Thoughts matter: a theory of motivated preference

Abstract

This paper develops a model of individual decision-making under bounded rationality in which discretionary cognitive adjustment creates a durable stock that complements choice of action. While it increases utility, adjustment also entails a cost, because focusing attention optimally is effortful and mental resources are scarce. Associated behavioral phenomena are categorized based on whether the operative motivation in adjusting is forward-looking utility maximization or justification of prior action. The theory is in line with prior conceptions of cognitive dissonance, but also offers a more empirically consistent explanation of the endowment effect, persuasive advertising, and sunk-cost effects than existing accounts.

Appendix

Proof of Theorem 1

The optimization problem is the same every period, so fix $t=\tau$ and consider the problem for self $\tau$ of solving (3). This player recognizes that his move $\left( x_{\tau },T_{\tau }\right)$ will influence the subsequent moves of future selves. Specifically, his choice of $T_{\tau }$ will influence future decisions, because it affects the future stock of adjustment which in turn influences the preferences of future selves; his choice of $x_{\tau }$ has no bearing on future selves’ preferences or decisions. A subgame perfect strategy takes account of how the choice of $T_{\tau }$ affects responses in all $t>\tau$ by treating $T_{t}$ as a function $T_{t}\left( T_{\tau }\right)$ for all $t>\tau$ when choosing the optimal $T_{\tau }$.

Observe from (2) that self $\tau$’s objective function nests the objective function of each $t>\tau$. One may rewrite (2) accordingly as

$$\begin{aligned} U_{\tau }\equiv \mu u\left( x_{\tau },T_{\tau }\right) +K-x_{\tau }-T_{\tau }+\delta U_{\tau +1} \end{aligned}$$

(6)

whence the nesting of future selves’ objectives for $t>\tau +1$ follows by induction. Observe also that $T_{t}$ for $t>\tau$ appears in self $\tau$’s objective function only within $U_{t}$. It follows that one may write

$$\begin{aligned} \frac{\partial U_{\tau }}{\partial T_{\tau }}=\left. \frac{\partial U_{\tau }}{\partial T_{\tau }}\right| _{D}+\sum _{t=1}^{\infty }\frac{\partial U_{\tau }}{\partial U_{\tau +t}}\frac{\partial U_{\tau +t}}{\partial T_{\tau +t}}\frac{\partial T_{\tau +t}}{\partial T_{\tau }} \end{aligned}$$

(7)

where the “D” indicates the direct effect, not through the choice of T by another self.

For an interior solution $\frac{\partial U_{\tau +t}}{\partial T_{\tau +t}}=0$ for all $t\ge 1$ in (7). Thus by the envelope theorem $\frac{\partial U_{\tau }}{\partial T_{\tau }}=\left. \frac{\partial U_{\tau }}{\partial T_{\tau }}\right| _{D}$, that is, the overall effect of $T_{\tau }$ on utility is equal to the direct effect, ignoring effects of $T_{\tau }$ through the $T_{\tau +t}$ that are already being optimized by future selves.

With this in mind, the first-order conditions of (3) may be written

$$\begin{aligned} \mu u_{x}=1; \mu u_{y}=1. \end{aligned}$$

(8)

As the problem is the same in every period, the first-order conditions yield the same $\left( x^{*}\left( \mu \right),y^{*}\left( \mu \right) \right)$ interior solution in all periods $t\ge 0$. (An interior solution is assured by the assumption of $\mu$ large enough that $\mu u_{x}\left( 0,0\right) >1$ and $\mu u_{y}\left( 0,0\right) >1$.) It follows that $T_{0}=y^{*}\left( \mu \right)$ and $T_{t}=\sigma y^{*}\left( \mu \right)$ for all $t>0$. Note that the constant marginal cost of adjustment implies the consumer engages in adjustment in each period t until the level $y^{*}\left( \mu \right)$ is reached, regardless of $y_{t-1}$. It is, moreover, easy to see from the first-order conditions that $x_{\mu }^{*},y_{\mu }^{*}>0$.

One now must show that this interior solution is the unique maximum. The derivatives of (6) needed to evaluate the Hessian with respect to this objective, reflecting the relevant values of $\left( x,y\right)$ at which they are evaluated, are

$$\begin{aligned}&\frac{\partial U_{\tau }}{\partial x_{\tau }}=\mu u_{x}\left( x_{\tau },y_{\tau }\right) -1 ;\quad \frac{\partial ^{2}U_{\tau }}{\partial x_{\tau }^{2}}=\mu u_{xx}\left( x_{\tau },y_{\tau }\right) \\ &\frac{\partial ^{2}U_{\tau }}{\partial x_{\tau }\partial T_{\tau }}=\mu u_{xy}\left( x_{\tau },y_{\tau }\right) ;\quad \frac{\partial U_{\tau }}{\partial T_{\tau }}=\mu u_{y}\left( x_{\tau },y_{\tau }\right) +\delta \left( 1-\sigma \right) \mu u_{y}\left( x_{\tau +1},y_{\tau +1}\right) +\cdots -1 \\ &\frac{\partial ^{2}U_{\tau }}{\partial T_{\tau }^{2}} = \mu u_{yy}\left( x_{\tau },y_{\tau }\right) +\delta \left( 1-\sigma \right) ^{2}\mu u_{yy}\left( x_{\tau +1},y_{\tau +1}\right) +\cdots \end{aligned}$$

(9)

To simplify notation, I drop the arguments for $t=\tau$ and show only arguments for $t>\tau$. The Hessian is

$$\begin{aligned} \left| H\right|= & {} \mu ^{2}\left| \begin{array}{cc} u_{xx} &{} u_{xy}\\ u_{xy} &{} u_{yy}+\delta \left( 1-\sigma \right) ^{2}u_{yy}\left( x_{\tau +1},y_{\tau +1}\right) +\cdots \end{array}\right| \\= & {} u_{xx}u_{yy}-u_{xy}^{2}+u_{xx}\delta \left( 1-\sigma \right) ^{2}u_{yy}\left( x_{\tau +1},y_{\tau +1}\right) +\cdots>u_{xx}u_{yy}-u_{xy}^{2}>0 \end{aligned}$$

(10)

which follows from the strict concavity of $u\left( .\right)$. Since $\frac{\partial ^{2}U}{\partial x_{\tau }^{2}}<0$ and $\frac{\partial ^{2}U}{\partial T_{\tau }^{2}}<0$, $U_{\tau }\left( x_{\tau },T_{\tau }\right)$ is strictly concave. This establishes the solution $\left( x_{\tau }^{*},y_{\tau }^{*}\right)$ to the first-order conditions as the unique interior absolute maximum for $\tau$. By induction, this is true for all $\tau \ge 0$.

