<p>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 $G'$. </p>

<p>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^*)) &amp; &amp; \forall \sigma_1 \in \Delta (A_1) \\ \mathbb{E}(u_2(\sigma_1^*, \sigma_2^*)) \geq  \mathbb{E}(u_2(\sigma_1^*, \sigma_2)) &amp; &amp; \forall \sigma_2 \in \Delta (A_2) \end{eqnarray*} </p>

<p>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 &amp; &amp; \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 &amp; &amp; \forall \sigma_2 \in \Delta (A_2) \end{eqnarray*} </p>

<p>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) &amp; &amp; \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) &amp; &amp; \forall \sigma_2 \in \Delta (A_2) \end{eqnarray*} </p>

<p>By definition of $v_1$ and $v_2$,</p>

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

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

CC BY-SA 3.0
2017-02-16T02:51:34.590Z
2017-02-16T02:51:34.590Z
Amit
<p>Let me borrow from my answer <a href="https://math.stackexchange.com/questions/2146263/zero-sum-games-constant-sum-game/2146433#2146433">here</a>.</p>

<p>The short answer is that adding constants values, even ones that differ across players, will not change the set of Nash equilibria.</p>

<p>This is easy to see from the definition of Nash equilibrium. </p>

<p>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 $i$,</p>

<p>$$ u_i (\sigma_i , \sigma_{-i}) \ge u_i (\sigma_i^\prime ,\sigma_{-i}) $$</p>

<p>for any other strategy $\sigma_i^\prime$ of player $i$.</p>

<p>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</p>

<p>$$ v_i (\sigma_i , \sigma_{-i}) \ge v_i (\sigma_i^\prime ,\sigma_{-i}) $$</p>

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

<p>In fact, it is a standard assumption in game theory that payoff functions represent preferences that satisfy the <a href="https://en.wikipedia.org/wiki/Von_Neumann%E2%80%93Morgenstern_utility_theorem" rel="nofollow noreferrer">von Neumann-Morgenstern axioms</a>. 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.)</p>

CC BY-SA 3.0
2017-05-27T06:44:44.680Z
Why answer the [exact same question](https://math.stackexchange.com/questions/2146263/zero-sum-games-constant-sum-game/2146433) on two SE sites, instead of flagging one for being crossposted?
Giskard
CC BY-SA 3.0
2017-02-16T00:47:09.063Z
2017-02-16T00:47:09.063Z
2017-04-13T12:19:37.213Z
Theoretical Economist
<p>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.</p>

CC BY-SA 3.0
2017-02-16T05:57:42.313Z
2017-05-27T06:42:44.207Z
2017-05-27T06:42:44.207Z
Giskard
<p>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?</p>

<game-theory><nash-equilibrium>
Zero sum game, constant sum game
Lauren Solis
# Zero sum game, constant sum game

(Asked by: username_0 on 2017-02-16)

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?

**Tags:** game-theory, nash-equilibrium

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## 3 answers

### Answer by username_1 on 2017-02-16. Score: 5

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 $G'$.


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### Answer by username_2 on 2017-02-16. Score: 3

Let me borrow from my answer [here](https://math.stackexchange.com/questions/2146263/zero-sum-games-constant-sum-game/2146433#2146433).

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](https://en.wikipedia.org/wiki/Von_Neumann%E2%80%93Morgenstern_utility_theorem). 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.)

> **Comments:**
> **username_3**: Why answer the [exact same question](https://math.stackexchange.com/questions/2146263/zero-sum-games-constant-sum-game/2146433) on two SE sites, instead of flagging one for being crossposted?  

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### Answer by username_3 on 2017-02-16. Score: 3

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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