Delhi School of Economics Course 001: Microeconomic Theory
Solutions to Problem Set 5
1. Suppose the given preference ordering can be represented by a VNM functionu(l), wherel2 fA; B; C; Dg. Observe that in preference orderings 1 and 3, B C, which implies that
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4u(100) +1
4u(20)> 1
4u(100) +3
4u(20),u(100)> u(20)
Since VNM functions are invariant to linear transformations (which gives us two free parameters), we can, without loss of generality, choose: u(100) = 1andu(20) = 0 for prefernce orderings 1 and 3. Let u(50) =x.
Preference Ordering 1: The following inequalities must be satis…ed:
B D)x < 3 4 D A)x > 1 2 A C)x > 1
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These inequalities can be satis…ed (e.g.,x= 23) simultaneously, so ordering 1 has a VNM representation.
Prefernce Ordering 3: The following inequalities must be satis…ed:
B D)x < 3 4 D C)x > 1 4 C A)x < 1
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Clearly, these preferences cannot be satis…ed simultaneously, so these are not VNM preferences.
Preference Ordering 2: Here,u(100)< u(20), so the assumption of more money is better than less is violated straight away (see above). Let u(100) = 0 andu(20) = 1. Then we have
C A)x < 5 4 A D)x < 1 2 D B)x > 1 4
Clearly, these can be simultaneously satis…ed (e.g., x= 13). Therefore, these are VNM preferences, even though money is not necessarily a “good”.
2. (a) The coe¢ cient of absolute risk aversion is
u00(y) u0(y) = 1
y
which is decreasing in income. The certainty equivalent is such that
logCE=1
2log 100 +1
2log 25 = log 50)CE= 50
Note that the expected monetary value of the lottery is Rs 62.50, so the risk premium is Rs 12.50.
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(b) The farmer solves
maxx
1
2log(100 px) +1
2log(25 + (1 p)x) The FOC is
p
100 px = 1 p
25 + (1 p)x Solving, we get
x(p) =25(4 5p) 2p(1 p) (c) The zero pro…t condition is
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2(2p 1)x(p) = 5 Substitute the expression forx(p)derived above and solve.
3. (a) The agent solves
maxx
pw x+ 2x+ (1 )p
w x
First-order condition (when the solution is interior):
pw+x= 1 pw x
which yields
x= 2 1
2+ (1 )2 w
The term in brackets must lie between 0 and 1. It is always weakly less than 1 and is greater than 0 if > 12.
(b) x= 0for 12. x=wif only if = 1.
(c) Zero pro…t implies
z (1 )pz = 0)p= 1 Fix somex. Thenzis chosen to solve
maxz
pw+x z+ (1 )p
w x+pz
which gives the …rst order condition
pw+x z = (1 )p pw x+pz which yields
z= 2(1 )x Now maximize with respect to x:
maxx [w+x 2(1 )x] + (1 ) [w x+ 2 x]
maxx w+ (2 1)x
Note that this objective function being linear, there will always be a corner solution. If < 12, x = 0and if >12,x =w.
Intuition: by betting in favor of failure, the agent can hedge against risk. It allows him to smooth out consumption between high and low states, and capture the expected return of the asset with certainty. Whenever this expected return is positive, he will want to put all his money on the risky asset, since it is no longer risky in e¤ect.
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4. (a) There are four possible outcomes, each equally likely: (i) both assets yield positive returns (c= 2x+ 3(w x) = 3w x) (ii) only assetAyields positive returns (c= 2x) (iii) only assetB yields positive returns (c = 3(w x)) (iv) none of the assets yields positive returns (c = 0). Optimal portfolio is given by
x = 1
4arg max
x [ln(3w x) + ln 2x+ ln 3(w x) + ln 0]
The FOC is
1 3w x+1
x 1 w x = 0 This yields the quadratic
2x2 8wx+ 3w2= 0 and the solution is
x = 2 r5
2
! w
(b) The reason to invest in both assets is portfolio diversi…cation (“don’t put all your eggs in one basket”). Risk averse agents will accept a lower average return on their portfolio in return for a lower variance in earnings.
5. Let us assume the following. Monty knows where the car is and he always opens an empty door among the ones that have not been chosen. You could think Monty is a robot in which this behaviour has been programmed. The correct answer in that case is that you should switch to the unopened door.
This will give you a 23 probability of winning as opposed to 13 if you stick to your initial choice.
One common response is that switching should not a¤ect your chances. Since one of the doors has been opened and shown to be empty, the prize lies behind the other two doors with equal probability.
One of them is your initial choice, the other is the one you are thinking of switching to. It should not matter whether you stick with your choice or change it. This reasoning is incorrect, however.
One way to understand why switching will improve your chances is as follows. There is a probability
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3 that your initial choice is correct and probability 23 that it is wrong. If it is correct, you will lose if you switch. If your initial choice was incorrect, on the other hand, you willwin if you switch (because the remaining unopened door is the one which contains the prize). Therefore, switching gives you a 23 probability of winning.
Consider an extended case. There are a hundred doors and there is a car behind only one of them.
After you pick a door, Monty will open 98 of the other doors and show them to be empty. Intuition suggests that the car is very likely to be behind the door he didn’t open rather than the random door you had picked in the very beginning.
This example shows that sometimes calculating correct probabilities in decision problems involving risk and uncertainty can be very tricky. To see a chequered history of the Monty Hall problem, with a prominent role played by the world’s smartest woman, see the Wikipedia entry.
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