<p>A couple hints.</p>
<ol>
<li>Regarding the lower bound on <span class="math-container">$\epsilon$</span>: What happens if deviation occurs at stage <span class="math-container">$T$</span>? In other words, there is no opportunity for your so-called &quot;punishment stages&quot;.</li>
<li>Regarding the upper bound on <span class="math-container">$\epsilon$</span>: Suppose player 2 deviates at stage <span class="math-container">$T-1$</span> but player 1 does not. What must be true about <span class="math-container">$\epsilon$</span> in order for player 1 to have an incentive to switch to <span class="math-container">$D$</span> at the very last stage?</li>
</ol>
<hr />
<p>In general, to show that the given strategy profile is subgame perfect, you need to argue that playing <span class="math-container">$C$</span> after <span class="math-container">$(C,C)$</span> and <span class="math-container">$D$</span> after either <span class="math-container">$(C,D),(D,C),(D,D)$</span> are optimal.</p>

CC BY-SA 4.0
2021-01-17T19:48:26.430Z
Correct me if I'm wrong. First, in the last stage, $T$, if either player deviates it will yield the deviating player $4-\varepsilon$ which is less that 3 for $\varepsilon > 1$. Thus, it will not be a profitable deviation in the last stage. Second, if player 2 deviates at $T-1$ then he'll gain $4-\epsilon$ and $2$ at $T$ i.e. the total payoff of the last two stages for player 2 is $4-\varepsilon + 2$. On the other hand, if player 2 deviates such that $(C,D)$ is played at time $T-1$, then player 1 would want to deviate in the last stage, $T$. Thus, he'll gain $0+2-\varepsilon$.
Justin Malik
CC BY-SA 4.0
2021-01-18T00:33:51.190Z
@JustinMalik: Yes, you are on the right track. To ensure that player 1 switches to $D$ after player 2's deviation at $T-1$, you'd need $0+2-\varepsilon > 0+0$, the payoff from not switching. Of course, to conclude that the given strategy profile is subgame perfect, you'd still have to argue that any one-shot deviation at earlier stages, either on or off equilibrium path, is not profitable.
Herr K.
CC BY-SA 4.0
2021-01-18T16:06:55.910Z
Thank you so much. I think I understand now. I've updated the post. Would you have a look at the solution I proposed to see if I indeed understood it the right way?
Justin Malik
CC BY-SA 4.0
2021-01-17T04:29:13.757Z
2021-01-17T04:29:13.757Z
Herr K.
<p><a href="https://i.sstatic.net/vmLRR.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/vmLRR.png" alt="enter image description here" /></a></p>
<p>I'm doing this finitely repeated Prisoner's dilemma with switching costs but I have trouble showing the fact that <span class="math-container">$\varepsilon$</span> had to be <span class="math-container">$1 &lt; \varepsilon &lt; 2$</span>. I do see why and  that it is a subgame-perfect Nash equilibrium, but how do I show it.</p>
<p><strong>My approach:</strong>
Consider deviations by player 1, given that player 2 sticks to his strategies in his subgames following histories that end in each of the four outcomes of the game. (By symmetry the below would also account for deviations by player 2 if player 1 sticks to his strategies).</p>
<p>We know that this game is played <span class="math-container">$T$</span> times. Let <span class="math-container">$m$</span> denote the number of cooperation stages (i.e. number of times the strategy profile <span class="math-container">$(C,C)$</span> has been played), let <span class="math-container">$n$</span> denote the number of punishment stages (number of times the strategy profile <span class="math-container">$(D,D)$</span> has been played), and let <span class="math-container">$T = m + n +1$</span>. If the players cooperate <span class="math-container">$T$</span> times they get <span class="math-container">$3$</span> in payoffs. Thus, we can write
<span class="math-container">$$3T=3(m+n+1)$$</span>
If player 1 deviates at any stage of the game he'll get a payoff of
<span class="math-container">$$3 \cdot m + (4 - \varepsilon) + 2 \cdot n$$</span>
For the deviation to be profitable for player 1 the deviation payoff has to be greater than or equal to the cooperative payoff. That is,
<span class="math-container">$$3 \cdot m + (4 - \varepsilon) + 2 \cdot n \geq 3m + 3n + 3$$</span>
We can simplify this
<span class="math-container">$$n \leq 1 - \varepsilon$$</span></p>
<p>And this is where I'm stuck. What am I supposed to interpret from this? That when <span class="math-container">$\varepsilon \in (1,2)$</span> then the number of punishment stages is negative?</p>
<p>Another question I have is: Assume that the question above got answered. Is that enough? Or do I need to show something more? Because I'm well aware that playing <span class="math-container">$(C,C)$</span> <span class="math-container">$T$</span> times is a SPNE and playing <span class="math-container">$(D,D)$</span> <span class="math-container">$T$</span> times is also a SPNE and either way there is no profitable deviations.</p>
<p>So, to summarize: I would like to know if I'm approaching this problem in the right way. I would like to get some help on how to move forward from where I got stuck. And if the above would be enough to show what the problem is asking for.</p>
<p>Many thanks in advance!</p>
<p>[Update]----------------
Below is my updated attempt on a solution with some inspiration from M.J. Osbourne and the feedback I got from @Herr K..
<a href="https://i.sstatic.net/rGSZG.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/rGSZG.png" alt="enter image description here" /></a>
<a href="https://i.sstatic.net/svhd3.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/svhd3.png" alt="enter image description here" /></a>
<a href="https://i.sstatic.net/j4XqS.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/j4XqS.png" alt="enter image description here" /></a></p>