To dispense with the possibility of a corner solution: it was noted previously that the unique interior solution is $\left( x_{t}^{*},y_{t}^{*}\right) =\left( x^{*}\left( \mu \right),y^{*}\left( \mu \right) \right)$ for all $t\ge 0$, thus $T_{t}^{*}=\sigma y^{*}\left( \mu \right)$ for all $t>0$. A corner solution of $T_{1}=0$ could result if and only if $\left( 1-\sigma \right) y_{0}^{*}\left( \mu \right) >y^{*}\left( \mu \right)$. But this would imply $y_{0}^{*}\left( \mu \right) >y^{*}\left( \mu \right)$, which is a contradiction of the unique interior solution for $t=0$. Thus no corner solution will result in $t=1$. By similar argument, a corner solution in any $t=\tau$ requires $y_{\tau -1}^{*}\left( \mu \right) >y^{*}\left( \mu \right)$, a contradiction of the interior solution for the prior period. By induction, the interior solution is the only solution for all periods.

This confirms $\left( x_{0},T_{0}\right) =\left( x^{*}\left( \mu \right),y^{*}\left( \mu \right) \right)$ and $\left( x_{t},T_{t}\right) =\left( x^{*}\left( \mu \right),\sigma y^{*}\left( \mu \right) \right)$ for all $t>0$ constitutes the unique subgame perfect equilibrium strategy for the infinite game.

Proof of Corollary 1

Without varying the setup for Theorem 1 substantively, one may treat self 0 as two selves, one ($0_{x}$) who sets $x_{0}$ and another ($0_{T}$) who subsequently sets $T_{0}$, both with $\mu$ known. Both receive the same payoff, given by (2) for $\tau =0$, subsequent to $0_{T}$’s move. It is obvious that the theorem holds for this trivial variation. Relative to this, an endowment constitutes an out-of-equilibrium deviation for self $0_{x}$. The subgame that follows this deviation has a unique subgame perfect equilibrium strategy, per Theorem 1, by definition of a subgame perfect equilibrium.

Proof of Lemma 1

The derivatives of (4) needed to evaluate the Hessian with respect to this objective are

$$\begin{aligned}&\frac{\partial U^{R}}{\partial x_{\tau }}=\left( 1-\beta _{\tau }\right) \left( \mu u_{x}\left( x_{\tau },y_{\tau }\right) -1\right) ;\quad \frac{\partial ^{2}U^{R}}{\partial x_{\tau }^{2}}=\left( 1-\beta _{\tau }\right) \mu u_{xx}\left( x_{\tau },y_{\tau }\right) \\ &\frac{\partial ^{2}U^{R}}{\partial x_{\tau }\partial T_{\tau }}=\left( 1-\beta _{\tau }\right) \mu u_{xy}\left( x_{\tau },y_{\tau }\right) \\ &\frac{\partial U^{R}}{\partial T_{\tau }}=\left( 1-\beta _{\tau }\right) \left( \mu u_{y}\left( x_{\tau },y_{\tau }\right) -1+\delta \left\{ \left[ \left( 1-\sigma \right) +T_{\tau +1}^{\prime }\left( T_{\tau }\right) \right] \mu u_{y}\left( x_{\tau +1},y_{\tau +1}\right) -T_{\tau +1}^{\prime }\left( T_{\tau }\right) \right\} \right. \\ &\quad \left. +\delta ^{2}\left\{ \left[ \left( 1-\sigma \right) ^{2}+\left( 1-\sigma \right) T_{\tau +1}^{\prime }\left( T_{\tau }\right) +T_{\tau +2}^{\prime }\left( T_{\tau }\right) \right] \mu u_{y}\left( x_{\tau +2},y_{\tau +2}\right) -T_{\tau +2}^{\prime }\left( T_{\tau }\right) \right\} +\cdots \right) \\ &\quad +\beta _{\tau }\left( \mu u_{y}\left( \bar{x}_{0},y_{\tau }\right) -\mu u_{y}\left( x_{0}^{*}\left( T_{\tau }\right),y_{\tau }\right) \right) \\ &\frac{\partial ^{2}U^{R}}{\partial T_{\tau }^{2}}=\left( 1-\beta _{\tau }\right) \left( \mu u_{yy}\left( x_{\tau },y_{\tau }\right) +\delta \left\{ \left[ \left( 1-\sigma \right) +T_{\tau +1}^{\prime }\right] ^{2}\mu u_{yy}\left( x_{\tau +1},y_{\tau +1}\right) \right\} \right. \\ &\quad \left. +\delta ^{2}\left\{ \left[ \left( 1-\sigma \right) ^{2}+\left( 1-\sigma \right) T_{\tau +1}^{\prime }+T_{\tau +2}^{\prime }\right] ^{2}\mu u_{yy}\left( x_{\tau +2},y_{\tau +2}\right) \right\} +\cdots \right) \\ &\quad +\beta _{\tau }\left[ \mu u_{yy}\left( \bar{x}_{0},y_{\tau }\right) -\mu u_{yy}\left( x_{0}^{*}\left( T_{\tau }\right),y_{\tau }\right) -\mu u_{xy}\left( x_{0}^{*}\left( T_{\tau }\right),y_{\tau }\right) x_{T}^{*}\right] \end{aligned}$$

(11)

where $x_{0}^{*}\left( T_{0}\right)$ arises implicitly as the solution to $\mu u_{x}\left( x_{0}^{*},y_{0}\right) =1$. (Thus terms $\mu u_{x}\left( x_{0}^{*}\left( T_{0}\right),y_{0}\right) x_{T}^{*}-x_{T}^{*}$ drop out of the parentheses on the bottom line of $\frac{\partial U^{R}}{\partial T_{\tau }}$.)

