Show that in each Nash equilibrium of the game, each player receives the same (and maximum) payo↵ from any two strategies that are played with positive probabilities (under, of course, the assumption that his opponents play the Nash equilibrium strategies)
I tried it and got the answer but I am not sure this is right could you please check whether it is right? To prove ui(si, sigma-i)= ui(s’i,sigma-i). suppose not let ui(si, sigma-i)>ui(s’i,sigma-i) since this is true,any player would increase the probability of si since it is giving more payoff than si’. Then, we would have new strategy sigma’i and sigma’i(si)=sigmai(si)+(si’)>0 and sigma’i(s’i)=0 For this new strategy we have ui(si,sigma-i)>ui(s’i,sigma-i) thus sigma is not a NE which is a contradiction
Suppose $\sigma_i$ to be a mixed strategy for the player $i$ beeing part of a Nash equilibrium $\sigma$. Then every pure strategy in the support of $\sigma_i$, i.e. pure strategy with non-zero probability to be played under $\sigma_i$, has the same utility.
Proof.
Suppose $s_1$ and $s_2$ to be two pure strategies with positive probability $p_1$ and $p_2$ to be played under $\sigma_i$. If their utility is different, for example $\mu(s_1)>\mu(s_2)$, then another strategy $\sigma_i'$ better than $\sigma_i$ exists for the player $i$ that can be obtained reassigning the probability of $s_2$ to $s_1$, infact $\mu(\sigma_i')=\mu(\sigma_i) + p_2(\mu(s_1)-\mu(s_2))> \mu(\sigma_i)$. Hence we have a contraddiciton: $\sigma$ is not a Nash equilibrium (a player can increase his utility by moving alone).