I am trying to understand the following exersice from the solutions of my professor and I really don't understand what she is doing. The exersice is the following:
Suppose the matrix below is a pay off matrix of a two person non cooperative game where the left pay offs are A's pay offs and right pay offs are B's pay offs and A has the horizontal strategies and B the vertical. The question is two find all the equilibrium pairs of randomized strategies.
$\begin{bmatrix}(1,2) & (4,3)\\(3,1) & (4,2)\end{bmatrix}$
The solution is the following. We first plot the pay-off set which is the whole of the convex set determined by the four points representing pairs of pure strategies. Then we note that the pairs (ai, b2)(i = 1, 2) are
both equilibrium pairs and so therefore is any pair of the form (α, b2) where α is a randomized strategy .Then the solution says that by considering x = −2pq − q + 4 and y = p − q + 2 we must show that there are no other strategies in equilibrium. This is the part I don't understand... How can I use this x and y to show that there is no strategy in equilibrium? Any help would be highly appreciated
Thanks in advance for any help
First, let's see directly from the bimatrix why there are no other equilibria. Notice that the right column strictly dominates the left column, since for the top row we have $3>2$ and for the bottom row we have $2>1$. Thus, player 2 only plays the right column in equilibrium.
Since our reasoning can deal only with player 2's payoff, let's now derive in detail the payoff to player 2 when both players play general mixed strategies. Let $p$ be the probability for player 1 playing the top row, and $q$ be the probability for player 2 playing the left column. Then, when the players play $((p,1-p),(q,1-q))$, the resulting payoff $y$ for player 2 is:
$$ \begin{align} y = & \quad p \cdot (q\cdot2 +(1-q)\cdot3) + (1-p) \cdot (q\cdot 1 + (1-q) \cdot 2) \\ = & \quad 2pq + 3p - 3pq + q +2 - 2q - pq - 2p + 2pq \\ = & \quad p - q + 2 \end{align} $$
So, how does this equation for player 2's payoff tell us that 2 never plays anything except the right column in equilibrium? Well it reads as follows:
For any mixed strategy of player 1 (that is for all $p$), the payoff of player 2 is decreasing in $q$, the amount of probability that player 1 places on the left column, since the only term that depends on $q$ is $-q$. Thus, for all $p$, player 2's payoff is maximized when $q=0$, i.e., when player 2 plays the right column.