# Sustainability of Collusion in a finite Bertrand competition with N > 2 firms

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?

• 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? Jan 26 at 19:47
• @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. Jan 26 at 22:49
• @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. Jan 26 at 22:51