Commitment and anticipated utilitarianism

Abstract

Much empirical evidence demonstrates that individual preferences may not be consistent. This leads to an important question: how should societal preferences be formulated when individuals behave inconsistently? This paper, restricted to a class of preferences, addresses this question by (1) proposing a new method to rationalize individual preferences; (2) introducing a new version of Pareto principle with respect to the rationalized preferences; and (3) characterizing the societal preferences which respect this principle.

Appendix 1: Proofs

1.1 Proof of Proposition 1

We first show necessity and suppose that $v(x)\ge v(y)$. We consider three cases.

• Case 1. $x\succ y$. That is $u(x)>u(y)$. Then

  $$\begin{aligned} U(\{x,y\})=u(x)=U(\{x\}). \end{aligned}$$

• Case 2. $y\succ x$. That is $u(y)>u(x)$. Then

  $$\begin{aligned} U(\{x,y\})=\left\{ \begin{array}{ll} u(y)+v(y)-v(x) &{} \quad \text {if }\;u(y)+v(y)\ge u(x)+v(x), \\ u(x)+v(x)-v(x) &{} \quad \text {otherwise}. \end{array} \right. \end{aligned}$$

  Therefore, $U(\{y\})=u(y)>U(\{x,y\})$.

• Case 3. $x\sim y\succsim z$. That is $u(x)=u(y)\ge u(z)$. We have

  $$\begin{aligned} U(\{x,z\})=\left\{ \begin{array}{ll} u(x) &{} \quad \text {if }\;v(x)\ge v(z),\\ u(x)+v(x)-v(z) &{} \quad \text {otherwise}. \end{array} \right. \end{aligned}$$

  and

  $$\begin{aligned} U(\{y,z\})=\left\{ \begin{array}{ll} u(y) &{} \quad \text {if }\;v(y)\ge v(z),\\ u(y)+v(y)-v(z) &{} \quad \text {otherwise}. \end{array} \right. \end{aligned}$$

  If $v(z)\ge v(x)\ge v(y)$, then

  $$\begin{aligned} U(\{x,z\})=u(x)+v(x)-v(z)>u(y)+v(y)-v(z)=U(\{y,z\}. \end{aligned}$$

  If $v(x)>v(z)\ge v(y)$, then

  $$\begin{aligned} U(\{x,z\})=u(x)\ge u(y)+v(y)-v(z)=U(\{y,z\}). \end{aligned}$$

  If $v(x)>v(y)>v(z)$, then

  $$\begin{aligned} U(\{x,z\})=u(x)=u(y)=U(\{y,z\}). \end{aligned}$$

We now show sufficiency and suppose that x is more tempting than y. We consider three cases.

• Case 1. $x\not \sim y$ and $x\sim \{x,y\}$. If $u(x)+v(x)\ge u(y)+v(y)$, then $v(x)\ge v(y)$. If $u(y)+v(y)>u(x)+v(x)$, suppose $v(y)>v(x)$. Then $u(x)=u(y)$, which contradicts the assumption.

• Case 2. $x\not \sim y$ and $y\succsim \{x,y\}$. If $u(y)+v(y)\ge u(x)+v(x)$, then $v(x)\ge v(y)$. If $u(x)+v(x)>u(y)+v(y)$, suppose $v(y)>v(x)$. Then $u(x)+v(x)-v(y)>u(y)$, which contradicts the assumption.

• Case 3. $x\sim y$ and $z=\alpha x+(1-\alpha )y$ for $\alpha \in (0,1)$. Assume that $v(y)>v(x)$. Then $u(x)=u(y)=u(z)$ and $v(x)<v(z)<v(y)$. Therefore,

  $$\begin{aligned} U(\{y,z\})=u(y)>u(x)+v(x)-v(z)=U(\{x,z\}), \end{aligned}$$

  which contradicts $\{x,z\}\succsim \{y,z\}$.

1.2 Proof of Theorem

The proof that Anticipation Pareto Principle is necessary is straightforward. We only show that Anticipation Pareto Principle is sufficient. The proof consists of three lemmas. Define $W=(W_0,W_1,\ldots ,W_I): \mathscr {A}\rightarrow \mathbb {R}^{I+1}$, where $W_i(A)=u_i(x_i^A)$ for all $A\in \mathscr {A}$ and all $i\in \mathcal {I}$. One may suggest that we can apply the Theorem of Meyer and Mongin (1995), which generalizes the Theorems of Harsanyi (1955) and Anscombe and Aumann (1963). Notice that to apply the theorem there, we need to show that function W is convex.

Lemma 1

$W(\mathscr {A})$ is convex.

