0
$\begingroup$

Prove that in a two-person general sum game, the expected payoff of any player at any Strategic Equilibrium (mixed or pure) can not be smaller than the safety level of this player.

How do I prove this?

$\endgroup$
  • $\begingroup$ You want to prove that expected payoff can not be less than lowest expected payoff, in simple terms? What does that mean? The lowest expected payoff is the value that the expected payoff can not go below and could be 0 or even negative. $\endgroup$ – ghosts_in_the_code Nov 4 '14 at 8:08
0
$\begingroup$

If player I's safety level is $s$, then by definition there exists a (possibly mixed) strategy $x$ such that whatever player II does, player I gets at least $s$ by playing $x$.

Now suppose $(x',y)$ is a Nash equilibrium in which player I's payoff is $p$ with $p < s$. By definition, since $(x',y)$ is a Nash equilibrium $x'$ must be a best response against $y$. Suppose that $x$ achieves payoff $q$ against $y$. Then we have $q \ge s > p$, which contradicts the fact that $x'$ is a best response. Thus $(x',y)$ is not a Nash equilibrium. This proves the claim.

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.