Dependence and Uniqueness in Bayesian Games

Proof of Theorem 1

The game satisfies the assumptions of Van Zandt and Vives (2007). It follows that there are least and greatest equilibria and these are monotone. Since the game is symmetric these must be symmetric. For if, say is the least equilibrium in cutoff strategies, then must also be the least equilibrium by symmetry. Hence, . To prove the result it is therefore sufficient to show that any symmetric equilibrium in cutoff strategies has the desired properties.

The interim payoff to player i at cutoff k from choosing action 1 rather than 0 is

[30]

On the other hand, since and are strictly increasing and continuous, so uniformly continuous on , for any , one can find so that

[31]

From Assumption 4, an equilibrium cutoff k must belong to the compact interval in the interior of . Applying 2, proved below, is uniformly close to for all for large enough , so applying this to [31], one obtains that for all greater than some for

[32]

Hence, for any symmetric equilibrium cutoff must belong to , which proves the result.

Proof of Lemma 2

The proof follows the lines indicated in the text. By letting u be its mid-point, any open interval in the interior of can be written as . It will be shown that for any such interval

[33]

Since converges to , the associated measures converge in distribution, so the measure under of any set whose boundary has measure zero under M converges to that under M (see Billingsley (1995, Theorem 29.1)). In particular, the areas of the rectangles and converge to their areas under M, since M is concentrated on the diagonal and is atomless. On the other hand, , so as argued in the text, converges to , which, together with symmetry, establishes [33].

Suppose now that does not converge pointwise to for some u in the interior of . If, say, , then one can find a subsequence , with with . In particular, given with , for all n greater than some N. Since is increasing for all if . This, however, contradicts [33]: consider an interval , – from what has been shown on the left hand-side of [33] is bounded above by for all . Similarly, if , then for all for all sufficiently large , which leads to a contradiction of [33]. It follows that . A similar argument shows that , so indeed converges to .

To prove uniform convergence, let be a closed interval in the interior of . Since converges to at u, given one can find such that for all . Similarly, given , one can find such that for all . Let . Since is increasing, for all if . This implies the stated uniform convergence.²¹

Proof of Theorem 2

The proof is almost identical to that of Theorem 1 with replacing . Since converges uniformly to , given , one can find such that [31] holds for for all large enough . Similarly, since Assumption 4 holds for , one can find a compact interval T in the interior of such that for all large enough , equilibrium cutoffs must belong to T. The remainder of the proof is as before with replacing .

Proof of Theorem 3

By Assumption 13, interim payoffs can be uniformly approximated by [30], replacing by . The proof then proceeds as in the proof of Theorem 1 replacing by .

Proof of Lemma 4

[34]

Now

[35]

as since for any x, , and as F is a joint distribution function.²² In the case (or mutatis mutandis with inequalities reversed) note that and as is the marginal distribution of the first component.

So, by the dominated convergence theorem

[36]

The last equality holds since G is increasing.

Now

[37]

So, by a similar argument as above

[38]

Now, G is strictly increasing on its support, as g is strictly positive, and so it follows that is continuous. Hence, it follows from van der Vaart (1998, 305, Lemma 21.2) that

[39]

where is defined as in [5]. Since from [4]

[40]

[36] and [39] imply that

[41]

as was to be shown.

Proof of Lemma 5

Parts (a) and (b) were proven in the text. For part (c), note from the discussion in Sections 4 and 5 that part (iii) of Assumption 5 is equivalent to the assertion that the distribution of dominates that of and in the likelihood ratio order.

Now

[42]

under [25].

The density function of is

[43]

where h and H are the density and distribution functions of the (assumed identically distributed). clearly dominates each in the likelihood ratio order.

According to Theorem 2.1(d) of Keilson and Sumita (1982), if and are two random variables such that dominates in the likelihood ratio order and has a log-concave density function and is independent of and , then dominates in the likelihood ratio order. Taking and , proves the result since and from [42], .