<p>Finally, I believe it is easy to answer it. Well, in case where some player will not cooperate and the triger strategy is enabled, we have the following soloutions:
Soppuse that p2 does not cooperate in the first round so, he will gain a payoof of 1, but in the second round p1 plays <span class="math-container">$a_2$</span> and he won't cooperate with p2 in perpetual. Thus, we have that p2 will play <span class="math-container">$a_1$</span>, because by playing <span class="math-container">$a_2$</span> he increases his looses and the value function is:
<span class="math-container">$$V_2^{NC}=1-1\cdot\delta-1\cdot\delta^2-1\cdot\delta^3+...=1-\delta\sum_{j=1}^{+\infty}\delta^{j}=1-\dfrac{\delta}{1-\delta}$$</span>
Hence, they will cooperate if:
<span class="math-container">$$0&gt;1-\dfrac{\delta}{1-\delta}\Rightarrow \delta&gt;\dfrac{1}{2}$$</span>
Differently, if both players choose trigger strategies, so p2 also chooses <span class="math-container">$a_2$</span>, then both will make looses of -2 and the value function is:
<span class="math-container">$$V_2^{NC}=1-2\cdot\delta-2\cdot\delta^2-2\cdot\delta^3+...=1-2\delta\sum_{j=1}^{+\infty}\delta^{j}=1-\dfrac{2\delta}{1-\delta}$$</span>
and hence they will cooperate if:
<span class="math-container">$$0&gt;1-\dfrac{2\delta}{1-\delta}\Rightarrow \delta&gt;\dfrac{1}{3}$$</span></p>

CC BY-SA 4.0
2020-01-28T21:00:29.627Z
2020-01-29T14:32:19.813Z
2020-01-29T14:32:19.813Z
Nav89
<p>We have the following <a href="https://link.springer.com/chapter/10.1007/978-1-349-20181-5_29" rel="nofollow noreferrer">super-game</a>:
<span class="math-container">\begin{array}{cc}
  &amp; a_1 &amp; a_2  \\
a_1 &amp; 0,0 &amp; -1,1 \\
a_2 &amp; 1,-1 &amp; -2,-2 
\end{array}</span>
I want to show that both the <a href="https://en.wikipedia.org/wiki/Trigger_strategy" rel="nofollow noreferrer">trigger</a> (or <a href="https://en.wikipedia.org/wiki/Grim_trigger" rel="nofollow noreferrer">grim</a>) strategy and the mutual punishment, give the cooperative solution <span class="math-container">$(a_1,a_1)=(0,0)$</span> for the discount factor <span class="math-container">$\delta&gt;\dfrac{1}{2}$</span>. In the mutual punishment case, if players have observed <span class="math-container">$(a_1,a_1)$</span> or <span class="math-container">$(a_2,a_2)$</span> in the preivous round, they choose <span class="math-container">$a_1$</span>. In case they have observed <span class="math-container">$(a_1,a_2)$</span> or <span class="math-container">$(a_2,a_1)$</span>, they choose <span class="math-container">$a_2$</span> for one round and then we asumme that they cooperate again.</p>

<p>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:
<span class="math-container">$$V_1^{C}= 0 + 0 \cdot \delta + 0 \cdot \delta^{2} + ... = 0$$</span>
Suppose that p2, deviates unilaterally from the cooperation strategy in round 1 by playing <span class="math-container">$a_2$</span>, so both players in the second round play the mutual punishment strategy, that is, <span class="math-container">$-2$</span> looses for everyone. From then onward, we assume that they both play according to the cooperative strategy, namely, the value function is:
<span class="math-container">$$V_2^{NC}=1 - 2 \cdot \delta + 0 \cdot \delta + 0 \cdot \delta^{2} + ... = 1   - 2 \cdot \delta$$</span>
They will cooperate if:
<span class="math-container">$$V_1^{C}&gt;V_2^{NC}\Rightarrow \delta&gt;\dfrac{1}{2}$$</span>
In the case of grim/trigger strategy, the players stop to cooperate in perpetuity, how does the solution differs? </p>

<p>Note: The game is the <a href="https://link.springer.com/chapter/10.1007%2FBFb0055028" rel="nofollow noreferrer">hawk-dove</a> type super-game.</p>

<game-theory><repeated-games><cooperative-game-theory>
Simple hawk-dove type super-game
Nav89
# Simple hawk-dove type super-game

(Asked by: username_0 on 2020-01-28)

We have the following [super-game](https://link.springer.com/chapter/10.1007/978-1-349-20181-5_29):
\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](https://en.wikipedia.org/wiki/Trigger_strategy) (or [grim](https://en.wikipedia.org/wiki/Grim_trigger)) 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](https://link.springer.com/chapter/10.1007%2FBFb0055028) type super-game.

**Tags:** game-theory, repeated-games, cooperative-game-theory

---

## 1 answers

### Answer by username_0 on 2020-01-28. Score: 0 | Accepted answer

Finally, I believe it is easy to answer it. Well, in case where some player will not cooperate and the triger strategy is enabled, we have the following soloutions:
Soppuse that p2 does not cooperate in the first round so, he will gain a payoof of 1, but in the second round p1 plays $a\_2$ and he won't cooperate with p2 in perpetual. Thus, we have that p2 will play $a\_1$, because by playing $a\_2$ he increases his looses and the value function is:
$$V\_2^{NC}=1-1\cdot\delta-1\cdot\delta^2-1\cdot\delta^3+...=1-\delta\sum\_{j=1}^{+\infty}\delta^{j}=1-\dfrac{\delta}{1-\delta}$$
Hence, they will cooperate if:
$$0>1-\dfrac{\delta}{1-\delta}\Rightarrow \delta>\dfrac{1}{2}$$
Differently, if both players choose trigger strategies, so p2 also chooses $a\_2$, then both will make looses of -2 and the value function is:
$$V\_2^{NC}=1-2\cdot\delta-2\cdot\delta^2-2\cdot\delta^3+...=1-2\delta\sum\_{j=1}^{+\infty}\delta^{j}=1-\dfrac{2\delta}{1-\delta}$$
and hence they will cooperate if:
$$0>1-\dfrac{2\delta}{1-\delta}\Rightarrow \delta>\dfrac{1}{3}$$


---