Proof

We want to show that for all $A,B\in \mathscr {A}$ and all $\alpha \in [0,1]$, $\alpha W(A)+(1-\alpha )W(B)\in W(\mathscr {A})$. Since function $W_0$ is linear, it suffices to show that for all $i\in \mathcal {I}$,

$$\begin{aligned} W_i(\alpha A+(1-\alpha )B)=\alpha W_i(A)+(1-\alpha )W_i(B). \end{aligned}$$

Equivalently, we will show for all $i\in \mathcal {I}$

$$\begin{aligned} u_i(x_i^{\alpha A+(1-\alpha )B})=\alpha u_i(x_i^A)+(1-\alpha )u_i(x_i^B). \end{aligned}$$

Let $x_i^{\alpha A+(1-\alpha )B}=\alpha \hat{x}_i^A+(1-\alpha )\hat{x}_i^B$ for some $\hat{x}_i^A\in A$ and $\hat{x}_i^B\in B$. According to the definition of $x_i^A$ and $x_i^B$, we have

$$\begin{aligned} u_i(x_i^A)+v_i(x_i^A)\ge & {} u_i(\hat{x}_i^A)+v_i(\hat{x}_i^A) \\ u_i(x_i^B)+v_i(x_i^B)\ge & {} u_i(\hat{x}_i^B)+v_i(\hat{x}_i^B) \end{aligned}$$

Suppose that $u_i(x_i^A)+v_i(x_i^A)> u_i(\hat{x}_i^A)+v_i(\hat{x}_i^A)$. Then by linearity of $u_i$ and $v_i$,

$$\begin{aligned}&u_i(\alpha x_i^A+(1-\alpha )x_i^B)+v_i(\alpha x_i^A+(1-\alpha )x_i^B)>u_i(\alpha \hat{x}_i^A \\&\quad +(1-\alpha )\hat{x}_i^B)+v_i(\alpha \hat{x}_i^A+(1-\alpha )\hat{x}_i^B), \end{aligned}$$

which contradicts that $x_i^{\alpha A+(1-\alpha ) B}$ is an anticipated consumption in $\alpha A+(1-\alpha ) B$. Therefore, we must have

$$\begin{aligned} u_i(x_i^A)+v_i(x_i^A)= u_i(\hat{x}_i^A)+v_i(\hat{x}_i^A). \end{aligned}$$

Similarly,

$$\begin{aligned} u_i(x_i^B)+v_i(x_i^B)= u_i(\hat{x}_i^B)+v_i(\hat{x}_i^B). \end{aligned}$$

Hence, definition of $x_i^A,x_i^B$ and $x_i^{\alpha A+(1-\alpha )B}$ imply

$$\begin{aligned} \alpha u_i(\hat{x}_i^A)+(1-\alpha )u_i(\hat{x}_i^B)=\alpha u_i(x_i^A)+(1-\alpha )u_i(x_i^B). \end{aligned}$$

By linearity of $u_i$, we have $u_i(x_i^{\alpha A+(1-\alpha )B})=\alpha u_i(x_i^A)+(1-\alpha )u_i(x_i^B)$. Thus, $W(\mathscr {A})$ is convex. $\square$

Define $L=\{l\in \mathbb {R}^{I+1}: l_0<-1\text { and }\ l_i=0, \quad i=1,\ldots ,I\}$ and let $\bar{L}$ denote the closure of L. Recall that a subset of $\mathbb {R}^{I+1}$ is said to be polyhedral if it is the set of solutions of a finite system of linear weak inequalities. Clearly $\bar{L}$ is polyhedral.

Define $M=\{w-w': w,w'\in W(\mathscr {A})\}$. By Lemma 1, M is convex. The span of M, write ${{\mathrm{span}}}(M)$, is defined as the set of all finite linear combinations of elements in M.

Lemma 2

$\bar{L}\cap {span}(M)=\emptyset$.

Proof

We begin by noticing that Anticipation Pareto Principle is equivalent to: for all $A,B\in \mathscr {A}$,

$$\begin{aligned} W_i(A)=W_i(B),\quad \text {for all } \; i=1,\ldots , I\Longrightarrow W_0(A)=W_0(B). \end{aligned}$$

Therefore, $\bar{L}\cap M=\emptyset$. Suppose that $\bar{L}\cap {{\mathrm{span}}}(M)\ne \emptyset$. Due to the symmetry property of M, there exists a sequence of $m_1,\ldots ,m_k$ in M and a sequence of positive numbers $\alpha _1,\ldots ,\alpha _k$ such that $\sum _{i=1}^k \alpha _i m_i\in \bar{L}$. The convexity of M implies that $\left( \sum _{j=1}^k\right) ^{-1}\sum _{i=1}^k \alpha _i m_i\in M$, which contradicts the fact that $\bar{L}\cap M=\emptyset$. $\square$

Lemma 3

There are real numbers $\lambda _1,\ldots ,\lambda _I$ and $\mu$ such that for all $A\in \mathscr {A}$,

$$\begin{aligned} W_0(A)=\sum ^I_{i=1}\lambda _iW_i(A)+\mu . \end{aligned}$$

Proof

According to Lemma 2, we can invoke the standard separating hyperplane theorem. Therefore, there is $\lambda =(\lambda _0,\lambda _1,\ldots ,\lambda _I)$ such that for all $l\in \bar{L}$ and all $m\in {{\mathrm{span}}}(M)$,

$$\begin{aligned} \left<\lambda ,l\right>>\left<\lambda ,m\right>. \end{aligned}$$

The property of ${{\mathrm{span}}}(M)$ implies that $\left<\lambda ,m\right>=0$. Hence, we have $\lambda _0<0$ and for all $A,B\in \mathscr {A}$,

$$\begin{aligned} W_0(A)-W_0(B)=\sum ^I_{i=1}\frac{\lambda _i}{-\lambda _0}(W_i(A)-W_i(B)). \end{aligned}$$

By renormalizing, we have for all $A\in \mathscr {A}$, $W_0(A)=\sum ^I_{i=1}\lambda _iW_i(A)$. $\square$

The sufficient result follows from the definition of W.