Econ 431: David Reiley

Due: Wednesday, 4 April 2007

In this problem set, you will encounter a couple of quadratic equations. If you need help, follow the link to a nice description of how to solve them. I recommend memorizing the quadratic formula, if you don't know it by heart already.

- (10 points) DS, question 12.2.
- (20 points) DS, question 12.3.
- (20 points) DS, question 12.6.
- (15 points) DS, question 12.7.
- (20 points) DS, question 12.8.
- (Extra credit, 20 points) In question 12.3 you found an asymmetric pure-strategy
Nash equilibrium. Now I would like you to find a symmetric mixed-strategy
Nash equilibrium to the same game. Suppose each player chooses to answer
Question 5 with probability
*P*, and not to answer it with probability 1-*P*. What is the equilibrium value of*P*?

To find this, you will need to follow a procedure similar to that followed in section 12.6 of the text, showing that an individual student is indifferent between answering the question and not answering the question. The present problem is a bit trickier, however. The expected payoff to answering the question depends on the probability that 0 out of 29 of the other students decide to answer the question, that 1 out of 29 of the other students decide to answer the question, that 2 out of 29 of the other students decide to answer the question, and so on. These probabilities are given by the binomial distribution, which you can look up in a probability textbook if you have forgotten it.

You will use these probabilities to construct an indifference equation that can be solved for the value of*P*. However, because this eq uation includes a very high-order polynomial, it's not worth trying to solve it analytically. Instead, I recommend finding a numeric solution. This can be done quite easily in Microsoft Excel by guessing a value of P in one cell, then setting up a formula in another cell that gives the expected value of answering the question. By changing the value in the first cell, you'll be able to zero in on the correct value of P. (If you prefer, you can solve the equation instead in another piece of software, such as Matlab or Mathematica.)

For full credit, please show the indifference equation you are trying to solve, then describe briefly how you solved it, and what answer you got.

- (Extra credit, 20 points) In this problem, we will find a mixed-strategy
equilibrium for a game in continuous strategies. (This game will be described
in Section 16.5 as an "all-pay auction" for a fixed prize.) Suppose
Professor Reiley auctions off a $20 bill to N students. Each student submits
a sealed bid for the $20 bill. The winner gets the $20 bill, but each student
must pay the amount of her bid. That is, the losers pay as well as the winner.
In case two or more players tie for the highest bid, then those players
split the $20 equally. (Each of the winners pays the amount of her bid,
as do each of the losers.) For most of this problem, let's assume for simplicity
that N=2, so there are just two players.

- First, let's show that there is no equilibrium in pure strategies
for this two-player, continuous-strategy game. To do this, first note
that a bid of more than $20 is a strictly dominated strategy (explain
why). Then assume that player 1 bids x
_{1}while player 2 bids x_{2}. There are then three possible cases: (i) x_{1}< x_{2}<20, (ii) x_{1}= x_{2}<20, and (i) x_{2}< x_{1}<20. Show that for each of these three cases, one of the players would have an incentive to deviate. Therefore, there is no pure-strategy equilibrium to this game.

- Now let's find the mixed-strategy equilibrium to this game. Since
this is a game with continuous strategies, let's look for a cumulative
probability distribution for each player. For each player (i=1,2), let
F
_{i}(x) = Pr(x_{i}<x), the probability that player i's bid x_{i}is less than some value x. What we want to find is the functional form of F_{i}(x).

Since this is a symmetric game, let's assume that each player has the same strategy in equilibrium: F_{1}(x)=F_{2}(x)=F(x). Let's suppose that player i's equilibrium mixture puts positive probability density f(x)>0 on all bids x between some lower bound a and some upper bound b. That means F(x)=0 for all x<a, F(x)=1 for all x>b, and F(x) is strictly increasing for all x in the interval [a,b]. We will need to determine the values of a and b.

- Write down the expected profit π
_{2}(x) that results for player 2 when she bids x. This will depend on F(x), the probability that player 1's bid is below x. There are three different regimes to consider: (1) x<a, (2) a≤x≤b, and (3) x>b. - Remember that in equilibrium, I must be indifferent between each
of the strategies I mix over. (Just as in a discrete game, we use
indifference between player 2's payoffs to figure out player 1's
mixing probabilities.) Suppose that π
_{2}(x)=c for all x in the interval [a,b]. Find an equation for F(x) that must be true for all x in the interval [a,b]. - Remembering that F(a) = 0, compute π
_{2}(a). Now remember that there can't be any gain to deviating from the equilibrium strategy. In particular, we must have π_{2}(0)≤π_{2}(a), because a is one of the bids in the equilibrium mixture. And note that bids in an auction must be nonnegative, so a≥0. Use this information to figure out the value of a. - Remembering that F(b)=1, compute π
_{2}(b). Remember that player 2 must be indifferent between playing x=a and playing x=b. Use this information to figure out the value of b. - Using your results to the previous two parts of the problem, find the value of c. Use this to find a simple expression for F(x).
- Now we know the values of a and b, and we have an expression for F(x). Check that when both players use this probability distribution, it is a mixed-strategy equilibrium.
- Finally, if you can, generalize this result to an all-pay auction with N≥2 players.

- Write down the expected profit π

- First, let's show that there is no equilibrium in pure strategies
for this two-player, continuous-strategy game. To do this, first note
that a bid of more than $20 is a strictly dominated strategy (explain
why). Then assume that player 1 bids x