Higher-order risk vulnerability

Abstract

We add an independent unfair background risk to higher-order risk-taking models in the current literature and examine its interaction with the main risk under consideration. Parallel to the well-known concept of risk vulnerability, which is defined by Gollier and Pratt (Econometrica 64:1109–1123, 1996), an agent is said to have a type of higher-order risk vulnerability if adding an independent unfair background risk to wealth raises his level of this type of higher-order risk aversion. We derive necessary and sufficient conditions for all types of higher-order risk vulnerabilities and explain their behavioral implications. We find that as in the case of risk vulnerability, all familiar HARA utility functions have all types of higher-order risk vulnerabilities except for a type of third-order risk vulnerability corresponding to a downside risk aversion measure called the Schwarzian derivative.

Appendices

Appendix 1: Proof of Theorem 1

To prove the theorem, we need the following lemma. Let $\varOmega _0$ be the set of all zero-mean binary random variables. Let $\varOmega$ be any set of zero-mean random variables which contains $\varOmega _0$. Let f(x), g(x), and h(x) be three functions such that $\forall ~{\tilde{\epsilon }}\in \varOmega$, $Ef({\tilde{\epsilon }})\ge 0$, $Eg({\tilde{\epsilon }})\ge 0$, and $Eh({\tilde{\epsilon }})\ge 0$.

Lemma 2

$\forall ~{\tilde{\epsilon }}\in \varOmega$, $Eg({\tilde{\epsilon }})Eh({\tilde{\epsilon }})\ge [Ef({\tilde{\epsilon }})]^2$ if and only if the inequality is true for all ${\tilde{\epsilon }}\in \varOmega _0$.

Proof We first prove the following statement: $\forall ~{\tilde{\epsilon }}\in \varOmega$, $Eg({\tilde{\epsilon }})Eh({\tilde{\epsilon }})\ge [Ef({\tilde{\epsilon }})]^2$, if and only if $\forall ~{\tilde{\epsilon }}_1\in \varOmega$ and $\forall ~{\tilde{\epsilon }}_2\in \varOmega$, $Eg({\tilde{\epsilon }}_1) Eh({\tilde{\epsilon }}_2) + Eg({\tilde{\epsilon }}_2)Eh({\tilde{\epsilon }}_1)\ge 2Ef({\tilde{\epsilon }}_1) Ef({\tilde{\epsilon }}_2)$. This is proved as follows. To prove the “if” part, we let ${\tilde{\epsilon }}_1={\tilde{\epsilon }}_2$ in the latter inequality, while to prove the “only if” part, we need only note that

$$\begin{aligned} Eg({\tilde{\epsilon }}_1)Eh({\tilde{\epsilon }}_2) + Eg({\tilde{\epsilon }}_2) Eh({\tilde{\epsilon }}_1)\ge 2\sqrt{Eg({\tilde{\epsilon }}_1) Eh({\tilde{\epsilon }}_1)Eh({\tilde{\epsilon }}_2) Eg({\tilde{\epsilon }}_2)}. \end{aligned}$$

On the other hand, if we fix either of the two random variables, the expression $Eg({\tilde{\epsilon }}_1) Eh({\tilde{\epsilon }}_2) + Eg({\tilde{\epsilon }}_2)Eh({\tilde{\epsilon }}_1)- 2Ef({\tilde{\epsilon }}_1) Ef({\tilde{\epsilon }}_2)$ is linear in the distribution of the other (in the probabilities of its possible values). This implies that $\forall ~{\tilde{\epsilon }}_1\in \varOmega$ and $\forall ~{\tilde{\epsilon }}_2\in \varOmega$, $Eg({\tilde{\epsilon }}_1) Eh({\tilde{\epsilon }}_2) + Eg({\tilde{\epsilon }}_2)Eh({\tilde{\epsilon }}_1)\ge 2Ef({\tilde{\epsilon }}_1)Ef({\tilde{\epsilon }}_2)$ if and only if $\forall ~{\tilde{\epsilon }}_1\in \varOmega _0$ and $\forall ~{\tilde{\epsilon }}_2\in \varOmega _0$, the inequality is true. Now applying the preceding statement for the case where $\varOmega =\varOmega _0$, we immediately obtain the lemma. $\square$

