Costly self-control and limited willpower

Abstract

In Gul and Pesendorfer (Econometrica 69(6):1403–1435, 2001), a decision-maker, when facing a choice among menus, evaluates each menu in terms of the maximum value of its commitment utility net of self-control costs. This paper extends the model such that this maximum is constrained by the condition that the cost of self-control cannot exceed the decision-maker’s stock of willpower w. Four of the five axioms of our characterization are as in their Theorem 3 except that the independence axiom is restricted to a subset of menus. We add one new axiom to regulate willpower as a limited (cognitive) resource in which the available “stock” does not vary across menus. In our characterization, choices within menus that satisfy WARP reveal a constant trade-off between commitment and temptation utilities. However, it is the discontinuity of preferences over menus (along with violations of WARP for choices within menus) that reveals w (measured in units of temptation utility), allowing for a behaviorally meaningful comparative measure of self-control across individuals.

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Notes

  1. 1.

    Throughout we refer to the representation result in Theorem 3 of GP (p1413), rather than Theorem 1 (p1409) which is the one usually cited in the literature.

  2. 2.

    In his characterization of finite additive utility representations for preferences over menus, Kopylov (2009) provides further clarification of the role of the mixture space theorem.

  3. 3.

    See, for example, Chatterjee and Vijay Krishna (2009), Dekel et al. (2009), Stovall (2010), Kopylov (2009) and Kopylov (2012).

  4. 4.

    For a discussion about the hot–cold empathy gap in decision-making, see Loewenstein (2000). The possibility of learning leading to perfect foresight of this gap is considered by Bénabou and Tirole (2004) and Ali (2011). For an extension allowing for uncertainty about the subjective state of temptation, see Dekel et al. (2009) and Stovall (2010).

  5. 5.

    It is this second channel that “erodes” the self-control of an individual in Gul and Pesendorfer (2007) for their model of addiction via compulsive consumption.

  6. 6.

    For the case where the stock of willpower is sufficiently large so that it never binds (respectively, the case where the stock of willpower is zero) the choice function describing the choice the DM makes from a menu is the one generated by the utility function \(u\left( x\right) +v\left( x\right) \) (respectively, \(v\left( x\right) \) with \(u\left( x\right) \) only used to break ties).

  7. 7.

    In particular, we have for any \(\alpha \), \(\beta \in \left( 0,1\right) \), and any A, \(B\in {\mathscr {M}}\left( \succsim \right) \), \(h_{\alpha }\left( h_{\beta }\left( A,B\right) ,B\right) =h_{\alpha \beta }\left( A,B\right) \).

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Correspondence to Simon Grant.

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Liang would like to acknowledge financial support from Taiwan’s Ministry of Science and Technology under Grant MOST 105-2410-H-001-012.

Appendices

Appendix A: Preliminary results for representation

We begin our derivation by noting it follows from Rader (1963), the ordering axiom and upper semi-continuity ensure the preferences over compact menus admit a utility representation.

Lemma A.1

If Axioms 1 and 2a hold, then there exists a function \(U:{\mathscr {A}}\rightarrow \mathbb {R}\) that represents \(\succsim \).

Note that unlike GP’s Lemma 1 (p1421), we do not adopt the independence axiom to establish the existence of a linear representation for the above lemma. Instead, we use our Axiom 3 in conjunction with Axiom 4, set betweenness to obtain a linear representation over \({\mathscr {M}}\left( \succsim \right) \) in Lemma A.5. As pointed out by GP (p 1413), upper semi-continuity and set betweenness together are enough to extend the representation from finite menus to arbitrary compact sets. (Proof appears in GP, lemma 8, p. 1430.) Hence, we only need to establish our results for finite menus. The next lemma is identical to GP’s lemma 2, which enables us to identify the utility of any finite set with an appropriately chosen two-element subset.

Lemma A.2

(GP, Lemma 2, p. 1422). Let U be a function that represents some \(\succsim \) satisfying Axiom 4. If \(A\in {\mathscr {A}}\) is a finite set, then

$$\begin{aligned} U\left( A\right) =\max _{x\in A}\min _{y\in A}U\left( \left\{ x,y\right\} \right) =\min _{y\in A}\max _{x\in A}U\left( \left\{ x,y\right\} \right) . \end{aligned}$$

Moreover, there is an \(x^{*}\), \(y^{*}\) such that \(\left( x^{*},y^{*}\right) \) solves the maxmin problem and \(\left( y^{*},x^{*}\right) \) solves the minmax problem.

We use Lemma A.2 to prove a result analogous to Lemma 3 in GP (p1422). But unlike GP we establish this result without assuming that the function representing \(\succsim \) is linear. Instead the proof invokes Axiom 3 (restricted independence) directly.

Lemma A.3

Let U be a function that represents some \(\succsim \) that satisfy Axioms 3 and 4 and \(A=\alpha \left\{ x,y\right\} +\left( 1-\alpha \right) \left\{ a,b\right\} \).

$$\begin{aligned} \left\{ x,y\right\} \succ \left\{ y\right\} \text { and }\left\{ a,b\right\} \succ \left\{ b\right\} \text { implies }U\left( A\right) =\min _{y^{\prime }\in A}U\left( \left\{ \alpha x+\left( 1-\alpha \right) a,y^{\prime }\right\} \right) , \end{aligned}$$

and

$$\begin{aligned}&\left\{ x\right\} \succ \left\{ x,y\right\} \text {, }\left\{ a\right\} \succ \left\{ a,b\right\} \text {, }y\notin I\left( x\right) \text { and }b\notin I\left( a\right) \text { }\\&\quad \text {implies }U\left( A\right) =\max _{x^{\prime }\in A}U\left( \left\{ x^{\prime },\alpha y+\left( 1-\alpha \right) b\right\} \right) . \end{aligned}$$

Proof

By Lemma A.2, there exists \((x^{*},y^{*})\) such that \(A\sim \{x^{*},y^{*}\}\) and \((x^{*},y^{*})\) solves the maxmin problem. First, we show that \(\{x,y\} \succ \{y\}\) and \(\{a,b\} \succ \{b\}\) implies \(x^{*}\)\(=\alpha x+(1-\alpha )a.\) By Axiom 3, we have

$$\begin{aligned} A&\succ \alpha \{y\}+(1-\alpha )\{a,b\},\\ A&\succ \alpha \{x,y\}+(1-\alpha )\{b\}. \end{aligned}$$

Suppose \(x^{*}=\alpha x+(1-\alpha )b.\) Then, since \(A\sim \{x^{*},y^{*}\}\) and it solves the maxmin problem, we have

$$\begin{aligned} A\succ \alpha \{x,y\}+(1-\alpha )\{b\}=\{ \alpha x+(1-\alpha )b,\alpha y+(1-\alpha )b\} \succsim A, \end{aligned}$$

which yields a contradiction. Similarly, if \(x^{*}=\alpha y+(1-\alpha )a\), then

$$\begin{aligned} A\succ \alpha \{y\}+(1-\alpha )\{a,b\}=\{ \alpha y+(1-\alpha )b,\alpha y+(1-\alpha )a\} \succsim A. \end{aligned}$$

