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.
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Notes
I thank the editor to suggest this term to distinguish from social preferences, which are used to express other-regarding preferences in the literature.
The formal definition of commitment utility is provided in Sect. 2.
See page 245 of Kelley (1975) for technical details.
Abe (2012) shows that x GP-tempts y if and only if \(u(y)>u(x)\) and \(v(x)>v(y)\).
The tie-breaking rule we suggest here for period 2 is not critical for our main results.
References
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Acknowledgments
I would like to thank David Ahn, Antoine Billot, Itzhak Gilboa, Ani Guerdjikova and Marcus Pivato. I am especially grateful to the associate editor and two anonymous referees for detailed comments which led to substantial improvements in the paper.
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This research has been conducted as part of the project Labex MME-DII (ANR11-LBX-0023-01).
Appendix 1: Proofs
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.
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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 \(i\in \mathcal {I}\),
Equivalently, we will show for all \(i\in \mathcal {I}\)
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
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 \(v_i\),
which contradicts that \(x_i^{\alpha A+(1-\alpha ) B}\) is an anticipated consumption in \(\alpha A+(1-\alpha ) B\). Therefore, we must have
Similarly,
Hence, definition of \(x_i^A,x_i^B\) and \(x_i^{\alpha A+(1-\alpha )B}\) imply
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 \(A,B\in \mathscr {A}\),
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 \(A\in \mathscr {A}\),
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 \(m\in {{\mathrm{span}}}(M)\),
The property of \({{\mathrm{span}}}(M)\) implies that \(\left<\lambda ,m\right>=0\). Hence, we have \(\lambda _0<0\) and for all \(A,B\in \mathscr {A}\),
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.
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Qu, X. Commitment and anticipated utilitarianism. Soc Choice Welf 47, 349–358 (2016). https://doi.org/10.1007/s00355-016-0965-0
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DOI: https://doi.org/10.1007/s00355-016-0965-0