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This is a question I encountered during my undergraduate studies in Industrial Economics. So the question goes like this:

Consider two identical firms with constant marginal cost $c$ which compete in quantities in each of an infinite number of periods. The quantities chosen are observed by both firms before the next play begins. Inverse demand is given by $p = 1 – q_{1} – q_{2}$. The firms use ‘trigger strategies’ and they revert to static Cournot behaviour if cooperation breaks down.

  1. What is the lowest value of the discount factor $\delta$ such that the firms can sustain the monopoly output level?

This part I can answer and the value of $\delta$ needed to sustain the monopoly output level turns out to be $\frac{9}{17}$.

Now onto the second part of the question:

  1. Suppose $\delta$ is too small to sustain the monopoly output. In particular, suppose $\delta = \frac{1}{2}$. What is the most profitable subgameperfect equilibrium that can be sustained using trigger strategies? (Assume $c = 0$ for simplicity.)

So my line of thinking for this is, to sustain collusion the profit from colluding has to exceed the profit from defecting one period plus the punishment profit in subsequent periods.

$$ \pi^{C}\frac{1}{(1-\delta)} \geq \pi^{D} + \pi^{P}\frac{\delta}{(1-\delta)} \tag{1} $$

Having from secion 1 of the question:

$$ \pi^{C} = \frac{(1-c)^2}{8},\quad \pi^{D} = \frac{9(1-c)^2}{64},\quad \pi^{P} = \frac{(1-c)^2}{9} \tag{2} $$ Take into account that $\pi^{C}$ is the profit each firm obtains if they collude, i.e. monopoly profit divided by 2.

So, to me section 2 of the question is asking about $\pi^{C}$ (in condition $(1)$) when $\delta=\frac{1}{2}$. I know $\pi^P$ remains the same since it is the Cournot profit but I guess $\pi^D$ changes because it was calculated assuming the other firm would produce the monopoly level of output and this is not the case anymore since $\delta$ is lower than what is needed to maintain that level of output.

You can see I am a bit confused. Can anyone help me?

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  • $\begingroup$ Could you transform the hint an answer to your question? $\endgroup$ – Bayesian Jan 20 '17 at 19:10
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I do not want to do all the algebra, but I give you a hint. You are correct: You are asked about $\pi^C$ in your Condition (1) and you are also right that $\pi^D$ changes.

In your parameter setting with $c=0$: $$\pi^C := \pi (q^C, q^C) = (1- 2 q^C) q^C,$$ where $q^C$ is the quantity they coordinate on.

Then $\pi^D$ is the profit from best-responding to $q^C$ with some $q^D$: $$ \pi^D := \pi (q^D, q^C) = (1- q^C-q^D)q^D,$$ i.e., $q^D = \arg max \quad \pi^D$. Now you solve $q^D$ as a function of $q^C$ and plug that into $\pi^D$. Having Condition (1) hold with equality gives you some $q^C$ that you can plug into $\pi^C$ again.

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  • $\begingroup$ First of all thanks for taking the time to answer my question. I got an answer, I don't know if the right one. In the end I got a quadratic equation for $q^C$ and two postive solutions: $\frac{1}{3}$ and $\frac{13}{51}$ and plugging those into the $\pi^C$ equation I get a higher profit for the $\frac{13}{51}$ quantity, namely $\pi^C=\frac{325}{2601}$. Did you solve it yourself? If so, what did you get? $\endgroup$ – soltzu Jan 21 '17 at 17:39
  • $\begingroup$ Looks good to me. $\endgroup$ – Bayesian Jan 23 '17 at 10:08

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