<p>For the first part: correct, any NE in the stage game is a SPNE in an repeated game. In fact, it is the <strong>only</strong> SPNE if the game is repeated finitely many times.</p>
<p>For the second part: to check that a strategy is SPNE you can use the one-shot deviation principle. That is, check that for any strategy there is no profitable deviation only in one particular stage. Provided there is none, then you are fine.</p>

CC BY-SA 4.0
2021-06-24T10:41:10.100Z
Hi Pedro. Thanks for the comment. Would I have to calculate the sums, i.e. what would happen if one player chose to deviate? I just have trouble with understanding how to. If you could show me how to it would be very much appreciated.
Mathias
CC BY-SA 4.0
2021-06-24T15:49:26.257Z
Exactly: take one stage and suppose the player deviates and plays another action. Assume the other player sticks to the strategy. For the following periods they will both stick to the strategy. For the case of a grim trigger strategy it would be like this: 1) player 1 plays D instead of C 2) player 2 still plays D because that's what his strategy says 3) this triggers both of them to play D on the subsequent periods. Compute the payoff of such deviation. If it is not profitable then you're fine.
Pedro Ignacio Martinez Bruera
CC BY-SA 4.0
2021-06-26T12:57:16.243Z
Hi Pedro. Would you check if my answer is alright? I posted it.
Mathias
CC BY-SA 4.0
2021-06-23T22:37:31.243Z
2021-06-23T22:37:31.243Z
Pedro Ignacio Martinez Bruera
<p>Consider the down below<a href="https://i.sstatic.net/MKKiv.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/MKKiv.png" alt="enter image description here" /></a></p>
<p>which I have trouble with solving.</p>
<p>For part 1) I have said that a possible outcome path is to play <span class="math-container">$(D,D)$</span> in the first round and for all rounds following until <span class="math-container">$i \leq 298$</span>. Afterwards play <span class="math-container">$(C,C)$</span> forever. If anyone deviates then play <span class="math-container">$(D,D)$</span> forever. This is since we have that</p>
<p><span class="math-container">$$0.01 ( \sum_{i=1}^n 0.99^{i-1} + 9 \sum_{i=n+1}^{\infty} 0.99^{i-1} ) = 1.4 \Leftrightarrow n \leq 298$$</span></p>
<p>Is this allright? However, then to show that it is a NE in 2) I have trouble with. I thought I could use a Proposition in my book saying that because <span class="math-container">$(D,D)$</span> is ne in stage game, then the strategy must be a SPNE. Is this alright?</p>
<p>For 3) I know what do just following the same as in 1) but I have no idea how to prove if it is a SPNE or not. Can you help me with that? The strategy is to play <span class="math-container">$(D,C)$</span> for <span class="math-container">$i \leq 92$</span> and <span class="math-container">$(C,C)$</span> aftwards. If anyone devaites, then play <span class="math-container">$(D,D)$</span> forever.</p>
<p>TIA.</p>

<microeconomics><game-theory><repeated-games>
How to show that a strategy is a SPNE in repeated games
Mathias
# How to show that a strategy is a SPNE in repeated games

(Asked by: username_0 on 2021-06-23)

Consider the down below[![enter image description here](https://i.sstatic.net/MKKiv.png)](https://i.sstatic.net/MKKiv.png)

which I have trouble with solving.

For part 1) I have said that a possible outcome path is to play $(D,D)$ in the first round and for all rounds following until $i \leq 298$. Afterwards play $(C,C)$ forever. If anyone deviates then play $(D,D)$ forever. This is since we have that

$$0.01 ( \sum\_{i=1}^n 0.99^{i-1} + 9 \sum\_{i=n+1}^{\infty} 0.99^{i-1} ) = 1.4 \Leftrightarrow n \leq 298$$

Is this allright? However, then to show that it is a NE in 2) I have trouble with. I thought I could use a Proposition in my book saying that because $(D,D)$ is ne in stage game, then the strategy must be a SPNE. Is this alright?

For 3) I know what do just following the same as in 1) but I have no idea how to prove if it is a SPNE or not. Can you help me with that? The strategy is to play $(D,C)$ for $i \leq 92$ and $(C,C)$ aftwards. If anyone devaites, then play $(D,D)$ forever.

TIA.

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

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## 1 answers

### Answer by username_1 on 2021-06-23. Score: 1

For the first part: correct, any NE in the stage game is a SPNE in an repeated game. In fact, it is the **only** SPNE if the game is repeated finitely many times.

For the second part: to check that a strategy is SPNE you can use the one-shot deviation principle. That is, check that for any strategy there is no profitable deviation only in one particular stage. Provided there is none, then you are fine.

> **Comments:**
> **username_0**: Hi Pedro. Thanks for the comment. Would I have to calculate the sums, i.e. what would happen if one player chose to deviate? I just have trouble with understanding how to. If you could show me how to it would be very much appreciated.  
> **username_1**: Exactly: take one stage and suppose the player deviates and plays another action. Assume the other player sticks to the strategy. For the following periods they will both stick to the strategy. For the case of a grim trigger strategy it would be like this: 1) player 1 plays D instead of C 2) player 2 still plays D because that's what his strategy says 3) this triggers both of them to play D on the subsequent periods. Compute the payoff of such deviation. If it is not profitable then you're fine.  
> **username_0**: Hi Pedro. Would you check if my answer is alright? I posted it.  

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