The curse of hope

Abstract

In Kőszegi and Rabin’s (Q J Econ 1133–1165, 2006, Am Econ Rev 97:1047–1073, 2007) reference-dependent model of preferences, the chance of obtaining a better outcome can reduce an agent’s expected utility through an increase in the stochastic reference point. This means that individuals may prefer stochastically dominated lotteries. In this sense, hope, understood as a small probability of a better outcome, can be a curse. While Kőszegi and Rabin focus on a linear specification of the utility function, we show that this effect occurs more broadly. Using fairly plausible assumptions and parameter values, we specify the conditions under which it occurs, as well as the type of lotteries in which this should be expected. We then show that while a simple subjective transformation of probability into weights of the reference point may in some cases mitigate the issue, in others, it can intensify it or even generate new ones. Finally, we extend the model by adding the individual’s current reference point (status quo) to the stochastic reference point. We show that this modification can reconcile Kőszegi and Rabin’s model with the apparent empirical infrequency of stochastically dominated choices while maintaining its main qualitative results.

Appendix

1.1 Proof of Proposition 2

$$\begin{aligned} \frac{\partial U(L_p|L_p)}{\partial p}=\Delta m+(1-2p)\mu ^-+(1-2p)\mu ^+ \end{aligned}$$

It is positive if $p<\frac{\Delta m +\Delta \mu }{\Delta \mu }$, that is if $p<\frac{\bar{p}}{2}$. Proposition 2 ensues.

We complement the proof by some additional elements on the behavior of $U(L_p|L_p)$.

$$\begin{aligned} \frac{\partial ^2 U(L_p|L_p)}{\partial p^2}=-2[\mu ^-+\mu ^+] \end{aligned}$$

Because of loss aversion, for any $t>0$ we have

$$\begin{aligned}{}[\mu (-t)+\mu (t)]<0 \end{aligned}$$

Hence

$$\begin{aligned} \frac{\partial ^2 U}{\partial p^2}=-2[\mu ^-+\mu ^+]>0 \end{aligned}$$

So U(p) is convex. Since $\frac{\partial U(L_p|L_p)}{\partial p}(\frac{\bar{p}}{2})=0$, we have $U(L_p|L_p)$ decreasing in p up to $\bar{p}/2$ and increasing afterwards, reaching 0 at $\bar{p}$. This means that $p\le \bar{p}$ not only implies $L_p \succ x$ but that the two are equivalent.

1.2 Proof of Proposition 3

For infinitesimal differences between x and y, the condition becomes

$$\begin{aligned} \lim _{\Delta m \rightarrow 0^+} \frac{-\Delta \mu }{\Delta m}>1 \end{aligned}$$

That gives

$$\begin{aligned} \lim _{\Delta m \rightarrow 0^+} \frac{-\mu ^+ - \mu ^-}{\Delta m}>1 \end{aligned}$$

Hence

$$\begin{aligned} -\mu '_{+}(0)+\mu '_{-}(0)>1 \end{aligned}$$

That is

$$\begin{aligned}\mu '_{+}(0)>\frac{1}{\lambda -1}\end{aligned}$$

The limit of $\bar{p}$ is given by

$$\begin{aligned} \lim _{\Delta m \rightarrow 0^+} \bar{p}= & {} \lim _{\Delta m \rightarrow 0^+} \left( 1+\frac{\Delta m}{\Delta \mu }\right) =\lim _{\Delta m \rightarrow 0^+} \left( 1+\frac{1}{1/(\Delta \mu /\Delta m)}\right) \\= & {} 1+\frac{1}{(1-\lambda )\mu '_{+}(0)} \end{aligned}$$

1.3 Proof of Proposition 4

$m(r+\delta )-m(r)$ is less than $m(r'+\delta )-m(r')$ because $m''<0$. Hence by continuity there exists $\delta '>\delta$ with $m(r'+\delta ')-m(r')=m(r+\delta )-m(r)$. Given that the condition for the curse of hope to occur at r and $r+\delta$ is that $\mu (m(r+\delta )-m(r))+\mu (m(r)-m(r+\delta ))>m(r+\delta )-m(r)$, it holds that $\mu (m(r'+\delta ')-m(r'))+\mu (m(r')-m(r'+\delta '))>m(r'+\delta ')-m(r')$; hence the curse of hope occurs at $r'$ and $r'+\delta '$.

1.4 Proof and conditions of Proposition 5

$x\succ L_{p}$ iff $m(x)-U(L_{p} |R_{\pi })$, that is, using Eq. (4) and rearranging, if

$$\begin{aligned} -(\pi ({(1-p)/ p} )\mu ^{-} +(1-\pi )\mu ^{+} )>\Delta m \end{aligned}$$

In the absence of loss aversion, $-\mu ^{-} =\mu ^{+}$ and the inequality boils down to ${({(\pi -p)/ p} )\mu ^{+} >\Delta m}$, which is satisfied for some values of the parameters, in particular when $\pi$ is large enough.

