Solutions: DSE entrance March 17, DSE Super20

Transcription

1 Solutions: entrance 2015 March 17, 2016 Super 20 March 17,

2 Super 20 Solutions : entrance 2015 Problem 1 There are two individuals, 1 and 2. Suppose, they are offered a lottery that gives Rs 160 or Rs 80 each with probability equal to 1/2. The alternative to the lottery is a fixed amount of money given to the individual. Assume that individuals are expected utility maximizers. Suppose, individual 1 will prefer to get Rs 110 with certainty over the lottery. However, Individual 2 is happy receiving a sure sum of Rs 90 rather than facing the lottery. Which of the following statements is correct? 1. both individuals are risk averse 2. 2 is risk averse but 1 loves risk 3. 1 is risk averse but 2 loves risk 4. none of the above A risk averse (risk loving) individual would have lower (higher) expected utility from a risky proposition than a risk neutral individual. The risk neutral individual will calculate his/ her expected utility as a probability-weighted average of the utilities from various outcomes. In this case, EU = = 120. Since both individuals value the risky proposition at values below 120, both are risk averse. Consider an exchange economy with agents 1 and 2 and goods x and y. The agents preferences over x and y are given. If it rains, 1 s endowment is (10, 0) and 2 s endowment is (0, 10). If it shines, 1 s endowment is Problem 2 (0, 10) and 2 s endowment is (10, 0). 1. the set of Pareto efficient allocations is independent of whether it rains or shines 2. the set of Pareto efficient allocations will depend on the weather 3. the set of Pareto efficient allocations may depend on the weather 4. whether the set of Pareto efficient allocations varies with the weather depends on the preferences of the agents The set of Pareto efficient allocations depends only on the preferences of the agents, and is independent of their endowments. The endowments only decide the reservation utilities of the agents, and hence determine the efficient allocations that both agents would willingly move to. Problem 3 Deadweight loss is a measure of 1. change in consumer welfare 2. change in producer welfare 3. change in social welfare 4. change in social inequality Deadweight loss refers to the loss in social welfare, which is the sum of consumer and producer welfare, as a result of some form of inefficiency. 2

3 Super 20 Solutions : entrance 2015 Problem 3 Problem 4 To regulate a natural monopolist with cost function C(q) = a + bq, the government has to subsidize the monopolist under 1. average cost pricing 2. marginal cost pricing 3. non-linear pricing 4. all of the above Subsidy is necessary under marginal cost pricing because marginal cost is less than average cost, leading to negative profits if price is capped at marginal cost. Problem 5 Suppose an economic agent lexicographically prefers x to y, then her indifference curves are 1. straight lines parallel to the x axis 2. straight lines parallel to the y axis 3. convex sets 4. L shaped curves Agents with lexicographic preferences don t have indifference curves, in the strictest sense of the word. The set of point(s), at all of which the agent is indifferent is a Singleton, hence Convex. This and the next question are based on the function f : R R given by for (x, y) R 2 Problem 6 f(x, y) = { 0, if (x, y) = 0 xy x 2 +y 2 if (x, y) 0 Which of the following statements is correct? 1. f is continuous and has partial derivatives at all points 2. f is discontinuous but has partial derivatives at all points 3. f is continuous but does not have partial derivatives at all points 4. f is discontinuous and does not have partial derivatives at all points Problem 6 continued on next page

4 Super 20 Solutions : entrance 2015 Problem 6 (continued) First, lets show that f is not continuous at (0,0). Consider the following 2 sequence of points. (x k, y k ) = ( 1, 0), k = 1, 2,... k (x k, y k) = ( 1 k, 1 ), k = 1, 2,... k Then as k, both sequence of points tend to (0,0) and yet in the former case, f(x k, y k ) 0 as k where as in the latter case f(x k, y k ) 1 2 as k Thus lim x f(x, y) does not exist, hence it is not continuous. To show that f x (0, 0) and f y (0, 0) exist, lets use the definitions of the partial derivatives f x (x, y) = lim h 0 f(x + h, y) f(x, y) h f y (x, y) = lim k 0 f(x, y + k) f(x, y) k Now using the above definitions, it is easy to see that f x (0, 0) = 0 and f y (0, 0) = 0. Which of the following statements is correct? Problem 7 1. f is continuous and differentiable 2. f is discontinuous and differentiable 3. f is continuous but not differentiable 4. f is discontinuous and non-differentiable Follows from the fact that f(x, y) is discontinuous at (0,0) Problem 8 Consider the following system of equations: x + 2y + 2z s + 2t = 0 x + 2y + 3z + s + t = 0 3x + 6y + 8z + s + 4t = 0 The dimension of the solution space of this system of equations is : Problem 8 continued on next page

