# subgame perfect nash equilibrium for war of attrition

the question is as follow:

suppose that two players are playing war of attrition,

that means both of them could choose either to fight or quit, if either one of them quit, the game ends, and if both of them fight, the game proceed to the next period and continue.

The reward is $10 meaning that if one of them fight and the other quit, the fighter gets the reward, and the quiter gets nothing, and if both of them quit, both of them get nothing. The cost for them is different: player 1 has a per-period cost of$6, but player 2 has a per-period cost of 2 dollars

a) Derive a SPNE where each player uses a stationary randomizing strategy.

Firstly, i don't really know what is meant by stationary randomizing strategy, i assume that means both of them would choose to mix their strategy for every game.

Thus from my calculation, the SPNE is that player 1 would play fight and quit with probability 5/6 and 1/6 respectively and player 2 would play fight and quit with probability 5/8 and 3/8 respectively in every period.

Am i correct in doing this question?

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If I understand the question correctly, the scenario is this. Both players (say R and C) pay their per-period cost(6 for R, 2 for C). Then if both players choose fight, we go on to the next round; if one chooses fight and the other quit, the one who chooses fight collects 10 and the game stops ; if both choose quit, the game stops. If $v_r$ and $v_c$ are the values of the game for players $R$ and $C$ respectively, the payoff matrices for the two players are as follows: $$\text{R's matrix }\ \pmatrix{ & F & Q\cr F & -6 + v_r & 4 \cr Q & -6 & -6\cr} \ \text{C's matrix }\ \pmatrix{ & F & Q\cr F & -2 + v_c & -2\cr Q & 8 & -2\cr}$$
In a mixed strategy Nash equilibrium, R's strategy $(x_F, x_Q)$ must make C's expected payoff equal to $v_c$ for both of C's possible moves. Thus \eqalign{ x_F + x_Q &= 1\cr (-2 + v_c) x_F + 8 x_Q &= v_c\cr -2 x_F - 2 x_Q &= v_c\cr} From the first and third equations, we get $v_c = -2$, and then $x_F = 5/6$, $x_Q = 1/6$.
Similarly, C's mixed strategy must make R's expected payoff equal to $v_r$ for both of R's possible moves, and we get $v_r = -6$, $x_F = 5/8$, $x_Q=3/8$.