# 2 player Nash equilibrium game

I am having difficulty solving this question and would be glad if someone can suggest an approach to this problem.

Consider 2 players playing a simultaneous game where each player chooses $x_i>0,i= 1,2$. Each player receives a payoff of $\pi_i(x_1,x_2)=2x_i +2x_1x_2-x_i^2$.

What are the pure strategy Nash equilibria? If the set of strategies is limited to [0,N], $N >0$, what are the pure strategy Nash equilibria now?

I tried to differentiate $\pi_i$ wrt. $x_i$ and $x_2$ and solve the 2 equations simultaneously but I obtained $x_2=x_2+2$

$$\dfrac{\partial}{\partial x_1} \pi_1(x_1, x_2) = 2 + 2 x_2 - 2 x_1$$ Thus player $1$ has an incentive to increase his pure strategy whenever $x_1 - x_2 < 1$. Similarly, player $2$ has an incentive to increase his pure strategy whenever $x_1 - x_2 > -1$. Since for every $(x_1, x_2) > 0$ at least one of these is the case, there are no pure strategy Nash equilibria.
In the modified game, if $x_i = N$ player $i$ can't increase his pure strategy. Thus we have a pure strategy Nash equilibrium with $x_1 = x_2 = N$.
• Just to clarify the solution to the modified game -- the reason why player $i$ can't increase his pure strategy if $x_i=N$ is because N is the maximum value and cannot increase further? Also, why do we reach a Nash equilibrium of $x_1=x_2=N$ ? Is it because both players will choose the max value of N to maximise their payoff? Nov 1 '17 at 6:51