1.5 Proof of Proposition 7

We first note that

$$\begin{aligned} U(x|R(w,x))= q U(x|w)+(1-q)U(x|x)=m(x)+q\mu (m(x)-m(w)) \end{aligned}$$

and

$$\begin{aligned} U(L_p|R(w,L_p))= & {} qU(L_p|w)+(1-q)U(L_p|L_p)\\= & {} q[p(m(y)+\mu (m(y)-m(w)) )\\&+\,(1-p)(m(x)+\mu (m(x)-m(w)) ) ]\\&+\,(1-q)[m(x)+p\Delta m+p(1-p)(\mu ^+ +\mu ^-)]\\= & {} m(x)+p \Delta m+qp\Delta \mu _w+q \mu (m(x)-m(w))\\&+\,(1-q)p(1-p)\Delta \mu , \end{aligned}$$

where $\Delta \mu _w=\mu (m(y)-m(w))-\mu (m(x)-m(w))\ge 0$, since $y>x$.

A curse of hope still exists if $U(x|R(w,x))>U(L_p|R(w,L_p))$, that is, after replacement:

$$\begin{aligned} - p \Delta m - qp \Delta \mu _w-(1-q)p(1-p)\Delta \mu >0 \end{aligned}$$

That is

$$\begin{aligned} -\Delta \mu> & {} \frac{ \Delta m + q \Delta \mu _w}{(1-q)(1-p)}\\> & {} \frac{ \Delta m}{1-p} +\frac{q \Delta m}{(1-q)(1-p)}+ \frac{q \Delta \mu _w}{(1-q)(1-p)}\\> & {} \frac{ \Delta m}{1-p} + \frac{q (\Delta m+\Delta \mu _w)}{(1-q)(1-p)} \end{aligned}$$

Set $A(q)=\frac{q (\Delta m+\Delta \mu _w)}{(1-q)(1-p)}$, $A(0)=0$ and $A'(q)>0$.

Now compare the value of the threshold probability below which the curse of hope occurs in the absence of inertia ($\bar{p}$) and the threshold probability for some value of q, denoted $\bar{p}_q$. As already noted, $\bar{p}=\frac{\Delta m + \Delta \mu }{\Delta \mu }$. For $q>0$, we have

$$\begin{aligned} - \Delta m - q \Delta \mu _w-(1-q)(1-\bar{p}_q)\Delta \mu =0 \end{aligned}$$

After rearrangement:

$$\begin{aligned} \bar{p}_q\Delta \mu =\Delta \mu +\frac{\Delta m + q \Delta \mu _w}{1-q} \end{aligned}$$

That is,

$$\begin{aligned} \bar{p}_q\Delta \mu =\frac{(1-q)\Delta \mu +(1-q)\Delta m +q\Delta m+ q \Delta \mu _w}{1-q} \end{aligned}$$

$$\begin{aligned} \bar{p}_q=\frac{\Delta \mu +\Delta m}{\Delta \mu }+\frac{q}{1-q}\frac{\Delta m+ \Delta \mu _w}{\Delta \mu } \end{aligned}$$

In the end,

$$\begin{aligned} \bar{p}_q=\bar{p}+\frac{q}{1-q}\frac{\Delta m+ \Delta \mu _w}{\Delta \mu } \end{aligned}$$

Note that $\frac{\Delta m +\Delta \mu _w}{\Delta \mu }<0$ and $\frac{q}{1-q}$ is increasing in q and is unbounded for q arbitrarily close to 1. Hence, $\bar{p}_q$ decreases in q and the interval $[0, \bar{p}_q]$ on which the curse of hope occurs shrinks as q increases. Moreover, for any $\mu _w$, i.e. for any w, exists $q<1$ such that $\bar{p}_q=0$.

1.6 Proof of Proposition 8

We have:

$$\begin{aligned} U(L_p|R(w,L_p))-U(x|R(w,x))>p \Delta m +(1-q)[p(1-p)\mu ^+ +p(1-p)\mu ^-] \end{aligned}$$

The right-hand side is of the same sign as

$$\begin{aligned} \Delta m +(1-q)[(1-p)\mu ^+ +(1-p)\mu ^-] \end{aligned}$$

A sufficient condition for this to be positive for any p is that

$$\begin{aligned} \Delta m +(1-q)[\mu ^+ +\mu ^-]>0 \end{aligned}$$

Define f on ${\mathbb {R}}_{+}$ by $f(\Delta m)=\Delta m +(1-q)[\mu ^+ +\mu ^-]$; then its derivative is given by

$$\begin{aligned} f'(\Delta m)=1+(1-q)[\mu '(\Delta m)-\mu '(-\Delta )m] \end{aligned}$$

Because $\mu '$ is bounded, exists $b>0$ such that $\mu '(\Delta m)-\mu '(-\Delta )>-b$. That is

$$\begin{aligned} f'(\Delta m)>1-(1-q)b \end{aligned}$$

Taking $q>1-\frac{1}{b}$ ensures that $f'(\Delta m)>0$ for all $\Delta m\ge 0$. Moreover, $f(0)=0$; hence f has non-negative values.

