The name is absent



S  Δ0

(1 —S2) ^BA + b

SS⅛-2za  ' SS⅛-2za

1


for
for


b ∈ [0,


S∆B0A ]
’ (1-S2) ]
SΔ3'0
sδba


(1-S2)


and


^70-1(^) = S

S    Δ30-1

(1—S2) δab    ∣__a

SSm—(>o+1)ZB + SSm-(J0+1)ZB


for
for


a ∈ [0,


S Δ30
s δab


where


(1-S2


S ∆A0B


(1-S2


b

S   Δ0-1

(1 —S2) δab

1


for b ∈ [0, 'AB ]

,       , S δA0b 1

for b > T-aBt •


Δ⅛λ = [⅛m-*>Zb - Si°Zλ] and ∆⅛1 = Ao1Z, - δm-l*0~-ZB].


(12)

(13)

(14)

(15)


Note first that this equilibrium candidate has the properties described in
Proposition 1. Players’ continuation values can be stated as functions of the
respective state j as follows:

(           & Zλ           for j<jα - 1

vλ(j) = ^  (⅛'-`    Zλ - r-(70-1)Zb]  for j = jo - 1      (16)

I                0                 for    j ≥ jo

and

(       δm- Zb        for   j>jo

vb(j)={ (⅛) Г-0 Zb - ^ Zλ] for j = jo         (17)

I             0              for j ≤ jo - 1∙

These constitute the payoffs stated in the proposition. For 0 < j < jo - 1,
player A wins the next j battles without any effort. This takes j periods and
explains why the value of the final prize must be discounted to 57Zλ. Also,
B does not expend effort in these j battles and finally loses after j battles.
Hence, B’s payoff is equal to zero. For m > j > jo, players A and B simply
switch roles.

Turn now to the states jo - 1 and jo as in Figure 2. We call these states
"tipping states", because of their pivotal role in determining the outcome
of the contest. Consider jo - 1. From there, if A wins, the game moves to
jo - 2 with continuation values vλi∕o - 2) = 570-2Zλ and vb(jo - 2) = 0.

13



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