### Section - A

1. Answer all of the following six parts in about 100 words each.

Marks: 5 × 7 = 35

Solution Video:

#### (b) The cost-minimizing demand for labour is: $L = \frac{Q}{{50}}\sqrt {\frac{r}{w}}$ and that for capital is: $K = \frac{Q}{{50}}\sqrt {\frac{w}{r}}$ where w and r denote wage and prices of capital respectively. Find the production function.

Let: $v = \sqrt {\frac{r}{w}}$ Now, the equations in the questions can be written as: $L = \frac{{Qv}}{{50}}$ $v = \frac{{50L}}{Q}$ and $K = \frac{Q}{{50v}}$ $v = \frac{Q}{{50K}}$ Equating results derived above: $\frac{Q}{{50K}} = \frac{{50L}}{Q}$ ${Q^2} = 2500LK$ $Q = \sqrt {2500LK}$ $Q = 50\sqrt {LK}$ $Q = 50{L^{\frac{1}{2}}}{K^{\frac{1}{2}}}$ This is the production function we want know.

#### (c) Explain the principle of average cost pricing in the contect of natural monopoly.

See Chapter - 7, Section - VI. Government-Regulated Monopoly, P. No. - 200, Microeconomics by A. Koutsoyinnis. The book is available in the Google Drive Folder.

#### (d) Find the monopolist's demand function when labour market is perfectly competitive.

See P. No. 456 Section: (c) The market demand for and supply of labour, Microeconomics by A. Koutsoyinnis. You can use the figure given therein.

#### (e) Explain the concept of external economies in context of marginal social benefits and marginal social costs.

External economies (positive externalities) occurs when marginal social costs are less than marginal social benefits. For example, vaccination save an individual from viruses but a vaccinated person breaks the chain of infection so as others are also safe from her. The cost of vaccine is marinal social cost and also private cost here. Safety from viruses is private benefit for which one pays for. Breaking the chain of infection is social benefit which is additional benefit for the society.

#### (f) Suppose that Leontief input-output coefficient matrix is: $A = \left[ {\begin{array}{*{20}{c}}{0.1}&{0.4}\\{0.2}&{0.5}\end{array}} \right]$ and the final demand vector is: $\left[ {\begin{array}{*{20}{c}}1\\1\end{array}} \right]$ Find the total direct and indirect requirement of the second input to satisfy the final demand.

Let requirements of inputs be $${x_1^*}$$ and $${x_2^*}$$: ${x^*} = {\left( {I - A} \right)^{ - 1}}d$ $\left[ {\begin{array}{*{20}{c}}{x_1^*}\\{x_2^*}\end{array}} \right] = {\left[ {\begin{array}{*{20}{c}}{1 - 0.1}&{0 - 0.4}\\{0 - 0.2}&{1 - 0.5}\end{array}} \right]^{ - 1}}\left[ {\begin{array}{*{20}{c}}1\\1\end{array}} \right]$ $\left[ {\begin{array}{*{20}{c}}{x_1^*}\\{x_2^*}\end{array}} \right] = {\left[ {\begin{array}{*{20}{c}}{0.9}&{ - 0.4}\\{ - 0.2}&{0.5}\end{array}} \right]^{ - 1}}\left[ {\begin{array}{*{20}{c}}1\\1\end{array}} \right]$ $\left[ {\begin{array}{*{20}{c}}{x_1^*}\\{x_2^*}\end{array}} \right] = \frac{1}{{0.37}}\left[ {\begin{array}{*{20}{c}}{0.5}&{0.4}\\{0.2}&{0.9}\end{array}} \right]\left[ {\begin{array}{*{20}{c}}1\\1\end{array}} \right]$ $\left[ {\begin{array}{*{20}{c}}{x_1^*}\\{x_2^*}\end{array}} \right] = \frac{1}{{0.37}}\left[ {\begin{array}{*{20}{c}}{0.5(1) + 0.4(1)}\\{0.2(1) + 0.9(1)}\end{array}} \right]$ $\left[ {\begin{array}{*{20}{c}}{x_1^*}\\{x_2^*}\end{array}} \right] = \frac{1}{{0.37}}\left[ {\begin{array}{*{20}{c}}{0.9}\\{1.1}\end{array}} \right]$ $x_2^* = \frac{{1.1}}{{0.37}} = 2.97$

#### (g) Show that in the regression model $${Y_i} = \alpha + \beta {X_i} + {U_i};i = 1,2,...,n$$, the covariance between the regressor and the error term is zero under ordinary least squares method of estimation.

${\mathop{\rm cov}} \left( {{X_i},{U_i}} \right) = E\left\{ {\left[ {{X_i} - E\left( {{X_i}} \right)} \right]\left[ {{U_i} - E\left( {{U_i}} \right)} \right]} \right\}$ By assumption of OLS: $E\left( {{U_i}} \right) = 0$ Now, we can write: ${\mathop{\rm cov}} \left( {{X_i},{U_i}} \right) = E\left\{ {\left[ {{X_i} - E\left( {{X_i}} \right)} \right]{U_i}} \right\}$ ${\mathop{\rm cov}} \left( {{X_i},{U_i}} \right) = E\left[ {{X_i}{U_i} - E\left( {{X_i}} \right){U_i}} \right]$ ${\mathop{\rm cov}} \left( {{X_i},{U_i}} \right) = E\left( {{X_i}{U_i}} \right) - E\left( {{X_i}} \right)E\left( {{U_i}} \right)$ ${\mathop{\rm cov}} \left( {{X_i},{U_i}} \right) = {X_i}E\left( {{U_i}} \right)$ Since Xi are fixed values: ${\mathop{\rm cov}} \left( {{X_i},{U_i}} \right) = {X_i}E\left( {{U_i}} \right)$ ${\mathop{\rm cov}} \left( {{X_i},{U_i}} \right) = 0$

### Section - B

Answer any five out of the following seven questions in about 200 words each.

Marks: 18 × 5 = 90

Solution Video:

#### 2 (b). If the price of good X decreases to 5 find the compensating variation in income in order to maintain her level of satisfaction in part (a)

Compensating variation in income refers how much income the consumer shhould be given after price change so that she can achieve the level of satisfaction without price change. This can be both negative or positive. We need to solve dual of the problem in part (a), that is, expenditure minimization when utility level will be fixed at what we calculated in part (a). Problem can be mathematically stated as follows: Minimize: $I = 5X + 10Y$ Subject to: $2500 = XY$ Taking Lagrangian: $L = 5X + 10Y + \lambda \left( {2500 - XY} \right)$ Applying the first order condition: $\frac{{\partial L}}{{\partial X}} = 0$ $5 - \lambda Y = 0$ $\lambda = \frac{5}{Y}$ and $\frac{{\partial L}}{{\partial Y}} = 0$ $10 - \lambda X = 0$ $\lambda = \frac{{10}}{X}$ Equating both results: $\frac{5}{Y} = \frac{{10}}{X}$ $\frac{X}{Y} = 2$ $X = 2Y$ Substituting this value of X in the constraint: $2500 = \left( {2Y} \right)Y$ $2500 = 2{Y^2}$ $Y = \sqrt {\frac{{2500}}{2}}$ $Y = 25\sqrt 2$ Again substituting this value of Y in the constraint: $2500 = X\left( {25\sqrt 2 } \right)$ $X = \frac{{2500}}{{25\sqrt 2 }}$ $X = 50\sqrt 2$ Substituting both values in the objective function: $I = 5\left( {50\sqrt 2 } \right) + 10\left( {25\sqrt 2 } \right)$ $I = 250\sqrt 2 + 250\sqrt 2$ $I = 500\sqrt 2$ $I = 707(aprrox.)$ Consumer needed ₹1000 to obtain 2500 units utility before decrese in the price. Now, she needs ₹707 (approx.). Therefore, compensating variation (CV) is: $CV = 1000 - 707 = 293$

#### 2 (c). An individual buys two goods X and Y at prices PX and PY. Check whether her behaviour satidfies the Weak Axiom of Revealed Prference, given the following information: When (PX, PY) = (1, 2), (X, Y) = (1, 2) When (PX, PY) = (2, 1), (X, Y) = (2, 1)

See Chapter - 7, Section - 7.5 Checking for WARP, P. No. - 125 in Intermediate Microeconomics by Hal & Varian. Same problem has been explained there. The book is available in the Google Drive Folder.

