As stated in Tim Roughgarden's book "twenty lectures on algorithmic game theory" (on page 176), "the mixed nash equilibrium of a game correspond to the correlated equilibrium that are product distributions" .
Put it mathematically.
Consider a cost-minimization game and a product distribution $\sigma=\sigma_1 \times \cdots \times \sigma_k$ over the game’s outcomes, where $\sigma_k$ is a mixed strategy for agent $i$. Prove that $\sigma$ is a correlated equilibrium of the game if and only if $\sigma_1, \cdots, \sigma_k$ form a mixed Nash equilibrium of the game.
I can prove the direction from CE to MNP given the product distribution condition. Specifically, from the CE definition, given $\sigma$ as a CE, we have \begin{equation} \label{eq1} \mathbf{E}_{\bar{s} \sim \sigma }[C_i(\bar{s}) | s_i] \leq \mathbf{E}_{\bar{s} \sim \sigma }[C_i(s^\prime_i, \bar{s}_{-i}) | s_i], \forall s_i, s^\prime_i \end{equation} where $s_i$ denotes the strategy of the $i$th player, $\bar{s}$ denotes the joint strategy, $ s^\prime_i$ denotes a deviation of $i$th player from the CE.
Given the product distribution condition $\sigma=\sigma_1 \times \cdots \times \sigma_k$, the above equation can be expressed as $$ \mathbf{E}_{\bar{s}_{-i} \sim \sigma_{-i}}[C_i(s_i, \bar{s}_{-i})] \leq \mathbf{E}_{\bar{s}_{-i} \sim \sigma_{-i}}[C_i(s^\prime_i, \bar{s}_{-i})],\forall s_i, s^\prime_i $$ where $\sigma_{-i}$ denotes $\sigma_1 \times \cdots\sigma_{i-1} \times \sigma_{i+1} \cdots \sigma_k$.
It follows that $$ \sum_{s_i}\sigma_i\mathbf{E}_{\bar{s}_{-i} \sim \sigma_{-i}}[C_i(s_i, \bar{s}_{-i})] \leq \sum_{s_i}\sigma_i \mathbf{E}_{\bar{s}_{-i} \sim \sigma_{-i}}[C_i(s^\prime_i, \bar{s}_{-i})] , \bar{s}_{-i})],\forall s^\prime_i $$ and $$ \mathbf{E}_{\bar{s} \sim \sigma}[C_i(s_i, \bar{s}_{-i})] \leq \mathbf{E}_{\bar{s} \sim \sigma}[C_i(s^\prime_i, \bar{s}_{-i})] , \bar{s}_{-i})], \forall s^\prime_i , $$ which is the definition of MNP.
However, I failed to prove the inverse direction, i.e., from MNP to CP. Could anyone help me out.
Thank you in advance.