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

enter image description here

  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.

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  • $\begingroup$ 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 $L$. $\endgroup$
    – Herr K.
    May 28, 2017 at 20:24
  • $\begingroup$ 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. $\endgroup$
    – thisisme
    May 28, 2017 at 22:41
  • $\begingroup$ 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? $\endgroup$
    – Herr K.
    May 29, 2017 at 0:00
  • $\begingroup$ 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? $\endgroup$
    – thisisme
    May 29, 2017 at 1:13
  • $\begingroup$ 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. $\endgroup$
    – thisisme
    May 29, 2017 at 1:19

1 Answer 1

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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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