Question

Consider this discrete Bertrand model, but assume that each student has a constant cost of 5 that is deducted from all payoffs. So whoever has the low number wins their number, minus 5. Whoever has the high number loses 5 total. In the event of a tie, each student wins an amount equal to their number divided by two, then minus five. Find any Nash equilibria in this game. Explain your reasoning. Hint: It is perfectly fine for both players to have losses in equilibrium! There are more than 1 Nash equilibria.

Answer 1

In this game, actions are restricted to the integers 0,...c−1,c,c + 1,c + 2,...

Let’s consider the possible cases:

• p1 = p2 = c. Both players are making zero profit. If player i lowered its costs to below c, it would make a negative profit. If it raised its costs to above c, it would not get any customers and hence make zero profit. As in the original game, there is no incentive to deviate, therefore (c,c) is a Nash equilibrium.
• pi < c for either player. The player with the lowest cost is making negative profit, and can improve to zero profit by setting pi = c. Not a Nash equilibrium.
• pi = p2 = c + 1. Both players are making positive profit. If player i lowered its cost, it would make zero or negative profit. If it raises its cost, it makes zero profit. Therefore, (c + 1,c + 1) is also a Nash equilibrium.
• pi = c,pj > c. Player i is making zero profit, can improve by raising its cost to c + 1. Not a Nash equilibrium.
• pi > pj ≥ c+1. Player i is making zero profit, can improve by lowering cost to equal pj. Not a Nash equilibrium.

As compared to the original formulation of Bertrand duopoly, there is an additional Nash equilibrium because we have eliminated some possible actions that would give an incentive to deviate.