Higher order risk attitudes and prevention under different timings of loss

Abstract

This paper provides experimental evidence of the role of higher order risk attitudes—especially prudence—in prevention behavior. Prudence, under an expected utility framework, increases (decreases) self-protection effort compared to the risk neutral level when the risk of losing part of an income exists in a future (the same) period. Motivated by these predictions that give the exact test on prudence, an experiment was designed where subjects go through higher order risk attitude elicitation and make a self-protection decision. In contrast to the expected utility theory, the observed efforts are less than the risk neutral level, regardless of the timing of loss. This violation of expected utility predictions can be explained by probability weighting.

Appendix A. Proofs

Proof of Proposition 3

(1) Future loss. Take any $\upgamma \in (0,1]$. An $\upgamma$-Prelec player chooses effort e to maximize

$$\,\,\,\,y - e + w^{\upgamma } (q)z + (1 - w^{\upgamma } (q))(z - d) \equiv P^{f,\upgamma } (e),$$

where $q = q(e) = 1 - p(e) = ke/(1 + ke)$. First, we check the first order condition.

$$\frac{{dP^{f,\upgamma } }}{de} = - 1 + d\frac{{dw^{\upgamma } }}{dq}\frac{dq}{de}.$$

By substituting $dw^{\upgamma } /dq = (\upgamma ( - \ln q)^{\upgamma - 1} /q) \cdot w^{\upgamma } (q)$, $p(e_{n} ) = 1/2,$ and $- p^{\prime}(e_{n} ) = 1/d,$

$$\begin{aligned} \frac{{dP^{f,\upgamma } }}{de}|_{{e = e_{n} }} = \,\, - 1 + d \cdot \upgamma \frac{{( - \ln (1/2))^{\upgamma - 1} }}{1/2} \cdot w^{\upgamma } (1/2) \cdot \frac{1}{d} \hfill \\ \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = - 1 + \cdot 2\upgamma (\ln 2)^{\upgamma - 1} \cdot \exp ( - (\ln 2)^{\upgamma } ) \hfill \\ \end{aligned}$$

Since

$$\frac{d}{d\upgamma }\{ \upgamma (\ln 2)^{\upgamma - 1} \cdot \exp ( - (\ln 2)^{\upgamma - 1} \} = { \exp }\left( { - (\ln 2)^{\upgamma } } \right)(\ln 2)^{ - 1 + \upgamma } \left( {2 - 2\upgamma \left( { - 1 + (\ln 2)^{\upgamma } } \right)\ln (\ln 2)} \right) > 0$$

by $0 < \upgamma \le 1,0 < (\ln 2)^{\upgamma } \le 1\,{\text{and - 1}} < \ln (\ln 2) < 0$, we have

$$\frac{{dP^{f,\upgamma } }}{de}|_{{e = e_{n} }} \le \frac{{dP^{f,1} }}{de}|_{{e = e_{n} }} = 0\,{\rm for\,any}\, \upgamma \in (0,1)$$

Second, we check the second order condition. Since

$$\frac{{d^{2} P^{f,\upgamma } }}{{de^{2} }} = \frac{{d \cdot { \exp }\left( { - \left( { - \ln q} \right)^{\upgamma } } \right)\upgamma \left( { - \ln q} \right)^{ - 2 + \upgamma } \left( {1 + \upgamma \left( { - 1 + \left( { - \ln q} \right)^{\upgamma } } \right) + \left( {1 + 2ke} \right)\ln q} \right)}}{{e^{2} \left( {1 + ek} \right)^{2} }}$$

and ke = q/(1-q), it suffices to show $h(q,\upgamma ) = 1 + \upgamma \left( { - 1 + \left( { - \ln q} \right)^{\upgamma } } \right) + \frac{1 + q}{1 - q}\ln q \le 0$ for all $q \in (0,1]$ and $\upgamma \in (0,1]$.

Let e be Napier’s constant.

