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Chapter 1

Mathematics and Statistics for Economics

TU MA past exam questions

Past questions and answers

91 questions set from this chapter. Most repeated first.

  • 2026 · Mathematics Economics (Econ. 556-2) · 20 marks

Explain Samuelson dynamic process of income determination model. Derive the income difference equation and discuss the roles of MPC and the accelerator along with conditions for stability and cycles.

Answer

Samuelson's model (1939) combines the multiplier and the accelerator to show how a once-for-all disturbance can generate a self-sustained or damped business cycle. Income in each period is the sum of consumption, induced investment and autonomous (government) spending.

Assumptions

  • Consumption depends on last period's income: Ct=a+bYt−1C_t = a + bY_{t-1}, with 0<b<10<b<1 (bb = MPC).
  • Induced investment depends on the change in consumption (the accelerator): It=v(Ct−Ct−1)=vb(Yt−1−Yt−2)I_t = v(C_t - C_{t-1}) = vb(Y_{t-1}-Y_{t-2}), with v>0v>0 the accelerator coefficient. (Writing β=vb\beta = vb gives the usual textbook form.)
  • Autonomous spending Gt=G0G_t = G_0 is constant.

Derivation of the income difference equation

Yt=Ct+It+G0=a+bYt−1+vb (Yt−1−Yt−2)+G0Y_t = C_t + I_t + G_0 = a + bY_{t-1} + vb\,(Y_{t-1}-Y_{t-2}) + G_0

Yt−b(1+v) Yt−1+bv Yt−2=a+G0\boxed{Y_t - b(1+v)\,Y_{t-1} + bv\,Y_{t-2} = a + G_0}

Equilibrium (set Yt=Yt−1=Yt−2=Y∗Y_t=Y_{t-1}=Y_{t-2}=Y^*):

Y∗=a+G01−bY^* = \frac{a+G_0}{1-b}

which is the simple multiplier solution; the accelerator does not change the equilibrium level, only the path towards it.

Roles of MPC and accelerator

  • The MPC (bb) drives the multiplier: a rise in income raises consumption, which raises income again in the next period.
  • The accelerator (vv) makes investment respond to the change in demand. A rising income raises investment sharply, which amplifies the boom; a slowdown cuts investment, which deepens the slump.
  • Multiplier alone gives monotonic adjustment; only the interaction of the two can give cycles.

Solution and conditions

The characteristic equation is

λ2−b(1+v)λ+bv=0,λ1,2=b(1+v)±b2(1+v)2−4bv2\lambda^2 - b(1+v)\lambda + bv = 0,\qquad \lambda_{1,2}=\frac{b(1+v)\pm\sqrt{b^2(1+v)^2-4bv}}{2}

with Yt=Y∗+A1λ1t+A2λ2tY_t = Y^* + A_1\lambda_1^t + A_2\lambda_2^t.

  • Stability: both roots must lie inside the unit circle. For this equation the conditions reduce to bv<1bv<1 (the others, 1−b(1+v)+bv=1−b>01-b(1+v)+bv=1-b>0 and 1+b(1+v)+bv>01+b(1+v)+bv>0, always hold). Time path is convergent if bv<1bv<1, uniform if bv=1bv=1, explosive if bv>1bv>1.
  • Cycles: the roots are complex when the discriminant is negative, b2(1+v)2<4bvb^2(1+v)^2<4bv, i.e. b(1+v)2<4vb(1+v)^2<4v. Then the path oscillates with modulus bv\sqrt{bv}.

Five cases (with bb, vv plotted in the plane)

RegionDiscriminantbvbvTime path
A≥0\ge 0<1<1Non-oscillatory, convergent
B<0<0<1<1Damped cycles
C<0<0=1=1Uniform (constant-amplitude) cycles
D<0<0>1>1Explosive cycles
E≥0\ge 0>1>1Explosive, non-oscillatory

The boundary between cyclical and non-cyclical regions is the curve b(1+v)2=4vb(1+v)^2=4v.

Numerical illustration: b=0.5, v=1b=0.5,\ v=1 gives λ2−λ+0.5=0\lambda^2 - \lambda + 0.5=0, so λ=0.5±0.5i\lambda = 0.5\pm 0.5i, modulus 0.5≈0.71<1\sqrt{0.5}\approx0.71<1: damped cycles around Y∗Y^*.

Conclusion: A high MPC and a high accelerator together make the economy cyclical and potentially explosive; stability requires bv<1bv<1.

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  • 2026 · Mathematics Economics (Econ. 556-2) · 20 marks

Derive the Slutsky equation considering U=f(x,y)U = f(x, y) be a utility function and B=pxx+pyyB = p_x x + p_y y be a budget constraint. Express the equation in elasticity form, and analyze the possible signs of price, substitution and income effects for normal, inferior and Giffen goods.

Answer

Setup. The consumer maximises U=f(x,y)U=f(x,y) subject to B=pxx+pyyB=p_xx+p_yy. The first-order conditions fx=λpx, fy=λpyf_x=\lambda p_x,\ f_y=\lambda p_y and the budget constraint give the Marshallian demand x=x(px,py,B)x=x(p_x,p_y,B).

Derivation. Let hx(px,py,Uˉ)h_x(p_x,p_y,\bar U) be the Hicksian (compensated) demand, the demand that minimises expenditure for a given utility Uˉ\bar U, with expenditure function e(px,py,Uˉ)e(p_x,p_y,\bar U). At the optimum, B=e(px,py,Uˉ)B=e(p_x,p_y,\bar U) and

hx(px,py,Uˉ)=x(px,py,e(px,py,Uˉ)).h_x(p_x,p_y,\bar U)=x\big(p_x,p_y,e(p_x,p_y,\bar U)\big).

Differentiate with respect to pxp_x, holding pyp_y and Uˉ\bar U fixed:

∂hx∂px=∂x∂px+∂x∂B⋅∂e∂px.\frac{\partial h_x}{\partial p_x}=\frac{\partial x}{\partial p_x}+\frac{\partial x}{\partial B}\cdot\frac{\partial e}{\partial p_x}.

By Shephard's lemma, ∂e/∂px=hx=x\partial e/\partial p_x=h_x=x. Rearranging gives the Slutsky equation

∂x∂px=∂hx∂px⏟substitution effect−x ∂x∂B⏟income effect\boxed{\frac{\partial x}{\partial p_x}=\underbrace{\frac{\partial h_x}{\partial p_x}}_{\text{substitution effect}}-\underbrace{x\,\frac{\partial x}{\partial B}}_{\text{income effect}}}

The substitution effect is always ≤0\le 0 (compensated demand slopes downward). The income effect is the loss of real income caused by the price rise, x dpxx\,dp_x, multiplied by the income response ∂x/∂B\partial x/\partial B.

Elasticity form. Multiply by px/xp_x/x and write the last term as pxxB⋅∂x∂BBx\dfrac{p_xx}{B}\cdot\dfrac{\partial x}{\partial B}\dfrac{B}{x}:

exx=exxs−sx exBe_{xx}=e^{s}_{xx}-s_x\,e_{xB}

where exxe_{xx} is the own-price elasticity, exxs=∂hx∂pxpxx≤0e^{s}_{xx}=\dfrac{\partial h_x}{\partial p_x}\dfrac{p_x}{x}\le0 the compensated (substitution) elasticity, exBe_{xB} the income elasticity and sx=pxx/Bs_x=p_xx/B the budget share.

Signs of the effects (for a price rise)

GoodSubstitution effectIncome effect −x ∂x/∂B-x\,\partial x/\partial BTotal effect
Normal (exB>0e_{xB}>0)NegativeNegative (reinforces)Negative; law of demand holds
Inferior (exB<0e_{xB}<0)NegativePositive (offsets)Negative if ∣SE∣>∣IE∣\lvert SE\rvert>\lvert IE\rvert
Giffen (strongly inferior)NegativePositive and ∣IE∣>∣SE∣\lvert IE\rvert>\lvert SE\rvertPositive; upward-sloping demand

For a Giffen good, exx>0e_{xx}>0 requires −sxexB>∣exxs∣-s_xe_{xB}>\lvert e^{s}_{xx}\rvert, so the good must have a large budget share and a strongly negative income elasticity. Staple foods for very poor households are the usual example.

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  • 2026 · Mathematics Economics (Econ. 556-2) · 20 marks

Given A=(341010321010)A = \begin{pmatrix} 3 & 4 \\ 10 & 10 \\ 3 & 2 \\ 10 & 10 \end{pmatrix} and d=(12101210)d = \begin{pmatrix} 12 \\ 10 \\ 12 \\ 10 \end{pmatrix} for the continuous time output adjustment input-output model. Find
a) the particular integral, b) the complementary function and c) the definite time path assuming initial conditions x1,0=18738x_{1,0} = \frac{187}{38} and x2,0=7213x_{2,0} = \frac{72}{13}

Answer

Reading the data as A=(3/104/103/102/10)A=\begin{pmatrix}3/10&4/10\\3/10&2/10\end{pmatrix} and d=(12/1012/10)d=\begin{pmatrix}12/10\\12/10\end{pmatrix}, the continuous-time output adjustment model is

x˙=Ax+d−x=−(I−A)x+d,I−A=(0.7−0.4−0.30.8), ∣I−A∣=0.44.\dot x = Ax+d-x=-(I-A)x+d,\qquad I-A=\begin{pmatrix}0.7&-0.4\\-0.3&0.8\end{pmatrix},\ |I-A|=0.44.

(a) Particular integral (the equilibrium xˉ\bar x with x˙=0\dot x=0):

xˉ=(I−A)−1d=10.44(0.80.40.30.7)(1.21.2)=10.44(1.441.20)=(36/1130/11)≈(3.272.73)\bar x=(I-A)^{-1}d=\frac{1}{0.44}\begin{pmatrix}0.8&0.4\\0.3&0.7\end{pmatrix}\begin{pmatrix}1.2\\1.2\end{pmatrix}=\frac1{0.44}\begin{pmatrix}1.44\\1.20\end{pmatrix}=\begin{pmatrix}36/11\\30/11\end{pmatrix}\approx\begin{pmatrix}3.27\\2.73\end{pmatrix}

(b) Complementary function. Solve x˙=(A−I)x\dot x=(A-I)x with A−I=(−0.70.40.3−0.8)A-I=\begin{pmatrix}-0.7&0.4\\0.3&-0.8\end{pmatrix}.

Characteristic equation: λ2+1.5λ+0.44=0⇒(λ+0.4)(λ+1.1)=0\lambda^2+1.5\lambda+0.44=0\Rightarrow(\lambda+0.4)(\lambda+1.1)=0, so λ1=−0.4, λ2=−1.1\lambda_1=-0.4,\ \lambda_2=-1.1.

Eigenvectors: for λ1=−0.4\lambda_1=-0.4, −0.3v1+0.4v2=0⇒v=(4,3)′-0.3v_1+0.4v_2=0\Rightarrow v=(4,3)'; for λ2=−1.1\lambda_2=-1.1, 0.4v1+0.4v2=0⇒v=(1,−1)′0.4v_1+0.4v_2=0\Rightarrow v=(1,-1)'.

xc=c1(43)e−0.4t+c2(1−1)e−1.1tx_c=c_1\begin{pmatrix}4\\3\end{pmatrix}e^{-0.4t}+c_2\begin{pmatrix}1\\-1\end{pmatrix}e^{-1.1t}

(c) Definite time path. The general solution is x(t)=xˉ+xcx(t)=\bar x+x_c. At t=0t=0:

4c1+c2=18738−3611=689418,3c1−c2=7213−3011=4021434c_1+c_2=\frac{187}{38}-\frac{36}{11}=\frac{689}{418},\qquad 3c_1-c_2=\frac{72}{13}-\frac{30}{11}=\frac{402}{143}

Adding: 7c1=689418+4021437c_1=\dfrac{689}{418}+\dfrac{402}{143}, so c1≈0.637c_1\approx0.637 and c2≈−0.900c_2\approx-0.900.

x1(t)≈3.27+2.55 e−0.4t−0.90 e−1.1t,x2(t)≈2.73+1.91 e−0.4t+0.90 e−1.1tx_1(t)\approx3.27+2.55\,e^{-0.4t}-0.90\,e^{-1.1t},\qquad x_2(t)\approx2.73+1.91\,e^{-0.4t}+0.90\,e^{-1.1t}

Both roots are negative, so the system is stable and converges monotonically to xˉ=(36/11, 30/11)′\bar x=(36/11,\,30/11)', with the slower root −0.4-0.4 dominating in the long run.

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  • 2026 · Mathematics Economics (Econ. 556-2) · 5 marks

Derive shepherd's lemma.

Answer

Shephard's Lemma

Statement: If the expenditure function is e(p,u)e(p, u), then the Hicksian (compensated) demand for good ii is obtained simply by partially differentiating the expenditure function with respect to the price of that good:

xih(p,u)=∂e(p,u)∂pix_i^h(p, u) = \frac{\partial e(p, u)}{\partial p_i}

Derivation:

Let the consumer minimise expenditure subject to attaining a fixed utility level uu:

min⁡x  p⋅xsubject tou(x)=u\min_{x} \; p \cdot x \quad \text{subject to} \quad u(x) = u

The expenditure function is the value of this minimisation problem:

e(p,u)=min⁡x{p⋅x:u(x)=u}=p⋅xh(p,u)e(p, u) = \min_{x} \{ p \cdot x : u(x) = u \} = p \cdot x^h(p, u)

where xh(p,u)x^h(p, u) is the vector of Hicksian demand functions.

Now differentiate e(p,u)e(p, u) with respect to pip_i using the chain rule. Since e(p,u)=∑jpjxjh(p,u)e(p, u) = \sum_j p_j x_j^h(p, u):

∂e(p,u)∂pi=xih(p,u)+∑jpj∂xjh(p,u)∂pi\frac{\partial e(p, u)}{\partial p_i} = x_i^h(p, u) + \sum_j p_j \frac{\partial x_j^h(p, u)}{\partial p_i}

The second term vanishes by the envelope theorem. At the optimum, the first-order conditions of the expenditure-minimisation problem are:

pj=λ∂u∂xjfor all jp_j = \lambda \frac{\partial u}{\partial x_j} \quad \text{for all } j

where λ\lambda is the Lagrange multiplier on the utility constraint. Substituting these into the second term:

∑jpj∂xjh∂pi=λ∑j∂u∂xj⋅∂xjh∂pi=λddpi[u(xh(p,u))]\sum_j p_j \frac{\partial x_j^h}{\partial p_i} = \lambda \sum_j \frac{\partial u}{\partial x_j} \cdot \frac{\partial x_j^h}{\partial p_i} = \lambda \frac{d}{dp_i}\big[ u(x^h(p, u)) \big]

But along the Hicksian demand curve utility is held constant at uu, so u(xh(p,u))=uu(x^h(p, u)) = u for all pp. Hence its derivative with respect to pip_i is zero:

λ⋅0=0\lambda \cdot 0 = 0

Therefore the second term vanishes and we are left with:

∂e(p,u)∂pi=xih(p,u)\boxed{\frac{\partial e(p, u)}{\partial p_i} = x_i^h(p, u)}

Interpretation: The partial derivative of the expenditure function with respect to a price gives the compensated demand for that good. It tells us how much extra money the consumer needs to maintain the same utility level when the price of good ii rises by one unit — which is exactly the quantity of good ii she is buying. This is why the result is also called the "compensated demand = price derivative of expenditure" property.

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  • 2026 · Mathematics Economics (Econ. 556-2) · 5 marks

Define Cobb-Douglas production function and find its productivities (MP) and productivity function.

Answer

Cobb–Douglas production function

The Cobb–Douglas production function is a special form of production function that shows the technological relationship between inputs and output. For two inputs, labour (L)(L) and capital (K)(K), it is written as:

Q=ALαKβQ = A L^{\alpha} K^{\beta}

where:

  • QQ = total output,
  • AA = efficiency/technology parameter (A>0)(A>0),
  • α\alpha = output elasticity of labour,
  • β\beta = output elasticity of capital,
  • α,β>0\alpha,\beta > 0.

It is homogeneous of degree α+β\alpha+\beta. Thus:

  • if α+β=1\alpha+\beta=1, constant returns to scale,
  • if α+β>1\alpha+\beta>1, increasing returns to scale,
  • if α+β<1\alpha+\beta<1, decreasing returns to scale.

1. Marginal productivities (MP)

Marginal product of labour (MPL)(MP_L):

MPL=∂Q∂L=αALα−1Kβ=αQL=αAPLMP_L = \frac{\partial Q}{\partial L} = \alpha A L^{\alpha-1} K^{\beta} = \alpha \frac{Q}{L} = \alpha AP_L

Marginal product of capital (MPK)(MP_K):

MPK=∂Q∂K=βALαKβ−1=βQK=βAPKMP_K = \frac{\partial Q}{\partial K} = \beta A L^{\alpha} K^{\beta-1} = \beta \frac{Q}{K} = \beta AP_K

2. Productivity function (average productivity)

Average productivity of labour:

APL=QL=ALα−1KβAP_L = \frac{Q}{L} = A L^{\alpha-1} K^{\beta}

Average productivity of capital:

APK=QK=ALαKβ−1AP_K = \frac{Q}{K} = A L^{\alpha} K^{\beta-1}

3. Output elasticity

EL=∂Q∂L⋅LQ=αE_L = \frac{\partial Q}{\partial L}\cdot \frac{L}{Q} = \alpha EK=∂Q∂K⋅KQ=βE_K = \frac{\partial Q}{\partial K}\cdot \frac{K}{Q} = \beta

So the sum of output elasticities is:

EL+EK=α+βE_L + E_K = \alpha+\beta

which also gives the degree of homogeneity and hence the returns to scale.

Conclusion: For Q=ALαKβQ = A L^{\alpha} K^{\beta}, the marginal products are MPL=αQ/LMP_L=\alpha Q/L and MPK=βQ/KMP_K=\beta Q/K, while the average productivity functions are APL=Q/LAP_L=Q/L and APK=Q/KAP_K=Q/K.

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  • 2026 · Mathematics Economics (Econ. 556-2) · 5 marks

Solve the following system of simultaneous difference equation: xt+1−3xt+2yt=12x_{t+1} - 3x_t + 2y_t = 12 yt+1−2xt+yt=6y_{t+1} - 2x_t + y_t = 6 Given the initial condition x(0)=4,y(0)=3x(0) = 4, y(0) = 3

Answer

Let
[ u_t=\begin{pmatrix}x_t\ y_t\end{pmatrix},\qquad A=\begin{pmatrix}3&-2\2&-1\end{pmatrix},\qquad c=\begin{pmatrix}12\6\end{pmatrix}. ] Then the system is [ u_{t+1}=Au_t+c. ]

The characteristic equation of (A) is [ |A-\lambda I|= \begin{vmatrix} 3-\lambda&-2\ 2&-1-\lambda \end{vmatrix} =(3-\lambda)(-1-\lambda)+4 =\lambda^2-2\lambda+1=(\lambda-1)^2. ] So (\lambda=1) is a repeated eigenvalue.

Now [ N=A-I= \begin{pmatrix} 2&-2\ 2&-2 \end{pmatrix}, \qquad N^2=0. ] Hence [ A^t=(I+N)^t=I+tN= \begin{pmatrix} 1+2t&-2t\ 2t&1-2t \end{pmatrix}. ]

The general solution is [ u_t=A^t u_0+\sum_{j=0}^{t-1}A^j c. ]

First, [ A^t u_0= \begin{pmatrix} 1+2t&-2t\ 2t&1-2t \end{pmatrix} \begin{pmatrix} 4\3 \end{pmatrix}

\begin{pmatrix} 4+2t\ 3+2t \end{pmatrix}. ]

Also, [ \sum_{j=0}^{t-1}A^j c

\left(tI+\frac{t(t-1)}2N\right)c. ] Since [ Nc= \begin{pmatrix} 2&-2\ 2&-2 \end{pmatrix} \begin{pmatrix} 12\6 \end{pmatrix}

\begin{pmatrix} 12\12 \end{pmatrix}, ] we get [ \sum_{j=0}^{t-1}A^j c

t\begin{pmatrix}12\6\end{pmatrix} + \frac{t(t-1)}2\begin{pmatrix}12\12\end{pmatrix}

\begin{pmatrix} 6t^2+6t\ 6t^2 \end{pmatrix}. ]

Therefore, [ u_t= \begin{pmatrix} 4+2t\ 3+2t \end{pmatrix} + \begin{pmatrix} 6t^2+6t\ 6t^2 \end{pmatrix}. ]

So the required solution is [ \boxed{x_t=6t^2+8t+4} ] [ \boxed{y_t=6t^2+2t+3} ]

Verification for (t=0): [ x_0=4,\qquad y_0=3, ] which satisfies the given initial conditions.

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  • 2026 · Mathematics Economics (Econ. 556-2) · 5 marks

Using the Hessian determinant method, examine the conditions for relative maxima and minima of a two-variable function Z=f(X,Y)Z=f(X, Y). Show the necessary derivations and interpretation.

Answer

1. Setup. Let Z=f(X,Y)Z=f(X,Y) be twice continuously differentiable, with first partials fX,fYf_X,f_Y and second partials fXX,fYY,fXY=fYXf_{XX},f_{YY},f_{XY}=f_{YX} (Young's theorem).

2. First-order (necessary) condition. At an interior extremum the total differential must vanish for arbitrary dX,dYdX,dY:

dZ=fX dX+fY dY=0 ⇒ fX=0, fY=0.dZ=f_X\,dX+f_Y\,dY=0\ \Rightarrow\ f_X=0,\ f_Y=0.

This locates stationary points but cannot tell a maximum from a minimum or a saddle.

3. Second-order condition and the Hessian. The second differential is the quadratic form

d2Z=fXX(dX)2+2fXY dX dY+fYY(dY)2,d^2Z=f_{XX}(dX)^2+2f_{XY}\,dX\,dY+f_{YY}(dY)^2,

with Hessian matrix and determinant

H=(fXXfXYfYXfYY),∣H∣=fXXfYY−(fXY)2.H=\begin{pmatrix}f_{XX}&f_{XY}\\f_{YX}&f_{YY}\end{pmatrix},\qquad |H|=f_{XX}f_{YY}-(f_{XY})^2.

4. Derivation of the sign conditions. Completing the square (for fXX≠0f_{XX}\ne0):

d2Z=1fXX(fXX dX+fXY dY)2+fXXfYY−(fXY)2fXX (dY)2.d^2Z=\frac{1}{f_{XX}}\big(f_{XX}\,dX+f_{XY}\,dY\big)^2+\frac{f_{XX}f_{YY}-(f_{XY})^2}{f_{XX}}\,(dY)^2.

The first term is a perfect square, so its sign follows the sign of fXXf_{XX}. The second term's sign follows ∣H∣/fXX|H|/f_{XX}. Hence:

  • Relative maximum: d2Z<0d^2Z<0 for all (dX,dY)≠0(dX,dY)\ne0 requires fXX<0f_{XX}<0 and ∣H∣>0|H|>0.
  • Relative minimum: d2Z>0d^2Z>0 requires fXX>0f_{XX}>0 and ∣H∣>0|H|>0.
  • Saddle point: ∣H∣<0|H|<0; the two terms have opposite signs, so d2Zd^2Z changes sign with direction.
  • Inconclusive: ∣H∣=0|H|=0; higher-order terms must be examined.

5. Summary

Relative maximumRelative minimum
First orderfX=fY=0f_X=f_Y=0fX=fY=0f_X=f_Y=0
Second order$f_{XX}<0,\H

Equivalently, the leading principal minors alternate in sign for a maximum (∣H1∣<0, ∣H2∣>0|H_1|<0,\ |H_2|>0) and are all positive for a minimum (∣H1∣>0, ∣H2∣>0|H_1|>0,\ |H_2|>0). Note that ∣H∣>0|H|>0 forces fXXf_{XX} and fYYf_{YY} to have the same sign.

6. Interpretation. ∣H∣|H| measures the curvature of the surface at the stationary point. If ∣H∣>0|H|>0 the surface curves the same way in every direction: dome-shaped (peak) when fXX<0f_{XX}<0, bowl-shaped (trough) when fXX>0f_{XX}>0. If ∣H∣<0|H|<0 it curves up in one direction and down in another, giving a saddle.

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  • 2026 · Mathematics Economics (Econ. 556-2) · 5 marks

Show that Dorse growth model is highly unstable.

Answer

The model meant here is Domar's growth model (the Harrod–Domar framework). Its instability is the famous "knife-edge" result: the economy can grow in balance only at one particular rate, and any deviation from it is magnified, not corrected.

Assumptions: saving is a constant fraction of income, S=sYS=sY; the capacity-output relation is fixed, so each unit of investment raises potential output by σ=1/v\sigma=1/v (the capital-output ratio vv is constant); no technical progress and no depreciation.

Demand side (multiplier). Equilibrium needs I=S=sYI=S=sY. If investment grows, income must grow by ΔY=ΔIs\Delta Y=\dfrac{\Delta I}{s}, so required income growth is

ΔYY=ΔII⋅IsY=ΔII(since I=sY).\frac{\Delta Y}{Y}=\frac{\Delta I}{I}\cdot\frac{I}{sY}=\frac{\Delta I}{I}\quad(\text{since }I=sY).

Supply side (capacity). Investment II raises capacity by ΔK=I\Delta K=I, i.e. potential output increases by σI\sigma I.

Condition for balanced growth. Demand growth must equal capacity growth:

ΔY=σI ⇒ ΔYY=σs=sv.\Delta Y=\sigma I\ \Rightarrow\ \frac{\Delta Y}{Y}=\sigma s=\frac{s}{v}.

This is the warranted rate Gw=s/vG_w=s/v. For example, with s=0.20s=0.20 and v=4v=4, Gw=5%G_w=5\%.

Why it is unstable (Harrod's argument). Suppose actual growth GG differs from GwG_w.

  • If G>GwG>G_w: actual output grows faster than planned, so the actual capital-output ratio is below the desired one. Firms see capital as scarce and raise investment further, so GG rises even more above GwG_w. The result is cumulative inflationary boom.
  • If G<GwG<G_w: capital is in excess, so firms cut investment, income growth falls further below GwG_w. The result is cumulative slump and unemployment.

Formally, the deviation G−GwG-G_w grows over time (a positive feedback), so the equilibrium is unstable. The economy "grows on a razor's edge".

Second knife-edge. Even if G=GwG=G_w, steady full employment also needs Gw=Gn=nG_w=G_n=n (the natural rate). Since ss, vv and nn are determined independently, equality is accidental; if Gw>GnG_w>G_n there is secular stagnation, and if Gw<GnG_w<G_n there is chronic unemployment.

Conclusion: because balanced growth requires G=Gw=GnG=G_w=G_n and any departure is self-reinforcing, the Domar/Harrod–Domar model is highly unstable. This is why later models (Solow–Swan) allow vv to vary to restore stability.

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  • 2026 · Mathematics Economics (Econ. 556-2) · 5 marks

Solve the following optimal control theory: Maximize ∫01(4x−3y2)dt\int_0^1 (4x - 3y^2)dt Subject to x˙=8y\dot{x} = 8y x(0)=5x(1)=freex(0) = 5 \quad x(1) = \text{free}

Answer

Given: Maximize [ J=\int_0^1 (4x-3y^2),dt ] subject to [ \dot{x}=8y,\qquad x(0)=5,\qquad x(1)\ \text{free}. ]

Step 1: Form the Hamiltonian

For a maximization problem, [ H=4x-3y^2+\lambda(8y). ]

Step 2: Pontryagin necessary conditions

Control condition: [ \frac{\partial H}{\partial y}=-6y+8\lambda=0 ] [ \Rightarrow y=\frac{4}{3}\lambda. ]

Costate equation: [ \dot{\lambda}=-\frac{\partial H}{\partial x}=-4. ]

Transversality condition since (x(1)) is free: [ \lambda(1)=0. ]

Step 3: Solve the costate equation

[ \dot{\lambda}=-4,\qquad \lambda(1)=0. ] Integrating, [ \lambda(t)=-4t+C. ] Using (\lambda(1)=0), [ -4+C=0 \Rightarrow C=4. ] Thus, [ \lambda(t)=4(1-t). ]

Step 4: Find the optimal control

[ y(t)=\frac{4}{3}\lambda(t) =\frac{4}{3}\cdot 4(1-t) =\frac{16}{3}(1-t). ]

So, [ \boxed{y^*(t)=\frac{16}{3}(1-t)}. ]

Step 5: Find the state trajectory

[ \dot{x}=8y=\frac{128}{3}(1-t). ] Integrating, [ x(t)=x(0)+\int_0^t \frac{128}{3}(1-s),ds. ] Since (x(0)=5), [ x(t)=5+\frac{128}{3}\left(t-\frac{t^2}{2}\right). ] Therefore, [ \boxed{x^*(t)=5+\frac{128}{3}t-\frac{64}{3}t^2}. ]

At (t=1), [ x(1)=5+\frac{128}{3}\cdot \frac12 =5+\frac{64}{3} =\frac{79}{3}. ]

Step 6: Maximum value of the objective functional

[ J^=\int_0^1 \left(4x^-3y^{2}\right)dt. ] Substituting, [ 4x^-3y^{2} =20+\frac{512}{3}t-\frac{256}{3}t^2 -\frac{256}{3}(1-2t+t^2) ] [ =-\frac{196}{3}+\frac{1024}{3}t-\frac{512}{3}t^2. ] Thus, [ J^=\int_0^1\left(-\frac{196}{3}+\frac{1024}{3}t-\frac{512}{3}t^2\right)dt ] [ =-\frac{196}{3}+\frac{1024}{3}\cdot\frac12-\frac{512}{3}\cdot\frac13 ] [ =-\frac{196}{3}+\frac{512}{3}-\frac{512}{9} =\frac{436}{9}. ]

Hence, [ \boxed{J^*=\frac{436}{9}} ]

Final answer: [ \boxed{y^(t)=\frac{16}{3}(1-t)} ] [ \boxed{x^(t)=5+\frac{128}{3}t-\frac{64}{3}t^2} ] [ \boxed{J^*=\frac{436}{9}} ]

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  • 2026 · Mathematics Economics (Econ. 556-2) · 5 marks

Define CGE model with important components.

Answer

CGE Model (Computable General Equilibrium Model)

A CGE model is a numerical, economy-wide model that applies general equilibrium theory to actual data of an economy to compute the equilibrium prices and quantities of all markets simultaneously. It converts Walrasian general equilibrium into an empirically solvable system, using a Social Accounting Matrix (SAM) as its database. It is widely used to analyse the effects of policy shocks such as tax reform, trade liberalisation, subsidies, and public expenditure changes on output, prices, income distribution, and welfare.

Important components:

  1. Social Accounting Matrix (SAM) — the data base that records all transactions among production activities, commodities, factors of production (labour, capital, land), households, government, and the rest of the world in a consistent double-entry format.

  2. Production block — production functions (usually CES or Cobb–Douglas) describing how firms combine intermediate inputs and primary factors to produce output, along with the zero-profit condition (price = unit cost) under perfect competition.

  3. Factor markets — supply and demand for factors of production, with factor prices determined endogenously; full employment or unemployment (e.g., efficiency-wage or fixed-wage) closures may be assumed.

  4. Household block — household income from factor payments and transfers, and consumption demand derived from utility maximisation (e.g., Stone–Geary or CES utility functions) subject to the budget constraint.

  5. Government block — tax and transfer instruments (direct and indirect taxes, tariffs, subsidies) and government revenue–expenditure accounts, with a fiscal closure rule.

  6. Foreign trade block — exports and imports modelled through the Armington assumption (imperfect substitutability between domestic and foreign goods) and the small-country or large-country trade closure.

  7. Equilibrium conditions and closure rules — market-clearing conditions for all commodity and factor markets, plus macro closures for the government budget, savings–investment balance, and the current account.

  8. Solution algorithm — the system of nonlinear equations is solved numerically (e.g., GAMS with a solver such as GAMS/MPSGE) to obtain counterfactual equilibrium, which is then compared with the base-run equilibrium.

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  • 2026 · Mathematics Economics (Econ. 556-2) · 5 marks

If a>ba > b, then prove ac>bca^c > b^c if c>0c > 0 and a,b∈Ra, b \in \mathbb{R}

Answer

The statement cannot hold for all real a,ba,b (it is not true for negative numbers), but it is true when a>b>0a>b>0, which is the case to prove.

Why the general case fails. Take a=1, b=−2, c=2a=1,\ b=-2,\ c=2. Then a>ba>b and c>0c>0, but ac=1<4=bca^c=1<4=b^c. Also, for non-integer cc, bcb^c may not even be real when b<0b<0. So the result needs a>b>0a>b>0 (or at least a>b≥0a>b\ge0).

Proof for a>b>0, c>0a>b>0,\ c>0. Let f(x)=xcf(x)=x^c for x>0x>0. Then

f′(x)=c xc−1>0,f'(x)=c\,x^{c-1}>0,

because c>0c>0 and xc−1>0x^{c-1}>0 for x>0x>0. Hence ff is strictly increasing on (0,∞)(0,\infty). Since a>b>0a>b>0, f(a)>f(b)f(a)>f(b), that is

ac>bc.a^c>b^c.

Alternative proof (without calculus). Write ac=ecln⁡aa^c=e^{c\ln a} and bc=ecln⁡bb^c=e^{c\ln b}. Since ln⁡\ln is increasing, a>b>0a>b>0 gives ln⁡a>ln⁡b\ln a>\ln b. Multiplying by c>0c>0 keeps the inequality, and exe^{x} is increasing, so ecln⁡a>ecln⁡be^{c\ln a}>e^{c\ln b}. ■\blacksquare

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  • 2026 · Mathematics Economics (Econ. 556-2) · 5 marks

Explain the Solow growth model, highlighting the roles of capital accumulation, labor growth and technological progress in determining long-run economic growth.

Answer

The Solow growth model (Solow–Swan model, 1956) is the foundational neoclassical model of long-run economic growth. It explains how output per worker is determined by capital accumulation, labour growth and technological progress, and shows that only technological progress can sustain long-run growth in living standards.

Assumptions

  • Production function with constant returns to scale: Y=F(K,L)=Kα(AL)1−αY = F(K, L) = K^{\alpha}(AL)^{1-\alpha} where (Y) is output, (K) capital, (L) labour and (A) the level of technology (labour-augmenting).
  • Saving is a constant fraction (s) of income: (S = sY); investment (I = sY).
  • Capital depreciates at a constant rate (\delta).
  • Labour force grows at rate (n) and technology at rate (g).
  • Perfect competition, so factors are paid their marginal products.

Capital accumulation

In intensive (per effective worker) form, with (k = K/AL) and (y = Y/AL = k^{\alpha}):

k˙=skα−(n+g+δ)k\dot{k} = s k^{\alpha} - (n + g + \delta)k

  • (s k^{\alpha}) is actual investment per effective worker.
  • ((n + g + \delta)k) is break-even investment — the investment needed to keep (k) constant as capital is diluted by population growth, technological progress and depreciation.

At the steady state, (\dot{k} = 0), giving:

k∗=(sn+g+δ)11−α,y∗=(k∗)αk^{*} = \left(\frac{s}{n + g + \delta}\right)^{\frac{1}{1-\alpha}}, \qquad y^{*} = (k^{*})^{\alpha}

Because of diminishing returns to capital, a higher saving rate raises (k^{}) and (y^{}) only in levels, not in the long-run growth rate. Once (k = k^{*}), output per effective worker stops growing. This is the model's central prediction of conditional convergence: poorer countries with lower (k) grow faster and catch up to their own steady states.

Labour growth

A rise in (n) steepens the break-even investment line, lowering (k^{}) and (y^{}). Higher population growth therefore reduces steady-state income per worker (a "dilution" effect), though it raises the growth rate of total output (Y) by (n + g). This helps explain why many developing countries with high (n) remain poor.

Technological progress

Technology (g) is the only source of sustained growth in output per worker. In steady state:

  • (k) and (y) are constant (per effective worker),
  • capital per worker (K/L) and output per worker (Y/L) grow at rate (g),
  • total output (Y) grows at rate (n + g).

Thus long-run living standards rise only through technological progress; capital accumulation alone cannot do so because of diminishing returns.

Policy implications

  • Raising (s) or lowering (\delta) increases steady-state income but not its long-run growth rate.
  • Lowering (n) (e.g. through family planning, education) raises income per worker.
  • The only route to permanently rising per-capita income is continuous technological progress through R&D, education and innovation.

Limitation: the model treats (g) as exogenous and does not explain what drives technology — a gap later addressed by endogenous growth theory (Romer, Lucas).

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  • 2026 · Mathematics Economics (Econ. 556-2) · 5 marks

Write short note on any two a) Optimum control theory with Hamiltonian condition b) Market inventory model c) Homogeneous production function

Answer

a) Optimal control theory with the Hamiltonian condition

Optimal control theory finds the time path of a control variable that optimises an objective functional subject to a differential equation governing the state variable. It is used in growth theory, resource economics and optimal saving.

Problem: maximise J=∫0Tf(t,y,u) dtJ=\displaystyle\int_0^T f(t,y,u)\,dt subject to y˙=g(t,y,u)\dot y=g(t,y,u), y(0)=y0y(0)=y_0, where yy is the state variable and uu the control.

Hamiltonian (present-value form):

H(t,y,u,λ)=f(t,y,u)+λ(t) g(t,y,u)H(t,y,u,\lambda)=f(t,y,u)+\lambda(t)\,g(t,y,u)

with λ(t)\lambda(t) the costate variable (shadow price of the state).

Necessary conditions (Pontryagin's maximum principle):

  1. Control: max⁡uH\max_u H at every tt; for an interior solution ∂H/∂u=0\partial H/\partial u=0.
  2. Costate: λ˙=−∂H/∂y\dot\lambda=-\partial H/\partial y.
  3. State: y˙=∂H/∂λ=g\dot y=\partial H/\partial\lambda=g.
  4. Transversality: λ(T)=0\lambda(T)=0 if y(T)y(T) is free (or H(T)=0H(T)=0 if TT is free).

