Download Solutions: DSE entrance 2015

Solutions: DSE entrance 2015
March 17, 2016
DSE
Super20
DSE Super 20
March 17, 2016
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DSE Super 20
Solutions : DSE 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 = 12 ·160+ 21 ·80 = 120.
Since both individuals value the risky proposition at values below 120, both are risk averse.
Problem 2
DSE
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
(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
Super20
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.
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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
DSE
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
(
0,
if (x, y) = 0
f (x, y) =
xy
if
(x, y) 6= 0
x2 +y 2
for (x, y) ∈ R2
Super20
Problem 6
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. . .
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Problem 6 (continued)
First, lets show that f is not continuous at (0,0). Consider the following 2 sequence of points.
1
(xk , yk ) = ( , 0), k = 1, 2, ...
k
1 1
(x0k , yk0 ) = ( , ), k = 1, 2, ...
k k
Then as k → ∞ , both sequence of points tend to (0,0) and yet in the former case, f (xk , yk ) → 0 as k → ∞
where as in the latter case f (x0k , yk0 ) → 21 as k → ∞ Thus limx→∞ f (x, y) does not exist, hence it is not
continuous.
To show that fx (0, 0) and fy (0, 0) exist, lets use the definitions of the partial derivatives
f (x + h, y) − f (x, y)
h→0
h
fx (x, y) = lim
fy (x, y) = lim
k→0
f (x, y + k) − f (x, y)
k
Now using the above definitions, it is easy to see that fx (0, 0) = 0 and fy (0, 0) = 0.
Problem 7
DSE
Which of the following statements is correct?
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)
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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 :
1. 1
2. 2
3. 3
4. 4
Problem 8 continued on next page. . .
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Problem 8 (continued)
First represent the system of linear equations in matrix form.
Ax = 0
where

1 2 2 −1
A= 1 2 3 1
3 6 8 1



2


1  x=


4
x
y
z
s
t







Reducing A to row-echelon form, we obtain

1
A= 0
0
2
0
0
2
1
0
−1
2
0

2
−1 
0
Upon solving the system of equations, by taking y = α, s = β, t = γ we have,
 






−2
5
−4
x
 0 
 0 
 y   1 
 






 






