Exercise 6.3.7

If we have a diagonal matrix $\mathrm{\Sigma }$, then the optimal strategy for $X$ will be: On each row, the worst payoff will be ${\sigma }_i$, among them, the minimum is ${\sigma }_n$, so he’ll choose the last row.

For $Y$, on each column, the minimum payoff will always be 0, so he can choose any column.

So if $X$ always choose the last row, $Y$ will choose the last column to maximize his payoff. So $X$ is better of to have probabilities in choosing the rows to make his worst payoff the same regardless of which row he picks.

Assume the probabilities of picking row $i$ is $x_i$, and we have a mixed row of ${\sigma }_1{x}_1,{\sigma }_2{x}_2,\dots ,{\sigma }_n{x}_n$, solve $x_i$ for ${\sigma }_1{x}_1={\sigma }_2{x}_2=\cdots ={\sigma }_n{x}_n$, we have $x_i=\frac{1}{{\sigma }_i}\frac{1}{{\sum <!--\; FUNCTION\; APPLICATION\; -->}_k\frac{1}{{\sigma }_k}}$, so now $X$’s worst payoff will be $\frac{1}{{\sum <!--\; FUNCTION\; APPLICATION\; -->}_k\frac{1}{{\sigma }_k}}$ regardless of his row pick. This payoff is clearly less than the ${\sigma }_n$ if he always picks the last row.

This is also $Y$’s probabilities to pick the columns.