To conclude, $q>1-\frac{1}{b}$ implies that $\Delta m +(1-q)[\mu ^+ +\mu ^-]>0$, which implies ${U(L_p|R(w,L_p))-U(x|R(w,x))\ge 0}$.

1.7 Proof of Proposition 9

We have for any binary lottery

$$\begin{aligned} U(L_p|R(w,L_p))= & {} pm(y)+(1-p)m(x)\\&+\,q[p \mu (m(y)-m(w))+(1-p) \mu (m(x)-m(w))]\\&+\, (1-q)p(1-p)[\mu (m(x)-m(y))+\mu (m(y)-m(x))] \end{aligned}$$

Then its derivative is given by

$$\begin{aligned} \frac{\partial U(L_p|R(w,L_p))}{\partial p}= & {} m(y)-m(x)+ q[ \mu (m(y)-m(w))- \mu (m(x)-m(w))]\\&+\,(1-q)(1-2p)[\mu (m(x)-m(y))+\mu (m(y)-m(x))] \end{aligned}$$

And

$$\begin{aligned} \frac{\partial ^2 U(L_p|R(w,L_p))}{\partial p^2}= -2q[\mu (m(x)-m(y))+\mu (m(y)-m(x))]>0 \end{aligned}$$

Hence $U(L_p|R(w,L_p))$ is convex.

The sign of the first-order derivative is given by the sign of

$$\begin{aligned}&\Delta m + q[ \mu (m(y)-m(w))- \mu (m(x)-m(w))]\\&\quad +\,(1-q)(1-2p)[\mu (\Delta m)+\mu (-\Delta m)] \end{aligned}$$

Given that $\mu (m(y)-m(w))- \mu (m(x)-m(w))>0$, this is positive for any p if

$$\begin{aligned} \Delta m +(1-q)[\mu (\Delta m)+\mu (-\Delta m)]>0 \end{aligned}$$

Which is the exact same condition studied in the proof of Proposition 8.

1.8 A generalization to non-binary lotteries

First suppose the effect exists for some $L_p=(p,y; 1-p, x)$ with $0<p<1$, i.e. the condition of Proposition 1 is satisfied. We then proceed by induction.

Suppose the effect occurs for n fixed. That is, there exists

$$\begin{aligned} L_n=(p_1, p_2, \ldots , p_n; x_1, x_2, \ldots x_n) \end{aligned}$$

with $x_1<x_2<x_2<\cdots <x_n$ and for all $k=1,\ldots , n$, $p_k>0$. $L_n$ is such that $U(x_1|x_1)>U(L_n|L_n)$. We want to show that there exists

$$\begin{aligned} L_{n+1}=(q_1, q_2, \ldots , q_n, q_{n+1}; x_1, x_2, \ldots , x_{n+1}) \end{aligned}$$

with for all $k=1,\ldots ,n+1$, $q_k>0$ and $x_{n+1}$ different from any $x_k$ for $k=1,\ldots , n$.

For $\epsilon <p_1$, denote $L_\epsilon =L_{n+1}=(p_1-\epsilon , p_2, p_3, \ldots , p_n, \epsilon ; x_1, x_2, \ldots ,x_n, x_{x+1})$. It is straightforward that the mapping f defined for any $\epsilon \in [0,p_1)$ by $f(\epsilon )=U(L_\epsilon |L_\epsilon )$ is continuous because of the continuity of U with respect to probabilities, and that $f(0)=U(L_n|L_n)$. This is true independently of $x_{n+1}$. Given that $f(0)=U(L_n|L_n)<U(x_1|x_1)$, by continuity exists $\epsilon >0$ small enough so that

$$\begin{aligned} f(\epsilon )=U(L_\epsilon |L_\epsilon )<U(x_1|x_1) \end{aligned}$$

Hence the existence of $L_\epsilon$ implies the existence of $L_{n+1}$. It obtains: if exists $L_n$ such that $U(L_n|L_n)<U(x_1|x_1)$, then there exists $L_{n+1}$ with $U(L_{n+1}|L_{n+1})<U(x_1|x_1)$.

By assumption and Proposition 1, $L_2$ exists. Hence exists $L_k$ for any k with $U(L_{k+1}|L_{k+1})<U(x_1|x_1)$.