The consumer doesn't satifies the weak axiom of revealed preference because when she revealed prefered the first bundel at intial price second bundle was also affordable. After change in price, the first bundle is still affordable so she should choose first bundle even after price change to be consistent with weak axiom of revealed preference.

#### 3 (a). Show that $$q = \gamma {\left[ {\delta {L^{ - \alpha }} + \left( {1 - \delta } \right){K^{ - \alpha }}} \right]^{ - \frac{1}{\alpha }}}$$ is a production function that represents the average of two inputs L and K for different values of $$\alpha$$, given that $$\gamma \prec 0$$ and $$0 \prec \delta \prec 1$$.

(Note - This question is quite confusing. I seems that want to ask elasticity of substitution $$\left( \sigma \right)$$ for different value of $$\alpha$$.)

The formula for elasticity of substitution is: $\sigma = \frac{{d\left( {\frac{K}{L}} \right)}}{{\frac{K}{L}}} \div \frac{{d\left( {\frac{w}{r}} \right)}}{{\frac{w}{r}}}$ $\sigma = \frac{{d\left( {\frac{K}{L}} \right)}}{{d\left( {\frac{w}{r}} \right)}} \times \frac{{\frac{w}{r}}}{{\frac{K}{L}}}$ where w and r are wage and rent respectively. At level of equilibrium: $MRT{S_{L,K}} = \frac{w}{r}$ $\frac{{M{P_L}}}{{M{P_K}}} = \frac{w}{r}$ $\frac{{\frac{{\partial q}}{{\partial L}}}}{{\frac{{\partial q}}{{\partial K}}}} = \frac{w}{r}$ Where, $\frac{{\partial q}}{{\partial L}} = - \frac{1}{\alpha }\gamma {\left[ {\delta {L^{ - \alpha }} + \left( {1 - \delta } \right){L^{ - \alpha }}} \right]^{ - \frac{1}{\alpha } - 1}}\left( { - \alpha \delta {L^{ - \alpha - 1}}} \right)$ $\frac{{\partial q}}{{\partial L}} = \gamma {\left[ {\delta {L^{ - \alpha }} + \left( {1 - \delta } \right){L^{ - \alpha }}} \right]^{ - \frac{1}{\alpha } - 1}}\left( {\delta {L^{ - \alpha - 1}}} \right)$ and, $\frac{{\partial q}}{{\partial K}} = - \frac{1}{\alpha }\gamma {\left[ {\delta {L^{ - \alpha }} + \left( {1 - \delta } \right){K^{ - \alpha }}} \right]^{ - \frac{1}{\alpha } - 1}}\left( { - \alpha \left( {1 - \delta } \right){K^{ - \alpha - 1}}} \right)$ $\frac{{\partial q}}{{\partial K}} = \gamma {\left[ {\delta {L^{ - \alpha }} + \left( {1 - \delta } \right){K^{ - \alpha }}} \right]^{ - \frac{1}{\alpha } - 1}}\left\{ {\left( {1 - \delta } \right){K^{ - \alpha - 1}}} \right\}$ Now, it can be written: $\frac{{M{P_L}}}{{M{P_K}}} = \frac{{\gamma {{\left[ {\delta {L^{ - \alpha }} + \left( {1 - \delta } \right){L^{ - \alpha }}} \right]}^{ - \frac{1}{\alpha } - 1}}\left( {\delta {L^{ - \alpha - 1}}} \right)}}{{\gamma {{\left[ {\delta {L^{ - \alpha }} + \left( {1 - \delta } \right){K^{ - \alpha }}} \right]}^{ - \frac{1}{\alpha } - 1}}\left\{ {\left( {1 - \delta } \right){K^{ - \alpha - 1}}} \right\}}}$ $\frac{{M{P_L}}}{{M{P_K}}} = \frac{{\delta {L^{ - \alpha - 1}}}}{{\left( {1 - \delta } \right){K^{ - \alpha - 1}}}}$ $\frac{{M{P_L}}}{{M{P_K}}} = \frac{\delta }{{1 - \delta }} \cdot {\left( {\frac{K}{L}} \right)^{1 + \alpha }}$ Now, equilibrium condition is: $\frac{\delta }{{1 - \delta }} \cdot {\left( {\frac{K}{L}} \right)^{1 + \alpha }} = \frac{w}{r}$ $\frac{K}{L} = {\left( {\frac{{1 - \delta }}{\delta }} \right)^{\frac{1}{{1 + \alpha }}}} \cdot {\left( {\frac{w}{r}} \right)^{\frac{1}{{1 + \alpha }}}}$ Taking derivating with repect to input price ratio: $\frac{{d\left( {\frac{K}{L}} \right)}}{{d\left( {\frac{w}{r}} \right)}} = \frac{1}{{1 + \alpha }} \cdot {\left( {\frac{{1 - \delta }}{\delta }} \right)^{\frac{1}{{1 + \alpha }}}} \cdot {\left( {\frac{w}{r}} \right)^{\frac{1}{{1 + \alpha }} - 1}}$ Now, elasticity of substitution is: $\sigma = \frac{1}{{1 + \alpha }} \cdot {\left( {\frac{{1 - \delta }}{\delta }} \right)^{\frac{1}{{1 + \alpha }}}}{\left( {\frac{w}{r}} \right)^{\frac{1}{{1 + \alpha }} - 1}} \times \frac{{\frac{w}{r}}}{{{{\left( {\frac{{1 - \delta }}{\delta }} \right)}^{\frac{1}{{1 + \alpha }}}} \cdot {{\left( {\frac{w}{r}} \right)}^{\frac{1}{{1 + \alpha }}}}}}$ $\sigma = \frac{1}{{1 + \alpha }}$

$$\sigma$$ is constant and its magnitude depends on the value of $$\alpha$$

Reference: See page no. 396 to 399 of Mathematical Economics book by A. C. Chiang available in the google drive folder.