If \(x^{*}=\alpha y+(1-\alpha )b\), then

$$\begin{aligned} A\succ \alpha \{y\}+(1-\alpha )\{a,b\} \succ \{ \alpha y+(1-\alpha )b\}=\{ \alpha y+(1-\alpha )b,\alpha y+(1-\alpha )b\} \succsim A. \end{aligned}$$

Hence, \(x^{*}\)\(=\alpha x+(1-\alpha )a.\) Suppose that we have \(\{x\} \succ \{x,y\}\) and \(\{a\} \succ \{a,b\}\) with \(y\not \in I\left( x\right) \) and \(b\not \in I\left( a\right) \). Then we can apply Axiom 3 and obtain

$$\begin{aligned} \alpha \{x\}+(1-\alpha )\{a,b\}&\succ A,\\ \alpha \{x,y\}+(1-\alpha )\{a\}&\succ A. \end{aligned}$$

Then since \(A\sim \{y^{*},x^{*}\}\) and it solves the minmax problem, we can use a similar argument as above to show \(y^{*}=\alpha y+(1-\alpha )b\). \(\square \)

Lemma A.3 enables us to define a mixture operation for the space \({\mathscr {M}}\left( \succsim \right) \), which we recall is comprised of the set of singleton menus and two-element menus in which there is a tempting alternative for which the DM can exert costly self-control. Since any A in \({\mathscr {M}}\left( \succsim \right) \) has at most two elements, it follows that for any pair of menus A and B in \({\mathscr {M}}\left( \succsim \right) \) and any \(\alpha \) in \(\left( 0,1\right) \), the menu \(\alpha A+\left( 1-\alpha \right) B\) has either one, two or four elements. So consider the following (set-)mixture operator which we denote by \(h_{\alpha }\left( \cdot ,\cdot \right) \). If \(A=\{a,b\}\) and \(B=\{x,y\}\) with \(\{a\} \succ \{a,b\} \succ \{b\}\) and \(\{x\} \succ \{x,y\} \succ \{y\}\), then the \(\left( \alpha ,1-\alpha \right) \)-(set-)mixing of A and B consists of taking the \(\left( \alpha ,1-\alpha \right) \)-convex combination of the two better alternatives from each set and the \(\left( \alpha ,1-\alpha \right) \)-convex combination of the two worse alternatives from each set. Thus, the resulting “mixture” set still contains only two elements. For all other possible configurations, the standard operation leads to at most two elements anyway, so no modification is required in these cases. More formally, we have for any A and B in \({\mathscr {M}}\left( \succsim \right) \) and any \(\alpha \) in \(\left( 0,1\right) \),

$$\begin{aligned} h_{\alpha }\left( A,B\right) :=\left\{ \begin{array}[c]{cl} \left\{ \alpha a+\left( 1-\alpha \right) x,\alpha b+\left( 1-\alpha \right) y\right\} &{} \quad \text {if } \begin{array}{l} A=\{a,b\}, b\in S\left( a\right) ,\\ B=\{x,y\} \text {, }y\in S\left( x\right) , \end{array}\\ \alpha A+\left( 1-\alpha \right) B &{}\quad \text {otherwise.} \end{array} \right. \end{aligned}$$

Lemma A.4

If a preference ordering \(\succsim \) satisfies Axioms 3 and 4, then \(\left( {\mathscr {M}}\left( \succsim \right) ,\{h_{\alpha }\}_{\alpha \in \left[ 0,1\right] }\right) \) is a mixture space as defined in Kreps (1988, p. 52).Footnote 7

Proof

First, we will show that \(h_{\alpha }\left( A,B\right) \in {\mathscr {M}}\left( \succsim \right) \) for any \(A,B\in {\mathscr {M}}\left( \succsim \right) \). From Lemma A.3, it is known that \(h_{\alpha }\) is either a singleton set or a two-element set. If \(h_{\alpha }\) is a singleton set, then obviously it is in \({\mathscr {M}}\left( \succsim \right) \). If \(h_{\alpha }\) has two elements, then it only takes one of the two possible forms, either \(h_{\alpha }(\{a,b\},\{x\}),\) or \(h_{\alpha }(\{a,b\},\{x,y\})\) with \(b\in S\left( a\right) \) and \(y\in S\left( x\right) .\) By Axiom 3, we have

$$\begin{aligned} \{ \alpha a+\left( 1-\alpha \right) x\}&=\alpha \{a\}+(1-\alpha )\{x\} \succ \{ \alpha a+\left( 1-\alpha \right) x,\alpha b+\left( 1-\alpha \right) x)\}\\&\succ \alpha \{b\}+(1-\alpha )\{x\}=\{ \alpha b+\left( 1-\alpha \right) x\} \end{aligned}$$

Hence, \(h_{\alpha }(\{a,b\},\{x\})\in {\mathscr {M}}\left( \succsim \right) \). By Axiom 3, we also have

$$\begin{aligned} \{ \alpha a+\left( 1-\alpha \right) x\}&\succ \alpha \{a,b\}+(1-\alpha )\{x\} \succ \alpha \{a,b\}+(1-\alpha )\{x,y\}\\&\succ \alpha \{a,b\}+(1-\alpha )\{y\} \succ \alpha \{b\}+(1-\alpha )\{y\}=\{ \alpha b+\left( 1-\alpha \right) y\} \end{aligned}$$

Hence, \(h_{\alpha }(\{a,b\},\{x,y\})\in {\mathscr {M}}\left( \succsim \right) \) as well.

Next we will show that \(h_{\alpha }(\left( h_{\beta }\left( A,B\right) ,B\right) =h_{\alpha \beta }(A,B)\) for any \(A,B\in {\mathscr {M}}\left( \succsim \right) \). We only deal with the case when \(A=\{x,y\}\) and \(B=\{a,b\}\) with \(y\in S\left( x\right) \) and \(b\in S\left( a\right) \) because for the rest cases, the argument is similar but easier.

$$\begin{aligned}&h_{\alpha }\left( h_{\beta }\left( \{x,y\},\{a,b\} \right) ,\{a,b\} \right) \\&\quad =h_{\alpha }(\{ \beta x+\left( 1-\beta \right) a,~\ \beta y+(1-\beta )b\},\{a,b\})\\&\quad =\{ \alpha \left( \beta x+\left( 1-\beta \right) a\right) +\left( 1-\alpha \right) a,\alpha (\beta y+(1-\beta )b)+\left( 1-\alpha \right) b\}\\&\quad =h_{\alpha \beta }(\{x,y\},\{a,b\}). \end{aligned}$$

\(\square \)

Since it follows from Lemma A.3 that for any AB in \({\mathscr {M}}\left( \succsim \right) \), \(h_{\alpha }\left( A,B\right) \sim \alpha A+\left( 1-\alpha \right) B\), as a consequence of Lemma A.4 we can apply the mixture space theorem Kreps (1988, Theorem 5.11, p. 54) to obtain the following representation of \(\succsim \) restricted to \({\mathscr {M}}\left( \succsim \right) \).

