Consider the game shown in Figure 4.1 , and suppose that player X plays Abate with probability px ∈ [0,1] and that player Y plays Abate with probability py ∈ [0,1]. Of course, since both players face a binary choice, X and Y must play Pollute with probability (1 − px) and (1 − py), respectively.
Player X's ex ante payoff isor(4A.1) By inspection, you can see X will want to set px = 0 if (1 − 2py) is negative and px = 1 if (1 − 2py) is positive. If py = 0.5, X will not care how it chooses px.
Country Y's ex ante payoff can be shown to be(4A.2)As above, it is easy to see that Y will want to set py = 0 if (1 − 2px) is negative and py = 1 if this term is positive. If px = 0.5, Y will not care how it chooses py.
Suppose Y plays py = 0. Then from (4A.1) we know that X will want to set px as large as possible; that is, px = 1. From (4A.2), we can see that if px = 1, then Y will want to set py as small as possible; that is, px = 0. Hence, (px = 1, py = 0) is an equilibrium. Equivalently, (Abate, Pollute) is an equilibrium. Using the same logic, it is easy to show that (Pollute, Abate) is also an equilibrium. These are the two equilibria in pure strategies.
Now let us derive the mixed strategy equilibrium. I noted that if py = 0.5, then X will not care how it chooses px. However, suppose X sets px = 0.75. Then, (4A.2) tells us that Y will want to set py as low as possible; that is, py = 0. However, (4A.1) tells us that if py = 0, then X will not want to set px = 0.75. Plainly, apart from the corner solutions, only if px = 0.5 and py = 0.5 will neither player be able to gain by deviating unilaterally. Hence, (px = 0.5, py = 0.5) is the equilibrium in mixed strategies.