With the help of Lemma 2, we are now ready to prove the theorem. The result where $i=2$ and 3 is a special case of Statement 1 of Theorem 5 which is proved in the last appendix; thus, we need only prove the result for the other three cases. Note that as an unfair risk can be decomposed into a certain reduction in wealth and a zero-mean risk, a necessary [sufficient] condition for all independent zero-mean background risks to increase $\tau _i(x)$ combined with the condition that $\tau _i(x)$ is monotone decreasing is a necessary [sufficient] condition for all independent unfair background risks to increase $\tau _i(x)$. Thus, we need only prove the result for the case of zero-mean background risks.

To prove the result for $\tau _1(x)=C(x)$, note that the inequality ${\hat{C}}(x)\ge C(x)$ is equivalent to

$$\begin{aligned} \frac{u^\prime (x)u^{\prime \prime \prime }(x)}{u^{\prime \prime 2}(x)} [Eu^{\prime \prime }(x+{\tilde{\epsilon }})]^2\le Eu^\prime (x+{\tilde{\epsilon }})Eu^{\prime \prime \prime }(x+{\tilde{\epsilon }}). \end{aligned}$$

(13)

Thus, the problem is to characterize the utility function u(x) which satisfies the following condition

$$\begin{aligned} E{\tilde{\epsilon }}=0\Rightarrow \frac{u^\prime (x)u^{\prime \prime \prime }(x) }{u^{\prime \prime 2}(x)}[Eu^{\prime \prime }(x+{\tilde{\epsilon }})]^2\le Eu^\prime (x+{\tilde{\epsilon }})Eu^{\prime \prime \prime }(x+{\tilde{\epsilon }}). \end{aligned}$$

It is straightforward to see that Lemma 2 is applicable to this case. From this lemma, it is clear that the inequality ${\hat{C}}(x)\ge C(x)$ is true for all zero-mean risks if and only if it is true for all zero-mean binary risks, which is equivalent to Inequality (1) as $C(x)=(\frac{1}{R(x)})^\prime$. This proves the result for $\tau _1(x)$.

To prove the result for $\tau _4(x)=-R^\prime (x)$, we have

$$\begin{aligned} -{\hat{R}}^\prime (x) = \frac{Eu^{\prime \prime \prime } (x+{\tilde{\epsilon }})Eu^\prime (x+{\tilde{\epsilon }}) - [Eu^{\prime \prime }(x+{\tilde{\epsilon }})]^2}{[Eu^{\prime } (x+{\tilde{\epsilon }})]^2}. \end{aligned}$$

Thus, the inequality $-{\hat{R}}^\prime (x)\ge -R^\prime (x)$ is equivalent to

$$\begin{aligned} E[u^{\prime \prime \prime } (x+{\tilde{\epsilon }}) + R^\prime (x)u^{\prime } (x+{\tilde{\epsilon }})]Eu^\prime (x+{\tilde{\epsilon }}) \ge [Eu^{\prime \prime }(x+{\tilde{\epsilon }})]^2. \end{aligned}$$

It follows that the problem is to characterize the utility function u(x) which satisfies the following condition

$$\begin{aligned} E{\tilde{\epsilon }} = 0\Rightarrow [Eu^{\prime \prime }(x+{\tilde{\epsilon }})]^2 \le E[u^{\prime \prime \prime } (x+{\tilde{\epsilon }}) + R^\prime (x)u^{\prime }(x+{\tilde{\epsilon }})]Eu^\prime (x+{\tilde{\epsilon }}). \end{aligned}$$