 z  =  0  α +  −2  β +  1  γ
 






 1 
 0 
 s   0 
0
0
1
t
DSE
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
Pn
The vectors v0 , v1 , ..., vn in Rm are said to be affinely independent if with scalars c0 , c1 , ..., cn , i=0 ci vi = 0
Pn
and i=0 ci = 0 implies ci = 0 for i = 0, 1, ..., n. For such an affinely independent set of vectors, which of
the following is an implication:
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I v0 , v1 , ..., vn are linearly independent
II (v1 − v0 ), (v2 − v0 ), ..., (vn − v0 ) 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 R2 : u0 = e1 , u1 = 2e1 and u2 = e1 + e2 . 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
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Problem 9
Problem 10
R2 → R2 be a linear mapping (i.e., for every pair of vectors (x1 , x2 ), (y1 , y2 ) and scalars c1 , c2 , F (c1 (x1 , x2 ) +
c2 (y1 , y2 )) = c1 F (x1 , x2 ) + c2 F (y1 , y2 ).) Suppose F (1, 2) = (2, 3) and F (0, 1) = (1, 4). Then in general,
F (x1 , x2 ) equals :
1. (x2 , 4x2 )
2. (x2 , x1 + x2 )
3. (1 + x1 , 4x2 )
4. (x2 , −5x1 + 4x2 )
Using the property of linear mapping F , calculate F (1, 0) which comes out to be (0,-5) and use the fact
that F (x1 , x2 ) = x1 F (1, 0) + x2 F (0, 1)
Problem 11
A correlation coefficient of 0.2 between Savings and Income implies that:
DSE
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 20
4. If we plot Savings against Income, the points would lie more or less on a straight line
Consider a regression specification Yi = β1 + β2 Xi + 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 β2 . All that’s given in
the question is the value of ρ and the relation between ρ and β2 is given by
Sx
Sy
Super20
ρ = β2
Sx and Sy being variances of x and y respectively. Thus without knowing Sx and Sy , one cannot say β2
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.
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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
2
(b−a)
Mean of the uniform distribution is a+b
2 where as the variance is
12 . The length of the interval is b − a.
If the interval is slightly lengthened, say by on both ends, i.e. [a − , b + ], 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
b(c) =
−1
infF ([c, 1]),
if c ∈ (0, 1]
Problem 14
DSE
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)
4. b is constant over (a, c)
Super20
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.
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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 (Yt ) is demand-determined. Aggregate demand on the other hand has
two components: consumption demand (Ct ) and investment demand (It ). Both consumption and investment
demands depend on agents expectation about period t output (Yte ) in the following way:
Ct = αYte ; 0 < α < 1
It = γ(Yte )2 ; γ > 0
———————
Problem 16
DSE
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
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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
4. none of the above
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 6=
run the actual output in this economy
1−α
γ ,
in the long
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 Y0 > or <
Problem 17 continued on next page. . .
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1−α
γ
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Problem 17 (continued)
e
Under static expectations, Yt+1
= Yt . Substituting into the given expression for Ct+1 and It+1 , we get
2
Yt+1 = αYt + γ(Yt ) which is convex and cuts the 45◦ line at 0, and then from below at Yt = 1−α
γ . Below
this value, Yt+1 < Yt 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.
Problem 19
DSE
Under rational expectations, 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 finite positive value given by
1−α
γ
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4. will go to zero or infinity depending on agents expectations
Rational expectations dictate that Yt = Et−1 Yt + µt where µ)t is a stochastic error term. Substituting this
into the expression for Yt and taking expectations again, we get Et−1 Yt (1 − α − γEt−1 Yt ) which means
that either Et−1 Yt = 0 or Et−1 Yt = 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 Yt 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. . .
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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 u1 (·) = xy, u2 (·) = x3 y and u3 (·) = 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. pX /pY = 1 is an equilibrium price ratio
3. pX /pY > 1 is an equilibrium price ratio
4. pX /pY < 1 is an equilibrium price ratio
Let px = p, py = 1. Maximizing u1 (·) = x1 y1 , u2 (·) = x32 y2 and u3 (·) = x3 y32 subject to px1 + y1 = 2p,
px2 + y2 = 12, and px3 + y3 = 12p, we get x1 = 1, y1 = p, x2 = 9/p, y2 = 3, x3 = 4, and y3 = 8p. Using
the conditions that x1 + x2 + x3 = 14 and y1 + y2 + y3 = 12, we get p = 1.
Problem 22
DSE
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
Super20
3. pX /pY < 1 is an equilibrium price ratio
4. pX /pY > 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 pX /pY > 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 ui (q, t) = θi ln(1 + q) − t, where q is electricity consumption and t is electricity tariff.