Suppose first that $\left( 1-\sigma \right) y_{\tau -1}<y_{\tau }^{*}$, such that the solution arising from the first-order condition is an interior solution and candidate unique solution. Note, in this case, that all the $T_{t}^{\prime }$ terms in (11) are constants that either take a value of zero or $-\left( 1-\sigma \right) ^{t-\tau }$. The value taken by each of these terms follows from the form of the state equation (1) and depends specifically on the size of the base value of $T_{\tau }$, which determines in turn whether each $T_{t}$ is a corner or interior solution. Suppose first that $T_{\tau }$ is relatively small; then the depreciated value of adjustment $\left( 1-\sigma \right) y_{\tau }$ will be less than the optimizing value of $y_{\tau +1}$, whereby there will be an interior solution for $T_{\tau +1}$ such that $T_{\tau +1}^{\prime }\left( T_{\tau }\right) =-\left( 1-\sigma \right)$. It follows that $T_{\tau +t}^{\prime }\left( T_{\tau }\right) =0$ for all $t>1$. If, however, $T_{\tau }$ is a bit larger but not too large, $\left( 1-\sigma \right) y_{\tau }$ will be large enough to push $T_{\tau +1}$ to a corner solution at zero; $T_{\tau }$ would then be large enough that changes in its value would influence the choice of T in the following period, $\tau +2$, whence $T_{\tau +2}^{\prime }\left( T_{\tau }\right) =-\left( 1-\sigma \right) ^{2}$ and $T_{\tau +t}^{\prime }\left( T_{\tau }\right) =0$ for all $t>2$ . And so on. It can be shown, therefore, that each squared expression in square brackets in $\frac{\partial ^{2}U^{R}}{\partial T_{\tau }^{2}}$ is non-negative; more precisely, they take the value $\left( 1-\sigma \right) ^{2t}$ up to a threshold t and zero thereafter. Overall, the sum of the curly bracketed terms within the non-regret component of $\frac{\partial ^{2}U^{R}}{\partial T_{\tau }^{2}}$ (i.e., the portion that is weighted by $1-\beta _{\tau }$), which I shall refer to as “NR” for convenience, is unambiguously negative. The regret component (weighted by $\beta _{\tau }$), which I shall refer to as “R”, is ambiguously signed.

Dropping the arguments for $t=\tau$ to simplify notation, the Hessian may be written

$$\begin{aligned} \left| H\right|&=\mu ^{2}\left| \begin{array}{cc} \left( 1-\beta _{\tau }\right) u_{xx} &{} \left( 1-\beta _{\tau }\right) u_{xy}\\ \left( 1-\beta _{\tau }\right) u_{xy} &{} \left( 1-\beta _{\tau }\right) \left( u_{yy}+NR\right) +\beta _{\tau }R \end{array}\right| \\&= \mu ^{2}\left( 1-\beta _{\tau }\right) ^{2}\left[ u_{xx}u_{yy}-u_{xy}^{2}+u_{xx}NR\right] +\mu ^{2}\left( 1-\beta _{\tau }\right) \beta _{\tau }u_{xx}R. \end{aligned}$$

This takes the sign of

$$\begin{aligned}&\left( 1-\beta _{\tau }\right) \left( u_{xx}u_{yy}-u_{xy}^{2}+u_{xx}NR\right) +\beta _{\tau }u_{xx}R. \end{aligned}$$

(12)

Given the strict concavity of $u\left( .\right)$, $u_{xx}u_{yy}-u_{xy}^{2}>0$, whereby it follows that the first term in (12) is positive. It is, therefore, sufficient for signing the entire expression positive, given $\beta _{\tau }$ sufficiently small, that R be bounded above.

Proof of Proposition 1

Begin by writing the version of (4) for $\tau =0$:

$$\begin{aligned}&\max _{T_{0}\left| \bar{x}_{0}\right. }U^{R}\left( \bar{x}_{0},T_{0}\right) =\left( 1-\beta _{0}\right) \left\{ \mu u\left( \bar{x}_{0},T_{0}\right) +K-\bar{x}_{0}-T_{0}+\sum _{t=1}^{\infty }\delta ^{t}\left[ \mu u\left( x_{t},T_{t}\right) +K-x_{t}-T_{t}\right] \right\} \\ &\quad +\beta _{0}\left\{ \left[ \mu u\left( \bar{x}_{0},T_{0}\right) +K-\bar{x}_{0}-T_{0}\right] -\left[ \mu u\left( x_{0}^{*}\left( T_{0}\right),T_{0}\right) +K-x_{0}^{*}\left( T_{0}\right) -T_{0}\right] \right\} . \end{aligned}$$

(13)

Take the first-order condition with respect to $T_{0}$, using (9):

$$\begin{aligned}&\left( 1-\beta _{0}\right) \left( \mu u_{y}\left( \bar{x}_{0},y_{0}\right) -1+\delta \left\{ \left[ \left( 1-\sigma \right) +T_{1}^{\prime }\left( T_{0}\right) \right] \mu u_{y}\left( x_{1},y_{1}\right) -T_{1}^{\prime }\left( T_{0}\right) \right\} \right. \\ &\quad \left. +\delta ^{2}\left\{ \left[ \left( 1-\sigma \right) ^{2}+\left( 1-\sigma \right) T_{1}^{\prime }\left( T_{0}\right) +T_{2}^{\prime }\left( T_{0}\right) \right] \mu u_{y}\left( x_{2},y_{2}\right) -T_{2}^{\prime }\left( T_{0}\right) \right\} +\cdots \right) \\ &\quad +\beta _{0}\left\{ \mu u_{y}\left( \bar{x}_{0},y_{0}\right) -\mu u_{y}\left( x_{0}^{*}\left( T_{0}\right),y_{0}\right) \right\} =0 \end{aligned}$$

