Nash equilibrium of a Bertrand game with different marginal costs Consider the following game of Bertrand (price competition):


There are two players, $1$ and $2$. Each has a publicly known marginal cost, $c_i$.
A strategy is a price, $p_i\in\mathbb{R}$.
Player $i$'s payoff (profit) is $\pi(p_i,p_j)=p_i-c_i$ if $p_i&lt;p_j$, $\pi(p_i,p_j)=\frac{p_i-c}{2}$ if $p_i=p_j$ and $\pi(p_i,p_j)=0$ if $p_i&gt;p_j$.




If $c_i=c_j=c$ the result is the straightforward Bertrand Nash equilibrium: both firms set $p_i=c$ and make zero profit. A higher price results in zero demand/profit; a lower price results in negative profits. There is therefore no profitable deviation.




  Now suppose $c_1&lt;c_2$. What is the Nash equilibrium?


I have previously contented myself with the intuition that the equilibrium is for firm 1 to take the whole market at a price 'slightly below' $c_2$? But looking at the technical details raises doubts in my mind:


We can't have an equilibrium with $p_1=c_2$. The best response for $2$ would be $p_2=c_2$, but then $1$'s profit is $(p_1-c_2)/2$ and $1 $ can profitably deviate to $p_1=c_2-\epsilon$ for some small $\epsilon$.
It seems like we can't have an equilibrium with $p_1=c_2-\epsilon&lt; c_2\leq p_2$ because firm 1 could do better with $p_1=c_2-(\epsilon/2)$.


Do we conclude that the only equilibrium of this game is in mixed strategies?
Yes, there is no equilibrium in pure strategies. For any price charged by firm 2 above $c_1$, firm one could only best respond by charging the largest price that is strictly smaller. which is impossible. If both firms charge at most $c_1$, one of these firms must make a loss, which cannot be a best response. So there is no Nash equilibrium in pure strategies.

There are, however, equilibria in mixed strategies. Consider an equilibrium in which firm 1 chooses a price of $c_2$, while firm two randomizes uniformly over the interval $[c_2,c_2+\epsilon]$ for some $\epsilon&gt;0$. For $\epsilon&lt;c_2-c_1$, this is a Nash equilibrium and, moreover, it uses no weakly dominated strategies. Firm 1 is making a profit of $c_2-c_1$ and firm 2 a profit of $0$. Clearly, firm 2 has no profitable deviation. To see that firm $1$ has no profitable deviation, recall that there can be no profitable deviation in mixed strategies if there is no profitable deviation in pure strategies (a very general game-theoretic fact), and observe that firm $1$ cannot profit from deviating to a price strictly below $c_2$ or weakly above $c_2+\epsilon$. So take any $\delta$ satisfying $0&lt;\delta&lt;\epsilon$ and assume that firm 1 charges $c_2+\delta$ (note that $\delta=0$ would be no deviation.) The probability that firm 2 charges a lower price is $\delta/\epsilon$, so the expected profit of firm 1 is
$$(1-\delta/\epsilon)(c_2-c_1+\delta)+\delta/\epsilon~ 0.$$
Write $K$ for $c_2-c_1$. We have a profitable deviation if
$$(1-\delta/\epsilon)(K+\delta)&gt;K,$$
which we can rearrange to get
$$\delta(1-K/\epsilon-\delta/\epsilon)&gt;0,$$
which is equivalent to
$$K/\epsilon+\delta/\epsilon&lt;1.$$ If $\epsilon&lt; c_2-c_1=K$, this can never be the case, so we obtain a Nash equilibrium if $\epsilon&gt;c_2-c_1$. 

One can actually use the same construction to construct equilibria in which firm 1 charges less than $c_2$, but in the resulting equilibrium, firm 2 must play a weakly dominated strategy. 

The construction in this answer is taken from the following short paper:

Blume, Andreas. "Bertrand without fudge." Economics Letters 78.2
  (2003): 167-168.