High type customers are more energy efficient, that is, θH > θL ; moreover θL > c.
Problem 23
Suppose the supplier can observe type of the consumer, i.e., whether θ = θH or θ = θL . If the supplier
decides to sell package (qH , tH ) to those for whom θ = θH and (qL , tL ) to those for whom θ = θL , then profit
maximizing tariffs will be
1. tH = c ln θcH and tL = c ln θcL
2. tH = θH ln θcH and tL = 0
Problem 23 continued on next page. . .
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3. tH = θH ln
4. tH = tL = c ln
θH
c
θL
c
and tL = θL ln
θH +θL
Problem 23 (continued)
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 ti = θi ln(1 + qi ) =⇒ qi = eti /θi − 1. Then, profit from
buyer of type i, πi = ti − cqi = ti − ceti /θi − 1 is maximized at ti = θi ln θci .
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
DSE
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.
Super20
1. Both will locate at
2. One will locate at
1
4
1
2.
and the other at 43 .
3. One will locate at 0 and the other at 1.
4. One will locate at
1
3
and the other at 32 .
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 .
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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 12 .
2. One will locate at
1
4
and the other at
3
4.
3. One will locate at 0 and the other at 1.
4. One will locate at
1
3
and the other at 32 .
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 12 (Options 1 or 3), or 13 (Option 4), leaving the first two players with expected shares of 14 or
1
1
3 . On the other hand, Option 2 forces Player 3 to settle for a share of 4 . giving players 1 and 2 expected
3
shares of 8 .
Problem 27
DSE
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 r1 and r2 . It cannot verify the truthfulness of the reports. It decides
to build the park if r1 + r2 ≥ 100, in which case, citizen 1 will pay the tax 100 − r2 and citizen 2 will pay
the tax 100 − r1 . If the park is not built, then no taxes are imposed. The reported valuations will be
1. r1 < 30 and r2 > 60
2. r1 > 30 and r2 < 60
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3. r1 = 60 and r2 = 30
4. r1 = 30 and r2 = 60
This is an example of a truth-telling equilibrium. Note that r1 + r2 = 30 + 60 = 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 10.
2. The park will be built and result in a government budget deficit of Rs 10.
3. 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. . .
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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 100 − 90 = 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
DSE
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.
Super20
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, px /py , for this economy
1. can be any positive number
2. is greater than 1
3. is less than 1
4. does not exist
Let px = p, py = 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
10p = (p + 1)q. In other words, he demands q = p+1
of each good. For the market in good x to clear,
10p
p
10
1
p+1 + p = 10 =⇒ p + p+1 = 1, which does not have a solution. Thus, no competitive equilibrium
allocation is possible.
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Problem 30
Problem 31
Consider a strictly increasing, differentiable function u : R2 → R and the equations:
D1 u(x1 , x2 )
p1
= and
D2 u(x1 , x2 )
p2
p1 x1 + p2 x2 = w
where p1 , p2 , w are strictly positive. What additional assumptions will guarantee the existence of continuously
differentiable functions x1 (p1 , p2 , w) and x2 (p1 , p2 , w) that will solve these equations for all strictly positive
p1 , p2 , w?
1. u is injective
2. u is bijective
3. u is twice continuously differentiable
4. u is twice continuously differentiable and
D11 u(x1 , x2 , p2 ) − D12 u(x1 , x2 , p2 ) D12 u(x1 , x2 , p2 ) − D22 u(x1 , x2 , p2 )
p1
p2
is nonsingular
DSE
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.
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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
n
The given sequence is a sum of two component sequences, un = (−1)n and vn = (−1)
n . Its easy to see
that vn converges to 0 where as un takes values +1 and −1 ad infinitum. Thus the given sequence doesn’t
converge
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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 In = [ n1 , 1 − n1 ]. Then the countable union of these closed intervals, represented by
∞ [
1
1
,1 −
n
n
n=1
is the open interval (0, 1)
Problem 34
The set [0, 1] can be expressed as
DSE
1. the union of a finite family of open intervals
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 In = (− n1 , 1 + n1 ). Then the countable intersection of these open intervals, represented by
∞ \
1
1
− ,1 +
n
n
n=1
Super20
is the closed interval [0, 1]
The following information is used in the next Two questions. Consider a linear transformation
P : Rn → Rn . Let R(P ) = {P x|x ∈ Rn } and N (P ) = {x ∈ Rn |P x = 0}. P is said to be a projector if
a . every x ∈ Rn 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. . .
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Problem 35 (continued)
4. Both 1 and 2