Case 1

$q \in (0,{\mathbf{e}}^{ - 1} ]$

Since

$$\frac{\partial h}{\partial \upgamma }(q,\upgamma ) = - 1 + \left( { - \ln q} \right)^{\upgamma } + \upgamma \left( { - \ln q} \right)^{\upgamma } \ln \left( { - \ln q} \right) \ge 0\quad{\text{for}}\,{\text{all}}\quad q \in (0,q*],$$

$h(q,\upgamma ) \le h(q,1) = 2q\ln q/(1 - q)$ for all $q \in (0,{\mathbf{e}}^{ - 1} ]$ and $\upgamma \in (0,1]$. Hence,$h(q,\upgamma ) \le h(q,1) < \mathop {\lim }\limits_{q \to 0} h(q,1) = \mathop {\lim }\limits_{q \to 0} \frac{2q\ln q}{1 - q} = 2 \cdot 1 \cdot \mathop {\lim }\limits_{q \to 0} q\ln q = 2 \cdot \left( { - \mathop {\lim }\limits_{r \to \infty } \frac{\ln r}{r}} \right) = 0\quad {\text{for}}\,{\text{all}}\,$$q \in (0,q*]$ and $\upgamma \in (0,1]$.

Case 2

$q \in ({\mathbf{e}}^{ - 1} ,1]$

Since

$$\frac{\partial h}{\partial \upgamma }(q,\upgamma ) < 0 \quad {\rm for\,all } q \in [q*,1],$$

$h(q,\upgamma ) \le h(q,0) = 1 + \frac{1 + q}{1 - q}\ln q$ for all $q \in ({\mathbf{e}}^{ - 1} ,1]$ and $\upgamma \in (0,1]$. Hence,

$$h(q,\upgamma ) \le h(q,0) \le \mathop {\lim }\limits_{q \to 1} h(q,0) = 1 + \mathop {\lim }\limits_{q \to 1} \frac{1 + q}{1 - q}\ln q = 1 - \mathop {\lim }\limits_{q \to 1} (1 + q)\mathop {\lim }\limits_{s \to 0} \frac{\ln (s + 1)}{s} = 1 - 2 = - 1.$$

Hence, the objective function is concave.

(2) Current loss. The same argument holds by linearity of valuation function of money. ||

Proof of observation

Take any a, b, z with a > b. $L = (a,.50;b + z,.25;b - z),R = (a + z,.25;a - z,.25;b).$

Suppose a > b + z.

$$\begin{aligned} & V(R) = w(.25)(a + z) + \{ w(.5) - w(.25)\} (a - z) + \{ w(1) - w(.5)\} b \\ & V(L) = \,w(.5)a + \{ w(.75) - w(5)\} (b + z) + \{ w(1) - w(.75)\} (b - z) \end{aligned}$$

Then

$$\begin{aligned} V(R) - V(L) &= w(.25)\{ (a + z) - (a - z)\} + w(.5)\{ (a - z) - b - a + (b + z)\} \\ &\quad + w(.75)\,\{ (b - z) - (b + z)\} + w(1)\{ b - (b - z)\} \\ & = (2w(.25) - 2w(.75)\, + 1)z \ge 0. \end{aligned}$$

The last inequality holds by $w(.75) - w(.25)\, \le 0.5$ for any Prelec function.

Suppose next a < b + z.

$$\begin{aligned} & V(R) = w(.25)(a + z) + \{ w(.75) - w(.25)\} b + \{ w(1) - w(.75)\} (a - z) \\ & V(L) = \,w(.25)(b + z) + \{ w(.75) - w(.25)\} a + \{ w(1) - w(.75)\} (b - z) \end{aligned}$$

Then

$$\begin{aligned} V(R) - V(L) &= w(.25)\{ (a + z) - (b + z) - (b - a)\} + w(.75)\{ b - (a - z) - a + (b - z)\} \\&\quad + w(1)\{ (a - z) - (b - z)\} \\ &= (2w(.25) - 2w(.75) + 1)(a - b) \ge 0.\end{aligned}$$

The last inequality holds by $w(.75) - w(.25)\, \le 0.5$ for any Prelec function.||