If for some zero-mean ${\tilde{\epsilon }}$, $E[u^{\prime \prime \prime } (x+{\tilde{\epsilon }}) + R^\prime (x)u^{\prime }(x+{\tilde{\epsilon }})]<0$, then there must exist a zero-mean binary ${\tilde{\epsilon }}$ such that $E[u^{\prime \prime \prime } (x+{\tilde{\epsilon }}) + R^\prime (x)u^{\prime }(x+{\tilde{\epsilon }})]<0$. Thus, we need only consider the case where for all zero-mean risks, $E[u^{\prime \prime \prime } (x+{\tilde{\epsilon }}) + R^\prime (x)u^{\prime } (x+{\tilde{\epsilon }})]\ge 0$. Now Lemma 2 is applicable to this case, and the inequality $-{\hat{R}}^\prime (x)\ge -R^\prime (x)$ is true for all zero-mean risks if and only if it is true for all binary zero-mean risks, which is equivalent to Inequality (4). This proves the result for $\tau _4(x)$.

To prove the result for $\tau _5(x)=S(x)$, we have

$$\begin{aligned} {\hat{S}}(x) = \frac{Eu^{\prime \prime \prime }(x+{\tilde{\epsilon }})}{Eu^\prime (x+{\tilde{\epsilon }})}-\frac{3}{2} \left[ \frac{Eu^{\prime \prime } (x+{\tilde{\epsilon }})}{Eu^\prime (x+{\tilde{\epsilon }})}\right] ^2. \end{aligned}$$

Thus, the inequality ${\hat{S}}(x)\ge S(x)$ is equivalent to

$$\begin{aligned} E \left[ u^{\prime \prime \prime }(x+{\tilde{\epsilon }}) - S(x)u^\prime (x+{\tilde{\epsilon }})\right] Eu^\prime (x+{\tilde{\epsilon }}) \ge \frac{3}{2}\left[ Eu^{\prime \prime }(x+{\tilde{\epsilon }})\right] ^2. \end{aligned}$$

(14)

It follows that the problem is to characterize the utility function u(x) which satisfies the following condition

$$\begin{aligned} E{\tilde{\epsilon }}=0\Rightarrow \frac{3}{2}[Eu^{\prime \prime }(x+{\tilde{\epsilon }})]^2\le E[u^{\prime \prime \prime }(x+{\tilde{\epsilon }}) - S(x)u^\prime (x+{\tilde{\epsilon }})]Eu^\prime (x+{\tilde{\epsilon }}). \end{aligned}$$

Similar to the case of $\tau _4(x)$, we need only consider the case where for all zero-mean risks, $E[u^{\prime \prime \prime }(x+{\tilde{\epsilon }}) - S(x)u^\prime (x+{\tilde{\epsilon }})]\ge 0$. Again, Lemma 2 is applicable to this case, and Inequality (14) is true for all zero-mean risks if and only if it is true for all binary zero-mean risks, which is equivalent to Inequality (5) as $S(x)=-R^\prime (x)-0.5R^2(x)$. This proves the result for $\tau _5(x)$. $\square$

Appendix 2: Proof of Theorem 2

As was explained at the beginning of the proof of Theorem 1, we need only prove the result for the case of zero-mean background risks. The result where $i=1$ follows from Theorem 1 in Hara et al. (2011), and its proof can be found there. The result where $i=2$ and 3 is a special case of Statement 2 of Theorem 5 which is proved in the last appendix; thus, we need only prove the result for the other two cases.

To prove the result where $i=4$, we have ${\hat{R}}(x)=E(\frac{u^\prime }{Eu^\prime }R)$.¹³ Differentiating the equation with respect to x, we obtain

$$\begin{aligned} {\hat{R}}^\prime (x) = E \left[ \left( \frac{u^\prime }{Eu^\prime }\right) ^\prime R\right] + E \left( \frac{u^\prime }{Eu^\prime }R^\prime \right) . \end{aligned}$$

(15)

