Value of a zero-sum matrix games, if one matrix has smaller entries than another Suppose that we have two zero-sum games represented by matrices $A = (a_{ij})$ and $B = (b_{ij})$. I am curious to know that if we assume $a_{ij} \leq b_{ij}$ for all $i,j$, does this imply that the value of nash equilibria in game $A$ has a smaller value than equilibria in game $B$?
This seems intuitive to me but I have not been able to find any information, nor prove or disprove the statement.
 A: For any arbitrarily fixed $x\geq 0$ and $y\geq 0$, we can say that $x^TAy\leq x^TBy$. 
$\Rightarrow$ $\min_y x^TAy\leq \min_y x^TBy $ (as the above inequality is true for all $y$)
$\Rightarrow$ $\max_x \min_y x^TAy\leq \max_x\min_y x^TBy$. Thus the nash equilibrium of the first game is always smaller than that of the last game. It seems this is true because of specific properties that $x$ and $y$ carry, namely $1^Tx=1, 1^Ty=1, x,y\geq 0$. Hope this helps.
A: We can write $B=A+E$ where $E$ is a matrix of non-negative numbers. If Player 1 (rows) plays the optimal strategy of the game $A$ (namely $x_A^*$) in the game $B$, and player 2 plays $y$, the payoff is:
$$x_A^*By=x_A^*(A+E)y=x_A^*Ay+x_A^*Ey\geq v_A + (x_A^*Ey)$$
where $v_A$ is the value of the game $A$. The second term is non-negative, since all entries in $E,x_A^*$ and $y$ are non-negative.
Thus, we found a strategy of Player 1 that garantuees at least $v_A$ in the game $B$. The optimal strategy in $B$ should get at least that, so $v_B\geq v_A$.