CC BY-SA 4.0
2021-01-18T18:00:21.720Z
This doesn't change your conclusion qualitatively, but I'd suggest being more explicit about the steam of payoffs after a deviation. For example, suppose player 1 deviates at time $t$, and the outcome at $t-1$ was $(C,C)$. Then the continuation payoff from $t$ onwards would be $4-\epsilon+2(T-t)$. Compared to the continuation payoff from not deviating at $t$, which is $3(T-t+1)$, deviation is not profitable for any $t\le T$ as long as $\epsilon >1$.
Herr K.
CC BY-SA 4.0
2021-01-20T16:33:48.830Z
Ah I think I understand. If let's say I would apply the same method you used to for example ($C,D$). If we suppose player 1 deviates at time $t$, and the outcome at time $t-1$ was ($C,D$). Then the continuation payoff from $t$ onwards would be $2-\varepsilon + 2(T-t)$. If he doesn't deviate it would be a stream of 0's from time from time $t$, which is $0(T-t+1)$. Thus, we have that deviation is profitable if $2-\varepsilon 2(T-t) \geq 0T \quad \Rightarrow \quad 0 \leq 2 - \varepsilon + 2T - 2t$. That is, a deviation will be profitable for any $t \leq T$ as long as $\varepsilon < 2$.
Justin Malik
CC BY-SA 4.0
2021-01-20T22:11:33.323Z
Well, if the outcome at $t-1$ is $(C,D)$, then the given strategy prescribes playing $D$ at $t$ for player 1. So playing $C$ is deviating, whereas playing $D$ is not. Also, since we only need to consider one-shot deviations, if player 1 deviates to playing $C$ at $t$, he should revert back to the original strategy from $t+1$ onwards, i.e. playing $D$ from $t+1$ till $T$.
Herr K.
CC BY-SA 4.0
2021-01-20T22:26:46.973Z
It might be a dumb question but doesn’t the outcome $(C,D)$ at $t-1$ count as a deviation? If the play was $(C,D)$ at time $t-1$ and again at time $t$ wouldn’t it be two deviations then?
Justin Malik
CC BY-SA 4.0
2021-01-20T23:22:58.847Z
The one stage deviation is for *every subgame*. So in any subgame that comes after the outcome $(C,D)$, we verify that there is no profitable one-shot deviation within that subgame, even though the subgame in question is clearly off-equilibrium path.
Herr K.
CC BY-SA 4.0
2021-01-20T23:27:34.717Z
Aaah okay, that was what was bothering and confusing me. Thank you so much. Your help is very much appriciated.
Justin Malik
<game-theory><self-study><nash-equilibrium><dynamic-games><repeated-games>
Finitely repeated Prisoner’s Dilemma with switching cost
Justin Malik
# Finitely repeated Prisoner’s Dilemma with switching cost

(Asked by: username_0 on 2021-01-16)

[![enter image description here](https://i.sstatic.net/vmLRR.png)](https://i.sstatic.net/vmLRR.png)

I'm doing this finitely repeated Prisoner's dilemma with switching costs but I have trouble showing the fact that $\varepsilon$ had to be $1 < \varepsilon < 2$. I do see why and that it is a subgame-perfect Nash equilibrium, but how do I show it.

**My approach:**
Consider deviations by player 1, given that player 2 sticks to his strategies in his subgames following histories that end in each of the four outcomes of the game. (By symmetry the below would also account for deviations by player 2 if player 1 sticks to his strategies).

We know that this game is played $T$ times. Let $m$ denote the number of cooperation stages (i.e. number of times the strategy profile $(C,C)$ has been played), let $n$ denote the number of punishment stages (number of times the strategy profile $(D,D)$ has been played), and let $T = m + n +1$. If the players cooperate $T$ times they get $3$ in payoffs. Thus, we can write
$$3T=3(m+n+1)$$
If player 1 deviates at any stage of the game he'll get a payoff of
$$3 \cdot m + (4 - \varepsilon) + 2 \cdot n$$
For the deviation to be profitable for player 1 the deviation payoff has to be greater than or equal to the cooperative payoff. That is,
$$3 \cdot m + (4 - \varepsilon) + 2 \cdot n \geq 3m + 3n + 3$$
We can simplify this
$$n \leq 1 - \varepsilon$$

And this is where I'm stuck. What am I supposed to interpret from this? That when $\varepsilon \in (1,2)$ then the number of punishment stages is negative?