Lemma A.5

A preference relation satisfies Axioms 14 if and only if there exists a linear function \(U:{\mathscr {M}}\left( \succsim \right) \rightarrow \mathbb {R}\), such that for any AB in \({\mathscr {M}}\left( \succsim \right) \), \(U\left( A\right) \geqslant U\left( B\right) \Leftrightarrow A\succsim B\). Moreover, U in the representation is unique up to a positive affine transformation and its restriction to singleton sets is continuous.

Now to extend the representation obtained in Lemma A.5, notice first it follows from Axiom 4 (set betweenness) that for any two-element menu either the menu is indifferent to a singleton menu that consists of just one element from that menu or that menu lies in preference terms strictly between the two singleton menus formed from its two elements. That is, \(\{a\} \sim \{a,b\}\) or \(\{a,b\} \sim \{b\}\) or \(\{a\} \succ \{a,b\} \succ \{b\}\). For the third case, since \(\{a,b\}\) is in \({\mathscr {M}}\left( \succsim \right) \), \(U\left( \{a,b\} \right) \) is already defined. For the other two cases, we can simply set \(U\left( \{a,b\} \right) \) either to \(U\left( \{a\} \right) \) or to \(U\left( \{b\} \right) \). This provides the unique extension of the function \(U\left( \cdot \right) \) from Lemma A.5 to extend the representation of \(\succsim \) to all two-element sets.

It remains to extend the representation to all menus. Our first step in this task is to define, as do GP, the linear (commitment utility) function \(u:\Delta \left( Z\right) \rightarrow \mathbb {R}\), by setting \(u\left( x\right) :=U\left( \{x\} \right) \). Next, for any two lotteries ab and any \(\gamma \in \left( 0,1\right) \), such that \(\{a,b\} \in {\mathscr {M}}\left( \succsim \right) \) and \(\{a,\left( 1-\gamma \right) b+\gamma x\} \in {\mathscr {M}}\left( \succsim \right) \) for all \(x\in \Delta \left( Z\right) \), we define the (temptation utility) function \(v:\Delta \left( Z\right) \rightarrow \mathbb {R}\), as follows:

$$\begin{aligned} v\left( x;a,b,\gamma \right) :=\frac{U\left( \{a,b\} \right) -U\left( \left\{ a,\left( 1-\gamma \right) b+\gamma x\right\} \right) }{\gamma }. \end{aligned}$$

We begin by noting a result that is analogous to GP’s Lemma 4 (p1423) holds in our setting even though the domain of the \(U\left( \cdot \right) \) in its statement is \({\mathscr {M}}\left( \succsim \right) \) rather than the unrestricted domain \({\mathscr {A}}\). GP’s proof is still valid in our setting since all two-element sets used in their proof are in \({\mathscr {M}}\left( \succsim \right) \).

Lemma A.6

Let U be a linear function that represents the restriction of some \(\succsim \) to \({\mathscr {M}}\left( \succsim \right) \). Suppose that \(\left( 1-\gamma \right) b+\gamma x\in S\left( a\right) \) for all \(x\in \Delta \left( Z\right) \). Then:

\(\left( i\right) \):

\(\forall \ x\) such that \(x\in S\left( a\right) \), \(v\left( x;a,b,\gamma \right) =U\left( \{a,b\} \right) -U\left( \left\{ a,x\right\} \right) \).

\(\left( ii\right) \):

\(v\left( a;a,b,\gamma \right) =U\left( \{a,b\} \right) -U\left( \left\{ a\right\} \right) \).

\(\left( iii\right) \):

\(v\left( \alpha x+\left( 1-\alpha \right) x^{\prime };a,b,\gamma \right) =\alpha v\left( x;a,b,\gamma \right) +\left( 1-\alpha \right) v\left( x^{\prime };a,b,\gamma \right) \).

\(\left( iv\right) \):

\(v\left( x;a,b,\gamma ^{\prime }\right) =v\left( x;a,b,\gamma \right) \), for all \(\gamma ^{\prime }\in \left( 0,\gamma \right) \).

\(\left( v\right) \):

Suppose that \(\left( 1-\gamma \right) b^{\prime }+\gamma x\in S\left( a^{\prime }\right) \), for all \(x\in \Delta \left( Z\right) \). Then \(v\left( x;a,b,\gamma \right) =v\left( x;a^{\prime },b^{\prime },\gamma \right) +v\left( b^{\prime };a,b,\gamma \right) \).

Although U is linear on \({\mathscr {M}}\left( \succsim \right) \), we have not established that it is linear on \({\mathscr {B}}\left( \succsim \right) \). However, using an argument similar to the proof of Lemma 5.6 in Kreps (1988), we obtain the following weaker version of linearity.

Lemma A.7

Let U be a function that restricted to \({\mathscr {M}} \left( \succsim \right) \) is linear and represents some \(\succsim \) satisfying Axioms 14. If \(\left\{ x,y\right\} \in {\mathscr {B}}\left( \succsim \right) \), then for any \(A\in {\mathscr {M}}\left( \succsim \right) \), and any \(\alpha \in \left( 0,1\right) \),

$$\begin{aligned} U\left( \alpha \left\{ x,y\right\} +\left( 1-\alpha \right) A\right) =\alpha U\left( \left\{ x,y\right\} \right) +\left( 1-\alpha \right) U\left( A\right) . \end{aligned}$$

Proof

If \(\{x,y\}\) satisfies \(\{x\} \succ \{x,y\} \succ \{y\}\), the linearity is already proven in lemma A.4. We have to deal with \(\{x,y\}\) with \(\{x\} \sim \{x,y\} \succ \{y\}\) or \(\{x\} \succ \{x,y\} \sim \{y\}\). If \(\{x,y\}\) satisfies \(\{x\} \succ \{x,y\} \sim \{y\}\), we claim that \(\alpha \{x,y\}+\left( 1-\alpha \right) A\sim \alpha \{y\}+\left( 1-\alpha \right) A\) for all \(A\in {\mathscr {M}}\left( \succsim \right) \). By Axiom 3, \(\{x\} \succ \{y\}\) implies \(\alpha \{x\}+\left( 1-\alpha \right) A\succ \alpha \{y\}+\left( 1-\alpha \right) A\), and Axiom 4 further implies \(\alpha \{x\}+\left( 1-\alpha \right) A\succsim \alpha \{x,y\}+\left( 1-\alpha \right) A\succsim \alpha \{y\}+\left( 1-\alpha \right) A\). In this case, we only have to show that \(\alpha \{x,y\}+\left( 1-\alpha \right) A\succ \alpha \{y\}+\left( 1-\alpha \right) A\) will lead to a contradiction. Let us take \(A=\{a,b\}\) with \(\{a\} \succ \{a,b\} \succ \{b\}\). Suppose that \(\alpha \{x,y\}+\left( 1-\alpha \right) A\succ \alpha \{y\}+\left( 1-\alpha \right) A\). In this case, we have \(\{x\} \succ \{x,y\} \sim \{y\}\). Since \(\{x\} \succ \{y\}\), letting \(\alpha ,\beta \in (0,1)\) and applying Axiom 3, we obtain