#### 3 (b). Find the marginal rate of technical substitution for the production function given in part (a).

It has already been calculated in the answer of part (a). $MRT{S_{L,K}} = \frac{{M{P_L}}}{{M{P_K}}}$ $MRT{S_{L,K}} = \frac{{\frac{{\partial q}}{{\partial L}}}}{{\frac{{\partial q}}{{\partial K}}}}$ where, $\frac{{\partial q}}{{\partial L}} = - \frac{1}{\alpha }\gamma {\left[ {\delta {L^{ - \alpha }} + \left( {1 - \delta } \right){L^{ - \alpha }}} \right]^{ - \frac{1}{\alpha } - 1}}\left( { - \alpha \delta {L^{ - \alpha - 1}}} \right)$ $\frac{{\partial q}}{{\partial L}} = \gamma {\left[ {\delta {L^{ - \alpha }} + \left( {1 - \delta } \right){L^{ - \alpha }}} \right]^{ - \frac{1}{\alpha } - 1}}\left( {\delta {L^{ - \alpha - 1}}} \right)$ and $\frac{{\partial q}}{{\partial K}} = - \frac{1}{\alpha }\gamma {\left[ {\delta {L^{ - \alpha }} + \left( {1 - \delta } \right){K^{ - \alpha }}} \right]^{ - \frac{1}{\alpha } - 1}}\left( { - \alpha \left( {1 - \delta } \right){K^{ - \alpha - 1}}} \right)$ $\frac{{\partial q}}{{\partial K}} = \gamma {\left[ {\delta {L^{ - \alpha }} + \left( {1 - \delta } \right){K^{ - \alpha }}} \right]^{ - \frac{1}{\alpha } - 1}}\left\{ {\left( {1 - \delta } \right){K^{ - \alpha - 1}}} \right\}$ Now, we can write: $MRT{S_{L,K}} = \frac{{\gamma {{\left[ {\delta {L^{ - \alpha }} + \left( {1 - \delta } \right){L^{ - \alpha }}} \right]}^{ - \frac{1}{\alpha } - 1}}\left( {\delta {L^{ - \alpha - 1}}} \right)}}{{\gamma {{\left[ {\delta {L^{ - \alpha }} + \left( {1 - \delta } \right){K^{ - \alpha }}} \right]}^{ - \frac{1}{\alpha } - 1}}\left\{ {\left( {1 - \delta } \right){K^{ - \alpha - 1}}} \right\}}}$ $MRT{S_{L,K}} = \frac{{\delta {L^{ - \alpha - 1}}}}{{\left( {1 - \delta } \right){K^{ - \alpha - 1}}}}$ $MRT{S_{L,K}} = \frac{\delta }{{1 - \delta }} \cdot {\left( {\frac{K}{L}} \right)^{1 + \alpha }}$

#### 4 (a). A firm with market power faces the demand curve given by: $P = 100 - 3Q + 4\sqrt A$ where P, Q and A denote price, quantity and expenditure on advertising respectively. The total cost is given as: $C = 4{Q^2} + 10Q + A$ Find the firm's profit-maximizing price. (Marks - 8)

$TR = PQ$ $TR = Q\left( {100 - 3Q + 4\sqrt A } \right)$ $TR = 100Q - 3{Q^2} + 4Q\sqrt A$ It is known that: $\Pr ofit\left( \pi \right) = TR - C$ $\pi = 100Q - 3{Q^2} + 4Q\sqrt A - \left( {4{Q^2} + 10Q + A} \right)$ $\pi = 90Q - 7{Q^2} + 4Q\sqrt A - A$ Applying the first order condition: $\frac{{\partial \pi }}{{\partial Q}} = 0$ $90 - 14Q + 4\sqrt A = 0$ and $\frac{{\partial \pi }}{{\partial A}} = 0$ $\frac{{2Q}}{{\sqrt A }} - 1 = 0$ $\frac{{2Q}}{{\sqrt A }} = 1$ $\sqrt A = 2Q$ $A = {\left( {2Q} \right)^2}$ $A = 4{Q^2}$ Substituting the value of A : $90 - 14Q + 4\sqrt A = 0$ $90 - 14Q + 4\sqrt {4{Q^2}} = 0$ $90 - 14Q + 8Q = 0$ $6Q = 90$ $Q = \frac{{90}}{6}$ $Q = 15$ Substituting the value of Q: $A = 4{Q^2}$ $A = 4{\left( {15} \right)^2}$ $A = 900$ Substituting the value of Q and A in the inverse demand function: $P = 100 - 3Q + 4\sqrt A$ $P = 100 - 3 \times 15 + 4\sqrt {900}$ $P = 175$

#### 4 (b). Suppose that the demand and supply function in a market are given as: ${Q_D} = 100 - P$ $and$ ${Q_S} = 200 - 5P$ Analyse whether the equilibrium is Walrasian stable or Marshallian stable. (Marks -10)

It can easility be obseved from the equations that both demand and supply are negatively sloped. Taking Q on vertical axis and P on horizontal axis, it can also be observed that absolute slope of supply curve is 5 and that of the demand curve is 1. As a thumb rule, equilibrium is Walrasian unstable but Marshallian stableif absolute slope of supply curve is greater than the absolute slope of demand curve when both demand and supply curve are negatively sloped (Q should be on the vertical axis and P on horizontal axis).

Equilibrium price and quantity for a market is as follows: $100 - P = 200 - 5P$ $5P - P = 200 - 100$ $P = 25$ Substituting this value in either demand or supply equation, we will get: ${Q_D} = {Q_S} = 75$

Walrasian Stability Analysis: In the Walrasian system, if there is excess demand then price increses. As a result of increase in price, demands falls and supply increases. This simultaneous process brings equilibrium in the market price. We can analyse the equations in the same framework. Let price be 30. At this price, demand and quantity will be: ${Q_D} = 100 - 30 = 70$ ${Q_S} = 200 - 5 \times 30 = 50$ As it can be observed that there is excess demand at price 30. So, the prices should increase but the equilibrium price is 25 so increase in price will result in further disequilibrium. It means that equilibrium cannot be achieved in Walrasian syatem for this model. In other words, equilibrium is Walrasian unstable for this model.

Marshallian Stability Analysis: In the Marshallian system, if demand price is greater than supply price for a given quantity then the sopplier will increase the supply. Due to incresed supply prices will fall. As a result of fall in prices, demand will increse. This will bring equilibrium in the market. Let quantity be 70. At this quantity level, demand price is: $70 = 100 - P$ $P = 30$ and the supply price is: $70 = 200 - 5P$ $P = \frac{{200 - 70}}{5}$ $P = 26$ It can be observed that the demand price is greater than supply price. So, supply should increase. As we have calculated, if supply is incresed to 75, market will be at the equilibrium level with price 25. It means equilibrium can be achieved in Marshallian system for this model. In other words, this equilibrium is Marshallian stable.

Also See: Stability of Equilibrium

#### 5 (a) and (b) [See the question paper from Google Drive Folder]

See P. No. - 458, Section: Model B. The firm bas monopolistic power in the commodity market and monopsonistic power in the factor market, Microeconomics by A. Koutsoyinnis. the book is available in the Google Drive Folder. Both question has been answered in this section.

#### 6 (a). [See the question paper from Google Drive Folder]

Following Section of Microeconomics by A Koutsoyiannis ( P. N0. - 497) exactly explains the answer.

• 2. STATIC PROPERTIES OF A GENERAL EQUILIBRIUM STATE (CONFIGURATION)
• (a) Equilibrium of productioo (efficiency in factor substitution)
• (b) EquiHbrium of consumption (efficiency in distribution of commodities): You can also call it. Equilibrium in exchange
• (c) Simultaneous equilibrium of production and consumption (efficiency in product-mix)

#### 6 (b). [See the question paper from Google Drive Folder]

Private Optimal Price and quantity $Q = 24 - P$ $P = 24 - Q$ Private equilibrium condition is: $P = MC$ $24 - Q = 2 + Q$ $2Q = 22$ $Q = \frac{{22}}{2} = 11$ This is the private optimal quantity. Since $P = 24 - Q$ So $P = 24 - 11 = 13$ This is the private optimal price.