5 Super 20 Solutions : entrance 2015 Problem 8 (continued) First represent the system of linear equations in matrix form. where A = Reducing A to row-echelon form, we obtain Ax = A = x = Upon solving the system of equations, by taking y = α, s = β, t = γ we have, x y z = 0 α + 2 β + 1 γ s t x y z s t Its easy to see that coefficient vectors of α, β and γ are linearly independent and thus form the basis for the solution space whose dimension hence is 3 Problem 9 The vectors v 0, v 1,..., v n in R m are said to be affinely independent if with scalars c 0, c 1,..., c n, n i=0 c iv i = 0 and n i=0 c i = 0 implies c i = 0 for i = 0, 1,..., n. For such an affinely independent set of vectors, which of the following is an implication: I v 0, v 1,..., v n are linearly independent II (v 1 v 0 ), (v 2 v 0 ),..., (v n v 0 ) are linearly independent. III n m 1. Only I and II are true 2. Only I and III are true 3. Only II is true 4. Only II and III are true Linear independence implies affine independence, but not vice versa. Thus I is not true. Consider the following three vectors in R 2 : u 0 = e 1, u 1 = 2e 1 and u 2 = e 1 + e 2. The three vectors are affinely independent but not linearly independent. Thus we can have n = m. We can of course have n < m. III is true II can easily be shown to be true using the definition of affine independence and linear independence 5

6 Super 20 Solutions : entrance 2015 Problem 9 Problem 10 R 2 R 2 be a linear mapping (i.e., for every pair of vectors (x 1, x 2 ), (y 1, y 2 ) and scalars c 1, c 2, F (c 1 (x 1, x 2 ) + c 2 (y 1, y 2 )) = c 1 F (x 1, x 2 ) + c 2 F (y 1, y 2 ).) Suppose F (1, 2) = (2, 3) and F (0, 1) = (1, 4). Then in general, F (x 1, x 2 ) equals : 1. (x 2, 4x 2 ) 2. (x 2, x 1 + x 2 ) 3. (1 + x 1, 4x 2 ) 4. (x 2, 5x 1 + 4x 2 ) Using the property of linear mapping F, calculate F (1, 0) which comes out to be (0,-5) and use the fact that F (x 1, x 2 ) = x 1 F (1, 0) + x 2 F (0, 1) Problem 11 A correlation coefficient of 0.2 between Savings and Income implies that: 1. A unit change in Income leads to a less than 20 percent increase in Savings 2. A unit change in Income leads to a 20 percent increase in Savings 3. A unit change in Income may cause Savings to increase by less than or more than If we plot Savings against Income, the points would lie more or less on a straight line Consider a regression specification Y i = β 1 + β 2 X i + ɛ i, where Y be Savings and X be Income. To be able to say how much Y changes with a unit change in X, one must know the value of β 2. All that s given in the question is the value of ρ and the relation between ρ and β 2 is given by ρ = β 2 S x S x and S y being variances of x and y respectively. Thus without knowing S x and S y, one cannot say β 2 Problem 12 In a simple regression model estimated using OLS, the covariance between the estimated errors and the regressors is zero by construction. This statement is: 1. True only if the regression model contains an intercept term 2. True only if the regression model does not contain an intercept term 3. True irrespective of whether the regression model contains an intercept term 4. False If there is no intercept term, the mean of the residuals is no longer zero, and the covariance need not be 0 any longer. S y 6

7 Super 20 Solutions : entrance 2015 Problem 12 Problem 13 Consider the uniform distribution over the interval [a, b]. 1. The mean of this distribution depends on the length of the interval, but the variance does not 2. The mean of this distribution does not depend on the length of the interval, but the variance does 3. Neither the mean, nor the variance, of this distribution depends on the length of the interval 4. The mean and the variance of this distribution depend on the length of the interval Mean of the uniform distribution is a+b 2 where as the variance is (b a)2 12. The length of the interval is b a. If the interval is slightly lengthened, say by ɛ on both ends, i.e. [a ɛ, b + ɛ], while the mean would remain the same, but the variance would increase. The next two questions are based on the following information Let F : R R be a (cumulative) distribution function. Define b : [0, 1] R by { 0, if c = 0 Problem 14 b(c) = If F has a jump at x, say c = F (x) > a F (x ), then 1. b has a jump at c 2. b has a jump at a 3. b is strictly increasing over (a, c) inff 1 ([c, 1]), if c (0, 1] 4. b is constant over (a, c) b( ), which is a generalised inverse of F, the CDF, represents the Quantile function. b(p) represents the minimum value of x at which the cumulative probability ( i.e. the value of CDF) is at least p. Now suppose there is a jump in the CDF at x, where c = F (x) > a F (x ), then for any value of p [a, c], inf(f 1 )( ) remains the same. Thus the graph of b( ) remains flat in that interval. Refer the figure for more clarity. 7

8 Super 20 Solutions : entrance 2015 Problem 14 Problem 15 If F is constant over (x, y) with F (z) < F (x) for every z < x, then 1. b has a jump at y 2. b has a jump at x 3. b is continuous at F (x) 4. b is decreasing over [0, F (x)] Since F and b are generalised inverses of one another, jump in one would mean a flat portion in the other. The following set of information is relevant for the next Four questions. Consider a closed economy where at any period t the actual output (Y t ) is demand-determined. Aggregate demand on the other hand has two components: consumption demand (C t ) and investment demand (I t ). Both consumption and investment demands depend on agents expectation about period t output (Y e t ) in the following way: Problem 16 C t = αyt e ; 0 < α < 1 I t = γ(yt e ) 2 ; γ > 0 Suppose agents have static expectations. Static expectation implies that 1. in every period agents expect the previous periods actual value to prevail 2. in every period agents adjust their expected value by a constant positive fraction of the expectational error made in the previous period 4. none of the above 3. in every period agents use all the information available in that period so that the expected value can dier from the actual value if and only if there is a stochastic element present By definition. (b) is Adaptive Expectations, and (c) is Rational Expectations. Problem 17 Under static expectations, starting from any given initial level of actual output Y 0 1 α γ, in the long run the actual output in this economy 1. will always go to zero 2. will always go to infinity 3. will always go to a nite positive value given by 1 α γ 4. will go to zero or infinity depending on whether Y 0 > or < 1 α γ Problem 17 continued on next page