As $E(\frac{u^\prime }{Eu^\prime })^\prime =0$, we have $E[(\frac{u^\prime }{Eu^\prime })^\prime R]=E[(\frac{u^\prime }{Eu^\prime })^\prime (R-{\hat{R}}(x))]$. Moreover, as $(\frac{u^\prime }{Eu^\prime })^\prime = \frac{u^\prime }{Eu^\prime }({\hat{R}}(x)-R)$, we obtain

$$\begin{aligned} E \left[ \left( \frac{u^\prime }{Eu^\prime }\right) ^\prime R \right] = -E \left[ \frac{u^\prime }{Eu^\prime }\left( {\hat{R}}(x)-R\right) ^2 \right] . \end{aligned}$$

(16)

This and Eq. (15) lead to

$$\begin{aligned} -{\hat{R}}^\prime (x)Eu^\prime = -E(u^\prime R^\prime ) + E \left[ u^\prime \left( {\hat{R}}(x)-R\right) ^2\right] . \end{aligned}$$

(17)

Adding $R^\prime (x)Eu^\prime$ to both sides, we obtain

$$\begin{aligned} \left( -{\hat{R}}^\prime (x)+R^\prime (x)\right) Eu^\prime = R^\prime (x)Eu^\prime -E(u^\prime R^\prime ) + E \left[ u^\prime \left( {\hat{R}}(x)-R\right) ^2\right] . \end{aligned}$$

(18)

Meanwhile, we have $(Eu^\prime )^2 E(u^\prime ({\hat{R}}(x)-R)^2) = E[u^\prime (-Eu^{\prime \prime }-REu^\prime )^2] = E[u^\prime (R^\prime (x)u^\prime (x)\epsilon +O(\epsilon ^2))^2 ]=R^{\prime 2}(x)u^{\prime 3}(x)\sigma _\epsilon ^2 +O(\sigma _\epsilon ^3)$. This implies that

$$\begin{aligned} E \left( u^\prime \left( {\hat{R}}(x)-R\right) ^2 \right) = u^\prime (x) R^{\prime 2}(x)\sigma _\epsilon ^2+O(\sigma _\epsilon ^3). \end{aligned}$$

(19)

We also have

$$\begin{aligned} R^\prime (x)Eu^\prime -E(u^\prime R^\prime )=0.5[R^\prime (x)u^{\prime \prime \prime }(x)-(u^\prime (x) R^\prime (x))^{\prime \prime }]\sigma _\epsilon ^2 +O(\sigma _\epsilon ^3). \end{aligned}$$

The last two equation, together with Eq. (18), imply that $(-{\hat{R}}^\prime (x)+R^\prime (x))Eu^\prime$ is equal to

$$\begin{aligned} 0.5(R^\prime (x)u^{\prime \prime \prime }(x)-(u^\prime (x) R^\prime (x))^{\prime \prime })\sigma _\epsilon ^2 +u^\prime (x) R^{\prime 2}(x)\sigma _\epsilon ^2+O(\sigma _\epsilon ^3). \end{aligned}$$

Thus for arbitrarily small $\sigma _\epsilon$, $-{\hat{R}}^\prime (x)\ge -R^\prime (x)$ only if $R^\prime (x)u^{\prime \prime \prime }(x)+2u^\prime (x)$ $R^{\prime 2}(x)- (u^\prime (x) R^\prime (x))^{\prime \prime }\ge 0$, which is equivalent to $-R^{\prime \prime \prime }(x)+2R(x)R^{\prime \prime }(x)+2R^{\prime 2}(x) \ge 0$.

To prove the result where $i=5$, we have

$$\begin{aligned} {\hat{S}}(x)=-{\hat{R}}^\prime (x)-0.5{\hat{R}}^2(x) = - \left[ E \left( \frac{u^\prime }{Eu^\prime }R\right) \right] ^\prime -0.5 \left[ E \left( \frac{u^\prime }{Eu^\prime }R\right) \right] ^2, \end{aligned}$$

where again for brevity, we have omitted the argument $(x+{\tilde{\epsilon }})$ of the functions under the expectation operators.