Social OPtimal Quatity and Price: Let Marginal Externality Cost be MECE: $MEC = \frac{d}{{dQ}}\left( { - 2Q + \frac{{{Q^2}}}{2}} \right)$ $MEC = - 2 + Q$ Let Marginal Social Cost be: $MSC = MC + MEC$ $MSC = 2 + Q - 2 + Q$ $MSC = 2Q$ Private equilibrium condition is: $P = MSC$ $13 = 2Q$ $Q = \frac{{13}}{2} = 6.5$ This is the Social Optimal Quantity. Since $P = 24 - Q$ So $P = 24 - 6.5 = 17.5$ This is the social optimal quantity.

Comparision: Social optimal quantitity is less than private optimal quantity and the social optimal price is greater than private optimal price.

For its theoretical explanation, see Microeconomics by A. Koutsoyiannis, P. No. - 542, Section: A. Externalities in production

#### 7. Consider the following data on two variables:

 Y Y -5 -4 -3 -2 -1 0 1 2 3 4 5 25 16 9 4 1 0 1 4 9 16 25

Solution Video:

#### 8 (a). [See the question paper from Google Drive Folder] By Reidpath - The original file was on WikiMedia Commons (<a class="external free" href="https://en.wikipedia.org/wiki/File:Economics_Gini_coefficient.svg">http://en.wikipedia.org/wiki/File:Economics_Gini_coefficient.svg</a>). I have edited the file., Public Domain, Link

Gini Coefficient is the ration of area A and the total area of the rectangle. A is above the Lorez curve and B is below the Lorez curve. Let G represent the Gini coefficient. Now, the formula for Gini coefficient is: $G = \frac{A}{{A + B}}$ It is assumed that A + B = 0.5, that is half of the area of square of area unit 1. So, we can write: $G = \frac{{0.5 - B}}{{0.5}}$ $G = \frac{{0.5}}{{0.5}} - \frac{B}{{0.5}}$ $G = 1 - 2B$ Which is exactly:
Gini Coefficient = 1 - 2 $$\times$$ Area below the Lorenz curve.

#### 8 (b). [See the question paper from Google Drive Folder]

If θ amount is added to all the person then Gini Coefficient will decrease meaning decline in inequality. θ amount will result in greater percentage of increase in lower income group as compared to higher income group. For example, if A earns ₹10,000 and B earns ₹100,000 and both are given addition ₹1000 then this ₹1000 is 10% for A but only 1% for B.

#### 8 (c). [See the question paper from Google Drive Folder]

Calculation of Gini coefficients entails arranging population in accending order in terms of income and see how much percentage of income is received by how much percentage of population. it has nothing to do with the distribution of population. Therefore, Gini Coefficient is distribution insensitive.
(Note: Mathematical proof is very lenghty so there is no possibility that it will be asked in any examination.)

### Section - C

Answer any three of the following five questions in about 300 words each.

Marks: 25 × 3 = 75

#### 9 (a). [See the question paper from Google Drive Folder]

See Equation 5.40, Pg. No. - 161 Microeconomics by Nicholson and Snyder. The book is available in Google Drive Folder.
Slutsky Equation in elasticity form: ${e_{x,{p_x}}} = {e_{{x^c},{p_x}}} - {s_x} \cdot {e_{x,I}}$ [Ordinary Demand Elasticity] = [Compensated Demand Elasticity] - [Share of Expenditure on X][Income Elasticity]
Ordinary Demand Elasticity and Compensated Demand Elasticity will have negative sign for Normal Goods. All negative sign can be cancelled after calculation. After cancelling negative sign it can be noticed that Ordinary Demand Elasticity is addition of Compensated Demand Elasticity and Share of Expenditure multiplied by Income elasticity. It simply means that ordinary demand elasticity is greater than compensated demand elasticity.

For inferior goods, income elasticity of demand has negative sign so it should be subtracted from compensated demand elasticity. It means compensated demand elasticity is greater than ordinary demand elasticity in case of inferior goods.

#### 9 (b). [See the question paper from Google Drive Folder]

$MC = \frac{{d\left( {TC} \right)}}{{dQ}}$ $MC = \frac{d}{{dQ}}\left( {5Q + 20} \right)$ $MC = 5$ Let Marginal Revenue in the first market be MR1 and in the second market be MR2: ${Q_1} = 55 - {P_1}$ ${P_1} = 55 - {Q_1}$ $T{R_1} = {P_1}{Q_1} = 55{Q_1} - Q_1^2$ $\therefore M{R_1} = 55 - 2{Q_1}$ and ${Q_2} = 70 - 2{P_2}$ $2{P_2} = 70 - {Q_2}$ ${P_2} = 35 - \frac{{{Q_2}}}{2}$ $T{R_2} = {P_2}{Q_2} = 35{Q_2} - \frac{{Q_2^2}}{2}$ $\therefore M{R_2} = 35 - {Q_2}$

(i) Equilibrium condition in the first market: $M{R_1} = MC$ $55 - 2{Q_1} = 5$ $2{Q_1} = 55 - 5$ ${Q_1} = \frac{{50}}{2}$ ${Q_1} = 25$ This is the equilibrium quantity for the first market.

Equilibrium condition in the second market: $M{R_2} = MC$ $35 - {Q_2} = 5$ ${Q_2} = 35 - 5$ ${Q_2} = 30$ This is the equilibrium quantity for the second market.

(ii) Equilibrium price in the first market: ${P_1} = 55 - {Q_1}$ ${P_1} = 55 - 25$ ${P_1} = 30$ This is the equilibrium price for the first market.

Equilibrium price in the second market: ${P_2} = 35 - \frac{{{Q_2}}}{2}$ ${P_2} = 35 - \frac{{30}}{2}$ ${P_2} = 35 - 15 = 20$ This is the equilibrium price for the second market.

(iii) Elasticity of demand in the first market at the point of equilibrium: ${e_{{Q_1},{P_1}}} = \frac{{d{Q_1}}}{{d{P_1}}} \times \frac{{{P_1}}}{{{Q_1}}}$ ${e_{{Q_1},{P_1}}} = - 1 \times \frac{{30}}{{25}}$ ${e_{{Q_1},{P_1}}} = - 1.2$ $\left| {{e_{{Q_1},{P_1}}}} \right| = 1.2$

${e_{{Q_2},{P_2}}} = \frac{{d{Q_2}}}{{d{P_2}}} \times \frac{{{P_2}}}{{{Q_2}}}$ ${e_{{Q_2},{P_2}}} = - 2 \times \frac{{20}}{{30}}$ ${e_{{Q_2},{P_2}}} = - 1.67$ $\left| {{e_{{Q_2},{P_2}}}} \right| = 1.67$ The monopolist charges higher price in the market where elasticity of demand in absolute term is lower and lower price in the market where elasticity of demand is low.

