# Nash equilibrium of a game

I've a doubt about the following game and I'm new with Game-Theory. We have two road managers that aim to maximize their profit. Let be $$i=1,2$$ the players and respectively $$\omega_1,\omega_2\in [0,1]$$ possible action. Let be $$u_i$$, profit of player $$i$$:

1. $$u_1(\omega_1,\omega_2)=\omega_1(1-\frac{2}{3}\max(0,\omega_1-\omega_2))$$.
2. $$u_2(\omega_1,\omega_2)=\omega_2(\frac{2}{3}\max(0,\omega_1-\omega_2))$$.

Where $$\omega_j$$ represents toll on player $$i$$'s road.

I've got to find Nash Equilibria of the game. I've found that $$[1,1/2]$$ is the unique Nash equilibrium, is that correct?

• If player 1 plays 1, isn't player 2's best response to play $1\over2$? Commented Jan 9, 2023 at 7:42
• sure, I wrote wrong number, I've corrected it Commented Jan 9, 2023 at 18:24

If $$\omega_1\le\omega_2$$, we have $$\max(0,\omega_1-\omega_2)=0$$ and thus $$u_1=\omega_1$$, $$u_2=0$$. If $$\omega_1\ge\omega_2$$, we have $$\max(0,\omega_1-\omega_2)=\omega_1-\omega_2$$ and thus $$u_1=\omega_1(1-\frac23(\omega_1-\omega_2))$$ and $$u_2=\frac23\omega_2(\omega_1-\omega_2)$$.

Thus, the payoff for player $$1$$ increases with $$\omega_1$$ until $$\omega_1=\omega_2$$, so there can be no Nash equilibrium with $$\omega_1\lt\omega_2$$. The derivatives for $$\omega_1\ge\omega_2$$ are

$$\frac{\partial u_1}{\partial\omega_1}=1-\frac43\omega_1+\frac23\omega_2$$

and

$$\frac{\partial u_2}{\partial\omega_2}=\frac23(\omega_1-2\omega_2)\;.$$

For a Nash equilibrium in the interior, $$0\lt\omega_2\lt\omega_1\lt1$$, these must both vanish. Solving the system of linear equations yields $$\omega_1=1$$ and $$\omega_2=\frac12$$. This doesn’t lie in the interior, but it’s a Nash equilibrium; both strategies are best responses to each other. The payoffs are $$\frac23$$ and $$\frac16$$, respectively.

Now we need to check the remaining boundary cases. Since we’ve excluded $$\omega_1\lt\omega_2$$ and we already found $$\omega_1=1$$ in solving the interior case, the only remaining boundary cases are $$\omega_2=0$$ and $$\omega_2=\omega_1$$. Neither of these yields a Nash equilibrium, so it seems you’ve correctly identified the one Nash equilibrium of the game.