Zero sum game, constant sum game 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?
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^*)) &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*} 

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*} 

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*} 

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^*)) &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*} 

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