#### 10 (a). [See the question paper from Google Drive Folder]

This question is easy. Google it.
Public goods have two main charateristics - (i) Non-rivalry and (ii) Non-excudability. Non-ravalry means consumption by one individual will not reduce its quantity for others while Non-excludability means one cannot be excluded from consuming it. For example, air is both non-rival and non-excludable in consumption

#### 10 (b). [See the question paper from Google Drive Folder]

Willingness to pay of individual - 1: ${P_1} = 100 - Q$ Willingness to pay of individual - 1: ${P_2} = 200 - Q$ Total willing to pay after horizontal summation: $WTP = 300 - 2Q$

(i) Optimal Provision for public good when the Marginal Cost is 240: $WTP = MC$ $300 - 2Q = 240$ $Q = 30$ When the price is zero, consumption will be 300 units and the provider of public goods, say, government need to spend 300(240) = 72,000. However, government can limit the quantity at 30 units through taxation for which individual 1 is willing to pay 70(30) = 2100 and individual 2 is willing to pay 170(30) = 5100, The cost of government will be 240(30)= 7200 which is exactly equal to 2100 + 5100 = 7200 which two individuals are collectly ready to pay.

(ii) Optimal Provision for public good when the Marginal Cost is 240: $WTP = MC$ $300 - 2Q = 50$ $Q = 125$ Analysise in the same way as above. In this case, individual 1 is not willing to pay anything. Instead, he need a compensation of 25 per unit of over production.

#### 10 (c). [See the question paper from Google Drive Folder]

You can simply Google the definition of Coase theorem. It's commonly available.

(i) If the factory has right to pollute then the fisherman has two options:

1. He can accept the damage of ₹1 lakh. In this case, he do not need to pay anything. His loss is ₹1 lakh.
2. He can pay to the factory ₹2 lakh to construct a pollution treatment plant. Thus, he willl avoid the damage of ₹1 lakh. However, his loss is still ₹1 lakh because he has paid ₹2 lakhs to avoid the damage of ₹1 lakh.

(i) If the fisherman has right to clean water then the factory has two options:

1. Factory can construct the pollution treatment plant so that the fisherman avoid the damage of ₹1lakh. The benefit of the fisherman is ₹1 lakh after construction of the plant.
2. The factory can pay the fisherman ₹2 lakhs and do not construct the pollution treatment plant. Thus, the fisherman will accept the damage of ₹1lakhs because he is being paid ₹2 lakh. His benefit is again ₹1 lakh.

#### 11 (a). Consider the multiple regression model: ${Y_i} = \alpha + {\beta _2}{X_{2i}} + {\beta _3}{X_{3i}} + {U_i}$ $i = 1,2,...,n$ where Uis are independent and normally distributed with mean zero and variance $${\sigma ^2}$$. Also consider the auxiliary regressions: ${X_{2i}} = \hat a + \hat b{X_{3i}} + {{\hat V}_{2i}}$ ${X_{3i}} = \hat c + \hat d{X_{2i}} + {{\hat V}_{3i}}$ where $${{\hat V}_{2}}$$ and $${{\hat V}_{2i}}$$ are error terms. Show that $${{\hat \beta }_2}$$, the ordinary least squares estimate of $${\beta _2}$$, can be interpreted as a simple regression of Y on $${{\hat V}_{2}}$$. (Marks - 8)

We can write the first auxiliary regression equation in devition form as follows: ${x_{2i}} = \hat b{x_{3i}} + {{\hat v}_{2i}}$ This equation implies that there exists multicollinearity in the original regression equation because $${X_{2i}}$$ is the function of $${X_{3i}}$$.
The formula for the least square estimator in absense of multicollinearity is given as follows: ${{\hat \beta }_2} = \frac{{{{\sum {{y_i}x} }_{2i}}\sum {x_{3i}^2} - \sum {{y_i}{x_{3i}}} \sum {{x_{2i}}{x_{3i}}} }}{{\sum {x_{2i}^2} \sum {x_{3i}^2} - {{\left( {\sum {{x_{2i}}{x_{3i}}} } \right)}^2}}}$ Now, we can substitute the value of deviation form of multicollinear variable, that is, $${x_{2i}}$$: ${{\hat \beta }_2} = \frac{{\left\{ {\sum {{y_i}\left( {\hat b{x_{3i}} + {{\hat v}_{2i}}} \right)} } \right\}\sum {x_{3i}^2} - \sum {{y_i}{x_{3i}}} \sum {{x_{3i}}\left( {\hat b{x_{3i}} + {{\hat v}_{2i}}} \right)} }}{{{{\sum {\left( {\hat b{x_{3i}} + {{\hat v}_{2i}}} \right)} }^2}\sum {x_{3i}^2} - {{\left\{ {\sum {{x_{3i}}\left( {\hat b{x_{3i}} + {{\hat v}_{2i}}} \right)} } \right\}}^2}}}$ ${{\hat \beta }_2} = \frac{{\hat b\sum {{y_i}{x_{3i}}} \sum {x_{3i}^2} + \sum {{y_i}{{\hat v}_{2i}}} \sum {x_{3i}^2} - \hat b\sum {{y_i}{x_{3i}}} \sum {x_{3i}^2} - \sum {{y_i}{x_{3i}}} \sum {{x_{3i}}{{\hat v}_{2i}}} }}{{{{\left( {\hat b\sum {x_{3i}^2} } \right)}^2} + \sum {\hat v_{2i}^2} \sum {x_{3i}^2} - {{\left( {\hat b\sum {x_{3i}^2} } \right)}^2} - {{\left( {\sum {{x_{3i}}{{\hat v}_{2i}}} } \right)}^2}}}$ By the assumption of no serial correlation or autocorrelation in the auxiliary regressions: $\sum {{x_{3i}}{{\hat V}_{2i}}} = 0$ As a result: $\sum {{y_i}{x_{3i}}} \sum {{x_{3i}}{{\hat v}_{2i}}} = 0$ ${\left( {\sum {{x_{3i}}{{\hat v}_{2i}}} } \right)^2} = 0$ Now, we are remained with: ${{\hat \beta }_2} = \frac{{\hat b\sum {{y_i}{x_{3i}}} \sum {x_{3i}^2} + \sum {{y_i}{{\hat v}_{2i}}} \sum {x_{3i}^2} - \hat b\sum {{y_i}{x_{3i}}} \sum {x_{3i}^2} }}{{{{\left( {\hat b\sum {x_{3i}^2} } \right)}^2} + \sum {\hat v_{2i}^2} \sum {x_{3i}^2} - {{\left( {\hat b\sum {x_{3i}^2} } \right)}^2}}}$ Subtracting similar terms: ${{\hat \beta }_2} = \frac{{\sum {{y_i}{{\hat v}_{2i}}} \sum {x_{3i}^2} }}{{\sum {\hat v_{2i}^2} \sum {x_{3i}^2} }}$ Removing similar terms from nominator and denominator: ${{\hat \beta }_2} = \frac{{\sum {{y_i}{{\hat v}_{2i}}} }}{{\sum {\hat v_{2i}^2} }}$ This is the similar result which is found in simple linear regression model. Therefore, we can say that $${{\hat \beta }_2}$$, the ordinary least squares estimate of $${\beta _2}$$, is a simple regression of Y on $${{\hat V}_{2}}$$.

For reference, See Chapter - 10, Section - 10.3 of Basic Econometrics by Gujarati.

