discontinuous games
1725
are strictly increasing in own signal, the proof of Proposition 1 in Maskin and
Riley (2000) is easily adapted to show that there is no loss in assuming the
bidding strategies b1ꢂb2 are monotone, in the sense that if tiꢁ > ti, then every bid
in the support of biꢄtiꢁ is at least as large as every bid in the support of biꢄti . It
follows immediately that there is an at most countable set of signals ti for which
the support of biꢄti is not a singleton. For such ti, replace bi by the infimum of
the support of biꢄti . It is easily checked that the modified bid functions b1ꢂb2
again constitute an equilibrium. Thus we have an equilibrium in monotone, pure
behavioral strategies. Altering bids following signals 0, 1 if necessary, there is no
loss in assuming that b1ꢂb2 are continuous at 0, 1.
Let b = maxꢈb1ꢄ0 ꢂb2ꢄ0 ꢉ, and for each i, let ꢊi = supꢈtꢀbiꢄt ≤ bꢉ. Suppose
that ꢊ2 = 0. Then, b1ꢄ0 wins with probability 0, and so earns 0. But, a bid of
b+1 by 1 wins with positive probability, and for 1 small, does so only when t2 ꢅ 0
(using that ꢊ2 = 0) so that v1 ꢅ 5+0−4ꢄ0 = 5. For there not to be a profitable
deviation of this form, it must thus be that b ≥ 5, and hence that the winning
bid is always at least 5. But, the average value of the object, even if allocated
optimally to the player with larger t, is 5+2/3−4ꢄ1/3 < 5, since 2/3 and 1/3 are
the expected higher and lower values of two draws from the uniform distribution.
So, someone is losing money on average, and would be better off to bid 0 always.
This is a contradiction, and so ꢊ2 > 0. Arguing symmetrically, ꢊ1 > 0.
Assume both players use b with positive probability, and assume that ties at b
are broken with probability p ∈ ꢄ0ꢂ1 in favor of player 1. Let tꢁ and tꢁꢁꢂtꢁ < tꢁꢁ, be
two values of t for which b1ꢄt = b. It follows that 5+tꢁ −4Eꢄt2ꢀb2ꢄt2 = b ≥ b,
else 1 would be better to bid b−1 with tꢁ. But then 5+tꢁꢁ −4Eꢄt2ꢀb2ꢄt2 = b > b,
and so 1 should deviate to b +1 with tꢁꢁ. This is a contradiction. There are thus
two remaining possibilities.
(i) One player, w.l.o.g. player 2, does not use b with positive probability (since
ꢊ2 > 0, this implies b2ꢄ0 < b, and also that player 1 bids b with positive probability
since ꢊ1 > 0, and by definition of b). Now, with t = ꢊ2 −1, player 2 never wins, but
by bidding b +1 wins with positive probability for an expected value of 5 +ꢊ2 −
1−4ꢄꢊ1/2 . So, for 2 not to want to deviate, it must be that b ≥ 5+ꢊ2 −4ꢄꢊ1/2 >
5+ꢊ2 −4ꢊ1.
(ii) Both players use b with positive probability. Then, by the above, one
player, again w.l.o.g. player 2, always has ties at b decided against him. Let
ꢄ51ꢂꢊ1 be the (nonempty) interval over which player 1 bids b. Then, with
t = ꢊ2 −1, player 2 wins only when t1 < 51, while by bidding 1 more, he can also
win when t1 ∈ ꢄ51ꢂꢊ1 . For this not to be a profitable deviation, it must be that
b ≥ 5+ꢊ2 −4ꢄ51 +ꢊ1 /2 > 5+ꢊ2 −4ꢊ1.
Assume that ꢊ2 < 1. Pick t = ꢊ2 + 1, and consider replacing b2ꢄt (which is
by definition greater than b) by any bid in ꢄbꢂb2ꢄt . This bid pays less in the
(positive probability, since ꢊ1 > 0) event that it still wins, and when it changes a
win into a loss, t1 ≥ ꢊ1, and hence v2 is at best 5 +ꢊ2 +1−4ꢊ1 < b. So, this is a
profitable deviation, a contradiction. Thus, ꢊ2 = 1.
Since ꢊ2 = 1, it follows that 1 wins with probability 1 (since 2 does not win when
t2 < 1), and hence that he always bids b (he bids at least this by definition, and