9 Super 20 Solutions : entrance 2015 Problem 17 (continued) Under static expectations, Yt+1 e = Y t. Substituting into the given expression for C t+1 and I t+1, we get Y t+1 = αy t + γ(y t ) 2 which is convex and cuts the 45 line at 0, and then from below at Y t = 1 α γ. Below this value, Y t+1 < Y t and the output goes to zero. Above this value, the reverse inequality holds, and the output goes to infinity. Problem 18 Suppose now agents have rational expectations. Rational expectation implies that 1. in every period agents expect the previous periods actual value to prevail 2. in every period agents adjust their expected value by a constant positive fraction of the expectational error made in the previous period 3. in every period agents use all the information available in that period so that the expected value can differ from the actual value if and only if there is a stochastic element present 4. none of the above By definition. Under rational expectations, in the long run the actual output in this economy Problem will always go to zero 2. will always go to infinity 3. will always go to a finite positive value given by 1 α γ 4. will go to zero or infinity depending on agents expectations Rational expectations dictate that Y t = E t 1 Y t + µ t where µ)t is a stochastic error term. Substituting this into the expression for Y t and taking expectations again, we get E t 1 Y t (1 α γe t 1 Y t ) which means that either E t 1 Y t = 0 or E t 1 Y t = 1 α γ, of which zero is not a reasonable expectation because at that point mean of random shocks can not be zero, as negative values of Y t are not possible. Problem 20 Suppose we conduct n independent Bernoulli trials, each with probability of success p. If k is such that the probability of k successes is equal to the probability of k + 1 successes, then 1. (n + 1)p = n(1 + p) 2. np = (n 1)(1 + p) 3. np is a positive integer 4. (n + 1)p is a positive integer Problem 20 continued on next page

10 Super 20 Solutions : entrance 2015 Problem 20 (continued) The random variable representing number of successes in n independent Bernoulli trials, say X follows Binomial distribution. i.e. X B(n, p). Now equate P r[x = k] = P r[x = k + 1] to get the result. The next Two questions are based on the following. Consider a pure exchange economy with three persons, 1, 2, 3, and two goods, x and y. The utilities are given by u 1 ( ) = xy, u 2 ( ) = x 3 y and u 3 ( ) = xy 2, respectively. Problem 21 If the endowments are (2,0), (0,12) and (12,0), respectively, then 1. an equilibrium price ratio does not exist 2. p X /p Y = 1 is an equilibrium price ratio 3. p X /p Y > 1 is an equilibrium price ratio 4. p X /p Y < 1 is an equilibrium price ratio Let p x = p, p y = 1. Maximizing u 1 ( ) = x 1 y 1, u 2 ( ) = x 3 2y 2 and u 3 ( ) = x 3 y3 2 subject to px 1 + y 1 = 2p, px 2 + y 2 = 12, and px 3 + y 3 = 12p, we get x 1 = 1, y 1 = p, x 2 = 9/p, y 2 = 3, x 3 = 4, and y 3 = 8p. Using the conditions that x 1 + x 2 + x 3 = 14 and y 1 + y 2 + y 3 = 12, we get p = 1. Problem 22 If the endowments are (0,2), (12,0) and (0,12), respectively, then 1. an equilibrium price ratio does not exist 2. equilibrium price ratio is the same as in the above question 3. p X /p Y < 1 is an equilibrium price ratio 4. p X /p Y > 1 is an equilibrium price ratio Proceeding as above, we see that for markets in both x andy to clear, we have p = 5/3. Thus p X /p Y > 1 The next Two questions are based on the following information. A city has a single electricity supplier. Electricity production cost is Rs. c per unit. There are two types of customers. Utility function for type i is given by u i (q, t) = θ i ln(1 + q) t, where q is electricity consumption and t is electricity tariff. High type customers are more energy efficient, that is, θ H > θ L ; moreover θ L > c. Problem 23 Suppose the supplier can observe type of the consumer, i.e., whether θ = θ H or θ = θ L. If the supplier decides to sell package (q H, t H ) to those for whom θ = θ H and (q L, t L ) to those for whom θ = θ L, then profit maximizing tariffs will be 1. t H = c ln ( θ Hc ) and tl = c ln ( θ Lc ) 2. t H = θ H ln ( θ Hc ) and tl = 0 Problem 23 continued on next page

