The name is absent

1) = 0. Hence, this deviation reduces B’s payoff by b'(j) compared to
b(j) = 0. A deviation b⁰(j) > 0 at j > j₀ does not change the state in t + 1
compared to b(j) = 0 in the candidate equilibrium, due to the tiebreaking rule
employed. The deviation reduces B’s payoff by b⁰(j) compared to b(j) = 0.
An equivalent logic applies for α(j) at states j ^ {j₀ — 1,j₀}.

Turn now to the state j₀. In the candidate equilibrium, in state j₀ con-
testant A randomizes on the support [0, ^Д₂) /V^{o 1}Z_a — ^^ (^{jo 1}'Z_b]]. All
actions in the equilibrium support for A at j₀ yield the same expected payoff
equal to Gj₀ (x)⅛_j(j₀ — 1) + (1 — Gj₀(x))0 — x = 0. A possible one-stage
deviation for A at j₀ is an α⁰(j₀) > (_1-f^₂) [^^j01Z_a — Z" ^{( jo k}Z_b]. Compared
to the action α(j₀) = (_1-f^₂) ∕l^{jo 1}Z_a — Z" (^{jo 1}'Z_b] that is inside A’s equilib-
rium support, this also leads to state j₀ — 1, but costs the additional amount
α⁰(j₀) — a(j₀) > 0. The deviation is therefore not profitable for A. The same
type of argument applies for b(j₀).

A similar argument applies to the state j₀ — 1. In the candidate equilib-
rium, in state j₀ — 1 contestant A randomizes on the support [0, (_1-f^₂) Z”' ^jo Z_b—
Z^oZ_a)]. All actions in the equilibrium support for A at j₀ — 1 yield the
same expected payoff equal to G_j∙₀-₁ (x)⅛_j(j₀ — 2) + (1 — G_j∙_o-₁(x))0 — x =
ɪɪa[Z^o1Z_a — Z" (^{jo 1}'Z_b] = u_A(j₀ — 1).¹¹ A possible one-stage deviation
for A at j₀ — 1 is an α⁰(j₀ — 1) > (_1-f^₂) Z" ^jo Z_b — 5^jo Z_a]. Compared to the
action α(j₀ — 1) = (_1-f^₂) [5^m~^joZ_b — 5^joZ_a] that is the upper bound of A’s
equilibrium support, this also leads to state j₀ — 2, but costs the additional
amount α⁰(j₀ — 1) — α(j₀ — 1) > 0. The deviation is not profitable for A. The
same type of argument applies for b(j₀ — 1). ■

Intuitively, outside of the states j₀ — 1 and j₀, one of the players is indif-
ferent between winning and losing the component contest. For instance, in
the state j₀ — 2, the best that player B could achieve by winning the next
component contest is to enter the state j₀ — 1 at which B’s continuation value
is still zero and smaller than player A’s continuation value. As B does not
gain anything from reaching j₀ — 1, B should not spend any effort trying to
reach this state. But if B does not spend effort to win, it is easy for A to
win.

The states j₀ — 1 and j₀ are different. Battle victory or defeat at one of

¹¹More formally, all actions in the support of A’s equilibrium local strategy that are
not mass points of B’s local strategy yield the same expected payoff. Since B has a mass
point at zero, this does not hold at a = 0, but for every a in a neigborhood above zero.

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