Game Theory- Second-Price Sealed Auction- 2 Players 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?
 A: 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.          
