Question

Game Theory class

5. HOTELLING COMPETITION (16 POINTS) Consumers are uniformly distributed along a boardwalk that is 1 mile long. They all like ice cream the same and dislike walking the same. Prices are regulated and equal for every vendor. The cost of producing ice cream is zero. If more than one vendor is at the same location, they split the business evenly (similarly, if two vendors are at the same distance, the consumer goes to each of them with the same probability). Assume that at the regulated prices the maximum distance that a consumer is willing to walk is 1 mile. 1. (4 points) Consider a game in which two ice-cream vendors pick their locations simulta- ncously. Write down the utility function of cach vendor. Find the pure strategy NE of the game.

Answer 1

Let us assume x1 to be the location of vendor 1 and x2 to be the location of vendor 2

Now We can associate a strategy for player i with xi ∈ [0, 1]. So the first step is to find the payoff function for each of the vendors. Also,Since the price of the ice cream is regulated, profit of each vendor can be identified with the number of customers he gets.

Suppose x1 < x2. In such case, all consumers located to the left of (x1 + x2)/2 will purchase from vendor 1, while all consumers located to the right of (x1 + x2)/2 will purchase from vendor 2.

Also, u1(x1, x2) = (x1 + x2)/ 2 u2(x1, x2) = 1 − (x1 + x2)/2

Similarly, for x1 > x2, u1(x1, x2) = 1 − (x1 + x2)/ 2 u2(x1, x2) = (x1 + x2)/ 2

Now, if x1 = x2, the vendors will split the business such that u1(x1, x2) = u2(x1, x2) = 1/2.

It is then easily visible to check that x1 = x2 = 1/2 constitutes a Nash equilibrium. No vendor can do better by deviating.

To show uniqueness, suppose first that x1 = x2 < 1/2. Then any vendor can do better by moving ε > 0 to the right, since he will sell almost 1 − x1 > 1/2 units rather than 1/2 units.

Similarly, it can be shown that x1 = x2 > 1/2 does not constitute a Nash equilibrium.

Suppose now let us suppose that x1 < x2. in such case, vendor 1 can be in a  better position by moving to x2−ε, with ε > 0, hence this cannot be a Nash equilibrium. Similarly, it can be shown that x1 > x2 does not constitute to be a Nash equilibrium