Given P is a projector transformation. Take any x ∈ Rn . 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 : Rm → Rn 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 ∈ Rm . Let Q(x) = y ∈ R(Q) = R(P ). Thus y ∈ R(P ) ⇒ P (Q(x)) = P (y) = y = Q(x)
Problem 37
DSE
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:
I g(a) and g(b) have opposite signs.
II f 0 (x) = 0 for some x ∈ (a, b)
Of these statements,
1. Both I and II are true.
Super20
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 D2 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. . .
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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 − kex , k being the constant of integration. Substitute back for z(x) and solve
again to get f (x) = x − kex + c, c is the constant of integration for second step. Now using the given fact
x
1
−1
. Therefore f (x) = x − ee−1
which is non negative in
f (0) = 0, we get k = c and f (1) = 0 we get c = e−1
the interval [0,1].
Problem 39
Suppose x1 , ..., xn are positive and λ1 , ..., λn are non-negative with
Pn
λ1
λn
1.
i=0 λi xi ≥ x1 · · · xn
Pn
λ1
λn
2.
i=0 λi xi < x1 · · · xn
q
Pn
xλ1 1 · · · xλnn
3.
λ
x
≤
i=0 i i
Pn
i=0
λi = 1. Then
4. None of the above is necessarily true.
DSE
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
Super20
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 < xn > and so is the case with bijection
between N and set Y , say < yn >. 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
Wi = 300 − 50 Di + ui
(20)
(10)
where Wi is the wage in Rs per day of the ith individual, Di = 1 if individual i is male, and 0 otherwise, ui
is a classical error term, and the figures in parentheses are standard errors.
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Problem 40
Problem 41
What is the average wage in the sample?
1. Rs. 250 per day
2. Rs. 275 per day
3. Rs. 262.50 per day
4. Rs. 267.50 per day
150 men receive, on average, Rs. 250 each, and 50 women receive, on average, Rs. 300 each, for an average
of Rs. 262.50.
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
DSE
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.
Super20
Problem 43
The explained (regression) sum of squares in this case is:
1. 93750
2. 1406.25
3. 15000
4. This cannot be calculated from the information given
ESS = 150 · 12.52 + 50 · 37.52 = 93750
Problem 44
A researcher estimate the following two models using OLS
Model A: yi = β0 + β1 Si + β2 Ai + i
Model B: yi = β0 + β1 Si + i
Problem 44 continued on next page. . .
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Problem 44 (continued)
where yi refers to the marks (out of 100) that a student i gets on an exam, Si refers to the number of hours
spent studying for the exam by the student, and Ai is an index of innate ability (varying continuously from
a low ability score of 1 to a high ability score of 10). i is the usual classical error term. The estimated
β1 coefficient is 7.1 for Model A, but 2.1 for Model B; both are statistically significant. The estimated β2
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 Si decreases on introducing Ai indicates some amount of collinearity
between the two. Let Ai = α0 +α1 Si +ui . Then the first model is yi = (β0 +β2 α0 )+(β1 +α1 β2 )Si +(β2 u+).
Comparing to Model B, we see that α1 must be negative.
Problem 45
An analyst estimates the model Y = β0 + β1 X1 + β2 X2 + β3 X3 + u using OLS. But the true β3 = 0. In this
case, by including X3
DSE
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
b3 = 0. However, non-usage of the information (that β3 = 0) leads to inefficiency (greater standard errors).
Super20
Problem 46
Let β̂ be the OLS estimator of the slope coefficient in a regression of Y on X1 . Let β̃ be the OLS estimator
of the coefficient on X1 on a regression of Y on X1 and X2 . Which of the following is true:
1. V ar(β̂) < V ar(β̃)
2. V ar(β̂) > V ar(β̃)
3. V ar(β̂) < or > V ar(β̃)
4. V ar(β̂) = V ar(β̃)
Either of the variances could be larger, depending on which model is correct. The variance will be lower
for the correctly specified model.
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Problem 46
Problem 47
You estimate the multiple regression Y = a + b1 X1 + b2 X2 + u with a large sample. Let t1 be the test
statistic for testing the null hypothesis b1 = 0 and t2 be the test statistic for testing the null hypothesis
b2 = 0. Suppose you test the joint null hypothesis that b1 = b2 = 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 b1 = b2 = 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 0.05 + 0.05 − 0.05 · 0.05 = 0.0975 = 9.75%.
Problem 48
DSE
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:
1.
1
4
2.
1
3
3.
1
27
4.
1
81
Super20
The only way a total score of 4 is obtained when each tester ranks all three brands 1, the probability of
1
which is 13 independently, thus the answer 81
Problem 49
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 0.999. If the person
does not have TB, the test will erroneously indicate that s/he does with probability 0.002. For a randomly
selected individual, the test shows that s/he has TB. What is the probability that this person actually has
TB?
1.
0.002
0.999
2.
1
1000
3.
1
3
4.
2
3
Problem 49 continued on next page. . .
