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In a infinitely repeated Bertrand competition, collusion is sustainable if, and only, if, the following inequality is satisfied,

$\frac{\pi}{N(1-\delta)}\geq\pi$

Where $\pi$ is equilibrium profits, and $0 < \delta < 1$ is the discount factor. If the game is finite, I make the following claim:

Collusion is not sustainable, as each firm has an incentive to deviate from collusion in the last stage of the game. However, how would I go ahead and prove this claim?

My initial attempt, was to setup the following inequality,

$\frac{1-\delta^T}{1-\delta}$ $\frac{\pi}{N}$+$\pi_{\varepsilon}$ $\delta^{T-1}$ $\geq$ $\pi$,

where $\pi_{\varepsilon} < \pi$, and is given to the deviating firm. Is this a valid approach, or should I change my approach?

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You're already on the right track. Just need to reorganize your argument a little bit.

Hint: Prove by contradiction. Suppose that collusion is sustainable in an equilibrium. Show that the collusive strategy (i.e. choosing the monopoly price in every stage) is not optimal given that everybody else is doing the same. Then conclude that collusion cannot be an equilibrium, hence a contradiction.

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  • $\begingroup$ Thank you for this hint. So if I understood you correctly, I should stick with the inequality, and assume some optimal value of $\delta$ and show that this is not satisfied? $\endgroup$ – TiredStudent Jan 26 at 19:47
  • $\begingroup$ @TiredStudent: I don't quite follow your notations, particularly the difference between $\pi_\varepsilon$ and $\pi$. Either way, your inequality is probably incorrect. If $\pi$ is the monopoly profit in a particular stage, then it is also approximately the one-time profit that a firm gets if it deviates. $\endgroup$ – Herr K. Jan 26 at 22:49
  • $\begingroup$ @TiredStudent: Given that all the other $N-1$ firms adopt the collusive strategy, you should compare the profit of the remaining firm if (a) it also adopts the collusive strategy, and (b) it colludes for the first $T-1$ periods and deviates in the last period. In scenario (a), the firm gets $\pi/N$ in all $T$ periods; in scenario (b), the firm gets $\pi/N$ in the first $T-1$ periods, and $\pi$ in the last. $\endgroup$ – Herr K. Jan 26 at 22:51
  • $\begingroup$ I guess Im living up to my alias. I thank you sincerely for your invested time. $\endgroup$ – TiredStudent Jan 27 at 12:05

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