Auction Design without Commitment



A Appendix

Proof of Lemma 4

First assume that φ = g h is deterministic, i.e., φ(θ) = g(h(θ)) is singleton
for all
θ. We show that g(h(θ)) = φE [p](θ), for all θ Θ. Denote, for notational
simplicity,
Yi = {θ Θ : w(θ) = i} for all i.

Since g h is not ex post dominated by φE [p(s, h)] for any s h(supp(p)),
g(h(θ))
has to allocate the good to the same buyer as φE [p] (θ) does, for all θ.
Thus the partition
{Yi} specifies the winner under g h. Since, by EXP-IR, a non-
winner cannot be imposed a strictly positive monetary transfer, the allocation of
g(h(θ)) may differ from φE [p] (θ) only in the winner’s monetary transfer. Denote
the winner’s monetary transfer under
g(h(θ)) by mi(θ). Our task reduces to
showing that
mi(θ) = mE(i, θ-i,p), for all θ Yi, for all i.

Fix i. Since Θi is discrete and bounded below, we can order its elements by
θi0 <...<θik < .... We prove by induction that mi ik-i)=mE(i, θ-i, p), for
all
θ-i such that ik , θ-i) Yi , for allk =0, 1,  Assume that the induction
hypothesis holds until step
k - 1, i.e.,

mili, θ-i) = mE(i, θ-i, p), for all li, θ-i) Yi supp(p), for all l =0, ..., k - 1.

(9)

We show that (9) holds also for step k.

Take any s h(supp(p)). Since φ does not leave surplus to the winner that
could be extracted by
φE (p((x, s), φ)),

miik, θ-i) mE(i, θ-i,p(s, h)), for all ik, θ-i) supp(p(s, h)).

Since supp(p(s, h)) supp(p),

mE (i, θ-i, p(s, h)) mE (p, θ-i), for all ik, θ-i) supp(p(s, h)).     (10)

Noting that (10) holds for all s h(supp(p)), it follows from the above two con-
ditions that

miik, θ-i) mE(i, θ-i, p), for all ik, θ-i) supp(p).            (11)

It remains to be shown that the weak inequality in (11) holds as equality.

By (6),

mE(i, θ-i, p) = θik for all ik, θ-i) Yi supp(p) such that ik-1 , θ-i) 6∈ Yi. (12)

This has two implications. First,

P   [θk-mE(i,θ-i,p)]p(θik-i)=     P    [θk-mE(i,θ-i,p)]p(θik-i).

(θik-i,)Yi                                                       (θik-1-i)Yi

(13)

18



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