The name is absent



vA = δj0 2 ZA

vB = 0


vA =r?


vB = 0


vA = 0

vB = ?


vA = 0

vB = δmj0+1)ZB


j о -2

j 0+1


Figure 2:

If В wins, the game moves to j with continuation values vA(j0) = 0 and
vb(j0). Assuming that Zo 2Zλvb(j0) (which can be confirmed later),
and applying the results on the standard all-pay auction, the continuation
values are

vA(j0 1) = zA(j0 1) zB (j0 1) = 5[57'0 2ZA vB(j0)]     (18)

and vb(j0 1) = 0, where zA(j0 1) and zb(j0 1) denote the prizes that
A and В respectively attribute to winning the battle at j
0 1, given the
continuation of the game as described in the candidate equilibrium. Similarly,
at j
0, if A wins, the game moves to j0 1 with continuation values vA(j0 1)
as in (18) and
vb(j0 1) = 0. If В wins, the game moves to j0 + 1 with
continuation values v
A(j0 + 1) = 0 and vb(j0 + 1) = δm~(∙io+11 Zb. This yields
a continuation value for player В of

Vb (J0) = Zb (j>) — ∙    = ⅛r~0i+1) Zbva(J01)],      (19)

and va(J0) = 0. The solution to this system of equations yields the pos-
itive equilibrium values in the middle lines of (16) and (17), and the zero
continuation value in the respective state for the other player.

It remains to be shown that the choices described in the candidate equilib-
rium indeed describe equilibrium behavior. The one-stage deviation principle
applies here.10 The continuation values (16) and (17) can be used to consider
one-stage deviations for A and for В.

A deviation b0(j ) > 0 at a state 0 < j < j0 1 changes the path from
moving to j
1 in the next period to j + 1. However, vb (j 1) = vb (j +

10To confirm this it is sufficient to show that the condition of continuity at infinity is
fulfilled for this game. We may then apply Theorem 4.2 in Fudenberg and Tirole (1993).
This condition requires that the supremum of the payoff difference that can emerge from
strategies that differ after period
t converges to zero as t → ∞. However, a supremum for
this is
δt[Zi + iɪjK] for i = A, B, and this converges to zero as t → ∞.

14



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