Hence, we obtain

$$\begin{aligned} {\hat{S}}(x)= & {} -E \left[ \left( \frac{u^\prime }{Eu^\prime }\right) ^\prime R\right] + E \left[ \frac{u^\prime }{Eu^\prime }(-R^\prime -0.5R^2)\right] \\&+\,0.5E \left( \frac{u^\prime }{Eu^\prime }R^2\right) - 0.5 \left( E\frac{u^\prime }{Eu^\prime }R\right) ^2. \end{aligned}$$

In the meantime, from Eq. (16), we have $-E[(\frac{u^\prime }{Eu^\prime })^\prime R]=E[\frac{u^\prime }{Eu^\prime }({\hat{R}}(x)-R)^2]$, while $E(\frac{u^\prime }{Eu^\prime }R^2) - (E\frac{u^\prime }{Eu^\prime }R)^2 = E[\frac{u^\prime }{Eu^\prime }({\hat{R}}(x)-R)^2]$. Hence from the preceding equation, we have

$$\begin{aligned} {\hat{S}}(x) = E \left[ \frac{u^\prime }{Eu^\prime }S\right] + \frac{3}{2}E \left[ \frac{u^\prime }{Eu^\prime } \left( {\hat{R}}(x)-R\right) ^2\right] . \end{aligned}$$

(20)

Rewrite it as

$$\begin{aligned} \left( {\hat{S}}(x)-S(x)\right) Eu^\prime = -S(x)Eu^\prime + E(u^\prime S) + \frac{3}{2} E \left[ u^\prime \left( {\hat{R}}(x)-R\right) ^2\right] . \end{aligned}$$

As we have $Eu^\prime = u^\prime (x)+u^{\prime \prime \prime }(x) \sigma _\epsilon ^2+O(\sigma _\epsilon ^3)$, $E(u^\prime S)=u^\prime (x)S(x)+(u^\prime (x)S(x))^{\prime \prime }$ $\sigma _\epsilon ^2+O(\sigma _\epsilon ^3)$, and Eq. (19), it follows that $({\hat{S}}(x)-S(x))Eu^\prime$ is equal to

$$\begin{aligned} \left[ -0.5S(x)u^{\prime \prime \prime }(x) + 0.5(u^\prime (x)S(x))^{\prime \prime } + \frac{3}{2}u^\prime (x) R^{\prime 2}(x)\right] \sigma _\epsilon ^2+O(\sigma _\epsilon ^3). \end{aligned}$$

Thus for arbitrarily small $\sigma _\epsilon$, ${\hat{S}}(x)\ge S(x)$, only if $-S(x)u^{\prime \prime \prime }(x)+(u^\prime (x)S(x))^{\prime \prime }$ $+3u^\prime (x) R^{\prime 2}(x)\ge 0$. As $S(x)=-R^\prime (x)-0.5R^2(x)$, it can be rewritten as $S(x)(u^\prime (x)R(x))^\prime - (u^\prime (x)R(x)S(x))^\prime + (u^\prime (x)S^\prime (x))^\prime +3u^\prime (x) (S(x)$ $+0.5R^2(x))^2\ge 0$. Simplifying it, we obtain $-2R(x)S^\prime (x)+S^{\prime \prime }(x)+3 (S(x)+0.5R^2(x))^2\ge 0$. $\square$

Appendix 3: Proof of Theorem 3

The result where $i=2$ and 3 is a special case of Statement 1 of Theorem 5 which is proved in the last appendix; thus, we need only prove the result for the other three cases. As was explained at the beginning of the proof of Theorem 1 in 1, we need only prove the result for the case of zero-mean risks. We first prove the following lemma.

Lemma 3

For $i\in \{1, 4, 5\}$, $\tau _i(x)$ is increased by a zero-mean background risk ${\tilde{\epsilon }}$ if

$$\begin{aligned} E(\tau _i(x+{\tilde{\epsilon }})h_i(x+{\tilde{\epsilon }}))\ge \tau _i(x)Eh_i(x+{\tilde{\epsilon }}), \end{aligned}$$

(21)

where $h_i(x)$ is defined in Theorem 3.