(14)

where as mentioned in the proof of Lemma 1, $x_{0}^{*}\left( T_{0}\right)$ arises implicitly as the solution to $\mu u_{x}\left( x_{0}^{*},y_{0}\right) =1$. (Thus terms $\mu u_{x}\left( x_{0}^{*}\left( T_{0}\right),y_{0}\right) x_{T}^{*}-x_{T}^{*}$ drop out of the last line.) Recall also from the proof of the lemma that all terms $T_{\tau }^{\prime }\left( T_{0}\right)$ are constants. Totally differentiating (14) and using $D_{1}$, $D_{2}$, $N_{11}$, $N_{12}$, and $N_{2}$ to represent respective curly bracketed terms yields

$$\begin{aligned}&\left[ \left( 1-\beta _{0}\right) \left\{ \mu u_{yy}\left( \bar{x}_{0},y_{0}\right) +\delta \left[ \left( 1-\sigma \right) +T_{1}^{\prime }\left( T_{0}\right) \right] ^{2}\mu u_{yy}\left( x_{1},y_{1}\right) +\cdots \right\} \right. \\ &\qquad \left. \left. +\beta _{0}\left\{ \mu u_{yy}\left( \bar{x}_{0},y_{0}\right) \right. -\mu u_{yy}\left( x_{0}^{*}\left( T_{0}\right),y_{0}\right) -\mu u_{xy}\left( x_{0}^{*}\left( T_{0}\right),y_{0}\right) x_{T}^{*}\right\} \right] dT_{0} \\ &\quad =-\left[ \left( 1-\beta _{0}\right) \left\{ u_{y}\left( \bar{x}_{0},y_{0}\right) +\delta \left[ \left( 1-\sigma \right) +T_{1}^{\prime }\left( T_{0}\right) \right] u_{y}\left( x_{1},y_{1}\right) +\cdots \right\} \right. \\ &\qquad \left. +\beta _{0}\left\{ u_{y}\left( \bar{x}_{0},y_{0}\right) -u_{y}\left( x_{0}^{*}\left( T_{0}\right),y_{0}\right) \right\} \right] d\mu -\left[ \left( 1-\beta _{0}\right) \left\{ \mu u_{xy}\left( \bar{x}_{0},y_{0}\right) \right\} +\beta _{0}\left\{ \mu u_{xy}\left( \bar{x}_{0},y_{0}\right) \right\} \right] dx_{0} \\ &\quad \Longleftrightarrow \left[ \left( 1-\beta _{0}\right) D_{1}+\beta _{0}D_{2}\right] dT_{0}=\left[ -\left( 1-\beta _{0}\right) N_{11}-\beta _{0}N_{12}\right] d\mu -N_{2}dx_{0}. \end{aligned}$$

(15)

This yields, using Cramer’s rule, for $\beta _{0}=0$:

$$\begin{aligned} \frac{\partial T_{0}}{\partial \mu }= & {} -\frac{N_{11}}{D_{1}}=-\frac{u_{y}\left( \bar{x}_{0},y_{0}\right) +\delta \left[ \left( 1-\sigma \right) +T_{1}^{\prime }\left( T_{0}\right) \right] u_{y}\left( x_{1},y_{1}\right) +\cdots }{\mu u_{yy}\left( \bar{x}_{0},y_{0}\right) +\delta \left[ \left( 1-\sigma \right) +T_{1}^{\prime }\left( T_{0}\right) \right] ^{2}\mu u_{yy}\left( x_{1},y_{1}\right) +\cdots }>0. \end{aligned}$$

(16)

For $\beta _{0}\approx 1$—the case of extreme regret in the initial period—$y_{0}^{*}$ is set for given $\bar{x}_{0}$ by implicit solution to $\mu u_{x}\left( \bar{x}_{0},y_{0}^{*}\right) =1$ as the “justifying” value of $y_{0}$. One performs comparative static analysis for this case by totally differentiating $\mu u_{x}\left( \bar{x}_{0},y_{0}^{*}\right) =1$:

$$\begin{aligned} \mu u_{xy}\left( \bar{x}_{0},y_{0}^{*}\right) dT_{0}=-u_{x}\left( \bar{x}_{0},y_{0}^{*}\right) d\mu -u_{xx}\left( \bar{x}_{0},y_{0}^{*}\right) d\bar{x}_{0}. \end{aligned}$$

(17)

This yields

$$\begin{aligned} \frac{\partial T_{0}}{\partial \mu }=-\frac{u_{x}\left( \bar{x}_{0},y_{0}^{*}\right) }{\mu u_{xy}\left( \bar{x}_{0},y_{0}^{*}\right) }<0. \end{aligned}$$

To understand what happens as $\beta _{0}$ varies between 0 and 1, one applies Cramer’s rule to (15) for general $\beta _{0}$ and then differentiates with respect to $\beta _{0}$:

$$\begin{aligned} \frac{\partial T_{0}}{\partial \mu }= & {} -\frac{\left( 1-\beta _{0}\right) N_{11}+\beta _{0}N_{12}}{\left( 1-\beta _{0}\right) D_{1}+\beta _{0}D_{2}} \\ \rightarrow \frac{\partial ^{2}T_{0}}{\partial \mu \partial \beta _{0}}= & {} -\tfrac{\left[ \left( 1-\beta _{0}\right) D_{1}+\beta _{0}D_{2}\right] \left( N_{12}-N_{11}\right) -\left[ \left( 1-\beta _{0}\right) N_{11}+ \beta _{0}N_{12}\right] \left( D_{2}-D_{1}\right) }{\left[ \left( 1- \beta _{0}\right) D_{1}+\beta _{0}D_{2}\right] ^{2}}. \end{aligned}$$