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Problem 49 (continued)
Lets define two random variables, X, Y , former to represent the outcome of the TB test {P os, N eg} 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 ] = 0.002. Apply Bayes rule
P r[Y = T |X = P os] =
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 ]
Substitute and the get the answer 13 .
Problem 50
There exists a random variable X with mean µx and variance σx2 for which P [µx 2σx ≤ X ≤ µx + 2σx ] = 0.6.
This statement is:
1. True for any distribution for appropriate choices of µx and σx2 .
2. True only for the uniform distribution defined over an appropriate interval
3. True only for the normal distribution for appropriate choices of µx and σx2 .
4. False
DSE
According to Chebyshev Inequality,
P r[µx − 2σx ≤ X ≤ µx + 2σx ] = P r[|X − µx | ≤ 2σx ] ≥ 0.75
Problem 51
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:
1.
1
9
2.
3
11
3.
11
3
4.
3
4
Super20
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 ) =
11
4 8
1
3 − 3 · 3 = 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
5
n−5
5−(n−5)
1.
(1/2)
(1/2)
n−5
Problem 52 continued on next page. . .
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DSE Super 20
2.
3.
4.
Solutions : DSE entrance 2015
10
n − 10
5
n−5
10
n − 10
(1/4)
n−10
(1/4)
(1/2)
n−5
n−10
(3/4)
n
(3/4)
(1/2)
Problem 52 (continued)
10−n
n
Probability with which 2 dice are rolled is 14 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 34 . 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 + Σ5i=0 Xi
where Xi 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
5
n−5
10−n
P r[N = n] = P r[N − 5 = n − 5] =
(1/4)
(3/4)
n−5
Problem 53
DSE
Suppose S is a set with n ≥ 1 elements and A1 , ..., Am are subsets of S with the following property: if
x, y ∈ S and x 6= y, then there exists i ∈ 1, ..., m such that, either x ∈ Ai and y ∈
/ Ai , or y ∈ Ai and x ∈
/ Ai .
Then the following necessarily holds.
1. n = 2m
2. n ≤ 2m
3. n > 2m
Super20
4. None of the above
Given there are m subsets Ai , there are 2m possible sets of subsets. Call the set of these 2m sets S. Given
any i ∈ S, exactly one element A ∈ S represents the list of all subsets that contain i. If n > 2m , then
∃j ∈ S; j 6= 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 (Kt ) and aggregate labour force (Lt ) in the following way: Yt = (Kt )α (Lt )(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 sw proportion
of their wage earnings and sr 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.
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Problem 53
Problem 54
Let sw = 0 and sr = 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)ktα and interest rate r = αktα−1 where kt is
capital labour ratio at time t. Savings are given by St = sw wLt + sr rKt . Capital augmentation equation is
Kt+1 = Kt + St . Dividing by Lt and using population growth equation Lt+1 = Lt (1 + n), we get dynamic
equation for capital labour ratio. kt+1 (1 + n) = kt + ktα [(1 − α)sw + αsr ]. In steady state kt+1 = kt = k ∗ .
Putting sw = 0 and sr = 1 then differentiating k ∗ with respect to α we get positive expression.
Problem 55
Let sr = 0 and 0 < sw < 1. An increase in the parameter value n
DSE
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 < sw < 1 and sr = 0 then differentiating k ∗ with respect to n we get a negative expression.
Super20
Problem 56
Now let both sw and sr be positive fractions such that sw < sr . 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−α)sw +αsr
n
αsw +(1−α)sr
n
1
1−α
α1
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.
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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 sw proportion of their net (posttax) wage earnings and sr 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
DSE
Let us now go back to case where both sw and sr are positive fractions such that sw < sr 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:
(
Akt for kt < k̄;
1 dL
=
L dt
0
for kt >> k̄
where k̄ >
(1−α)sw +αsr
A
1
2−α
is a given constant. In the long run, the capital-labour ratio in this economy
1. approaches zero
Super20
2. approaches infinity
3. approaches a constant value given by
αsw +(1−α)sr
A
α1
4. approaches infinity or a constant value given by
(1−α)sw +αsr
A
1
2−α
depending on whether
the initial k0 > or < k̄
If k0 >> k̄, 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 k̄ 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. . .
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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 k0 < k̄
4. leaves the long run steady state value of the capital-labour ratio unchanged
if k 0 < k̄, then only steady state is k ∗ = 0, which is independent of A, hence k ∗ can not unambiguously
depend upon A. However, if k 0 > k̄, then steady state depends upon A positively.
Problem 60
2
A profit maximizing firm owns two production plants with cost functions c1 (q) = q2 and c2 (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
0 2
DSE
c(q, q 0 ) = (q2) + (q − q 0 )2 is minimized at q 0 = 2(q − q 0 ) =⇒ q 0 =
c0 (q) = 23 q, which lies below the 45 degree line for all q > 0.
2q
3 .
Then, c(q) = 29 q 2 +
q2
9
=
q2
3
=⇒
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