Belief-consistent Pareto dominance

Abstract

The classic Pareto criterion claims that all voluntary trades, even on the grounds of heterogeneous beliefs, should be encouraged. I argue that a trade without hope for Pareto improvement remains controversial. I introduce and characterize a notion of belief-consistent Pareto dominance to formalize this argument, which, in addition to unanimity of preferences, requires all rankings in a trade to be supported by some common beliefs that must coincide with the agents’ beliefs about the events on which all agents agree.

Appendix: Proofs

Proof of Theorem

We know that a finite partition $\{E_1,\ldots ,E_N\}$ generates the set of unanimous events $\Lambda$. We also know that probability p on $\Lambda$ satisfies that $p(E)=p_i(E)$ for $i\in \mathcal {I}$ and $E\in \Lambda$.

Pick a pair of allocations f and g. Since both allocations are simple and measurable, there is also a finite partition of S, $(A_j)_{j\le J}$, such that both f and g are constant over each $A_j$. For each $n\le N$ and $j\le J$, we define

$$\begin{aligned} B_{nj}=E_n\cup A_j. \end{aligned}$$

Therefore, $(B_{nj})_{n\le N; j\le J}$ consists a coarser finite partition of S. Thus, we use $f(B_{nj})$ and $g(B_{nj})$ to denote the outcomes in X that f and g, respectively, assume over event $B_{nj}$ for each $n\le N$ and $j\le J$. The theorem characterizes the definition of Belief-consistent Pareto dominance, namely, there exists a probability measure $p^*\in \Delta (\Lambda )$ such that for each $i\in \mathcal {I}(f,g)$,

$$\begin{aligned} \int _S u_i(f)\mathrm {d}p^*>\int _S g_i(f)\mathrm {d}p^* \end{aligned}$$

(1)

It is equivalent to the existence of a probability vector $(p^*(n,j))_{n\le N; j\le J}\in \Delta (N\times J)$ such that

$$\begin{aligned}&\sum _{\begin{array}{c} n\le N \\ j\le J \end{array}}p^*(n,j)u_i(f(B_{nj}))>\sum _{\begin{array}{c} n\le N\\ j\le J \end{array}}p^*(n,j)u_i(g(B_{nj}))\quad \text {and }\nonumber \\&\quad \sum _{j\le J}p^*(n,j)=p(E_n),\quad \text {for all }n\le N. \end{aligned}$$

(2)

Notice that if a probability measure $p^*\in \Delta (\Lambda )$ exists, it can reduce to a probability vector $p^*(n,j)$ satisfying Eq. (2). Conversely, if a probability vector $p^*(n,j)$ exists, we can extend it to a unanimous probability measure on $(S,\Sigma )$, which satisfies Eq. (1).

Condition (i) in Theorem 1 is equivalent to the existence of $p^*(n,j)$ satisfying Eq. (2). To our end, we will show that the existence of $p^*(n,j)$ is equivalent to Condition (ii) in Theorem 1. To see such equivalence, we construct a two-person zero-sum game. In such constructed game, we will show that $p^*(n,j)$ satisfying Eq. (2) is equivalent to Condition (ii) according to minmax theorem for two-person zero-sum games.

For each $n\le N$, let $B^n=\{B_{n1},\ldots ,B_{nJ}\}$. The strategy for player I is

$$\begin{aligned} (b^1,\ldots ,b^N)\in B=B^1\times \cdots \times B^N. \end{aligned}$$

The strategy for player II is $i\in \mathcal {I}(f,g)$. Given a strategy profile (b, i), the payoff for player I is defined as

$$\begin{aligned} U(b,i)=\displaystyle \sum ^N_{n=1}p(E_n)[u_i(f(b^n))-u_i(g(b^n))]. \end{aligned}$$

