Mixed nash equilibrium equilibrium $2\times4$ players

I have to solve this exercise, but I don't know how to find mixed strategies nash equilibria with $2\times4$ matrix, I've only done it with $2\times2$ ones. Also I don't know how I could prove that it is a correlated nash equilibrium apart from saying it's symmetric ?

Thanks for your help.

Game metrix as follows

[1]: https://i.stack.imgur.com/KlKE9.png


The game has two pure strategy equilibria, $(U,LL)$ and $(D,R)$. Lets consider mixed strategy equilibria. In a mixed strategy equilibrium both players have to be indifferent between all strategies that they choose with positive probability. For a mixed strategy equilibrium, make the following observation: Player 2 mixes at most between two of their strategies. Why is that? Suppose, Player 2 mixes between three strategies, say LL, L, and R. In this case, the indifference conditions that the payoff from playing LL, L, and R have to be the same result in two equations, but only one variable $x_U$, the probability with which Player 1 plays $U$ (the probality for playing $D$ has to be $1-x_U$). Unless, two of the strategies of Player 2 have the same payoffs, such a system of equations has no solution. Similar argument can be made for Player 2 mixing between all 4 strategies. Thus, Player 2 mixes between at most two strategies.

Observe that there is no mixed equilibrium in which Player 2 mixes between $L$ and $M$, $L$ and $R$, or $M$ and $R$. Why is that? To all those strategies the best reply of Player 1 is $D$. Thus, Player 1 would never mix and thus cannot make Player 2 indifferent. Overall, we are left with three candidates for mixed equilibrium. Player 1 mixes between $U$ and $D$ and Player 2 mixes either between $LL$ and $L$, $LL$ and $M$, and $LL$ and $R$. From here you can figure out the mixed equilibria using the techniques you said you know for 2x2 games.

With respect to the correlated equilibrium. Every correlated equilibrium is a Nash equilibrium, as correlated equilibrium is an extension of Nash equilibrium. Nevertheless, lets make a direct proof for the case at hand. Take a mixed strategy equilibrium of your game. Player 1 plays $U$ with probability $0.5$ and $D$ with probability $0.5$. Player 2 plays $LL$ with probability $0.5$ and $R$ with probability $0.5$. Consider the following correlated equilibrium. The randomization device recommends the following strategies with probability 0.25 each $(U,LL)$, $(U,R)$, $(D,LL)$, $(D,R)$. Suppose Player 1 observes the recommendation $U$. She believes, that Player 2 received the recommendation $LL$ with probability $0.5$ and the recommendation $R$ with probability $0.5$. Thus, Player 1 is indifferent between $U$ and $D$ and therefore $U$ is a best reply. The obedience with respect to the other recommendations can be checked in exactly the same manner. The probabilities that players choose their respective strategies are the same as in the mixed strategy equilibrium. Thus, the proposed recommendations and randomization device constitute a correlated equilibrium.