#### 11 (b). Consider the regression: ${Y_i} = \gamma + {{\hat \delta }_2}{{\hat V}_{2i}} + {{\hat \delta }_3}{{\hat V}_{3i}} + {W_i}$ where $${{\hat V}_{2}}$$ and $${{\hat V}_{3}}$$ are as defined in part (a). Find the relationship between the ordinary least squares estimate $${{\hat \delta }_2}$$ and the estimate $${{\hat \beta }_2}$$ as defined in part (a). (Marks - 5)

$${{\hat \beta }_2}$$ explains regression of Y on only one variable, that is, V2 while $${{\hat \delta }_2}$$ will explain regression of Y on both V2 and V3. So, the relationship can be explained by specification bias.
One of the normal equation derived from the multiple regression model in this question is: $\sum {{y_i}{{\hat v}_{2i}}} = {{\hat \delta }_2}\sum {\hat v_{2i}^2} + {{\hat \delta }_3}\sum {{{\hat v}_{2i}}{{\hat v}_{3i}}}$ Dividing both sides by $${\sum {\hat v_{2i}^2} }$$, we get: $\frac{{\sum {{y_i}{{\hat v}_{2i}}} }}{{\sum {\hat v_{2i}^2} }} = {{\hat \delta }_2} + {{\hat \delta }_3}\frac{{\sum {{{\hat v}_{2i}}{{\hat v}_{3i}}} }}{{\sum {\hat v_{2i}^2} }}$ We have already calculated that: $\frac{{\sum {{y_i}{{\hat v}_{2i}}} }}{{\sum {\hat v_{2i}^2} }} = {{\hat \beta }_2}$ $$\frac{{\sum {{{\hat v}_{2i}}{{\hat v}_{3i}}} }}{{\sum {\hat v_{2i}^2} }}$$ is simple regression of V3 on V2. Let us denote it by a : $\frac{{\sum {{{\hat v}_{2i}}{{\hat v}_{3i}}} }}{{\sum {\hat v_{2i}^2} }} = a$ Now, we can write: ${{\hat \beta }_2} = {{\hat \delta }_2} + {{\hat \delta }_3} \cdot a$ or ${{\hat \beta }_2} - {{\hat \delta }_2} = {{\hat \delta }_3} \cdot a$ where, $${{\hat \beta }_2} - {{\hat \delta }_2}$$ is the specification bias. This specification bias is due to removal of one variable in $${{\hat \beta }_2}$$. There is no reason to assume that $${{\hat \delta }_3}$$ is zero. a is zero if there is perfect multicollinearity. So, $${{\hat \delta }_3} \cdot a$$ is positive. It implies that $${{\hat \beta }_2} \succ {{\hat \delta }_2}$$ if multicollinearity is imperfect and $${{\hat \beta }_2} = {{\hat \delta }_2}$$ if multicollinearity is perfect.

#### 11 (c): (i) - In the regression model of part (a), what will be the variance of ordianary least squares estimate $${{\hat \beta }_2}$$? (ii) - Develop the F-test statistic for the goodness of fit of the regression model of part (a).

(i) Variance in multiple regression setup is: ${\mathop{\rm var}} \left( {{{\hat \beta }_2}} \right) = \sigma _u^2\frac{{\sum {x_{3i}^2} }}{{\sum {x_{2i}^2} \sum {x_{3i}^2} - {{\left( {\sum {{x_{2i}}{x_{3i}}} } \right)}^2}}}$ But we know that the model is multicollinear where, ${x_{2i}} = \hat b{x_{3i}} + {{\hat v}_{2i}}$ So, by substituting the value of above equation we get: ${\mathop{\rm var}} \left( {{{\hat \beta }_2}} \right) = \sigma _u^2\frac{{\sum {x_{3i}^2} }}{{{{\sum {\left( {\hat b{x_{3i}} + {{\hat v}_{2i}}} \right)} }^2}\sum {x_{3i}^2} - {{\left\{ {\sum {{x_{3i}}\left( {\hat b{x_{3i}} + {{\hat v}_{2i}}} \right)} } \right\}}^2}}}$ ${\mathop{\rm var}} \left( {{{\hat \beta }_2}} \right) = \sigma _u^2\frac{{\sum {x_{3i}^2} }}{{{{\left( {\hat b\sum {x_{3i}^2} } \right)}^2} + \sum {\hat v_{2i}^2} \sum {x_{3i}^2} - {{\left( {\hat b\sum {x_{3i}^2} } \right)}^2} - {{\left( {\sum {{x_{3i}}{{\hat v}_{2i}}} } \right)}^2}}}$ ${\mathop{\rm var}} \left( {{{\hat \beta }_2}} \right) = \sigma _u^2\frac{{\sum {x_{3i}^2} }}{{{{\left( {\hat b\sum {x_{3i}^2} } \right)}^2} + \sum {\hat v_{2i}^2} \sum {x_{3i}^2} - {{\left( {\hat b\sum {x_{3i}^2} } \right)}^2}}}$ ${\mathop{\rm var}} \left( {{{\hat \beta }_2}} \right) = \sigma _u^2\frac{{\sum {x_{3i}^2} }}{{\sum {\hat v_{2i}^2} \sum {x_{3i}^2} }}$ ${\mathop{\rm var}} \left( {{{\hat \beta }_2}} \right) = \frac{{\sigma _u^2}}{{\sum {\hat v_{2i}^2} }}$

We have reduced multiple regression in part (a) in a simple linear regression model which can be expressed as follow: ${y_i} = {{\hat \beta }_2}{{\hat v}_{2i}} + {e_i}$ ${y_i} = {{\hat y}_i} + {e_i}$ where, ${{\hat y}_i} = {{\hat \beta }_2}{{\hat v}_{2i}}$ For this equation, the measure of goodness of fit R2 is given by: ${R^2} = \frac{{\sum {\hat y_i^2} }}{{\sum {y_i^2} }}$ Summing up and squaring the second equation we get, $\sum {y_i^2} = \sum {\hat y_i^2} + \sum {e_i^2}$ Dividing both sides by $$\sum {y_i^2}$$: $1 = \frac{{\sum {\hat y_i^2} }}{{\sum {y_i^2} }} + \frac{{\sum {e_i^2} }}{{\sum {y_i^2} }}$ $1 = {R^2} + \frac{{\sum {e_i^2} }}{{\sum {y_i^2} }}$ $\frac{{\sum {e_i^2} }}{{\sum {y_i^2} }} = 1 - {R^2}$ F-statistic for a two variable regression is: $F = \left( {N - 2} \right)\frac{{\sum {\hat y_i^2} }}{{\sum {e_i^2} }}$ Dividing both nominator and denominator by $${\sum {y_i^2} }$$, $F = \left( {N - 2} \right)\frac{{\frac{{\sum {\hat y_i^2} }}{{\sum {y_i^2} }}}}{{\frac{{\sum {e_i^2} }}{{\sum {y_i^2} }}}}$ $F = \left( {N - 2} \right)\frac{{{R^2}}}{{1 - {R^2}}}$ The is the F-statistic of goodness of fit for the multicollinear regression model of part (a).

