Social Welfare and Pareto Optimality in a Bayesian Game

In a normal form game with two players, we call a joint strategy $$s=(s_1,s_2)$$ a Pareto optimal outcome if for no joint strategy $$s'$$, for all players $$i\in\{1,2\}$$, we have $$u_i(s')\geq u_i(s)\qquad(1)$$ and also at least for one player $$i$$ $$u_i(s')>u_i(s).\qquad(2)$$

The social welfare of $$s$$ is defined as $$\sum_{i=1}^{2}u_i(s).\qquad(3)$$ If the social welfare of $$s$$ is maximal, then the joint strategy $$s$$ is a social optimum.

I want to use these definitions in a Bayesian game where the utility functions are defined as $$u_i:A\times\Theta\to\mathbb{R}$$, where $$A$$ is the action set and $$\Theta$$ is the type set.

Can we simply replace $$u_i(s)$$ in $$(1),(2),(3)$$ with $$u_i(a,\theta)$$? Or we have to consider the expected utility of a player?

How should we write the definitions if the game is in the ex-ante stage?

• Feb 12 '19 at 6:14
• @HerrK.This seems to be implementable only to bayesian mechanism design. In my case the bayesian game is finite and I don't consider incentive compatible allocation rules. Feb 12 '19 at 23:28

While it is a bit unusual to describe a strategy profile as being Pareto optimal, especially in the context of Bayesian games, I guess you can still define Pareto optimality in different stages of such games as follows. Recall that in a Bayesian game, a pure strategy is a function $$s_i:\Theta_i\to A_i$$.
A strategy profile $$s(\theta)=(s_1(\theta_1),\dots,s_n(\theta_n))$$ is ex ante Pareto optimal if $$\begin{equation} \mathbb E_\theta [u_i(s(\theta);\theta)]\ge \mathbb E_\theta[u_i(s'(\theta);\theta)],\quad\text{\forall i and \forall s'(\theta)} \end{equation}$$ where the above inequality is strict for some $$i$$. [$$\mathbb E_\theta$$ means taking expectation over the state vector $$\theta$$.]
A strategy profile $$s(\theta)=(s_1(\theta_1),\dots,s_n(\theta_n))$$ is ex post Pareto optimal if, given a particular realization of states $$\bar\theta=(\bar\theta_1,\dots,\bar\theta_n)$$, $$\begin{equation} u_i(s(\bar\theta);\bar\theta)\ge u_i(s'(\bar\theta);\bar\theta),\quad\text{\forall i and \forall s'(\bar\theta)} \end{equation}$$ where the above inequality is strict for some $$i$$.