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A row $r$ of the payoff matrix is said to dominate a row $s$ if $a_{rj}\geq a_{sj}$ for all $j$ = 1,2,......,$n$. Similarly, a column $r$ of the payoff matrix is said to dominate a column $s$ if $a_{ir}\geq a_{is}$ for all $i$ = 1,2,......,$m$.

Prove:

$(i)$ If a row $r$ is dominated by another row, then the row player has at least one optimal strategy $x^{*}$ in which $x_r^{*}=0$. In particular, if row $r$ is deleted from the payoff matrix, then the value of the game does not change.

$(ii)$ If a column $s$ is dominated by another column, then the column player has at least one optimal strategy $y^{*}$ in which $y_s^{*}=0$. In particular, if column $s$ is deleted from the payoff matrix, then the value of the game does not change.

I do not know how to start on this proof, can anyone give me some hints?

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  • $\begingroup$ Do $x_r^{*}=0$ and $y_s^{*}=0$ refer to payoffs? If yes, what you are required to prove does not hold in general - it depends on specific assumptions on the payoffs of the game, that are not stated in your question. Otherwise, please clarify. $\endgroup$ – Alecos Papadopoulos Oct 27 '13 at 2:29
  • $\begingroup$ $x_r^{*}=0$ means the probability of row player choose row $r$ as the strategy is $0$, $y_s^{*}=0$ means the probability of column player choose column $s$ as the strategy is 0. $\endgroup$ – user59036 Oct 27 '13 at 5:50
  • $\begingroup$ In that case, it might also be helpful to define "optimal strategy." In this case, you're looking for mixed-strategy Nash equilibria. $\endgroup$ – jmbejara Oct 27 '13 at 6:23
  • $\begingroup$ "Optimal strategy" is the vector $x$= $[x_1,x_2,....x_r,....x_i]$, which will maximize $z$, the payoff to the row player. And similar for $y$, which will minimize the loss of column player. $\endgroup$ – user59036 Oct 27 '13 at 8:11
  • $\begingroup$ Prove it by reductio ad absurdum: Assume that even though $r$ is strictly dominated then all optimal strategies are characterized by $x_r^{*}\neq 0$. Can they be optimal? $\endgroup$ – Alecos Papadopoulos Oct 27 '13 at 11:19
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As a hint, I think the best way to proceed is to think about each player's maximization problem. Player A maximizes his/her payoff given player B's strategy. For the row player, given that row $r$ always gives equal or higher payoffs, her/she can do just as well by playing $r$ instead of $s$. Thus, if you deleted the row, the payoff in the equilibrium strategy could not be less.

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