#### 12 (a). Suppose that the relationship to be estimated is: ${Y_i} = {b_0} + {b_1}{X_{1i}} + {b_2}{X_{2i}} + {U_i}$ and X1 and X2 are related with the exact relation ${X_2} = k{X_1}$ where k is the arbitrary constant. Show that the estimates of the coefficients are indeterminate and standard errors of these estimates become infinitely large. (Marks - 10)

1. The estimates of the coefficients are indeterminate.
Proof: Let $${{\mathord{\buildrel{\lower3pt\hbox{\scriptscriptstyle\frown}} \over b} }_0}$$, $${{\mathord{\buildrel{\lower3pt\hbox{\scriptscriptstyle\frown}} \over b} }_1}$$ and $${{\mathord{\buildrel{\lower3pt\hbox{\scriptscriptstyle\frown}} \over b} }_2}$$ be the estimates of $${b_0}$$, $${b_1}$$ and $${b_2}$$ respectively and it is known that th formulae for estimation of coefficients are a follows: ${{\mathord{\buildrel{\lower3pt\hbox{\scriptscriptstyle\frown}} \over b} }_0} = \bar Y - {{\mathord{\buildrel{\lower3pt\hbox{\scriptscriptstyle\frown}} \over b} }_1}{{\bar X}_1} - {{\mathord{\buildrel{\lower3pt\hbox{\scriptscriptstyle\frown}} \over b} }_2}{{\bar X}_2}$ ${{\mathord{\buildrel{\lower3pt\hbox{\scriptscriptstyle\frown}} \over b} }_1} = \frac{{\left( {\sum {{x_{1i}}{y_i}} } \right)\left( {\sum {x_{2i}^2} } \right) - \left( {\sum {{x_{2i}}{y_i}} } \right)\left( {\sum {{x_{1i}}{x_{2i}}} } \right)}}{{\left( {\sum {x_{1i}^2} } \right)\left( {\sum {x_{2i}^2} } \right) - {{\left( {\sum {{x_{1i}}{x_{2i}}} } \right)}^2}}}$ ${{\mathord{\buildrel{\lower3pt\hbox{\scriptscriptstyle\frown}} \over b} }_2} = \frac{{\left( {\sum {{x_{2i}}{y_i}} } \right)\left( {\sum {x_{1i}^2} } \right) - \left( {\sum {{x_{1i}}{y_i}} } \right)\left( {\sum {{x_{1i}}{x_{2i}}} } \right)}}{{\left( {\sum {x_{1i}^2} } \right)\left( {\sum {x_{2i}^2} } \right) - {{\left( {\sum {{x_{1i}}{x_{2i}}} } \right)}^2}}}$ Substituting kX1 for X2 in $${{\mathord{\buildrel{\lower3pt\hbox{\scriptscriptstyle\frown}} \over b} }_1}$$, we get: ${{\mathord{\buildrel{\lower3pt\hbox{\scriptscriptstyle\frown}} \over b} }_1} = \frac{{\left( {\sum {{x_{1i}}{y_i}} } \right)\left\{ {\sum {{{\left( {k{x_{1i}}} \right)}^2}} } \right\} - \left\{ {\sum {\left( {k{x_{1i}}} \right){y_i}} } \right\}\left\{ {\sum {{x_{1i}}\left( {k{x_{1i}}} \right)} } \right\}}}{{\left( {\sum {x_{1i}^2} } \right)\left\{ {\sum {{{\left( {k{x_{1i}}} \right)}^2}} } \right\} - {{\left\{ {\sum {{x_{1i}}\left( {k{x_{1i}}} \right)} } \right\}}^2}}}$ Taking k2 as common: ${{\mathord{\buildrel{\lower3pt\hbox{\scriptscriptstyle\frown}} \over b} }_1} = \frac{{{k^2}\left( {\sum {{x_{1i}}{y_i}} } \right)\left( {\sum {x_{1i}^2} } \right) - {k^2}\left( {\sum {{x_{1i}}{y_i}} } \right)\left( {\sum {x_{_{1i}}^2} } \right)}}{{{k^2}{{\left( {\sum {x_{1i}^2} } \right)}^2} - {k^2}{{\left( {\sum {x_{_{1i}}^2} } \right)}^2}}}$ ${{\mathord{\buildrel{\lower3pt\hbox{\scriptscriptstyle\frown}} \over b} }_1} = \frac{0}{0}$ Simularly, for $${{\mathord{\buildrel{\lower3pt\hbox{\scriptscriptstyle\frown}} \over b} }_2}$$, we obtain: ${{\mathord{\buildrel{\lower3pt\hbox{\scriptscriptstyle\frown}} \over b} }_2} = \frac{{k\left( {\sum {{x_{1i}}{y_i}} } \right)\left( {\sum {x_{1i}^2} } \right) - k\left( {\sum {{x_{1i}}{y_i}} } \right)\left( {\sum {x_{_{1i}}^2} } \right)}}{{{k^2}{{\left( {\sum {x_{1i}^2} } \right)}^2} - {k^2}{{\left( {\sum {x_{_{1i}}^2} } \right)}^2}}}$ ${{\mathord{\buildrel{\lower3pt\hbox{\scriptscriptstyle\frown}} \over b} }_2} = \frac{0}{0}$ Thus, $${{\mathord{\buildrel{\lower3pt\hbox{\scriptscriptstyle\frown}} \over b} }_1}$$ and $${{\mathord{\buildrel{\lower3pt\hbox{\scriptscriptstyle\frown}} \over b} }_2}$$ are indeterminate. Since $${{\mathord{\buildrel{\lower3pt\hbox{\scriptscriptstyle\frown}} \over b} }_0}$$ depends on the values of $${{\mathord{\buildrel{\lower3pt\hbox{\scriptscriptstyle\frown}} \over b} }_1}$$ and $${{\mathord{\buildrel{\lower3pt\hbox{\scriptscriptstyle\frown}} \over b} }_2}$$ , it is also indeterminate.

2. Standard errors of these estimates are infinitely large.
Proof: Variances of the estimates are given as follows: ${\mathop{\rm var}} \left( {{{\mathord{\buildrel{\lower3pt\hbox{\scriptscriptstyle\frown}} \over b} }_1}} \right) = \sigma _u^2\frac{{\sum {x_{2i}^2} }}{{\left( {\sum {x_{1i}^2} } \right)\left( {\sum {x_{2i}^2} } \right) - {{\left( {\sum {{x_{1i}}{x_{2i}}} } \right)}^2}}}$ ${\mathop{\rm var}} \left( {{{\mathord{\buildrel{\lower3pt\hbox{\scriptscriptstyle\frown}} \over b} }_2}} \right) = \sigma _u^2\frac{{\sum {x_{1i}^2} }}{{\left( {\sum {x_{1i}^2} } \right)\left( {\sum {x_{2i}^2} } \right) - {{\left( {\sum {{x_{1i}}{x_{2i}}} } \right)}^2}}}$ Substituting kX1 for X2 in $${\mathop{\rm var}} \left( {{{\mathord{\buildrel{\lower3pt\hbox{\scriptscriptstyle\frown}} \over b} }_1}} \right)$$: ${\mathop{\rm var}} \left( {{{\mathord{\buildrel{\lower3pt\hbox{\scriptscriptstyle\frown}} \over b} }_1}} \right) = \sigma _u^2\frac{{{k^2}\sum {x_{_{1i}}^2} }}{{{k^2}{{\left( {\sum {x_{1i}^2} } \right)}^2} - {k^2}{{\left( {\sum {x_{_{1i}}^2} } \right)}^2}}}$ ${\mathop{\rm var}} \left( {{{\mathord{\buildrel{\lower3pt\hbox{\scriptscriptstyle\frown}} \over b} }_1}} \right) = \frac{{\sigma _u^2\sum {x_{_{1i}}^2} }}{0}$ Similarly for $${\mathop{\rm var}} \left( {{{\mathord{\buildrel{\lower3pt\hbox{\scriptscriptstyle\frown}} \over b} }_2}} \right)$$: ${\mathop{\rm var}} \left( {{{\mathord{\buildrel{\lower3pt\hbox{\scriptscriptstyle\frown}} \over b} }_2}} \right) = \frac{{\sigma _u^2\sum {x_{_{1i}}^2} }}{0}$ Any positive number devided by zero is infinitely large. Therefore, variances of both estimates are infinitely large.