(18)

Recognizing that (18) takes the sign of the numerator, let us evaluate the numerator. This can be simplified as follows, prior to substitutions:

$$\begin{aligned}&-\left[ \left( 1-\beta _{0}\right) D_{1}+\beta _{0}D_{2}\right] \left( N_{12}-N_{11}\right) +\left[ \left( 1-\beta _{0}\right) N_{11}+\beta _{0}N_{12}\right] \left( D_{2}-D_{1}\right) \\ &\quad =-\left( 1-\beta _{0}\right) D_{1}N_{12}+\left( 1-\beta _{0}\right) D_{1}N_{11}-\beta _{0}D_{2}N_{12}+\beta _{0}D_{2}N_{11}\\ &\qquad -\left( 1-\beta _{0}\right) D_{1}N_{11}+\left( 1-\beta _{0}\right) D_{2}N_{11}-\beta _{0}D_{1}N_{12}+\beta _{0}D_{2}N_{12}\\ &\quad =-\left( 1-\beta _{0}\right) D_{1}N_{12}+\beta _{0}D_{2}N_{11}+\left( 1-\beta _{0}\right) D_{2}N_{11}-\beta _{0}D_{1}N_{12}\\ &\quad =-D_{1}N_{12}+D_{2}N_{11}. \end{aligned}$$

Substitution back yields

$$\begin{aligned}&-\left\{ \mu u_{yy}\left( \bar{x}_{0},y_{0}\right) +\delta \left[ \left( 1-\sigma \right) +T_{1}^{\prime }\left( T_{0}\right) \right] ^{2}\mu u_{yy}\left( x_{1},y_{1}\right) +\cdots \right\} \left\{ u_{y}\left( \bar{x}_{0},y_{0}\right) -u_{y}\left( x_{0}^{*}\left( T_{0}\right),y_{0}\right) \right\} \\ &\quad +\left\{ \mu u_{yy}\left( \bar{x}_{0},y_{0}\right) -\mu u_{yy}\left( x_{0}^{*}\left( T_{0}\right),y_{0}\right) -\mu u_{xy}\left( x_{0}^{*}\left( T_{0}\right),y_{0}\right) x_{T}^{*}\right\} \\ &\quad \times \left\{ u_{z}z_{y}\left( \bar{x}_{0},y_{0}\right) +\delta \left[ \left( 1-\sigma \right) +T_{1}^{\prime }\left( T_{0}\right) \right] u_{z}z_{y}\left( x_{1},y_{1}\right) +\cdots \right\} . \end{aligned}$$

Now evaluate this expression at $\mu =\bar{\mu }$, the value of $\mu$ at which the endowment $\bar{x}_{0}$ would have resulted if $\left( \bar{x}_{0},T_{0}\right)$ were chosen simultaneously with $\mu$ known, that is, the value of $\mu$ for which $x_{0}^{*}\left( T_{0}\right) =\bar{x}_{0}$. This is the appropriate value at which to evaluate $\nicefrac {\partial ^{2}T_{0}}{\partial \mu \partial \beta _{0}}$, as the expression is then interpreted as showing how regret influences the individual’s response to a quality surprise relative to endowment-based expectations. At this value, the second term in curly brackets is zero. This leaves

$$\begin{aligned} \mu \left\{ u_{yy}\left( \bar{x}_{0},y_{0}\right) -u_{yy}\left( x_{0}^{*}\left( T_{0}\right),y_{0}\right) -u_{xy}\left( x_{0}^{*}\left( T_{0}\right),y_{0}\right) x_{T}^{*}\right\} \left\{ u_{y}\left( \bar{x}_{0},y_{0}\right) +\delta \left[ \left( 1-\sigma \right) +T_{1}^{\prime }\left( T_{0}\right) \right] u_{y}\left( x_{1},y_{1}\right) +\cdots \right\} \end{aligned}$$

The first and second terms cancel at $x_{0}^{*}\left( T_{0}\right) =\bar{x}_{0}$, leaving

$$\begin{aligned} -\mu u_{xy}\left( x_{0}^{*}\left( T_{0}\right),y_{0}\right) x_{T}^{*}\cdot \left\{ u_{y}\left( \bar{x}_{0},y_{0}\right) +\delta \left[ \left( 1-\sigma \right) +T_{1}^{\prime }\left( T_{0}\right) \right] u_{y}\left( x_{1},y_{1}\right) +\cdots \right\} \end{aligned}$$

(19)

An expression for $x_{T}^{*}$ comes from totally differentiating $\mu u_{x}\left( x_{0}^{*},y_{0}\right) =1$:

$$\begin{aligned} \mu u_{xx}\left( x_{0}^{*},y_{0}\right) dx_{0}= & {} -\mu u_{xy}\left( x_{0}^{*},y_{0}\right) dT_{0} \\ \rightarrow x_{T}^{*}= & {} -\frac{u_{xy}\left( x_{0}^{*},y_{0}\right) }{u_{xx}\left( x_{0}^{*},y_{0}\right) }>0. \end{aligned}$$

(20)

Thus, one can sign (19) as negative. It follows that $\nicefrac {\partial ^{2}T_{0}}{\partial \mu \partial \beta _{0}}<0$.

