From Talk:Dollar auction:
I am the guy who posted the mixed strategy equilibrium claim. I would also like to apologize for rude comment insinuating that my being an economics prof gives me authority. I have never seen a published proof of the existence of the mixed strategy equilibrium, but I will sketch a simplified version of it below. Also, I want to say that this is not the unique equilibrium, of course (bid a dollar, no other bid is an alternative, as are nash equilibria that aren't subgame perfect as described above). This equilibrium is, I believe, the unique mixed strategy equilibrium, but I haven't proved uniqueness.
Let the bid increment be B and the amount to be auctioned be NB, with N an integer greater than 1. Let t denote the period.
Player 1 bids in odd periods and player 2 bids in even periods. For simplicity, if player 1 does not bid in period 1, player 2 receives the prize (this assumption isn't necessary, but it makes the sketch below complete).
Let Vi(t) be the expected payoff to type i in period t, assuming the game has reached that point. We search for mixed strategy equilibria. This implies that the payoff from bidding and not bidding must be equal.
Let the probability of a bid at time t be Pt. Let's say that t is odd and greater than 1, so that it is player 1's turn to bid and there are already bids on the table.
The current bid is (t-1)*B. Therefore, if t is odd, the payoff to player 1 not bidding is -(t-2)B (she loses her current bid).
Payoff to bidding must equal that amount for her to be willing to randomize. Therefore, for t odd, V1(t)=-(t-2)B.
Now let t+1 be even. Then player 1's payoff at time t+1 depends on whether player 2 bids. Her expected payoff is the probability of no bid times her payoff net of her bid plus the probability of a bid times her continuation payoff at time t+2: V1(t+1)=(1-Pt+1)(N-t)B+(Pt+1)*V1(t+2).
Plugging in for above, we get V1(t+1)=(1-Pt+1)(N-t)B+(Pt+1)*V1(t+2)=(1-Pt+1)(N-t)B-(Pt+1)*(t+2-2)B.
The payoff from continuing the game for player 1 if t is odd equals -(t-2)B whether or not she bids, so we can set V1(t+1)=-(t-2)B.
Then we get (1-Pt+1)(N-t)B-(Pt+1)*(t+2-2)B=-(t-2)B.
Rearranging yields Pt+1=(N-2)/N for all t.
This argument works for player 2 as well. The randomization probability at any time t is (N-2)/N.
This argument follows for every t except t=1, since the assumption above is that by not bidding the amount (t-2)B is lost for good.
In period 1 this doesn't follow, obviously. Dealing with the initial bid is a hassle I won't get into here, but I'm sure any of you could adapt this argument to period t=1. —Preceding unsigned comment added by 66.57.254.60 (talk) 13:52, 11 September 2008 (UTC)