<p>Converting my comments into an answer... </p>

<p>In the infinitely repeated game, if the players want to collude on $(L,L)$, then there needs to be an incentive for each to stick to the plan, i.e. a punishment if one of them deviates from playing $L$.</p>

<p>As you have figured out, this amounts comparing the discounted payoffs from playing $(L,L)$ with a one-shot deviation of playing $H$ against $L$, and then playing $(H,H)$ forever. </p>

CC BY-SA 3.0
2017-05-29T06:06:30.963Z
2017-05-29T06:06:30.963Z
Herr K.
<p>Two firms can produce either low (L), medium (M) or high (H) quantity. The payoff matrix is given by:</p>

<p><a href="https://i.sstatic.net/gKHNw.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/gKHNw.png" alt="enter image description here"></a></p>

<ol>
<li>What is the outcome of this game if firms play only once?</li>
<li>Suppose the game is played for infinite periods and $\delta = 1/2$. Is it possible for firms to collude and choose $(L, L)$? Is it possible for firms to collude and choose $(M, M)$?</li>
</ol>

<p>From the picture we can see that if the firms play once then the outcome is $(H,H)$. However I don't really know how to solve the second part and would really appreciate some help. What I do know in repeated infinite games that may be useful for this problem is the following.</p>

<p>Lets take the case of deviating from $(H,H)$ to $(M,M)$ for firm 1. </p>

<p>Equilibrium payoff = $3+3\delta+3\delta^2...=\frac{3}{1-\delta}$</p>

<p>Deviation payoff (I think that if firm 1 deviated and played $M$ then firm 2 would play L since they have higher payoff $(10)$ and so we would have)= $5+2\delta+2\delta^2... = 5+\frac{2\delta}{(1-\delta) }  $</p>

<p>Where $\delta$ is the interest rate after the first round. I feel that this portion is irrelevant, however, since we are talking about collusion and should consider both firms, instead of just firm 1's strategy. </p>

CC BY-SA 3.0
2017-05-28T20:24:03.937Z
Hint: in the infinitely repeated game, if the players want to collude on $(L,L)$, then there needs to be an incentive for each to stick to the plan, i.e. a punishment if one of them deviates from playing $L$.
Herr K.
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2017-05-28T22:41:34.903Z
If P1 plays L P2 will respond by playing H. If P2 plays L P1 will respond by playing H. There is no incentive for either player to stick to (L,L) as there are higher payoffs. Similarly, if P1 plays M P2 will respond by playing H. If P2 plays M P1 will respond by playing H. Again, there is no incentive for either player to stick to (M,M) as there are alternate greater payoffs. So does that mean that they will never collude to chose (L,L) or (M,M)? I don't think this is fully correct, since I did not implement $\delta=\frac{1}{2}$ in my reasoning.
thisisme
CC BY-SA 3.0
2017-05-29T00:00:07.450Z
You're reasoning as if they are playing the one-shot game. What if they play the game repeatedly? What is the payoff from playing $(L,L)$ every period, and what is the payoff if one deviates from this?
Herr K.
CC BY-SA 3.0
2017-05-29T01:13:08.497Z
I see. To play (L,L) repeatedly results in $10+10\delta+10\delta^2...=\frac{10}{1-\delta}$. Now for deviating, suppose P1 plays L but P2 deviates and plays H then the long run payoff will be $20+0\delta+0\delta^2...=20$ We know that $\delta=\frac{1}{2}$ and so equilibrium will be 20 and deviating will be 20. So does that mean P2 is indifferent between replying by L or deviating?
thisisme
CC BY-SA 3.0
2017-05-29T01:19:30.607Z
I made a mistake above. I think the following is correct. I see. To play (L,L) repeatedly results in $10+10\delta+10\delta^2...=\frac{10}{1-\delta}$. Now for deviating, suppose P1 plays L but P2 deviates and plays H then the long run payoff will be $20+3\delta+3\delta^2...=20+\frac{6\delta}{(1-\delta)}$ We know that $\delta=\frac{1}{2}$ and so equilibrium will be 20 and deviating will be 9. As we can see if P1 chooses L then P2 will not deviate since the payoff of 20 is greater than 9.
thisisme
CC BY-SA 3.0
2017-05-29T02:44:45.483Z
Your deviation payoff doesn't seem right. I got $20+3\frac{\delta}{1-\delta}=23>20=10\frac{1}{1-\delta}$
Herr K.
CC BY-SA 3.0
2017-05-29T02:53:38.510Z
Ah I made calculation errors, but I now understand how to finish. Thank you for the help
thisisme
<game-theory><industrial-organisation><repeated-games>
Collusion, deviation from equilibrium
thisisme
# Collusion, deviation from equilibrium

