# Expected payoff calculation difficulty in the Bayesian environment of Bergemman and Morris

Suppose that we have two states of the world equally likely to occur, and say $$\psi$$ is the common prior of the state $$\theta\in\Theta=\{G,B\}$$. The types of the players are given by the following matrices in each state of the world

$$\begin{pmatrix}{G} & t_1 & t_2 \\ t_1 & 1/4 & 1/4 \\ t_2 & 1/4 & 1/4 \end{pmatrix}\quad\begin{pmatrix}{B} & t_1 & t_2 \\ t_1 & 1/3 & 1/6 \\ t_2 & 1/3 & 1/6 \end{pmatrix}$$

and the payoff matrices are

$$\begin{pmatrix}{G} & a_2 & p_2 \\ a_1 & 8,8 & 3,10 \\ p_1 & 10,3 & 0,0 \end{pmatrix}\quad \begin{pmatrix}{B} & a_2 & p_2 \\ a_1 & 6,6 & 2,7 \\ p_1 & 7,2 & 0,0 \end{pmatrix}$$

As in Bergemann and Morris in the paper [of 2016][1] the value function of the palyer i={1,2} is given by

$$V(a,t,\theta)=\sum_{a_{-i},t_{-i},\theta}\psi(\theta)\pi((t_i,t_{-i})|\theta)\sigma((a_i,a_{-i})|(t_i,t_{-i}),\theta)u_i((a_i,a_{-i}),\theta)$$

where $$\pi:\Theta\to\Delta(T)$$ is the probability distribution of the types with respect to the states of the world.

If we solve for the N.E of each game not taking into account the types of the players and the a-priori distributions, we know that the first matrix game gives two pure Nash Equilibrium that is $$(a_1,p_2)$$ and $$(a_2,p_1)$$ and the same holds for the second game while each game has a mixed Nash equilibrium that is $$\sigma_1^G(a_1,p_1),\sigma_2^G(a_2,p_2)=((3/5,2/5),(3/5,2/5))$$ in the good state and $$\sigma_1^B(a_1,p_1),\sigma_2^B(a_2,p_2)=((2/3,1/3),(2/3,1/3))$$ in the bad state respectively.

By taking into account the types of the players, where for example, if player $$1$$ is of type $$t_1$$ and player $$2$$ is of type $$t_1$$ in the good state, then each of them plays $$a_1$$ and $$a_2$$ as their actions in the game. If player $$1$$ is of type $$t_1$$ and player $$2$$ is of type $$t_2$$ in the good state, then each of them plays $$a_1$$ and $$p_2$$ as their actions in the game and so on. In this case the mixed strategies are a little different and the new solution is given by $$(\sigma_1^{G,T}(a_1,p_1),\sigma_2^{G,T}(a_2,p_2))=((3/5,2/5),(3/5,2/5))$$ in the good state and $$(\sigma_1^{B,T}(a_1,p_1),\sigma_2^{B,T}(a_2,p_2))=((2/3,1/3),(1/2,1/2))$$ in the bad state respectively. Due to the update of the beliefs with respect to $$\pi$$, we have a different strategy in the case of $$\sigma_2^{B,T}(a_2,p_2)$$ with respect ot the case where the types were not taken into account. If we make any changes in the a-priori distributions about the types, then the mixed strategy equilibria change a lot and I have done some calculation son my own.

My problem is the following:

When I try to calculate the expected payoof of the players in the pure strategy N.E. I am doing the following calcultion. I show the results only for player $$1$$ and not for player $$2$$, that is

$$V((a_1,p_2),(t_1,t_2),\theta)=\underbrace{\psi(G)}_{1/2}\times\left(\underbrace{\pi((t_1,t_2)|G)}_{1/2}\underbrace{\sigma((a_1,p_2)|(t_1,t_2),G)}_{=1\quad\text{pure strategies}}\underbrace{u_1((a_1,p_2),G)}_{3}\right)+\underbrace{\psi(B)}_{1/2}\times\left(\underbrace{\pi((t_1,t_2)|B)}_{1/3}\underbrace{\sigma((a_1,p_2)|(t_1,t_2),B)}_{=1\quad\text{pure strategies}}\underbrace{u_1((a_1,p_2),B)}_{2}\right)=13/12$$

