I'm solving part (b) of the following exercise in general-sum game:
My attempt:
Assume $\sigma^1 = (x,y,z)$ and $\sigma^2 = (a,b,c)$. Then $g_1(\sigma^1, \sigma^2) = 3xa+4yb+2zc$ and $g_2(\sigma^1, \sigma^2) = 4xa+3yb+5zc$.
It follows from $\sigma^1$ is a best reply against $\sigma^2$ that $3xa+4yb+2zc \ge 3x'a+4y'b+2z'c$ for any action profile $(x',y',z')$ of Player $1$. Similarly, $4xa+3yb+5zc \ge 4xa'+3yb'+5zc'$ for any action profile $(a',b',c')$ of Player $2$.
If $x=0$ then $$\begin{cases} 4yb+2zc \ge 3x'a+4y'b+2z'c \\ 3yb+5zc \ge \quad \quad \quad \, 3yb'+5zc'\end{cases}$$ for any action profile $(x',y',z')$ of Player $1$ and any action profile $(a',b',c')$ of Player $2$.
Then I'm stuck at proving $a=0$. How can I proceed to finish the proof?
Thank you so much!
You should first argue why player $1$ must have a strictly positive payoff in the given equilibrium. Moreover, every action which $\sigma^{1}$ plays with positive probability must yield the same payoff. Therefore, every action taken by $\sigma^{1}$ must yield a strictly positive payoff. This immediately implies $a = 0$ if $x = 0$, since $3 x $ is the payoff from playing $a$.
This suggests that there would be a total of seven equilibria: Three where both agents play the same action with probability one, three were both mix over the same two actions, and one where both mix over all three actions. I suspect that in these mixed equilibria the probabilites are uniquely determined as the actions all yield different payoffs (i.e. payoffs vary along the diagonal for each player). If that's indeed the case, then there would be exactly one equilibrium where both play the first two actions, exactly one where both play the second and third action, and so on.