Another question I have is: Assume that the question above got answered. Is that enough? Or do I need to show something more? Because I'm well aware that playing $(C,C)$ $T$ times is a SPNE and playing $(D,D)$ $T$ times is also a SPNE and either way there is no profitable deviations.

So, to summarize: I would like to know if I'm approaching this problem in the right way. I would like to get some help on how to move forward from where I got stuck. And if the above would be enough to show what the problem is asking for.

Many thanks in advance!

[Update]----------------
Below is my updated attempt on a solution with some inspiration from M.J. Osbourne and the feedback I got from @username_1.
[![enter image description here](https://i.sstatic.net/rGSZG.png)](https://i.sstatic.net/rGSZG.png)
[![enter image description here](https://i.sstatic.net/svhd3.png)](https://i.sstatic.net/svhd3.png)
[![enter image description here](https://i.sstatic.net/j4XqS.png)](https://i.sstatic.net/j4XqS.png)

**Tags:** game-theory, self-study, nash-equilibrium, dynamic-games, repeated-games

> **Comments:**
> **username_1**: This doesn't change your conclusion qualitatively, but I'd suggest being more explicit about the steam of payoffs after a deviation. For example, suppose player 1 deviates at time $t$, and the outcome at $t-1$ was $(C,C)$. Then the continuation payoff from $t$ onwards would be $4-\epsilon+2(T-t)$. Compared to the continuation payoff from not deviating at $t$, which is $3(T-t+1)$, deviation is not profitable for any $t\le T$ as long as $\epsilon >1$.  
> **username_0**: Ah I think I understand. If let's say I would apply the same method you used to for example ($C,D$). If we suppose player 1 deviates at time $t$, and the outcome at time $t-1$ was ($C,D$). Then the continuation payoff from $t$ onwards would be $2-\varepsilon + 2(T-t)$. If he doesn't deviate it would be a stream of 0's from time from time $t$, which is $0(T-t+1)$. Thus, we have that deviation is profitable if $2-\varepsilon 2(T-t) \geq 0T \quad \Rightarrow \quad 0 \leq 2 - \varepsilon + 2T - 2t$. That is, a deviation will be profitable for any $t \leq T$ as long as $\varepsilon < 2$.  
> **username_1**: Well, if the outcome at $t-1$ is $(C,D)$, then the given strategy prescribes playing $D$ at $t$ for player 1. So playing $C$ is deviating, whereas playing $D$ is not. Also, since we only need to consider one-shot deviations, if player 1 deviates to playing $C$ at $t$, he should revert back to the original strategy from $t+1$ onwards, i.e. playing $D$ from $t+1$ till $T$.  
> **username_0**: It might be a dumb question but doesn’t the outcome $(C,D)$ at $t-1$ count as a deviation? If the play was $(C,D)$ at time $t-1$ and again at time $t$ wouldn’t it be two deviations then?  
> **username_1**: The one stage deviation is for *every subgame*. So in any subgame that comes after the outcome $(C,D)$, we verify that there is no profitable one-shot deviation within that subgame, even though the subgame in question is clearly off-equilibrium path.  
> **username_0**: Aaah okay, that was what was bothering and confusing me. Thank you so much. Your help is very much appriciated.  

---

## 1 answers

### Answer by username_1 on 2021-01-17. Score: 6 | Accepted answer

A couple hints.

1. Regarding the lower bound on $\epsilon$: What happens if deviation occurs at stage $T$? In other words, there is no opportunity for your so-called "punishment stages".
2. Regarding the upper bound on $\epsilon$: Suppose player 2 deviates at stage $T-1$ but player 1 does not. What must be true about $\epsilon$ in order for player 1 to have an incentive to switch to $D$ at the very last stage?

---

In general, to show that the given strategy profile is subgame perfect, you need to argue that playing $C$ after $(C,C)$ and $D$ after either $(C,D),(D,C),(D,D)$ are optimal.

> **Comments:**
> **username_0**: Correct me if I'm wrong. First, in the last stage, $T$, if either player deviates it will yield the deviating player $4-\varepsilon$ which is less that 3 for $\varepsilon > 1$. Thus, it will not be a profitable deviation in the last stage. Second, if player 2 deviates at $T-1$ then he'll gain $4-\epsilon$ and $2$ at $T$ i.e. the total payoff of the last two stages for player 2 is $4-\varepsilon + 2$. On the other hand, if player 2 deviates such that $(C,D)$ is played at time $T-1$, then player 1 would want to deviate in the last stage, $T$. Thus, he'll gain $0+2-\varepsilon$.  
> **username_1**: @JustinMalik: Yes, you are on the right track. To ensure that player 1 switches to $D$ after player 2's deviation at $T-1$, you'd need $0+2-\varepsilon > 0+0$, the payoff from not switching. Of course, to conclude that the given strategy profile is subgame perfect, you'd still have to argue that any one-shot deviation at earlier stages, either on or off equilibrium path, is not profitable.  
> **username_0**: Thank you so much. I think I understand now. I've updated the post. Would you have a look at the solution I proposed to see if I indeed understood it the right way?  

---