$$\begin{aligned} \beta \{x\}+(1-\beta )\{y\} \succ \beta \{y\}+(1-\beta )\{y\}=\{y\} \sim \{x,y\}, \end{aligned}$$

and

$$\begin{aligned} \alpha \{x^{\prime }\}+(1-\alpha )\{a,b\} \succ \alpha \{x,y\}+(1-\alpha )\{a,b\}, \end{aligned}$$

where \(\{x^{\prime }\}=\beta \{x\}+(1-\beta )\{y\}.\)Since \(\alpha \{x^{\prime }\}+\left( 1-\alpha \right) \{a,b\} \succ \alpha \{x,y\}+\left( 1-\alpha \right) \{a,b\} \succ \alpha \{y\}+\left( 1-\alpha \right) \{a,b\}\), von Neumann–Morgenstern continuity implies there exists some \(\gamma \in (0,1)\) such that

$$\begin{aligned}&\alpha \{x\}+\left( 1-\alpha \right) \{a,b\}\\&\quad \succ \gamma \left( \alpha \{x^{\prime }\}+\left( 1-\alpha \right) \{a,b\} \right) +\left( 1-\gamma \right) \left( \alpha \{x^{\prime }\}+\left( 1-\alpha \right) \{a,b\} \right) . \end{aligned}$$

We only deal with the case when \(A=\{a,b\}\) with \(\{a\} \succ \{a,b\} \succ \{b\}\) because when A is a singleton set, the argument is similar but easier. Since \(\alpha \{x^{\prime }\}+\left( 1-\alpha \right) \{a,b\}\) and \(\alpha \{y\}+\left( 1-\alpha \right) \{a,b\}\) are both in \({\mathscr {M}}\left( \succsim \right) \), we use Lemma A.3 to obtain the first “\(\sim \)” below

$$\begin{aligned} \alpha \{x,y\}{+}\left( 1{-}\alpha \right) \{a,b\}&\succ \gamma \left( \alpha \{x^{\prime }\}{+}\left( 1{-}\alpha \right) \{a,b\} \right) {+}\left( 1{-}\gamma \right) \left( \alpha \{y\}{+}\left( 1{-}\alpha \right) \{a,b\} \right) \\&\sim h_{\gamma }\left( \alpha \{x^{\prime }\}+\left( 1-\alpha \right) \{a,b\},\alpha \{y\}+\left( 1-\alpha \right) \{a,b\} \right) \\&\sim \alpha \left( \gamma \{x^{\prime }\}+\left( 1-\gamma \right) \{y\} \right) +\left( 1-\alpha \right) \{a,b\}\\&\succ \alpha \{x,y\}+\left( 1-\alpha \right) \{a,b\} , \end{aligned}$$

which yields a contradiction. The last “\(\succ \)” uses the fact that \(\{x^{\prime }\} \succ \{y\}\), \(\gamma \{x^{\prime }\}+\left( 1-\alpha \right) \{y\} \succ \{y\} \sim \{x,y\}\) and Axiom 4. Since \(\alpha \{x,y\}+\left( 1-\alpha \right) A\sim \alpha \{y\}+\left( 1-\alpha \right) A\), we have \(U(\alpha \{x,y\}+\left( 1-\alpha \right) A)=U\left( \alpha \{y\}+\left( 1-\alpha \right) A\right) \). Using Lemma A.5, we have \(U\left( \alpha \{y\}+\left( 1-\alpha \right) A\right) =\alpha U\left( \{y\} \right) +\left( 1-\alpha \right) U\left( A\right) =\alpha U\left( \{x,y\} \right) +\left( 1-\alpha \right) U\left( A\right) \) for all \(A\in {\mathscr {M}}\left( \succsim \right) \). For \(\{x,y\}\) with \(\{x\} \sim \{x,y\} \succ \{y\}\), Axioms 3 and 4 imply \(\alpha \{x\}+\left( 1-\alpha \right) \{a,b\} \succsim \alpha \{x,y\}+\left( 1-\alpha \right) \{a,b\}\) in the previous discussion. By using the fact that \(\{x\} \succ \{y\}\), and applying a similar argument, we can rule out the possibility that \(\alpha \{x\}+\left( 1-\alpha \right) \{a,b\} \succ \alpha \{x,y\}+\left( 1-\alpha \right) \{a,b\}\) and further obtain \(U(\alpha \{x,y\}+\left( 1-\alpha \right) A)=U(\alpha \{x\}+\left( 1-\alpha \right) A)=\alpha U\left( \{x\} \right) +\left( 1-\alpha \right) U\left( A\right) =\alpha U\left( \{x,y\} \right) +\left( 1-\alpha \right) U\left( A\right) \) for all \(A\in {\mathscr {M}}\left( \succsim \right) \)\(\square \)

Next we adapt Lemma 5 of GP(p1424) to our framework and establish, given a certain condition holds, there exists a costly self-control representation over any two-element menu in which the willpower constraint never binds. This certain condition requires that for any lottery a for which \(S\left( a\right) \ne \varnothing \), we can find a lottery \(b\in S\left( a\right) \) and a \(\gamma >0\), such that \((1-\gamma )b+\gamma x\in S\left( a\right) \) for all \(x\in \Delta \left( Z\right) \).

Lemma A.8

Let U be a function over menus. Suppose its restriction to \({\mathscr {M}}\left( \succsim \right) \) is linear and it represents a preference relation \(\succsim \) satisfying Axioms 14. Consider a lottery \(a\in \Delta \left( Z\right) \) for which \(I\left( a\right) =\varnothing \). Suppose there exist lottery \(b\in \Delta \left( Z\right) \) and \(\gamma \in \left( 0,1\right) \), such that \(\left( 1-\gamma \right) b+\gamma x\in S\left( a\right) \), for all \(x\in \Delta \left( Z\right) \). Then for any lottery \(y\in \Delta \left( Z\right) \), such that \(U\left( \left\{ y\right\} \right) \leqslant U\left( \left\{ a\right\} \right) :\)

$$\begin{aligned} U\left( \left\{ a,y\right\} \right) =\max _{x\in \left\{ a,y\right\} }\left\{ u\left( x\right) +v\left( x;a,b,\gamma \right) \right\} -\max _{x^{\prime }\in \left\{ a,y\right\} }v\left( x^{\prime };a,b,\gamma \right) . \end{aligned}$$

