Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Assume Player 1 has valuation 400. Player 2 has valuation on the interval [0, 100]. What would the equlibrium be? Assume Player 2 knows Player 1's valuation, but Player 1 knows only the probability distribution of Player 2.

I know that this is functionally an all-pay auction, and so it incentivizes players to bid aggressively. I also know that Player 1 won't bid more than 100 because he knows player 2 won't do that either. I also believe there isn't a PSNE ... what what the equilibrium be?

share|cite|improve this question
I think some explanation might be helpful to the non-specialists here. For example, what is PSNE? My guess is that the NE stands for Nash equilibrium, but I have no idea what PS is. – Robert Israel May 10 '12 at 2:34
Pure Strategy Nash Equlibrium – Parseltongue May 10 '12 at 3:23
Who wins if both bid 100? – Andrew May 10 '12 at 4:41
Also, does player 2 know that player 1 knows their probability distribution? – Andrew May 10 '12 at 4:43
up vote 1 down vote accepted

Please correct me if my assumptions are incorrect. I assume players $1$ and $2$ each secretly choose a bid ($X_1$ and $X_2$ respectively). To avoid technical difficulties I'll suppose these must be integers. The highest bid wins the auction. I'll suppose that if $X_1 = X_2$ a fair coin-flip decides who wins. The winning bidder then pays his bid to gain the asset, whose worth to him is his evaluation: $400$ to player $1$, $V$ to player $2$, where $V$ is the value of a random variable with uniform (continuous) distribution on $[0,100]$. Player $2$ knows the actual value of $V$ before making his bid, so $X_2$ can be a function of $V$. Each player wants to maximize his expected payoff. If $W_1(X_1, X_2) = 1$ when $X_1 > X_2(V)$, $1/2$ when $X_1 = X_2(V)$ and $0$ when $X_1 < X_2(V)$, the expected payoff to player $1$ is $E(W_1) (400 - X_1)$ and the expected payoff to player $2$ is $E[(1-W_1)(V - X_2(V))]$. A Nash equilibrium will maximize each player's expected payoff given the other player's (mixed) strategy. Player $2$ will choose $X_2(V)$ to maximize $E[(1-W_1)(V - X_2(V)) | V]$ for each individual value of $V$.

If player $2$ knows $X_1$ (presumably some number in the interval $[0,100]$, there are basically three rational choices: $<X_1$, $X_1$ or $X_1 + 1$. With valuation $V$, his corresponding payoff is $0$, $(V - X_1)/2$, or $V - X_1 - 1$ respectively. The the optimal choice is $X_2(V) <X_1$ if $V < X_1$, $=X_1$ if $X_1 < V < X_1 + 2$, $=X_1 + 1$ if $V > X_1$. I'll suppose $X_2(V) = \lfloor V\rfloor$ if $V < X_1$: although that doesn't change player 2's payoff, it does allow him to take advantage of player $1$ varying from his optimal strategy.

Now if player $2$ is using this strategy, player $1$'s payoff is $400 - X_1$ when $V < X_1$ (which has probability $X_1/100$), $(400 - X_1)/2$ if $X_1 < V < X_1 + 2$ (which has probability $2/100$ assuming $X_1 \le 98$, $1/100$ if $X_1 = 99$, $0$ if $X_1 = 100$), $0$ if $V < X_1 + 2$. It is easy to see that the expected payoff is maximized (with a value of $300$) if $X_1 = 100$, which guarantees player $1$ a return of $300$.

Could this pair of strategies be a Nash equilibrium? Player $2$ certainly can't profit by varying his strategy. Player 1, however, might. Thus if player 1 bids $x \le 99 $ instead of $100$, he gains $100 - x$ when player 2's bid is less than $x$ (which has probability $x/100$, $-100-x/2$ when player 2's bid is also $x$ (probability $1/100$), $-300$ when player 2's bid is more than $x$ (probability $(99-x)/100$). The net expected gain is $-x^2/100 + 799 x/200 - 298$. But it is easily seen that this is always negative for $x \le 99$. So in fact we do have a Nash equilibrium.

share|cite|improve this answer
perfect. Thanks a lot! – Parseltongue May 17 '12 at 10:57

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.