Aversion to risk of regret and preference for positively skewed risks

Abstract

We assume that the ex post utility of a regret-sensitive agent facing a menu of lotteries depends upon the payoff of the chosen lottery in the realized state together with the forgone best payoff available in the menu in that state. If two lotteries in the menu have the same distribution of payoffs, the one whose payoff is less statistically concordant with the best alternative yields a larger risk of regret, which should be rejected by regret-risk-averse individuals. This means that regret-risk aversion is equivalent to the supermodularity of the bivariate utility function. We define regret-risk aversion in the small and in the large. We also show that regret-risk aversion tends to induce a bias in favor of the risky act in any one-risky-one-safe menu, in particular when the payoff of the risky choice is highly positively skewed. This is compatible with the “possibility effect” that is well documented in prospect theory. Symmetrically, we define the aversion to rejoicing-risk that can prevail when the ex post utility is alternatively sensitive to the forgone worst payoff. We show that rejoicing-risk-seeking is compatible with the “certainty effect”. We finally show that regret-risk-averse and rejoicing-risk-seeking people behave as if they had rank-dependent utility preferences with an inverse-S shaped probability weighting function that reproduces estimates existing in the literature.

Appendices

Appendix 1: Proof of Proposition 5

Eliminating w from (27)–(28) yields

$$\begin{aligned} F(c)= & {} \frac{p_+v(x_+)}{(1-p_+)v(c)+p_+v(x_+)}u(x_+) \\ \nonumber&\quad +\frac{(1-p_+)v(c)}{(1-p_+)v(c)+p_+v(x_+)}u(x_-)-u(c)=0 \end{aligned}$$

(36)

We have that

$$\begin{aligned} \ F(x_-)=\frac{p_+v(x_+)}{(1-p_+)v(x_-)+p_+v(x_+)} \left( u(x_+)-u(x_-)\right) \ge 0 \end{aligned}$$

(37)

and

$$\begin{aligned} F(x_+)=-(1-p_+)\left( u(x_-)-u(x_+)\right) \le 0. \end{aligned}$$

(38)

Because F is continuous, there must exists a real $c\in [x_-,x_+]$ such that $F(c)=0$. This must be the solution of system (27)–(28). Because c is less than $x_+$, its is immediate from (28) that $w(p_+)$ is larger than $p_+$ if v is increasing.

We now examine the shape of the probability weighting function. To do this, we fully differentiate system (27)–(28) with respect to $p_-$. It yields (we simplify the notation by replacing $p_+=p$)

$$\begin{aligned} \frac{dc}{dp}=\frac{u(x_+)-u(x_-)}{u'(c)}\frac{dw}{dp} \end{aligned}$$

(39)

and

$$\begin{aligned} \frac{dw}{dp}=\frac{w v(c)+(1-w)v(x_+)}{(1-p)v(c) +p v(x_+)+(1-p)w v'(c)\frac{u(x_+)-u(x_-)}{u'(c)}}. \end{aligned}$$

(40)

Let us first examine the case $p=1$. The above equations imply that $c=x_+$ and $w=1$, and $w'(1)=1$. Moreover, differentiating the above equation around $p=1$ yields

$$\begin{aligned} \left. \frac{d^2w}{dp^2}\right| _{p=1}=\frac{2v'(x_+)(u(x_+)-u(x_-))}{u'(x_+)v(x_+)}\ge 0. \end{aligned}$$

(41)

Let us alternatively consider the case $p=0$, which implies that $c=x_-$, $w=0$, $c'(0)=(u(x_-)-u(x_+)v(x_+)/u'(x_-)v(x_-)$ and $w'(0)=v(x_+)/v(x_-)$. Finally, differentiating equation (40) in the neighborhood of $p=0$ yields

$$\begin{aligned} \left. \frac{d^2w}{dp^2}\right| _{p=0}= -\frac{2 v(x_+)^2v'(x_-)(u(x_+)-u(x_-))}{u'(x_-)v(x_-)^3}-\frac{2 v(x_+)(v(x_+)-v(x_-))}{v(x_-)^2}\le 0. \end{aligned}$$

(42)