Proof

For the case where \(U\left( \{a\} \right)>U(\{a,y\})>U\left( \{y\} \right) \), by GP(p 1424) we know \(v\left( y;a,b,\gamma \right) \geqslant v\left( a;a,b,\gamma \right) \) and \(u\left( a\right) +v\left( a;a,b,\gamma \right) -v\left( y;a,b,\gamma \right) >u\left( y\right) +v\left( y;a,b,\gamma \right) -v\left( y;a,b,\gamma \right) \). Let \(A=(1-\gamma )\{a,b\}+\gamma \{a,y\}\). Since \(\{a,b\} \in {\mathscr {M}}\left( \succsim \right) \), by Lemma A.7 we have \(U(A)=\left( 1-\gamma \right) U\left( \{a,b\} \right) +\gamma U\left( \{a,y\} \right) \) for \(\{a,y\} \in {\mathscr {B}}\left( \succsim \right) \). For the case where \(U\left( \{a\} \right) =\)\(U(\{a,y\})>U\left( \{y\} \right) ,\) the first part of Lemma A.3 establishes that \(U\left( A\right) =\min _{x^{\prime }\in A}U\left( \{a,x^{\prime }\} \right) .\) Hence, we have \(v\left( a;a,b,\gamma \right) \geqslant v\left( y;a,b,\gamma \right) \) by the same argument in GP. For the case where \(U\left( \{a\} \right)>\)\(U(\{a,y\})=U\left( \{y\} \right) \) and \(y\in D(a)\), we will show \(v\left( y;a,b,\gamma \right) \geqslant v\left( a;a,b,\gamma \right) +u\left( a\right) -u\left( y\right) \). From GP this is equivalent to show that

$$\begin{aligned} U\left( \{a,(1-\gamma \right) b+\gamma y\})\leqslant \left( 1-\gamma \right) U\left( \{a,b\} \right) +\gamma U\left( \{a,y\} \right) =U(A) \end{aligned}$$

The above inequality holds because of the second part of Lemma A.3\(U\left( A\right) =\max _{w\in A}\left( \{w,(1-\gamma )b+\gamma y\} \right) \). \(\square \)

With these preliminary results in hand, we are in a position to characterize the family of preferences over menus for which either the willpower constraint is never binding or temptation is overwhelming, that is, the stock of willpower is zero. As we noted in the previous section, this is GP’s Theorem 3 (p1413).

Lemma A.9

Suppose \(I\left( a\right) =\phi \) for all \(a\in \Delta (Z)\). A preference relation \(\succsim \) satisfies Axioms 14 if and only if the preference relation \(\succsim \) admits a representation of the form in expression (1), where w is either sufficiently large so the constraint is never binding or \(w=0\).

Appendix B: Proof of Theorem 1

For the case that there exists a lottery a for which \(I\left( a\right) \ne \varnothing \), we establish two lemmas before we prove Theorem 1. The first lemma extends the representation result from Lemma A.8 to include two-element menus in which the willpower constraint binds.

Lemma B.10

Let U be a function over menus. Suppose its restriction to \({\mathscr {M}}\left( \succsim \right) \) is linear and it represents a preference relation \(\succsim \) satisfying Axioms 15. Consider a lottery a for which \(I\left( a\right) \ne \varnothing \). Suppose there exist lottery \(b\in \Delta \left( Z\right) \) and \(\gamma \in \left( 0,1\right) \), such that \(\left( 1-\gamma \right) b+\gamma x\in S\left( a\right) \), for all \(x\in \Delta \left( Z\right) \). Then for any lottery \(y\in \Delta \left( Z\right) \), such that \(U\left( \left\{ y\right\} \right) \leqslant U\left( \left\{ a\right\} \right) :\)

$$\begin{aligned} U\left( \left\{ a,y\right\} \right)&=\max _{x\in \left\{ a,y\right\} }\left\{ u\left( x\right) +v\left( x;a,b,\gamma \right) \right\} -\max _{x^{\prime }\in \left\{ a,y\right\} }v\left( x^{\prime };a,b,\gamma \right) ,\\&\mathrm{s.t. }\max _{x^{\prime }\in \{a,y\}}v\left( x^{\prime };a,b,\gamma \right) -v\left( x;a,b,\gamma \right) \leqslant w\left( a\right) \end{aligned}$$

where \(w\left( a\right) =\max _{x^{\prime }\in S\left( a\right) }v\left( x^{\prime };a,b,\gamma \right) -v\left( a;a,b,\gamma \right) \).

Proof

For the case where \(y\in S\left( a\right) \), we have \(v(y;a,b,\gamma )\leqslant \max _{z\in S\left( a\right) }v\left( z;a,b,\gamma \right) \). Hence, by Lemma A.8, we have the desired result for all \(\{a,y\} \in {\mathscr {B}}\left( \succsim \right) \). The remaining part of the proof is to show that if \(y\in I\left( a\right) \), then we must have \(v(y;a,b,\gamma )-v(a;a,b,\gamma )\geqslant w(a)\), which is equivalent to show that \(v\left( y;a,b,\gamma \right) >\max _{z\in S\left( a\right) }v\left( z;a,b,\gamma \right) \). Since \(y\in I\left( a\right) \), \(\exists \bar{\gamma }\in (0,1)\) such that \(\gamma ^{\prime }y+\left( 1-\gamma ^{\prime }\right) a\in S\left( a\right) \) if \(\gamma ^{\prime }\leqslant \bar{\gamma }\) and \(\gamma ^{\prime }y+\left( 1-\gamma ^{\prime }\right) a\in I\left( a\right) \) if \(\gamma ^{\prime }>\bar{\gamma }\). Let \(\bar{y}=\bar{\gamma }y+\left( 1-\bar{\gamma }\right) a\). Since \(\bar{y}\in S(a)\), by Lemma A.6 (iii), we have \(\bar{\gamma }v\left( y;a,b,\gamma \right) +\left( 1-\bar{\gamma }\right) v\left( a;a,b,\gamma \right) =v\left( \bar{y};a,b,\gamma \right) \geqslant v\left( a;a,b,\gamma \right) \). Hence, \(v\left( y;a,b,\gamma \right) \geqslant v\left( a;a,b,\gamma \right) \). Moreover, \(u\left( a\right) +v\left( a;a,b,\gamma \right) >u\left( \bar{y}\right) +v\left( \bar{y};a,b,\gamma \right) =\bar{\gamma }(u\left( y\right) +v\left( y;a,b,\gamma \right) )+\left( 1-\bar{\gamma }\right) (u\left( a\right) +v\left( a;a,b,\gamma \right) )\). Hence, \(u\left( a\right) +v\left( a;a,b,\gamma \right) >u\left( y\right) +v\left( y;a,b,\gamma \right) \). We claim that \(\max _{z\in S\left( a\right) }v\left( z;a,b,\gamma \right) =v(\bar{y};a,b,\gamma )\). For any \(b^{\prime }\in S\left( a\right) \), from Axiom 5, we have \(\frac{1}{2}\{a\}+\frac{1}{2}\{a,b^{\prime }\} \succsim \frac{1}{2}\{a\}+\frac{1}{2}\{a,\bar{y}\}\). Since \(\{a,b^{\prime }\}\) and \(\{a,\bar{y}\}\) are in \({\mathscr {M}}\left( \succsim \right) \), we have \(\frac{1}{2}u\left( a\right) +\frac{1}{2}U\left( \{a,b^{\prime }\} \right) \geqslant \frac{1}{2} u(a)+\frac{1}{2}U(\{a,\bar{y}\})\), \(U\left( \{a,b^{\prime }\} \right) =u\left( a\right) +v\left( a;a,b,\gamma \right) -v\left( b^{\prime };a,b,\gamma \right) \) and \(U\left( \{a,\bar{y}\} \right) =u\left( a\right) +v\left( a;a,b,\gamma \right) -v\left( \bar{y};a,b,\gamma \right) \). Hence, \(v\left( b^{\prime };a,b,\gamma \right) \leqslant v\left( \bar{y};a,b,\gamma \right) \). \(\square \)

In order to be able to apply the above lemmas, we need to establish that for any lottery a in which \(S\left( a\right) \ne \varnothing \), we can find a lottery \(b\in S\left( a\right) \) and a \(\gamma >0\), such that \((1-\gamma )b+\gamma x\in S\left( a\right) \) for all \(x\in \Delta \left( Z\right) \). This is the import of GP’s claim 1 (p1426) but their proof does not work when there exists a lottery a for which \(I\left( a\right) \ne \varnothing \). So we prove the following directly.

