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