If the objective has a discount factor, f=e−ρtF(⋅)f=e^{-\rho t}F(\cdot), the current-value Hamiltonian Hc=F+μgH_c=F+\mu g (with μ=λeρt\mu=\lambda e^{\rho t}) is often used. Its conditions become μ˙=ρμ−∂Hc/∂y\dot\mu=\rho\mu-\partial H_c/\partial y and the transversality condition lim⁡t→∞e−ρtμ(t)y(t)=0\lim_{t\to\infty}e^{-\rho t}\mu(t)y(t)=0 (or μ(T)e−ρT=0\mu(T)e^{-\rho T}=0 for free y(T)y(T)).

Interpretation: HH balances the current gain ff against the future value of the change in the state, λg\lambda g. Applications include the Ramsey optimal-saving model and Hotelling's rule for exhaustible resources.

b) Market inventory model

Producers hold inventories as a buffer between production and sales, so price adjusts according to the stock rather than clearing the market instantly.

Let Qd=a−bPQ_d=a-bP (demand), Qs=−c+dPQ_s=-c+dP (supply), and let inventory change be I˙=Qs−Qd\dot I=Q_s-Q_d. If price adjusts in the direction opposite to the inventory stock, P˙=−k (I−Iˉ)\dot P=-k\,(I-\bar I) with k>0k>0 and Iˉ\bar I the desired stock, then excess supply accumulates as stock and price falls until the stock returns to the desired level.

In equilibrium I˙=0\dot I=0 (so Qs=QdQ_s=Q_d) and I=IˉI=\bar I. The inventory buffer smooths price fluctuations: excess supply is absorbed into stock instead of causing an immediate price crash. The model is relevant for storable goods such as grains, and underlies buffer-stock policies.

c) Homogeneous production function

A production function Q=f(L,K)Q=f(L,K) is homogeneous of degree kk if f(tL,tK)=tkf(L,K)f(tL,tK)=t^k f(L,K) for all t>0t>0.

  • k=1k=1: constant returns to scale (linear homogeneity).
  • k>1k>1: increasing returns to scale.
  • k<1k<1: decreasing returns to scale.

Euler's theorem: L fL+K fK=k fL\,f_L+K\,f_K=k\,f. For k=1k=1 the total product is exactly exhausted when each factor is paid its marginal product (product exhaustion), the basis of the marginal productivity theory of distribution.

Cobb–Douglas Q=ALαKβQ=AL^{\alpha}K^{\beta} is homogeneous of degree α+β\alpha+\beta. For linearly homogeneous functions, average and marginal products depend only on the capital–labour ratio K/LK/L, which is why they are central in growth theory.

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  • 2025 · Research Methodology (Econ. 561) · 10 marks

Suppose you are planning to formulate a research problem in economics relating to an area of your interest. What are the steps you would follow in formulating the research problem? Elucidate with examples.

Answer

Formulating a Research Problem in Economics

A research problem is a clear, specific statement of the gap between what is known and what needs to be known in a chosen area. Formulating it is the first and most decisive step of research, because a badly framed problem wastes the entire study. The steps I would follow are set out below, with examples drawn from an area of my interest — labour economics in Nepal, specifically youth out-migration and its effect on agricultural productivity.

Step 1: Identify a broad area of interest

I begin with a field I know and care about. Interest sustains the long effort a study demands. My area: the economics of migration and rural labour markets in Nepal.

Step 2: Preliminary review of literature and data

Before narrowing, I read existing studies (journal articles, dissertations, government reports such as the Nepal Labour Force Survey, NLSS, Census, and Nepal Rastra Bank surveys) to learn what has already been established. This tells me where knowledge is thin. I find that much work exists on remittances and consumption, but little on how the absence of young workers affects farm output at the household level.

Step 3: Narrow the area into a specific topic

A broad area cannot be researched; it must be narrowed by place, time, population, and variable. From "migration and agriculture" I narrow to: the effect of youth out-migration on agricultural productivity in Nawalparasi district, 2015–2023.

Step 4: State the problem precisely (problem statement)

I convert the topic into a question that research can answer. For example:

"Has the out-migration of male youth aged 20–40 reduced household agricultural output per hectare in Nawalparasi, and through which channel — labour shortage, shift to less labour-intensive crops, or land abandonment?"

This states the gap: existing studies measure remittance income but not the labour-loss effect on productivity.

Step 5: Define objectives

Objectives must be specific, measurable, and few. For my study:

  1. To measure the change in agricultural productivity of migrant-sending households.
  2. To identify the channels through which labour loss affects output.
  3. To suggest policy measures for labour-substituting technology.

Step 6: Formulate research questions and hypotheses

Each objective becomes a question, and each question a testable hypothesis:

  • H₀: Out-migration has no significant effect on output per hectare.
  • H₁: Out-migration significantly reduces output per hectare.

Step 7: Define scope and delimitation

I state clearly what is included and excluded — one district, one crop season, households with at least one migrant, the period 2015–2023 — so the findings are not over-generalised.

Step 8: Assess feasibility (data, time, cost, ethics)

I check whether data are obtainable. If secondary data are insufficient, I plan a primary survey of, say, 200 households using a structured questionnaire. I confirm the budget, time, and ethical clearance for interviewing migrants' families.

Step 9: Define key concepts and variables

Operational definitions prevent confusion:

  • Out-migration = a household member aged 20–40 working abroad for 6+ months.
  • Agricultural productivity = output (kg) per hectare per season.
  • Independent variables: number of migrants, remittance received, land size, irrigation, mechanisation.

Step 10: State the significance of the study

I explain who benefits — policymakers designing labour-substitution programmes, local governments, and researchers — and how the study fills the identified gap.

Step 11: Finalise the title

A good title is short, specific, and reflects variables, place, and time:

"Youth Out-Migration and Agricultural Productivity: A Household-Level Study of Nawalparasi District, Nepal (2015–2023)."

Conclusion

The sequence — interest → literature → narrowing → problem statement → objectives → hypotheses → scope → feasibility → definitions → significance → title — converts a vague interest into a researchable problem. Each step disciplines the next, and together they ensure the study is original, feasible, and useful. For my chosen area, this process yields a problem that is economically meaningful: whether Nepal's remittance economy is quietly eroding its own food-producing base.

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  • 2025 · Research Methodology (Econ. 561) · 10 marks

What are the different types of sampling techniques? If your sampling technique you adopt in an experimental study design? Give reasons. [6+4]

Answer

Types of Sampling Techniques

Sampling is the process of selecting a representative subset (sample) from a population for study. The techniques are broadly divided into two categories:

1. Probability (Random) Sampling

In this method, every unit of the population has a known, non-zero chance of being selected. It is based on the principle of randomisation and allows the calculation of sampling error.

(a) Simple Random Sampling Every unit has an equal probability of selection. The sample is drawn by lottery method or random number tables. It is simple and unbiased but requires a complete sampling frame and is costly for large, scattered populations.

(b) Systematic Sampling Units are selected at a fixed interval (k = N/n) after a random start. It is convenient and quick, but may introduce bias if the population has a periodic pattern.

(c) Stratified Random Sampling The population is divided into homogeneous strata (e.g., income groups, regions), and random samples are drawn from each stratum. It increases precision and ensures representation of all groups. It is widely used in socio-economic surveys.

(d) Cluster Sampling The population is divided into clusters (e.g., wards, villages), and entire clusters are randomly selected and studied. It is cost-effective for geographically dispersed populations but has higher sampling error.

(e) Multistage Sampling Sampling is done in successive stages (e.g., district → ward → household). It is practical for large national surveys.

2. Non-Probability (Non-Random) Sampling

Here, selection depends on the researcher's judgement or convenience, and the probability of selection is unknown.

(a) Convenience Sampling — Units are chosen because they are easily available. Quick and cheap but highly biased.

(b) Purposive (Judgement) Sampling — The researcher deliberately selects units believed to be representative or relevant.

(c) Quota Sampling — The population is divided into categories and a fixed quota is filled from each, but selection within quota is non-random.

(d) Snowball Sampling — Existing respondents refer further respondents; useful for hidden or rare populations.

(e) Voluntary Sampling — Respondents self-select, as in online polls.


Sampling Technique Adopted in an Experimental Study Design

In an experimental study design, the most appropriate sampling technique is random probability sampling — specifically simple random sampling (or stratified random sampling where heterogeneity exists) for the selection of subjects, followed by random assignment of subjects to treatment and control groups.

Reasons

1. Minimises selection bias Random selection gives every unit an equal chance of being included, so the sample is representative of the population. Systematic differences between subjects are avoided.

2. Ensures internal validity Random assignment (randomisation) balances both known and unknown confounding variables across treatment and control groups. This is the core requirement of a true experiment, as it allows any observed difference in the outcome to be attributed to the treatment rather than to pre-existing differences.

3. Permits statistical inference Probability sampling provides a known sampling distribution, so sampling error can be estimated. This makes it valid to apply t-tests, ANOVA, or regression and to generalise results from the sample to the population with a measurable degree of confidence.

4. Satisfies the assumption of independence Random selection and assignment ensure that observations are independent of one another, which is a basic assumption of most parametric statistical tests.

5. Controls extraneous variation If the population is heterogeneous (e.g., by income, gender, or location), stratified random sampling is used so that each stratum is represented proportionally. This reduces variance and increases the precision (power) of the experiment to detect a true treatment effect.

6. Enhances external validity (generalizability) Because the sample is drawn randomly from a defined population, the causal findings of the experiment can be generalized beyond the sample to the wider population.

Conclusion

Thus, in an experimental study design, random probability sampling (simple or stratified) combined with random assignment is adopted, because it eliminates selection bias, controls confounding factors, satisfies the assumptions of statistical inference, and strengthens both the internal and external validity of the causal conclusions drawn from the experiment.

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  • 2025 · Research Methodology (Econ. 561) · 10 marks

Define and distinguish descriptive and inferential data analysis.

Answer

Descriptive and Inferential Data Analysis

Descriptive Data Analysis

Definition: Descriptive data analysis is the branch of statistics concerned with collecting, organizing, summarizing, and presenting data in a meaningful form so that the basic features of a dataset become clear. It describes what is in the data without drawing conclusions beyond the data itself. It uses measures such as frequency distributions, percentages, measures of central tendency (mean, median, mode), measures of dispersion (range, variance, standard deviation, coefficient of variation), and graphical tools such as bar charts, histograms, pie charts, and scatter diagrams.

Purpose: To reduce a large mass of raw data into a few understandable summary values and to reveal patterns, trends, and comparisons within the observed data.

Inferential Data Analysis

Definition: Inferential data analysis is the branch of statistics concerned with drawing conclusions, predictions, or generalizations about a population on the basis of information contained in a sample drawn from that population. It uses probability theory to measure the reliability of those conclusions. Its main tools are estimation (point and interval estimation) and hypothesis testing (z-test, t-test, chi-square test, F-test, ANOVA, regression analysis).

Purpose: To generalize sample findings to the whole population, to test theories or claims, and to support decision-making under uncertainty.

Distinction between Descriptive and Inferential Data Analysis

BasisDescriptive AnalysisInferential Analysis
MeaningSummarizes and describes the characteristics of the available data.Draws conclusions about a population from sample data.
ScopeLimited to the data at hand; no generalization.Extends beyond the data to the whole population.
Data usedUsually the entire set of collected observations (census or full dataset).A representative sample drawn from the population.
Tools usedMean, median, mode, range, variance, standard deviation, tables, graphs.Estimation, hypothesis testing, confidence intervals, t-test, z-test, chi-square, ANOVA.
Role of probabilityNot required; results are certain for the given data.Essential; results are probabilistic and carry a margin of error.
ResultGives summary values and visual descriptions.Gives inferences, estimates, and decisions about population parameters.
ExamplesAverage income of 500 surveyed households; a pie chart of expenditure shares.Testing whether average national income exceeds Rs 30,000 using a sample of 500 households.

Conclusion

The two are complementary rather than competing. Descriptive analysis is the first step that organizes and simplifies data, while inferential analysis builds on these summaries to make generalizations about the population. In economic research, descriptive statistics describe the sample or economy, and inferential statistics test hypotheses and forecast economic behaviour.

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  • 2025 · Research Methodology (Econ. 561) · 5 marks

"Research design ensures that the procedures required to complete the study are adequate to obtain valid, objective, and accurate answers to the research question." Explain.

Answer

Research design is the overall plan or blueprint that a researcher prepares before actually carrying out a study. It specifies the type of study, the population and sample, the methods of data collection, the tools of analysis, and the time and cost involved. Its central purpose is to link the research question to the data and the conclusions in a logically sound way. The statement that research design "ensures that the procedures required to complete the study are adequate to obtain valid, objective, and accurate answers to the research question" can be explained as follows.

1. It makes the procedures adequate to the research question. A research question may be descriptive, explanatory, or exploratory. Each type demands a different design. A descriptive question ("What is the level of financial literacy among rural households?") needs a survey design with a representative sample; an explanatory question ("Does microcredit reduce poverty?") needs a causal or quasi-experimental design with control groups. By choosing the design that matches the question, the researcher ensures that the procedures adopted are capable of answering that particular question rather than some other question. Without a design, the researcher may collect data that are interesting but irrelevant to the objective.

2. It ensures validity. Validity means that the study measures what it claims to measure and that the conclusions follow from the evidence. Research design safeguards validity in several ways:

  • Internal validity: Random sampling and random assignment, use of control groups, and control of extraneous variables reduce bias and confounding, so that an observed effect can be attributed to the treatment rather than to other factors.
  • External validity: A clearly defined population and an appropriate sampling frame allow the findings to be generalized beyond the sample.
  • Construct validity: Clear operational definitions of concepts (for example, defining "poverty" by a consumption expenditure line) ensure that the indicators actually represent the theoretical construct.
  • Statistical conclusion validity: Choosing an adequate sample size and the correct statistical test reduces Type I and Type II errors.

3. It ensures objectivity. Objectivity means that the findings do not depend on the personal bias, convenience, or preconceptions of the researcher. A well-prepared design achieves this by pre-specifying the sampling method (probability sampling rather than convenience sampling), standardising the instruments (structured questionnaires, uniform instructions to enumerators), pre-testing the tools, and fixing the hypotheses and analysis plan in advance. When procedures are predetermined and applied uniformly to all units, the personal judgement of the investigator plays a minimal role, and another researcher repeating the study should reach similar results.

4. It ensures accuracy. Accuracy refers to the closeness of the estimates to the true population values and the reliability of the measurements. Design contributes to accuracy through:

  • correct definition of population, unit of analysis, and variables;
  • use of reliable and pre-tested instruments;
  • adequate sample size and low sampling error;
  • proper training and supervision of data collectors to reduce non-sampling error;
  • appropriate data processing and use of suitable analytical techniques.

5. It provides a basis for control and economy. By deciding in advance what data are needed, from whom, how, and when, the design prevents the collection of superfluous data. This saves time, money, and effort, and keeps the study within feasible limits — an important consideration in economics research where surveys are costly.

6. It guides the entire research process. The design acts as a roadmap: it tells the researcher how to proceed at each stage, from problem formulation to sampling, measurement, analysis, and interpretation. This reduces confusion, prevents mid-course changes that could introduce bias, and makes the study replicable and comparable with other studies.

Conclusion. A well-constructed research design is therefore not a mere formality but the foundation of good research. It aligns the research question with suitable procedures, and by doing so it secures the three essential qualities of scientific enquiry — validity, objectivity, and accuracy. A defective design, however carefully the data are later collected and analysed, cannot be repaired at the analysis stage; hence the saying that the whole research stands or falls on its design.

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  • 2025 · Research Methodology (Econ. 561) · 5 marks

How do you evaluate the scientific literature and write a literature review? Discuss.

Answer

Evaluating Scientific Literature and Writing a Literature Review

A literature review is a critical, systematic summary and synthesis of existing knowledge on a research question. It is not a list of summaries but an argument that shows what is known, what is contested, and what remains unstudied. It therefore requires both evaluation of individual studies and organisation of the evidence as a whole.

A. Evaluating Scientific Literature

Each source is judged on the following criteria:

  1. Relevance – Does the study address the research question, the same population, period and context (e.g., a study of inflation in Nepal vs. one in the USA)?
  2. Authority and credibility – Is the author/institution recognised? Is the source peer-reviewed, or a working paper, report or blog?
  3. Methodology – Is the design appropriate (experiment, survey, panel data, RCT)? Is the sample adequate and representative? Are the variables measured correctly? Is there a control group or counterfactual?
  4. Validity and reliability – Internal validity (causal inference, omitted variable bias, endogeneity), external validity (generalisability), and consistency of results across studies.
  5. Objectivity and bias – Funding source, ideological slant, selective reporting, publication bias.
  6. Currency – Recent enough for the topic; classic works are retained only for theory.
  7. Contribution – Does it add new evidence, method or theory, or merely repeat earlier work?

B. Steps in Writing a Literature Review

  1. Define the question and scope – specify topic, time frame, geography and key concepts.
  2. Search the literature – use databases (JSTOR, Google Scholar, NepJOL and EconLit), keywords, and backward/forward citation tracking.
  3. Screen and select – apply inclusion/exclusion criteria; keep only relevant, quality studies.
  4. Read and record – prepare a matrix of author, year, objective, method, data, findings and limitations.
  5. Critically appraise – apply the criteria in Part A; note agreements and contradictions.
  6. Synthesise thematically – group studies by theme, theory, method or finding rather than by author; compare and contrast; explain why results differ.
  7. Identify the research gap – the unresolved issue that justifies the new study.
  8. Write the review – structure it as: introduction (topic, purpose, roadmap), body (themes with evidence and critique), conclusion (summary of knowledge, gap, implications).
  9. Cite and reference properly – follow APA style consistently; avoid plagiarism.

Conclusion

Good evaluation asks whether a study is relevant, credible, methodologically sound and current; good reviewing goes further by synthesising the evidence into themes and exposing the gap. A review that merely describes studies is weak; one that analyses and integrates them provides the theoretical and empirical foundation for the research.

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  • 2025 · Research Methodology (Econ. 561) · 5 marks

What are the differences between Concept and Variables? How do you convert concepts into variables?

Answer

Differences between Concept and Variables

BasisConceptVariable
MeaningA general, abstract idea or mental image of a phenomenonA measurable characteristic that takes different values
NatureAbstract and not directly measurableConcrete and measurable
ValuesHas no numerical valueTakes two or more values
ExamplePoverty, intelligence, development, satisfactionPoverty line income, IQ score, GDP per capita, satisfaction score (1–5)
Role in researchDefines what is to be studiedDefines how it is measured and analysed
LevelConceptual/theoretical levelOperational/empirical level

In short, a concept is an idea; a variable is that idea made measurable so that data can be collected and tested.

Conversion of Concepts into Variables

A concept is converted into a variable through operationalisation — the process of defining an abstract concept in terms of measurable indicators. The steps are:

  1. Define the concept clearly — state exactly what is meant (e.g., poverty as inability to meet basic needs).
  2. Identify dimensions — break the concept into its components (e.g., income, consumption, health, education).
  3. Select indicators — choose observable measures for each dimension (e.g., per capita income, calorie intake, school enrolment).
  4. Assign measurement scale — decide whether the variable is nominal, ordinal, interval or ratio.
  5. Fix units and categories — e.g., income in Rs per month, or poor/non-poor classification.
  6. Pilot test and refine — check reliability and validity of the measure before final use.

Example: The concept "poverty" → dimension: economic deprivation → indicator: monthly per capita consumption expenditure → variable: per capita consumption expenditure in Rs per month (ratio scale), with a poverty line to classify households as poor or non-poor.

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  • 2025 · Research Methodology (Econ. 561) · 5 marks

What are the methods of quantitative data collection? Which method would you follow and why? [2+3]

Answer

Methods of Quantitative Data Collection

Quantitative data (numerical, measurable information) is collected mainly through the following methods:

  1. Survey method — structured questionnaires or interviews administered to a sample (mail, telephone, online, face-to-face).
  2. Observation method — systematic recording of countable behaviour or events, often with checklists or rating scales.
  3. Experiment method — controlled manipulation of variables to measure cause–effect relationships (lab or field experiments).
  4. Secondary data method — use of already-published numerical records: government statistics, central bank reports, census, surveys of national/international agencies.
  5. Administrative/by-product records — data generated from routine operations (school registers, hospital records, tax files).

Method I Would Follow and Why

For a study of a socio-economic issue such as household income, consumption or employment, I would follow the sample survey method using a structured questionnaire.

Reasons:

  • Representativeness: A properly drawn random sample allows valid generalization to the whole population, which observation or case records cannot give.
  • Standardization and comparability: Fixed questions and response options produce uniform numerical data that can be coded, tabulated and statistically analysed (mean, variance, regression).
  • Cost and time efficiency: It covers a large population at far lower cost and time than a census or a controlled experiment.
  • Flexibility of administration: It can be conducted face-to-face, by phone or online, and can include both closed questions (for quantitative measurement) and a few open questions for context.
  • High response quality when designed well: With trained enumerators and pre-testing, problems of non-response and measurement error can be minimized.

Thus, the sample survey is the most practical, reliable and widely used method for collecting quantitative data in economics, provided the sampling design and questionnaire are carefully prepared.

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  • 2025 · Research Methodology (Econ. 561) · 5 marks

What is Participatory Rural Appraisal? How does it work? [2+3]

Answer

Participatory Rural Appraisal (PRA)

Meaning (2 marks)

Participatory Rural Appraisal is a family of approaches and methods that enable rural people to share, enhance and analyse their own knowledge of life and conditions, to plan and to act. It emerged in the late 1980s and early 1990s, growing out of Rapid Rural Appraisal (RRA) and the participatory research traditions of Robert Chambers and others. Its defining feature is that the outside professional acts as a facilitator and learner rather than as an extractor of data: local people are treated as the primary analysts of their own situation, and the visual and verbal methods used are ones they can handle themselves. It is thus both a research approach and a planning/empowerment tool, widely used in Nepal by NGOs, donor projects, community forestry and local government planning.

How PRA works (3 marks)

  1. Entry and rapport building. The team first visits the community, meets key informants, explains the purpose and seeks consent. Trust and rapport are prerequisites; the team avoids a "surveyor" posture and behaves as a guest.

  2. Selection of methods and tools. A mix of flexible, visual, group-based tools is chosen according to the issue. Common tools include:

    • Social mapping — villagers draw their settlement, showing households, caste/ethnic clusters, roads, water points and services.
    • Resource mapping — land use, forest, irrigation and cropping patterns.
    • Transect walk — a systematic walk across the village with local guides to observe and discuss zones, soils and problems.
    • Seasonal calendar — cropping, labour, migration, food availability and disease by month.
    • Wealth ranking / well-being ranking — households classified by local criteria of poverty and wealth.
    • Venn diagram (institutional diagram) — local institutions and their importance shown as overlapping circles.
    • Matrix ranking and scoring — comparing options (crops, tree species, technologies) by locally chosen criteria.
    • Focus group discussions, key informant interviews, timelines and trend analysis.
  3. Triangulation and iteration. Findings from one tool or group are cross-checked against others (different methods, different informants, different sites) to reduce bias. The process is iterative — hypotheses are revised as understanding deepens.

  4. Analysis by the community. The outputs (maps, charts, rankings) remain with the community and are discussed in group meetings and plenary sessions, where villagers themselves identify problems, priorities and possible solutions. The outsider's role is to ask probing questions, not to supply answers.

  5. Planning, action and follow-up. The analysis feeds into a Participatory Rural Appraisal plan / community action plan, which is implemented with local institutions, monitored and reviewed. The process is meant to be continuous, with the community retaining ownership of both the information and the decisions.

Key principles: reversal of learning (learning from, not about, local people), offsetting biases (spatial, wealth, gender, seasonal, professional), optimal ignorance (not collecting more than is needed), triangulation, seeking diversity rather than averages, and facilitating local analysis.

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  • 2025 · Research Methodology (Econ. 561) · 5 marks

Explain Measurement scales.

Answer

Measurement Scales

Measurement scales are the ways in which variables (numbers or symbols) are assigned to objects or events according to a set of rules. They determine what statistical operations can legitimately be performed on the data. There are four levels of measurement, arranged from the weakest to the strongest: nominal, ordinal, interval, and ratio.

1. Nominal Scale This is the simplest and weakest level. Numbers or labels are used only to name, identify, or classify objects into mutually exclusive and collectively exhaustive categories. There is no order or ranking among the categories. The numbers have no quantitative meaning — they are merely tags. Examples: gender (male = 1, female = 2), religion, marital status, blood group, district of birth. Only the mode and frequency counts (chi-square, contingency tests) can be used; arithmetic operations like addition or averaging are meaningless.

2. Ordinal Scale Here the categories are ranked or ordered according to some criterion, but the intervals between ranks are not equal or known. We can say one item is greater or better than another, but not by how much. Examples: socio-economic status (low, middle, high), Likert-scale responses (strongly disagree to strongly agree), education level, ranking of preferences. The median and mode, along with non-parametric tests such as the Mann–Whitney U or Spearman's rank correlation, are appropriate.

3. Interval Scale This scale has ordered categories with equal intervals between consecutive points, so differences between values are meaningful. However, it lacks a true or absolute zero, so ratios are not meaningful. Examples: temperature in Celsius or Fahrenheit, IQ scores, calendar years. We can say 40°C is 10°C warmer than 30°C, but not that 40°C is twice as hot as 20°C. The mean, standard deviation, correlation, and t-tests can be applied.

4. Ratio Scale This is the highest and most complete level. It possesses all the properties of the interval scale plus a true, meaningful absolute zero, which indicates the complete absence of the attribute. Hence all arithmetic operations, including multiplication and division (ratios), are valid. Examples: income, weight, height, age, number of children, GDP. Statements such as "A earns twice as much as B" are meaningful. Almost all statistical techniques are applicable.

Summary: The nominal scale only classifies, the ordinal scale classifies and ranks, the interval scale adds equal distances, and the ratio scale further adds a true zero. The choice of scale determines which statistical tools — from simple frequency counts to advanced parametric tests — can be validly used in economic and social research.

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  • 2025 · Research Methodology (Econ. 561) · 5 marks

What is a research proposal? Discuss three key functions of a research proposal.

Answer

Research proposal — A research proposal is a formal, written document that a researcher prepares before conducting a study. It outlines the research problem, objectives, research questions, hypotheses, methodology, scope, time frame and budget of the proposed study. It serves as a plan or blueprint that guides the researcher and seeks approval or funding from a supervisor, department or funding agency. In economics, a proposal typically states the topic (e.g., "Determinants of remittance use in Nepal"), the rationale, the data and the econometric method to be used.

Three key functions of a research proposal

  1. Planning and direction (blueprint function): It forces the researcher to think through the problem, objectives, hypotheses and methodology in advance. It specifies what data will be collected, from where, and how it will be analysed, so the study proceeds in a systematic way and avoids confusion and wasted effort.

  2. Communication and justification (persuasive function): It communicates the research idea to supervisors, departments, funding agencies or ethical review committees. It justifies why the study is worth doing — its significance, originality and contribution to economic knowledge or policy — and thereby helps secure approval, funding and cooperation.

  3. Evaluation and control (monitoring function): It provides a basis for evaluating the feasibility and quality of the study before it begins, and later for monitoring progress. Supervisors and funding bodies can judge whether the objectives, methods, budget and time schedule are realistic, and can check whether the completed work matches what was proposed.

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  • 2025 · Research Methodology (Econ. 561) · 5 marks

Distinguish between Preliminaries, Main Body and Appendices of a research report with suitable examples.

Answer

Preliminaries, Main Body and Appendices of a Research Report

A research report is organised into three broad divisions, each serving a distinct purpose.

1. Preliminaries (prefatory parts) These appear before the actual text and introduce the report to the reader. They include the title page, letter of transmittal, table of contents, list of tables and figures, and acknowledgements. They are not numbered as chapters and are written briefly. Example: In a study on "Determinants of Household Saving in Kathmandu Valley," the title page, table of contents and list of tables form the preliminaries.

2. Main Body (text) This is the core of the report, containing the actual research work in a logical sequence — introduction, statement of the problem, objectives, review of literature, methodology, data presentation and analysis, findings, conclusions and recommendations. It carries the main argument and evidence, and is divided into numbered chapters or sections. Example: The same study's chapters on research methodology, regression analysis of saving behaviour, and conclusions constitute the main body.

3. Appendices (supplementary parts) These come after the main body and contain supporting material that is too detailed or bulky to include in the text but is useful for verification. They include the questionnaire, raw data tables, computer outputs, mathematical derivations and the bibliography or references. Example: The sample questionnaire, SPSS regression output and the list of references attached at the end of the saving study form the appendices.

Key difference: Preliminaries help the reader locate and understand the report, the main body presents and analyses the research findings, and the appendices support and authenticate the findings with detailed evidence.

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  • 2025 · Research Methodology (Econ. 561) · 5 marks

"The collected data is not yet ready for analysis." Discuss the key steps a researcher should undertake to clean the data and make it suitable for analysis.

Answer

Data cleaning: key steps to make collected data ready for analysis

Raw data collected from the field is almost never ready for analysis. It usually contains errors, missing values, inconsistencies and outliers. Data cleaning (also called data preparation or pre-processing) is the process of detecting, correcting or removing these problems so that the data becomes accurate, complete, consistent and usable. The key steps are as follows:

1. Data editing and validation The first step is to check the raw data for completeness and accuracy. Editing may be field editing (done on the spot while collecting data) or central editing (done at the office after collection). The researcher checks whether all questions have been answered, whether entries are legible, and whether the recorded values fall within a reasonable range. Invalid entries — for example, an age of 250 years or a sex code of "5" — are identified and corrected or removed.

2. Handling missing values Missing data must be dealt with because most statistical techniques cannot handle blanks. The researcher first examines the pattern of missingness: whether it is random or systematic. Common remedies are:

  • Deletion: dropping the incomplete case (list-wise) or the variable (column-wise) if the loss is small.
  • Imputation: replacing the missing value with the mean, median or mode; with the value from a similar respondent (hot-deck); or with a model-based estimate (regression imputation). The choice depends on the amount and nature of missingness; careless imputation can bias results.

3. Removing duplicates and correcting inconsistencies Duplicate records (common in survey and secondary data) are detected and removed. Inconsistencies across related variables are also corrected — for example, a respondent recorded as "unmarried" but having a spouse, or age and date of birth that do not match. This ensures internal consistency of the data set.

4. Identifying and treating outliers Outliers are extreme values that may be genuine or due to data-entry mistakes. They are detected using box plots, z-scores or the interquartile range rule. Genuine outliers may be retained or handled by winsorizing/transformation; erroneous ones are corrected or deleted. Outliers matter because they distort means, variances and regression estimates.

5. Data coding and transformation Responses are converted into numerical codes suitable for software (for example, male = 1, female = 0). New variables may be created through transformation — recoding into categories, computing ratios, indices or scales, and standardising units (for example, converting all income figures to Rs per month). This step makes the data machine-readable and comparable.

6. Checking data entry and structure If data were entered manually, a verification or double-entry check is done to catch typing errors. The researcher also checks the structure of the data set — variable names, data types (numeric, string, date), and the correct number of observations — so that it can be imported into software such as SPSS, STATA or R without error.

7. Final screening and documentation Finally, the cleaned data set is screened once more using descriptive statistics (frequencies, means, ranges) to confirm that all values are plausible. Every cleaning decision — deletions, imputations, recodes — should be documented in a codebook or cleaning log so that the process is transparent and reproducible.

Conclusion Data cleaning is not a single act but a systematic sequence of editing, handling missing values, removing duplicates, treating outliers, coding, transforming and verifying the data. Only after these steps is the data set accurate, consistent and complete enough for meaningful statistical analysis; skipping them risks biased and unreliable conclusions.

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  • 2025 · Research Methodology (Econ. 561) · 5 marks

Write short notes on (any two) a) Footnotes and Endnotes b) Research in Economics c) Univariate Data Analysis

Answer

a) Footnotes and Endnotes

Footnotes and endnotes are supplementary notes used in research writing to provide additional information, explanations, citations, or comments without interrupting the main text.

Footnotes appear at the bottom of the same page on which the reference is made. They are separated from the main text by a short horizontal line. The reader can glance down the page and immediately see the note, so footnotes are convenient when the note is closely tied to the discussion on that page.

Endnotes appear together at the end of the chapter, article, or the whole document, usually under the heading "Notes." They keep the pages of the main text clean and uncluttered, but the reader has to turn to the end of the document to consult them.

Key points of difference:

BasisFootnotesEndnotes
LocationBottom of the same pageEnd of chapter/document
Reading easeImmediate, on the same pageRequires turning pages
Effect on textPage looks crowdedMain text looks clean
Best suited forDetailed page-specific commentsLong lists of citations

Both are indicated in the text by a superscript number (or symbol such as *, †) placed after the relevant word or sentence. In economics research, footnotes are commonly used to explain data sources, definitions of variables, and assumptions of a model, while endnotes are used when references are numerous. Their main advantage is that they keep the argument of the main text continuous and readable while still giving the reader the supporting detail.


b) Research in Economics

Research in Economics is a systematic and scientific process of inquiry aimed at discovering, explaining, and predicting economic behaviour and economic phenomena. It involves the collection, organisation, analysis, and interpretation of economic data to answer a well-defined question or to test a hypothesis.

Objectives:

  1. To describe economic phenomena (what is happening).
  2. To explain causal relationships among economic variables (why it happens).
  3. To predict future economic trends and outcomes.
  4. To provide a basis for policy formulation and decision-making.

Types:

  • Pure (basic) research — develops economic theory, e.g. testing the permanent income hypothesis.
  • Applied research — solves practical problems, e.g. studying the impact of a subsidy on rice production.
  • Descriptive vs. analytical, and quantitative vs. qualitative research are other common classifications.

Steps in economic research:

  1. Formulating the research problem.
  2. Reviewing the literature.
  3. Formulating hypotheses.
  4. Designing the research (sampling, data sources).
  5. Collecting data (primary or secondary).
  6. Analysing data using statistical and econometric tools.
  7. Interpreting results and drawing conclusions.
  8. Writing the research report.

Importance: It helps in testing economic theories, measuring relationships such as elasticity, demand, inflation and growth, evaluating government policies, and guiding planning and development in countries like Nepal. Its main limitations are the non-experimental nature of economic data, measurement errors, and the difficulty of controlling all variables.


c) Univariate Data Analysis

Univariate data analysis is the statistical analysis of a single variable at a time. The word "uni" means one and "variate" means variable, so the analysis deals with one characteristic of a set of observations — for example, the monthly income of 100 households, or the marks of 50 students in economics.

Purpose: To describe the central tendency, dispersion, and shape of the distribution of that one variable, and to detect outliers.

Main tools:

  1. Measures of central tendency — mean, median, mode. They show the typical or average value of the variable.
  2. Measures of dispersion — range, quartile deviation, mean deviation, variance, standard deviation, coefficient of variation. They show how spread out the values are.
  3. Measures of shape — skewness and kurtosis, which describe asymmetry and peakedness of the distribution.
  4. Graphical presentation — frequency tables, histograms, bar diagrams, pie charts, box plots, stem-and-leaf plots.
  5. Distribution fitting and testing — checking whether the data follow a normal or other theoretical distribution, using tests such as chi-square goodness of fit.

Uses in economics: describing income distribution, poverty and inequality (Lorenz curve, Gini coefficient), analysing price and consumption patterns, summarising survey data, and preparing the ground for bivariate and multivariate analysis.

Limitations: It ignores relationships between variables. A univariate analysis of income alone cannot show how income depends on education or family size; for that, bivariate or multivariate techniques are needed.

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  • 2015 · Research Methodology · 10 marks

Define the concept of "measurement" in research. Explain the types of measurement scales with suitable example.

Answer

Measurement in Research

Measurement is the process of assigning numerals or other symbols to the properties or characteristics of objects, persons, or events according to a set of rules. In research, measurement provides the quantitative or qualitative description of variables so that they can be analysed, compared, and interpreted systematically. For example, assigning the number "1" to male and "2" to female respondents is a measurement of the variable "gender". A good measurement must be valid (it measures what it is intended to measure) and reliable (it gives consistent results on repeated application).

Measurement in social science research is more difficult than in physical sciences because many variables (attitudes, satisfaction, motivation) are abstract and cannot be measured directly; they are measured through indicators or proxy variables.


Types of Measurement Scales

The four levels of measurement, arranged from the lowest to the highest order of precision, are as follows:

1. Nominal Scale

This is the simplest and most basic scale, in which numbers or labels are used only to name, identify or classify objects into mutually exclusive and collectively exhaustive categories. The numbers have no quantitative meaning — no order, no distance, no origin. Only the mode can be computed, and the appropriate statistics are frequency, percentage, and chi-square.

Example: Classification of respondents by religion — 1 = Hindu, 2 = Muslim, 3 = Buddhist, 4 = Christian; or by gender — 1 = Male, 2 = Female. Here "2" is not greater than "1"; the numerals are mere labels.

2. Ordinal Scale

This scale ranks or orders the objects according to the degree to which they possess a particular attribute. It tells us whether one item is greater or smaller than another, but not by how much. The intervals between ranks are unequal and unknown. Median, mode, percentile, and rank-order correlation can be used.

Example: Socio-economic status of households ranked as 1 = Low, 2 = Middle, 3 = High; or a Likert-scale response — Strongly Disagree, Disagree, Neutral, Agree, Strongly Agree. We know the order but not the exact distance between "Low" and "Middle".

3. Interval Scale

This scale has the properties of order and equal distance between consecutive points, so that the difference between any two values is meaningful. However, it has an arbitrary (non-absolute) zero; therefore ratios cannot be interpreted meaningfully. Arithmetic mean, standard deviation, and correlation can be computed.

Example: Measurement of temperature in Celsius or Fahrenheit — the difference between 20°C and 30°C equals that between 30°C and 40°C, but 40°C is not "twice as hot" as 20°C because 0°C does not mean the absence of temperature. Similarly, IQ scores and calendar years are interval data.

4. Ratio Scale

This is the highest level of measurement. It possesses all the properties of the interval scale plus an absolute or true zero, which represents the complete absence of the attribute. Hence all arithmetic operations, including multiplication and division, are meaningful. Geometric mean and coefficient of variation can also be used.

Example: Measurement of income, age, weight, height, or number of children. A person with an income of Rs 60,000 earns exactly twice the income of a person earning Rs 30,000, and zero income means no income at all.