Lemma B.11

If there exists some pair \(x,y^{\prime }\in \Delta \left( Z\right) \) such that \(y^{\prime }\in S\left( x\right) \), then there is a \(\gamma >0\) such that \((1-\gamma )y+\gamma a\in S\left( x\right) \) for all \(a\in \Delta \left( Z\right) \) and \(y=\frac{x+y^{\prime }}{2}.\)

Proof

From Lemma A.4 and \(\{x,y\}=\frac{1}{2}\{x\}+\frac{1}{2}\{x,y^{\prime }\}\), we know \(y\in S\left( x\right) \). Suppose there is a \(\gamma _{z}>0\) such that \((1-\gamma _{z})y+\gamma _{z}{\mathscr {[}}z]\in S\left( x\right) \) for all \(z\in Z\). Since Z is finite, letting \(\gamma =\min _{z\in Z}\{ \gamma _{z}\}\) and applying Lemma A.4, we can obtain the desired result. Hence, for any \(z\in Z,\) let \(y_{i} :=(1-\gamma _{i})y+\gamma _{i}{\mathscr {[}}z]\), and \(x_{i}:=(1-\gamma _{i} )x+\gamma _{i}{\mathscr {[}}z]\) and \(\gamma _{i}\rightarrow 0\). We will show that \(y_{i}\in S\left( x\right) \) for sufficiently large i. Suppose to the contrary that we can find a subsequence \(y_{i^{\prime }}\) from \(y_{i}\), such that \(y_{i^{\prime }}\not \in S\left( x\right) \). We show all the possible alternatives will lead to a contradiction.

  • Case 1. Suppose we have \(\{x,y_{i^{\prime }}\} \sim \{x\}\) for all \(i^{\prime }\). By Axiom 2a, we have \(\{x,y\} \succsim \{x\}\), which contradicting \(y\in S\left( x\right) \).

  • Case 2. Suppose we have \(y_{i^{\prime }}\in I\left( x\right) \) for all \(i^{\prime }\). By the definition of \(I\left( x\right) \), there exists \(\alpha \in (0,1)\) such that \((1-\alpha )x+\alpha y_{i^{\prime }}\in S\left( x\right) \). Let \(a=\frac{1-\alpha }{2-\alpha }y^{\prime }+(1-\frac{1-\alpha }{2-\alpha })((1-\alpha )x+\alpha y_{i^{\prime }})\). By Lemma A.4, we have a \(\in S\left( x\right) .\) However, we can rewrite \(a=\left( 1-\frac{\alpha \gamma _{i}}{2-\alpha }\right) y+\frac{\alpha \gamma _{i}}{2-\alpha }{\mathscr {[}}z]\). Hence, for all \(\gamma _{j}\leqslant \frac{\alpha \gamma _{i}}{2-\alpha }\), we have \(y_{j}\in S\left( x\right) \), which yields a contradiction.

  • Case 3. Suppose we have \(y_{i^{\prime }}\in D\left( x\right) \) and \(\ y\in T\left( x_{i^{\prime }}\right) \) for all \(i^{\prime }\). By Axiom 4, we have

    $$\begin{aligned} U(\{x,y_{i^{\prime }},x_{i^{\prime }},y\})\leqslant \max \{U(\{x,y_{i^{\prime } }\}),U\left( \{x_{i^{\prime }},y\right) \}=\max \{U\left( \{y_{i^{\prime }}\} \right) ,U\left( \{y\} \right) \} \end{aligned}$$

    and

    $$\begin{aligned} U(\{x,y_{i^{\prime }},x_{i^{\prime }},y\})\geqslant \min \{U(\{x,y\}),U\left( \{x_{i^{\prime }},y_{i^{\prime }}\} \right) \}. \end{aligned}$$

    Note that \(\{x_{i^{\prime }},y_{i^{\prime }}\}=(1-\gamma _{i^{\prime } })\{x,y\}+\gamma _{i^{\prime }}\{ {\mathscr {[}}z]\}\). By Lemma A.7, we have \(U\left( \{x_{i^{\prime }},y_{i^{\prime }}\} \right) =(1-\gamma _{i^{\prime }})U\left( \{x,y\} \right) +\gamma _{i^{\prime }}U(\{ {\mathscr {[}}z]\}).\) Since \(U(\{x,y\})>U\left( \{y\} \right) \), we obtain a contradiction.

  • Case 4. Suppose we have \(y_{i^{\prime }}\in D\left( x\right) \) and \(y\in S\left( x_{i^{\prime }}\right) \) for all \(i^{\prime }\). We apply Lemma A.7 to obtain

    $$\begin{aligned} \frac{1}{2}U(\{x,y_{i^{\prime }}\})+\frac{1}{2}U(\{x_{i^{\prime }},y\})=U\left( \frac{1}{2}\{x,y_{i^{\prime }}\}+\frac{1}{2}\{x_{i^{\prime }},y\} \right) ; \end{aligned}$$

therefore, the same argument in GP (Claim 1, p. 1426) follows. \(\square \)

Now we are ready to prove our main theorem.

Proof of Theorem 1

We show that the axioms imply that the relation \(\succsim \) admits a representation of the form given by the function U as defined in Theorem 1. Since there exists a in \(\Delta (Z)\) such that \(I\left( a\right) \not =\varnothing \), we have \(S\left( a\right) \not =\varnothing .\) By Lemma B.11, there exists \(b\in S(a),\)\(\gamma \in (0,1)\) satisfy \((1-\gamma )b+\gamma a^{\prime }\in S\left( a\right) \) for all \(a^{\prime }\in \Delta (Z)\). By Lemma B.10, we can let \(u\left( a^{\prime }\right) :=U\left( \{a^{\prime }\} \right) \), \(v\left( a^{\prime }\right) :=v\left( a^{\prime };a,b,\gamma \right) \) for all \(a^{\prime }\in \Delta (Z)\) and \(w=v\left( \bar{a};a,b,\gamma \right) -v\left( a;a,b,\gamma \right) ,\) where \(\bar{a} \in \overline{I\left( a\right) }\cap S(a)\). By the first part of the proof of Lemma A.8, we know \(u\left( a\right) +v\left( a\right) >u\left( b\right) +v\left( b\right) \) and \(v(b)>v\left( a\right) \). Hence, neither u nor v is constant and \(v\left( \cdot \right) \not =-\alpha u\left( \cdot \right) +\beta \) for some \(\alpha \in (-\infty ,0]\cup [1,\infty )\) and \(\beta \in R\).

