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?


3 Answers 3


Suppose $(\sigma_1^*, \sigma_2^*)$ is the Nash Equilibrium of the game $G$ consisting of players $\{1,2\}$ having strategy sets $A_1$ and $A_2$ and the payoff functions $u_1:A_1\times A_2 \rightarrow \mathbb{R}$ and $u_2:A_1\times A_2 \rightarrow \mathbb{R}$. Consider the game $G'$ in which everything else is same as $G$ except that the payoff functions are $v_1:A_1\times A_2 \rightarrow \mathbb{R}$ and $v_2:A_1\times A_2 \rightarrow \mathbb{R}$ defined as: $v_1(a_1, a_2) = u_1(a_1, a_2) + d_1$ and $v_2(a_1, a_2) = u_2(a_1, a_2) + d_2$. We will now show that $(\sigma_1^*, \sigma_2^*)$ is also the Nash Equilibrium of $G'$.

Since $(\sigma_1^*, \sigma_2^*)$ is the Nash Equilibrium of $G$, \begin{eqnarray*} \mathbb{E}(u_1(\sigma_1^*, \sigma_2^*)) \geq \mathbb{E}(u_1(\sigma_1, \sigma_2^*)) & & \forall \sigma_1 \in \Delta (A_1) \\ \mathbb{E}(u_2(\sigma_1^*, \sigma_2^*)) \geq \mathbb{E}(u_2(\sigma_1^*, \sigma_2)) & & \forall \sigma_2 \in \Delta (A_2) \end{eqnarray*}

The above can also be rewritten as \begin{eqnarray*} \mathbb{E}(u_1(\sigma_1^*, \sigma_2^*)) + d_1 \geq \mathbb{E}(u_1(\sigma_1, \sigma_2^*)) + d_1 & & \forall \sigma_1 \in \Delta (A_1) \\ \mathbb{E}(u_2(\sigma_1^*, \sigma_2^*)) + d_2 \geq \mathbb{E}(u_2(\sigma_1^*, \sigma_2)) + d_2 & & \forall \sigma_2 \in \Delta (A_2) \end{eqnarray*}

Because $d_1$ and $d_2$ are constants, we can take them inside the expectation and rewrite the above inequalities as: \begin{eqnarray*} \mathbb{E}(u_1(\sigma_1^*, \sigma_2^*) + d_1) \geq \mathbb{E}(u_1(\sigma_1, \sigma_2^*) + d_1) & & \forall \sigma_1 \in \Delta (A_1) \\ \mathbb{E}(u_2(\sigma_1^*, \sigma_2^*) + d_2) \geq \mathbb{E}(u_2(\sigma_1^*, \sigma_2) + d_2) & & \forall \sigma_2 \in \Delta (A_2) \end{eqnarray*}

By definition of $v_1$ and $v_2$,

\begin{eqnarray*} \mathbb{E}(v_1(\sigma_1^*, \sigma_2^*)) \geq \mathbb{E}(v_1(\sigma_1, \sigma_2^*)) & & \forall \sigma_1 \in \Delta (A_1) \\ \mathbb{E}(v_2(\sigma_1^*, \sigma_2^*)) \geq \mathbb{E}(v_2(\sigma_1^*, \sigma_2)) & & \forall \sigma_2 \in \Delta (A_2) \end{eqnarray*}

Therefore, $(\sigma_1^*, \sigma_2^*)$ is also the Nash Equilibrium of $G'$.


Let me borrow from my answer here.

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.)

  • $\begingroup$ Why answer the exact same question on two SE sites, instead of flagging one for being crossposted? $\endgroup$
    – Giskard
    May 27, 2017 at 6:44

Other answers took the game theoretic approach, I will take the utility approach. Adding a real number $d_i$ to all payoffs of a player is an affine transformation of said payoffs. This does not change the player's preferences even when dealing with expected payoffs. Thus the desirability of outcomes is unchanged from the player's perspective hence best responses remain the same.


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