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?

  • 1
    $\begingroup$ See Bayesian efficiency $\endgroup$ – Herr K. Feb 12 at 6:14
  • $\begingroup$ @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. $\endgroup$ – johnny09 Feb 12 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 $i$.


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