Now consider a set \(A=\{x,y^{\prime }\}\), where x and \(y^{\prime }\) are in the relative interior of \(\Delta (Z)\). Assume without loss of generality, that \(u\left( x\right) \geqslant u\left( y^{\prime }\right) \). Since x is in the interior of \(\Delta (Z)\) and \(b\in S\left( a\right) \), we can select \(\alpha \)\(\in (0,1)\) and \(x^{\prime }\)\(\in \Delta (Z)\) such that \(\{x,y\}=\alpha \{x^{\prime }\}+(1-\alpha )\{a,b\} \in {\mathscr {M}}\left( \succsim \right) \). Hence, by Lemma B.11, there exists \(\gamma ^{\prime } \in (0,1)\) such that \(\left( 1-\gamma ^{\prime }\right) y+\gamma ^{\prime }a^{\prime }\in S\left( x\right) \) for all \(a^{\prime }\in \Delta (Z)\). If \(I\left( x\right) =\varnothing \), then \(\{x,y^{\prime }\} \in {\mathscr {B}} \left( \succsim \right) \). Hence, we can apply Lemma A.8 and obtain

$$\begin{aligned} U\left( \{x,y^{\prime }\} \right) =\max _{x^{\prime \prime }\in \{x,y^{\prime } \}}\{u\left( x^{\prime \prime }\right) +v\left( x^{\prime \prime } ;x,y,\gamma ^{\prime }\right) \}-\max _{x^{\prime \prime }\in \{x,y^{\prime } \}}v\left( x^{\prime \prime };x,y,\gamma ^{\prime }\right) . \end{aligned}$$

Note that the willpower constraint is relevant only when \(y^{\prime }\in S\left( x\right) \). Hence, we need to show that \(v\left( y^{\prime };x,y,\gamma ^{\prime }\right) -v\left( x;x,y,\gamma ^{\prime }\right) \leqslant w\) if \(y^{\prime }\in S\left( x\right) \). Since \(y^{\prime }\in S\left( x\right) \), by Axiom 5, we have \(\frac{1}{2}\{a\}+\frac{1}{2}\{x,y^{\prime }\} \succsim \frac{1}{2}\{x\}+\frac{1}{2}\{a,a^{*}\}\). By Lemma A.7, we then have \(\frac{1}{2}u\left( a\right) +\frac{1}{2}U\left( \{x,y^{\prime }\} \right) \geqslant \frac{1}{2} u(x)+\frac{1}{2}U(\{a,\bar{a}\})\), where \(U\left( \{x,y^{\prime }\} \right) =u\left( x\right) +v\left( x;x,y,\gamma ^{\prime }\right) -v\left( y^{\prime };x,y,\gamma ^{\prime }\right) \) and \(U\left( \{a,\bar{a}\} \right) =u\left( a\right) -w\). Hence, we have \(v\left( y^{\prime } ;x,y,\gamma ^{\prime }\right) -v\left( x;x,y,\gamma ^{\prime }\right) \leqslant w\). Let \(\gamma ^{*}=\min \{ \gamma ,\gamma ^{\prime }\}\). By Lemma A.6 (iv), \(v\left( \cdot ;a,b,\gamma ^{*}\right) =v\left( \cdot ;a,b,\gamma \right) \) and \(v\left( \cdot ;x,y,\gamma ^{*}\right) =v\left( \cdot ;x,y,\gamma ^{\prime }\right) \). By Lemma A.6 (v), for an appropriate constant k\(v\left( \cdot ;a,b,\gamma ^{*}\right) =v\left( \cdot ;x,y,\gamma ^{*}\right) +k\) and hence it follows that

$$\begin{aligned} U\left( \{x,y^{\prime }\} \right)&=\max _{x^{\prime \prime }\in \{x,y^{\prime }\}}\{u\left( x^{\prime \prime }\right) +v\left( x^{\prime \prime }\right) \}-\max _{y^{\prime \prime }\in \{x,y^{\prime }\}}v\left( y^{\prime \prime }\right) \text { }\\&\mathrm{s.t.}\max _{y^{\prime \prime }\in \{x,y^{\prime }\}}v\left( y^{\prime \prime }\right) -v\left( x^{\prime \prime }\right) \leqslant w \end{aligned}$$

If \(I\left( x\right) \not =\varnothing ,\) we can apply Lemma B.10,

$$\begin{aligned} U\left( \{x,y^{\prime }\} \right)&=\max _{x^{\prime \prime }\in \{x,y^{\prime }\}}\{u\left( x^{\prime \prime }\right) +v\left( x^{\prime \prime };x,y,\gamma ^{\prime }\right) \}-\max _{y^{\prime \prime }\in \{x,y^{\prime }\}}v\left( y^{\prime \prime };x,y,\gamma ^{\prime }\right) ,\\&\mathrm{s.t. }\max _{y^{\prime \prime }\in \{x,y^{\prime }\}}v(y^{\prime \prime };x,y,\gamma ^{\prime })-v(x^{\prime \prime };x,y,\gamma ^{\prime })\leqslant w(x)\text {, } \end{aligned}$$

where \(w\left( x\right) =\max _{y^{\prime \prime }\in S\left( x\right) }v\left( y^{\prime \prime };x,y,\gamma ^{\prime }\right) -v(x;x,y,\gamma ^{\prime }).\) Take \(\bar{x}\in \)\(\overline{I\left( x\right) }\cap S\left( x\right) .\) By Axiom 5, we have \(\frac{1}{2}\{a\}+\frac{1}{2}\{x,\bar{x}\} \sim \frac{1}{2}\{x\}+\frac{1}{2}\{a,\bar{a}\}\). By Lemma A.7, we then have \(\frac{1}{2}u\left( a\right) +\frac{1}{2}U\left( \{x,\bar{x}\} \right) =\frac{1}{2}u(x)+\frac{1}{2}U(\{a,\bar{a}\})\), where \(U\left( \{x,\bar{x}\} \right) =u\left( x\right) +v\left( x;x,y,\gamma ^{\prime }\right) -v\left( \bar{x};x,y,\gamma ^{\prime }\right) \) and \(U\left( \{a,\bar{a}\} \right) =u\left( a\right) -w\). Hence, we have \(w\left( x\right) =v\left( \bar{x};x,y,\gamma ^{\prime }\right) -v\left( x;x,y,\gamma ^{\prime }\right) =w.\) Follow the same argument as above, we have

$$\begin{aligned} U\left( \{x,y^{\prime }\} \right)&=\max _{x^{\prime \prime }\in \{x,y^{\prime }\}}\{u\left( x^{\prime \prime }\right) +v\left( x^{\prime \prime }\right) \}-\max _{y^{\prime \prime }\in \{x,y^{\prime }\}}v\left( y^{\prime \prime }\right) \text { }\\&\mathrm{s.t. }\max _{y^{\prime \prime }\in \{x,y^{\prime }\}}v\left( y^{\prime \prime }\right) -v\left( x^{\prime \prime }\right) \leqslant w. \end{aligned}$$