Proof The results where $i=4$ and 5 immediately follow from Eqs. (17) and (20), respectively; thus, we need only prove the case where $i=1$. To prove this case, we have

$$\begin{aligned} {\hat{C}}(x)+1=\frac{ Eu^\prime (x+{\tilde{\epsilon }}) Eu^{\prime \prime \prime }(x+{\tilde{\epsilon }} )}{(Eu^{\prime \prime }(x+{\tilde{\epsilon }}))^2}. \end{aligned}$$

Some simple calculations lead to

$$\begin{aligned} {\hat{C}}(x) +1= \frac{ Eu^\prime (x+{\tilde{\epsilon }})}{(Eu^{\prime \prime }(x+{\tilde{\epsilon }}))^2} E \left[ (C(x+{\tilde{\epsilon }})+1) \frac{(u^{\prime \prime } (x+{\tilde{\epsilon }}))^2}{u^\prime (x+{\tilde{\epsilon }})}\right] . \end{aligned}$$

(22)

Meanwhile, from the Cauchy–Schwarz inequality ($E{\tilde{\epsilon }}_1^2 E{\tilde{\epsilon }}_2^2 \ge [E({\tilde{\epsilon }}_1{\tilde{\epsilon }}_2)]^2$), we have

$$\begin{aligned} Eu^\prime (x+{\tilde{\epsilon }})E\frac{(u^{\prime \prime } (x+{\tilde{\epsilon }}))^2}{u^\prime (x+{\tilde{\epsilon }})}\ge (Eu^{\prime \prime }(x+{\tilde{\epsilon }}))^2. \end{aligned}$$

Rewrite it as

$$\begin{aligned} \frac{ Eu^\prime (x+{\tilde{\epsilon }})}{(Eu^{\prime \prime }(x+{\tilde{\epsilon }}))^2}\ge \frac{1}{E\frac{(u^{\prime \prime }(x+{\tilde{\epsilon }}))^2}{u^\prime (x+{\tilde{\epsilon }})}}. \end{aligned}$$

This, together with Eq. (22), implies that

$$\begin{aligned} {\hat{C}}(x) +1\ge & {} \frac{1}{E\frac{(u^{\prime \prime }(x+{\tilde{\epsilon }}))^2}{u^\prime (x+{\tilde{\epsilon }})}} E \left[ (C(x+{\tilde{\epsilon }})+1) \frac{(u^{\prime \prime }(x+{\tilde{\epsilon }}))^2}{u^\prime (x+{\tilde{\epsilon }})}\right] \\= & {} \frac{1}{E\frac{(u^{\prime \prime }(x+{\tilde{\epsilon }}))^2}{u^\prime (x+{\tilde{\epsilon }})}} E\left[ C(x+{\tilde{\epsilon }}) \frac{(u^{\prime \prime }(x+{\tilde{\epsilon }}))^2}{u^\prime (x+{\tilde{\epsilon }})}\right] +1. \end{aligned}$$

Hence, ${\hat{C}}(x)\ge C(x)$ if $E[C(x+{\tilde{\epsilon }})h_1(x+{\tilde{\epsilon }}) ]\ge C(x)Eh_1(x+{\tilde{\epsilon }})$, where $h_1(x)=\frac{(u^{\prime \prime }(x))^2}{u^\prime (x)}$. $\square$