(Asked by: username_0 on 2017-05-28)

Two firms can produce either low (L), medium (M) or high (H) quantity. The payoff matrix is given by:

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

1. What is the outcome of this game if firms play only once?
2. Suppose the game is played for infinite periods and $\delta = 1/2$. Is it possible for firms to collude and choose $(L, L)$? Is it possible for firms to collude and choose $(M, M)$?

From the picture we can see that if the firms play once then the outcome is $(H,H)$. However I don't really know how to solve the second part and would really appreciate some help. What I do know in repeated infinite games that may be useful for this problem is the following.

Lets take the case of deviating from $(H,H)$ to $(M,M)$ for firm 1.

Equilibrium payoff = $3+3\delta+3\delta^2...=\frac{3}{1-\delta}$

Deviation payoff (I think that if firm 1 deviated and played $M$ then firm 2 would play L since they have higher payoff $(10)$ and so we would have)= $5+2\delta+2\delta^2... = 5+\frac{2\delta}{(1-\delta) } $

Where $\delta$ is the interest rate after the first round. I feel that this portion is irrelevant, however, since we are talking about collusion and should consider both firms, instead of just firm 1's strategy.

**Tags:** game-theory, industrial-organisation, repeated-games

> **Comments:**
> **username_1**: Hint: in the infinitely repeated game, if the players want to collude on $(L,L)$, then there needs to be an incentive for each to stick to the plan, i.e. a punishment if one of them deviates from playing $L$.  
> **username_0**: If P1 plays L P2 will respond by playing H. If P2 plays L P1 will respond by playing H. There is no incentive for either player to stick to (L,L) as there are higher payoffs. Similarly, if P1 plays M P2 will respond by playing H. If P2 plays M P1 will respond by playing H. Again, there is no incentive for either player to stick to (M,M) as there are alternate greater payoffs. So does that mean that they will never collude to chose (L,L) or (M,M)? I don't think this is fully correct, since I did not implement $\delta=\frac{1}{2}$ in my reasoning.  
> **username_1**: You're reasoning as if they are playing the one-shot game. What if they play the game repeatedly? What is the payoff from playing $(L,L)$ every period, and what is the payoff if one deviates from this?  
> **username_0**: I see. To play (L,L) repeatedly results in $10+10\delta+10\delta^2...=\frac{10}{1-\delta}$. Now for deviating, suppose P1 plays L but P2 deviates and plays H then the long run payoff will be $20+0\delta+0\delta^2...=20$ We know that $\delta=\frac{1}{2}$ and so equilibrium will be 20 and deviating will be 20. So does that mean P2 is indifferent between replying by L or deviating?  
> **username_0**: I made a mistake above. I think the following is correct. I see. To play (L,L) repeatedly results in $10+10\delta+10\delta^2...=\frac{10}{1-\delta}$. Now for deviating, suppose P1 plays L but P2 deviates and plays H then the long run payoff will be $20+3\delta+3\delta^2...=20+\frac{6\delta}{(1-\delta)}$ We know that $\delta=\frac{1}{2}$ and so equilibrium will be 20 and deviating will be 9. As we can see if P1 chooses L then P2 will not deviate since the payoff of 20 is greater than 9.  
> **username_1**: Your deviation payoff doesn't seem right. I got $20+3\frac{\delta}{1-\delta}=23>20=10\frac{1}{1-\delta}$  
> **username_0**: Ah I made calculation errors, but I now understand how to finish. Thank you for the help  

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## 1 answers

### Answer by username_1 on 2017-05-29. Score: 1 | Accepted answer

Converting my comments into an answer...

In the infinitely repeated game, if the players want to collude on $(L,L)$, then there needs to be an incentive for each to stick to the plan, i.e. a punishment if one of them deviates from playing $L$.

As you have figured out, this amounts comparing the discounted payoffs from playing $(L,L)$ with a one-shot deviation of playing $H$ against $L$, and then playing $(H,H)$ forever.


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