Summary Table

ScalePropertiesMeaningful StatisticsExample
NominalClassification onlyFrequency, percentage, modeGender, religion
OrdinalClassification + orderMedian, percentile, rank correlationSocio-economic rank, Likert scale
IntervalOrder + equal intervalMean, SD, correlationTemperature (°C), IQ
RatioOrder + interval + true zeroAll statistics incl. geometric meanIncome, age, weight

Conclusion: The choice of measurement scale determines the statistical techniques applicable to the data. A researcher must therefore select the appropriate scale in accordance with the nature of the variable and the objectives of the study.

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  • 2015 · Research Methodology · 10 marks

Describe the major components of a research proposal.

Answer

Major Components of a Research Proposal

A research proposal is a written plan that outlines what the researcher intends to study, why it is worth studying, and how it will be carried out. In economics, it serves both as a guide for the researcher and as a document for obtaining approval, funding, or academic supervision. The major components are described below.

1. Title of the Study

The title should be clear, specific, and self-explanatory. It should indicate the subject matter, the variables involved, and the scope of the study. A good title is concise and reflects the core research problem.

2. Introduction and Background

This section presents the context of the study. It explains the broader economic setting, the historical or theoretical background of the problem, and the relevance of the topic. It gradually narrows down from the general context to the specific area of inquiry.

3. Statement of the Problem

Here the researcher states precisely what problem is being investigated. It identifies the gap between the existing situation and the desired situation, points out deficiencies in existing knowledge, and formulates the issue in clear, researchable terms. A well-stated problem clarifies the direction of the entire study.

4. Objectives of the Study

The objectives state what the researcher intends to achieve. They should be specific, measurable, and consistent with the problem. A proposal usually has one general objective and a few specific objectives that break it into operational parts.

5. Research Questions or Hypotheses

Research questions are the specific queries the study seeks to answer. Where the study is explanatory or theory-testing, these are converted into testable hypotheses—statements about the expected relationship between variables that can be accepted or rejected on the basis of evidence.

6. Rationale and Significance of the Study

This section explains why the study matters. It states the academic contribution (filling a knowledge gap), the policy relevance (guiding decision-makers), and the practical usefulness for institutions, planners, or society at large.

7. Literature Review

The proposal reviews relevant theoretical and empirical literature. It summarises earlier studies, identifies their findings, methods, and limitations, and shows how the present study differs from or extends them. This establishes the theoretical foundation and avoids duplication.

8. Conceptual and Theoretical Framework

The researcher presents the concepts, variables, and their expected relationships, often in the form of a diagram or model. This framework links the research problem to established economic theory and guides the selection of variables and methods.

9. Research Methodology

This is the core of the proposal and typically includes:

  • Research design — descriptive, analytical, explanatory, or exploratory.
  • Nature and sources of data — primary (survey, interview) or secondary (CBS, NRB, World Bank reports).
  • Sampling design — population, sample size, and sampling technique.
  • Methods of data collection — questionnaire, interview schedule, observation, or documentary sources.
  • Methods of data analysis — descriptive statistics, regression, correlation, or other econometric tools.
  • Model specification — the functional form of the model and definition of variables.

10. Scope and Limitations of the Study

The scope defines the geographical area, time period, and subject coverage. The limitations acknowledge constraints such as time, budget, data availability, and the assumptions under which the findings will hold.

11. Chapter Plan or Organisation of the Study

This outlines how the final report will be structured—introduction, literature review, methodology, results and discussion, and conclusion with recommendations.

12. Work Schedule (Time Schedule)

A time frame, often presented in a Gantt chart, showing the sequence and duration of each activity from proposal writing to final report submission.

13. Budget Estimate

An estimate of the financial requirements—stationery, travel, data collection, data processing, and miscellaneous costs—usually presented in a table.

14. References

A list of all sources cited in the proposal, arranged alphabetically and formatted in a standard style such as APA.

Conclusion

These components together make a proposal systematic, feasible, and academically sound. A well-prepared proposal reduces uncertainty during research, ensures efficient use of resources, and provides a clear basis for evaluating the completed study.

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  • 2015 · Research Methodology · 10 marks

Describe the descriptive method of data analysis.

Answer

Descriptive Method of Data Analysis

Meaning. The descriptive method of data analysis is the set of statistical techniques used to organise, summarise and present data in a compact and meaningful form so that the basic features of the data can be understood without drawing inferences beyond the collected data. It answers the question "what is in the data?" rather than "what can we conclude about the population?" It is the first and essential step of any economic analysis, because data must be cleaned and summarised before inferential or causal analysis is attempted.

Objectives.

  1. To reduce a large mass of raw observations into a few meaningful figures.
  2. To reveal the central tendency, dispersion, shape and interrelationships present in the data.
  3. To make comparison between groups, regions or time periods possible.
  4. To present the findings in tables, graphs and summary numbers for policy use.

Main tools of the descriptive method.

(a) Measures of central tendency. These locate the typical value of the series:

  • Arithmetic mean, Xˉ=∑XN\bar{X} = \dfrac{\sum X}{N} (or ∑fX∑f\dfrac{\sum fX}{\sum f} for grouped data)
  • Median — the positional middle value, useful for skewed distributions
  • Mode — the most frequently occurring value, useful for qualitative and discrete data
  • Geometric and harmonic means — used for growth rates and index numbers

(b) Measures of dispersion. These show how far the observations spread around the average:

  • Range and quartile deviation
  • Mean deviation
  • Variance, σ2=∑(X−Xˉ)2N\sigma^2 = \dfrac{\sum (X-\bar{X})^2}{N}
  • Standard deviation, σ=∑(X−Xˉ)2N\sigma = \sqrt{\dfrac{\sum (X-\bar{X})^2}{N}}
  • Coefficient of variation, CV=σXˉ×100CV = \dfrac{\sigma}{\bar{X}} \times 100, used to compare variability across groups with different means

(c) Measures of shape. Skewness (asymmetry of the distribution) and kurtosis (peakedness) describe the form of the frequency curve. Karl Pearson's coefficient of skewness is Sk=Xˉ−MoσSk = \dfrac{\bar{X} - M_o}{\sigma}.

(d) Measures of relationship. Simple correlation coefficient, r=N∑XY−(∑X)(∑Y)[N∑X2−(∑X)2][N∑Y2−(∑Y)2]r = \dfrac{N\sum XY - (\sum X)(\sum Y)}{\sqrt{[N\sum X^2 - (\sum X)^2][N\sum Y^2 - (\sum Y)^2]}}, and simple linear regression measure the degree and direction of association between two variables.

(d) Tabular and graphical presentation. Frequency distributions, cross-tabulations, bar diagrams, pie charts, histograms, frequency polygons, ogives and time-series line graphs present the data visually. In economics, Lorenz curves and scatter diagrams are also used.

(e) Index numbers and growth rates. Wholesale and consumer price indices, GDP deflators and compound growth rates summarise changes over time in a single figure.

Steps in the descriptive method.

  1. Data collection — from primary sources (survey, field observation) or secondary sources (CBS, NRB, World Bank).
  2. Editing and cleaning — checking for omissions, inconsistencies and outliers.
  3. Classification and tabulation — arranging data by attribute, time, or class interval.
  4. Computation of summary measures — averages, dispersion, correlation, index numbers.
  5. Presentation — tables, diagrams and graphs.
  6. Interpretation — drawing simple conclusions about the pattern, level and variability of the data.

Uses in economics.

  • Measuring national income, inflation, unemployment and poverty levels.
  • Comparing per capita income, HDI or consumption across provinces and countries.
  • Studying the trend and seasonal pattern of prices, remittances, exports and agricultural output.
  • Providing the factual base for planning, budgeting and policy formulation.
  • Supplying the input data for inferential statistics (estimation and hypothesis testing).

Limitations.

  1. It describes only the collected sample or series; it cannot generalise to the population.
  2. It cannot establish cause and effect — correlation is not causation.
  3. Results depend heavily on the quality, completeness and representativeness of the data.
  4. Summary measures may hide important details and are sensitive to extreme values.
  5. It is not suitable for testing hypotheses or for forecasting.

Conclusion. The descriptive method is the foundation of economic data analysis. By converting raw data into averages, measures of dispersion, tables and graphs, it gives a clear picture of the level, variability and trend of economic variables. It must, however, be supplemented by inferential and econometric methods when the objective is generalisation, hypothesis testing or causal explanation.

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  • 2015 · Research Methodology · 5 marks

Account the significance of research in Economics.

Answer

Significance of Research in Economics

Research occupies a central place in economics, both as a scientific discipline and as a guide to policy. Its significance can be accounted under the following points:

1. Development of economic theory. Economic theories are not speculative constructions; they are built and tested through research. The classical, Keynesian, and monetarist revolutions in economic thought each emerged from systematic observation and analysis of real economic phenomena. Research thus supplies the empirical foundation on which theory stands or falls.

2. Formulation of economic policy. Sound policy requires evidence. Whether the issue is inflation control, taxation reform, employment generation, or poverty reduction, planners and policymakers depend on research findings to choose instruments, estimate their effects, and avoid unintended consequences. Research converts political intentions into workable economic measures.

3. Solution of economic problems. Problems such as unemployment, inequality, balance-of-payments deficits, food insecurity, and energy crisis are complex and multidimensional. Research identifies their causes, measures their magnitude, and evaluates alternative remedies, thereby enabling rational rather than ad hoc decision-making.

4. Measurement of economic variables and performance. National income accounting, price indices, poverty lines, human development indices, and growth rates all rest on research-based methodology. Reliable measurement is a precondition for comparison over time and across countries, and for judging whether an economy is progressing or regressing.

5. Formulation and evaluation of development plans. In a developing economy like Nepal, periodic plans, budget allocations, and donor-supported programmes require feasibility studies, baseline surveys, and impact evaluations. Research informs project selection, resource allocation, and mid-course correction.

6. Business and managerial decision-making. Firms use market research, demand forecasting, cost analysis, and consumer surveys to decide on pricing, production, investment, and marketing. Research reduces business uncertainty and improves competitiveness.

7. Testing and refinement of economic hypotheses. Through quantitative techniques such as regression, time-series analysis, and econometric modelling, research verifies whether postulated relationships hold in practice. This makes economics a progressive, self-correcting science rather than a body of fixed dogma.

8. Understanding of economic behaviour. Research in microeconomics, behavioural economics, and institutional economics explains how consumers, producers, households, and institutions actually behave, which is essential for designing effective incentives and regulations.

9. International comparison and learning. Comparative research allows countries to learn from one another's successes and failures, and provides the evidence base for international trade negotiations, foreign aid, and global economic cooperation.

10. Academic and professional development. Research generates new knowledge, trains economists in analytical skills, and supplies the literature on which teaching, further study, and professional practice depend.

Conclusion. Research is therefore not an optional academic luxury but the backbone of economic science and economic policy. It provides the facts without which theory is empty and policy is blind, and it is indispensable for a developing economy seeking sustained, inclusive, and evidence-based growth.

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  • 2015 · Research Methodology · 5 marks

Describe the major types of research design used in social research.

Answer

Major Types of Research Design in Social Research

A research design is the overall plan or blueprint that guides the collection, measurement, and analysis of data. In social research, designs are broadly grouped into three major types according to the purpose of the study.

1. Exploratory (Formulative) Research Design Used when the problem is new, vague, or little understood. Its aim is to gain familiarity with the phenomenon and generate hypotheses rather than test them. Common methods include literature review, case studies, key informant interviews, focus group discussions, and pilot surveys. It is flexible, unstructured, and relies mainly on qualitative data. Example: an exploratory study on the livelihood strategies of climate migrants in Karnali.

2. Descriptive Research Design Used to describe the characteristics of a population, situation, or event as it exists, answering the questions what, who, where, when, and how much. It involves careful definition of the population, sample, and variables, and uses surveys, censuses, correlational studies, and comparative studies. It does not manipulate variables, so it cannot establish causation. Example: a survey describing the level of financial literacy among rural households in Nepal.

3. Explanatory (Causal or Experimental) Research Design Used to explain why relationships exist and to test causal hypotheses. It is of two kinds:

  • Experimental design — the researcher manipulates the independent variable and controls extraneous factors, using control and treatment groups with random assignment (e.g., a randomised field experiment on the effect of a cash transfer on school enrolment).
  • Quasi-experimental and non-experimental causal designs — used when randomisation is not possible; causal inference is drawn through statistical control, panel data, instrumental variables, or difference-in-differences (e.g., impact evaluation of a microcredit programme).

Supplementary classification

  • Exploratory, descriptive, and explanatory — based on purpose.
  • Cross-sectional and longitudinal — based on the time dimension; longitudinal designs include trend, cohort, and panel studies.
  • Quantitative and qualitative — based on the nature of data; mixed-method designs combine both.

In practice, these types form a continuum: exploratory work generates hypotheses, descriptive work measures them, and explanatory work tests them. The choice depends on the research question, objectives, available time and resources, and the nature of the problem being studied.

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  • 2015 · Research Methodology · 5 marks

Discuss the procedure for x2x^2 test.

Answer

Procedure for the χ2\chi^2 (Chi-square) Test

The χ2\chi^2 test is a non-parametric test used to examine whether observed frequencies differ significantly from expected frequencies. Its procedure involves the following steps:

1. Set up the hypotheses

  • Null hypothesis (H0H_0): There is no significant difference between observed and expected frequencies (the variables are independent / the data fits the claimed distribution).
  • Alternative hypothesis (H1H_1): There is a significant difference between observed and expected frequencies.

2. Choose the level of significance (α\alpha) Usually α=0.05\alpha = 0.05 or 0.010.01 is selected, along with the appropriate degrees of freedom.

3. Calculate the expected frequencies (EE) Expected frequency for each cell is obtained as:

E=(Row total)×(Column total)Grand totalE = \frac{(\text{Row total}) \times (\text{Column total})}{\text{Grand total}}

4. Compute the value of χ2\chi^2 Using the formula:

χ2=∑(O−E)2E\chi^2 = \sum \frac{(O - E)^2}{E}

where OO = observed frequency and EE = expected frequency. For each cell, the deviation (O−E)(O-E) is squared, divided by EE, and all such values are summed.

5. Determine the degrees of freedom (df)

  • For a goodness-of-fit test: df=n−1df = n - 1 (or n−k−1n - k - 1 when parameters are estimated).
  • For a contingency table: df=(r−1)(c−1)df = (r - 1)(c - 1), where rr = number of rows and cc = number of columns.

6. Find the critical (table) value of χ2\chi^2 From the χ2\chi^2 table, at the chosen α\alpha and computed dfdf, the critical value χα,df2\chi^2_{\alpha, df} is located.

7. Compare and decide

  • If calculated χ2>\chi^2 > table χ2\chi^2, reject H0H_0 (significant difference exists).
  • If calculated χ2<\chi^2 < table χ2\chi^2, accept H0H_0 (no significant difference).

8. Conclusion The result is interpreted in the context of the problem, stating whether the observed data are consistent with the expected pattern or whether the variables are associated.

Note: The test assumes that expected frequencies are not too small (generally E≥5E \geq 5 for each cell); otherwise, adjacent categories should be pooled.

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  • 2015 · Research Methodology · 5 marks

Mention different factors a researcher should consider while using Focus Group Discussion (FGD) for qualitative data collection.

Answer

Factors to Consider While Using Focus Group Discussion (FGD) for Qualitative Data Collection

  1. Purpose and research question — FGD is suitable when the aim is to explore shared experiences, group norms, perceptions, and attitudes rather than individual biographies. It should not be used where sensitive personal or confidential matters are involved.

  2. Composition of the group (homogeneity vs. heterogeneity) — Participants should be similar enough (age, sex, caste/ethnicity, occupation, education) to feel comfortable, but different enough to generate diverse views. Mixing power levels (e.g., a boss with subordinates) suppresses free expression.

  3. Group size — Ideally 6–10 participants. Too few limits discussion; too many prevents everyone from speaking and makes moderation difficult.

  4. Number of groups (saturation) — Usually 3–6 FGDs per category of participants, continued until new themes stop emerging (theoretical saturation).

  5. Selection and recruitment of participants — Purposive or snowball sampling is used. Participants must be knowledgeable about the topic and willing to talk; incentives and timing of recruitment matter.

  6. Skill and role of the moderator — The moderator must be neutral, familiar with the topic, able to probe, control dominant speakers, encourage quiet ones, and avoid leading questions. A separate note-taker is desirable.

  7. Venue and physical setting — A neutral, accessible, quiet place where participants sit in a circle; seating should not create hierarchy. Travel cost and convenience affect attendance.

  8. Timing and duration — Sessions should suit participants' daily schedule (farm/office hours, festivals) and last about 60–90 minutes without fatigue.

  9. Language and cultural sensitivity — Use the local language and culturally acceptable terms; respect taboos, gender norms, and local power structures.

  10. Discussion guide (topic guide) — A semi-structured guide with open-ended questions and probes keeps the discussion focused while allowing flexibility.

  11. Recording and documentation — Audio/video recording with prior consent, plus field notes on non-verbal behaviour; ethical approval and confidentiality must be ensured.

  12. Ethical issues — Informed consent, voluntary participation, anonymity, and the right to withdraw; special care when discussing sensitive topics.

  13. Data management and analysis — Transcription, translation, coding, and thematic analysis; group interaction itself is data, so notes on agreement/disagreement are important.

  14. Resources and logistics — Budget, time, transport, refreshments, and manpower required for conducting and transcribing multiple sessions.

  15. Researcher bias and reflexivity — Awareness of the researcher's own assumptions, and avoidance of dominating the discussion or steering responses.

Considering these factors ensures that FGD yields valid, reliable, and contextually rich qualitative data.

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  • 2015 · Research Methodology · 5 marks

Describe the significance of "Formatting" for writing a good research report.

Answer

Significance of Formatting in a Research Report (5 marks)

Formatting refers to the systematic arrangement of the content of a research report — its title page, headings, sub-headings, tables, figures, citations, references, spacing, and pagination — according to an accepted style (APA, MLA, Chicago, etc.). Its significance for a good research report can be described as follows:

  1. Ensures clarity and readability. Proper headings, sub-headings, margins, and line spacing guide the reader logically from one section to another. A well-formatted report is easy to read and understand, whereas a poorly formatted one confuses the reader even if the content is sound.

  2. Maintains logical organisation and coherence. Formatting arranges the report into standard parts — preliminary pages (title page, certificate, acknowledgements, table of contents, list of tables/figures), main body (introduction, literature review, methodology, results, discussion, conclusion), and supplementary pages (references, appendices). This structure helps the reader locate information quickly and see the relationship among ideas.

  3. Establishes academic credibility and professionalism. A neat, consistent format reflects the researcher's care, discipline, and seriousness. It creates a good impression on examiners, supervisors, and readers, and increases the acceptability of the report in academic and professional circles.

  4. Facilitates proper citation and avoids plagiarism. Formatting prescribes standard rules for in-text citations and the reference list. This gives due credit to original authors, protects the researcher from charges of plagiarism, and allows other scholars to trace the sources used.

  5. Aids effective presentation of data. Tables, graphs, charts, and figures are numbered, titled, and placed systematically under formatting rules. This makes quantitative findings attractive, comparable, and self-explanatory, strengthening the evidence base of the report.

  6. Ensures uniformity and comparability. When reports follow a common style, different studies can be compared, evaluated, and reused by other researchers. It also makes the work acceptable to journals, universities, and funding agencies that insist on a prescribed format.

Conclusion: Formatting is not mere decoration; it is an essential requirement that gives a research report order, clarity, credibility, and communicative effectiveness. A report with excellent content but poor formatting fails to create the impact it deserves.

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  • 2015 · Research Methodology · 5 marks

Highlight the importance of literature review in scientific research.

Answer

Importance of Literature Review in Scientific Research

A literature review is a systematic survey of existing knowledge on a topic. In scientific research it is not merely a formality but the foundation on which the entire study rests. Its importance can be highlighted as follows:

  1. Identification of the research gap. By reviewing what has already been studied, the researcher discovers what remains unexplored. This gap becomes the justification and the problem statement of the new study.

  2. Formulation of research objectives and hypotheses. Existing theories and findings guide the framing of precise objectives, research questions and testable hypotheses. In economics, for example, a review of consumption theories helps in specifying the variables and the expected relationships.

  3. Theoretical and conceptual framework. The review supplies the concepts, definitions and theories (e.g., Keynesian, monetarist, new classical) that give the study its analytical base and help interpret results.

  4. Selection of appropriate methodology. Earlier studies reveal which research designs, sampling techniques, models and econometric tools have worked, and which have failed. This saves the researcher from repeating known mistakes and justifies the chosen method.

  5. Avoidance of duplication. It prevents the wasteful repetition of research already conducted and directs effort towards new contributions.

  6. Comparison and validation of findings. The results of the present study can be compared with those of previous studies to confirm, contradict or extend existing knowledge, thereby adding to the cumulative body of science.

  7. Identification of variables and data sources. The review points to the relevant independent and dependent variables, their measurement, and reliable sources of secondary data such as CBS reports, NRB publications and international databases.

  8. Identification of inconsistencies and debates. Conflicting findings in the literature expose weaknesses in earlier work and open the door to refinement and new research questions.

  9. Justification of the study's significance. A well-documented review shows that the research is timely, relevant and useful to policy makers, academics and practitioners.

  10. Improvement of research quality and credibility. A thorough review demonstrates the researcher's command over the field, sharpens the arguments, and increases the acceptability of the study in the scientific community.

Conclusion. Thus, the literature review provides direction, theoretical grounding, methodological guidance and scholarly credibility to scientific research. Without it, research becomes blind, repetitive and weak in both design and contribution.

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  • 2015 · Research Methodology · 5 marks

Construct a contingency table showing the relationship between two variables with hypothetical data.

Answer

Contingency table: education level and employment status

A contingency table (cross-tabulation) shows the joint frequency of two categorical variables. Suppose 200 persons are surveyed, with XX = education level (Graduate, Postgraduate) and YY = employment status (Employed, Unemployed).

Education levelEmployedUnemployedRow total
Graduate7050120
Postgraduate602080
Column total13070200

Marginals. Row totals: 120 graduates and 80 postgraduates. Column totals: 130 employed and 70 unemployed.

Row percentages (conditional distribution).

Education levelEmployed (%)Unemployed (%)
Graduate70/120 = 58.350/120 = 41.7
Postgraduate60/80 = 75.020/80 = 25.0

Relationship. Postgraduates have a higher employment rate (75.0%) than graduates (58.3%). The unemployment rate of graduates (41.7%) is about 1.7 times that of postgraduates (25.0%). The data therefore suggest a positive association between higher education and employment.

Test of independence. Whether the association is statistically significant can be checked with the chi-square test:

χ2=∑(Oij−Eij)2Eij,Eij=(row total)i(column total)jN\chi^2 = \sum \frac{(O_{ij}-E_{ij})^2}{E_{ij}}, \qquad E_{ij} = \frac{(\text{row total})_i(\text{column total})_j}{N}

For Graduate-Employed, E11=120×130200=78E_{11} = \dfrac{120 \times 130}{200} = 78, against an observed 70. The other expected values are 42 (Graduate-Unemployed), 52 (Postgraduate-Employed) and 28 (Postgraduate-Unemployed). Then

χ2=(70−78)278+(50−42)242+(60−52)252+(20−28)228≈0.82+1.52+1.23+2.29≈5.86\chi^2 = \frac{(70-78)^2}{78}+\frac{(50-42)^2}{42}+\frac{(60-52)^2}{52}+\frac{(20-28)^2}{28} \approx 0.82+1.52+1.23+2.29 \approx 5.86

With (2−1)(2−1)=1(2-1)(2-1)=1 degree of freedom, the 5% critical value is 3.84. Since 5.86>3.845.86 > 3.84, we reject independence: education and employment status are associated in this hypothetical sample.

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  • 2015 · Research Methodology · 5 marks

What is sampling? Distinguish between probability and non-probability sampling.

Answer

Sampling

Sampling is the statistical process of selecting a subset of individuals (called a sample) from a larger population for the purpose of estimating characteristics (parameters) of that whole population. Instead of studying every unit in the population (a census), the researcher studies a manageable part and draws inferences about the whole. A good sample must be representative, unbiased, and of adequate size so that the estimates are reliable and generalisable.

Probability vs. Non-probability Sampling

BasisProbability SamplingNon-probability Sampling
DefinitionEvery unit of the population has a known, non-zero probability of being selected.The probability of selection of units is not known; selection depends on the researcher's judgement or convenience.
SelectionSelection is done by a random (mechanical) process.Selection is done by personal judgement, convenience, or quota.
RepresentativenessSample is generally representative of the population.Sample may or may not be representative.
Sampling errorSampling error can be measured and controlled.Sampling error cannot be measured.
BiasFree from selection bias.Highly prone to selection bias.
InferenceResults can be generalised to the population with a known degree of confidence.Results cannot be safely generalised beyond the sample.
Cost and timeRelatively costly and time-consuming.Cheaper, quicker, and simpler.
ExamplesSimple random sampling, systematic sampling, stratified sampling, cluster sampling.Convenience sampling, purposive (judgement) sampling, quota sampling, snowball sampling.

Conclusion: Probability sampling is preferred when statistical inference and generalisation are required, while non-probability sampling is used in exploratory or pilot studies where cost, time, or the unavailability of a sampling frame makes random selection impractical.

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  • 2015 · Research Methodology · 5 marks

Highlight the importance of "Editing" in Data Processing.

Answer

Importance of Editing in Data Processing

Editing is the stage of data processing that lies between data collection and coding/classification. It consists of inspecting the raw data (questionnaires, schedules, records) to detect and correct errors, omissions, inconsistencies and irrelevant entries before the data are tabulated and analysed. Its importance can be highlighted as follows:

  1. Ensures accuracy and reliability. Editing removes wrong entries, misrecorded figures and transcription mistakes, so the final results reflect the true situation of the respondents. Without editing, conclusions drawn from the data would be misleading.

  2. Ensures completeness. It detects missing responses and blank items. Wherever possible, the editor supplies the missing information by recalling the respondent or by inference, so that the number of usable observations does not shrink unnecessarily.

  3. Ensures consistency and comparability. Editing checks whether answers to related questions agree with one another — for example, age and marital status, income and expenditure, occupation and education. It also brings data to a uniform unit, definition and time reference, which makes different schedules comparable.

  4. Detects and eliminates irrelevant and duplicate data. Entries outside the scope of the enquiry, and repeated or double-counted records, are removed, keeping the data set homogeneous.

  5. Improves quality of tabulation and analysis. Clean, consistent data can be coded, classified and tabulated without difficulty. Errors that escape editing get multiplied during tabulation and distort averages, ratios and measures of dispersion.

  6. Saves time and cost at later stages. Correcting errors at the editing stage is far cheaper and simpler than tracing them back after tabulation or during interpretation. It prevents the need to repeat an entire survey.

  7. Increases the validity of statistical inference. Since sampling and estimation procedures assume correct observations, editing protects the validity of estimates, tests of hypothesis and forecasts based on the data.

  8. Makes data suitable for storage and future use. Edited and cleaned data can be safely stored in data banks and reused for comparison over time and space.

Conclusion. Editing is thus a quality-control step in data processing. It converts raw, defective information into a usable, accurate and consistent data set, and thereby forms the foundation on which the entire edifice of tabulation, analysis and interpretation rests.

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  • 2015 · Research Methodology · 5 marks

Distinguish between closed ended and open ended questions with suitable examples.

Answer

Closed-ended vs Open-ended Questions

Closed-ended questions are questions that restrict the respondent to a limited set of predetermined answers — typically "Yes/No", "True/False", or a choice from a fixed list of options. The respondent cannot answer beyond the given alternatives. They are also called structured or fixed-alternative questions.

Open-ended questions are questions that allow the respondent to answer freely in his or her own words. No alternatives are supplied; the respondent has full freedom to express opinions, reasons, or experiences in detail. They are also called unstructured or free-response questions.

Points of distinction

BasisClosed-ended QuestionsOpen-ended Questions
Form of answerFixed alternatives are givenFree answer in respondent's own words
Freedom to respondentRestricted to given optionsComplete freedom
Ease of answeringEasy; only tick or selectDifficult; respondent must frame the reply
Ease of coding and analysisEasy; answers are pre-codedDifficult; requires content analysis
Cost and timeLow; can be administered to manyHigh; needs probing and transcription
Depth of informationShallow but preciseDeep, rich, and qualitative
ReliabilityHigh; uniform for all respondentsLower; varies with respondent's expression
Risk of biasResponse bias due to forced choicesInterviewer bias and irrelevant answers
Suitable forLarge-scale surveys, factual dataExploratory studies, sensitive or complex issues

Suitable examples

Closed-ended:

  1. "Do you own a smartphone?" — (a) Yes (b) No
  2. "How often do you visit a bank?" — (a) Weekly (b) Monthly (c) Rarely (d) Never
  3. "Rate your satisfaction with the service on a 5-point scale."

Open-ended:

  1. "What do you think is the main cause of unemployment in Nepal?"
  2. "How has inflation affected your household expenditure?"
  3. "What suggestions do you have for improving the quality of higher education?"

Conclusion

Closed-ended questions are best when the researcher knows the possible responses in advance and needs quantifiable, comparable data from a large sample. Open-ended questions are best in exploratory research where the range of answers is unknown and the researcher seeks insight, depth, and new ideas. In practice, most questionnaires combine both types — closed-ended items for measurable facts and open-ended items for opinions and explanations.

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  • 2015 · Econometrics · 10 marks

Show that OLS estimators are best linear unbiased estimators.

Answer

Answer (10 marks)

Setup. Consider the classical linear regression model

Yi=β1+β2Xi+ui,i=1,2,…,nY_i=\beta_1+\beta_2X_i+u_i,\qquad i=1,2,\dots,n

with the classical assumptions: E(ui)=0E(u_i)=0, Var⁡(ui)=σ2\operatorname{Var}(u_i)=\sigma^2 (homoscedasticity), Cov⁡(ui,uj)=0\operatorname{Cov}(u_i,u_j)=0 for i≠ji\neq j, XiX_i non-random (fixed), and ∑Xi2\sum X_i^2 finite with ∑xi2>0\sum x_i^2>0, where xi=Xi−Xˉx_i=X_i-\bar X.

The OLS estimator of the slope is

β^2=∑xiyi∑xi2=∑wiYi,wi=xi∑xi2.\hat\beta_2=\frac{\sum x_iy_i}{\sum x_i^2}=\sum w_iY_i,\qquad w_i=\frac{x_i}{\sum x_i^2}.

We must prove that β^2\hat\beta_2 is linear, unbiased, and has minimum variance among all linear unbiased estimators (the Gauss–Markov theorem).


1. Linearity. Since β^2=∑wiYi\hat\beta_2=\sum w_iY_i, it is a linear function of the sample values YiY_i. Hence it is a linear estimator. ✓\checkmark


2. Unbiasedness. Substituting Yi=β1+β2Xi+uiY_i=\beta_1+\beta_2X_i+u_i:

β^2=∑wi(β1+β2Xi+ui)=β1∑wi+β2∑wiXi+∑wiui.\hat\beta_2=\sum w_i(\beta_1+\beta_2X_i+u_i)=\beta_1\sum w_i+\beta_2\sum w_iX_i+\sum w_iu_i.

Now ∑wi=∑xi∑xi2=0\sum w_i=\dfrac{\sum x_i}{\sum x_i^2}=0 (since ∑xi=0\sum x_i=0), and

∑wiXi=∑xiXi∑xi2=∑xi(Xi−Xˉ)∑xi2=∑xi2∑xi2=1.\sum w_iX_i=\frac{\sum x_iX_i}{\sum x_i^2}=\frac{\sum x_i(X_i-\bar X)}{\sum x_i^2}=\frac{\sum x_i^2}{\sum x_i^2}=1.

Therefore

β^2=β2+∑wiui.\hat\beta_2=\beta_2+\sum w_iu_i .

Taking expectations and using E(ui)=0E(u_i)=0:

E(β^2)=β2+∑wiE(ui)=β2.E(\hat\beta_2)=\beta_2+\sum w_iE(u_i)=\beta_2.

So β^2\hat\beta_2 is unbiased. ✓\checkmark


3. Minimum variance among linear unbiased estimators.

Let any other linear estimator of β2\beta_2 be

β~2=∑ciYi,\tilde\beta_2=\sum c_iY_i,

where the cic_i are fixed weights (functions of XX only). For unbiasedness we require

E(β~2)=β1∑ci+β2∑ciXi=β2for all β1,β2,E(\tilde\beta_2)=\beta_1\sum c_i+\beta_2\sum c_iX_i=\beta_2 \quad\text{for all }\beta_1,\beta_2,

which gives the two restrictions

∑ci=0,∑ciXi=1.\sum c_i=0,\qquad \sum c_iX_i=1.

Its variance is

Var⁡(β~2)=σ2∑ci2.\operatorname{Var}(\tilde\beta_2)=\sigma^2\sum c_i^2.

Write ci=wi+dic_i=w_i+d_i, where di=ci−wid_i=c_i-w_i. Then

∑di=∑ci−∑wi=0−0=0,\sum d_i=\sum c_i-\sum w_i=0-0=0, ∑diXi=∑ciXi−∑wiXi=1−1=0.\sum d_iX_i=\sum c_iX_i-\sum w_iX_i=1-1=0.

Now

∑ci2=∑(wi+di)2=∑wi2+2∑widi+∑di2.\sum c_i^2=\sum (w_i+d_i)^2=\sum w_i^2+2\sum w_id_i+\sum d_i^2.

The cross term vanishes:

∑widi=∑xidi∑xi2=∑xi(ci−wi)∑xi2=∑ciXi−Xˉ∑ci−∑wiXi1=1−0−1=0.\sum w_id_i=\frac{\sum x_id_i}{\sum x_i^2}=\frac{\sum x_i(c_i-w_i)}{\sum x_i^2}=\frac{\sum c_iX_i-\bar X\sum c_i-\sum w_iX_i}{1}=1-0-1=0.

Hence

Var⁡(β~2)=σ2∑ci2=σ2∑wi2+σ2∑di2=Var⁡(β^2)+σ2∑di2.\operatorname{Var}(\tilde\beta_2)=\sigma^2\sum c_i^2=\sigma^2\sum w_i^2+\sigma^2\sum d_i^2=\operatorname{Var}(\hat\beta_2)+\sigma^2\sum d_i^2.

Since σ2∑di2≥0\sigma^2\sum d_i^2\ge 0, we get

Var⁡(β~2) ≥ Var⁡(β^2),\operatorname{Var}(\tilde\beta_2)\ \ge\ \operatorname{Var}(\hat\beta_2),

with equality only when di=0d_i=0 for all ii, i.e. ci=wic_i=w_i and β~2=β^2\tilde\beta_2=\hat\beta_2. Thus no other linear unbiased estimator has smaller variance; β^2\hat\beta_2 is the best (minimum-variance) linear unbiased estimator. ✓\checkmark


4. The intercept. Similarly, β^1=Yˉ−β^2Xˉ\hat\beta_1=\bar Y-\hat\beta_2\bar X is linear in YiY_i with weights (1n−Xˉwi)\left(\frac1n-\bar X w_i\right); these weights sum to 11 and satisfy ∑(1n−Xˉwi)Xi=0\sum\left(\frac1n-\bar Xw_i\right)X_i=0, so β^1\hat\beta_1 is also unbiased. The same ci=wi+dic_i=w_i+d_i argument gives Var⁡(β~1)≥Var⁡(β^1)\operatorname{Var}(\tilde\beta_1)\ge\operatorname{Var}(\hat\beta_1).


Conclusion. The OLS estimators β^1\hat\beta_1 and β^2\hat\beta_2 are linear functions of YiY_i, are unbiased, and possess the least variance in the class of all linear unbiased estimators. Therefore, by the Gauss–Markov theorem, the OLS estimators are BLUE — Best Linear Unbiased Estimators.

Answer checked by Claude Haiku · Oct 21, 2026

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  • 2015 · Econometrics · 10 marks

What is autocorrelation? Describe its consequences, detection and remedial measures.

Answer

Autocorrelation

Autocorrelation (also called serial correlation) is the correlation of a time series with its own lagged values. In the context of the classical linear regression model, it means that the error terms of different observations are correlated with one another:

E(utus)≠0for t≠sE(u_t u_s) \neq 0 \quad \text{for } t \neq s

The classical model assumes no autocorrelation, i.e. E(utus)=0E(u_t u_s) = 0 for all t≠st \neq s. When this assumption is violated, the model suffers from autocorrelation. It is mainly a problem of time-series data, though it can also arise in cross-section data (e.g. grouped or spatially ordered data).

Causes:

  1. Inertia (sluggishness) — economic time series such as GDP, prices, employment move slowly and smoothly, so successive error terms carry over from one period to the next.
  2. Omitted variables — if a relevant variable is left out, its influence gets absorbed into the error term, which then shows a systematic pattern.
  3. Wrong functional form — using a linear form when the true relation is non-linear produces serially correlated errors.
  4. Cobweb phenomenon — supply in period tt depends on price in period t−1t-1, generating a pattern in residuals.
  5. Data manipulation — interpolation, extrapolation or smoothing of data creates artificial serial correlation.
  6. Lagged dependent variable — including Yt−1Y_{t-1} as a regressor with autocorrelated errors.

Consequences of Autocorrelation

  1. OLS estimators remain unbiased and consistent — as long as XX is non-stochastic or uncorrelated with uu, the Gauss–Markov unbiasedness result still holds. Autocorrelation does not destroy unbiasedness.

  2. OLS estimators are no longer efficient (not BLUE) — the Gauss–Markov theorem requires E(utus)=0E(u_t u_s) = 0. With autocorrelation, OLS is no longer the best linear unbiased estimator; another estimator (e.g. GLS) has smaller variance.

  3. Underestimation of standard errors — the usual OLS formula Var(β^)=σ2(X′X)−1\text{Var}(\hat\beta) = \sigma^2 (X'X)^{-1} is wrong. In the presence of positive autocorrelation, the true variance is larger, so the computed standard errors are biased downward.

  4. Inflation of tt and FF statistics — because standard errors are understated, the tt-ratios are overstated. This leads to spurious significance: variables may appear statistically significant when they are not. The R2R^2 may also be inflated.

