# Bertrand Duopoly Equilibrium for Discrete Prices

• There are two identical firms, $$1$$ and $$2$$, with zero marginal costs. They produce homogenous product, which is demanded by a unit mass of identical consumers, each of which has inelastic unit demand with a reservation price of $$2$$. Prices are constrained to take only integer values. Using standard game theoretic reasoning, determine whether or not each possible pair $$(p_1,p_2)$$ can be regarded as Bertrand Nash equilibrium.

Would the Nash equilibria be $$(p_{1}^{*},p_{2}^{*})=$$ $$\{(0,0),(1,1)\}$$. For symmetric prices, $$\{(2,2)\}$$ since the demand would be $$0$$, I can see how unilateral deviation to $$1$$ could increase profits for a firm. However for prices $$\{(0,0),(1,1)\}$$, no unilateral profits are possible. For other non symmetric price pairings, unilateral deviations are possible. Therefore, the Nash equilirium prices are $$\{(0,0),(1,1)\}$$. Is my reasoning correct?

The problem formulation admits the following Normal Form representation. We can reject any strategy involving price greater than 2, as demand falls to zero and such strategies are strictly dominated by those for which prices are either 1 or 2.

    0         1         2

0 [0,0]     [0,0]     [0,0]

1 [0,0]   [0.5,0.5]   [1,0]

2 [0,0]     [0,1]     [1,1]


With this setup you can see that there are three Nash equilibria: (0,0), (1,1) and (2,2). At (0,0), firms are indifferent to the outcome of a higher price, and will not deviate. At (1,1), Firm 1 will not deviate to 2, nor will Firm 2 deviate to 2, as this cedes the market to the opponent. At (2,2) firms are indifferent to reducing price to 1.

(0,1), (0,2), (1,0), and (2,0) are strictly dominated by deviating to a price-matching strategy, splitting the market between firms.

Similarly, (1,2) and (2,1) fail to be Nash equilibria because whichever firm opens with a price of 2 can capture half the market by deviating to a price of 1.

• What's wrong with weak dominance, why are you leaving out 0? – Giskard Sep 16 '19 at 19:01
• I edited my answer. I initially left it out because I thought the result wasn't interesting - but then your comment helped me realize I was leaving out the classic Bertrand equilibrium result. – heh Sep 16 '19 at 19:17
• I think 1 is weakly dominating 2 (not the other way around), but otherwise you are spot on. Perhaps you mean 2 is weakly dominated? – Giskard Sep 16 '19 at 19:40
• Yep - that is indeed what I meant. I literally joined and started posting because my game theory has been getting rusty - I had no idea by how much! – heh Sep 16 '19 at 19:59
• Pricing above 2 is only weakly, not strictly, dominated, since profit is zero if the other firm prices at zero, regardless of what your price is. – Herr K. Sep 16 '19 at 22:28

Firstly, at reservation prices, it's not compulsory that demand will be 0. The consumers will be indifferent between buying and not buying the good. Let q* be the demand by the mass when price is 2 or less. If you take out the profit function of both the firms, you will see that at p1=2, profit received by firm 2 will be the same if P2=1,2. Because costs are 0, it will share the market at p2=2 and gain profit equal to q* and at p2= 1 it will gain the entire market and make profit equal to q* again. When p1=1, firm 2's best response is p2=1. When p1=0, firm 2's best response is p2 is a member of N, where N is a natural number (0 is a natural number). This is because firm 2 will earn profits 0 for any price. The best response of firm 1 will be analogous because cost functions are the same. Therefore Nash Equilibriums are {(0,0),(1,1),(2,2)}. For the reservation price P1=2, suppose the demand is 0, then the best response by firm 2 will be p2 is a member of N again, because it will earn 0 profits at all N. Also, when p1>2, best response of firm 2 will be p2=2, but there will be no intersection of best responses beyond (2,2).

• Why do you think demand will be the same $q^*$ when price is 2 or less, given that consumers have inelastic unit demand? – Herr K. Sep 16 '19 at 18:53
• You're right, I mistakenly took it for perfectly inelastic demand under 2. – ZlatanIbrahimovic Sep 17 '19 at 4:06
• @Herr K. Don't you think that inelastic unit demand at reservation price 2 means that the consumers will demand 1 unit at price less than 2 and 0 at prices more than 2? Isn't that why 2,2 is also a Nash equilibrium because the firm will get same profits at p1=P2=2 and p1=2, P2=1? – ZlatanIbrahimovic Sep 21 '19 at 19:33
• Yes, (2,2) is a NE if we assume that all consumers (i.e. the entire unit mass of them) will purchase at price equal to 2. But when you use $q^*$, I thought you meant any $q^*\in[0,1]$, which is okay when consumers are indifferent when price is 2, but when price is 1 or 0, all consumers should make the purchase. – Herr K. Sep 21 '19 at 21:29
• K thank you so much – ZlatanIbrahimovic Sep 24 '19 at 4:15