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Given any bilateral zero-sum game G, show that strategy profile σ is a Nash equilibrium for G if, and only if, it is a Nash equilibrium for the constant-sum game G' obtained from G by adding any fixed amount "d" to the payoffs of both players. Is the conclusion affected if the fixed amount, call it now d_i for each i = 1 , 2 , differs between the two players?

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  • $\begingroup$ The actual statement is : Given any bilateral zero-sum game G, show that strategy profile σ is a Nash equilibrium for G if, and only if, it is a Nash equilibrium for the constant-sum game G' obtained from G by adding any fixed amount "d" to the payoffs of both players. Is the conclusion affected if the fixed amount, call it now d_i for each i = 1 , 2 , differs between the two players? $\endgroup$ – Lauren Solis Feb 15 '17 at 22:32
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The short answer is that adding constants values, even ones that differ across players, will not change the set of Nash equilibria.

This is easy to see from the definition of Nash equilibrium.

Let $u_i$ represent player $i$'s payoffs as a function of all the players' strategies. A strategy profile $\sigma= (\sigma_i , \sigma_{-i})$, where $\sigma_i$ is $i$'s strategy, and $\sigma_{-i}$ is a vector of all the other players' strategies, is a Nash equilibrium if, for each player $i$,

$$ u_i (\sigma_i , \sigma_{-i}) \ge u_i (\sigma_i^\prime ,\sigma_{-i}) $$

for any other strategy $\sigma_i^\prime$ of player $i$.

Let $v_i = u_i + d_i$, for each $i$, and let $\sigma$ be a Nash equilibrium strategy profile under $u_i$. Since adding a constant to both sides will preserve the inequality above, it must also be the case that

$$ v_i (\sigma_i , \sigma_{-i}) \ge v_i (\sigma_i^\prime ,\sigma_{-i}) $$

for all $i$ and $\sigma_i^\prime$, so $\sigma$ is a Nash equilibrium strategy profile under the modified set of payoff functions as well.

In fact, it is a standard assumption in game theory that payoff functions represent preferences that satisfy the von Neumann-Morgenstern axioms. This implies that preference orderings are preserved by affine transformations of utility functions. (In this context, an affine transformation is just multiplication by a positive number and adding arbitrary constants to payoff functions.)

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