Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

The problem is summarized as:

There are two players. Player 1's strategy is h. Player 2's strategy is w. Both of their strategy sets are within the range [0,500].

Player 1's payoff function is:

$ P_h(h, w) = 50h + 2hw-\frac{1}{2}(h)^2 $

Player 2's payoff function is:

$ P_w(h, w) = 50w + 2hw - \frac{1}{2}(w)^2 $

Find a Nash Equilibrium.

I was taught to solve these problems in the following way. Find the first derivative of Player 1's payoff function with respect to h, equate it to 0, then solve for h, and then repeat for Player 2 but with respect to w and solving for w instead. However, I found the first derivatives to be:

$ P_h(h, w)^\prime = 50 + 2w - h $

$ P_w(h, w)^\prime = 50 + 2h - w $

Now after equating these first derivatives to 0 and solving for h and w, we get that h = -50 and w = -50. The issue now is that these strategies aren't within the strategy set [0,500] as mentioned in the problem question. Where am I going wrong?

share|improve this question
1  
As with any optimization problem, you should check the boundary. The optimal strategy is (500,500), but the players would choose higher numbers if they were allowed to do so. (That is, as Paxinum pointed out, the derivatives are not 0 at (500,500)). –  Théophile Jul 24 '12 at 20:00
    
@Théophile For this particular problem, what do you mean by checking the boundary? Do you mean checking at 0 and at 500 for both players, or one for each, or am I completely missing the point here? (I believe it's the last one.) –  Kevin Jul 25 '12 at 7:30
1  
Almost, Kevin: don't just check the four corners, though, but all four edges. One of the edges, for example, is $h = 0$. The derivative for Player 1 is then $P_h(h,w)^\prime = 50 + 2w$, which is positive regardless of the value of $w$. In other words, Player 1 has no incentive to stay at $0$. As for Player 2, $P_w(h,w)^\prime = 50 - w$, which is $0$ when $w=50$. However, this latter calculation wasn't really necessary because Player 1's dissatisfaction here means that there are no equilibria along $h=0$. Try similar calculations along the other edges. Does that make sense? –  Théophile Jul 25 '12 at 16:18
add comment

1 Answer

up vote 1 down vote accepted

(I have not studied Nash equilibria before, but I'll take a stab at this anyway).

Ok, so say h is on the x axis, and w is on the y axis. Both players are trying to maximize the profit, i.e. they change their strategies according to the derivatives of the payout functions.

Plotting this stream plot, we get the graph below:

Strategies streamplot We see that in the entire region, at least one of the derivatives is always positive. Thus, at least one player gains on increasing w or h. (We see this because each arrow points either right or up or both).

From this, we can see that the point (h,w)=(500,500) is an equilibrium, and you can verify this by seeing that both the derivatives in this point are positive, and even more, $P_h(h,500)>0$ for all $h\in[0,500]$, and similarly $P_w(500,w)>0$ for all $w \in [0,500].$ Thus, no player would gain on changing the strategy if they are in (500,500).

share|improve this answer
    
Thanks for your help! This answer seems correct, but, if you don't mind, I'd like to wait to see if anyone else has a solution that doesn't require graphing (because I won't have the time to graph the functions during an exam). –  Kevin Jul 24 '12 at 16:00
    
Ah, well, I think the general way to solve it then, should be to look for points you mentioned. If there aren't any in the allowed region, you need to examine the edges of the region, and then the corners, just like in a 2-variable optimization problem on a compact set. –  Paxinum Jul 24 '12 at 20:20
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.