Player I may use mixed strategy and a simplex $\Delta (B)$ represents the set of mixed strategy for player I. Therefore, should player I chooses mixed strategy $\sigma$ and player II chooses i, the payoff for player I is

$$\begin{aligned} U(\sigma ,i)=\displaystyle \sum _{(b^1,\ldots ,b^N)\in B}\sigma (b^1,\ldots ,b^N)\left[ \displaystyle \sum ^N_{n=1}p(E_n)(u_i(f(b^n))-u_i(g(b^n)))\right] . \end{aligned}$$

For any event $B_{nj}$, we denote

$$\begin{aligned} \sigma (B_{nj})=\sum _{\begin{array}{c} b^n\ne B_{nj}\\ (b^1,\ldots ,b^N)\in B \end{array}}\sigma (b^1,\ldots ,b^N). \end{aligned}$$

Notice that $\sum _{j\le J}\sigma (B_{nj})=1$. Therefore, the payoff for player I given strategy profile $(\sigma ,i)$ can be written as

$$\begin{aligned} U(\sigma ,i)=\displaystyle \sum _{n\le N}p(E_n)\sum _{j\le J}\sigma (B_{nj})[u_i(f(B_{nj})-u_i(g(B_{nj})] \end{aligned}$$

If there exists $\sigma \in \Delta (B)$ such that, for every pure strategy of player II, $i\in \mathcal {I}(f,g)$,

$$\begin{aligned} U(\sigma ,i)>0, \end{aligned}$$

then we can claim that there exists $p^*(n,j)$ satisfying Eq. (2). (To see this, simply define $p^*(n,j)=p(E_n)\cdot \sigma (B_{nj})$. Then, $\sum _{j\le J}p^*(n,j)=p(E_n)\cdot \sum _{j\le J}\sigma (B_{nj})=p(E_n)$.)

Therefore, existence of $p^*(n,j)$ satisfying Eq. (2) is equivalent to the existence of $\sigma \in \Delta (B)$ such that, for every mixed strategy of play II, $\lambda \in \Delta (\mathcal {I}(f,g))$,

$$\begin{aligned} \displaystyle \sum _{i\in \mathcal {I}(f,g)}\lambda (i)\cdot U(\sigma , i)>0. \end{aligned}$$

In other words

$$\begin{aligned} \displaystyle \max _{\sigma \in \Delta (B)}\min _{\lambda \in \Delta (\mathcal {I}(f,g))}\sum _i\lambda (i)\cdot U(\sigma ,i)>0. \end{aligned}$$

According to the minmax theorem for two-person zero-sum games, we have

$$\begin{aligned} \displaystyle \min _{\lambda \in \Delta (\mathcal {I}(f,g))}\max _{\sigma \in \Delta (B)}\sum _i\lambda (i)\cdot U(\sigma ,i)>0. \end{aligned}$$

That is to say, there exists $p^*(n,j)$ satisfying Eq. (2) if and only if for any $\lambda$, there exists $\sigma \in \Delta (B)$ such that

$$\begin{aligned} \displaystyle \sum _{i\in \mathcal {I}(f,g)}\lambda (i)\displaystyle \sum _{(b^1,\ldots ,b^N)\in B}\sigma (b^1,\ldots ,b^N)\left[ \displaystyle \sum ^N_{n=1}p(E_n)(u_i(f(b^n))-u_i(g(b^n)))\right] >0. \end{aligned}$$

Notice that for each $\lambda$, such a $\sigma \in \Delta (B)$ exists if and only if there exists a unit vector, namely, there exists $(b^1,\ldots ,b^N)$ such that

$$\begin{aligned} \displaystyle \sum _{i\in \mathcal {I}(f,g)}\lambda (i) \displaystyle \sum ^N_{n=1}p(E_n)(u_i(f(b^n))-u_i(g(b^n)))>0. \end{aligned}$$

This is the same case that there exists $s_n\in E_n$ for each $n\le N$ such that

$$\begin{aligned} \displaystyle \sum _{i\in \mathcal {I}(f,g)}\lambda (i) \displaystyle \sum ^N_{n=1}p(E_n)(u_i(f(s_n))-u_i(g(s_n)))>0. \end{aligned}$$

$\square$