and in mixed strategies I am a little confused about how to calcualte the expected payoff

$$V((\sigma_1^{(\theta,t,*)},\sigma_2^{(\theta,t,*)}),t,\theta)=\underbrace{\psi(G)}_{1/2}\times\left(\pi((t_1,t_1)|G)\sigma((a_1,a_2)|(t_1,t_1),G)u_1((a_1,a_2),G)+\pi((t_1,t_2)|G)\sigma((a_1,p_2)|(t_1,t_2),G)u_1((a_1,p_2),G)+\pi((t_2,t_1)|G)\sigma((p_1,a_2)|(t_2,t_1),G)u_1((p_1,a_2),G)+\pi((t_2,t_2)|G)\sigma((p_1,p_2)|(t_2,t_2),G)u_1((p_1,p_2),G)\right)+\underbrace{\psi(B)}_{1/2}\times\left(\pi((t_1,t_1)|B)\sigma((a_1,a_2)|(t_1,t_1),B)u_1((a_1,a_2),B)+\pi((t_1,t_2)|B)\sigma((a_1,p_2)|(t_1,t_2),B)u_1((a_1,p_2),B)+\pi((t_2,t_1)|B)\sigma((p_1,a_2)|(t_2,t_1),B)u_1((p_1,a_2),B)+\pi((t_2,t_2)|B)\sigma((p_1,p_2)|(t_2,t_2),B)u_1((p_1,p_2),B)\right)=$$

Can anybody help me? Am I in the right direction? [1]: https://onlinelibrary.wiley.com/doi/epdf/10.3982/TE1808

$$\underbrace{\pi((t_1,t_1)|G)}_{1/2}\underbrace{\sigma((a_1,a_2)|(t_1,t_1),G)}_{3/5\times 3/5=9/25}\underbrace{u_1((a_1,a_2),G)}_{8}=36/25$$ $$\underbrace{\pi((t_1,t_2)|G)}_{1/2}\underbrace{\sigma((a_1,p_2)|(t_1,t_2),G)}_{3/5\times 2/5=6/25}\underbrace{u_1((a_1,p_2),G)}_{3}=9/25$$ $$\underbrace{\pi((t_2,t_1)|G)}_{1/2}\underbrace{\sigma((p_1,a_2)|(t_2,t_1),G)}_{2/5\times 3/5=6/25}\underbrace{u_1((p_1,a_2),G)}_{10}=30/25$$ $$\underbrace{\pi((t_2,t_2)|G)}_{1/2}\underbrace{\sigma((p_1,p_2)|(t_2,t_2),G)}_{2/5\times 2/5=4/25}\underbrace{u_1((p_1,p_2),G)}_{0}=0$$ and hence the first part if you sum everything up is $$75/25=3$$
Similarly $$\pi((t_1,t_1)|B)\sigma((a_1,a_2)|(t_1,t_1),B)u_1((a_1,a_2),B)=2/3\times(2/3\times 1/2)\times 6=4/3$$ $$\pi((t_1,t_2)|B)\sigma((a_1,p_2)|(t_1,t_2),B)u_1((a_1,p_2),B)=1/3\times(2/3\times 1/2)\times 2=2/9$$ $$\pi((t_2,t_1)|B)\sigma((p_1,a_2)|(t_2,t_1),B)u_1((p_1,a_2),B)=2/3\times(1/3\times 1/2)\times 7=7/9$$ $$\pi((t_2,t_2)|B)\sigma((p_1,p_2)|(t_2,t_2),B)u_1((p_1,p_2),B)=1/3\times(1/3\times 1/2)\times 0=0$$
and if you sum up the outcomes you have, $$21/9=7/3$$ and hence
$$V((\sigma_1^{(\theta,t,*)},\sigma_2^{(\theta,t,*)}),t,\theta)=1/2\times 3 + 1/2\times 7/3=16/6$$