  5. Confidence intervals and hypothesis tests become invalid — since the estimated variances are biased, confidence intervals are too narrow and tests of significance are unreliable.

  6. Forecasts may be inefficient — though forecasts can still be unbiased, they are not minimum-variance, so prediction intervals are unreliable.

  7. Loss of efficiency in prediction and policy conclusions — wrong standard errors can lead to wrong policy recommendations.


Detection of Autocorrelation

(a) Graphical method Plot the residuals u^t\hat u_t against time tt. If they show a systematic pattern (runs of positive and negative residuals), autocorrelation is suspected. A plot of u^t\hat u_t against u^t−1\hat u_{t-1} also reveals correlation.

(b) Runs test Count the number of runs (sequences of same-sign residuals). Compare with the expected number of runs under randomness. Too few runs indicate positive autocorrelation.

(c) Durbin–Watson (DW) test — the most widely used test.

d=∑t=2n(u^t−u^t−1)2∑t=1nu^t2d = \frac{\sum_{t=2}^{n}(\hat u_t - \hat u_{t-1})^2}{\sum_{t=1}^{n}\hat u_t^2}

  • dd ranges from 0 to 4; E(d)≈2(1−ρ^)E(d) \approx 2(1-\hat\rho).
  • d≈2d \approx 2 → no autocorrelation; d→0d \to 0 → positive autocorrelation; d→4d \to 4 → negative autocorrelation.
  • Compare dd with lower (dLd_L) and upper (dUd_U) critical values from Durbin–Watson tables:
    • If d<dLd < d_L → reject H0H_0 (positive autocorrelation present).
    • If d>dUd > d_U → do not reject H0H_0.
    • If dL≤d≤dUd_L \le d \le d_U → inconclusive.
  • Limitations: applies only to first-order autocorrelation, assumes an intercept term, cannot be used when a lagged dependent variable appears as a regressor, and may give inconclusive results.

(d) Durbin's hh test Used when the regressors include a lagged dependent variable Yt−1Y_{t-1}:

h=ρ^n1−n⋅Var(β^Yt−1)h = \hat\rho \sqrt{\frac{n}{1 - n \cdot \text{Var}(\hat\beta_{Y_{t-1}})}}

where ρ^=1−d/2\hat\rho = 1 - d/2. Under H0H_0, h∼N(0,1)h \sim N(0,1).

(e) Breusch–Godfrey (LM) test Regress u^t\hat u_t on the original regressors and lagged residuals u^t−1,u^t−2,…,u^t−p\hat u_{t-1}, \hat u_{t-2}, \dots, \hat u_{t-p}. Test statistic (n−p)R2∼χp2(n-p)R^2 \sim \chi^2_p. It detects higher-order autocorrelation and works even with lagged dependent variables.

(f) Box–Pierce / Ljung–Box QQ test Tests joint significance of several autocorrelations of residuals using the autocorrelation function (ACF) and partial autocorrelation function (PACF).


Remedial Measures

  1. Check for omitted variables and misspecification — if autocorrelation is due to a wrong functional form or a missing variable, respecify the model (add the omitted variable, use log or polynomial form). This often removes the problem at its source.

  2. Cochrane–Orcutt iterative procedure — for first-order autocorrelation ut=ρut−1+εtu_t = \rho u_{t-1} + \varepsilon_t:

    • Estimate ρ\rho from the DW statistic: ρ^=1−d/2\hat\rho = 1 - d/2 (or by regressing u^t\hat u_t on u^t−1\hat u_{t-1}).
    • Transform the variables: Yt∗=Yt−ρ^Yt−1Y_t^* = Y_t - \hat\rho Y_{t-1}, Xt∗=Xt−ρ^Xt−1X_t^* = X_t - \hat\rho X_{t-1}.
    • Run OLS on the transformed variables and iterate until ρ^\hat\rho converges.
  3. Prais–Winsten transformation — similar to Cochrane–Orcutt but preserves the first observation (using 1−ρ2\sqrt{1-\rho^2}) instead of dropping it, so no observation is lost.

  4. First-difference method — when ρ=1\rho = 1 (perfect positive autocorrelation), take first differences: ΔYt=β2ΔXt+Δut\Delta Y_t = \beta_2 \Delta X_t + \Delta u_t. This removes the trend and the autocorrelation.

  5. Generalized Least Squares (GLS) — the theoretically correct estimator when the form of the error covariance matrix Ω\Omega is known. GLS is BLUE. In practice, feasible GLS (FGLS) is used by estimating ρ\rho first.

  6. Newey–West procedure — when the exact form of autocorrelation is unknown, use Newey–West heteroskedasticity- and autocorrelation-consistent (HAC) standard errors. This keeps OLS estimates but corrects the standard errors, making tt and FF tests valid.

  7. Include lagged dependent variable — if autocorrelation arises from dynamics, adding Yt−1Y_{t-1} as a regressor may absorb the serial dependence.

  8. Use of ARIMA / dynamic models — in time-series econometrics, model the error process explicitly as AR(pp) or MA(qq).


Conclusion: Autocorrelation violates a key assumption of the classical linear regression model. Although OLS estimates remain unbiased and consistent, they lose efficiency and their standard errors are biased, making hypothesis tests and confidence intervals unreliable. It must therefore be detected through graphical methods, the Durbin–Watson test, Durbin's hh test, the Breusch–Godfrey LM test or the Ljung–Box QQ test, and corrected through respecification, the Cochrane–Orcutt/Prais–Winsten procedures, first differencing, GLS/FGLS or Newey–West standard errors.

Answer checked by Claude Haiku · Oct 21, 2026

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  • 2015 · Econometrics · 10 marks

Given the following information
N=9N = 9 ΣY=1052\Sigma Y = 1052 ΣX2=1,017\Sigma X_2 = 1,017 ΣX3=954\Sigma X_3 = 954 ΣYX2=110,750\Sigma Y X_2 = 110,750 ΣX2X3=111,433\Sigma X_2 X_3 = 111,433 ΣX2Y=107,606\Sigma X_2 Y = 107,606 ΣY2=124,288\Sigma Y^2 = 124,288 ΣX22=115,371\Sigma X_2^2 = 115,371 ΣX32=101,772\Sigma X_3^2 = 101,772
Then,
a. Estimate the linear relationship Y=β1+β2X2+β3X3+UY = \beta_1 + \beta_2 X_2 + \beta_3 X_3 + U b. How do you interpret the estimated coefficients? c. Estimate the variance covariance matrix of estimated β\beta d. Test the hypothesis that β2=0\beta_2 = 0

Answer

Note on the data. The question lists ΣYX2=110,750\Sigma YX_2 = 110{,}750 and ΣX2Y=107,606\Sigma X_2Y = 107{,}606, which are the same quantity written twice. The second is taken to be ΣYX3=107,606\Sigma YX_3 = 107{,}606. Even with this reading the data turn out to be internally inconsistent (shown below), so the numbers are a mechanical application of the formulas and not a valid fit.

(a) Estimation

With N=9N = 9, the deviation-form sums are:

Σy2=124288−105229=1320.89,Σx22=115371−101729=450,Σx32=101772−95429=648\Sigma y^2 = 124288 - \frac{1052^2}{9} = 1320.89,\quad \Sigma x_2^2 = 115371 - \frac{1017^2}{9} = 450,\quad \Sigma x_3^2 = 101772 - \frac{954^2}{9} = 648

Σx2x3=111433−1017×9549=3631,Σyx2=110750−1052×10179=−8126,Σyx3=107606−1052×9549=−3906\Sigma x_2x_3 = 111433 - \frac{1017\times954}{9} = 3631,\quad \Sigma yx_2 = 110750 - \frac{1052\times1017}{9} = -8126,\quad \Sigma yx_3 = 107606 - \frac{1052\times954}{9} = -3906

Normal equations:

450 β^2+3631 β^3=−8126,3631 β^2+648 β^3=−3906450\,\hat\beta_2 + 3631\,\hat\beta_3 = -8126,\qquad 3631\,\hat\beta_2 + 648\,\hat\beta_3 = -3906

The determinant is 450×648−36312=−12,890,161450\times648 - 3631^2 = -12{,}890{,}161, so

β^2=(−8126)(648)−(3631)(−3906)−12890161=−0.692,β^3=(450)(−3906)−(3631)(−8126)−12890161=−2.152\hat\beta_2 = \frac{(-8126)(648) - (3631)(-3906)}{-12890161} = -0.692,\qquad \hat\beta_3 = \frac{(450)(-3906) - (3631)(-8126)}{-12890161} = -2.152

With Yˉ=116.89\bar Y = 116.89, Xˉ2=113\bar X_2 = 113, Xˉ3=106\bar X_3 = 106:

β^1=116.89+0.692(113)+2.152(106)≈423.18\hat\beta_1 = 116.89 + 0.692(113) + 2.152(106) \approx 423.18

Y^=423.18−0.692X2−2.152X3\hat Y = 423.18 - 0.692X_2 - 2.152X_3

Why the data are inconsistent. The correlation between X2X_2 and X3X_3 would be 3631/450×648≈6.73631/\sqrt{450\times648} \approx 6.7, and the correlation between YY and X2X_2 would be −8126/450×1320.89≈−10.5-8126/\sqrt{450\times1320.89}\approx -10.5. A correlation cannot exceed 1 in absolute value. The matrix x′xx'x has a negative determinant, so it is not positive definite. Some of the printed sums must be misprinted.

(b) Interpretation

  • β^1\hat\beta_1: the value of YY when X2=X3=0X_2 = X_3 = 0; it has meaning only if zero lies in the sample range.
  • β^2=−0.692\hat\beta_2 = -0.692: holding X3X_3 constant, a one-unit rise in X2X_2 lowers YY by about 0.69 units on average.
  • β^3=−2.152\hat\beta_3 = -2.152: holding X2X_2 constant, a one-unit rise in X3X_3 lowers YY by about 2.15 units on average.

(c) Variance-covariance matrix

RSS=Σy2−β^2Σyx2−β^3Σyx3=1320.89−5620.6−8406.3≈−12,706RSS = \Sigma y^2 - \hat\beta_2\Sigma yx_2 - \hat\beta_3\Sigma yx_3 = 1320.89 - 5620.6 - 8406.3 \approx -12{,}706

A sum of squared residuals cannot be negative, which confirms the inconsistency. Following the formulas mechanically, σ^2=RSS/(N−3)=−2117.7\hat\sigma^2 = RSS/(N-3) = -2117.7 and

Var-Cov^(β^2,β^3)=σ^2(45036313631648)−1=(0.1064−0.5964−0.59640.0739)\widehat{\text{Var-Cov}}(\hat\beta_2,\hat\beta_3) = \hat\sigma^2\begin{pmatrix}450&3631\\3631&648\end{pmatrix}^{-1} = \begin{pmatrix}0.1064 & -0.5964\\ -0.5964 & 0.0739\end{pmatrix}

These entries come from multiplying two negative numbers, so they have no statistical meaning. With correct data, σ^2=RSS/6\hat\sigma^2 = RSS/6 would be positive and the matrix is computed in exactly this way.

(d) Test of H0:β2=0H_0:\beta_2 = 0 against H1:β2≠0H_1:\beta_2\neq0

t=β^2se(β^2)=−0.6920.1064≈−2.12t = \frac{\hat\beta_2}{\text{se}(\hat\beta_2)} = \frac{-0.692}{\sqrt{0.1064}} \approx -2.12

The critical value is t0.025,6=2.447t_{0.025,6} = 2.447. Formally ∣t∣=2.12<2.447|t| = 2.12 < 2.447, so H0H_0 would not be rejected at 5 percent. Because the variance was obtained from a negative RSS, this decision cannot be relied on. The conclusion is that the data must be corrected before any inference is drawn.

Answer checked by Claude Sonnet · Oct 21, 2026

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  • 2015 · Econometrics · 5 marks

What is heteroscedasticity? Explain its consequences.

Answer

Heteroscedasticity is the violation of the classical assumption of homoscedasticity — the assumption that the error term in a regression model has constant variance across all observations. Under heteroscedasticity, the variance of the disturbance term is not constant but varies with the values of the explanatory variable(s):

Var(ui)=σi2≠σ2(varies across i)\text{Var}(u_i) = \sigma_i^2 \neq \sigma^2 \quad \text{(varies across } i\text{)}

For example, in a study of household consumption on income, low-income households show small variation in consumption while high-income households show large variation — the error variance grows with income. This is a common problem in cross-sectional data (heteroscedasticity is usually absent in time-series data).

Consequences:

  1. OLS estimators remain unbiased and consistent — the Gauss–Markov theorem's unbiasedness result does not require homoscedasticity, so β^\hat{\beta} is still unbiased and consistent.

  2. OLS estimators are no longer BLUE (efficient) — they lose the minimum-variance property. The OLS estimator is no longer the best linear unbiased estimator; the generalized least squares (GLS) estimator is more efficient.

  3. Standard errors are biased and inconsistent — the usual OLS formula Var(β^)=σ2(X′X)−1\text{Var}(\hat{\beta}) = \sigma^2 (X'X)^{-1} is wrong, so tt and FF statistics are invalid. This leads to misleading hypothesis tests — we may wrongly accept or reject hypotheses (Type I or Type II errors).

  4. Confidence intervals become unreliable — since they are built on the biased standard errors, interval estimates lose their stated coverage probability.

  5. Forecasting and prediction may be inefficient — though predictions remain unbiased, they are not minimum-variance, so prediction intervals are unreliable.

  6. R2R^2 and goodness-of-fit measures may be misleading — the usual interpretation of R2R^2 as the proportion of variance explained becomes questionable.

Remedies (brief): use White's heteroscedasticity-consistent standard errors, weighted least squares (WLS)/GLS, or transform the variables (e.g., log transformation) to stabilize variance.

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  • 2015 · Econometrics · 5 marks

What is multicollinearity? Describe its remedial measures.

Answer

Multicollinearity is the situation in multiple regression where two or more explanatory variables are highly (but not perfectly) linearly related, so that one can be approximated as a linear combination of the others. If the relation is exact, it is perfect multicollinearity and OLS cannot be estimated at all; if it is near-exact, OLS estimates exist but become unreliable.

For the model Yi=β1+β2X2i+β3X3i+uiY_i = \beta_1 + \beta_2 X_{2i} + \beta_3 X_{3i} + u_i, multicollinearity means X2X_2 and X3X_3 move so closely together that the model cannot separate their individual effects.

Consequences

  • OLS estimators remain unbiased and BLUE, but their variances and standard errors become very large, so tt-ratios fall and variables appear insignificant.
  • Estimated coefficients become unstable — small changes in data or dropping one observation can change signs and magnitudes sharply.
  • Confidence intervals become wide; the overall FF-test may be significant while individual tt-tests are not.
  • The determinant of X′XX'X approaches zero, so (X′X)−1(X'X)^{-1} is ill-conditioned and rounding errors get magnified.

Detection

  • High pairwise correlation coefficients among regressors.
  • Variance Inflation Factor: VIFj=11−Rj2\text{VIF}_j = \dfrac{1}{1-R_j^2}, where Rj2R_j^2 is from regressing XjX_j on the other regressors. VIF>10\text{VIF} > 10 (or tolerance <0.1< 0.1) signals serious multicollinearity.
  • Large changes in coefficients when a variable is added or removed; insignificant tt-ratios with high R2R^2.

Remedial measures

  1. Drop one of the collinear variables — the simplest remedy when theory permits, since the retained variable captures the common effect. The cost is possible omitted-variable bias.
  2. Use extraneous information — prior estimates or theoretical restrictions on coefficients (e.g. constant returns to scale, β2+β3=1\beta_2 + \beta_3 = 1) reduce the number of free parameters.
  3. Increase the sample size — more observations raise the variation in XX and lower estimator variances; often the most practical long-run solution.
  4. Respecify the model — use first differences, ratios, or combine the collinear variables into an index (e.g. replace X2,X3X_2, X_3 by X2/X3X_2/X_3 or by their sum).
  5. Ridge regression — add a small positive constant kk to the diagonal of X′XX'X: β^ridge=(X′X+kI)−1X′Y\hat{\beta}_{\text{ridge}} = (X'X + kI)^{-1}X'Y. This trades a little bias for a large reduction in variance.
  6. Principal component analysis — replace the collinear set by a few orthogonal principal components and regress on these.
  7. Do nothing — if the purpose is prediction rather than measuring individual effects, and R2R^2 is high, the inflated variances may be tolerable.

In practice the choice depends on whether the study aims at prediction (multicollinearity is then harmless) or at estimating separate structural coefficients (then remedies 1–6 are needed).

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  • 2015 · Econometrics · 5 marks

Describe the dummy variable trap.

Answer

Dummy Variable Trap

The dummy variable trap is the problem of perfect multicollinearity that arises when a set of dummy variables used to represent a qualitative variable with (m) categories is included in a regression model without omitting one category.

The problem

If a qualitative variable has (m) categories, we create (m) dummy variables — one for each category. Each dummy takes the value 1 if the observation belongs to that category and 0 otherwise. Since every observation belongs to exactly one category, the sum of all (m) dummies equals 1 for every observation:

D1+D2+⋯+Dm=1D_1 + D_2 + \cdots + D_m = 1

If the model also contains an intercept term (a column of ones), then the intercept column is an exact linear combination of the dummy columns. The regressor matrix (X) therefore does not have full column rank, and (X'X) becomes singular (its determinant is zero). Consequently, ((X'X)^{-1}) does not exist, so the OLS estimator

β^=(X′X)−1X′Y\hat{\beta} = (X'X)^{-1}X'Y

cannot be computed. This is the dummy variable trap — a case of perfect multicollinearity.

The solution

The trap is avoided by dropping one dummy variable (the reference or base category). If there are (m) categories, we include only (m-1) dummies along with the intercept. The omitted category is captured by the intercept, and each estimated dummy coefficient then measures the difference in the mean value of the dependent variable between that category and the base category. This gives a unique set of OLS estimates.

Illustration

For a variable "education level" with three categories — Primary, Secondary, Tertiary — we define:

  • (D_1 = 1) if Primary, else 0
  • (D_2 = 1) if Secondary, else 0
  • (D_3 = 1) if Tertiary, else 0

Including all three plus an intercept gives (D_1 + D_2 + D_3 = 1), so the trap occurs. Dropping (D_3) (base = Tertiary) and estimating

Y=β0+β1D1+β2D2+uY = \beta_0 + \beta_1 D_1 + \beta_2 D_2 + u

avoids the trap. Here (\beta_0) is the mean for Tertiary, (\beta_0 + \beta_1) for Primary, and (\beta_0 + \beta_2) for Secondary.

Key points

  • The trap is perfect multicollinearity, not merely high correlation.
  • It occurs only when all (m) dummies are included together with an intercept.
  • If the model is estimated without an intercept, all (m) dummies may be included without the trap.
  • The standard remedy is to omit one category and interpret coefficients relative to that base.

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  • 2015 · Econometrics · 5 marks

Explain the Logit model and its use in economics.

Answer

Logit Model and Its Use in Economics

The logit model is a binary choice model used when the dependent variable is qualitative and takes only two values, usually 0 or 1. It estimates the probability that an event occurs as a function of one or more explanatory variables.

Specification. Let Y=1Y = 1 if an event occurs (e.g., a person is employed) and Y=0Y = 0 otherwise. The logit model assumes the probability of occurrence is given by the logistic (cumulative logistic) function:

Pi=P(Yi=1∣Xi)=11+e−(β1+β2Xi)=eZi1+eZi,Zi=β1+β2XiP_i = P(Y_i = 1 \mid X_i) = \frac{1}{1 + e^{-(\beta_1 + \beta_2 X_i)}} = \frac{e^{Z_i}}{1 + e^{Z_i}}, \quad Z_i = \beta_1 + \beta_2 X_i

The probability of non-occurrence is 1−Pi=11+eZi1 - P_i = \dfrac{1}{1 + e^{Z_i}}. Taking the ratio,

Pi1−Pi=eZi\frac{P_i}{1 - P_i} = e^{Z_i}

This ratio is the odds of the event occurring. Taking natural logs gives the logit or log-odds:

Li=ln⁡(Pi1−Pi)=Zi=β1+β2XiL_i = \ln\left(\frac{P_i}{1 - P_i}\right) = Z_i = \beta_1 + \beta_2 X_i

So the logit model is linear in parameters in the log-odds form, though non-linear in the probabilities. PiP_i lies between 0 and 1, while LiL_i ranges from −∞-\infty to +∞+\infty, which makes estimation convenient. Parameters are estimated by the maximum likelihood (ML) method rather than OLS, since OLS gives inefficient estimates and predicted probabilities outside [0, 1] for a linear probability model.

Interpretation. β2\beta_2 measures the change in the log-odds (logit) per unit change in XX. The odds change by the factor eβ2e^{\beta_2}; if β2>0\beta_2 > 0, XX raises the probability of Y=1Y = 1, and if β2<0\beta_2 < 0, it lowers it. The marginal effect on probability is β2Pi(1−Pi)\beta_2 P_i (1 - P_i), which varies with XiX_i.

Uses in economics.

  1. Labour economics — analysing labour force participation, employment status, union membership, and migration decisions.
  2. Consumer choice — brand choice, purchase/no-purchase decisions, and adoption of new products.
  3. Finance and banking — credit default or loan repayment prediction, where the dependent variable is default (1) or no default (0).
  4. Public economics — whether a household is below the poverty line, whether a person pays tax, or uses a particular public service.
  5. Health and education — school enrolment, dropout, contraceptive use, and health insurance coverage.
  6. Programme evaluation — estimating the probability of participation in a development programme.

Limitations. It assumes the error term follows the logistic distribution; results are sensitive to outliers and to the choice of explanatory variables; and coefficients are not directly readable as probability effects. Despite these, the logit model is widely used because it is simple, gives probabilities within the valid range, and permits meaningful odds-ratio interpretation.

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  • 2015 · Econometrics · 5 marks

Explain the Koyck distributed lag model. How do you estimate it?

Answer

Koyck Distributed Lag Model

The Model

The Koyck distributed lag model (Koyck, 1954) is a geometric (infinite) distributed lag model. It assumes that the effect of the explanatory variable XX on the dependent variable YY declines geometrically over time:

Yt=α+β0Xt+β1Xt−1+β2Xt−2+⋯+utY_t = \alpha + \beta_0 X_t + \beta_1 X_{t-1} + \beta_2 X_{t-2} + \cdots + u_t

Koyck assumed that all the β\beta coefficients have the same sign and decline at a constant rate λ\lambda (0<λ<10 < \lambda < 1):

βk=β0λkfor k=0,1,2,…\beta_k = \beta_0 \lambda^k \quad \text{for } k = 0, 1, 2, \dots

So the model becomes:

Yt=α+β0Xt+β0λXt−1+β0λ2Xt−2+⋯+utY_t = \alpha + \beta_0 X_t + \beta_0 \lambda X_{t-1} + \beta_0 \lambda^2 X_{t-2} + \cdots + u_t

Here λ\lambda is the geometric decay rate (or Koyck lag parameter). The closer λ\lambda is to 1, the slower the decay; the closer to 0, the faster the decay. The long-run (total) multiplier is:

∑k=0∞βk=β01−λ\sum_{k=0}^{\infty} \beta_k = \frac{\beta_0}{1-\lambda}

The Koyck Transformation (Estimation)

Because the model has infinitely many lagged terms and cannot be estimated directly, Koyck used a transformation to reduce it to a finite model.

Step 1: Lag the original model by one period and multiply by λ\lambda:

λYt−1=λα+β0λXt−1+β0λ2Xt−2+⋯+λut−1\lambda Y_{t-1} = \lambda\alpha + \beta_0 \lambda X_{t-1} + \beta_0 \lambda^2 X_{t-2} + \cdots + \lambda u_{t-1}

Step 2: Subtract this from the original model:

Yt−λYt−1=α(1−λ)+β0Xt+(ut−λut−1)Y_t - \lambda Y_{t-1} = \alpha(1-\lambda) + \beta_0 X_t + (u_t - \lambda u_{t-1})

Step 3: Rearranging gives the estimable Koyck model:

Yt=α(1−λ)+β0Xt+λYt−1+vtY_t = \alpha(1-\lambda) + \beta_0 X_t + \lambda Y_{t-1} + v_t

where vt=ut−λut−1v_t = u_t - \lambda u_{t-1} is a moving-average error term.

Estimation Procedure

The transformed model is estimated by OLS in the form:

Yt=α∗+β0∗Xt+λ∗Yt−1+vtY_t = \alpha^* + \beta_0^* X_t + \lambda^* Y_{t-1} + v_t

where α∗=α(1−λ)\alpha^* = \alpha(1-\lambda) and λ∗=λ\lambda^* = \lambda.

From the estimated coefficients we recover the structural parameters:

  • λ^=λ^∗\hat{\lambda} = \hat{\lambda}^* (the coefficient on Yt−1Y_{t-1})
  • α^=α^∗1−λ^\hat{\alpha} = \dfrac{\hat{\alpha}^*}{1 - \hat{\lambda}}
  • β^0\hat{\beta}_0 is read directly, and the long-run multiplier is β^01−λ^\dfrac{\hat{\beta}_0}{1-\hat{\lambda}}

Problems in Estimation

  1. Serial correlation: The error term vt=ut−λut−1v_t = u_t - \lambda u_{t-1} is autocorrelated even if utu_t is not, so OLS estimates are biased and inconsistent.
  2. Correlated regressor and error: Yt−1Y_{t-1} depends on ut−1u_{t-1}, which appears in vtv_t, so Yt−1Y_{t-1} is correlated with the error term — OLS is inconsistent.
  3. Loss of one observation and possible multicollinearity.

Remedies

  • Use instrumental variables (e.g., Xt−1X_{t-1} or Xt−2X_{t-2}) for Yt−1Y_{t-1}.
  • Apply the method of moments or maximum likelihood estimation.
  • Use the Liviatan or Wallis procedure to correct for autocorrelation.

Thus, the Koyck model converts an infinite distributed lag into a simple autoregressive form that can be estimated, but its estimates must be corrected for the induced autocorrelation and endogeneity.

Answer checked by Claude Haiku · Oct 21, 2026

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  • 2015 · Econometrics · 5 marks

What is linear probability model (LPM)? Show that normality assumption for disturbance term does not hold true in LPM.

Answer

Linear Probability Model (LPM)

The linear probability model is a model in which a binary (dummy) dependent variable is regressed on one or more explanatory variables by ordinary least squares. If the dependent variable YY takes the value 1 with probability pp (occurrence of an event) and 0 with probability 1−p1-p (non-occurrence), the LPM is written as

Yi=β1+β2Xi+uiY_i = \beta_1 + \beta_2 X_i + u_i

where E(Yi∣Xi)=β1+β2Xi=piE(Y_i \mid X_i) = \beta_1 + \beta_2 X_i = p_i is the conditional probability that the event occurs. Since YiY_i is a Bernoulli random variable, the model is also called the linear probability model because the probability of the event is a linear function of XX.

Normality assumption does not hold in LPM

The classical normal linear regression model assumes that the disturbance term uiu_i follows a normal distribution, ui∼N(0,σ2)u_i \sim N(0, \sigma^2). In the LPM this assumption fails. The reasoning is as follows:

  1. The disturbance term is defined as ui=Yi−(β1+β2Xi)u_i = Y_i - (\beta_1 + \beta_2 X_i), i.e. ui=Yi−piu_i = Y_i - p_i.

  2. Since YiY_i takes only two values, 1 and 0, the disturbance term also takes only two values:

    • When Yi=1Y_i = 1: ui=1−piu_i = 1 - p_i, which occurs with probability pip_i.
    • When Yi=0Y_i = 0: ui=−piu_i = -p_i, which occurs with probability 1−pi1 - p_i.
  3. Thus uiu_i is a discrete random variable with only two possible values; it is not continuous at all. A normal distribution is continuous and defined over the whole real line, so a two-point discrete distribution cannot be normal. Hence normality of the disturbance term is violated.

  4. The variance of the disturbance term is also not constant:

    Var(ui)=pi(1−pi)=(β1+β2Xi)(1−β1−β2Xi)\text{Var}(u_i) = p_i(1 - p_i) = (\beta_1 + \beta_2 X_i)(1 - \beta_1 - \beta_2 X_i)

    This depends on XiX_i, so the disturbances are heteroscedastic. Hence the usual OLS standard errors, tt and FF tests are not valid, and the usual maximum likelihood (normal) framework cannot be applied.

Conclusion: In the LPM the disturbance term is a binary (Bernoulli) variable taking only two values, so it cannot be normally distributed; the normality assumption of the classical model therefore does not hold true.

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  • 2015 · Econometrics · 5 marks

Given the following simple linear stochastic relationship between XiX_i and YiY_i variables
Yi=α+βXi+Ui;Ui∼NID(0,σ2)Y_i = \alpha + \beta X_i + U_i; \quad U_i \sim NID(0, \sigma^2)
Estimate the model by maximum likelihood estimation method.

Answer

Maximum Likelihood Estimation of the Simple Linear Regression Model

Model and Assumptions

Yi=α+βXi+Ui;Ui∼NID(0,σ2),i=1,2,…,nY_i = \alpha + \beta X_i + U_i; \quad U_i \sim NID(0, \sigma^2), \quad i = 1, 2, \dots, n

Since Ui∼N(0,σ2)U_i \sim N(0, \sigma^2) and XiX_i is non-stochastic, the dependent variable is normally distributed as

Yi∼N(α+βXi, σ2)Y_i \sim N(\alpha + \beta X_i,\ \sigma^2)

Likelihood Function

The probability density function of YiY_i is

f(Yi)=1σ2πexp⁡[−12σ2(Yi−α−βXi)2]f(Y_i) = \frac{1}{\sigma\sqrt{2\pi}} \exp\left[-\frac{1}{2\sigma^2}(Y_i - \alpha - \beta X_i)^2\right]

Because the YiY_i are independently distributed, the joint density (likelihood function) is the product of the individual densities:

L(α,β,σ2)=∏i=1n1σ2πexp⁡[−12σ2(Yi−α−βXi)2]L(\alpha, \beta, \sigma^2) = \prod_{i=1}^{n} \frac{1}{\sigma\sqrt{2\pi}} \exp\left[-\frac{1}{2\sigma^2}(Y_i - \alpha - \beta X_i)^2\right]

L=(1σ2π)nexp⁡[−12σ2∑i=1n(Yi−α−βXi)2]L = \left(\frac{1}{\sigma\sqrt{2\pi}}\right)^{n} \exp\left[-\frac{1}{2\sigma^2}\sum_{i=1}^{n}(Y_i - \alpha - \beta X_i)^2\right]

Log-Likelihood Function

Taking natural logarithms:

ln⁡L=−nln⁡σ−n2ln⁡(2π)−12σ2∑i=1n(Yi−α−βXi)2\ln L = -n\ln\sigma - \frac{n}{2}\ln(2\pi) - \frac{1}{2\sigma^2}\sum_{i=1}^{n}(Y_i - \alpha - \beta X_i)^2

First-Order Conditions (Normal Equations)

With respect to α\alpha:

∂ln⁡L∂α=1σ2∑i=1n(Yi−α−βXi)=0\frac{\partial \ln L}{\partial \alpha} = \frac{1}{\sigma^2}\sum_{i=1}^{n}(Y_i - \alpha - \beta X_i) = 0

⇒∑Yi=nα+β∑Xi⋯(1)\Rightarrow \sum Y_i = n\alpha + \beta \sum X_i \quad \cdots (1)

With respect to β\beta:

∂ln⁡L∂β=1σ2∑i=1nXi(Yi−α−βXi)=0\frac{\partial \ln L}{\partial \beta} = \frac{1}{\sigma^2}\sum_{i=1}^{n}X_i(Y_i - \alpha - \beta X_i) = 0

⇒∑XiYi=α∑Xi+β∑Xi2⋯(2)\Rightarrow \sum X_i Y_i = \alpha \sum X_i + \beta \sum X_i^2 \quad \cdots (2)

With respect to σ2\sigma^2:

∂ln⁡L∂σ2=−n2σ2+12σ4∑i=1n(Yi−α−βXi)2=0\frac{\partial \ln L}{\partial \sigma^2} = -\frac{n}{2\sigma^2} + \frac{1}{2\sigma^4}\sum_{i=1}^{n}(Y_i - \alpha - \beta X_i)^2 = 0

⇒σ2=1n∑i=1n(Yi−α−βXi)2⋯(3)\Rightarrow \sigma^2 = \frac{1}{n}\sum_{i=1}^{n}(Y_i - \alpha - \beta X_i)^2 \quad \cdots (3)

Solving the Normal Equations

Dividing equation (1) by nn:

Yˉ=α+βXˉ⇒α=Yˉ−βXˉ⋯(3′)\bar{Y} = \alpha + \beta \bar{X} \quad \Rightarrow \quad \alpha = \bar{Y} - \beta \bar{X} \quad \cdots (3')

Substituting α\alpha into equation (2):

∑XiYi=(Yˉ−βXˉ)∑Xi+β∑Xi2\sum X_i Y_i = (\bar{Y} - \beta \bar{X})\sum X_i + \beta \sum X_i^2

∑XiYi=Yˉ∑Xi−βXˉ∑Xi+β∑Xi2\sum X_i Y_i = \bar{Y}\sum X_i - \beta \bar{X}\sum X_i + \beta \sum X_i^2

Since ∑Xi=nXˉ\sum X_i = n\bar{X}:

∑XiYi=nXˉYˉ−nβXˉ2+β∑Xi2\sum X_i Y_i = n\bar{X}\bar{Y} - n\beta \bar{X}^2 + \beta \sum X_i^2

∑XiYi−nXˉYˉ=β(∑Xi2−nXˉ2)\sum X_i Y_i - n\bar{X}\bar{Y} = \beta\left(\sum X_i^2 - n\bar{X}^2\right)

Therefore, the ML estimators are:

β^ML=∑(Xi−Xˉ)(Yi−Yˉ)∑(Xi−Xˉ)2=∑xiyi∑xi2\boxed{\hat{\beta}_{ML} = \frac{\sum (X_i - \bar{X})(Y_i - \bar{Y})}{\sum (X_i - \bar{X})^2} = \frac{\sum x_i y_i}{\sum x_i^2}}

α^ML=Yˉ−β^MLXˉ\boxed{\hat{\alpha}_{ML} = \bar{Y} - \hat{\beta}_{ML}\bar{X}}

σ^ML2=1n∑(Yi−α^ML−β^MLXi)2=∑ei2n\boxed{\hat{\sigma}^2_{ML} = \frac{1}{n}\sum (Y_i - \hat{\alpha}_{ML} - \hat{\beta}_{ML}X_i)^2 = \frac{\sum e_i^2}{n}}

where xi=Xi−Xˉx_i = X_i - \bar{X} and yi=Yi−Yˉy_i = Y_i - \bar{Y}.

Conclusion

The ML estimators of α\alpha and β\beta are identical to the OLS estimators. However, the ML estimator of σ2\sigma^2 is biased because it divides the residual sum of squares by nn instead of n−2n-2; the OLS estimator σ^2=∑ei2/(n−2)\hat{\sigma}^2 = \sum e_i^2/(n-2) is unbiased. The ML estimators are consistent and asymptotically efficient.

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  • 2015 · Econometrics · 5 marks

Describe adaptive expectation model. How do you estimate it?

Answer

Adaptive expectations model

The model, due to Cagan and Friedman, says that people revise their expectations of a variable in proportion to the error they made last period. Let Xt∗X_t^* be the expected (unobservable) value and XtX_t the actual value:

Xt∗−Xt−1∗=λ (Xt−Xt−1∗),0<λ≤1X_t^* - X_{t-1}^* = \lambda\,(X_t - X_{t-1}^*),\qquad 0<\lambda\le1

λ\lambda is the coefficient of expectation. If λ=1\lambda = 1 expectations are fully realised (Xt∗=XtX_t^*=X_t); if λ→0\lambda\to0 they never change. Rearranging,

Xt∗=λXt+(1−λ)Xt−1∗X_t^* = \lambda X_t + (1-\lambda)X_{t-1}^*

and by repeated substitution Xt∗=λ∑i=0∞(1−λ)iXt−iX_t^* = \lambda\sum_{i=0}^{\infty}(1-\lambda)^iX_{t-i}, a geometrically weighted average of past actual values, with the most recent weighted most.

Estimation (Koyck transformation)

Suppose the behavioural equation depends on the expected value:

Yt=β0+β1Xt∗+utY_t = \beta_0 + \beta_1X_t^* + u_t

Step 1. Substitute the expectation equation:

Yt=β0+β1λXt+β1(1−λ)Xt−1∗+utY_t = \beta_0 + \beta_1\lambda X_t + \beta_1(1-\lambda)X_{t-1}^* + u_t

Step 2. Lag the behavioural equation by one period and multiply by (1−λ)(1-\lambda):

(1−λ)Yt−1=(1−λ)β0+β1(1−λ)Xt−1∗+(1−λ)ut−1(1-\lambda)Y_{t-1} = (1-\lambda)\beta_0 + \beta_1(1-\lambda)X_{t-1}^* + (1-\lambda)u_{t-1}

Step 3. Subtract Step 2 from Step 1, which eliminates the unobservable Xt−1∗X_{t-1}^*:

Yt=λβ0+λβ1Xt+(1−λ)Yt−1+vt,vt=ut−(1−λ)ut−1\boxed{Y_t = \lambda\beta_0 + \lambda\beta_1X_t + (1-\lambda)Y_{t-1} + v_t},\qquad v_t = u_t - (1-\lambda)u_{t-1}

This equation contains only observable variables. Writing it as Yt=a0+a1Xt+a2Yt−1+vtY_t = a_0 + a_1X_t + a_2Y_{t-1} + v_t, the parameters are recovered as

λ^=1−a^2,β^1=a^1λ^,β^0=a^0λ^\hat\lambda = 1-\hat a_2,\qquad \hat\beta_1 = \frac{\hat a_1}{\hat\lambda},\qquad \hat\beta_0 = \frac{\hat a_0}{\hat\lambda}

Estimation problems

  • Yt−1Y_{t-1} is correlated with vtv_t through ut−1u_{t-1}, and vtv_t follows an MA(1) process. Plain OLS is therefore biased and inconsistent.
  • A consistent estimate uses instrumental variables, for example Xt−1X_{t-1} as an instrument for Yt−1Y_{t-1}, or maximum likelihood.
  • Because Yt−1Y_{t-1} is a regressor, the Durbin-Watson statistic is biased towards 2, so Durbin's hh test is used to check for autocorrelation.