Proof of Proposition 2

Applying Cramer’s rule to (15) yields, for $\beta _{0}=0$:

$$\begin{aligned} \frac{\partial T_{0}}{\partial \bar{x}_{0}}= & {} -\frac{N_{2}}{D_{1}}=-\frac{\mu u_{xy}\left( \bar{x}_{0},y_{0}\right) }{\mu u_{yy}\left( \bar{x}_{0},y_{0}\right) +\delta \left[ \left( 1-\sigma \right) +T_{1}^{\prime }\left( T_{0}\right) \right] ^{2}\cdot \mu u_{yy}\left( x_{1},y_{1}\right) +\cdots }>0. \end{aligned}$$

For $\beta _{0}\approx 1$, one may apply Cramer’s rule to (17), yielding

$$\begin{aligned} \frac{\partial T_{0}}{\partial \bar{x}_{0}}= & {} -\frac{u_{xx}\left( \bar{x}_{0},y_{0}^{*}\right) }{\mu u_{xy}\left( \bar{x}_{0},y_{0}^{*}\right) }>0. \end{aligned}$$

To understand what happens as $\beta _{0}$ varies between 0 and 1, apply Cramer’s rule to (15) for general $\beta _{0}$ and then differentiate with respect to $\beta _{0}$:

$$\begin{aligned} \frac{\partial T_{0}}{\partial x_{0}}=-\frac{N_{2}}{\left( 1-\beta _{0}\right) D_{1}+\beta _{0}D_{2}}\rightarrow \frac{\partial ^{2}T_{0}}{\partial x_{0}\partial \beta _{0}}=\tfrac{N_{2}\left( D_{2}-D_{1}\right) }{\left[ \left( 1-\beta _{0}\right) D_{1}+\beta _{0}D_{2}\right] ^{2}}. \end{aligned}$$

(21)

The sign of (21) takes on the sign of the numerator. Substitution back yields

$$\begin{aligned}&\left\{ \mu u_{xy}\left( \bar{x}_{0},y_{0}\right) \right\} \cdot \left\{ \mu u_{yy}\left( \bar{x}_{0},y_{0}\right) -\mu u_{yy}\left( x_{0}^{*}\left( T_{0}\right),y_{0}\right) -\mu u_{xy}\left( x_{0}^{*}\left( T_{0}\right),y_{0}\right) x_{T}^{*}\right. \\ &\quad \left. -\mu u_{yy}\left( \bar{x}_{0},y_{0}\right) -\delta \left[ \left( 1-\sigma \right) +T_{1}^{\prime }\left( T_{0}\right) \right] ^{2}\mu u_{yy}\left( x_{1},y_{1}\right) -\cdots \right\} . \end{aligned}$$

The first bracketed term is clearly positive. The first and fourth terms in the second bracketed expression cancel. Evaluate the second and third, substituting (20):

$$\begin{aligned}&-\mu u_{yy}\left( x_{0}^{*}\left( T_{0}\right),y_{0}\right) - \mu u_{xy}\left( x_{0}^{*}\left( T_{0}\right),y_{0}\right) \left[ -\frac{u_{xy}\left( x_{0}^{*}\left( T_{0}\right),y_{0}\right) }{u_{xx}\left( x_{0}^{*}\left( T_{0}\right),y_{0}\right) }\right] \\ &\quad =-\mu u_{yy}\left( x_{0}^{*}\left( T_{0}\right),y_{0}\right) + \mu \frac{\left[ u_{xy}\left( x_{0}^{*}\left( T_{0}\right),y_{0}\right) \right] ^{2}}{u_{xx}\left( x_{0}^{*}\left( T_{0}\right),y_{0}\right) }. \end{aligned}$$

Multiply through by $-u_{xx}\left( x_{0}^{*}\left( T_{0}\right),y_{0}\right)$ and the sign is the same. Removing the arguments to save space:

$$\begin{aligned} -\mu u_{xy}^{2}+\mu u_{xx}u_{yy}=\mu \left( u_{xx}u_{yy}-u_{xy}^{2}\right) >0. \end{aligned}$$

Thus $\nicefrac {\partial ^{2}T_{0}}{\partial \bar{x}_{0}\partial \beta _{0}}>0$.

Proof of Proposition 3

Let $\left( x^{*},y^{*}\right)$ represent the steady-state optimum under the unique subgame perfect equilibrium with no endowment, and let $y_{0}\left( \bar{x}_{0}\right) =T_{0}\left( \bar{x}_{0}\right)$ represent the level of $y_{0}$ arising from the no-regret first-order condition for T based on $\bar{x}_{0}$. Observe that $\left( 1-\sigma \right) y_{0}\left( \bar{x}_{01}\right) =y^{*}$ defines implicitly the threshold level $\bar{x}_{01}$ of the endowment that leads to a corner solution for adjustive thinking in $t=1$. That is, it is the minimum level that induces the consumer to set $T_{1}=0$: for lower levels of $x_{0}$, $y_{1}<y^{*}$ unless $T_{1}>0$. At the threshold itself, the unique interior solution that obtains for $t>0$ remains unaffected, thus $\left( x_{t},y_{t}\right) =\left( x^{*},y^{*}\right)$ for $t\ge 1$ even when $\left( x_{0},y_{0}\right) =\left( \bar{x}_{01},\frac{y^{*}}{1-\sigma }\right)$. Because $y_{0}\left( \bar{x}_{01}\right)$ itself is defined based on the first-order condition for T, use that condition to write

$$\begin{aligned} \mu u_{y}\left( \bar{x}_{01},\frac{y^{*}}{1-\sigma }\right) +\delta \left( 1-\sigma \right) \mu u_{y}\left( x^{*},y^{*}\right) +\cdots= & {} 1 \end{aligned}$$

(22)

(Note that here, because endowments may create corner solutions in the optimization problem of future selves, the full first-order condition for T must be used, rather than the simplified version used in the proof of Theorem 1 based on application of the envelope theorem.) Using the fact that the optimum still obtains at the threshold, one may write

$$\begin{aligned} \mu u_{y}\left( \bar{x}_{01},\frac{y^{*}}{1-\sigma }\right)= & {} \mu u_{y}\left( x^{*},y^{*}\right) . \end{aligned}$$

(23)