11 Super 20 Solutions : entrance 2015 Problem 23 (continued) 3. t H = θ H ln ( θ Hc ) and tl = θ L ln ( θ Lc ) 4. t H = t L = c ln ( θ H +θ L c ) For a perfectly discriminating monopolist with constant marginal costs, the optimization problem can be separated into two separate optimization problems, one for each kind of buyer. Further, he can extract all surplus from the buyers of either type, and hence t i = θ i ln(1 + q i ) = q i = e ti/θi 1. Then, profit from buyer of type i, π i = t i cq i = t i ce ti/θi 1 is maximized at t i = θ i ln ( θ i c ). Problem 24 Now, assume that the supplier cannot observe type of the consumer. Suppose, he puts on offer both of the packages that he would offer in the above question. If consumers are free to choose any of the offered packages, then 1. Both types will earn zero utility 2. Only low type can earn positive utility 3. Only high type can earn positive utility 4. Both types can earn positive utility The high-type buyer can earn a positive utility by opting for the low-type package, and the low-type buyer will receive a negative utility if he opts for the high-type package. Choosing your own package naturally results in zero utility. Problem 25 Suppose buyers of ice-cream are uniformly distributed on the interval [0, 1]. Ice-cream sellers 1 and 2 simultaneously locate on the interval, each locating so to maximize her market share given the location of the rival. Each seller s market share corresponds to the proportion of buyers who are located closer to her location than to the rival s location. 1. Both will locate at One will locate at 1 4 and the other at One will locate at 0 and the other at One will locate at 1 3 and the other at 2 3. In any of the other cases, though the two players have equal market share, either can increase her own share by moving further towards the other - the area she leaves behind is still closer to her, whereas she is now reducing the common area in between that is being split between them - and thus, both will converge to

12 Super 20 Solutions : entrance 2015 Problem 25 Problem 26 In the context of previous question, suppose it is understood by all players that seller 3 will locate on [0, 1] after observing the simultaneous location choices of sellers 1 and 2. Seller 3 aims to maximize market share given the locations of 1 and 2. The locations of sellers 1 and 2 are as follows: 1. Both will locate at One will locate at 1 4 and the other at One will locate at 0 and the other at One will locate at 1 3 and the other at 2 3. In any of the other cases, the last Player will locate just next to one of the first two players, and will get a share of 1 2 (Options 1 or 3), or 1 3 (Option 4), leaving the first two players with expected shares of 1 4 or 1 3. On the other hand, Option 2 forces Player 3 to settle for a share of 1 4. giving players 1 and 2 expected shares of 3 8. Problem 27 Consider a government and two citizens. The government has to decide whether to create a public good, say a park, at cost Rs 100. The value of the park is Rs 30 to the citizen 1 and Rs 60 to citizen 2; each valuation is private information for the relevant citizen and not known to the government. The government asks the citizens to report their valuations, say r 1 and r 2. It cannot verify the truthfulness of the reports. It decides to build the park if r 1 + r 2 100, in which case, citizen 1 will pay the tax 100 r 2 and citizen 2 will pay the tax 100 r 1. If the park is not built, then no taxes are imposed. The reported valuations will be 1. r 1 < 30 and r 2 > r 1 > 30 and r 2 < r 1 = 60 and r 2 = r 1 = 30 and r 2 = 60 This is an example of a truth-telling equilibrium. Note that r 1 + r 2 = = 90 < 100, and the bridge will not be built. Reporting below their valuation will not change anything (the bridge will still not be built), and while they can report higher values and get the bridge built (say if citizen 1 reports 40 rather than 30), but this will result in negative utilities for the lying citizen, and is, hence, not preferred. Problem 28 In the context of the previous question, suppose the only change is that citizen 1 s valuation rises to 50 and the same procedure is followed, then 1. The park will be built and result in a government budget surplus of Rs The park will be built and result in a government budget deficit of Rs The park will be built and result in a government balanced budget. 4. The park will not be built. Problem 28 continued on next page

13 Super 20 Solutions : entrance 2015 Problem 28 (continued) Under the given scheme, by telling the truth, the bridge gets built, and the citizens both get a utility level of 50 (100 60) = 60 (100 50) = 10. If they report a valuation that is too low, the bridge might not be built (0 utility, which means they are worse off). If they report something higher, their utility remains unchanged. As a result, truth-telling is a weakly dominant strategy. The total revenue is (100 60) + (100 50) = 90, which means the government runs a budget deficit of = 10. Problem 29 Consider the following two games in which player 1 chooses a row and player 2 chooses a column. Analysis of these games shows 1. Having an extra option cannot hurt. 2. Having an extra option cannot hurt as long as it dominates other options. 3. Having an extra option can hurt if the other player is irrational. 4. Having an extra option can hurt if the other player is rational. In the absence of the extra option, the equilibrium payoffs are (0,6). When the new option is available, the equilibrium payoffs are (3,3). Player 2 is hurt by having a new option available. In strategic interactions, rational players can force you into less desirable positions by posing credible threats. This is more likely to happen when you have more options available. Problem 30 Consider an exchange economy with agents 1 and 2 and goods x and y. Agent 1 lexicographically prefers x to y. Agent 2 s utility function is min{x, y}. Agent 1 s endowment is (0, 10) and agent 2 s endowment is (10, 0). The competitive equilibrium price ratio, p x /p y, for this economy 1. can be any positive number 2. is greater than 1 3. is less than 1 4. does not exist Let p x = p, p y = 1. Since agent 1 has lexicographic preferences, he is willing to give up good y at any price to get more of good x. Thus, in equilibrium, he will want to spend his entire allocation on good x. Thus, the demand of agent 1 for good x will be 10 p. Since agent 2 s utility function is a min function, he will want to consume equal quantities, say q of each good. Then, his budget constraint says that 10p = (p + 1)q. In other words, he demands q = 10p p+1 of each good. For the market in good x to clear, 10p p p = 10 = 1 p + p p+1 = 1, which does not have a solution. Thus, no competitive equilibrium allocation is possible. 13

