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 $p_i<p_j$, $\pi(p_i,p_j)=\frac{p_i-c}{2}$ if $p_i=p_j$ and $\pi(p_i,p_j)=0$ if $p_i>p_j$.

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 $c_1<c_2$. 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:

1. 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 $\epsilon$.
2. It seems like we can't have an equilibrium with $p_1=c_2-\epsilon< c_2\leq p_2$ 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?

• Your reasoning is correct, there is no pure equilibrium. But are you also asking whether a mixed strategy equilibrium actually exists? (Nash's existence theorem does not apply in this case because of the discontinuity of the payoff functions.) – Giskard Oct 29 '15 at 14:23
• If there is a minimum increment in prices, e.g. a cent, then your previous intuition would be correct. – Herr K. Oct 29 '15 at 16:26
• @densep I think that existence of a mixed strategy equilibrium should be guaranteed by Theorem 5 here: web.mit.edu/~nstein/Public/Game%20Theory%20Papers/… – Ubiquitous Oct 29 '15 at 16:33
• @Ubiquitous I would have referred you to the same paper :) – Giskard Oct 29 '15 at 17:18
• Theorem 5 of Dasgupta and Maskin does not apply. The sum of the payoff functions, here aggregate profit, is not upper semicontinuous at points where both firms charge the same price. If the high-cost firm would slightly raise their price, aggregate profit would jump up. Indeed, the industry revenue would not change but the industry average cost would jump down. – Michael Greinecker Sep 25 '17 at 12:33

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 $\epsilon>0$. For $\epsilon<c_2-c_1$, 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 $0<\delta<\epsilon$ 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 $$(1-\delta/\epsilon)(K+\delta)>K,$$ which we can rearrange to get $$\delta(1-K/\epsilon-\delta/\epsilon)>0,$$ which is equivalent to $$K/\epsilon+\delta/\epsilon<1.$$ If $\epsilon< c_2-c_1=K$, this can never be the case, so we obtain a Nash equilibrium if $\epsilon>c_2-c_1$.

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.

It is my impression that this has been formalized under the $\varepsilon$-equilibrium concept ("epsilon-equilibrium"). It is even called "approximate-Nash" equilibrium.

Shamelessly copying from the relevant wikipedia article (which includes some literature references)

=== The standard definition ===

Given a game and a real non-negative parameter $\varepsilon$, a strategy profile is said to be an $\varepsilon$-equilibrium if it is not possible for any player to gain more than $\varepsilon$ in payoff dominant equilibrium|expected payoff by unilaterally deviating from his Strategy. Every Nash Equilibrium is equivalent to an $\varepsilon$-equilibrium where $\varepsilon = 0$.

Formally, let $G = (N, A=A_1 \times \dotsb \times A_N, u\colon A \to > R^N)$ be an $N$-player game with action sets $A_i<$ for each player i and utility function $u$. Let $u_i (s)$ denote the payoff to player $i$ when Strategy s is played. Let $\Delta_i$ be the space of probability distributions over $A_i$. A vector of strategies $\sigma \in \Delta = \Delta_1 \times \dotsb > \times \Delta_N$ is an $\varepsilon$-Nash Equilibrium for $G$ if :

$$u_i(\sigma)\geq u_i(\sigma_i^{'},\sigma_{-i})-\varepsilon\;\;\; \forall \sigma_i^{'} \in \Delta_i,\;\; i \in N$$

This is tied to a probabilistic setting, but it appears that it can be used more generally.

The article even mentions that

"In Economics, the concept of a Pure strategy epsilon-equilibrium is used when the mixed-strategy approach is seen as unrealistic. In a pure-strategy epsilon-equilibrium, each player chooses a pure-strategy that is within epsilon of its best pure-strategy."

To not kid ourselves, this is just a formalization that attempts to hide under the carpet the fact that "infinitesimal deviations" may bug mathematicians but they don't bug economists, which know that such infinitesimal deviations play no role in determining real world economic situations.

I think standard in Bertrand competition with different constant marginal cost is another assumption in case of equal prices. Instead of sharing demand equally, you could assume that in case of equal prices the more efficient firms supplies the entire demand. As a result, all price pairs $$p_1=p_2=p$$ with $$p \in [c_1,c_2]$$ constitute an equilibrium. All prices $$p \in [c_1,c_2)$$ are a bit weird because firm 2 prices below marginal cost. However, that does not matter, because --by assumption-- firm 1 serves the entire demand. As an alternative, you can discretize the price space (e.g., prices are a multiple of, say, $$\varepsilon =1$$cent). Then you can construct similar equilibria in which only firm 1 trade. That is, prices $$p_1 =p$$ and $$p_2=p+\varepsilon$$ for any $$p \in [c_1,c_2]$$ constitute an equilibrium.

Interesting question. You will have to make the assumption that the second firm is indifferent between not producing and earning 0 profit. Actually, the way you have set it up, the firm makes 0 profit anyway. If the case is that $c_{1}<c_{2}$ , then case 1) that you have mentioned is the Nash Equilibrium. This is similar to a second price sealed bid auction with known valuation- the player with the highest valuation bids the second highest player's valuation so that the second player is indifferent between bidding 0 and her own valuation.

Let me point out that if you don't make the indifference assumption, then many games may not have a Nash equilibrium given the always possible epsilon deviations.

• I don't see why the indifference assumption would yield a Nash-equilibrium. What would firm 1 bid? Aren't you also assuming the existence of entry decisions before the price decisions? – Giskard Oct 29 '15 at 14:24
• Because given a 0 profit (which would result if firm 1 were to set the price at firm 2's cost yielding firm 2's profit = c2-c2/2=0), firm 2 should be indifferent between pricing anything more than p2. Put differently, even if your original set up, isn't p1=c and p2>c a NE? Both firms make 0 profit in eqm and cannot profitably deviate. – ChinG Oct 29 '15 at 14:26
• So, a NE in the example with the first firm more efficient than the second can be p1=c2 and p2= x>c2. Firm 2 cannot deviate optimally. – ChinG Oct 29 '15 at 14:32
• @ChinG But suppose $p_1=c_2$ and $p_2=x>c_2$. Then firm 1 has a profitable deviation to $p_1\in(c_2,x)$ so I don't think that is an equilibrium. – Ubiquitous Oct 29 '15 at 16:32
• Yes, you are right. My apologies. Intuitively, I don't think there is even a NE in mixed strategies in this case. – ChinG Oct 29 '15 at 16:46