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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?

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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
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1 Answer

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.

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perfect. Thanks a lot! –  Parseltongue May 17 '12 at 10:57
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