14 Super 20 Solutions : entrance 2015 Problem 30 Problem 31 Consider a strictly increasing, differentiable function u : R 2 R and the equations: D 1 u(x 1, x 2 ) D 2 u(x 1, x 2 ) = p 1 p 2 and p 1 x 1 + p 2 x 2 = w where p 1, p 2, w are strictly positive. What additional assumptions will guarantee the existence of continuously differentiable functions x 1 (p 1, p 2, w) and x 2 (p 1, p 2, w) that will solve these equations for all strictly positive p 1, p 2, w? 1. u is injective 2. u is bijective 3. u is twice continuously differentiable 4. u is twice continuously differentiable and [ D11 u(x 1, x 2, p 2 ) D 12 u(x 1, x 2, p 2 ) D 12 u(x 1, x 2, p 2 ) D 22 u(x 1, x 2, p 2 ) ] is nonsingular p 1 p 2 The given set of equations are the ones encountered in a standard constrained utility maximization exercise. The first expression equates the ratio of the two products marginal rates of substitution to the price ratio, whereas the second equation gives the budget constraint. The Implicit Function Theorem ensures continuously differentiable Marshallian Demand Functions if the Jacobian Matrix of the given system of equations exists - i.e. the given expressions are differentiable, and hence u is twice continuously differentiable - and is nonsingular. Problem 32 As n, the sequence ( 1) n (1 + n 1 ) 1. converges to 1 2. converges to 1 3. converges to both 1 and 1 4. does not converge The given sequence is a sum of two component sequences, u n = ( 1) n and v n = ( 1)n n. Its easy to see that v n converges to 0 where as u n takes values +1 and 1 ad infinitum. Thus the given sequence doesn t converge 14

15 Super 20 Solutions : entrance 2015 Problem 32 Problem 33 The set (0, 1) can be expressed as 1. the union of a finite family of closed intervals 2. the intersection of a finite family of closed intervals 3. the union of an infinite family of closed intervals 4. the intersection of an infinite family of closed intervals Consider I n = [ 1 n, 1 1 n ]. Then the countable union of these closed intervals, represented by is the open interval (0, 1) n=1 [ 1 n, 1 1 ] n Problem the union of a finite family of open intervals The set [0, 1] can be expressed as 2. the intersection of a finite family of open intervals 3. the union of an infinite family of open intervals 4. the intersection of an infinite family of open intervals Consider I n = ( 1 n, n ). Then the countable intersection of these open intervals, represented by is the closed interval [0, 1] n=1 ( 1 n, ) n The following information is used in the next Two questions. Consider a linear transformation P : R n R n. Let R(P ) = {P x x R n } and N (P ) = {x R n P x = 0}. P is said to be a projector if a. every x R n can be uniquely written as x = y + z for some y R(P ) and z N (P ), and b. P (y + z) = y for all y R(P ) and z N (P ). Problem 35 If P is a projector, then 1. P 2 = I, where I is the identity mapping 2. P = P 1 3. P 2 = P Problem 35 continued on next page

16 Super 20 Solutions : entrance 2015 Problem 35 (continued) 4. Both 1 and 2 Given P is a projector transformation. Take any x R n. P 2 (x) = P (P (y + z)) = P (y) = y + 0 as y R(P ) and z, 0 N (P ) Problem 36 If P is a projector and Q : R m R n is a linear transformation such that R(P ) = R(Q), then 1. QP = P 2. P Q = Q 3. QP = I 4. P Q = I Take any x R m. Let Q(x) = y R(Q) = R(P ). Thus y R(P ) P (Q(x)) = P (y) = y = Q(x) Suppose that a and b are two consecutive roots of a polynomial function f, with a < b. Suppose a and b are non-repeated roots. Consequently, f(x) = (x a)(x b)g(x) for some polynomial function g. Consider the statements: Problem 37 I g(a) and g(b) have opposite signs. II f (x) = 0 for some x (a, b) Of these statements, 1. Both I and II are true. 2. Only I is true. 3. Only II is true. 4. Both I and II are false. f is continuous and differentiable, being a polynomial function. II follows from Mean value theorem. I is not true, because if g(a) and g(b) have different signs, then, by the Intermediate Value Theorem, g(x) = 0for somex (a, b) = f(x) = 0for somex (a, b), which is not possible since a and b are consecutive roots of f(x). For a counter example take, f(x) = (x 1)(x 2)(x 3), g(x) = x 3 g(1), g(2) < 0 Problem 38 Suppose f : [0, 1] R is a twice differentiable function that satisfies D 2 f(x)+df(x) = 1 for every x (0, 1) and f(0) = 0 = f(1). Then, 1. f does not attain positive values over (0, 1) 2. f does not attain negative values over (0,1) Problem 38 continued on next page

