Question

1. Bar A and Bar B are located across the street from each other, and are both deciding how much to charge for a pint of beer: $10 or S4. There are 100 potential customers Of these, are loyal customers of Bar A: they will each buy their pint of beer from Bar A, no matter what prices each bar charges. The remaining 100-x customers will buy their pint from whichever bar charges the lowest price. If both bars charge the same price, each of these 100 - x customers will choose randomly between the two bars so that, on expectation, half of them will end up at each bar. Payoffs for each bar are equal to expected revenues (a) Fill in the payoffs in the following table Bar B $10 $4 $10 Bar A $4 (a) True or false: "Both bars charging $10 is a Nash equilibrium for any 100-x-0." Provide a proof of your answer. (b) Fil in the blank, and show your work: "If a <both bars charging $4 is a Nash equilibrium

Answer 1

Solution:

a) Two players: Bar A and Bar B, Strategies for the game: Charge for a pint of beer: $10 or $4.

For any price x customers go to Bar A. So, x*p is a fixed revenue for Bar A, where p is the price charged (p = {4, 10})

If any one of the bars charges $4, while other charges $10, then $4 charging firm gets (100 - x)*4 (since, lower price charged, the bar gets an entire market)

Lastly, if both firms charge $4, or both firms charge $10, each bar gets half a market, thus receiving ((100 - x)*4)/2 or ((100 - x)*10)/2, respectively.

Accordingly, we can make the payoff matrix as follows:

		Bar B
		$10	$4
Bar A	$10	[(10x+((100-x)10)/2), (0*x+((100-x)10)/2] = [(5x+500), (500-5x)]	[(10x+0*(100-x)), (0x+((100-x)4)] = [(10x), (400 - 4x)]
	$4	[(4x + 4(100-x)), (0*x+0*(100-x))] = [(400), (0)]	[(4x+((100-x)4)/2), (0*x+((100-x)4)/2] = [(2x+200), (200-2x)]

where for any payoff combination [(a), (b)], a belongs to the expected payoff to Bar A, and b belongs to the expected payoff to Bar B.

b) We have to see when is $10 charged by both bars is a Nash equilibrium:

This means when Bar A charges $10, best response for bar B must be $10 as well. This further means that when bar A charges $10, expected payoff received by bar B from charging $10 must be at least as much as what it receives by charging $4. Thus, it must be that

500 - 5x >= 400 - 4x

Solving this inequality gives us: 100 >= x

Also, when Bar B charges $10, best response for bar A must be $10 as well. This further means that when bar B charges $10, expected payoff received by bar A from charging $10 must be at least as much as what it receives by charging $4. Thus, another condition to be satisfied simultaneously is:

5x + 500 >= 400

Solving this inequality gives us: x >= -20. Since x is number of customers, it cannot be negative. Thus, lower threshold that can be set for x is minimum possible non-negative integer, which is 0 itself. So, x >= 0

Combining the two solutions, we get that $10 charged by both bars is Nash equilibrium when: 0 <= x <= 100

Hence, the given statement is True.

c) Now, we have to see when is $4 charged by both bars is a Nash equilibrium:

Just like the solution in part (b), this means when Bar A charges $4, best response for bar B must be $4 as well; meaning that when bar A charges $4, expected payoff received by bar B from charging $4 must be at least as much as what it receives by charging $10. Thus, it must be that

0 <= 200 - 2x

Solving this inequality gives us: x <= 100

Also, when Bar B charges $4, best response for bar A must be $4 as well. So, when bar B charges $4, expected payoff received by bar A from charging $4 must be at least as much as what it receives by charging $10. Thus, another condition to be satisfied simultaneously is:

10x <= 2x + 200

Solving this inequality gives us: x <= 25.

Combining the two solutions, we get that $4 charged by both bars is Nash equilibrium when: x <= 25 (the common solution from both inequalities). So, if x <= 25, both bars charging $4 is a Nash equilibrium.