The name is absent

equilibrium characterized in the Proposition 1 is unique in the class of Markov
perfect equilibria.

We will demonstate the uniqueness of continuation values for every state
j. For given state-contingent continuation values we have already argued that
the problem reduces to a standard all-pay auction for both players at each in-
terior state. Hence, uniqueness results from the uniqueness of the equilibrium
in the standard two-player all-pay auction with complete information.

Our proof will start by assuming that m > 3 and j₀ ∈ {2,...m — 1}. (The
case of m = 2 corresponds trivially to the all-pay auction.) We claim the
following:

Claim 8 In any Markov perfect equilibrium, for all к ≤ j₀ — 1, v_b (к) = 0
and for all к ≤ j₀ — 2, va(^ = δ^kZ_a.

Proof. At к = 0 by construction va(0) = Za and v_b (0) = 0, so the claim
holds for к = 0. Since Z_a > Z_b and m > 3, from (9) evaluated at к = 2
it follows that J(0) = Z_a > max(S²Z_A,S^m-2Z_B) > v_a(2) + v_b(2) = J(2).
Hence from (i) z_a(1) > z_b(I) and from (3) and (4) v_a(1) = 5[Z_a — v_b(2)]
and v_b(1) = 0. It immediately follows that J(1) = 5[Z_a — v_b(2)]. This
implies that the claim holds when j₀ = 2, which by definition of j₀ implies
that m = 3.

So assume that j₀ ∈ {3,...m — 1}. We will now prove the claim by
induction on к. Suppose that for some к, 1 ≤ к ≤ j₀ — 2, v_b(Z) = 0 for all
I ≤ к and Va(Z) = S¹Z_a for all Z ≤ к — 1. (Note that the supposition holds
for к = 1) We claim that v_a^) = S^kZ_a and v_b(к + 1) = 0.

To demonstrate this observe that by (5)

va(^ = Sva^ + 1) + max(0,S[(vA^ —1)+ Vb (к —1)) — (va (к + 1)+ Vb (к + 1))])

Since Vb(к) = 0, by (6) Zb(к) — za(^ = S[(va^ + 1) + Vb(к + 1)) —
(v_A(k — 1) + v_b(к — 1))] ≤ 0, which implies by (iii') that

1>А(к) = %(к — 1) + Vb (к — 1)) — Vb (к + 1)] = 4'>^{l 1}Z _l. — Vb (к + 1)]

Moreover, v_b(к + 1) = Sv_b(к) + Smax(0, (v_a^ + 2)+ v_b(к + 2)) — ^_А(к) +
Vb(к))) = Smax(0, (va (к + 2) + Vb(к + 2)) — S[S^k-1ZA — Vb(к + 1)]).

Suppose by way of contradiction that v_b (к + 1) > 0. Then v_b (к + 1) =
S[v_a^ + 2) + v_b(к + 2) — S(S^k-1Z_a) + Sv_b(к + 1)] > 0,which implies that

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