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 (θik,θ-i)=mE(i, θ-i, p), for
all θ-i such that (θik , θ-i) ∈ Yi , for allk =0, 1, Assume that the induction
hypothesis holds until step k - 1, i.e.,
mi(θli, θ-i) = mE(i, θ-i, p), for all (θli, 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)),
mi(θik, θ-i) ≥ mE(i, θ-i,p(s, h)), for all (θik, 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 (θik, 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
mi(θik, θ-i) ≥ mE(i, θ-i, p), for all (θik, 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 (θik, θ-i) ∈ Yi ∩ supp(p) such that (θik-1 , θ-i) 6∈ Yi. (12)
This has two implications. First,
P [θk-mE(i,θ-i,p)]p(θik,θ-i)= P [θk-mE(i,θ-i,p)]p(θik,θ-i).
(θik,θ-i,)∈Yi (θik-1,θ-i)∈Yi
(13)
18
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