#### 12 (b). Assume three explanatory variables like X1, X2 and X3, which are found to be highly collinear. The following three principal components (PC) and their corresponding eigenvalues (λ) are reported as $P{C_1} = {a_{11}}{x_1} + {a_{12}}{x_2} + {a_{13}}{x_3}$ $P{C_2} = {a_{21}}{x_1} + {a_{22}}{x_2} + {a_{23}}{x_3}$ $P{C_3} = {a_{31}}{x_1} + {a_{32}}{x_2} + {a_{33}}{x_3}$ where aij be the factor loading of the ith principal component of jth factor, and λ1, λ2 and λ3 are the egenvalues of first, second and third principal components respectively. How are factor loadings (aij) related to egenvalues in each PC? Do you think that λ1 > λ2 > λ3? What is the sum of λs ? (Marks - 10)

Also see Q. No. 13 (b) General Economics - 1 Previous Year Paper Solution 2011.

Eigenvalue in each PC is the summation of square of factor loading of ith principal component. Here, the three egenvalues are as follows: ${\lambda _1} = \sum\limits_i^3 {a_{1i}^2} = a_{11}^2 + a_{12}^2 + a_{13}^2$ ${\lambda _2} = \sum\limits_i^3 {a_{2i}^2} = a_{21}^2 + a_{22}^2 + a_{23}^2$ ${\lambda _3} = \sum\limits_i^3 {a_{3i}^2} = a_{31}^2 + a_{32}^2 + a_{33}^2$

aij are a form of correlation coefficients. They are derived from a correlation table of the set of k variables. The formula is: ${a_{ij}} = \frac{{\sum\limits_j^k {{r_{{x_i}{x_j}}}} }}{{\sqrt {\sum\limits_i^k {\sum\limits_j^k {{r_{{x_i}{x_j}}}} } } }}$

By the design of the principal component method, the first principal components has the higher egeinvalue than the second; the second principal component has a higher egenvalue than the third and so on. To be precise, the values of the egenvalues become smaller and smaller for subsequent principal component.

The of the egenvalues of all principal components is eual to the number of independent variables Xs: $\sum\limits_i^k {{\lambda _i}} = k$ where k is the number of independent variables.

#### 12 (c). How do you use principal component analysis to tackle the problem of multicollinearity in regression analysis? (Marks - 5)

In the principal component analysis, the set of new variable, PCi is constructed out of theset of original variables, Xi. The set of PCi are called principal components and they are also linear combination of Xi. Xi are followed by new sets of coefficients, aij which are called factor loadings.
Factor loadings are chosen such that the constructed principal components are uncorrelated. Three such principal components has been given in the example. Now, regression of Y on these principal components can be done to find the coefficients of original regression equation. Since principal components have no multicollinearity, the coefficients are also free from the effects of multicollinearity.

#### 13 (a). [See the question paper from Google Drive Folder]

At p = 10: ${Q_1} = 50 - {p_1}$ ${Q_1} = 50 - 10$ ${Q_1} = 40$ At the point of intersection: ${Q_1} = {Q_2} = 40$ ${p_1} = {p_2} = 10$ According to the question at the point of equilibrium: $6{e_{{Q_1},{p_1}}} = {e_{{Q_2},{p_2}}}$ $6 \cdot \frac{{d{Q_1}}}{{d{p_1}}} \cdot \frac{{{p_1}}}{{{Q_1}}} = \frac{{d{Q_2}}}{{d{p_2}}} \cdot \frac{{{p_2}}}{{{Q_2}}}$ $6 \cdot \frac{d}{{d{p_1}}}\left( {50 - {p_1}} \right) \cdot \frac{{10}}{{40}} = \frac{{d{Q_2}}}{{d{p_2}}} \cdot \frac{{10}}{{40}}$ $\frac{{d{Q_2}}}{{d{p_2}}} = - 6$ Takinf anti-derivative: ${Q_2} = \int {\left( { - 6} \right)d} {p_2}$ ${Q_2} = - 6{p_2} + c$ At the point of intersection: $40 = - 6\left( {10} \right) + c$ $c = 60 + 40$ $c = 100$ Therefore, the demand function for Q2 is: ${Q_2} = 100 - 6{p_2}$

#### 13 (b). [See the question paper from Google Drive Folder]

$p = {\left( {6 - X} \right)^2}$ $p = 36 - 12X + {X^2}$ $TR = pX = {X^3} - 12{X^2} + 36X$ $MR = 3{X^2} - 24X + 36$ Equilibrium Condition for a monopolist is: $MR = MC$ $3{X^2} - 24X + 36 = 14 + X$ $3{X^2} - 25X + 22 = 0$ $3{X^2} - 3X - 22X + 22 = 0$ $3X\left( {X - 1} \right) - 22\left( {X - 1} \right) = 0$ $\left( {X - 1} \right)\left( {3X - 22} \right) = 0$ $X - 1 = 0$ $X = 1$ and $3X - 22 = 0$ $X = \frac{{22}}{3}$ Both value for X are positive so second derivative test is needed to check for the valid value of x. Here, the second derivative test is to check whether the second derive of profit function is negative or not. second derivative of profit function should be negative because at the maximum point of profit, negative second derivative means profit will decline after maximum. First derivative of profit function is: $\frac{{d\pi }}{{dX}} = MR - MC$ $\frac{{d\pi }}{{dX}} = 3{X^2} - 25X + 22$ Second derivate: $\frac{{{d^2}\pi }}{{d{X^2}}} = 6X - 25$ Value of second derivative for X = 1: $\frac{{{d^2}\pi }}{{d{X^2}}} = 6\left( 1 \right) - 25$ $\frac{{{d^2}\pi }}{{d{X^2}}} = - 19$
Value of second derivative for X = 22/3: $\frac{{{d^2}\pi }}{{d{X^2}}} = 6 \times \frac{{22}}{3} - 25$ $\frac{{{d^2}\pi }}{{d{X^2}}} = 44 - 25$ $\frac{{{d^2}\pi }}{{d{X^2}}} = 19$
Since second derivative has minus sign for X = 1, it is a valid value of X because it is the point of maximum. For X = 1, price is: $P = {\left( {6 - X} \right)^2}$ $P = {\left( {6 - 1} \right)^2}$ $P = 25$ Now, Consumer Surplus can be calculated using following formulas: $CS = \int_0^X {f\left( Q \right)dQ} - QP$ $CS = \int_0^1 {{{\left( {6 - X} \right)}^2}dQ} - 1 \times 25$ $CS = \int_0^1 {\left( {36 - 12X + {X^2}} \right)dQ} - 25$ $CS = \left[ {36X - 6{X^2} + \frac{{{X^3}}}{3}} \right]_0^1 - 25$ $CS = 36\left( 1 \right) - 6{\left( 1 \right)^2} + \frac{{{1^3}}}{3} - 25$ $CS = \frac{{16}}{3} = 5.33$

#### 13 (c). A firm is producing two goods A and B. It has two factories that jointly produce the two goods in the following quantities (per hour):

Factory 1Factory 2
Good A1020
Good B2525

Solution Video:

#### 1 comment:

1. Sir pls can you explain Question no. 4 (A) step jismein ye equation mili last ke ek step mein 90−14Q+8Q=0 ismein 8Q kaise aaya? 4Q ke upar jo square tha vo kaise chala gaya? underroot 4 ka 2 agya but what about the square exponent that was on Q???