The name is absent

state j ∈ {1, ...m

1} at any time t is

Mj) = 5v_A(j + 1) + max(0,ZA(j) - z_b(j)) = ôv_A(j + 1) +
+ max(0,5[(vA(j - 1) - v_A(j + 1)) - (vb(j + 1) - Vb(j - 1))])

and

Vb(j) = ⅛b(j - 1) + max(0,ZB(j) - z_a(J)) = ⅛b(j - 1) +
+ max(0,5[(vB(j + 1) - Vb(j - 1)) - (vA(j - 1) - v_A(j + 1))]).

Rearranging (3) and (4) we obtain

Va(j) = M(j + 1)+max(0,5[(vA(j -1)+ Vb(j -1)) - (vA(j + 1)+ Vb(j + 1))])

(5)

and

Vb ( j ) = Svb (j -1)+ max(0,5 [( Va ( j + 1)+ Vb ( j + 1)) - ( Va ( j -1)+ Vb ( j -1))])

(6)
Note that the first summand in (5) and (6) is the discounted value of losing
the contest at j and the second summand in each of these expressions is
the expected gain arising from the contest at j. For at least one player this
gain will be zero and for the other player it will be non-negative and strictly
positive as long as J(j - 1) = J(j + 1), where J(Z) ≡ v_a(Z) + v_b(Z) is the
joint present value of being in state Z.

Three immediate implications of the above construction are

(i) z_A(j) - z_b(j) ≥ 0 if and only if J(j - 1) - J(j + 1) ≥ 0 with strict
inequality in one if and only if in the other.

(ii) ZA(j )-zb (j) ≥ 0 if and only if vb (j) = ⅛b (j-1) and z_a(J)-Zb (j) ≤ 0
if and only if va(J) = ⅛a(J + 1).

(iii) If za(j) - zb(j) ≥ 0 then va(J) = 5[v_a(J - 1) + Vb(j - 1)) - Vb(j + 1)],
and if za(j)-zb(j) ≤ 0 then Vb(j) = 5[va(J + 1)+vb(j + 1))-va(J-1)].

By assumption 0 and m are terminal states so that v_a(0) = Z_a ≥ Z_b =
v_b (m) and v_A(m) = v_b (0) = 0. Moreover, since player A can only receive a
positive payoff in the state 0, player B can only receive a positive payoff in the
state m, and both players have available the opportunity to always expend
zero effort, in any Markov perfect equilibrium the following inequalities hold
for all j :

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