Limitations

The model is backward-looking: it ignores new information and policy announcements, and agents can make systematic errors. This weakness, evident in the 1970s, led to the rational expectations critique of Lucas and Muth.

Answer checked by Claude Sonnet · Oct 21, 2026

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  • 2015 · Econometrics · 5 marks

What is coefficient of determination? How do you interpret it?

Answer

Coefficient of determination ((R^2))

The coefficient of determination, denoted (R^2), is a statistical measure that indicates the proportion of the total variation in the dependent variable ((Y)) that is explained by the variation in the independent variable(s) ((X)) in a regression model. It is computed as the ratio of the explained sum of squares (ESS) to the total sum of squares (TSS):

R2=ESSTSS=∑(Y^i−Yˉ)2∑(Yi−Yˉ)2=1−RSSTSSR^2 = \frac{ESS}{TSS} = \frac{\sum(\hat{Y}_i - \bar{Y})^2}{\sum(Y_i - \bar{Y})^2} = 1 - \frac{RSS}{TSS}

where (RSS = \sum(Y_i - \hat{Y}_i)^2) is the residual (unexplained) sum of squares. In simple linear regression, (R^2) is also the square of the correlation coefficient between (X) and (Y), i.e. (R^2 = r^2).

Interpretation

  1. Range: (R^2) lies between 0 and 1 ((0 \le R^2 \le 1)).
  2. Proportion explained: An (R^2) of, say, 0.85 means 85% of the variation in (Y) is explained by the regressor(s), while the remaining 15% is due to factors not included in the model (captured by the error term).
  3. Goodness of fit: The closer (R^2) is to 1, the better the regression line fits the data; the closer to 0, the weaker the explanatory power of the model.
  4. Limitations: A high (R^2) does not prove causation, nor does it guarantee a good model. (R^2) never decreases when more regressors are added, so in multiple regression the adjusted (R^2) ((\bar{R}^2 = 1 - \frac{RSS/(n-k)}{TSS/(n-1)})) is preferred, as it penalises the addition of irrelevant variables. Also, a low (R^2) is not necessarily bad if the estimated coefficients are statistically significant and theoretically meaningful.

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  • 2015 · Econometrics · 5 marks

Distinguish between ex-ante and ex-post forecast.

Answer

Ex-ante vs Ex-post Forecast

Ex-ante forecast (Latin ex ante = "before the event") is a prediction made before the actual event or period occurs. It uses past data, assumptions and economic models to estimate future values — for example, forecasting next year's inflation rate or next quarter's GDP growth. Since the future is unknown, an ex-ante forecast is always subject to uncertainty and is revised as new information arrives. It is the type of forecast used for planning, budgeting and policy formulation.

Ex-post forecast (Latin ex post = "after the event") is a forecast made after the event has already occurred, or a comparison of a past forecast with the actual realised value. It is essentially a backward-looking evaluation — for example, checking how close last year's predicted inflation was to the actual inflation recorded. It is used to validate the accuracy of forecasting models, measure forecast errors (such as RMSE or MAPE) and improve future predictions.

Key points of difference:

BasisEx-ante forecastEx-post forecast
TimingMade before the event/periodMade after the event/period
PurposePlanning, policy, decision-makingEvaluation, model validation
Data usedPast data + assumptions about the futureActual realised data
UncertaintyHigh — outcome unknownLow — outcome already known
NaturePredictive / forward-lookingEvaluative / backward-looking
ExampleForecasting 2025 inflation in 2024Comparing 2024 forecast with actual 2024 inflation

In short, ex-ante forecasting looks forward to guide decisions, while ex-post forecasting looks backward to judge how good those decisions and forecasts were.

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  • 2082 · Statistics for Economics (Econ. 553) · 15 marks

The joint probability distribution of the two random variables X and Y is given as:
X \ Y01
00.150.15
10.200.25
20.100.15
a) Obtain the marginal probability distributions of the random variables X and Y. (2) b) Compute the mean (expected value) and variance of both X and Y. (3) c) State the condition under which two random variables X and Y are said to be independent. Using this condition, examine whether X and Y in the above table are independent. If they are not independent, compute the correlation coefficient between X and Y. (5)

Answer

(a) Marginal probability distributions

Marginal distribution of (X):

[ P_X(x)=\sum_y P(X=x,Y=y) ]

(x)(P_X(x))
0(0.15+0.15=0.30)
1(0.20+0.25=0.45)
2(0.10+0.15=0.25)

So, [ P_X(0)=0.30,\quad P_X(1)=0.45,\quad P_X(2)=0.25 ]

Marginal distribution of (Y):

[ P_Y(y)=\sum_x P(X=x,Y=y) ]

(y)(P_Y(y))
0(0.15+0.20+0.10=0.45)
1(0.15+0.25+0.15=0.55)

So, [ P_Y(0)=0.45,\quad P_Y(1)=0.55 ]


(b) Mean and variance of (X) and (Y)

For (X):

[ E(X)=\sum_x xP_X(x) ]

[ E(X)=0(0.30)+1(0.45)+2(0.25)=0+0.45+0.50=0.95 ]

[ E(X^2)=\sum_x x^2P_X(x) ]

[ E(X^2)=0^2(0.30)+1^2(0.45)+2^2(0.25)=0+0.45+1.00=1.45 ]

[ \operatorname{Var}(X)=E(X^2)-[E(X)]^2 ]

[ \operatorname{Var}(X)=1.45-(0.95)^2=1.45-0.9025=0.5475 ]

Thus, [ E(X)=0.95,\qquad \operatorname{Var}(X)=0.5475 ]

For (Y):

[ E(Y)=\sum_y yP_Y(y) ]

[ E(Y)=0(0.45)+1(0.55)=0.55 ]

[ E(Y^2)=\sum_y y^2P_Y(y) ]

[ E(Y^2)=0^2(0.45)+1^2(0.55)=0.55 ]

[ \operatorname{Var}(Y)=E(Y^2)-[E(Y)]^2 ]

[ \operatorname{Var}(Y)=0.55-(0.55)^2=0.55-0.3025=0.2475 ]

Thus, [ E(Y)=0.55,\qquad \operatorname{Var}(Y)=0.2475 ]


(c) Independence condition and correlation coefficient

Two random variables (X) and (Y) are independent if and only if, for all values (x) and (y),

[ P(X=x,Y=y)=P_X(x)P_Y(y) ]

Now check the given table.

For example, for (x=0,\ y=0),

[ P(X=0,Y=0)=0.15 ]

But,

[ P_X(0)P_Y(0)=0.30\times 0.45=0.135 ]

Since,

[ 0.15\ne 0.135 ]

the condition fails. Therefore, (X) and (Y) are not independent.

Now compute the correlation coefficient.

First,

[ E(XY)=\sum_x\sum_y xyP(X=x,Y=y) ]

Only the terms with (x=1,y=1) and (x=2,y=1) contribute:

[ E(XY)=1(1)(0.25)+2(1)(0.15)=0.25+0.30=0.55 ]

Covariance:

[ \operatorname{Cov}(X,Y)=E(XY)-E(X)E(Y) ]

[ \operatorname{Cov}(X,Y)=0.55-(0.95)(0.55) ]

[ =0.55-0.5225=0.0275 ]

Correlation coefficient:

[ r=\frac{\operatorname{Cov}(X,Y)}{\sqrt{\operatorname{Var}(X)\operatorname{Var}(Y)}} ]

[ r=\frac{0.0275}{\sqrt{0.5475\times 0.2475}} ]

[ r=\frac{0.0275}{0.3681}=0.0747 ]

Therefore,

[ \boxed{r=0.0747} ]

So (X) and (Y) are not independent, but their correlation coefficient is positive and very weak.

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  • 2082 · Statistics for Economics (Econ. 553) · 15 marks

The following intermediate result is related to the industrial output. (10+2+3)
n=40ΣX1=120ΣX2=160ΣX3=80n=40 \quad \Sigma X_1=120 \quad \Sigma X_2=160 \quad \Sigma X_3=80 ΣX12=5200ΣX22=4600ΣX32=4600\Sigma X_1^2=5200 \quad \Sigma X_2^2=4600 \quad \Sigma X_3^2=4600 ΣY=800ΣY2=25000ΣX1Y=3800ΣX2Y=4400ΣX3Y=3600\Sigma Y=800 \quad \Sigma Y^2=25000 \quad \Sigma X_1Y=3800 \quad \Sigma X_2Y=4400 \quad \Sigma X_3Y=3600 Σ(X1−X2)2=1500Σ(Y−Yˉ)2=2500\Sigma(X_1-X_2)^2=1500 \quad \Sigma(Y-\bar{Y})^2=2500
Where, Y be the industrial output, X₁ be the expenditure on labour and X₂ be the expenditure on machinery in thousand NRs.
a) Find the regression equation Y on X₁ and X₂. b) Calculate the coefficient of determination R² and interpret the result. c) Estimate the variance of parameters.

Answer

Note on the data. ΣY=800\Sigma Y=800 and ΣY2=25000\Sigma Y^2=25000 with n=40n=40 give Σ(Y−Yˉ)2=25000−8002/40=9000\Sigma(Y-\bar Y)^2 = 25000 - 800^2/40 = 9000, which conflicts with the stated 25002500. The problem states 25002500 separately for use in R2R^2, so it is used below, and the result with 90009000 is also given. ΣX3\Sigma X_3 and related sums are not needed.

Deviation sums (n=40n=40; Yˉ=20\bar Y=20, Xˉ1=3\bar X_1=3, Xˉ2=4\bar X_2=4):

Σx12=5200−120240=4840,Σx22=4600−160240=3960\Sigma x_1^2 = 5200 - \frac{120^2}{40} = 4840,\qquad \Sigma x_2^2 = 4600 - \frac{160^2}{40} = 3960

Since Σ(X1−X2)2=Σx12+Σx22−2Σx1x2\Sigma(X_1-X_2)^2 = \Sigma x_1^2 + \Sigma x_2^2 - 2\Sigma x_1x_2 is 15001500:

Σx1x2=4840+3960−15002=3650\Sigma x_1x_2 = \frac{4840+3960-1500}{2} = 3650

Σx1y=3800−120×80040=1400,Σx2y=4400−160×80040=1200\Sigma x_1y = 3800 - \frac{120\times800}{40} = 1400,\qquad \Sigma x_2y = 4400 - \frac{160\times800}{40} = 1200

(a) Regression equation

β^1=(1400)(3960)−(1200)(3650)(4840)(3960)−36502=1,164,0005,843,900=0.1992\hat\beta_1 = \frac{(1400)(3960) - (1200)(3650)}{(4840)(3960) - 3650^2} = \frac{1{,}164{,}000}{5{,}843{,}900} = 0.1992

β^2=(1200)(4840)−(1400)(3650)5,843,900=698,0005,843,900=0.1194\hat\beta_2 = \frac{(1200)(4840) - (1400)(3650)}{5{,}843{,}900} = \frac{698{,}000}{5{,}843{,}900} = 0.1194

β^0=20−0.1992(3)−0.1194(4)=18.92\hat\beta_0 = 20 - 0.1992(3) - 0.1194(4) = 18.92

Y^=18.92+0.1992X1+0.1194X2\boxed{\hat Y = 18.92 + 0.1992X_1 + 0.1194X_2}

(b) Coefficient of determination

ESS=β^1Σx1y+β^2Σx2y=0.1992(1400)+0.1194(1200)=422.2ESS = \hat\beta_1\Sigma x_1y + \hat\beta_2\Sigma x_2y = 0.1992(1400) + 0.1194(1200) = 422.2

R2=ESSΣ(Y−Yˉ)2=422.22500=0.169R^2 = \frac{ESS}{\Sigma(Y-\bar Y)^2} = \frac{422.2}{2500} = 0.169

About 16.9 percent of the variation in industrial output is explained jointly by expenditure on labour and on machinery; the remaining 83 percent is due to other factors and random error. The fit is weak. (With the value 90009000 implied by ΣY2\Sigma Y^2, R2=0.047R^2 = 0.047, which is weaker still.)

(c) Variances of the parameters

RSS=2500−422.2=2077.8,σ^2=RSSn−3=2077.837=56.16RSS = 2500 - 422.2 = 2077.8,\qquad \hat\sigma^2 = \frac{RSS}{n-3} = \frac{2077.8}{37} = 56.16

Var(β^1)=σ^2Σx22D=56.16×39605843900=0.0381\text{Var}(\hat\beta_1) = \hat\sigma^2\frac{\Sigma x_2^2}{D} = 56.16\times\frac{3960}{5843900} = 0.0381

Var(β^2)=σ^2Σx12D=56.16×48405843900=0.0465\text{Var}(\hat\beta_2) = \hat\sigma^2\frac{\Sigma x_1^2}{D} = 56.16\times\frac{4840}{5843900} = 0.0465

Cov(β^1,β^2)=−σ^2Σx1x2D=−0.0351\text{Cov}(\hat\beta_1,\hat\beta_2) = -\hat\sigma^2\frac{\Sigma x_1x_2}{D} = -0.0351

Var(β^0)=σ^2[1n+Xˉ12Σx22+Xˉ22Σx12−2Xˉ1Xˉ2Σx1x2D]≈1.65\text{Var}(\hat\beta_0) = \hat\sigma^2\left[\frac1n + \frac{\bar X_1^2\Sigma x_2^2 + \bar X_2^2\Sigma x_1^2 - 2\bar X_1\bar X_2\Sigma x_1x_2}{D}\right] \approx 1.65

Here D=4840×3960−36502=5,843,900D = 4840\times3960 - 3650^2 = 5{,}843{,}900. Standard errors are se(β^1)=0.195\text{se}(\hat\beta_1)=0.195, se(β^2)=0.216\text{se}(\hat\beta_2)=0.216 and se(β^0)=1.28\text{se}(\hat\beta_0)=1.28.

Answer checked by Claude Sonnet · Oct 21, 2026

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  • 2082 · Statistics for Economics (Econ. 553) · 5 marks

Explain the concept of Stratified Random Sampling. Under what circumstances is stratified random sampling preferred over simple random sampling? Also discuss the different approaches for allocating the total sample size say n among the various strata with suitable illustration. (3+3+9)

Answer

Stratified Random Sampling

Concept of Stratified Random Sampling

Stratified random sampling is a probability sampling technique in which the entire population is first divided into a number of non-overlapping, homogeneous subgroups called strata (singular: stratum), and then a simple random sample is drawn independently from each stratum. The strata are formed on the basis of some characteristic (e.g., income, age, sex, region, landholding size) that is believed to be related to the variable under study. The samples drawn from all strata are then pooled together to form the overall sample.

Procedure:

  1. Divide the population of size N into k strata of sizes N₁, N₂, …, Nₖ, such that N₁ + N₂ + … + Nₖ = N. The strata must be mutually exclusive and collectively exhaustive.
  2. Decide the sample size n to be drawn from the population.
  3. Allocate n among the strata as n₁, n₂, …, nₖ, such that n₁ + n₂ + … + nₖ = n.
  4. Draw a simple random sample of size nₕ from the h-th stratum independently (h = 1, 2, …, k).
  5. Combine all nₕ units to form the final sample of size n.

Estimate of population mean: If ȳₕ is the sample mean of the h-th stratum and Wₕ = Nₕ/N is the stratum weight, the stratified sample mean is

yˉst=∑h=1kWhyˉh\bar{y}_{st} = \sum_{h=1}^{k} W_h \bar{y}_h

with variance

V(yˉst)=∑h=1kWh2 V(yˉh)=∑h=1kWh2(Nh−nhNh)Sh2nhV(\bar{y}_{st}) = \sum_{h=1}^{k} W_h^2 \, V(\bar{y}_h) = \sum_{h=1}^{k} W_h^2 \left(\frac{N_h - n_h}{N_h}\right)\frac{S_h^2}{n_h}

where Sₕ² is the population variance within the h-th stratum.

Circumstances Under Which Stratified Random Sampling is Preferred Over Simple Random Sampling

Stratified random sampling is preferred over simple random sampling in the following situations:

  1. Heterogeneous population: When the population is highly heterogeneous with respect to the study variable, stratification divides it into homogeneous groups, thereby reducing within-group variability and increasing precision.

  2. Presence of distinct subgroups: When the population naturally consists of distinct groups (e.g., male/female, rural/urban, different ecological belts) whose characteristics differ markedly, stratification ensures representation of each group.

  3. Need for separate estimates: When estimates are required not only for the whole population but also for each subgroup (domain), stratified sampling provides reliable separate estimates for each stratum.

  4. Greater precision at lower cost: When the cost of surveying units differs across strata (e.g., remote areas costlier than accessible ones), stratification permits efficient allocation of effort and yields a more precise estimate for the same cost, or the same precision at lower cost.

  5. Administrative convenience: When the population is already divided into administrative units (districts, wards, schools), stratification simplifies field operations and supervision.

  6. Extreme values or skewed distribution: When a few very large or very small units dominate the population (e.g., large industries, big farms), isolating them into separate strata prevents distortion of the estimate.

  7. When a related auxiliary variable is known: When information on a variable highly correlated with the study variable is available for all units, stratification on that variable substantially reduces sampling error.

Approaches for Allocating the Total Sample Size n Among Strata

Let the population be divided into k strata with sizes N₁, N₂, …, Nₖ, stratum weights Wₕ = Nₕ/N, and within-stratum standard deviations S₁, S₂, …, Sₖ. The total sample size n is to be allocated as n₁, n₂, …, nₖ with ∑nₕ = n. The main allocation methods are:

1. Equal Allocation

The same number of units is taken from each stratum:

nh=nkn_h = \frac{n}{k}

This method is simple and is used when strata are of roughly equal size and have similar variability. It is rarely optimal when strata differ greatly in size or variance.

2. Proportional Allocation

The sample size in each stratum is proportional to the stratum size:

nh=n⋅NhN=nWhn_h = n \cdot \frac{N_h}{N} = n W_h

This is the most widely used method because it is simple, requires no information on Sₕ, and gives a self-weighting sample (the overall sample mean is simply the arithmetic mean of all sampled units). It is preferred when strata do not differ much in variability.

3. Neyman (Optimum) Allocation

When strata differ in variability, more units should be drawn from more variable strata. Neyman allocation minimizes the variance of the stratified mean for a fixed n:

nh=n⋅NhSh∑h=1kNhSh=n⋅WhSh∑WhShn_h = n \cdot \frac{N_h S_h}{\sum_{h=1}^{k} N_h S_h} = n \cdot \frac{W_h S_h}{\sum W_h S_h}

This is optimal when the cost per unit is the same in all strata. It gives greater precision than proportional allocation whenever Sₕ vary across strata.

4. Cost-Optimum (Economic) Allocation

When costs per unit differ across strata, allocation must balance both variability and cost. If cₕ is the cost per unit in the h-th stratum, the optimum allocation is:

nh=n⋅NhSh/ch∑h=1kNhSh/chn_h = n \cdot \frac{N_h S_h / \sqrt{c_h}}{\sum_{h=1}^{k} N_h S_h / \sqrt{c_h}}

This minimizes the total cost for a fixed variance (or minimizes variance for a fixed cost). Proportional and Neyman allocations are special cases of this when costs are equal.

5. Neyman Allocation with a Fixed Total Cost

If total cost C = c₀ + ∑ cₕ nₕ is fixed, the allocation becomes

nh=(C−c0) NhSh/ch∑NhShchn_h = \frac{(C - c_0) \, N_h S_h / \sqrt{c_h}}{\sum N_h S_h \sqrt{c_h}}

Illustration

Suppose a population of N = 1,000 farmers is divided into three strata by landholding size, and a sample of n = 100 is to be drawn. The relevant data are:

StratumNₕWₕ = Nₕ/NSₕ (ha)cₕ (Rs/unit)
Small5000.5024
Medium3000.3059
Large2000.201016
Total1,0001.00——

(a) Proportional Allocation (nₕ = nWₕ):

  • n₁ = 100 × 0.50 = 50
  • n₂ = 100 × 0.30 = 30
  • n₃ = 100 × 0.20 = 20

(b) Neyman Allocation (nₕ ∝ NₕSₕ):

Compute NₕSₕ: 500×2 = 1,000; 300×5 = 1,500; 200×10 = 2,000. Total = 4,500.

  • n₁ = 100 × (1,000/4,500) = 22.22 ≈ 22
  • n₂ = 100 × (1,500/4,500) = 33.33 ≈ 33
  • n₃ = 100 × (2,000/4,500) = 44.44 ≈ 45

(c) Cost-Optimum Allocation (nₕ ∝ NₕSₕ/√cₕ):

Compute NₕSₕ/√cₕ: 1,000/2 = 500; 1,500/3 = 500; 2,000/4 = 500. Total = 1,500.

  • n₁ = 100 × (500/1,500) = 33.33 ≈ 33
  • n₂ = 100 × (500/1,500) = 33.33 ≈ 33
  • n₃ = 100 × (500/1,500) = 33.33 ≈ 34

Interpretation: Proportional allocation gives the largest sample to the largest stratum. Neyman allocation shifts units toward the most variable stratum (large farmers). Cost-optimum allocation balances variability against cost; here, because the costlier strata are also the more variable ones, the three allocations converge to nearly equal shares. This shows that the choice of allocation depends on whether strata differ in size, variability, or cost.

Answer checked by Claude Haiku · Oct 21, 2026

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  • 2082 · Statistics for Economics (Econ. 553) · 10 marks

A researcher wants to examine whether there is an association between gender and employment status. A random sample of 200 individuals is classified according to gender and employment status as shown below:
GenderEmployedUnemployedTotal
Male7030100
Female5050100
Total12080200
Stating the null and alternative hypotheses, discuss the procedure for testing whether employment status is associated with gender.

Answer

Hypotheses

  • H0H_0: Gender and employment status are independent (no association).
  • H1H_1: Gender and employment status are associated.

Test statistic

Both variables are categorical and the data are frequencies, so the chi-square test of independence is used:

χ2=∑(Oij−Eij)2Eij,Eij=(row totali)(column totalj)N\chi^2 = \sum\frac{(O_{ij}-E_{ij})^2}{E_{ij}},\qquad E_{ij} = \frac{(\text{row total}_i)(\text{column total}_j)}{N}

Procedure

  1. Set up the 2×22\times2 contingency table and fix the significance level, say α=0.05\alpha = 0.05.
  2. Compute the expected frequencies under H0H_0:
GenderEmployedUnemployedTotal
Male100×120/200=60100\times120/200 = 60100×80/200=40100\times80/200 = 40100
Female100×120/200=60100\times120/200 = 60100×80/200=40100\times80/200 = 40100
Total12080200
  1. Compute the chi-square value:

χ2=(70−60)260+(30−40)240+(50−60)260+(50−40)240=1.667+2.5+1.667+2.5=8.333\chi^2 = \frac{(70-60)^2}{60} + \frac{(30-40)^2}{40} + \frac{(50-60)^2}{60} + \frac{(50-40)^2}{40} = 1.667 + 2.5 + 1.667 + 2.5 = 8.333

  1. Degrees of freedom: (r−1)(c−1)=(2−1)(2−1)=1(r-1)(c-1) = (2-1)(2-1) = 1. The critical value is χ0.05,12=3.841\chi^2_{0.05,1} = 3.841.
  2. Decision: since 8.333>3.8418.333 > 3.841, reject H0H_0.
  3. Conclusion: at the 5 percent level there is sufficient evidence that employment status is associated with gender. In this sample 70 percent of men but only 50 percent of women are employed.

For a 2×22\times2 table, Yates' continuity correction may be applied: χ2=∑(∣O−E∣−0.5)2/E≈7.52\chi^2 = \sum(|O-E|-0.5)^2/E \approx 7.52, which still exceeds 3.841, so the conclusion is unchanged.

Answer checked by Claude Sonnet · Oct 21, 2026

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  • 2082 · Statistics for Economics (Econ. 553) · 10 marks

Define and distinguish between estimator and estimate. Discuss the desirable small-sample properties of a good estimator. (2+8)

Answer

Estimator and Estimate

An estimator is a rule or formula (a statistic) that tells us how to calculate an approximate value of an unknown population parameter from sample data. It is a random variable because its value varies from sample to sample. For example, the sample mean Xˉ=1n∑i=1nXi\bar{X} = \frac{1}{n}\sum_{i=1}^{n} X_i is an estimator of the population mean μ\mu.

An estimate is the specific numerical value obtained when the estimator is applied to a particular sample. It is a fixed, non-random number. For example, if a sample of 5 households gives incomes 10, 12, 14, 16, 18 (in thousand Rs), then xˉ=14\bar{x} = 14 thousand Rs is an estimate of μ\mu.

Distinction

BasisEstimatorEstimate
NatureA rule, formula or statisticA numerical value
RandomnessRandom variable (varies across samples)Fixed constant for a given sample
Notationθ^\hat{\theta}, Xˉ\bar{X}, s2s^2θ^\hat{\theta}, xˉ\bar{x}, s2s^2 (computed value)
DependenceDepends on sample size and sampling designDepends on the particular sample drawn
PurposeUsed to construct the estimateUsed for inference about the parameter

Desirable small-sample properties of a good estimator

Small-sample (exact) properties are those that hold for any sample size nn, however small.

  1. Unbiasedness. An estimator θ^\hat{\theta} is unbiased if its expected value equals the true parameter: E(θ^)=θE(\hat{\theta}) = \theta The sampling distribution of θ^\hat{\theta} is then centred on θ\theta; there is no systematic over- or under-estimation. The sample mean Xˉ\bar{X} is an unbiased estimator of μ\mu, and s2=1n−1∑(Xi−Xˉ)2s^2 = \frac{1}{n-1}\sum (X_i - \bar{X})^2 is unbiased for σ2\sigma^2.

  2. Efficiency. Among all unbiased estimators of θ\theta, the one with the smallest variance is most efficient. If θ^1\hat{\theta}_1 and θ^2\hat{\theta}_2 are both unbiased, θ^1\hat{\theta}_1 is more efficient if Var(θ^1)<Var(θ^2)\text{Var}(\hat{\theta}_1) < \text{Var}(\hat{\theta}_2). The ratio Var(θ^2)Var(θ^1)\frac{\text{Var}(\hat{\theta}_2)}{\text{Var}(\hat{\theta}_1)} measures relative efficiency. An estimator attaining the Cramér–Rao lower bound is a minimum variance unbiased estimator (MVUE).

  3. Consistency. θ^\hat{\theta} is consistent if it converges in probability to θ\theta as n→∞n \to \infty: plimn→∞θ^=θ\text{plim}_{n\to\infty} \hat{\theta} = \theta Sufficient conditions are E(θ^)→θE(\hat{\theta}) \to \theta and Var(θ^)→0\text{Var}(\hat{\theta}) \to 0 as n→∞n \to \infty. Consistency is technically a large-sample property but is listed among the basic requirements of a good estimator.

  4. Sufficiency. An estimator is sufficient if it uses all the information in the sample relevant to θ\theta; no other statistic can add information about θ\theta. Formally, the conditional distribution of the sample given the statistic does not depend on θ\theta. The sample mean is sufficient for μ\mu in a normal population.

  5. Efficiency of the best linear unbiased estimator (BLUE). When attention is restricted to linear estimators, the Gauss–Markov property requires the estimator to be linear, unbiased, and of minimum variance among linear unbiased estimators — as in OLS.

  6. Minimum mean squared error. For biased estimators, the criterion is MSE(θ^)=Var(θ^)+[Bias(θ^)]2\text{MSE}(\hat{\theta}) = \text{Var}(\hat{\theta}) + [\text{Bias}(\hat{\theta})]^2 A good estimator keeps MSE small, trading a little bias for a large reduction in variance (e.g. shrinkage estimators).

  7. Invariance. If θ^\hat{\theta} is a good estimator of θ\theta, then g(θ^)g(\hat{\theta}) should be a good estimator of g(θ)g(\theta) for a suitable function gg (as with maximum likelihood invariance).

In practice no single estimator satisfies every property, so the choice depends on the criterion most relevant to the problem — unbiasedness and minimum variance being the most commonly used standards in small samples.

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  • 2082 · Statistics for Economics (Econ. 553) · 10 marks

Prove that total variation is the sum of explained and unexplained variation. (10)

Answer

Prove that total variation is the sum of explained and unexplained variation

Let the simple linear regression model be

yi=α+βxi+uiy_i=\alpha+\beta x_i+u_i

and the fitted sample regression line be

y^i=α^+β^xi\hat y_i=\hat\alpha+\hat\beta x_i

where:

  • (y_i) = actual value of dependent variable,
  • (\hat y_i) = estimated/fitted value,
  • (\bar y) = mean of actual (y_i),
  • (e_i=y_i-\hat y_i) = residual or unexplained part.

Definitions

Total variation in (y) is measured by the total sum of squares:

TSS=∑(yi−yˉ)2TSS=\sum (y_i-\bar y)^2

Explained variation is measured by the explained sum of squares:

ESS=∑(y^i−yˉ)2ESS=\sum (\hat y_i-\bar y)^2

Unexplained variation is measured by the residual sum of squares:

RSS=∑(yi−y^i)2=∑ei2RSS=\sum (y_i-\hat y_i)^2=\sum e_i^2

We have to prove:

TSS=ESS+RSSTSS=ESS+RSS

Proof

For each observation,

yi−yˉ=(y^i−yˉ)+(yi−y^i)y_i-\bar y=(\hat y_i-\bar y)+(y_i-\hat y_i)

Squaring both sides and summing over all observations:

∑(yi−yˉ)2=∑[(y^i−yˉ)+(yi−y^i)]2\sum (y_i-\bar y)^2 = \sum \left[(\hat y_i-\bar y)+(y_i-\hat y_i)\right]^2

Expanding the square:

∑(yi−yˉ)2=∑(y^i−yˉ)2+∑(yi−y^i)2+2∑(y^i−yˉ)(yi−y^i)\sum (y_i-\bar y)^2 = \sum (\hat y_i-\bar y)^2 + \sum (y_i-\hat y_i)^2 + 2\sum (\hat y_i-\bar y)(y_i-\hat y_i)

So,

TSS=ESS+RSS+2∑(y^i−yˉ)eiTSS=ESS+RSS+2\sum (\hat y_i-\bar y)e_i

Now we must show that the cross-product term is zero:

∑(y^i−yˉ)ei=0\sum (\hat y_i-\bar y)e_i=0

The OLS method minimizes (\sum e_i^2=\sum (y_i-\hat\alpha-\hat\beta x_i)^2). The first-order conditions are:

∂∑ei2∂α^=−2∑(yi−α^−β^xi)=0\frac{\partial \sum e_i^2}{\partial \hat\alpha} = -2\sum (y_i-\hat\alpha-\hat\beta x_i)=0 ∂∑ei2∂β^=−2∑xi(yi−α^−β^xi)=0\frac{\partial \sum e_i^2}{\partial \hat\beta} = -2\sum x_i(y_i-\hat\alpha-\hat\beta x_i)=0

Therefore,

∑ei=0\sum e_i=0

and

∑xiei=0\sum x_i e_i=0

Since (\hat y_i=\hat\alpha+\hat\beta x_i),

∑y^iei=α^∑ei+β^∑xiei=α^(0)+β^(0)=0\sum \hat y_i e_i = \hat\alpha \sum e_i+\hat\beta \sum x_i e_i = \hat\alpha(0)+\hat\beta(0)=0

Also,

yˉ∑ei=yˉ(0)=0\bar y \sum e_i=\bar y(0)=0

Hence,

∑(y^i−yˉ)ei=∑y^iei−yˉ∑ei=0−0=0\sum (\hat y_i-\bar y)e_i = \sum \hat y_i e_i-\bar y\sum e_i = 0-0=0

Thus the cross-product term vanishes:

2∑(y^i−yˉ)ei=02\sum (\hat y_i-\bar y)e_i=0

Therefore,

TSS=ESS+RSSTSS=ESS+RSS

That is,

Total variation=Explained variation+Unexplained variation\boxed{\text{Total variation} = \text{Explained variation} + \text{Unexplained variation}}

Interpretation

  • Total variation (\sum (y_i-\bar y)^2) measures the total fluctuation of (y) around its mean.
  • Explained variation (\sum (\hat y_i-\bar y)^2) is the part of total variation explained by the regression model, i.e. by changes in (x).
  • Unexplained variation (\sum (y_i-\hat y_i)^2) is the part left unexplained, due to random factors or omitted variables.

Thus, in OLS regression, the total variation in the dependent variable is exactly decomposed into explained and unexplained variation. This result also gives the coefficient of determination:

R2=ESSTSS=1−RSSTSSR^2=\frac{ESS}{TSS}=1-\frac{RSS}{TSS}

Answer checked by Claude Haiku · Oct 21, 2026

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  • 2082 · Statistics for Economics (Econ. 553) · 10 marks

Differentiate between positive and negative correlation with suitable example. Find the Spearman's rank correlation between the rainfall and paddy production in Western Nepal.
Rainfall (mm)787465607490856575
Production (mt)726878559170755880

Answer

Positive and negative correlation

Correlation measures the direction and strength of the linear relationship between two variables.

  • Positive correlation: the variables move in the same direction; r>0r>0. Example: as household income rises, household consumption expenditure rises.
  • Negative correlation: the variables move in opposite directions; r<0r<0. Example: as the price of a commodity rises, the quantity demanded falls.

Spearman's rank correlation

Rank both series (rank 1 for the smallest). Tied values receive the average of their ranks: rainfall 65 occurs twice (ranks 2 and 3, so 2.5 each), and 74 occurs twice (ranks 4 and 5, so 4.5 each).

RainfallProductionR1R_1R2R_2d=R1−R2d = R_1-R_2d2d^2
78727524
74684.531.52.25
65782.57-4.520.25
60551100
74914.59-4.520.25
907094525
85758624
65582.520.50.25
758068-24
Total80

Two tied pairs in the rainfall series require a correction of m3−m12=8−212=0.5\frac{m^3-m}{12} = \frac{8-2}{12} = 0.5 for each, a total of 11:

rs=1−6[Σd2+1]n(n2−1)=1−6(80+1)9(81−1)=1−486720=0.325r_s = 1 - \frac{6\left[\Sigma d^2 + 1\right]}{n(n^2-1)} = 1 - \frac{6(80+1)}{9(81-1)} = 1 - \frac{486}{720} = 0.325

Without the tie correction rs=1−480/720=0.333r_s = 1 - 480/720 = 0.333.

Interpretation. rs≈0.33r_s\approx0.33 indicates a weak positive correlation: higher rainfall in Western Nepal tends to go with higher paddy production, but the relationship is far from strong, so other factors such as irrigation, seed and fertiliser also matter.

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  • 2082 · Statistics for Economics (Econ. 553) · 10 marks

Attempt any two: (2x5=10) a) Sampling Error b) Properties of Standard Normal distribution c) What are dummy variables? Why is it incorporated in regression model?

Answer

a) Sampling Error

Meaning. Sampling error is the difference between a statistic computed from a sample and the true value of the corresponding population parameter. If xˉ\bar{x} is the sample mean and μ\mu the population mean, the sampling error is xˉ−μ\bar{x}-\mu; similarly for a proportion, p^−p\hat{p}-p.

Nature. It arises purely because observation is confined to a part of the population rather than the whole. It is not a mistake of the investigator; it is inherent in the sampling method. It can be reduced (by enlarging the sample size or adopting a better sampling design) but never completely eliminated unless a complete enumeration (census) is taken.

Standard error. Since the exact error is unknown, its magnitude is measured by the standard error of the statistic. For the sample mean from a simple random sample,

SE(xˉ)=σxˉ=σnSE(\bar{x}) = \sigma_{\bar{x}} = \frac{\sigma}{\sqrt{n}}

and, when σ\sigma is unknown,   SE(xˉ)=sn\;SE(\bar{x}) = \dfrac{s}{\sqrt{n}}. For a proportion, SE(p^)=pqnSE(\hat{p}) = \sqrt{\dfrac{pq}{n}}.

Determinants. Sampling error varies inversely with sample size (nn), directly with population variability (σ\sigma), and depends on the sampling design; it is smaller under stratified or systematic sampling than under simple random sampling for the same nn.

Distinction from non-sampling error. Non-sampling errors arise from faulty frame, non-response, wrong measurement, editing and tabulation mistakes; they may increase with sample size, whereas sampling error falls as nn rises.

Uses. Standard error is used to construct confidence intervals, e.g. xˉ±zα/2 σn\bar{x} \pm z_{\alpha/2}\,\dfrac{\sigma}{\sqrt{n}}, to test hypotheses about μ\mu, and to determine the sample size required for a desired precision, n=z2σ2E2n = \dfrac{z^2\sigma^2}{E^2}.


b) Properties of Standard Normal Distribution

A normal distribution with mean μ=0\mu = 0 and standard deviation σ=1\sigma = 1 is called the standard normal distribution, denoted Z∼N(0,1)Z \sim N(0,1), with density

f(z)=12πe−12z2,−∞<z<∞f(z)=\frac{1}{\sqrt{2\pi}}e^{-\frac{1}{2}z^{2}},\qquad -\infty<z<\infty

Its main properties are:

  1. Symmetry. The curve is perfectly symmetric about the mean z=0z=0; the left half is the mirror image of the right half. Hence skewness β1=0\beta_1 = 0.
  2. Mean, median and mode coincide. All three equal zero, so the curve has a single central peak at z=0z=0.
  3. Total area is one. The area under the curve and above the horizontal axis equals 1, so it represents a probability distribution.
  4. Mesokurtic. Kurtosis β2=3\beta_2 = 3 (excess kurtosis 0); the curve is neither flat nor peaked relative to the normal form.
  5. Asymptotic. The curve approaches but never touches the horizontal axis as z→±∞z \to \pm\infty; the range is unlimited.
  6. Points of inflection lie at z=±1z = \pm 1, i.e. one standard deviation from the mean.
  7. Standard deviation equals one, and the total area is distributed as: about 68.27% within ±1σ\pm 1\sigma, 95.45% within ±2σ\pm 2\sigma, and 99.73% within ±3σ\pm 3\sigma.
  8. Standardisation property. Any normal variable X∼N(μ,σ2)X \sim N(\mu,\sigma^2) can be converted into a standard normal variable by z=X−μσz = \dfrac{X-\mu}{\sigma}, so one table of ZZ serves all normal distributions.
  9. Linear transformation. A linear combination of independent normal variables is also normally distributed.
  10. Moment property. All odd central moments are zero and even central moments follow μ2r=(2r)!2r r! σ2r\mu_{2r} = \dfrac{(2r)!}{2^{r}\,r!}\,\sigma^{2r}; for the standard normal, μ2=1, μ4=3\mu_2=1,\ \mu_4=3.

c) Dummy Variables and Their Use in Regression

Meaning. A dummy variable (indicator or binary variable) is a qualitative variable quantified by taking only two values, usually 0 and 1. Value 1 indicates the presence of an attribute (e.g. male, urban, married, skilled) and 0 its absence. It is also called a binary, categorical or indicator variable.