We now use the lemma to prove the theorem. Applying the lemma, we need only prove that the inequality in Theorem 3 is necessary and sufficient for Inequality (21). This is shown as follows. It is well known that Inequality (21) is true for all zero-mean ${\tilde{\epsilon }}$ if and only if it is true for all zero-mean binary ${\tilde{\epsilon }}$, i.e., $p\tau _i(x+\epsilon _1)h_i(x+\epsilon _1)+(1-p) \tau _i(x+\epsilon _2) h_i(x+\epsilon _2)\ge \tau _i(x)[ph_i(x+\epsilon _1) + (1-p)h_i(x+\epsilon _2)]$, for all $p\in (0,1)$, $\epsilon _1$ and $\epsilon _2$ satisfying $p\epsilon _1+(1-p)\epsilon _2=0$. After the elimination of p, this is equivalent to the condition that there exists a scalar m(x) such that

$$\begin{aligned} h_i(x+\epsilon _1)\frac{\tau _i(x+\epsilon _1)-\tau _i(x)}{\epsilon _1}\ge m(x)\ge h_i(x+\epsilon _2)\frac{\tau _i(x+\epsilon _2)-\tau _i(x)}{\epsilon _2}, \end{aligned}$$

for any $\epsilon _1>0>\epsilon _2$. By symmetry, the only candidate for m(x) is $h_i(x)\tau _i^\prime (x)$. The above condition is thus equivalent to $h_i(y)(\tau _i(y)-\tau _i(x))- h_i(x)\tau _i^\prime (x)(y-x)\ge 0$. $\square$

Appendix 4: Proof of Theorem 4

As was explained at the beginning of the proof of Theorem 1 in 1, we need only prove the result for the case of zero-mean risks. We first prove the following lemma. The proof uses a technique that is very close to the one used by Gollier and Pratt (1996) in the proof of their Proposition 3.

Lemma 4

Assume that for a given w, $g(w)> 0$, and for all x, $f(x)>0$, $f^\prime (x)\le 0$, and $(\frac{g(x)}{f(x)})^\prime \le 0$. If for all x, $(\frac{g(x)}{f(x)})^{\prime \prime }\ge -(\frac{g(x)}{f(x)})^\prime \frac{f^\prime (x)}{f(x)}$ then, for any unfair ${\tilde{\epsilon }}, ~\frac{Eg(w+{\tilde{\epsilon }})}{Ef(w+{\tilde{\epsilon }})}\ge \frac{g(w)}{f(w)}$.

Proof Note that as for all x, $f(x)>0$ and $(\frac{g(x)}{f(x)})^{\prime \prime }\ge -(\frac{g(x)}{f(x)})^\prime \frac{f^\prime (x)}{f(x)}$, we have $[(\frac{g(x)}{f(x)})^\prime f(x)]^\prime =(\frac{g(x)}{f(x)})^{\prime \prime } f(x)+ (\frac{g(x)}{f(x)})^\prime f^\prime (x)\ge 0$; i.e., $(\frac{g(x)}{f(x)})^\prime f(x)=g^\prime (x)-\frac{g(x)f^\prime (x)}{f(x)}$ is increasing. Thus as $g(w)> 0$ and for all x, $f^\prime (x)\le 0$, $\frac{g(x)}{f(x)}$ is decreasing, and $g^\prime (x)-\frac{g(x)f^\prime (x)}{f(x)}$ is increasing, y has the same sign as

$$\begin{aligned}&\frac{1}{g(w)} \left[ \left( g^\prime (w+y) - \frac{g(w+y)f^\prime (w+y)}{f(w+y)}\right) - \left( g^\prime (w)-\frac{g(w)f^\prime (w)}{f(w)}\right) \right] \\&\qquad +\,\frac{f^\prime (w+y)}{g(w)} \left( \frac{g(w+y)}{f(w+y)}-\frac{g(w)}{f(w)}\right) , \end{aligned}$$

which is equal to

$$\begin{aligned} \frac{g^\prime (w+y)}{g(w)} - \frac{f^\prime (w+y)}{f(w)} - \left( \frac{g^\prime (w)}{g(w)}-\frac{f^\prime (w)}{f(w)}\right) . \end{aligned}$$

