microeconomics & mathematics

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1. Consider a homeowner with Von-Neumann and Morgenstern utility function u, where u (x) = 1 − e−x for wealth level x, measured in million US dollars. His entire wealth is his house. The value of a house is 1 (million US dollars), but the house can be destroyed by a flood, reducing its value to 0, with probability π ∈ (0, 1). What is the largest premium P is the homeowner is willing to pay for a full insurance? (He pays the premium P and gets back 1 in case of a flood, making his wealth 1 − P regardless of the flood.)

The homeowners utility from getting 1-P is

u(1-P) = 1-e-(1-P)

Without insurance the homeowner faces the following expected loss

π(u(0)) +(1- π)(u(1)) = π (1-e-(0))-(1- π )( 1-e-(1))=(1- π) ( 1-e-(1))

Thus the largest premium he would be willing to pay is the premuim that sets the utility from insurance just equal to the utility from facing the risk.

1-e-(1-P) =(1- π) ( 1-e-(1))

1− P = −ln(1 − (1 − π)(1 − e −1))

P = 1+ ln(1− (1 − π)(1 − e −1))

2. Consider the following game

Find all Nash Equilibrium

(z,c) is the only NASH

3. Find all pure and mixed strategy Nash Equilibriums in the following games.

There is no pure strategy Nash Equilibrium.

Recall, any mixed strategy NE has to involve Player 2 mixing, because otherwise Player 1 would not mix either.

Let pA, pB, and pC be the probabilities that 1 plays A, B, and C, respectively. Let q and 1−q be the probabilities with which 2 plays L and R, respectively. For Player 2 to be willing to mix, it is necessary that pB = pC, because 2 strictly prefers L if C is more likely and prefers R if B is more likely.

Case 1 Suppose pB = pC > 0. Player 1 will only be willing to mix B and C if L and R are equally likely (q = 1/2), but this would mean that A is preferred to both. There is no NE in this case.

Case 2 Suppose pB = pC = 0. A is preferred to both B and C for 1/3≤ q ≤ 2/3. In this case, pA = 1, to which all q∈ [0,1] is a best response for 2. Thus (A,q) is a NE for 1/3≤ q ≤ 2/3

These are all of the NE of this game.

Again, there is no pure strategy NE. Note that A is dominated by an even mix between B and C.

Let p be the probability with which Player 1 plays B and let q be the probability that 2 plays L.

For mixing to be sustained in equilibrium, each player’s strategy must make the other indiﬀerent between the two pure strategies in that person’s mix.

Player 2 is indiﬀerent when 2(1−p) = 3p, or p = 2/5. Player 1 is indiﬀerent (between B and C) when q = 1/2. So the unique NE involves 1 playing (0,2/5,3/5) and 2 playing (1/2,1/2).

4. Consider the following extensive form game

(a) List the strategies available to each player in the game

P1: (L1, L2), (L1, R2), (R1)

P2: (l1, l3), (l1,r3), (r1), (l2), (r2)

(b) Construct payoffs (a, b, c, d, e, f, g, h, i, j, k, l) such that the terminal node r3, with payoffs (k, l), will be the unique backwards-induction outcome.

We will solve this in class. There are many possible answers.