Need. Regression analysis requires quantitative data. Variables such as sex, caste, region, religion, season, marital status or level of education are qualitative and cannot be entered directly. Dummy variables convert such attributes into numerical form so that their influence on the dependent variable can be measured and tested.

Nature of the coefficient. In the model

Yi=β1+β2Di+uiY_i = \beta_1 + \beta_2 D_i + u_i

where Di=1D_i = 1 for the category of interest and 00 otherwise, the conditional means are:

E(Yi∣Di=0)=β1,E(Yi∣Di=1)=β1+β2E(Y_i \mid D_i = 0) = \beta_1, \qquad E(Y_i \mid D_i = 1) = \beta_1 + \beta_2

Thus β1\beta_1 is the intercept (mean value of the reference or base category) and β2\beta_2 is the differential intercept, showing by how much the mean of the dummy category differs from that of the base category. If β2\beta_2 is statistically significant, the qualitative attribute has a real effect.

Reasons for incorporation in the regression model.

  1. To include qualitative or categorical information that cannot otherwise enter a quantitative model.
  2. To estimate and test the effect of an attribute on the dependent variable, holding other regressors constant.
  3. To allow the intercept (and, through interaction terms, the slope) to differ between groups, i.e. to capture structural differences.
  4. To avoid specification bias that would arise if a relevant qualitative variable were omitted.
  5. To handle seasonal or structural shifts, as in seasonal dummies in demand and time-series models.

Dummy variable trap and rule. If a qualitative variable has mm categories, use only m−1m-1 dummies and keep one category as the base; including all mm dummies together with a common intercept makes the regressor matrix singular (perfect collinearity), which is the dummy variable trap. The base category's effect is absorbed in the intercept. For example, for three regions — Eastern, Central and Western — use two dummies, D1=1D_1 = 1 for Eastern and D2=1D_2 = 1 for Central, taking Western as the base:

Yi=β1+β2D1i+β3D2i+uiY_i = \beta_1 + \beta_2 D_{1i} + \beta_3 D_{2i} + u_i

where β1\beta_1 is the mean for Western, β1+β2\beta_1+\beta_2 for Eastern and β1+β3\beta_1+\beta_3 for Central. Dummy variables may also be interacted with quantitative regressors (e.g. DiXiD_i X_i) to test whether slopes differ across groups.

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  • 2082 · Statistics for Economics (Econ 553) · 15 marks

A discrete random variable X has the following probability mass function (pmf) P(X=x)=c(2x+1), for x=0,1,2,3.P(X = x) = c(2x + 1), \text{ for } x = 0,1,2,3.
a. Find the value of constant c such that above function serves as a pmf. (2) b. Show the probability distribution in the tabular form. (1) c. Compute the mean E(X) and Variance V(X) of the distribution. (3+3) d. Find the probability P(0≤X<3)P(0 \le X < 3). (2) e. Find E(5X+3)E(5X+3) and Var(7X)Var(7X) based on the above distribution. (2+2)

Answer

Given pmf

[ P(X=x)=c(2x+1),\qquad x=0,1,2,3 ]


(a) Find (c)

For a valid pmf, [ \sum_{x=0}^{3}P(X=x)=1 ] So, [ c[(2(0)+1)+(2(1)+1)+(2(2)+1)+(2(3)+1)]=1 ] [ c[1+3+5+7]=1 ] [ 16c=1 ] [ \boxed{c=\frac{1}{16}} ]


(b) Probability distribution in tabular form

[ P(X=x)=\frac{1}{16}(2x+1) ]

(x)(P(X=x))
0(\frac{1}{16})
1(\frac{3}{16})
2(\frac{5}{16})
3(\frac{7}{16})
Total(1)

(c) Mean (E(X)) and Variance (V(X))

Mean

[ E(X)=\sum xP(X=x) ] [ E(X)=0\cdot\frac{1}{16}+1\cdot\frac{3}{16}+2\cdot\frac{5}{16}+3\cdot\frac{7}{16} ] [ E(X)=\frac{0+3+10+21}{16}=\frac{34}{16} ] [ \boxed{E(X)=\frac{17}{8}=2.125} ]

Variance

First compute (E(X^2)): [ E(X^2)=\sum x^2P(X=x) ] [ E(X^2)=0^2\cdot\frac{1}{16}+1^2\cdot\frac{3}{16}+2^2\cdot\frac{5}{16}+3^2\cdot\frac{7}{16} ] [ E(X^2)=\frac{0+3+20+63}{16}=\frac{86}{16}=\frac{43}{8} ]

Now, [ V(X)=E(X^2)-[E(X)]^2 ] [ V(X)=\frac{43}{8}-\left(\frac{17}{8}\right)^2 ] [ V(X)=\frac{43}{8}-\frac{289}{64} ] [ V(X)=\frac{344}{64}-\frac{289}{64}=\frac{55}{64} ] [ \boxed{V(X)=\frac{55}{64}=0.859375} ]


(d) Find (P(0\le X<3))

[ P(0\le X<3)=P(X=0)+P(X=1)+P(X=2) ] [ =\frac{1}{16}+\frac{3}{16}+\frac{5}{16} ] [ =\frac{9}{16} ] [ \boxed{P(0\le X<3)=\frac{9}{16}=0.5625} ]


(e) Find (E(5X+3)) and (Var(7X))

(E(5X+3))

[ E(5X+3)=5E(X)+3 ] [ =5\left(\frac{17}{8}\right)+3 ] [ =\frac{85}{8}+\frac{24}{8}=\frac{109}{8} ] [ \boxed{E(5X+3)=\frac{109}{8}=13.625} ]

(Var(7X))

[ Var(7X)=7^2Var(X) ] [ =49\left(\frac{55}{64}\right) ] [ =\frac{2695}{64} ] [ \boxed{Var(7X)=\frac{2695}{64}=42.109375} ]


Final Answers

[ \boxed{c=\frac{1}{16}} ] [ \boxed{E(X)=\frac{17}{8},\quad V(X)=\frac{55}{64}} ] [ \boxed{P(0\le X<3)=\frac{9}{16}} ] [ \boxed{E(5X+3)=\frac{109}{8},\quad Var(7X)=\frac{2695}{64}} ]

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  • 2082 · Statistics for Economics (Econ 553) · 15 marks

The following results were obtained from a sample of 40 firms on their output (Y), Labour input (X1X_1) and Capital input (X2X_2). ΣY=56ΣX1=320ΣX2=16\Sigma Y = 56 \quad \Sigma X_1 = 320 \quad \Sigma X_2 = 16 ΣY2=882ΣX12=290ΣX22=94\Sigma Y^2 = 882 \quad \Sigma X_1^2 = 290 \quad \Sigma X_2^2 = 94 ΣX1Y=505ΣX2Y=276ΣX1X2=159\Sigma X_1Y = 505 \quad \Sigma X_2Y = 276 \quad \Sigma X_1X_2 = 159
a. Find the least square equation of Y on X1X_1 and X2X_2 (10) b. Interpret the meaning of the coefficients estimated. (2) c. Find adjusted RR-square and interpret it. (3)

Answer

Note on the data. The sums are impossible as printed. For n=40n=40 and ΣX1=320\Sigma X_1 = 320, we must have ΣX12≥3202/40=2560\Sigma X_1^2 \ge 320^2/40 = 2560, but ΣX12=290\Sigma X_1^2 = 290 is given, so Σx12=290−2560=−2270\Sigma x_1^2 = 290 - 2560 = -2270, which is negative. The formulas can still be applied mechanically, as below, but the estimates are not a valid fit and the question should be treated as having misprinted data.

(a) Least squares equation

Deviation sums (n=40n = 40):

Σy2=882−56240=803.6,Σx12=−2270,Σx22=94−16240=87.6\Sigma y^2 = 882 - \frac{56^2}{40} = 803.6,\qquad \Sigma x_1^2 = -2270,\qquad \Sigma x_2^2 = 94 - \frac{16^2}{40} = 87.6

Σx1x2=159−320×1640=31,Σx1y=505−320×5640=57,Σx2y=276−16×5640=253.6\Sigma x_1x_2 = 159 - \frac{320\times16}{40} = 31,\quad \Sigma x_1y = 505 - \frac{320\times56}{40} = 57,\quad \Sigma x_2y = 276 - \frac{16\times56}{40} = 253.6

β^1=(57)(87.6)−(253.6)(31)(−2270)(87.6)−312=−2868.4−199,813=0.0144\hat\beta_1 = \frac{(57)(87.6) - (253.6)(31)}{(-2270)(87.6) - 31^2} = \frac{-2868.4}{-199{,}813} = 0.0144

β^2=(253.6)(−2270)−(57)(31)−199813=−577,439−199,813=2.890\hat\beta_2 = \frac{(253.6)(-2270) - (57)(31)}{-199813} = \frac{-577{,}439}{-199{,}813} = 2.890

With Yˉ=1.4\bar Y = 1.4, Xˉ1=8\bar X_1 = 8, Xˉ2=0.4\bar X_2 = 0.4: β^0=1.4−0.0144(8)−2.890(0.4)=0.129\hat\beta_0 = 1.4 - 0.0144(8) - 2.890(0.4) = 0.129.

Y^=0.129+0.0144X1+2.890X2\boxed{\hat Y = 0.129 + 0.0144X_1 + 2.890X_2}

(b) Interpretation

  • β^1=0.0144\hat\beta_1 = 0.0144: holding capital constant, one more unit of labour input raises output by about 0.014 units.
  • β^2=2.890\hat\beta_2 = 2.890: holding labour constant, one more unit of capital input raises output by about 2.89 units.
  • β^0=0.129\hat\beta_0 = 0.129: output when both inputs are zero; it has no practical meaning.

(c) Adjusted R2R^2

ESS=β^1Σx1y+β^2Σx2y=0.0144(57)+2.890(253.6)=733.7ESS = \hat\beta_1\Sigma x_1y + \hat\beta_2\Sigma x_2y = 0.0144(57) + 2.890(253.6) = 733.7

R2=733.7803.6=0.913,Rˉ2=1−(1−R2)n−1n−k=1−0.087×3937=0.908R^2 = \frac{733.7}{803.6} = 0.913,\qquad \bar R^2 = 1 - (1-R^2)\frac{n-1}{n-k} = 1 - 0.087\times\frac{39}{37} = 0.908

Formally, about 91 percent of the variation in output would be explained by the two inputs after adjusting for degrees of freedom. Almost all of this comes from capital (X2X_2), as labour has a tiny coefficient. Since ΣX12\Sigma X_1^2 is impossible, this figure is only indicative until the data are corrected.

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  • 2025 · Statistics Methods (Econ. 557) · 10 marks

Define the sampling distribution of the sample mean. Consider a population consisting of elements {2,4,6,8,10}\{2, 4, 6, 8, 10\} and a sample of size 4 is to be drawn. Illustrate the sampling distribution of the sample mean when sample are drawn without replacement. Compute the mean and variance of the sampling distribution so obtained. Also, compare the mean and variance of the sampling distribution with the population mean and variance. Are the equal? Why or why not? [1+1+3+2+2+1]

Answer

1. Sampling distribution of the sample mean

The sampling distribution of the sample mean is the probability distribution of all possible values of the sample mean (\bar X), obtained from all possible samples of a fixed size (n) drawn from a given population. It shows each possible value of (\bar X) and its probability of occurrence.


2. Population parameters

Population:
[ X={2,4,6,8,10}, \qquad N=5 ]

Population mean: [ \mu=\frac{\sum X}{N}=\frac{2+4+6+8+10}{5}=\frac{30}{5}=6 ]

Population variance: [ \sigma^2=\frac{\sum (X-\mu)^2}{N} ]

Deviations from mean (6): [ -4,\ -2,\ 0,\ 2,\ 4 ]

Squares: [ 16,\ 4,\ 0,\ 4,\ 16 ]

So, [ \sigma^2=\frac{16+4+0+4+16}{5}=\frac{40}{5}=8 ]

Thus, [ \mu=6,\qquad \sigma^2=8 ]


3. Sampling distribution of (\bar X) for (n=4) without replacement

Since (N=5) and (n=4), the number of possible samples is: [ \binom{5}{4}=5 ]

Each sample is obtained by leaving out one element from the population. Each sample has probability (\frac15).

Sample no.Sample elementsSample mean (\bar X)Probability (P(\bar X))
1(4,6,8,10)(\frac{28}{4}=7)(\frac15)
2(2,6,8,10)(\frac{26}{4}=6.5)(\frac15)
3(2,4,8,10)(\frac{24}{4}=6)(\frac15)
4(2,4,6,10)(\frac{22}{4}=5.5)(\frac15)
5(2,4,6,8)(\frac{20}{4}=5)(\frac15)

So the sampling distribution is:

(\bar X)55.566.57
(P(\bar X))(\frac15)(\frac15)(\frac15)(\frac15)(\frac15)

4. Mean of the sampling distribution

[ E(\bar X)=\sum \bar X P(\bar X) ]

[ E(\bar X)=\frac15(5+5.5+6+6.5+7) ]

[ E(\bar X)=\frac{30}{5}=6 ]

Therefore, [ \boxed{E(\bar X)=\mu=6} ]


5. Variance of the sampling distribution

[ \operatorname{Var}(\bar X)=E(\bar X^2)-[E(\bar X)]^2 ]

First, [ E(\bar X^2)=\frac15(5^2+5.5^2+6^2+6.5^2+7^2) ]

[ =\frac15(25+30.25+36+42.25+49) ]

[ =\frac{182.5}{5}=36.5 ]

Thus, [ \operatorname{Var}(\bar X)=36.5-(6)^2 ]

[ =36.5-36=0.5 ]

So, [ \boxed{\operatorname{Var}(\bar X)=0.5} ]

Using the finite population correction formula: [ \operatorname{Var}(\bar X)=\frac{\sigma^2}{n}\cdot \frac{N-n}{N-1} ]

[ =\frac{8}{4}\cdot \frac{5-4}{5-1} ]

[ =2\cdot \frac14=0.5 ]

This confirms the result.


6. Comparison with population mean and variance

MeasurePopulationSampling distribution of (\bar X)
Mean(\mu=6)(E(\bar X)=6)
Variance(\sigma^2=8)(\operatorname{Var}(\bar X)=0.5)

Conclusion

  • The mean of the sampling distribution is equal to the population mean: [ E(\bar X)=\mu=6 ] So the sample mean is an unbiased estimator of the population mean.

  • The variance of the sampling distribution is not equal to the population variance: [ \operatorname{Var}(\bar X)=0.5 \neq 8=\sigma^2 ]

Why are the variances not equal?

Because the sample is drawn without replacement from a finite population. In this case, the observations in the sample are not independent; there is a finite population correction factor: [ \frac{N-n}{N-1} ]

Here, [ \frac{N-n}{N-1}=\frac{5-4}{5-1}=\frac14 ]

So the variance of the sample mean becomes much smaller than the population variance. If sampling were done with replacement, the variance would be: [ \operatorname{Var}(\bar X)=\frac{\sigma^2}{n}=\frac{8}{4}=2 ]

But because sampling is without replacement from a finite population, the correct variance is: [ \boxed{\operatorname{Var}(\bar X)=0.5} ]

Thus, the mean is equal, but the variance is not equal due to the finite population correction under without-replacement sampling.

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  • 2025 · Statistics Methods (Econ. 557) · 5 marks

The following data relating to the number of Washing Machine sold by different companies in three years 2022, 2023 and 2024.
YearsLGSamsungWhirlpoolSkyworth
202248503940
202350484548
202458564850
Is there a significance difference in the sales made (a) by four companies? (b) by three different years? [2+3]

Answer

Two-way ANOVA (without replication)

Let rows = years and columns = companies.

  • (r = 3) years, (c = 4) companies, (N = 12)
  • Total sales (T = 580)
  • Correction factor: [ CF=\frac{T^2}{N}=\frac{580^2}{12}=28033.3333 ]

Sums of squares

[ SS_{\text{Total}}=\sum x^2-CF=28362-28033.3333=328.6667 ]

Year sums: (177,\ 191,\ 212)

[ SS_{\text{Years}}=\frac{177^2+191^2+212^2}{4}-CF =28188.5-28033.3333=155.1667 ]

Company sums: (156,\ 154,\ 132,\ 138)

[ SS_{\text{Companies}}=\frac{156^2+154^2+132^2+138^2}{3}-CF =\frac{84520}{3}-28033.3333=140.0000 ]

[ SS_{\text{Error}}=SS_{\text{Total}}-SS_{\text{Years}}-SS_{\text{Companies}} =328.6667-155.1667-140=33.5 ]

ANOVA table

SourceSSdfMSF
Years155.1667277.5833(13.8955)
Companies140.0000346.6667(8.3582)
Error33.500065.5833
Total328.666711

Critical values at 5% level

[ F_{0.05}(2,6)=5.14,\qquad F_{0.05}(3,6)=4.76 ]

(a) Difference among four companies

[ F_{\text{Companies}}=8.3582 > 4.76 ]

So (H_0) is rejected. There is a significant difference in sales among the four companies.

(b) Difference among three years

[ F_{\text{Years}}=13.8955 > 5.14 ]

So (H_0) is rejected. There is a significant difference in sales among the three years.

Conclusion: At the 5% level of significance, sales differ significantly both by company and by year. At the 1% level, the year effect remains significant, while the company effect is not significant.

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  • 2025 · Statistics Methods (Econ. 557) · 10 marks

Suppose a probability density function (pdf) of the variable is given by f(x)=kx2f(x) = kx^2. For 0<x<20<x<2, where k is constant. For what value of k, the function will serve as a valid pdf? Compute E(X) and V(X) and also V(2X+4) and V(3X+4) for the given pdf, where symbols have their usual meanings. [1+4+5]

Answer

Given pdf: [ f(x)=kx^2,\qquad 0<x<2 ]

1. Value of (k)

For a valid pdf, [ \int_0^2 f(x),dx=1 ] [ \int_0^2 kx^2,dx=1 ] [ k\left[\frac{x^3}{3}\right]_0^2=1 ] [ k\left(\frac{8}{3}\right)=1 ] [ k=\frac{3}{8} ]

So, [ \boxed{k=\frac{3}{8}} ]

Thus, [ f(x)=\frac{3}{8}x^2,\qquad 0<x<2 ]


2. (E(X))

[ E(X)=\int_0^2 x f(x),dx ] [ =\int_0^2 x\left(\frac{3}{8}x^2\right)dx ] [ =\frac{3}{8}\int_0^2 x^3,dx ] [ =\frac{3}{8}\left[\frac{x^4}{4}\right]_0^2 ] [ =\frac{3}{8}\cdot \frac{16}{4} ] [ =\frac{3}{8}\cdot 4 ] [ =\frac{3}{2} ]

[ \boxed{E(X)=\frac{3}{2}=1.5} ]


3. (V(X))

First compute (E(X^2)): [ E(X^2)=\int_0^2 x^2 f(x),dx ] [ =\int_0^2 x^2\left(\frac{3}{8}x^2\right)dx ] [ =\frac{3}{8}\int_0^2 x^4,dx ] [ =\frac{3}{8}\left[\frac{x^5}{5}\right]_0^2 ] [ =\frac{3}{8}\cdot \frac{32}{5} ] [ =\frac{12}{5} ]

Now, [ V(X)=E(X^2)-[E(X)]^2 ] [ =\frac{12}{5}-\left(\frac{3}{2}\right)^2 ] [ =\frac{12}{5}-\frac{9}{4} ] [ =\frac{48-45}{20} ] [ =\frac{3}{20} ]

[ \boxed{V(X)=\frac{3}{20}=0.15} ]


4. (V(2X+4))

Using the property [ V(aX+b)=a^2V(X) ] we get [ V(2X+4)=2^2V(X) ] [ =4\cdot \frac{3}{20} ] [ =\frac{12}{20} ] [ =\frac{3}{5} ]

[ \boxed{V(2X+4)=\frac{3}{5}=0.6} ]


5. (V(3X+4))

Similarly, [ V(3X+4)=3^2V(X) ] [ =9\cdot \frac{3}{20} ] [ =\frac{27}{20} ]

[ \boxed{V(3X+4)=\frac{27}{20}=1.35} ]


Final Answers

[ \boxed{k=\frac{3}{8}} ]

[ \boxed{E(X)=\frac{3}{2}} ]

[ \boxed{V(X)=\frac{3}{20}} ]

[ \boxed{V(2X+4)=\frac{3}{5}} ]

[ \boxed{V(3X+4)=\frac{27}{20}} ]

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  • 2025 · Statistics Methods (Econ. 557) · 5 marks

A survey conducted by the Traffic Department of Kathmandu Valley finds that, on average, there are 5 first-vehicle accidents per day in Kathmandu valley. In a typical day, what is the probability that (i) no accident occurs (ii) exactly 3 accidents occur (iii) more than 2 accidents occur? [1+2+2]

Answer

Let (X) be the number of first-vehicle accidents per day in Kathmandu Valley. Since accidents occur randomly and independently at an average rate of 5 per day,

[ X \sim \text{Poisson}(\lambda = 5) ]

The Poisson probability mass function is

[ P(X=x)=\frac{e^{-\lambda}\lambda^x}{x!} ]

Here, (\lambda=5) and (e^{-5}\approx 0.0067379).


(i) Probability that no accident occurs

[ P(X=0)=\frac{e^{-5}5^0}{0!}=e^{-5} ]

[ P(X=0)\approx 0.0067379 ]

[ \boxed{P(X=0)\approx 0.0067} ]


(ii) Probability that exactly 3 accidents occur

[ P(X=3)=\frac{e^{-5}5^3}{3!} ]

[ =\frac{e^{-5}(125)}{6} ]

[ =20.8333 \times 0.0067379 ]

[ \boxed{P(X=3)\approx 0.1404} ]


(iii) Probability that more than 2 accidents occur

[ P(X>2)=1-P(X\le 2) ]

[ P(X\le 2)=P(X=0)+P(X=1)+P(X=2) ]

Now,

[ P(X=1)=\frac{e^{-5}5^1}{1!}=5e^{-5}\approx 0.03369 ]

[ P(X=2)=\frac{e^{-5}5^2}{2!}=\frac{25e^{-5}}{2}\approx 0.08422 ]

So,

[ P(X\le 2)=0.0067379+0.0336897+0.0842243 ]

[ =0.124652 ]

Therefore,

[ P(X>2)=1-0.124652 ]

[ \boxed{P(X>2)\approx 0.8753} ]


Final Answers

[ \boxed{P(X=0)\approx 0.0067} ]

[ \boxed{P(X=3)\approx 0.1404} ]

[ \boxed{P(X>2)\approx 0.8753} ]

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  • 2025 · Statistics Methods (Econ. 557) · 5 marks

Define binomial distribution of a random variable X. Provide the derivation of its mean and variance. [2+3]

Answer

Binomial Distribution

A random variable XX is said to follow a binomial distribution if it represents the number of successes in nn independent Bernoulli trials, where each trial has only two possible outcomes (success or failure), the probability of success pp remains constant from trial to trial, and the trials are statistically independent. Its probability mass function is

P(X=x)=(nx)pxqn−x,x=0,1,2,…,nP(X=x) = \binom{n}{x} p^x q^{n-x}, \quad x = 0, 1, 2, \dots, n

where q=1−pq = 1-p and nn is the number of trials. We write X∼B(n,p)X \sim B(n, p).

Derivation of Mean

By definition,

E(X)=∑x=0nx(nx)pxqn−xE(X) = \sum_{x=0}^{n} x \binom{n}{x} p^x q^{n-x}

Since the term for x=0x = 0 vanishes,

E(X)=∑x=1nx⋅n!x!(n−x)!pxqn−xE(X) = \sum_{x=1}^{n} x \cdot \frac{n!}{x!(n-x)!} p^x q^{n-x}

Using x(nx)=n(n−1x−1)x \binom{n}{x} = n \binom{n-1}{x-1},

E(X)=n∑x=1n(n−1x−1)pxqn−xE(X) = n \sum_{x=1}^{n} \binom{n-1}{x-1} p^x q^{n-x}

Put y=x−1y = x - 1, so that when x=1x = 1, y=0y = 0 and when x=nx = n, y=n−1y = n-1:

E(X)=np∑y=0n−1(n−1y)pyq(n−1)−yE(X) = n p \sum_{y=0}^{n-1} \binom{n-1}{y} p^y q^{(n-1)-y}

The summation is the total probability of a binomial distribution with parameters (n−1)(n-1) and pp, hence equals 1. Therefore,

E(X)=μ=np\boxed{E(X) = \mu = np}

Derivation of Variance

First compute E[X(X−1)]E[X(X-1)]:

E[X(X−1)]=∑x=0nx(x−1)(nx)pxqn−xE[X(X-1)] = \sum_{x=0}^{n} x(x-1) \binom{n}{x} p^x q^{n-x}

The terms for x=0x = 0 and x=1x = 1 vanish, so

E[X(X−1)]=∑x=2nx(x−1)n!x!(n−x)!pxqn−xE[X(X-1)] = \sum_{x=2}^{n} x(x-1) \frac{n!}{x!(n-x)!} p^x q^{n-x}

Using x(x−1)(nx)=n(n−1)(n−2x−2)x(x-1)\binom{n}{x} = n(n-1)\binom{n-2}{x-2},

E[X(X−1)]=n(n−1)∑x=2n(n−2x−2)pxqn−xE[X(X-1)] = n(n-1) \sum_{x=2}^{n} \binom{n-2}{x-2} p^x q^{n-x}

Put y=x−2y = x - 2:

E[X(X−1)]=n(n−1)p2∑y=0n−2(n−2y)pyq(n−2)−y=n(n−1)p2E[X(X-1)] = n(n-1) p^2 \sum_{y=0}^{n-2} \binom{n-2}{y} p^y q^{(n-2)-y} = n(n-1)p^2

Now,

Var(X)=E(X2)−[E(X)]2=E[X(X−1)]+E(X)−[E(X)]2\text{Var}(X) = E(X^2) - [E(X)]^2 = E[X(X-1)] + E(X) - [E(X)]^2

Substituting the results obtained:

Var(X)=n(n−1)p2+np−(np)2\text{Var}(X) = n(n-1)p^2 + np - (np)^2

=n2p2−np2+np−n2p2=np−np2=np(1−p)= n^2p^2 - np^2 + np - n^2p^2 = np - np^2 = np(1-p)

Var(X)=σ2=npq\boxed{\text{Var}(X) = \sigma^2 = npq}

Thus the mean of the binomial distribution is npnp and its variance is npqnpq, where q=1−pq = 1-p.

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  • 2025 · Statistics Methods (Econ. 557) · 5 marks

Define estimation. Describe the properties of good estimator. [2+3]

Answer

Estimation is the statistical procedure of using sample data to infer (estimate) the unknown value of a population parameter. The sample statistic used for this purpose is called an estimator, and the numerical value it takes in a particular sample is called an estimate. For example, the sample mean xˉ\bar{x} is an estimator of the population mean μ\mu.

A good estimator should possess the following properties:

  1. Unbiasedness: An estimator TT is unbiased if its expected value equals the true parameter, i.e. E(T)=θE(T) = \theta. It is correct on average over repeated sampling; there is no systematic over- or under-estimation.

  2. Efficiency: Among all unbiased estimators, the most efficient is the one with the smallest variance. If T1T_1 and T2T_2 are unbiased, T1T_1 is more efficient than T2T_2 when Var(T1)<Var(T2)\text{Var}(T_1) < \text{Var}(T_2). The sample mean is the most efficient estimator of μ\mu for a normal population.

  3. Consistency: An estimator is consistent if it approaches the true parameter value as the sample size increases indefinitely, i.e. plim Tn=θ\text{plim } T_n = \theta as n→∞n \to \infty. This requires both bias and variance to vanish in the limit.

  4. Sufficiency: An estimator is sufficient if it uses all the information in the sample relevant to the parameter, so that no other statistic can provide additional information about θ\theta. The statistic TT is sufficient if the conditional distribution of the sample given TT does not depend on θ\theta.

  5. Efficiency in small samples (Minimum Variance): Ideally the estimator should have minimum variance among unbiased estimators (MVUE), giving the most precise estimates for a given sample size.

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  • 2025 · Statistics Methods (Econ. 557) · 5 marks

What are dummy variables? Why it is incorporated in regression model? Illustrate with an example. [2+2+1]

Answer

Dummy Variables

1. Meaning of Dummy Variables (2 marks)

A dummy variable (also called an indicator, binary, or qualitative variable) is an artificial variable that takes only two values, usually 0 or 1, used to represent a qualitative or categorical attribute that cannot be measured numerically — such as sex, marital status, religion, region, or season.

  • 1 = the attribute is present (the category of interest)
  • 0 = the attribute is absent (the reference or base category)

For example, for the attribute "sex":

Di=1 if the person is female,Di=0 if the person is maleD_i = 1 \text{ if the person is female}, \quad D_i = 0 \text{ if the person is male}

2. Why Dummy Variables Are Incorporated in the Regression Model (2 marks)

  1. To handle qualitative/qualitative-type information: Many economic variables (e.g., gender, caste, urban–rural residence, before/after a policy) are qualitative. Regression requires numerical inputs, so such attributes must be quantified through dummies.
  2. To allow intercept (or slope) shifts: Dummies let the regression line shift up or down for different categories, capturing structural differences between groups that a purely quantitative model would ignore.
  3. To avoid omitted variable bias: Excluding a relevant qualitative factor pushes its effect into the error term, biasing the estimated coefficients. Including dummies reduces this bias.
  4. To test hypotheses about group differences: The significance of a dummy's coefficient directly tests whether the category differs significantly from the base group.
  5. To capture seasonal or structural breaks: Dummies are used for seasonal effects (quarterly/monthly data) and for structural change (pre-reform vs. post-reform periods).

3. Illustration with an Example (1 mark)

Suppose we study the starting salary of graduates as a function of years of experience and sex:

Salaryi=β1+β2Experiencei+β3Di+ui\text{Salary}_i = \beta_1 + \beta_2 \text{Experience}_i + \beta_3 D_i + u_i

where Di=1D_i = 1 for female, 00 for male.

  • Male: E(Salary)=β1+β2ExperienceE(\text{Salary}) = \beta_1 + \beta_2 \text{Experience}
  • Female: E(Salary)=(β1+β3)+β2ExperienceE(\text{Salary}) = (\beta_1 + \beta_3) + \beta_2 \text{Experience}

Here β2\beta_2 is the common return to experience, and β3\beta_3 measures the salary differential between females and males. If β3<0\beta_3 < 0 and significant, females earn less than males at any given experience level — evidence of discrimination. If β3=0\beta_3 = 0, sex has no effect on salary.

Note on the dummy-variable trap: For a category with mm groups, use only (m−1)(m-1) dummies plus the intercept; including all mm dummies causes perfect multicollinearity.

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  • 2025 · Statistics Methods (Econ. 557) · 5 marks

If the regression equation of Y on X is Y=3+0.4XY = 3 + 0.4X and Xˉ=200\bar{X} = 200, ∑X2=2000\sum X^2 = 2000, ∑Y2=3990\sum Y^2 = 3990, ∑XY=2700\sum X Y = 2700, Yˉ=1660\bar{Y} = 1660 a) Find the lines of least square regression. b) Find the coefficient of determination and interpret it. [3+2]

Answer

Note on the data. The data conflict. The line Y=3+0.4XY = 3 + 0.4X passes through the means, so Xˉ=200\bar X = 200 gives Yˉ=3+0.4(200)=83\bar Y = 3 + 0.4(200) = 83, not 16601660. Also ΣX2=2000\Sigma X^2 = 2000 is smaller than nXˉ2n\bar X^2 for any n≥1n\ge1. The given sums ΣX2\Sigma X^2, ΣY2\Sigma Y^2, ΣXY\Sigma XY are therefore treated as sums of squares and products about the means (deviation sums), and Yˉ=83\bar Y = 83 from the regression line is used.

(a) Lines of regression

Regression of YY on XX (given):   Y=3+0.4X\;Y = 3 + 0.4X, so byx=0.4b_{yx} = 0.4.

Regression of XX on YY: the coefficient is

bxy=ΣxyΣy2=27003990=0.677b_{xy} = \frac{\Sigma xy}{\Sigma y^2} = \frac{2700}{3990} = 0.677

X−Xˉ=bxy(Y−Yˉ)  ⇒  X−200=0.677(Y−83)  ⇒  X=143.8+0.677YX - \bar X = b_{xy}(Y-\bar Y)\;\Rightarrow\; X - 200 = 0.677(Y-83)\;\Rightarrow\; X = 143.8 + 0.677Y

(b) Coefficient of determination

r2=byx×bxy=0.4×0.677=0.271,r=+0.271=0.52r^2 = b_{yx}\times b_{xy} = 0.4\times0.677 = 0.271,\qquad r = +\sqrt{0.271} = 0.52

The sign is positive because both regression coefficients are positive. About 27 percent of the variation in YY is explained by its linear dependence on XX, and the remaining 73 percent is due to other factors. This is a moderate, not strong, fit.

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  • 2025 · Statistics Methods (Econ. 557) · 5 marks

A random sample of 21 values from a normal population showed a mean of 58 and sum of the squares of deviations from this mean equals 15. Can this sample be regarded as taken from the population having 61 as mean? Obtain 95% and 99% confidence limits of the mean. [2+3]

Answer

Given: n=21n = 21, xˉ=58\bar x = 58, Σ(x−xˉ)2=15\Sigma(x-\bar x)^2 = 15, μ0=61\mu_0 = 61.

Test of H0:μ=61H_0:\mu = 61 against H1:μ≠61H_1:\mu\neq61. The population variance is unknown and the sample is small, so the tt-test is used.

s2=1520=0.75,s=0.866,SE=sn=0.86621=0.189s^2 = \frac{15}{20} = 0.75,\qquad s = 0.866,\qquad SE = \frac{s}{\sqrt n} = \frac{0.866}{\sqrt{21}} = 0.189

t=58−610.189=−15.87t = \frac{58-61}{0.189} = -15.87

With df=20df = 20 and α=0.05\alpha = 0.05, t0.025,20=2.086t_{0.025,20} = 2.086. Since ∣t∣=15.87>2.086|t| = 15.87 > 2.086, reject H0H_0: the sample cannot be regarded as drawn from a population with mean 61.

Confidence limits for the mean

95 percent: 58±2.086(0.189)=58±0.39458 \pm 2.086(0.189) = 58 \pm 0.394, so the limits are (57.61, 58.39)(57.61,\ 58.39).

99 percent (t0.005,20=2.845t_{0.005,20} = 2.845): 58±2.845(0.189)=58±0.53858 \pm 2.845(0.189) = 58\pm0.538, so the limits are (57.46, 58.54)(57.46,\ 58.54).

Neither interval contains 61, which agrees with the rejection of H0H_0.

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  • 2025 · Statistics Methods (Econ. 557) · 5 marks

Write normal density function of a random variable. The SEE (Secondary Education Examination) results of students in Kathmandu follow a normal distribution with Mean score (μ\mu) = 65 and Standard deviation (σ\sigma) = 10. Find the probability that a randomly selected students (i) scores more than 80 marks (ii) scores between 60 and 75 marks. [1+2+2]

Answer

Normal density function

A continuous random variable XX is said to follow a normal distribution with mean μ\mu and standard deviation σ\sigma if its probability density function is

f(x)=1σ2πexp⁡[−(x−μ)22σ2],−∞<x<∞f(x)=\frac{1}{\sigma\sqrt{2\pi}}\exp\left[-\frac{(x-\mu)^2}{2\sigma^2}\right],\qquad -\infty<x<\infty

For the given problem, μ=65\mu=65 and σ=10\sigma=10, so

f(x)=1102πexp⁡[−(x−65)2200]f(x)=\frac{1}{10\sqrt{2\pi}}\exp\left[-\frac{(x-65)^2}{200}\right]

Let XX be the SEE score. Then X∼N(65,102)X\sim N(65,10^2).

The standard normal variable is

Z=X−μσ=X−6510Z=\frac{X-\mu}{\sigma}=\frac{X-65}{10}

(i) Probability of scoring more than 80 marks

P(X>80)=P(Z>80−6510)=P(Z>1.5)P(X>80)=P\left(Z>\frac{80-65}{10}\right)=P(Z>1.5)

From the standard normal table,

P(Z>1.5)=1−P(Z<1.5)=1−0.9332=0.0668P(Z>1.5)=1-P(Z<1.5)=1-0.9332=0.0668

P(X>80)=0.0668≈6.68%\boxed{P(X>80)=0.0668\approx 6.68\%}

(ii) Probability of scoring between 60 and 75 marks

P(60<X<75)=P(60−6510<Z<75−6510)P(60<X<75)=P\left(\frac{60-65}{10}<Z<\frac{75-65}{10}\right)

=P(−0.5<Z<1)=P(-0.5<Z<1)

=P(Z<1)−P(Z<−0.5)=P(Z<1)-P(Z<-0.5)

=0.8413−0.3085=0.5328=0.8413-0.3085=0.5328

P(60<X<75)=0.5328≈53.28%\boxed{P(60<X<75)=0.5328\approx 53.28\%}

Thus, the probability that a randomly selected student scores more than 80 marks is about 0.06680.0668, and the probability that the score lies between 60 and 75 marks is about 0.53280.5328.