Now consider that $\left( 1-\sigma \right) ^{2}y_{0}\left( \bar{x}_{02}\right) =y^{*}$ defines implicitly the threshold level $\bar{x}_{02}$ of the endowment that leads to a corner solution for adjustive thinking in $t=2$. Analogous to the above, one needs the first-order condition for T to obtain a precise expression for $\bar{x}_{02}$; however, in that expression it will be necessary to replace $\left( x_{1},y_{1}\right)$ with threshold values in place of the optimal values appearing in (23). Given the depreciation rate of adjustment, $y_{1}=\frac{y^{*}}{1-\sigma }$. Therefore, because the process is the same every period, there is an expression identical to (23) that defines a hypothetical $t=1$ threshold level of consumption $\bar{x}_{12}$ for adjusting thinking in $t=2$, whereby $\bar{x}_{12}=\bar{x}_{01}$. Using the first-order condition with respect to T then

$$\begin{aligned} \mu u_{y}\left( \bar{x}_{02},\frac{y^{*}}{\left( 1-\sigma \right) ^{2}}\right) +\delta \left( 1-\sigma \right) \mu u_{y}\left( \bar{x}_{01},\frac{y^{*}}{1-\sigma }\right) +\cdots= & {} 1 \end{aligned}$$

whereby substituting (23) yields

$$\begin{aligned} \mu u_{y}\left( \bar{x}_{02},\frac{y^{*}}{\left( 1-\sigma \right) ^{2}}\right) +\delta \left( 1-\sigma \right) \mu u_{y}\left( x^{*},y^{*}\right) +\cdots= & {} 1 \end{aligned}$$

whence, again using the fact that the optimum still obtains at the threshold, one may write

$$\begin{aligned} \mu u_{y}\left( \bar{x}_{02},\frac{y^{*}}{\left( 1-\sigma \right) ^{2}}\right)= & {} \mu u_{y}\left( x^{*},y^{*}\right) . \end{aligned}$$

By induction, the level of the endowment that corresponds to the threshold of a corner solution for adjustive thinking in $t=\bar{t}$ is defined implicitly by

$$\begin{aligned} \mu u_{y}\left( \bar{x}_{0\bar{t}},\frac{y^{*}}{\left( 1-\sigma \right) ^{\bar{t}}}\right)= & {} \mu u_{y}\left( x^{*},y^{*}\right) . \end{aligned}$$

For any $\bar{x}_{0}>\bar{x}_{0\bar{t}}$, $u_{xy}>0$ implies

$$\begin{aligned} \mu u_{y}\left( \bar{x}_{0},\frac{y^{*}}{\left( 1-\sigma \right) ^{\bar{t}}} \right)> & {} \mu u_{y}\left( x^{*},y^{*}\right) \end{aligned}$$

and, moreover, $y_{t}>y^{*}$ for all $t\le \bar{t}$; which in turn implies $x_{t}>x^{*}$, as well as an increased marginal valuation for $x_{t}$ at all values of $x_{t}$, which is what we sought to show.

Proof of Proposition 4

Let $\left( x^{*},y^{*}\right)$ represent the steady-state optimum under the unique subgame perfect equilibrium with no advertising, and let $\left( x\left( \tilde{y}_{01}\right),\tilde{y}_{01}\right)$ represent the consumption-adjustment pair corresponding to a level of advertising (write as $A_{01}$) in $t=0$ just sufficient to induce a corner solution for adjustive thinking in $t=1$. Using the full first-order condition for T as in the proof of Proposition 3, it follows that

$$\begin{aligned}&\mu u_{y}\left( x^{*},y^{*}\right) +\delta \left( 1-\sigma \right) \mu u_{y}\left( x^{*},y^{*}\right) +\cdots = 1\\ &\mu u_{y}\left( x\left( \tilde{y}_{01}\right),\tilde{y}_{01}\right) +\delta \left( 1-\sigma \right) \mu u_{y}\left( x^{*},y^{*}\right) +\cdots = 1-\phi \left( A_{01}\right) \end{aligned}$$

where the second condition follows, because one is dealing with the threshold value of the corner solution. Using the fact that all the terms on the left-hand side in these expressions are equal beyond the first, one may write

$$\begin{aligned} 1-\mu u_{y}\left( x^{*},y^{*}\right) &= {} 1-\phi \left( A_{01}\right) -\mu u_{y}\left( x\left( \tilde{y}_{01}\right),\tilde{y}_{01}\right) \\ \rightarrow \frac{\phi \left( A_{01}\right) }{\mu }&= u_{y}\left( x^{*},y^{*}\right) -u_{y}\left( x\left( \tilde{y}_{01}\right),\tilde{y}_{01}\right) \\&=u_{y}\left( x^{*},y^{*}\right) -u_{y}\left( x\left( \frac{y^{*}}{1-\sigma }\right),\frac{y^{*}}{1-\sigma }\right) \end{aligned}$$

where, for the last line, I again use the fact that $\tilde{y}_{01}$ just induces a corner solution in $t=1$.

Now consider the level of advertising in $t=0$, $A_{02}$, just sufficient to induce a corner solution for adjustive thinking in $t=2$, and consider the corresponding $\tilde{y}_{02}$. Using the first-order condition for T:

$$\begin{aligned} \mu u_{y}\left( x\left( \tilde{y}_{02}\right),\tilde{y}_{02}\right) +\delta \left( 1-\sigma \right) \mu u_{y}\left( x\left( \left( 1-\sigma \right) \tilde{y}_{02}\right),\left( 1-\sigma \right) \tilde{y}_{02}\right) +\cdots= & {} 1-\phi \left( A_{02}\right) . \end{aligned}$$

Here, because it reflects the threshold value of the corner solution at $t=2$, all terms on the left-hand side are equal to those in the steady-state first-order condition beyond the second; this means one can write

$$\begin{aligned}&1-\mu u_{y}\left( x^{*},y^{*}\right) -\delta \left( 1-\sigma \right) \mu u_{y}\left( x^{*},y^{*}\right) = 1-\phi \left( A_{02}\right) -\mu u_{y}\left( x\left( \tilde{y}_{02}\right),\tilde{y}_{02}\right) \\ &\qquad - \delta \left( 1-\sigma \right) \mu u_{y}\left( x\left( \left( 1-\sigma \right) \tilde{y}_{02}\right),\left( 1-\sigma \right) \tilde{y}_{02}\right) \\ &\quad \rightarrow \frac{\phi \left( A_{02}\right) }{\mu } = u_{y}\left( x^{*},y^{*}\right) -u_{y}\left( x\left( \frac{y^{*}}{\left( 1-\sigma \right) ^{2}}\right),\frac{y^{*}}{\left( 1-\sigma \right) ^{2}}\right) \\ &\qquad + \delta \left( 1-\sigma \right) u_{y}\left( x^{*},y^{*}\right) -\delta \left( 1-\sigma \right) u_{y}\left( x\left( \frac{y^{*}}{1-\sigma }\right),\frac{y^{*}}{1-\sigma }\right) . \end{aligned}$$

