Positive affine transformation doesn’t affect the pure strategy Nash equilibrium

Question

While studying a pure strategy Nash equilibrium, I read such a statement

“ the positive affine transformations of the payoffs functions do not change the set of pure strategy Nash equilibria”

How I show this property ?

What my view is as follows:

I construct two finite pure strategy

$\Gamma = (N,S,u)$ and $\Gamma’=(N, S,u’)$

where $s\in S$ is strategy profile.

And I define a postive affine function for all player $i\in N$,

$u’_i(s)=a * u_i(s)+b$ for $a>0$ and $b\in IR$

I need to show that $PNE(\Gamma)=PNE(\Gamma’)$ where PNE is all pure Nash equilibrium.

For that, I guess firstly show that $PNE(\Gamma)\subseteq PNE(\Gamma’)$ and similarly I need to show that $PNE(\Gamma’)\subseteq PNE(\Gamma)$

But since I am not mathematician, I could not proceed this proof explicitly. Please help me show that how can I do this proof?

Thank you.

Answer 1

It doesn't really require any crazy mathematics; it follows pretty immediately from the definition of a Nash equilibrium in pure strategies.

Consider a Nash equilibrium $s^* \in \text{PNE}(\Gamma)$, and consider some other arbitrary strategy profile $s^*_{-i} \in S$ which player $i$ might consider deviating to, which is the same as $s^*$ for all other players $j\neq i$, but differs only for player $i$. If $s^*$ is already a pure-strategy Nash equilibrium of $\Gamma$, then by definition, $u_i (s^*) \ge u_i (s^*_{-i})$.

Now just multiply both sides by $a$, which is positive and so does not flip the sign of the inequality, and add $b$, hence $au_i (s^*) + b \ge au_i (s^*_{-i}) + b$, hence $u'_i (s^*) \ge u'_i (s^*_{-i})$ by the definition of $u'$.

But if $u'_i (s^*) \ge u'_i (s^*_{-i})$, for any player $i$, and any strategy profile $s^*_{-i} \neq s^*$, then again by the definition, $s^*$ is also now a Nash equilibrium of $\Gamma'$.

This then shows that $s^* \in \text{PNE}(\Gamma) \Longrightarrow s^* \in \text{PNE}(\Gamma')$, i.e. $\text{PNE}(\Gamma) \subseteq \text{PNE}(\Gamma')$; it is now straightforward enough to simply go in the other direction (i.e. start with $u'$, subtract $b$ and divide by $a$) to show the converse implication, hence $\text{PNE}(\Gamma) = \text{PNE}(\Gamma')$.