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  • 2025 · Statistics Methods (Econ. 557) · 5 marks

Discuss the general procedure for Z-test. [1+2+2]

Answer

General procedure for testing a hypothesis using the Z-test

The Z-test is used to test a hypothesis about a population parameter when the sample size is large (n ≥ 30) or when the population standard deviation (σ) is known and the population is normal. The general procedure involves the following steps:

1. Setting up the hypotheses (1 mark) State the null hypothesis (H₀) and the alternative hypothesis (H₁). For example, for testing a population mean:

  • H₀: μ = μ₀ (no difference)
  • H₁: μ ≠ μ₀ (two-tailed) or μ > μ₀ / μ < μ₀ (one-tailed)

2. Specifying the test statistic and its distribution (2 marks) Choose the Z-statistic appropriate to the parameter being tested. For a sample mean from a large sample or known σ:

Z=Xˉ−μ0σ/nZ = \frac{\bar{X} - \mu_0}{\sigma/\sqrt{n}}

When σ is unknown but n is large, use the sample standard deviation s:

Z=Xˉ−μ0s/nZ = \frac{\bar{X} - \mu_0}{s/\sqrt{n}}

For testing a population proportion:

Z=p^−p0p0q0/nZ = \frac{\hat{p} - p_0}{\sqrt{p_0 q_0 / n}}

Under H₀, this Z-statistic follows the standard normal distribution with mean 0 and variance 1. The level of significance (α) is chosen, and the critical value Z_α (or Z_α/2 for a two-tailed test) is obtained from the standard normal table.

3. Making the decision (2 marks) Compute the value of Z from the sample data and compare it with the critical value:

  • If |Z| > Z_α/2 (two-tailed) or Z > Z_α (right-tailed) / Z < −Z_α (left-tailed), reject H₀.
  • Otherwise, do not reject H₀.

Alternatively, use the p-value approach: if the p-value is less than α, reject H₀; otherwise do not reject it. Finally, interpret the result in the context of the problem.

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  • 2025 · Statistics Methods (Econ. 557) · 5 marks

Find the correlation coefficient between age and playing habits of the following students.
Age151617181920
No. of students25020015012010080
Regular Players20015090483012

Answer

Let age be XX, and measure playing habit by the percentage of regular players, Y=regular playersno. of students×100Y = \dfrac{\text{regular players}}{\text{no. of students}}\times100.

Age XXStudentsRegular playersYY (%)dx=X−17.5dx = X-17.5dy=Y−50dy = Y-50dx2dx^2dy2dy^2dx dydx\,dy
1525020080-2.5306.25900-75
1620015075-1.5252.25625-37.5
171509060-0.5100.25100-5
1812048400.5-100.25100-5
1910030301.5-202.25400-30
208012152.5-356.251225-87.5
Total30017.53350-240

Xˉ=105/6=17.5\bar X = 105/6 = 17.5 and Yˉ=300/6=50\bar Y = 300/6 = 50.

r=Σdx dyΣdx2  Σdy2=−24017.5×3350=−240242.1≈−0.99r = \frac{\Sigma dx\,dy}{\sqrt{\Sigma dx^2\;\Sigma dy^2}} = \frac{-240}{\sqrt{17.5\times3350}} = \frac{-240}{242.1} \approx -0.99

There is a very high negative correlation between age and playing habit: as age increases, the proportion of regular players falls sharply.

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  • 2025 · Statistics Methods (Econ. 557) · 5 marks

Write short notes on any two a) Confidence Interval b) Control limit theorem c) Sample space

Answer

a) Confidence Interval

A confidence interval (CI) is a range of values, calculated from sample data, that is likely to contain the true (unknown) population parameter. Because a sample statistic (like xˉ\bar{x}) varies from sample to sample, we do not report a single value but an interval with a stated degree of confidence.

Formula (for population mean, σ\sigma known):

CI=xˉ±zα/2⋅σnCI = \bar{x} \pm z_{\alpha/2} \cdot \frac{\sigma}{\sqrt{n}}

where xˉ\bar{x} = sample mean, σ\sigma = population standard deviation, nn = sample size, and zα/2z_{\alpha/2} is the critical value from the standard normal table.

Interpretation: A 95% confidence interval means that if we repeatedly drew samples and built intervals the same way, about 95% of those intervals would contain the true parameter. It does not mean there is a 95% probability that the parameter lies in this particular interval.

Key points:

  • Higher confidence level (e.g., 99% vs 95%) gives a wider interval.
  • Larger sample size nn gives a narrower (more precise) interval.
  • When σ\sigma is unknown and nn is small, we use the tt-distribution: xˉ±tα/2⋅sn\bar{x} \pm t_{\alpha/2} \cdot \frac{s}{\sqrt{n}}.

Example: If xˉ=50\bar{x} = 50, σ=10\sigma = 10, n=100n = 100, then the 95% CI is 50±1.96(1)=(48.04, 51.96)50 \pm 1.96(1) = (48.04,\ 51.96).


b) Control Limit Theorem

Note: The standard term is the Central Limit Theorem (CLT); "control limit theorem" is often used loosely for the same idea in quality-control contexts.

The Central Limit Theorem states that if random samples of size nn are drawn from any population with mean μ\mu and finite variance σ2\sigma^2, then as nn becomes large, the sampling distribution of the sample mean xˉ\bar{x} approaches a normal distribution with:

μxˉ=μ,σxˉ=σn\mu_{\bar{x}} = \mu, \qquad \sigma_{\bar{x}} = \frac{\sigma}{\sqrt{n}}

Standardised form:

Z=xˉ−μσ/n→dN(0,1)as n→∞Z = \frac{\bar{x} - \mu}{\sigma/\sqrt{n}} \xrightarrow{d} N(0,1) \quad \text{as } n \to \infty

Key points:

  • The parent population need not be normal; the theorem holds for large nn (usually n≥30n \ge 30).
  • It explains why the normal distribution is so widely used in inference.
  • It forms the basis of confidence intervals, hypothesis testing and control charts.

Application in quality control: Control limits are set at μ±3σ/n\mu \pm 3\sigma/\sqrt{n}. By the CLT, sample means fall within these limits with very high probability (≈99.73%\approx 99.73\%) when the process is in control; points outside indicate the process is out of control.


c) Sample Space

A sample space is the set of all possible outcomes of a random experiment. It is usually denoted by SS (or Ω\Omega).

Examples:

  • Tossing a coin: S={H,T}S = \{H, T\}
  • Rolling a die: S={1,2,3,4,5,6}S = \{1, 2, 3, 4, 5, 6\}
  • Tossing two coins: S={HH,HT,TH,TT}S = \{HH, HT, TH, TT\}

Types:

  • Discrete (finite or countably infinite): e.g., number of defective items in a lot.
  • Continuous: e.g., the lifetime of a bulb, S={t:t≥0}S = \{t : t \ge 0\}.

Related terms:

  • Event: any subset of the sample space, e.g., A={2,4,6}A = \{2, 4, 6\} (even number on a die).
  • Simple event: an event with exactly one outcome.
  • Exhaustive events: events whose union equals SS.
  • Mutually exclusive events: events with no common outcome.

Probability: For equally likely outcomes, P(A)=n(A)n(S)P(A) = \dfrac{n(A)}{n(S)}, where n(A)n(A) is the number of favourable outcomes and n(S)n(S) the total number of outcomes. The sample space is thus the foundation on which all probability calculations are built.

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  • 2082 · Mathematics For Economics (Econ. 503) · 10 marks

If the consumer's utility function is U=x0.5y0.5U=x^{0.5}y^{0.5} and prices and x and y are Rs.25 and Rs.75, respectively. Find the optimum commodity purchase where the consumer's total income is Rs.1500. Verify that the second-order conditions for utility maximization are satisfied. Compute the maximum utility attained at the optimum consumption bundle and verify that the entire income is exhausted. [10+3+2]

Answer

Let the utility function be
[ U=x^{0.5}y^{0.5}=\sqrt{xy} ] with prices [ p_x=25,\qquad p_y=75 ] and income [ M=1500. ]

The consumer maximizes utility subject to the budget constraint: [ 25x+75y=1500. ]

1. Optimum commodity purchase

Use the Lagrange multiplier method: [ L=x^{0.5}y^{0.5}+\lambda(1500-25x-75y). ]

First-order conditions: [ \frac{\partial L}{\partial x}=0.5x^{-0.5}y^{0.5}-25\lambda=0 ] [ \frac{\partial L}{\partial y}=0.5x^{0.5}y^{-0.5}-75\lambda=0 ] [ \frac{\partial L}{\partial \lambda}=1500-25x-75y=0. ]

Dividing the first condition by the second: [ \frac{0.5x^{-0.5}y^{0.5}}{0.5x^{0.5}y^{-0.5}}=\frac{25}{75} ] [ \frac{y}{x}=\frac{1}{3} ] [ x=3y. ]

Substitute (x=3y) into the budget constraint: [ 25(3y)+75y=1500 ] [ 75y+75y=1500 ] [ 150y=1500 ] [ y=10. ]

Then, [ x=3y=30. ]

Therefore, the optimum consumption bundle is [ \boxed{x=30,\qquad y=10}. ]

2. Second-order condition

For utility maximization with two choice variables and one constraint, the bordered Hessian determinant must be positive: [ |\bar H|>0. ]

Here, [ U_{xx}=-0.25x^{-1.5}y^{0.5} ] [ U_{yy}=-0.25x^{0.5}y^{-1.5} ] [ U_{xy}=0.25x^{-0.5}y^{-0.5}. ]

At (x=30,\ y=10): [ U_{xx}=-\frac{1}{120\sqrt3},\qquad U_{yy}=-\frac{\sqrt3}{40},\qquad U_{xy}=\frac{1}{40\sqrt3}. ]

The bordered Hessian is [ \bar H= \begin{bmatrix} 0 & 25 & 75\ 25 & U_{xx} & U_{xy}\ 75 & U_{xy} & U_{yy} \end{bmatrix}. ]

Its determinant is [ |\bar H| =2(25)(75)U_{xy}-(25)^2U_{yy}-(75)^2U_{xx}. ]

Substituting the values: [ |\bar H| =2(25)(75)\left(\frac{1}{40\sqrt3}\right) -(25)^2\left(-\frac{\sqrt3}{40}\right) -(75)^2\left(-\frac{1}{120\sqrt3}\right) ] [ =62.5\sqrt3>0. ]

Since [ \boxed{|\bar H|>0}, ] the second-order condition for utility maximization is satisfied.

3. Maximum utility

At the optimum bundle ((x,y)=(30,10)), [ U=30^{0.5}\cdot 10^{0.5} =\sqrt{30\cdot 10} =\sqrt{300} =10\sqrt3. ]

Thus, [ \boxed{U_{\max}=10\sqrt3\approx 17.32\text{ utils}}. ]

4. Verification that entire income is exhausted

Check the budget constraint: [ p_x x+p_y y=25(30)+75(10) ] [ =750+750=1500. ]

Hence, [ \boxed{25(30)+75(10)=1500}. ]

So the consumer spends the entire income of Rs.1500 at the optimum.

Final answer

[ \boxed{x=30,\quad y=10,\quad U_{\max}=10\sqrt3\approx 17.32} ] and the second-order condition is satisfied.

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  • 2082 · Mathematics For Economics (Econ. 503) · 5 marks

Define a difference equation. Using the demand function Qd=a−bpQ_d = a - bp and supply function Qs=−c+dpQ_s = -c + dp, show the economic application of the first order difference equation in the Cobweb model. Analyze the time path of adjustment with suitable diagram. [2+4+4]

Answer

Difference Equation

A difference equation is an equation that relates the value of a variable at one period to its value in one or more previous periods. If yty_t denotes the value of the variable yy at time tt, a general first-order difference equation is written as:

yt=f(yt−1,t)y_t = f(y_{t-1}, t)

When the relation is linear with constant coefficients, it takes the form:

yt=a+b yt−1y_t = a + b\,y_{t-1}

where aa and bb are constants. The order of a difference equation is the difference between the highest and the lowest time subscripts appearing in it; hence the above equation is of the first order, because only yty_t and yt−1y_{t-1} appear. Its solution (time path) is:

yt=yˉ+A b t,yˉ=a1−by_t = \bar{y} + A\,b^{\,t}, \qquad \bar{y} = \frac{a}{1-b}

where yˉ\bar{y} is the intertemporal equilibrium (steady-state) value and AA is determined by the initial condition y0y_0. The term Ab tA b^{\,t} is the complementary (transient) part. The behaviour of the time path depends entirely on bb:

  • ∣b∣<1|b| < 1: the path converges to yˉ\bar{y} (stable).
  • ∣b∣>1|b| > 1: the path diverges from yˉ\bar{y} (unstable).
  • b=1b = 1: there is no finite steady state and the path changes by the constant aa each period (linear drift).
  • b<0b < 0: the path oscillates (alternates above and below yˉ\bar{y}).

Economic Application: The Cobweb Model

The Cobweb model explains price fluctuations in an agricultural market where production decisions must be made one period in advance. Because the crop takes one period to grow, farmers decide how much to plant in period tt on the basis of the price prevailing in period t−1t-1, while demand responds to the current price.

Demand function:

Qd(t)=a−b Pt(a,b>0)Q_d(t) = a - b\,P_t \qquad (a, b > 0)

Supply function (lagged):

Qs(t)=−c+d Pt−1(c,d>0)Q_s(t) = -c + d\,P_{t-1} \qquad (c, d > 0)

Market clearing requires demand in period tt to equal supply in period tt:

a−b Pt=−c+d Pt−1a - b\,P_t = -c + d\,P_{t-1}

Solving for PtP_t:

b Pt=a+c−d Pt−1b\,P_t = a + c - d\,P_{t-1}

Pt=a+cb−db Pt−1…(1)P_t = \frac{a+c}{b} - \frac{d}{b}\,P_{t-1} \qquad \dots (1)

This is a first-order linear difference equation in price. Comparing with Pt=A0+A1Pt−1P_t = A_0 + A_1 P_{t-1}, we have A0=a+cbA_0 = \dfrac{a+c}{b} and A1=−dbA_1 = -\dfrac{d}{b}.

Intertemporal equilibrium price (setting Pt=Pt−1=PˉP_t = P_{t-1} = \bar{P}):

Pˉ=a+cb+d\bar{P} = \frac{a+c}{b+d}

Solution of the difference equation:

Pt=Pˉ+(P0−Pˉ)(−db)t…(2)P_t = \bar{P} + \left(P_0 - \bar{P}\right)\left(-\frac{d}{b}\right)^{t} \qquad \dots (2)

The term (−db)t\left(-\dfrac{d}{b}\right)^{t} is the oscillating component, and its magnitude governs the time path.


Time Path of Adjustment

Let k=dbk = \dfrac{d}{b} = ratio of the slopes of the supply and demand curves. The time path in equation (2) becomes:

Pt=Pˉ+(P0−Pˉ)(−k)tP_t = \bar{P} + (P_0 - \bar{P})(-k)^{t}

Case 1: k<1k < 1 (b>db > d)

Here ∣−k∣<1|{-k}| < 1, so (−k)t→0(-k)^t \to 0 as t→∞t \to \infty. Price oscillates with diminishing amplitude and settles at Pˉ\bar{P}. The market is dynamically stable (convergent cobweb). In the usual diagram with price on the vertical axis, dP/dQ=−1/bdP/dQ = -1/b for demand and +1/d+1/d for supply, so b>db > d means the supply curve is steeper than the demand curve: demand is more responsive to price than supply.

Case 2: k>1k > 1 (d>bd > b)

Here ∣−k∣>1|{-k}| > 1, so the oscillations explode and price moves further from Pˉ\bar{P} each period. The market is dynamically unstable (divergent cobweb). In the diagram the demand curve is steeper than the supply curve: supply reacts strongly to last period's price.

Case 3: k=1k = 1 (b=db = d)

Here (−k)t=±1(-k)^t = \pm 1, so price oscillates between two values with constant amplitude (closed rectangle in the diagram) and never reaches Pˉ\bar{P}.

Diagram (cobweb) and how to draw it: plot price PP on the vertical axis and quantity QQ on the horizontal axis, with the downward-sloping demand curve DD and the upward-sloping supply curve SS intersecting at (Qˉ,Pˉ)(\bar{Q}, \bar{P}). Start at an initial price P0P_0. Move horizontally to SS to read the quantity Q1Q_1 that farmers supply, then vertically to DD to find the price P1P_1 at which that quantity clears. Move horizontally back to SS for Q2Q_2, then vertically to DD for P2P_2, and so on. The stepwise path traces a cobweb around (Qˉ,Pˉ)(\bar{Q}, \bar{P}): it spirals inward if SS is steeper than DD (Case 1), outward if DD is steeper than SS (Case 2), and forms a closed rectangle if the slopes are equal (Case 3). A companion time-path graph of PtP_t against tt shows damped, explosive or constant zig-zag oscillations around Pˉ\bar{P} respectively.

Economic interpretation: Because producers respond to last period's price, price need not move smoothly to equilibrium. Stability depends on the relative responsiveness of supply and demand: the market converges when supply is less responsive to price than demand (d<bd < b), and diverges when supply is more responsive than demand (d>bd > b). This explains cyclical swings in agricultural prices (for example potato or vegetable prices in Nepal) and why price supports or buffer-stock policies may be used to stabilise them.

Answer checked by Claude Sonnet · Oct 21, 2026

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  • 2082 · Mathematics For Economics (Econ. 503) · 10 marks

A manufacturer makes chairs and tables. Processing these products are done by two types of machine A and B. A unit of chair making requires 2 hrs on Machine A and 6 hrs on machine B. A unit of Table requires 5 hrs. in Machine A and no time in machine B. There are 16 hrs. of time available on machine A and 30 hrs. in machine B. If profit from chairs and tables are Rs. 30 and 40 respectively. i) Formulate the problem as the Linear Programming (LP) ii) Find the optimal solution iii) Interpret the optimality iii) Interpret the optimality iv) Solve the LP using Simplex method v) Interpret the optimal solution and non-binding constraint to the given LPP. [4+4+1+2]

Answer

Let xx = number of chairs and yy = number of tables produced.

(i) Formulation of the LP

max⁡Z=30x+40y\max Z = 30x + 40y

subject to

2x+5y≤16(Machine A hours)2x + 5y \le 16 \quad \text{(Machine A hours)} 6x≤30(Machine B hours)6x \le 30 \quad \text{(Machine B hours)} x,y≥0x, y \ge 0

(ii) Graphical solution

Machine B constraint: x≤5x \le 5. Machine A constraint: 2x+5y≤162x + 5y \le 16, which cuts the axes at (8,0)(8, 0) and (0,3.2)(0, 3.2). The feasible region has corners:

CornerZ=30x+40yZ = 30x + 40y
O(0,0)O(0,0)0
(5,0)(5,0)150
(5,1.2)(5, 1.2)150+48=198150 + 48 = 198
(0,3.2)(0, 3.2)128

The corner (5,1.2)(5, 1.2) is where x=5x = 5 meets 2x+5y=162x + 5y = 16, since y=(16−10)/5=1.2y = (16-10)/5 = 1.2. The maximum is Z=Z = Rs 198 at x=5x = 5, y=1.2y = 1.2.

(iii) Interpretation of optimality

The firm maximises profit by making 5 chairs and 1.2 tables, earning Rs 198. At this point both machines are fully used. Any move along the boundary toward another corner lowers profit.

(iv) Simplex method

Add slack variables s1,s2s_1, s_2:

2x+5y+s1=16,6x+s2=302x + 5y + s_1 = 16, \qquad 6x + s_2 = 30

Table 1 (initial). Basis s1,s2s_1, s_2. CjC_j: 30, 40, 0, 0. Zj=0Z_j = 0, so Cj−Zj=30,40,0,0C_j - Z_j = 30, 40, 0, 0. The entering variable is yy (largest Cj−ZjC_j - Z_j = 40). Ratios: 16/5=3.216/5 = 3.2 for row 1; row 2 has no entry for yy. So s1s_1 leaves, and the pivot is 5.

Table 2. After pivoting: yy row is 0.4x+y+0.2s1=3.20.4x + y + 0.2 s_1 = 3.2; s2s_2 row is 6x+s2=306x + s_2 = 30. Z=128Z = 128. Cj−ZjC_j - Z_j: xx: 30−40(0.4)=1430 - 40(0.4) = 14; s1s_1: −8-8. The entering variable is xx. Ratios: 3.2/0.4=83.2/0.4 = 8 and 30/6=530/6 = 5, so s2s_2 leaves (pivot 6).

Table 3 (optimal).

Basisxxyys1s_1s2s_2RHS
yy (40)011/51/5−1/15-1/151.2
xx (30)1001/61/65
Cj−ZjC_j - Z_j00−8-8−7/3-7/3Z=198Z = 198

All Cj−Zj≤0C_j - Z_j \le 0, so the solution is optimal: x=5x = 5, y=1.2y = 1.2, Z=Z = Rs 198, the same as the graphical result.

(v) Interpretation of the optimal solution and non-binding constraints

  • Produce 5 chairs and 1.2 tables (1.2 tables means 6 tables per 5 production runs; if only whole units are allowed, the best integer plan is 5 chairs and 1 table, with profit Rs 190).
  • Both slack variables are zero, so both constraints are binding. Machine A (16 hours) and Machine B (30 hours) are fully used, so there is no idle machine time and no non-binding constraint in this problem.
  • The shadow price (the −(Cj−Zj)-(C_j - Z_j) entry for the slack variable) of Machine A is Rs 8 per hour, and of Machine B is Rs 7/3 = about Rs 2.33 per hour. One extra hour on Machine A would raise profit by Rs 8, so the firm should pay up to Rs 8 for it. Check: 16(8)+30(7/3)=19816(8) + 30(7/3) = 198.
  • A non-binding constraint would have positive slack and a shadow price of zero, so adding more of that resource would not raise profit.

Answer checked by Claude Sonnet · Oct 21, 2026

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  • 2082 · Mathematics For Economics (Econ. 503) · 5 marks

Consider the following first-order linear difference equation with constant coefficient and constant term: yt+1+ayt=by_{t+1} + ay_t = b where a is constant coefficient and b is a constant term. Derive the characteristic equation. Find the homogeneous, particular solutions and general solution and discuss the nature of time path of the system.

Answer

First-order linear difference equation

Given: [ y_{t+1}+ay_t=b ] where (a) is the constant coefficient and (b) is the constant term.


1. Characteristic equation

For the homogeneous equation, set (b=0): [ y_{t+1}+ay_t=0 ]

Assume a trial solution: [ y_t=A\lambda^t ]

Substitute: [ A\lambda^{t+1}+aA\lambda^t=0 ] [ A\lambda^t(\lambda+a)=0 ]

Since (A\lambda^t\neq 0), the characteristic equation is: [ \boxed{\lambda+a=0} ] so the characteristic root is: [ \boxed{\lambda=-a} ]


2. Homogeneous solution

The homogeneous solution is: [ \boxed{y_t^h=A(-a)^t} ] where (A) is an arbitrary constant determined by the initial condition.


3. Particular solution

Since the right-hand side is a constant (b), try a constant particular solution: [ y_t^p=\bar y ]

Substitute into the original equation: [ \bar y+a\bar y=b ] [ (1+a)\bar y=b ]

Thus, for (a\neq -1): [ \boxed{y_t^p=\frac{b}{1+a}} ]


4. General solution

The general solution is the sum of the homogeneous and particular solutions: [ \boxed{y_t=A(-a)^t+\frac{b}{1+a}} ]

The steady-state equilibrium is: [ \boxed{y^*=\frac{b}{1+a}} ]


5. Nature of the time path

The behaviour of (y_t) depends on the characteristic root (\lambda=-a).

ConditionNature of time path
(a
(0<a<1)Damped oscillation
(-1<a<0)Monotonic convergence
(a
(a>1)Explosive oscillation
(a<-1)Monotonic divergence
(a=1)Bounded oscillation, no convergence
(a=-1)Special case: (y_t=A+bt), linear trend if (b\neq 0)

Special case: (a=-1)

If (a=-1), the equation becomes: [ y_{t+1}-y_t=b ]

The homogeneous solution is: [ y_t^h=A ]

A particular solution is: [ y_t^p=bt ]

Hence the general solution is: [ \boxed{y_t=A+bt} ]

If (b=0), (y_t=A), a constant path.
If (b\neq 0), the path diverges linearly.

Answer checked by Claude Haiku · Oct 21, 2026

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  • 2082 · Mathematics For Economics (Econ. 503) · 5 marks

Solve the following problem using the maximum principle: Max∫0T(x−y2)dt\text{Max} \int_0^T (x - y^2)dt Subject to x′=y,x(0)=2x' = y, x(0) = 2

Answer

Treat xx as the state variable, yy as the control, and the horizon TT as fixed with the terminal state x(T)x(T) free (none is given).

Step 1: Hamiltonian

H=x−y2+λyH = x - y^2 + \lambda y

Step 2: Maximum principle conditions

  • Control: ∂H∂y=−2y+λ=0  ⇒  y=λ2\dfrac{\partial H}{\partial y} = -2y + \lambda = 0 \;\Rightarrow\; y = \dfrac{\lambda}{2}. Since ∂2H∂y2=−2<0\dfrac{\partial^2 H}{\partial y^2} = -2 < 0, this maximises HH.
  • Costate: λ′=−∂H∂x=−1  ⇒  λ(t)=c−t\lambda' = -\dfrac{\partial H}{\partial x} = -1 \;\Rightarrow\; \lambda(t) = c - t.
  • State: x′=yx' = y, with x(0)=2x(0) = 2.
  • Transversality (free x(T)x(T)): λ(T)=0  ⇒  c=T\lambda(T) = 0 \;\Rightarrow\; c = T.

Step 3: Solution

λ(t)=T−t,y∗(t)=T−t2\lambda(t) = T - t, \qquad y^*(t) = \frac{T - t}{2}

x′=T−t2  ⇒  x∗(t)=2+Tt2−t24x' = \frac{T - t}{2} \;\Rightarrow\; x^*(t) = 2 + \frac{Tt}{2} - \frac{t^2}{4}

So x(T)=2+T24x(T) = 2 + \dfrac{T^2}{4}.

Step 4: Sufficiency. HH is concave in (x,y)(x, y) (linear in xx, concave in yy), so these necessary conditions are also sufficient and give the maximum.

Optimal value

J∗=∫0T[x∗−(y∗)2]dt=2T+T34−T312−T312=2T+T312J^* = \int_0^T \left[x^* - (y^*)^2\right]dt = 2T + \frac{T^3}{4} - \frac{T^3}{12} - \frac{T^3}{12} = 2T + \frac{T^3}{12}

The optimal control y∗y^* is positive and declines linearly to zero at TT. The planner raises xx fast early on, since xx earns benefit for longer, and slows near the end because there is no value placed on terminal x(T)x(T).

Answer checked by Claude Sonnet · Oct 21, 2026

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  • 2082 · Mathematics For Economics (Econ. 503) · 5 marks

Discuss the Economic Interpretation of the Dual considering a standard Linear Programming Production Problem.

Answer

Economic Interpretation of the Dual in a Linear Programming Production Problem

1. The Primal Production Problem

A firm produces nn goods using mm scarce resources. Let:

  • xjx_j = quantity of good jj produced (j=1,2,…,nj = 1, 2, \dots, n)
  • cjc_j = profit (contribution) per unit of good jj
  • bib_i = available amount of resource ii (i=1,2,…,mi = 1, 2, \dots, m)
  • aija_{ij} = amount of resource ii required to produce one unit of good jj

The primal LP is:

max⁡Z=∑j=1ncjxj\max Z = \sum_{j=1}^{n} c_j x_j

subject to

∑j=1naijxj≤bi(i=1,…,m),xj≥0\sum_{j=1}^{n} a_{ij} x_j \le b_i \quad (i = 1, \dots, m), \qquad x_j \ge 0

This asks: how much of each good should be produced to maximise total profit, given limited resources?

2. The Dual Problem

Associating a dual variable yi≥0y_i \ge 0 with each resource constraint, the dual LP is:

min⁡W=∑i=1mbiyi\min W = \sum_{i=1}^{m} b_i y_i

subject to

∑i=1maijyi≥cj(j=1,…,n),yi≥0\sum_{i=1}^{m} a_{ij} y_i \ge c_j \quad (j = 1, \dots, n), \qquad y_i \ge 0

3. Economic Interpretation of the Dual Variables

The dual variable yiy_i is the imputed (shadow) price of resource ii — the marginal value of one additional unit of that resource.

  • Dual objective min⁡W=∑biyi\min W = \sum b_i y_i: the minimum total opportunity cost (or imputed value) the firm must assign to its resource endowment. It is the lowest valuation of resources consistent with the profit rates the firm actually earns.
  • Dual constraints ∑iaijyi≥cj\sum_i a_{ij} y_i \ge c_j: for every good actually produced, the total imputed cost of the resources used in producing one unit must be at least as great as the profit that unit earns. Otherwise the firm would be valuing its resources below what they can earn, and the valuation would not be sustainable.

Thus the dual asks: what minimum prices should be imputed to the scarce resources so that no product's resource cost falls short of its profit contribution?

4. Key Economic Results (Duality Theorems)

ResultEconomic meaning
Weak duality: Z≤WZ \le WNo feasible production plan can yield profit greater than the total imputed cost of resources.
Strong duality: at optimum Z∗=W∗Z^* = W^*Maximum profit exactly equals the minimum imputed value of the resource endowment.
Complementary slacknessIf a resource is not fully used (∑aijxj<bi\sum a_{ij}x_j < b_i), its shadow price is zero (yi=0y_i = 0) — a free (non-scarce) resource has no value. If yi>0y_i > 0, the resource is fully employed.
Complementary slackness (products)If a good is produced (xj>0x_j > 0), its dual constraint holds with equality (∑aijyi=cj\sum a_{ij}y_i = c_j) — its resource cost exactly equals its profit. If the resource cost exceeds the profit, the good is not produced (xj=0x_j = 0).

5. Shadow Price as Marginal Value

The shadow price yi∗=∂Z∗/∂biy_i^* = \partial Z^* / \partial b_i measures the increase in maximum profit from one extra unit of resource ii. Hence:

  • yi∗>0y_i^* > 0 → resource is scarce and binding; the firm would pay up to yi∗y_i^* to acquire more.
  • yi∗=0y_i^* = 0 → resource is abundant/slack; extra units add nothing to profit.

This guides managerial decisions on buying additional resources, renting out capacity, or subcontracting.

6. Illustrative Example

Maximise Z=40x1+30x2Z = 40x_1 + 30x_2 subject to 2x1+x2≤100 (labour),x1+x2≤80 (material),x1,x2≥02x_1 + x_2 \le 100 \text{ (labour)}, \qquad x_1 + x_2 \le 80 \text{ (material)}, \qquad x_1, x_2 \ge 0

Optimal primal: x1=20, x2=60, Z∗=2600x_1 = 20,\ x_2 = 60,\ Z^* = 2600.

Dual: min⁡W=100y1+80y2\min W = 100y_1 + 80y_2 s.t. 2y1+y2≥402y_1 + y_2 \ge 40, y1+y2≥30y_1 + y_2 \ge 30, y1,y2≥0y_1, y_2 \ge 0.

Solving gives y1∗=10, y2∗=20y_1^* = 10,\ y_2^* = 20, so W∗=100(10)+80(20)=2600=Z∗W^* = 100(10) + 80(20) = 2600 = Z^*.

Interpretation: an extra unit of labour raises profit by Rs 10; an extra unit of material raises profit by Rs 20. Both resources are fully used, so both shadow prices are positive. The firm should be willing to pay up to Rs 10 per extra labour unit and Rs 20 per extra material unit.

7. Conclusion

The dual converts a quantity problem (how much to produce) into a price/valuation problem (what the scarce resources are worth). Its variables are shadow prices reflecting marginal productivity of resources; strong duality shows that maximum profit equals the minimum imputed value of the resource base; and complementary slackness links scarcity to positive value. The dual is thus the theoretical foundation for resource pricing, opportunity-cost accounting, and optimal resource-allocation decisions in production economics.

Answer checked by Claude Haiku · Oct 21, 2026

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  • 2082 · Mathematics For Economics (Econ. 503) · 5 marks

Find the complementary function and the particular integral of the following second-order differential equation. Define the definite solution and discuss the stability and time path of the solution. y′′+2y′+3y=2t2−t+2y'' + 2y' + 3y = 2t^2 - t + 2, where y(0)=0y(0) = 0 and y′(0)=1y'(0) = 1

Answer

Given differential equation

[ y''+2y'+3y=2t^2-t+2,\qquad y(0)=0,\quad y'(0)=1 ]

1. Complementary function (CF)

The homogeneous equation is [ y''+2y'+3y=0. ] Its auxiliary equation is [ r^2+2r+3=0. ] Solving, [ r=\frac{-2\pm\sqrt{4-12}}{2} =\frac{-2\pm\sqrt{-8}}{2} =-1\pm i\sqrt2. ] Thus the complementary function is [ \boxed{y_c(t)=e^{-t}\left(A\cos(\sqrt2,t)+B\sin(\sqrt2,t)\right)} ]

2. Particular integral (PI)

Since the RHS is a polynomial of degree 2, try [ y_p=at^2+bt+c. ] Then [ y_p'=2at+b,\qquad y_p''=2a. ] Substituting into the differential equation: [ 2a+2(2at+b)+3(at^2+bt+c)=2t^2-t+2. ] Comparing coefficients: [ 3a=2 \implies a=\frac23, ] [ 4a+3b=-1 \implies \frac83+3b=-1 \implies b=-\frac{11}{9}, ] [ 2a+2b+3c=2 \implies \frac43-\frac{22}{9}+3c=2 \implies c=\frac{28}{27}. ] Hence [ \boxed{y_p(t)=\frac23t^2-\frac{11}{9}t+\frac{28}{27}} ]

3. General solution

[ y(t)=y_c(t)+y_p(t) ] [ \boxed{ y(t)=e^{-t}\left(A\cos(\sqrt2,t)+B\sin(\sqrt2,t)\right) +\frac23t^2-\frac{11}{9}t+\frac{28}{27} } ]

4. Definite solution

A definite solution is the particular solution obtained by using the given initial conditions to determine the arbitrary constants (A) and (B).

Using (y(0)=0): [ 0=A+\frac{28}{27} ] [ A=-\frac{28}{27}. ]

Now differentiate: [ y'(t)=e^{-t}\left[(-A+\sqrt2 B)\cos(\sqrt2,t) +(-B-\sqrt2 A)\sin(\sqrt2,t)\right] +\frac43t-\frac{11}{9}. ] Using (y'(0)=1): [ 1=-A+\sqrt2 B-\frac{11}{9}. ] Since (A=-\frac{28}{27}), [ 1=\frac{28}{27}+\sqrt2 B-\frac{11}{9}. ] [ \sqrt2 B=1+\frac{11}{9}-\frac{28}{27} =\frac{32}{27}. ] [ B=\frac{32}{27\sqrt2}=\frac{16\sqrt2}{27}. ]

Therefore the definite solution is [ \boxed{ y(t)=e^{-t}\left(-\frac{28}{27}\cos(\sqrt2,t) +\frac{16\sqrt2}{27}\sin(\sqrt2,t)\right) +\frac23t^2-\frac{11}{9}t+\frac{28}{27} } ]

5. Stability and time path

The characteristic roots are [ r=-1\pm i\sqrt2. ] Their real part is [ \alpha=-1<0. ] Therefore the complementary function satisfies [ y_c(t)\to 0 \quad \text{as } t\to\infty. ] So the solution is dynamically stable: any deviation from the particular integral dies out over time.

The time path consists of two parts:

  1. Transient part:
    [ e^{-t}\left(A\cos(\sqrt2,t)+B\sin(\sqrt2,t)\right) ] This is a damped oscillation. It oscillates with angular frequency (\sqrt2), so its period is [ T=\frac{2\pi}{\sqrt2}=\sqrt2,\pi, ] and its amplitude decays exponentially because of (e^{-t}).

  2. Long-run part:
    [ y_p(t)=\frac23t^2-\frac{11}{9}t+\frac{28}{27} ] As (t\to\infty), the solution approaches this quadratic path: [ y(t)\to \frac23t^2-\frac{11}{9}t+\frac{28}{27}. ]

Thus the definite solution is stable in the sense that the transient oscillatory component vanishes, but the long-run equilibrium path itself grows without bound because the forcing term is quadratic with positive leading coefficient.

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  • 2082 · Mathematics For Economics (Econ. 503) · 5 marks

Write short note on any TWO a) Application of differential equations in Economics b) Supremum and Infimum of Sets of Real Numbers c) Hamiltonian Function for an Investment Problem

Answer

a) Application of Differential Equations in Economics

A differential equation relates a function to its derivatives and is used in economics whenever a variable changes continuously with time or with another variable. The main applications are:

1. Growth and compound interest. If a capital stock KK grows continuously at rate rr,

dKdt=rK⇒K(t)=K0ert\frac{dK}{dt} = rK \quad \Rightarrow \quad K(t) = K_0 e^{rt}

This is the continuous-time version of compound interest and is the base of all growth models.

2. Domar's growth model. With saving proportional to income (S=sYS = sY) and investment I=dKdtI = \dfrac{dK}{dt},

dKdt=sYandK=vY⇒dYdt=svY\frac{dK}{dt} = sY \quad \text{and} \quad K = vY \quad \Rightarrow \quad \frac{dY}{dt} = \frac{s}{v}Y

so income grows exponentially at the rate s/vs/v. This gives the famous condition for full-employment growth.

3. Price adjustment (cobweb and market models). If price rises when demand exceeds supply,

dPdt=k(D(P)−S(P))\frac{dP}{dt} = k(D(P) - S(P))

With linear demand and supply this gives a first-order linear equation whose solution is stable only if the slope condition holds; otherwise price oscillates or diverges.

4. Solow model. The capital–labour ratio k=K/Lk = K/L follows

dkdt=sf(k)−(n+δ)k\frac{dk}{dt} = sf(k) - (n + \delta)k

The steady state k∗k^* is found by setting dkdt=0\dfrac{dk}{dt} = 0, i.e. sf(k∗)=(n+δ)k∗sf(k^*) = (n+\delta)k^*.