By induction

$$\begin{aligned} \frac{\phi \left( A_{0\bar{t}}\right) }{\mu }> & {} \sum _{i=0}^{\bar{t}-1}\delta ^{i}\left( 1-\sigma \right) ^{i}\left[ u_{y}\left( x^{*},y^{*}\right) -u_{y}\left( x\left( \frac{y^{*}}{\left( 1-\sigma \right) ^{\bar{t}-i}}\right),\frac{y^{*}}{\left( 1-\sigma \right) ^{\bar{t}-i}}\right) \right] \end{aligned}$$

(24)

defines the level of advertising $A_{0\bar{t}}$ in $t=0$ sufficiently large to induce $y_{t}>y^{*}$ and $x_{t}>x^{*}$ for all $t\le \bar{t}$. Note that existence of such a level of advertising depends on $\frac{\alpha }{\mu }$ being larger than the right-hand side of (24); specifically, this is a necessary condition for $\phi \left( A\right)$ being able to get large enough, given its limit, for (24) to hold.

The welfare results follow trivially: because the consumer adjusts in equilibrium, and advertising reduces the cost of adjustment, it relaxes the consumer’s constraint and so increases utility. The only additional requirement for the welfare effect of advertising to be unambiguously positive overall is that advertising be profitable for firms.

Proof of Proposition 5

The proof of Proposition 3 establishes that for the no-regret case there exists a threshold endowment level for every $\tau$ that is defined implicitly by $\left( 1-\sigma \right) ^{\tau }y_{0}\left( \bar{x}_{0\tau }\right) =y^{*}$, where $y_{0}\left( \bar{x}_{0\tau }\right) =T_{0}\left( \bar{x}_{0\tau }\right)$ is in turn defined by the first-order condition with respect to T, evaluated at $\bar{x}_{0\tau }$. Totally differentiating $\left( 1-\sigma \right) ^{\tau }T_{0}\left( \bar{x}_{0\tau }\right) =y^{*}$ yields

$$\begin{aligned} \left( 1-\sigma \right) ^{\tau }\frac{\partial T_{0}}{\partial \bar{x}_{0\tau }}d\bar{x}_{0\tau }= & {} -\left( 1-\sigma \right) ^{\tau }\frac{\partial T_{0}\left( \bar{x}_{0\tau }\right) }{\partial \beta _{0}}d\beta _{0} \end{aligned}$$

and, by Cramer’s rule,

$$\begin{aligned} \frac{\partial \bar{x}_{0\tau }\left( \beta _{0}\right) }{\partial \beta _{0}}= & {} -\frac{\nicefrac {\partial T_{0}\left( \bar{x}_{0\tau }\right) }{\partial \beta _{0}}}{\nicefrac {\partial T_{0}\left( \bar{x}_{0\tau }\right) }{\partial \bar{x}_{0\tau }}}. \end{aligned}$$

The denominator can be signed positive based on Proposition 2. Given $\nicefrac {\partial ^{2}T_{0}}{\partial \bar{x}_{0}\partial \beta _{0}}>0$—also from Proposition 2—the numerator is also positive for $\bar{x}_{0\tau }>x^{*}$. It follows that $\frac{\partial \bar{x}_{0\tau }\left( \beta _{0}\right) }{\partial \beta _{0}}<0$.

Proof of Proposition 6

Consider first the extreme regret case $\beta _{\tau }=1$ for some $\tau >0$. It is clear from analysis of (4) that $\bar{x}_{0}>x^{*}$ implies $y_{\tau }>y^{*}$. Now instead suppose $\beta _{\tau }\in \left( 0,1\right)$. We consider two cases. Suppose first that $\left( 1-\sigma \right) ^{\tau }y_{0}\left( \bar{x}_{0}\right) >y^{*}$. Then $y_{\tau }>y^{*}$ for $\beta _{\tau }=0$, whence by complementarity of x and y, it follows that $x_{\tau }>x^{*}$ for $\beta _{\tau }=0$. As it has already been established that $y_{\tau }>y^{*}$ in the extreme regret case, then $y_{\tau }>y^{*}$ also when $\beta _{\tau }\in \left( 0,1\right)$, and we are done. Suppose instead that $\left( 1-\sigma \right) ^{\tau }y_{0}\left( \bar{x}_{0}\right) \le y^{*}$. Then $y_{\tau }=y^{*}$ for $\beta _{\tau }=0$. However, since $y_{\tau }>y^{*}$ in the extreme regret case, it follows also that $y_{\tau }>y^{*}$ for any $\beta _{\tau }\in \left( 0,1\right)$, and so $x_{\tau }>x^{*}$ follows by complementarity of x and y.

Proof of Proposition 7

Consider first the extreme regret case $\beta _{\tau }=1$. It is clear from analysis of (4) that $\bar{x}_{0}<x^{*}$ implies $y_{\tau }<y^{*}$. Now instead suppose $\beta _{\tau }\in \left( 0,1\right)$. It is obvious that $\left( 1-\sigma \right) ^{\tau }y_{0}\left( \bar{x}_{0}\right) <y^{*}$. This means $y_{\tau }=y^{*}$ for $\beta _{\tau }=0$. However, since $y_{\tau }<y^{*}$ in the extreme regret case, it follows also that $y_{\tau }<y^{*}$ for any $\beta _{\tau }\in \left( 0,1\right)$, and so $x_{\tau }<x^{*}$ follows by complementarity of x and y.