This concludes the proof of Proposition 5. $\square$

Appendix 2: Proof of Lemma 2

Without loss of generality, let $\mu$ be zero. Let $S=m+n$, and let the risky lottery $x_1$ in menu M be

$$\begin{aligned} x_1=x_1^0\sim (a_1^-,p_1^-;\ldots ;a_m^-,p_m^-;a_1^+,p_1^+;\ldots ;a_n^+,p_n^+), \end{aligned}$$

(43)

with $a_1^-<\cdots<a_m^-< 0<a_1^+<\cdots <a_n^+$, $p_i^->0$, $i=1,\ldots ,m$, $p_j^+>0$, $j=1,\ldots ,n$, and $\sum _{i=1}^m p_i^- + \sum _{j=1}^n p_j^+ =1$. Assume that $Ex_1=0$. Define $p_{ij}$ as

$$\begin{aligned} p_{ij}=\frac{-a_i^-}{a_j^+-a_i^-}\in [0,1]. \end{aligned}$$

(44)

Define lottery $x_{ij}\sim (a_j^+,p_{ij};a_i^-,1-p_{ij})$. By construction, $Ex_{ij}=0$. Initialize the probability vector $Q=(q_1^{0-},\ldots ,q_m^{0-},q_1^{0+},\ldots ,q_n{0+})$ such that for all i, $q_i^{0,-}=p_i^-$ and for all j, $q_j^{0+}=p_j^+$. We also initialize two sets $I=J=\{\emptyset \}$.

We consider the following $n+m-2$ iterations. At the beginning of iteration k, $I\cup J$ contains the $k-1$ states of nature whose lottery’s initial payoff has been replaced by a binary zero-mean lottery.

Iteration k: Take an arbitrary pair (i, j), $i\in \{1,\ldots ,m\}/I$, $j\in \{1,\ldots ,n\}/J$. Consider two cases.

Case 1: Suppose that $q_i^{k-1-}<-q_j^{k-1+}a_j^+/a_i^-$. Then, define $\pi ^k=q_i^{k-1-}(a_j^+-a_i^-)/a_j^+$. Perform the following two operations on lottery $x^{k-1}$:

• Replace the atom $a_i^-$ by lottery $x_{ij}$, and raises the associated probability $q_i^{k-1-}$ up to $q_i^{k-}=\pi ^k$.

• Reduce the probability associated with $a_j^+$ from $q_j^{k-1+}$ to $q_j^{k+}=q_j^{k-1+}+(q_i^{k-1-}a_i^-/a_j^+)>0$.

Moreover, append state i into the set of negative states whose initial payoff $a_i^-$ as been replaced by a binary zero-mean lottery $x_{ij}$: $I^k=I^{k-1}\cup {i}$.

Case 2: Suppose that $q_i^{k-1-}\ge -q_j^{k-1+}a_j^+/a_i^-$. Then, define $\pi ^k=-q_j^{k-1+}(a_j^+-a_i^-)/a_i^-$. Perform the following two operations on lottery $x^{k-1}$:

• Replace the atom $a_j^+$ by lottery $x_{ij}$, and raises the associated probability $q_j^{k-1+}$ up to $q_j^{k+}=\pi ^k$.

• Reduce the probability associated with $a_i^-$ from $q_i^{k-1-}$ to $q_i^{k-}=q_i^{k-1-}+(q_j^{k-1+}a_j^+/a_i^-)\ge 0$.

Moreover, append state j into the set of positive states whose initial payoff $a_j^+$ as been replaced by a binary zero-mean lottery $x_{ij}$: $J^k=J^{k-1}\cup {j}$.

In both cases, this procedure yields a new lottery $x^k$ that has the same distribution of payoffs, but in which one payoff has been replaced by a binary zero-mean lottery. After $m+n-2$ iterations, all payoffs have replaced by such lotteries, expect two of them. By construction, since the $Ex^k=0$, the remaining two atoms $(a_i^-,a_j^+)$ must be opposite in sign and have remaining probabilities $q_i^{m+n-2-}$ and $q_j^{m+n-2+}$ such that

$$\begin{aligned} a_i^-q_i^{m+n-2-}+a_j^+q_j^{m+n-2+}=0. \end{aligned}$$

(45)

This concludes the proof of Lemma 2. $\square$