17 Super 20 Solutions : entrance 2015 Problem 38 (continued) 3. f attains positive and negative values over (0, 1) 4. f is constant over (0, 1) Convert the second order differential equation into first order by substituting Df(x) = z(x) to get Dz+z = 1 which solve to z(x) = 1 ke x, k being the constant of integration. Substitute back for z(x) and solve again to get f(x) = x ke x + c, c is the constant of integration for second step. Now using the given fact f(0) = 0, we get k = c and f(1) = 0 we get c = 1 e 1. Therefore f(x) = x ex 1 e 1 which is non negative in the interval [0,1]. Problem 39 Suppose x 1,..., x n are positive and λ 1,..., λ n are non-negative with n i=0 λ i = 1. Then n 1. i=0 λ ix i x λ1 1 xλn n 2. n i=0 λ ix i < x λ1 1 xλn n 3. n i=0 λ ix i x λ1 1 xλn n 4. None of the above is necessarily true. n-extension of the property, Arithmetic Mean Geometric Mean Problem 40 Let N = {1, 2, 3,...}. Suppose there is a bijection, i.e., a one-to-one correspondence (an into and onto mapping), between N and a set X. Suppose there is also a bijection between N and a set Y. Then, 1. there is a bijection between N and X Y 2. there is a bijection between N and X Y 3. there is no bijection between N and X Y 4. there is no bijection between N and X Y The bijection between N and set X is essentially a sequence < x n > and so is the case with bijection between N and set Y, say < y n >. Now we can define a new bijection from N to X Y in the following manner, map all odd naturals to elements in set X and even naturals to elements in set Y. Thus we have a bijection between N and X Y. The next Three questions pertain to the following: A simple linear regression of wages on gender, run on a sample of 200 individuals, 150 of whom are men, yields the following W i = 300 (20) 50 (10) D i + u i where W i is the wage in Rs per day of the i th individual, D i = 1 if individual i is male, and 0 otherwise, u i is a classical error term, and the figures in parentheses are standard errors. 17

18 Super 20 Solutions : entrance 2015 Problem 40 Problem 41 What is the average wage in the sample? 1. Rs. 250 per day 2. Rs. 275 per day 3. Rs per day 4. Rs per day 150 men receive, on average, Rs. 250 each, and 50 women receive, on average, Rs. 300 each, for an average of Rs Problem 42 The most precise estimate of the difference in wages between men and women would have been obtained if, among these 200 individuals, 1. There were an equal number (100) of men and women in the sample 2. The ratio of the number of men and women in the sample was the same as the ratio of their average wages 3. There were at least 30 men and 30 women; this is sufficient for estimation: precision does not depend on the distribution of the sample across men and women 4. None of the above Precision depends on sample size, and not on sample composition. Problem 43 The explained (regression) sum of squares in this case is: This cannot be calculated from the information given ESS = = Problem 44 A researcher estimate the following two models using OLS Model A: y i = β 0 + β 1 S i + β 2 A i + ɛ i Model B: y i = β 0 + β 1 S i + ɛ i Problem 44 continued on next page

19 Super 20 Solutions : entrance 2015 Problem 44 (continued) where y i refers to the marks (out of 100) that a student i gets on an exam, S i refers to the number of hours spent studying for the exam by the student, and A i is an index of innate ability (varying continuously from a low ability score of 1 to a high ability score of 10). ɛ i is the usual classical error term. The estimated β 1 coefficient is 7.1 for Model A, but 2.1 for Model B; both are statistically significant. The estimated β 2 coefficient is 1.9 and is also significantly different from zero. This suggests that: 1. Students with lower ability also spend fewer hours studying 2. Students with lower ability spend more time studying 3. There is no way that students of even high ability can get more than 40 marks 4. None of the above The fact that the coefficient on S i decreases on introducing A i indicates some amount of collinearity between the two. Let A i = α 0 +α 1 S i +u i. Then the first model is y i = (β 0 +β 2 α 0 )+(β 1 +α 1 β 2 )S i +(β 2 u+ɛ). Comparing to Model B, we see that α 1 must be negative. Problem 45 An analyst estimates the model Y = β 0 + β 1 X 1 + β 2 X 2 + β 3 X 3 + u using OLS. But the true β 3 = 0. In this case, by including X 3 1. there is no harm done as all the estimates would be unbiased and efficient 2. there is a problem because all the estimates would be biased and inconsistent 3. the estimates would be unbiased but would have larger standard errors 4. the estimates may be biased but they would still be efficient Addition of unwanted variables does not lead to bias. In this case, the estimation will give the estimate b 3 = 0. However, non-usage of the information (that β 3 = 0) leads to inefficiency (greater standard errors). Problem 46 Let ˆβ be the OLS estimator of the slope coefficient in a regression of Y on X 1. Let β be the OLS estimator of the coefficient on X 1 on a regression of Y on X 1 and X 2. Which of the following is true: 1. V ar( ˆβ) < V ar( β) 2. V ar( ˆβ) > V ar( β) 3. V ar( ˆβ) < or > V ar( β) 4. V ar( ˆβ) = V ar( β) Either of the variances could be larger, depending on which model is correct. The variance will be lower for the correctly specified model. 19