5. Inventory and consumption models. Optimal inventory and consumption paths are obtained by solving first-order differential equations subject to boundary conditions.

Conclusion. Differential equations convert static economic relations into dynamic ones, allowing economists to study growth, stability, cycles and convergence to equilibrium.


b) Supremum and Infimum of Sets of Real Numbers

Upper bound. A real number uu is an upper bound of a set S⊂RS \subset \mathbb{R} if x≤ux \le u for every x∈Sx \in S.

Supremum (least upper bound). The supremum of SS, written sup⁡S\sup S, is the smallest of all upper bounds. It satisfies:

  1. x≤sup⁡Sx \le \sup S for all x∈Sx \in S;
  2. for any ε>0\varepsilon > 0, there exists x∈Sx \in S with x>sup⁡S−εx > \sup S - \varepsilon.

Lower bound. A real number ll is a lower bound of SS if l≤xl \le x for every x∈Sx \in S.

Infimum (greatest lower bound). The infimum of SS, written inf⁡S\inf S, is the largest of all lower bounds. It satisfies:

  1. inf⁡S≤x\inf S \le x for all x∈Sx \in S;
  2. for any ε>0\varepsilon > 0, there exists x∈Sx \in S with x<inf⁡S+εx < \inf S + \varepsilon.

Completeness axiom. Every non-empty set of real numbers that is bounded above has a supremum in R\mathbb{R}; similarly, every non-empty set bounded below has an infimum. This property is what distinguishes R\mathbb{R} from Q\mathbb{Q}.

Examples.

Set SSinf⁡S\inf Ssup⁡S\sup SMax / Min
{1,2,3}\{1, 2, 3\}13max = 3, min = 1
(0,1)(0, 1)01no max, no min
{1n:n∈N}\left\{\dfrac{1}{n} : n \in \mathbb{N}\right\}01max = 1, no min
{x∈R:x2<2}\{x \in \mathbb{R} : x^2 < 2\}−2-\sqrt{2}2\sqrt{2}no max, no min

Key point. The supremum and infimum need not belong to the set. When they do belong to the set, they are the maximum and minimum respectively. For example, in (0,1)(0,1) the supremum is 1 and the infimum is 0, but neither is an element of the set.

Answer checked by Claude Haiku · Oct 21, 2026

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  • 2025 · Mathematics For Economics (Econ. 503) · 15 marks

Solve the particular solution and homogeneous equation by finding roots of characteristics of the following second order linear differential equation by classify the nature of roots and analyze the time path of the solution: d2ydt2+adydt+by=k\frac{d^2y}{dt^2} + a\frac{dy}{dt} + by = k, where aa and bb are constant coefficients, and kk is a constant term. [6+9]

Answer

Model Answer

Question: Solve the particular solution and homogeneous equation by finding roots of characteristics of the following second order linear differential equation by classify the nature of roots and analyze the time path of the solution: d2ydt2+adydt+by=k\frac{d^2y}{dt^2} + a\frac{dy}{dt} + by = k, where aa and bb are constant coefficients, and kk is a constant term. [6+9]


Solution

The given differential equation is a second-order linear non-homogeneous differential equation: d2ydt2+adydt+by=k…(1)\frac{d^2y}{dt^2} + a\frac{dy}{dt} + by = k \quad \dots (1)

The general solution y(t)y(t) is the sum of the homogeneous solution yh(t)y_h(t) and the particular solution yp(t)y_p(t): y(t)=yh(t)+yp(t)y(t) = y_h(t) + y_p(t)

Part 1: Homogeneous Equation and Roots Analysis (yhy_h) [Approx. 6 Marks]

The associated homogeneous equation is obtained by setting the non-homogeneous term (kk) to zero: d2ydt2+adydt+by=0…(2)\frac{d^2y}{dt^2} + a\frac{dy}{dt} + by = 0 \quad \dots (2)

We assume a solution of the form y=erty = e^{rt}. Substituting this into Equation (2) yields the characteristic equation: r2+ar+b=0…(3)r^2 + ar + b = 0 \quad \dots (3)

The roots of this quadratic equation are found using the quadratic formula: r=−a±a2−4b2r = \frac{-a \pm \sqrt{a^2 - 4b}}{2}

The nature of the roots depends on the discriminant, Δ=a2−4b\Delta = a^2 - 4b.

Case 1: Distinct Real Roots (Δ>0\Delta > 0) If a2−4b>0a^2 - 4b > 0, the roots r1r_1 and r2r_2 are real and distinct. r1=−a+a2−4b2,r2=−a−a2−4b2r_1 = \frac{-a + \sqrt{a^2 - 4b}}{2}, \quad r_2 = \frac{-a - \sqrt{a^2 - 4b}}{2} The homogeneous solution is: yh(t)=C1er1t+C2er2ty_h(t) = C_1 e^{r_1 t} + C_2 e^{r_2 t}

Case 2: Repeated Real Roots (Δ=0\Delta = 0) If a2−4b=0a^2 - 4b = 0, there is one repeated real root: r=−a2r = -\frac{a}{2} The homogeneous solution is: yh(t)=(C1+C2t)erty_h(t) = (C_1 + C_2 t) e^{rt}

Case 3: Complex Conjugate Roots (Δ<0\Delta < 0) If a2−4b<0a^2 - 4b < 0, the roots are complex conjugates. Let α=−a/2\alpha = -a/2 and β=4b−a22\beta = \frac{\sqrt{4b - a^2}}{2}. r=α±iβr = \alpha \pm i\beta The homogeneous solution is: yh(t)=eαt(C1cos⁡(βt)+C2sin⁡(βt))y_h(t) = e^{\alpha t} (C_1 \cos(\beta t) + C_2 \sin(\beta t))


Part 2: Particular Solution and Time Path Analysis (ypy_p and yy) [Approx. 9 Marks]

1. Finding the Particular Solution (ypy_p)

Since the non-homogeneous term is a constant (kk), we assume a particular solution of the form: yp=Ay_p = A Differentiating ypy_p: dypdt=0,d2ypdt2=0\frac{dy_p}{dt} = 0, \quad \frac{d^2y_p}{dt^2} = 0 Substituting these into the original equation (1): 0+a(0)+b(A)=k0 + a(0) + b(A) = k bA=kbA = k Assuming b≠0b \neq 0, the particular solution is: yp=kby_p = \frac{k}{b}

2. General Solution (yy)

The general solution is y(t)=yh(t)+yp(t)y(t) = y_h(t) + y_p(t).

  • Case 1 (Distinct Real Roots): y(t)=C1er1t+C2er2t+kby(t) = C_1 e^{r_1 t} + C_2 e^{r_2 t} + \frac{k}{b}
  • Case 2 (Repeated Real Roots): y(t)=(C1+C2t)ert+kby(t) = (C_1 + C_2 t) e^{rt} + \frac{k}{b}
  • Case 3 (Complex Roots): y(t)=eαt(C1cos⁡(βt)+C2sin⁡(βt))+kby(t) = e^{\alpha t} (C_1 \cos(\beta t) + C_2 \sin(\beta t)) + \frac{k}{b}

3. Analysis of Time Path (Stability and Behavior)

The time path of the solution is determined by the behavior of the homogeneous part (yhy_h) as t→∞t \to \infty, as the particular solution k/bk/b represents the steady-state (equilibrium) value.

Case 1: Distinct Real Roots (r1,r2r_1, r_2)

  • Stable Node: If both r1<0r_1 < 0 and r2<0r_2 < 0, the transient terms decay to zero, and the system approaches the steady state k/bk/b.
  • Unstable Node: If both r1>0r_1 > 0 and r2>0r_2 > 0, the solution diverges exponentially away from k/bk/b.
  • Saddle Point (unstable): If the roots have opposite signs, the positive root dominates as t→∞t\to\infty, so the path diverges from k/bk/b except along the one stable arm (CC on the growing term equal to zero).
  • Zero root: If one root is 00 (that is b=0b=0), the particular solution k/bk/b does not exist; the path does not settle at a fixed level and a different particular solution (such as yp=kt/ay_p = kt/a) is needed.

Case 2: Repeated Real Roots (rr)

  • Stable Node: If r<0r < 0, the solution decays exponentially to k/bk/b.
  • Unstable Node: If r>0r > 0, the solution grows exponentially away from k/bk/b.

Case 3: Complex Conjugate Roots (α±iβ\alpha \pm i\beta) The behavior is governed by the real part of the roots, α=−a/2\alpha = -a/2.

  • Stable Spiral (Damped Oscillation): If α<0\alpha < 0, the exponential term eαte^{\alpha t} decays, causing the oscillations to dampen and approach the steady state k/bk/b.
  • Unstable Spiral (Growing Oscillation): If α>0\alpha > 0, the exponential term eαte^{\alpha t} grows, causing the oscillations to increase indefinitely away from k/bk/b.
  • Center (Undamped Oscillation): If α=0\alpha = 0 (i.e., a=0a=0), the solution oscillates perpetually around the steady state k/bk/b without decaying or growing.

Summary Table

Nature of Roots (Δ=a2−4b\Delta = a^2 - 4b)Roots (rr)Homogeneous Solution (yhy_h)Steady State (ypy_p)Time Path Analysis
Δ>0\Delta > 0 (Distinct Real)r1,r2r_1, r_2C1er1t+C2er2tC_1 e^{r_1 t} + C_2 e^{r_2 t}k/bk/bStable node if r1,r2<0r_1, r_2 < 0; unstable node if both >0>0; saddle (unstable) if opposite signs.
Δ=0\Delta = 0 (Repeated Real)rr(C1+C2t)ert(C_1 + C_2 t) e^{rt}k/bk/bStable if r<0r < 0. Unstable if r>0r > 0.
Δ<0\Delta < 0 (Complex)α±iβ\alpha \pm i\betaeαt(C1cos⁡(βt)+C2sin⁡(βt))e^{\alpha t} (C_1 \cos(\beta t) + C_2 \sin(\beta t))k/bk/bStable Spiral if α<0\alpha < 0. Unstable Spiral if α>0\alpha > 0. Center if α=0\alpha = 0.

Answer checked by Claude Sonnet · Oct 21, 2026

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  • 2025 · Mathematics For Economics (Econ. 503) · 10 marks

Solve the following linear programming problem using the Simplex Method. Show all iterations and simplex tableaus clearly. Also, write the dual of the given problem and solve it to verify that the optimal values of the primal and dual are equal. [10+5] Minimize Z=3x1+2x2Z = 3x_1 + 2x_2 Subject to: 2x1+x2≥102x_1 + x_2 \ge 10 x1+3x2≥15x_1 + 3x_2 \ge 15 x1,x2≥0x_1, x_2 \ge 0

Answer

Primal: Minimize Z=3x1+2x2Z=3x_1+2x_2 s.t. 2x1+x2≥102x_1+x_2\ge10, x1+3x2≥15x_1+3x_2\ge15, x1,x2≥0x_1,x_2\ge0.

Simplex (dual simplex form). Write W=−ZW=-Z (maximise) and subtract surplus variables so the constraints become −2x1−x2+s1=−10-2x_1-x_2+s_1=-10 and −x1−3x2+s2=−15-x_1-3x_2+s_2=-15. The row W+3x1+2x2=0W+3x_1+2x_2=0 is the objective row.

Tableau 0

Basisx1x_1x2x_2s1s_1s2s_2RHS
s1s_1-2-110-10
s2s_2-1-301-15
WW32000

The most negative RHS is −15-15, so s2s_2 leaves. Ratios 31\frac{3}{1} and 23\frac{2}{3} give the minimum 23\frac23, so x2x_2 enters.

Tableau 1

Basisx1x_1x2x_2s1s_1s2s_2RHS
s1s_1−53-\frac5301−13-\frac13-5
x2x_213\frac1310−13-\frac135
WW73\frac730023\frac23-10

s1s_1 (RHS −5-5) leaves. Ratios 7/35/3=1.4\frac{7/3}{5/3}=1.4 and 2/31/3=2\frac{2/3}{1/3}=2, so x1x_1 enters.

Tableau 2 (optimal: all RHS ≥0\ge0, all objective coefficients ≥0\ge0)

Basisx1x_1x2x_2s1s_1s2s_2RHS
x1x_110−35-\frac3515\frac153
x2x_20115\frac15−25-\frac254
WW0075\frac7515\frac15-17

Optimal solution: x1=3,  x2=4,  Zmin⁡=3(3)+2(4)=17x_1=3,\;x_2=4,\;Z_{\min}=3(3)+2(4)=\mathbf{17}. (Check: 2(3)+4=102(3)+4=10 and 3+3(4)=153+3(4)=15, both binding.) The objective-row entries 75=1.4\frac75=1.4 and 15=0.2\frac15=0.2 are the dual values.

Dual: Maximize Z′=10y1+15y2Z'=10y_1+15y_2 s.t. 2y1+y2≤32y_1+y_2\le3, y1+3y2≤2y_1+3y_2\le2, y1,y2≥0y_1,y_2\ge0.

Basisy1y_1y2y_2s3s_3s4s_4RHS
s3s_321103
s4s_413012
Z′Z'-10-15000

y2y_2 enters (most negative), minimum ratio 23\frac23 in row s4s_4:

Basisy1y_1y2y_2s3s_3s4s_4RHS
s3s_353\frac5301−13-\frac1373\frac73
y2y_213\frac131013\frac1323\frac23
Z′Z'-500510

y1y_1 enters, minimum ratio 7/35/3=1.4\frac{7/3}{5/3}=1.4 in row s3s_3:

Basisy1y_1y2y_2s3s_3s4s_4RHS
y1y_11035\frac35−15-\frac1575\frac75
y2y_201−15-\frac1525\frac2515\frac15
Z′Z'003417

Dual optimum: y1=1.4,  y2=0.2,  Zmax⁡′=10(1.4)+15(0.2)=17y_1=1.4,\;y_2=0.2,\;Z'_{\max}=10(1.4)+15(0.2)=\mathbf{17}.

Verification. Primal minimum == dual maximum =17=17, as the strong duality theorem requires. The dual's objective-row entries under the slack columns (3 and 4) equal the primal x1=3x_1=3 and x2=4x_2=4.

Answer checked by Claude Sonnet · Oct 21, 2026

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  • 2025 · Mathematics For Economics (Econ. 503) · 10 marks

Define and highlight the implications of constraint optimization. Find the optimum commodity purchase when the consumer's total income is Rs.100. If the consumer's utility function is U=x2y2U=x^2y^2 and prices of x and y are Rs.1 and Rs.4 respectively. [2+2+1+5]

Answer

1. Definition of Constrained Optimization

Constrained optimization is the mathematical technique of maximizing or minimizing an objective function subject to one or more constraints.

In economics, the consumer’s problem is:

max⁡x,yU(x,y)\max_{x,y} U(x,y)

subject to the budget constraint:

Pxx+Pyy=MP_x x + P_y y = M

where UU is utility, xx and yy are quantities of two goods, PxP_x and PyP_y are their prices, and MM is money income.

The standard method used is the Lagrange multiplier method:

L=U(x,y)+λ(M−Pxx−Pyy)L = U(x,y) + \lambda (M - P_x x - P_y y)

where λ\lambda is the Lagrange multiplier.


2. Implications of Constrained Optimization

  1. Scarcity and choice
    The consumer cannot buy unlimited quantities. Income and prices restrict choice, so the optimum lies on the budget line.

  2. Opportunity cost
    The slope of the budget line, Px/PyP_x/P_y, shows how much of one good must be sacrificed to obtain one more unit of the other good.

  3. Tangency condition
    At the optimum, the marginal rate of substitution equals the price ratio:

    MRSxy=MUxMUy=PxPyMRS_{xy} = \frac{MU_x}{MU_y} = \frac{P_x}{P_y}

    Equivalently,

    MUxPx=MUyPy=λ\frac{MU_x}{P_x} = \frac{MU_y}{P_y} = \lambda

    This means the marginal utility per rupee spent on each good is equal.

  4. Shadow price of the constraint
    The Lagrange multiplier λ\lambda measures the marginal utility of income. It shows how much utility would increase if income increased by one rupee.

  5. Demand functions
    The solution gives ordinary demand functions:

    x∗=x(Px,Py,M),y∗=y(Px,Py,M)x^* = x(P_x, P_y, M), \qquad y^* = y(P_x, P_y, M)
  6. Comparative statics and welfare analysis
    Changes in income or prices change the optimum. This helps analyse income and substitution effects, consumer surplus, and effects of taxes, subsidies, and rationing.


3. Budget Constraint

Given:

M=Rs. 100,Px=Rs. 1,Py=Rs. 4M = \text{Rs. }100, \qquad P_x = \text{Rs. }1, \qquad P_y = \text{Rs. }4

The budget constraint is:

1x+4y=1001x + 4y = 100

or

x+4y=100x + 4y = 100

4. Optimum Commodity Purchase

The consumer maximizes:

U=x2y2U = x^2 y^2

subject to:

x+4y=100x + 4y = 100

Using the Lagrange multiplier method:

L=x2y2+λ(100−x−4y)L = x^2 y^2 + \lambda (100 - x - 4y)

First-order conditions:

∂L∂x=2xy2−λ=0\frac{\partial L}{\partial x} = 2xy^2 - \lambda = 0 ∂L∂y=2x2y−4λ=0\frac{\partial L}{\partial y} = 2x^2 y - 4\lambda = 0

From the first condition:

λ=2xy2\lambda = 2xy^2

From the second condition:

4λ=2x2y4\lambda = 2x^2 y λ=x2y2\lambda = \frac{x^2 y}{2}

Equating the two expressions for λ\lambda:

2xy2=x2y22xy^2 = \frac{x^2 y}{2}

Multiplying both sides by 2:

4xy2=x2y4xy^2 = x^2 y

Since x>0x>0 and y>0y>0, divide by xyxy:

4y=x4y = x

Thus:

x=4yx = 4y

Substitute x=4yx = 4y into the budget constraint:

4y+4y=1004y + 4y = 100 8y=1008y = 100 y=12.5y = 12.5

Then:

x=4y=4(12.5)=50x = 4y = 4(12.5) = 50

Therefore, the optimum commodity purchase is:

x∗=50 units of X\boxed{x^* = 50 \text{ units of } X} y∗=12.5 units of Y\boxed{y^* = 12.5 \text{ units of } Y}

5. Verification

Expenditure on XX:

Pxx∗=1×50=Rs. 50P_x x^* = 1 \times 50 = \text{Rs. }50

Expenditure on YY:

Pyy∗=4×12.5=Rs. 50P_y y^* = 4 \times 12.5 = \text{Rs. }50

Total expenditure:

Rs. 50+Rs. 50=Rs. 100\text{Rs. }50 + \text{Rs. }50 = \text{Rs. }100

Marginal utilities:

MUx=∂U∂x=2xy2=2(50)(12.5)2=15,625MU_x = \frac{\partial U}{\partial x} = 2xy^2 = 2(50)(12.5)^2 = 15{,}625 MUy=∂U∂y=2x2y=2(50)2(12.5)=62,500MU_y = \frac{\partial U}{\partial y} = 2x^2 y = 2(50)^2(12.5) = 62{,}500

Marginal utility per rupee:

MUxPx=15,6251=15,625\frac{MU_x}{P_x} = \frac{15{,}625}{1} = 15{,}625 MUyPy=62,5004=15,625\frac{MU_y}{P_y} = \frac{62{,}500}{4} = 15{,}625

Equal marginal utility per rupee confirms the first-order (tangency) condition.

Second-order condition (bordered Hessian). With Uxx=2y2=312.5U_{xx}=2y^2=312.5, Uyy=2x2=5000U_{yy}=2x^2=5000, Uxy=4xy=2500U_{xy}=4xy=2500 at the optimum:

∣Hˉ∣=∣0141312.52500425005000∣=−1(5000−10000)+4(2500−1250)=10,000>0|\bar H|=\begin{vmatrix}0&1&4\\1&312.5&2500\\4&2500&5000\end{vmatrix} =-1(5000-10000)+4(2500-1250)=10{,}000>0

For two variables and one constraint, ∣Hˉ∣>0|\bar H|>0 indicates a maximum, so the point is a utility maximum.

Maximum utility:

U=x2y2=(50)2(12.5)2=390,625U = x^2 y^2 = (50)^2 (12.5)^2 = 390{,}625

Final Answer

The consumer should purchase:

50 units of X and 12.5 units of Y\boxed{50 \text{ units of } X \text{ and } 12.5 \text{ units of } Y}

This gives maximum utility:

Umax⁡=390,625\boxed{U_{\max} = 390{,}625}

and the marginal utility of income is:

λ=15,625\boxed{\lambda = 15{,}625}

Answer checked by Claude Sonnet · Oct 21, 2026

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  • 2025 · Mathematics For Economics (Econ. 503) · 10 marks

Calculate the amount of money that an investment of Rs.1000 will generate in the following scenarios. In each instance, assume an annual interest rate of 10%. (a) At the end of one year, with semi-annual compounding, (b) At the end of five years, with annual compounding, (c) At the end of one year, with monthly compounding, (d) At the end of five years, with monthly compounding, and (e) At the end of one year, with continuous compounding. [2+2+2+2+2]

Answer

Let principal (P = \text{Rs. }1000), annual interest rate (r = 10% = 0.10).

(a) End of one year, semi-annual compounding

[ A=P\left(1+\frac{r}{n}\right)^{nt} ] Here (n=2,\ t=1). [ A=1000\left(1+\frac{0.10}{2}\right)^{2} =1000(1.05)^2 =1000(1.1025) =\text{Rs. }1102.50 ] Interest generated: [ 1102.50-1000=\text{Rs. }102.50 ]

(b) End of five years, annual compounding

Here (n=1,\ t=5). [ A=1000(1+0.10)^5 =1000(1.10)^5 =1000(1.61051) =\text{Rs. }1610.51 ] Interest generated: [ 1610.51-1000=\text{Rs. }610.51 ]

(c) End of one year, monthly compounding

Here (n=12,\ t=1). [ A=1000\left(1+\frac{0.10}{12}\right)^{12} =1000(1.008333\ldots)^{12} =\text{Rs. }1104.71 ] Interest generated: [ 1104.71-1000=\text{Rs. }104.71 ]

(d) End of five years, monthly compounding

Here (n=12,\ t=5), so (nt=60). [ A=1000\left(1+\frac{0.10}{12}\right)^{60} =\text{Rs. }1645.31 ] Interest generated: [ 1645.31-1000=\text{Rs. }645.31 ]

(e) End of one year, continuous compounding

[ A=Pe^{rt} ] Here (r=0.10,\ t=1). [ A=1000e^{0.10} =1000(1.105170\ldots) =\text{Rs. }1105.17 ] Interest generated: [ 1105.17-1000=\text{Rs. }105.17 ]

Final results

ScenarioTotal amountInterest generated
(a) Semi-annual, 1 yearRs. 1102.50Rs. 102.50
(b) Annual, 5 yearsRs. 1610.51Rs. 610.51
(c) Monthly, 1 yearRs. 1104.71Rs. 104.71
(d) Monthly, 5 yearsRs. 1645.31Rs. 645.31
(e) Continuous, 1 yearRs. 1105.17Rs. 105.17

Answer checked by Claude Haiku · Oct 21, 2026

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  • 2025 · Mathematics For Economics (Econ. 503) · 10 marks

Solve the following linear programming problem by simplex method. [10] Maximize Q=3x+5yQ = 3x + 5y Subject to: 2x+y≤202x + y \le 20 x+y≤12x + y \le 12 x≥0,y≥0x \ge 0, y \ge 0

Answer

Standard form

Introduce slack variables s1,s2≥0s_1, s_2 \ge 0:

Maximize
Q=3x+5y+0s1+0s2Q = 3x + 5y + 0s_1 + 0s_2

Subject to
2x+y+s1=202x + y + s_1 = 20 x+y+s2=12x + y + s_2 = 12 x,y,s1,s2≥0x, y, s_1, s_2 \ge 0


Initial simplex tableau

Basisxxyys1s_1s2s_2RHSRatio
s1s_121102020/1=2020/1=20
s2s_211011212/1=1212/1=12
ZjZ_j00000
Cj−ZjC_j-Z_j3500

Entering variable: yy because Cj−Zj=5C_j-Z_j=5 is largest positive.
Leaving variable: s2s_2 because ratio 1212 is minimum.
Pivot element: 11.


Iteration 1

New yy-row = old s2s_2-row:

(1, 1, 0, 1 ∣ 12)(1,\ 1,\ 0,\ 1\ |\ 12)

New s1s_1-row = old s1s_1-row −- old s2s_2-row:

(2,1,1,0∣20)−(1,1,0,1∣12)=(1,0,1,−1∣8)(2,1,1,0|20) - (1,1,0,1|12) = (1,0,1,-1|8)

New ZjZ_j row:

Zj=5×y-row+0×s1-row=(5,5,0,5∣60)Z_j = 5 \times y\text{-row} + 0 \times s_1\text{-row} = (5,5,0,5|60)

New Cj−ZjC_j-Z_j row:

(3−5, 5−5, 0−0, 0−5)=(−2,0,0,−5)(3-5,\ 5-5,\ 0-0,\ 0-5) = (-2,0,0,-5)


Final simplex tableau

Basisxxyys1s_1s2s_2RHS
yy110112
s1s_1101-18
ZjZ_j550560
Cj−ZjC_j-Z_j-200-5

Since all Cj−Zj≤0C_j-Z_j \le 0, the optimal solution is reached.

Basic variables:
y=12,s1=8y=12,\qquad s_1=8

Non-basic variables:
x=0,s2=0x=0,\qquad s_2=0

Hence the optimal solution is
x=0,y=12x=0,\qquad y=12

Maximum value:
Qmax⁡=3(0)+5(12)=60Q_{\max}=3(0)+5(12)=60

Check:
2(0)+12=12≤202(0)+12=12 \le 20 0+12=12≤120+12=12 \le 12

Therefore,
Qmax⁡=60 at (x,y)=(0,12)\boxed{Q_{\max}=60 \text{ at } (x,y)=(0,12)}

Answer checked by Claude Haiku · Oct 21, 2026

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  • 2025 · Mathematics For Economics (Econ. 503) · 10 marks

Solve the following dynamic optimization problem: Max ∫02(2x−3y2)dt\int_0^2 (2x - 3y^2)dt Subject to: x˙=y\dot{x}=y, x(0)=1x(0)=1, x(2)=10x(2)=10, and y(2)=0y(2)=0 Formulate the Hamiltonian to derive the necessary optimality conditions, solve the resulting system of differential equations to find the optimal state and control time paths, and compute the value of the integral. [4+4+2]

Answer

Hamiltonian and conditions

State xx, control yy, costate λ\lambda:

H=2x−3y2+λyH = 2x - 3y^2 + \lambda y

∂H∂y=−6y+λ=0  ⇒  y=λ6,λ˙=−∂H∂x=−2,x˙=y\frac{\partial H}{\partial y}=-6y+\lambda=0\;\Rightarrow\;y=\frac{\lambda}{6},\qquad \dot\lambda=-\frac{\partial H}{\partial x}=-2,\qquad \dot x=y

HH is concave in (x,y)(x,y), so these conditions are also sufficient.

Solving

λ(t)=c−2t,y(t)=c−2t6,x(t)=1+ct−t26    (using x(0)=1)\lambda(t)=c-2t,\qquad y(t)=\frac{c-2t}{6},\qquad x(t)=1+\frac{ct-t^2}{6}\;\;(\text{using }x(0)=1)

Conflict in the data. The problem gives three boundary conditions for a two-constant system (one constant cc). Using x(2)=10x(2)=10 gives 1+2c−46=101+\frac{2c-4}{6}=10, so c=29c=29. But using y(2)=0y(2)=0 means λ(2)=0\lambda(2)=0, so c=4c=4, and then x(2)=53≠10x(2)=\frac53\ne10. The conditions x(2)=10x(2)=10 and y(2)=0y(2)=0 cannot both hold, so one has to be dropped.

Case 1: keep x(2)=10x(2)=10 (fixed terminal state), drop y(2)=0y(2)=0. Here c=29c=29:

y∗(t)=29−2t6,x∗(t)=1+29t−t26y^*(t)=\frac{29-2t}{6},\qquad x^*(t)=1+\frac{29t-t^2}{6}

∫02(2x−3y2) dt=∫02[2+29t−t23−(29−2t)212]dt=4+1669−219118=−178718≈−99.3\int_0^2(2x-3y^2)\,dt=\int_0^2\left[2+\frac{29t-t^2}{3}-\frac{(29-2t)^2}{12}\right]dt=4+\frac{166}{9}-\frac{2191}{18}=-\frac{1787}{18}\approx-99.3

Case 2: keep y(2)=0y(2)=0 (free terminal state, transversality λ(2)=0\lambda(2)=0), drop x(2)=10x(2)=10. Here c=4c=4:

y∗(t)=2−t3,x∗(t)=1+4t−t26,x(2)=53y^*(t)=\frac{2-t}{3},\qquad x^*(t)=1+\frac{4t-t^2}{6},\qquad x(2)=\frac53

∫02(2x−3y2) dt=(4+169)−89=449≈4.89\int_0^2(2x-3y^2)\,dt=\left(4+\frac{16}{9}\right)-\frac{8}{9}=\frac{44}{9}\approx4.89

Because the stated terminal conditions are inconsistent, Case 1 (which respects the explicit state target x(2)=10x(2)=10) is the solution of the problem as posed, with value −1787/18-1787/18, and Case 2 is the result if the terminal state is left free.

Answer checked by Claude Sonnet · Oct 21, 2026

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  • 2025 · Mathematics For Economics (Econ. 503) · 8 marks

If the cobweb model is Qtd=21−2PtQ_t^d = 21 - 2P_t; Qts=−3+6Pt−1Q_t^s = -3 + 6P_{t-1} and P0=7.5P_0=7.5. Find the equilibrium price and comment on its stability. [8+2]

Answer

Given the cobweb model:

Qtd=21−2Pt,Qts=−3+6Pt−1,P0=7.5Q_t^d = 21 - 2P_t,\qquad Q_t^s = -3 + 6P_{t-1},\qquad P_0=7.5

1. Equilibrium price

At equilibrium, price is constant over time, so

Pt=Pt−1=PP_t=P_{t-1}=P

and quantity demanded equals quantity supplied:

21−2P=−3+6P21-2P=-3+6P

24=8P24=8P

P∗=3P^*=3

The equilibrium quantity is

Q∗=21−2(3)=15Q^*=21-2(3)=15

So the equilibrium price is Rs 3 and equilibrium quantity is 15 units.


2. Difference equation and time path

Equating demand and supply for any period:

21−2Pt=−3+6Pt−121-2P_t=-3+6P_{t-1}

24=2Pt+6Pt−124=2P_t+6P_{t-1}

12=Pt+3Pt−112=P_t+3P_{t-1}

Pt=12−3Pt−1P_t=12-3P_{t-1}

This is a first-order linear difference equation. Its particular (equilibrium) solution is

P∗=121+3=3P^*=\frac{12}{1+3}=3

The deviation from equilibrium follows

Pt−3=−3(Pt−1−3)P_t-3=-3(P_{t-1}-3)

Hence,

Pt=3+(P0−3)(−3)tP_t=3+(P_0-3)(-3)^t

Using (P_0=7.5),

Pt=3+(7.5−3)(−3)tP_t=3+(7.5-3)(-3)^t

Pt=3+4.5(−3)tP_t=3+4.5(-3)^t

3. Stability

For the difference equation

Pt=a+bPt−1P_t=a+bP_{t-1}

the equilibrium is stable if and only if

∣b∣<1|b|<1

Here,

b=−3b=-3

so

∣b∣=3>1|b|=3>1

Therefore, the equilibrium is unstable.

The price path is:

P0=7.5,P1=−10.5,P2=43.5,P3=−118.5,…P_0=7.5,\quad P_1=-10.5,\quad P_2=43.5,\quad P_3=-118.5,\dots

Thus the price oscillates above and below the equilibrium price (P^*=3), but the oscillations become larger and larger. The cobweb diverges explosively from equilibrium. In cobweb terminology, since

∣slope of supplyslope of demand∣=62=3>1\left|\frac{\text{slope of supply}}{\text{slope of demand}}\right|=\frac{6}{2}=3>1

the model is explosively unstable.

Answer checked by Claude Haiku · Oct 21, 2026

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  • 2025 · Mathematics For Economics (Econ. 503) · 10 marks

Write short notes on any two: [2×5=10] a) Let the cost function and demand function are C=100+2QC=100+2Q and Q=20−P2Q=20-\frac{P}{2}. Find the profit maximizing level of output and price. b) Cost minimization c) Hessian Determinant

Answer

a) Profit-maximizing output and price

Given: C=100+2QQ=20−P2C = 100 + 2Q \qquad Q = 20 - \frac{P}{2}

Step 1 — Invert the demand function: Q=20−P2  ⇒  P2=20−Q  ⇒  P=40−2QQ = 20 - \frac{P}{2} \;\Rightarrow\; \frac{P}{2} = 20 - Q \;\Rightarrow\; P = 40 - 2Q

Step 2 — Total revenue: TR=P⋅Q=(40−2Q)Q=40Q−2Q2TR = P \cdot Q = (40 - 2Q)Q = 40Q - 2Q^2 MR=dTRdQ=40−4QMR = \frac{dTR}{dQ} = 40 - 4Q

Step 3 — Marginal cost: MC=dCdQ=2MC = \frac{dC}{dQ} = 2

Step 4 — First-order condition (MR = MC): 40−4Q=2  ⇒  4Q=38  ⇒  Q∗=9.5 units40 - 4Q = 2 \;\Rightarrow\; 4Q = 38 \;\Rightarrow\; Q^* = 9.5 \text{ units}

Step 5 — Price: P∗=40−2(9.5)=40−19=Rs 21P^* = 40 - 2(9.5) = 40 - 19 = \text{Rs } 21

Step 6 — Second-order condition: d2πdQ2=dMRdQ−dMCdQ=−4−0=−4<0(maximum)\frac{d^2\pi}{dQ^2} = \frac{dMR}{dQ} - \frac{dMC}{dQ} = -4 - 0 = -4 < 0 \quad \text{(maximum)}

Step 7 — Maximum profit: π=TR−TC=(21×9.5)−(100+2×9.5)=199.5−119=Rs 80.5\pi = TR - TC = (21 \times 9.5) - (100 + 2 \times 9.5) = 199.5 - 119 = \text{Rs } 80.5

Answer: The firm maximizes profit at Q∗=9.5Q^* = 9.5 units and P∗=P^* = Rs 21, earning a maximum profit of Rs 80.5.


b) Cost minimization

Meaning. Cost minimization is the dual problem of profit maximization. It means producing a given level of output (Qˉ\bar{Q}) at the lowest possible total cost, by choosing the least-cost combination of inputs (labour LL and capital KK).

Condition. The firm minimizes C=wL+rKC = wL + rK subject to the production constraint Qˉ=f(L,K)\bar{Q} = f(L,K). The solution requires the isoquant to be tangent to the isocost line, i.e. the marginal rate of technical substitution equals the input price ratio: MRTSLK=MPLMPK=wrorMPLw=MPKr\text{MRTS}_{LK} = \frac{MP_L}{MP_K} = \frac{w}{r} \quad\text{or}\quad \frac{MP_L}{w} = \frac{MP_K}{r}

This means the last rupee spent on each input yields the same marginal product — the equi-marginal principle.

Lagrangian method. Form Z=wL+rK+λ[Qˉ−f(L,K)]Z = wL + rK + \lambda\big[\bar{Q} - f(L,K)\big] Setting the partial derivatives equal to zero: ∂Z∂L=w−λMPL=0,∂Z∂K=r−λMPK=0,∂Z∂λ=Qˉ−f(L,K)=0\frac{\partial Z}{\partial L} = w - \lambda MP_L = 0,\qquad \frac{\partial Z}{\partial K} = r - \lambda MP_K = 0,\qquad \frac{\partial Z}{\partial \lambda} = \bar{Q} - f(L,K) = 0 Dividing the first two gives MPLMPK=wr\dfrac{MP_L}{MP_K} = \dfrac{w}{r}, and λ\lambda is interpreted as marginal cost.

Second-order condition. The bordered Hessian determinant must be positive, ∣Hˉ∣>0\left|\bar{H}\right| > 0, ensuring the isoquant is convex to the origin and the tangency point is a true minimum.

Expansion path. Joining the tangency points for different output levels gives the expansion (scale) line, which shows how the least-cost input combination changes as output expands. For a homogeneous production function the expansion path is a straight line through the origin.

Significance. It explains the firm's optimal input choice, the derivation of the long-run cost curve, factor substitution, and the effect of a change in relative factor prices on the capital–labour ratio.


c) Hessian determinant (if chosen instead of b)

Meaning. The Hessian is the matrix of second-order partial derivatives of a function, used to test the second-order (sufficiency) condition in optimization.

For y=f(x1,x2)y = f(x_1, x_2): H=∣f11f12f21f22∣=f11f22−f12f21H = \begin{vmatrix} f_{11} & f_{12} \\ f_{21} & f_{22} \end{vmatrix} = f_{11}f_{22} - f_{12}f_{21}

Unconstrained optimization. For a maximum, f11<0f_{11} < 0 and ∣H∣>0|H| > 0 (Hessian negative definite); for a minimum, f11>0f_{11} > 0 and ∣H∣>0|H| > 0 (positive definite).

Constrained optimization. With a constraint g(x1,x2)=0g(x_1,x_2)=0, the bordered Hessian is used: Hˉ=∣0g1g2g1f11f12g2f21f22∣\bar{H} = \begin{vmatrix} 0 & g_1 & g_2 \\ g_1 & f_{11} & f_{12} \\ g_2 & f_{21} & f_{22} \end{vmatrix} For a maximum, ∣Hˉ∣>0\left|\bar{H}\right| > 0; for a minimum, ∣Hˉ∣<0\left|\bar{H}\right| < 0 (sign of the determinant of the bordered Hessian, with the relevant principal minors alternating in sign).

Use in economics. It verifies that a stationary point found from the first-order conditions is genuinely a maximum or minimum — e.g. in profit maximization, utility maximization and cost minimization — and thus confirms the stability of the equilibrium.

Answer checked by Claude Haiku · Oct 21, 2026

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