Say you’re at the Clippers-Lakers game at the Staples Center and you can choose

Question

Say you’re at the Clippers-Lakers game at the Staples Center and you can choose either to yell at the top of your lungs like a maniac or to sit quietly and watch the game. If you make n minutes of noise and watch the game for w minutes, your utility is given by u(n, w) = nw. Since the game is 60 minutes long, you have the constraint n + w = 60.

a. Say that you are the only fan in the audience. What is your optimal choice of n and w?

b. Now say that there are two fans in the audience. Now the total amount of noise n is given by n = n1+n2. Person 1’s utility is u1(n, w1) = nw1 and person 2’s utility is u2(n, w2) = nw2. Of course, we have the constraints n1 + w1 = 60 and n2 + w2 = 60. Model this as a strategic form game and find the (pure strategy) Nash equilibrium. Is the Nash equilibrium Pareto efficient (can you both become better off by not playing the Nash equilibrium)?

c. Now say that there are m identical fans in the audience (each with the same utility function given above). There is a Nash equilibrium in which everyone cheers an identical amount. Find it. Does the total amount of cheering approach 60 as m increases?

Explanation / Answer

a.
   utility = nw
utility = n*(60-n)

= 60n-(n^2)

We have to maximise it.

By differentiating the above we get,

0=60-2n
n=30 which implies w=30

Maximum utility = u(30,30) = 30*30=900

b. Maximum utility of person 1 does not vary based on person 2’s utility and the vice versa is true as well. Hence the nash equilibrium would be achieved when each maximises their own utility i.e. when n=30, w=30.

Yes, it is Pareto efficient as if person 1 tries to increase his utility it does not affect person 2’s utility.

c. Everyone will cheer for 30 minutes to maximise their utility. When m=2, total cheering will be 60 and it will cross 60 when m>2.

Related Questions

Navigate

• Browse (All)
• Browse S
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.