Now consider an arbitrary finite set A. We know that

Let \(y^{*}\)\(\in \arg \max _{y\in A}v\left( y\right) \). If, \(x\in A\) such that \(v\left( y^{*}\right) -v\left( x\right) >w\), then x does not solve the constraint maxminmax problem because for the pair \(\{x,y^{*}\}\) we would choose \(y^{*}\) instead of x. Hence, x will not survive after the second requirement, i.e., \(\min _{y\in A}\) when we take \(y=y^{*}\). Now consider any \(x\in A\) such that \(v\left( y^{*}\right) -v\left( x\right) \leqslant w\), then for any pair \(\{x,x^{\prime }\}\) where \(x^{\prime }\in A\), we have \(v\left( x^{\prime }\right) -v(\)\(x^{*})\leqslant w\). Hence, if \(v\left( y^{*}\right) -v\left( x^{\prime }\right) \leqslant w\), then we choose x over \(x^{\prime }\) only when \(u\left( x\right) +v\left( x\right) \geqslant u\left( x^{\prime }\right) +v\left( x^{\prime }\right) \). Hence, we have

$$\begin{aligned} U\left( A\right)&=\max _{x\in A}\{u\left( x\right) +v\left( x\right) \}-\max _{y\in A}\{v\left( y\right) \}\\&\text {s.t. }\max _{y\in A}v\left( y\right) -v\left( x\right) \leqslant w \end{aligned}$$

To show that the axioms are necessary, for any \(A\in {\mathscr {A}}\) let us denote the most tempting lottery \(y_{A}\in \arg \max _{y\in A}v\left( y\right) \), the admissible set \(D_{A}=\)\(\{x\in A:v\left( x\right) \geqslant w-v(y_{A})\}\) and the best admissible compromise lottery \(x_{A}\in \arg \max _{x\in D_{A} }\{u\left( x\right) +v\left( x\right) \}\). Note that u and v are both continuous functions and \(D_{A}\) is a compact set. Hence, Axiom 2a holds. For Axiom 2 to hold, note that \(\alpha y_{A}+\left( 1-\alpha \right) y_{C}\)\(\in \arg \max _{y\in \alpha A+\left( 1-\alpha \right) C}v\left( y\right) \) for all \(\alpha \in [0,1]\) and \(A,C\in {\mathscr {A}}\). Hence, we have \(\alpha x_{A}+\left( 1-\alpha \right) x_{C}\)\(\in D_{\alpha A+\left( 1-\alpha \right) C}\). By selecting \(\alpha >\frac{U\left( B\right) -U\left( C\right) }{U\left( A\right) -U\left( C\right) }\), we can verify that preferences represented by U satisfy Axiom 2b. For Axiom 3 to hold, if \(A_{1},A_{2}\in {\mathscr {B}}\left( \succsim \right) \), we claim that \(\alpha x_{A_{1}}+\left( 1-\alpha \right) x_{A_{2}}\in \arg \max _{x\in D_{\alpha A_{1}+\left( 1-\alpha \right) A_{2}}}\{u\left( x\right) +v\left( x\right) \}\). Hence, \(U\left( \alpha A_{1}+\left( 1-\alpha \right) A_{2}\right) =\alpha U\left( A_{1}\right) +\left( 1-\alpha \right) U\left( A_{2}\right) \), which yields the result. Suppose to the contrary that the above claim does not hold. Since \(\alpha x_{A_{1}}+\left( 1-\alpha \right) x_{A_{2}}\in D_{\alpha A_{1}+\left( 1-\alpha \right) A_{2}} \), we must have some \(x^{\prime }\in A_{i}\) such that \(u(x^{\prime })+v\left( x^{\prime }\right) >u\left( x_{i}\right) +v\left( x_{i}\right) \) for some \(i\in \{1,2\} \, \ \)and \(x^{\prime }\not \in D_{A_{i}}\). Since \(A_{i}\) has at most two elements and \(x^{\prime }\not =x_{A_{i}}\), we must have \(x_{A}=y_{A},\)\(u\left( x^{\prime }\right) >u\left( x_{i}\right) \) and \(U(A_{i})=u\left( x_{A}\right) \). Hence, \(x^{\prime }\in T\left( x_{A}\right) \). Since \(w>0\), we can find an \(\alpha \in (0,1)\) such that \(\alpha x^{\prime }+\left( 1-\alpha \right) x_{A_{i}}\in D_{A_{i}}\), which implies \(\alpha x^{\prime }+\left( 1-\alpha \right) x_{A_{i}}\in S\left( x_{A}\right) \). This contradicts \(A_{i}\in {\mathscr {B}}\left( \succsim \right) \). To prove Axiom 4, if \(v\left( y_{A}\right) \leqslant v\left( y_{A}\right) \), then we have \(y_{A\cup B}=y_{B}\). Hence,\(U\left( A\cup B\right) \leqslant \)\(u\left( x_{A}\right) +v\left( x_{A}\right) -v\left( y_{A}\right) \) and \(x_{B}\in D_{A\cup B}\), which implies \(A\succsim A\cup B\succsim B\). If \(v\left( y_{A}\right) >v\left( y_{B}\right) \), then \(y_{A\cup B}=y_{A}\). Hence, we have \(U(A)=U(A\cup B)\). To show that Axiom 5 holds, since \(b\in S\left( a\right) \) and \(y\in S\left( x\right) \), we have \(u\left( a\right)>U(\{a,b\})>u\left( b\right) \) and \(u\left( x\right)>U\left( \{x,y\} \right) >u\left( y\right) \). Hence, \(0<v\left( b\right) -v\left( a\right) \leqslant w\) and \(0<v\left( y\right) -v\left( x\right) \leqslant w\). Moreover, \(b\in \overline{I\left( a\right) }\) implies \(v\left( b\right) -v\left( a\right) =w\). Otherwise, we would have an open ball centered at b,  denoted as \(B\left( b\right) \), such that all \(b^{\prime }\in B\left( b\right) \) we have \(v\left( b^{\prime }\right) -v\left( a\right) <w\) and \(b^{\prime }\in S\left( a\right) \), which contradicts \(b\in \overline{I\left( a\right) }\). Hence, by straightforward computation we can conclude that \(U(\{ \frac{1}{2}a+\frac{1}{2}x,\frac{1}{2}a+\frac{1}{2}y\})\geqslant U\left( \{ \frac{1}{2}x+\frac{1}{2}a,\frac{1}{2}x+\frac{1}{2}b\} \right) \). \(\square \)

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Liang, MY., Grant, S. & Hsieh, SL. Costly self-control and limited willpower. Econ Theory 70, 607–632 (2020). https://doi.org/10.1007/s00199-019-01231-6

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Keywords

  • Temptation
  • Self-control
  • Willpower
  • Revealed preference

JEL Classifications

  • D81
  • D91
  • D11