20 Super 20 Solutions : entrance 2015 Problem 46 Problem 47 You estimate the multiple regression Y = a + b 1 X 1 + b 2 X 2 + u with a large sample. Let t1 be the test statistic for testing the null hypothesis b 1 = 0 and t2 be the test statistic for testing the null hypothesis b 2 = 0. Suppose you test the joint null hypothesis that b 1 = b 2 = 0 using the principle reject the null if either t1 or t2 exceeds 1.96 in absolute value, taking t1 and t2 to be independently distributed. 1. The probability of error Type 1 is 5 percent in this case 2. The probability of error Type 1 is less than 5 percent in this case 3. The probability of error Type 1 is more than 5 percent in this case 4. The probability of error Type 1 is either 5 percent or less than 5 percent in this case Each of t1 and t2 has a 5% chance to be greater than 1.96 in absolute value even if b 1 = b 2 = 0. Thus, the Type I error, which can occur as a result of either of these errors, has a chance of cropping up with a probability of = = 9.75%. Problem 48 Four taste testers are asked to independently rank three different brands of chocolate (A, B, C). The chocolate each tester likes best is given the rank 1, the next 2 and then 3. After this, the assigned ranks for each of the chocolates are summed across the testers. Assume that the testers cannot really discriminate between the chocolates, so that each is assigning her ranks at random. The probability that chocolate A receives a total score of 4 is given by: Problem The only way a total score of 4 is obtained when each tester ranks all three brands 1, the probability of which is 1 3 independently, thus the answer 1 81 Suppose 0.1 percent of all people in a town have tuberculosis (TB). A TB test is available but it is not completely accurate. If a person has TB, the test will indicate it with probability If the person does not have TB, the test will erroneously indicate that s/he does with probability For a randomly selected individual, the test shows that s/he has TB. What is the probability that this person actually has TB? Problem 49 continued on next page

21 Super 20 Solutions : entrance 2015 Problem 49 (continued) Lets define two random variables, X, Y, former to represent the outcome of the TB test {P os, Neg} and latter to represent the proportion of the TB patients in the population. P r[y = T ] = 0.001, P r[x = P os Y = T ] = 0.999, P r[x = P os Y = N] = Apply Bayes rule P r[y = T X = P os] = Substitute and the get the answer 1 3. P r[y = T ] P r[x = P os Y = T ]P r[y = T ] + P r[x = P os Y = N]P [Y = N] Problem 50 There exists a random variable X with mean µ x and variance σ 2 x for which P [µ x 2σ x X µ x + 2σ x ] = 0.6. This statement is: 1. True for any distribution for appropriate choices of µ x and σ 2 x. 2. True only for the uniform distribution defined over an appropriate interval 3. True only for the normal distribution for appropriate choices of µ x and σx False According to Chebyshev Inequality, Problem 51 P r[µ x 2σ x X µ x + 2σ x ] = P r[ X µ x 2σ x ] 0.75 Consider a sample size of 2 drawn without replacement from an urn containing three balls numbered 1, 2, and 3. Let X be the smaller of the two numbers drawn and Y the larger. The covariance between X and Y is given by: X can be 1 with probability 2/3 and 2 with probability 1/3. Y can be 2 with probability 1/3 and 3 with probability 2/3. XY can be 2, 3, or 6 with equal probability. Then, cov(xy ) = E(XY ) E(X)E(Y ) = = 1 9. Problem 52 Consider the square with vertices (0,0), (0,2), (2,0) and (2,2). Five points are independently and randomly chosen from the square. If a point (x,y) satisfies x + 2y 2, then a pair of dice are rolled. Otherwise, a single die is rolled. Let N be the total number of dice rolled. For 5 n 10, the probability that N = n is 1. ( 5 n 5 ) (1/2) n 5 (1/2) 5 (n 5) Problem 52 continued on next page

22 Super 20 Solutions : entrance 2015 Problem 52 (continued) ( 10 n 10 ( 5 n 5 ( 10 n 10 ) (1/4) n 10 (3/4) n ) (1/4) n 5 (3/4) 10 n ) (1/2) n 10 (1/2) n Probability with which 2 dice are rolled is 1 4 which is equal to the probability that point (x, y) satisfies x + 2y 2, given that (x, y) is a point in the square defined by 4 vertices in the question. Consequently probability of a single die being rolled is 3 4. N be the random variable representing the total number of dice that are rolled. One can see that, N is atleast 5, irrespective of where (x, y) lies. Further, N = 5 + Σ 5 i=0x i where X i are Bernoulli random variables, which are defined to be 1 if 2 dice are rolled and 0 if 1 die is rolled. Now N is just distributed binomially, because N 5 is binomially distributed. Thus Problem 53 ( 5 n 5 P r[n = n] = P r[n 5 = n 5] = ) (1/4) n 5 (3/4) 10 n Suppose S is a set with n 1 elements and A 1,..., A m are subsets of S with the following property: if x, y S and x y, then there exists i 1,..., m such that, either x A i and y / A i, or y A i and x / A i. Then the following necessarily holds. 1. n = 2 m 4. None of the above 2. n 2 m 3. n > 2 m Given there are m subsets A i, there are 2 m possible sets of subsets. Call the set of these 2 m sets S. Given any i S, exactly one element A S represents the list of all subsets that contain i. If n > 2 m, then j S; j i such that A also represents the list of all subsets that contain j. This is due to the Pigeonhole Principle in logic, which says that if you have n pigeonholes and n + 1 pigeons, at least one pigeonhole has more than one pigeon. The following set of information is relevant for the next Six questions. Consider the following version of the Solow growth model where the aggregate output at time t depends on the aggregate capital stock (K t ) and aggregate labour force (L t ) in the following way: Y t = (K t ) α (L t ) (1 α) ; 0 < α < 1. At every point of time there is full employment of both the factors and each factor is paid its marginal product. Total output is distributed equally to all the households in the form of wage earnings and interest earnings. Households propensity to save from the two types of earnings differ. In particular, they save s w proportion of their wage earnings and s r proportion of their interest earnings in every period. All savings are invested which augments the capital stock over time (dk/dt). There is no depreciation of capital. The aggregate labour force grows at a constant rate n. 22

