The mixed-strategy Nash equilibrium for the Battle of the Sexes game described in Example 10.4 may depend on the numerical values of the payoffs. To generalize this solution, assume that the payoff matrix for the game is given by where \(K \geq 1 .\) Show how the Nash equilibrium in mixed strategies for this game depends on the value of \(K\)

The Battle of the Sexes game is a two-player game with two possible strategies for each player. The players are often referred to as Player 1 (the row player) and Player 2 (the column player). The payoff matrix \(A\) for the row player, in this case, is given by: $$ A = \begin{pmatrix} K & 0 \\ 0 & 1 \end{pmatrix} $$ Let \(p\) be the probability of Player 1 choosing the first strategy, and \(1-p\) be the probability of Player 1 choosing the second strategy. Similarly, let \(q\) be the probability of Player 2 choosing the first strategy, and \(1-q\) be the probability of Player 2 choosing the second strategy. Our goal is to find the mixed-strategy Nash equilibrium by finding \(p^*\) and \(q^*\) that satisfy the equilibrium conditions.

In order to find the Nash equilibrium in mixed strategies, we need to find the best response functions for both players. These functions tell us what each player should optimally do, given the other player's strategy. In this case, we need to find Player 1's best response to Player 2's strategy \(q\), denoted by \(BR_1(q)\), and Player 2's best response to Player 1's strategy \(p\), denoted by \(BR_2(p)\). Given Player 2's strategy \(q\), Player 1's expected payoff from playing the first strategy is \(Kq\), and the expected payoff from playing the second strategy is \(1 - q\). The best response function for Player 1, \(BR_1(q)\), is given by: $$ BR_1(q) = \begin{cases} 1, & \text{if } Kq > 1 - q \\ 0, & \text{if } Kq < 1 - q \\ p, & \text{otherwise} \end{cases} $$ Similarly, given Player 1's strategy \(p\), Player 2's expected payoff from playing the first strategy is \(p\), and the expected payoff from playing the second strategy is \(1 - p\). The best response function for Player 2, \(BR_2(p)\), is given by: $$ BR_2(p) = \begin{cases} 1, & \text{if } p > 1 - p \\ 0, & \text{if } p < 1 - p \\ q, & \text{otherwise} \end{cases} $$

It can be seen from the best response functions for both players that a Nash equilibrium in mixed strategies occurs when neither player has an incentive to change their mixed strategy if the opponent keeps their strategy unchanged. This is true when both best response functions hold simultaneously, i.e., when \(BR_1(q^*) = p^*\) and \(BR_2(p^*) = q^*\). When we look at the best response functions, we can see the Nash equilibrium in mixed strategies occurs when: $$ Kq^* = 1 - q^* \text{ and } p^* = 1 - p^* $$ From these conditions, we can find the equilibrium strategies for both players: $$ q^* = \frac{1}{K+1} \text{ and } p^* = \frac{1}{2} $$

From our solution, we can see that the mixed-strategy Nash equilibrium depends on the value of \(K\) as follows: - For Player 1 (row player), their equilibrium probability of playing the first strategy, \(p^*\), is always \(\frac{1}{2}\), independent of the value of \(K\). - For Player 2 (column player), their equilibrium probability of playing the first strategy, \(q^*\), depends on \(K\) through the expression \(\frac{1}{K+1}\). As \(K\) increases, \(q^*\) decreases, making Player 2 more likely to choose the second strategy. In conclusion, the mixed-strategy Nash equilibrium in the Battle of the Sexes game depends on the value of \(K\). Specifically, the row player's equilibrium strategy is always to choose both strategies with equal probability, while the column player's equilibrium strategy depends on the value of \(K\). As \(K\) increases, the column player becomes more likely to choose the second strategy.