In this section, we will try to gain a greater understanding of equilibrium strategies in a game. In general, we call the pair of equilibrium strategies an equilibrium pair, while we call the specific payoff vector associated with an equilibrium pair an equilibrium point.
By experimenting with some examples, try to create an example of a game with two equilibrium points where those points have different values for one of the players. If you can successfully create such an example, you will have answered the question. But just because you can’t find an example, that doesn’t mean one doesn’t exist!
Activity2.5.3.Experimenting with different values.
Let’s see if we can create a \(2\times 2\) matrix for a zero-sum game that has two equilibrium points with different values. Let’s assume the two equilbrium are \((1, -1)\) and \((2, -2)\text{.}\)
Now place \((1, -1)\) and \((2, -2)\) diagonally in the matrix (different rows and columns). Try to fill in values for the other two places in the matrix so that \((1, -1)\) is an equilibrium. Is \((2, -2)\) an equilibrium in any of your examples?
Remember, if \((1, -1)\) is an equilibrium, then \(1\) must be the biggest value for Player 1 in the column and \(1\) is the smallest in the row, so that \(-1\) is the biggest for Player 2 in the row.
Do you think it is possible to have both \((1, -1)\) and \((2, -2)\) be equilibrium points in a \(2\times 2\) matrix? Explain your answer based on your examples.
After trying several examples, you might be beginning to believe that the answer to Question 2.5.1 is “no.” Now you are ready to try to prove the following theorem.
Let’s start with the \(2 \times 2\) case. We will use a proof by contradiction. We will assume the theorem is false and show that we get a logical contradiction. Once we reach a logical contradiction (a statement that is both true and false), we can conclude we were wrong to assume the theorem was false; hence, the theorem must be true. Make sure you are comfortable with the logic of this before moving on.
Since we want to assume the theorem is false, we assume we have a two-player zero-sum game with two different equilibrium values. Since we don’t have a specific example of such a game, we want to represent the game in a general form. In particular, we can represent the general game
Note that if \(a\) is negative, then \(-a\) is positive; thus, every possible set of values is represented by this matrix. We want to look at the possible cases for equilibria.
Think about the different cases, such as \(a\lt d\text{,}\)\(a>d\text{.}\) Can you show that in each case either \((a, -a)\) or \((d, -d)\) is NOT an equilibrium?
Generalize your answer to Activity 2.5.4 to explain what goes wrong if the two equilibria with different values are in the same column. Similarly, explain what happens if the two equilibria are in the same row.
From your last answer, you should see that we need to do more work to figure out what happens if the equilibria are diagonal. So let’s assume that the two equilibria are \((a, -a)\) and \((b, -b)\) with \(a \neq b\text{.}\) It might be helpful to draw the payoff matrix and circle the equilibria.
Activity2.5.7.A player prefers the value of an equilibrium.
Construct a system of inequalities using the fact that a player prefers an equilibrium point to another choice. For example, Player 1 prefers \(a\) to \(d\text{.}\) Thus, \(a > d\text{.}\) List all four inequalities you can get using this fact. You should get two for each player. Remember that Player 1 can only compare values in the same column since he has no ability to switch columns. If necessary, convert all inequalities to ones without negatives. (Algebra review: \(-5 \lt -2\) means \(5 > 2\text{.}\))
Now string your inequalities together. For example, if \(a \lt b\) and \(b \lt c\) then we can write \(a \lt b \lt c\text{.}\) Be careful, the inequalities must face the same way; we cannot write \(a> b \lt c\text{.}\)
Repeat the above argument (Activity 2.5.7, Activity 2.5.8, and Activity 2.5.9) for the case that the two equilibria are \((d, -d)\) and \((c, -c)\) with \(d\neq c\text{.}\)
Can you see how you might generalize to a larger game matrix? You do not need to write up a proof of the general case, just explain how the key ideas from the \(2 \times 2\) case would apply to a bigger game matrix.
We’ve seen that any two equlibrium points must have the same value. However, it is important to note that just because an outcome has the same value as an equilibrium point, that does not mean it is also an equilibrium point.
Working through the steps of a mathematical proof can be challenging. As you think about what we did in this section, first make sure you understand the argument for each step. Then work on understanding how the steps fit together to create the larger argument.
