We have the following super-game: \begin{array}{cc} & a_1 & a_2 \\ a_1 & 0,0 & -1,1 \\ a_2 & 1,-1 & -2,-2 \end{array} I want to show that both the trigger (or grim) strategy and the mutual punishment, give the cooperative solution $(a_1,a_1)=(0,0)$ for the discount factor $\delta>\dfrac{1}{2}$. In the mutual punishment case, if players have observed $(a_1,a_1)$ or $(a_2,a_2)$ in the preivous round, they choose $a_1$. In case they have observed $(a_1,a_2)$ or $(a_2,a_1)$, they choose $a_2$ for one round and then we asumme that they cooperate again.
My thoughts for the second is (player 1 (p1) is the line player and player 2 the column player (p2)): Strategy: For as long as we cooperate we both gain a zero payoff, so the value function is: $$V_1^{C}= 0 + 0 \cdot \delta + 0 \cdot \delta^{2} + ... = 0$$ Suppose that p2, deviates unilaterally from the cooperation strategy in round 1 by playing $a_2$, so both players in the second round play the mutual punishment strategy, that is, $-2$ looses for everyone. From then onward, we assume that they both play according to the cooperative strategy, namely, the value function is: $$V_2^{NC}=1 - 2 \cdot \delta + 0 \cdot \delta + 0 \cdot \delta^{2} + ... = 1 - 2 \cdot \delta$$ They will cooperate if: $$V_1^{C}>V_2^{NC}\Rightarrow \delta>\dfrac{1}{2}$$ In the case of grim/trigger strategy, the players stop to cooperate in perpetuity, how does the solution differs?
Note: The game is the hawk-dove type super-game.