The name is absent

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, v_A (к) = 0 and for all к ≥ J₀ + 1, v_B(к) = S^m-kZ_B.

Proof. By an argument similar to that above, v_B (m) = Z_b and v_A(m) =
0 imply J(m) > J(m-2), which in turn implies v_A(m-1) = 0+S max(0,J(m-
2) - Zb) = 0.

Suppose now that for some к, J₀ + 1 ≤ к < m, v_A(Z) = 0 for all I ≥ к and
vb (Z) = S^m-ZB for all Z ≥ к + 1. We will now demonstrate that this implies
that v_B(к) = S^m-kZ_b and v_A^ - 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, v_A(Z) = 0 for all Z ≥ к
and vb(Z) = S^m-ZB for all Z ≥ к + 1. Since va (к) = 0, we know that
v_a^ - 1) + v_B(к - 1) - v_a^ + 1) - v_B(к + 1) ≤ 0, so that

v_B(к) = Sv_B(к - 1) + S max(θ,vA(fc + 1) + v_B(к + 1) - v_A^ - 1) - v_B(к - 1))
= S[va^ + 1)+ vb(к + 1) - va(& - 1)]

= S[S^m-(k+1)ZB - va^ - 1)]

Moreover, v_A^-1) = Sv_A(^ + Smax(0,v_A^-2)+ v_B(к-2)-v_A(к)— v_B(к)).
Since va(^) = 0 by assumption and v_B(к) = S^m-kZ_B - Sv_A^ - 1), we have
v_a^ - 1) = S max(0,v_A^ - 2) + v_B(к - 2) - S^m-kZ_b + Sv_A^ - 1)). Suppose
by way of contradiction that va (к - 1) > 0. Then

va(& - 1) = S[va^ - 2) + vb(к - 2) - S^m-kZb + Sva^ - 1)] > 0.

or

va(£ - 1)[S^-1 - S] = va (к - 2) + vb (к - 2) - S^m-k Zb > 0.

26

<<   24   25   26   27   28   29   30   >>

More intriguing information