vA(m - 2)
Sm-2 Za - S2Zb
1 - S2
and
SZb - Za
vB (m - 1) =---------2---- > 0
1 - S
where both inequalities follow from J0 = m - 1.
This completes the case for J0 = m - 1. So suppose J0 ∈ {3,..., m - 2}
and look at к > J0.
Claim 9 In any subgame perfect equilibrium in stationary Markov strategies,
for all к ≥ J0, vA (к) = 0 and for all к ≥ J0 + 1, vB(к) = Sm-kZB.
Proof. By an argument similar to that above, vB (m) = Zb and vA(m) =
0 imply J(m) > J(m-2), which in turn implies vA(m-1) = 0+S max(0,J(m-
2) - Zb) = 0.
Suppose now that for some к, J0 + 1 ≤ к < m, vA(Z) = 0 for all I ≥ к and
vb (Z) = Sm-ZB for all Z ≥ к + 1. We will now demonstrate that this implies
that vB(к) = Sm-kZb and vA^ - 1) = 0. Since the supposition holds for
к = m - 1, this will then prove claim 2 by induction.
So assume that for some к, J0 + 1 ≤ к < m, vA(Z) = 0 for all Z ≥ к
and vb(Z) = Sm-ZB for all Z ≥ к + 1. Since va (к) = 0, we know that
va^ - 1) + vB(к - 1) - va^ + 1) - vB(к + 1) ≤ 0, so that
vB(к) = SvB(к - 1) + S max(θ,vA(fc + 1) + vB(к + 1) - vA^ - 1) - vB(к - 1))
= S[va^ + 1)+ vb(к + 1) - va(& - 1)]
= S[Sm-(k+1)ZB - va^ - 1)]
Moreover, vA^-1) = SvA(^ + Smax(0,vA^-2)+ vB(к-2)-vA(к)— vB(к)).
Since va(^) = 0 by assumption and vB(к) = Sm-kZB - SvA^ - 1), we have
va^ - 1) = S max(0,vA^ - 2) + vB(к - 2) - Sm-kZb + SvA^ - 1)). Suppose
by way of contradiction that va (к - 1) > 0. Then
va(& - 1) = S[va^ - 2) + vb(к - 2) - Sm-kZb + Sva^ - 1)] > 0.
or
va(£ - 1)[S-1 - S] = va (к - 2) + vb (к - 2) - Sm-k Zb > 0.
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