23 Super 20 Solutions : entrance 2015 Problem 53 Problem 54 Let s w = 0 and s r = 1. An increase in the parameter value α 1. unambiguously increases the long run steady state value of the capital-labour ratio 2. unambiguously decreases the long run steady state value of the capital-labour ratio 3. increases the long run steady state value of the capital-labour ratio if α > n 4. leaves the long run steady state value of the capital-labour ratio unchanged Given the aggregate output equation we get wages w = (1 α)kt α and interest rate r = αkt α 1 where k t is capital labour ratio at time t. Savings are given by S t = s w wl t +s r rk t. Capital augmentation equation is K t+1 = K t + S t. Dividing by L t and using population growth equation L t+1 = L t (1 + n), we get dynamic equation for capital labour ratio. k t+1 (1 + n) = k t + kt α [(1 α)s w + αs r ]. In steady state k t+1 = k t = k. Putting s w = 0 and s r = 1 then differentiating k with respect to α we get positive expression. Problem 55 Let s r = 0 and 0 < s w < 1. An increase in the parameter value n 1. unambiguously increases the long run steady state value of the capital-labour ratio 2. unambiguously decreases the long run steady state value of the capital-labour ratio 3. increases the long run steady state value of the capital-labour ratio if α > n 4. leaves the long run steady state value of the capital-labour ratio unchanged Putting 0 < s w < 1 and s r = 0 then differentiating k with respect to n we get a negative expression. Problem 56 Now let both s w and s r be positive fractions such that s w < s r. In the long run, the capital-labour ratio in this economy 1. approaches zero 2. approaches infinity 3. approaches a constant value given by 4. approaches a constant value given by ( ) 1 (1 α)sw+αs 1 α r n ( ) 1 αsw+(1 α)s α r n Putting steady state condition in the dynamic capital labour ratio equation as explained in question 54, we get two roots for k. The first root 0 is unstable but the second root as stated in option 3 is stable. 23

24 Super 20 Solutions : entrance 2015 Problem 56 Problem 57 Suppose now the government imposes a proportional tax on wage earnings at the rate τ and redistributes the tax revenue in the form of transfers to the capital-owners. People still save s w proportion of their net (posttax) wage earnings and s r proportion of their net (post-transfer) interest earnings. In the new equilibrium, an increase in the tax rate τ 1. unambiguously increases the long run steady state value of the capital-labour ratio 2. unambiguously decreases the long run steady state value of the capital-labour ratio 3. increases the long run steady state value of the capital-labour ratio if α > n 4. leaves the long run steady state value of the capital-labour ratio unchanged Taxing those with lower saving propensity and giving it to people with hgher saving propensity would increase the aggregate savings in the economy. And as we know steady state level of capital per capita increases with aggregate savings. Problem 58 Let us now go back to case where both s w and s r are positive fractions such that s w < s r but without the tax-transfer scheme. However, now let the growth rate of labour force be endogenous such that it depends on the economys capital-labour ratio in the following way: where k ( ) 1 > (1 α)sw+αs 2 α r A { 1 dl L dt = Akt for k t < k; 0 for k t >> k is a given constant. In the long run, the capital-labour ratio in this economy 1. approaches zero 2. approaches infinity 3. approaches a constant value given by ( ) 1 αsw+(1 α)s α r A 4. approaches infinity or a constant value given by the initial k 0 > or < k ( ) 1 (1 α)sw+αs 2 α r A depending on whether If k 0 >> k, then Labour growth rate is zero. using it in dynamic equation for capital labour ratio with steady state condition, we get only trivial solution at k = 0. and the curve always remains above 45 degree line. Capital stock can not begin from zero so in this case capital labour ratio becomes unbounded. In the other case k itself is one steady state. Given the dynamic equation is concave, it is one stable equilibrium. Problem 59 In the above question, an increase in the parameter value A 1. unambiguously increases the long run steady state value of the capital-labour ratio Problem 59 continued on next page

25 Super 20 Solutions : entrance 2015 Problem 59 (continued) 2. unambiguously decreases the long run steady state value of the capital-labour ratio 3. increases the long run steady state value of the capital-labour ratio if k 0 < k 4. leaves the long run steady state value of the capital-labour ratio unchanged if k 0 < k, then only steady state is k = 0, which is independent of A, hence k can not unambiguously depend upon A. However, if k 0 > k, then steady state depends upon A positively. Problem 60 A profit maximizing firm owns two production plants with cost functions c 1 (q) = q2 2 and c 2 (q) = q 2, respectively. The firm is free to use either just one or both of the plants to achieve any given level of output. For this firm, the marginal cost curve 1. lies above the 45 degree line through the origin, for all positive output levels 2. lies below the 45 degree line through the origin, for all positive output levels 3. is the 45 degree line through the origin 4. none of the above c(q, q ) = (q ) (q q ) 2 is minimized at q = 2(q q ) = q = 2q 3. Then, c(q) = 2 9 q2 + q2 9 = q2 c (q) = 2 3q, which lies below the 45 degree line for all q > 0. dsesuper20@gmail.com = 25