Therefore, since the direction of integration cancels out the sign, we have

$$\begin{aligned} \int _0^\epsilon \left[ \frac{g^\prime (w+y)}{g(w)} - \frac{f^\prime (w+y)}{f(w)} - \left( \frac{g^\prime (w)}{g(w)} - \frac{f^\prime (w)}{f(w)}\right) \right] dy\ge 0. \end{aligned}$$

Since this is true for any $\epsilon$, taking the expectation over a random variable ${\tilde{\epsilon }}$ yields

$$\begin{aligned} \frac{Eg(w+{\tilde{\epsilon }})-g(w)}{g(w)} - \frac{Ef(w+{\tilde{\epsilon }})-f(w)}{f(w)}\ge \left( \frac{g^\prime (w)}{g(w)}-\frac{f^\prime (w)}{f(w)}\right) E{\tilde{\epsilon }}. \end{aligned}$$

As f(w) and g(w) are both strictly positive and $\frac{g(w)}{f(w)}$ is decreasing, the right-hand side is nonnegative, and it follows that $\frac{Eg(w+{\tilde{\epsilon }})}{Ef(w+{\tilde{\epsilon }})}\ge \frac{g(w)}{f(w)}$. $\square$

We now use the lemma to prove Theorem 4. Let $f(x)=h_i(x)$ and $g(x)= (\tau _i(x)+\alpha _0)h_i(x)$, where $h_i(x)$ is defined in Theorem 3 and $\alpha _0>0$ is arbitrarily large. Note that for $i=1, \ldots , 5$, for all x, $f(x)=h_i(x)> 0$, and for any given w, as $\alpha _0>0$ is arbitrarily large, $g(w)=(\tau _i(w)+\alpha _0)h_i(w)>0$. Also note that for $i=1$, $C(x)\ge -0.5$ implies $h_1^\prime (x)=-h_1(x)R(x)(2C(x)+1)\le 0$, for $i=3$, $h_i^\prime (x)=-u^{\prime \prime \prime }(x)\le 0$, and for $i=2,~4,~5$, $h_i^\prime (x)=u^{\prime \prime }(x)\le 0$. Moreover, as $\tau _i(x)$ is decreasing, $\frac{g(x)}{f(x)}=\tau _i(x)+\alpha _0$ is also decreasing. Furthermore, we have $-(\ln h_i(x))^\prime =\zeta _i(x)$, where $\zeta _i(x)$ is defined in the theorem, and it follows from the given condition in the theorem that $(\frac{g(x)}{f(x)})^{\prime \prime } +(\frac{g(x)}{f(x)})^\prime \frac{f^\prime (x)}{f(x)}= \tau _i^{\prime \prime }(x)-\tau _i^\prime (x)\zeta _i(x)\ge 0$. Thus, f(x) and g(x) satisfy all the conditions in the lemma. Now applying the lemma, we obtain $\frac{Eg(w+{\tilde{\epsilon }})}{Ef(w+{\tilde{\epsilon }})}\ge \frac{g(w)}{f(w)}~\forall {\tilde{\epsilon }}$. This implies that $\forall {\tilde{\epsilon }}, ~\frac{E[(\tau _i(w+{\tilde{\epsilon }}) + \alpha _0) h_i(w+{\tilde{\epsilon }})]}{Eh_i(w+{\tilde{\epsilon }})}\ge \frac{(\tau _i(w)+\alpha _0)h_i(w)}{h_i(w)}$, i.e., $E[\tau _i(w+{\tilde{\epsilon }})h_i(w+{\tilde{\epsilon }})]\ge \tau _i(w)Eh_i(w+{\tilde{\epsilon }})$. Applying Lemma 3 of the last appendix, we immediately obtain the conclusion in the theorem. $\square$

Appendix 5: Proof of Theorem 5

As was explained at the beginning of the proof of Theorem 1 in 1, we need only prove the result for the case of zero-mean risks. We first prove Statement 1. The inequality $(-1)^{n-m}\frac{Eu^{(n)}(x+{\tilde{\epsilon }})}{Eu^{(m)}(x+{\tilde{\epsilon }})}\ge R_{n/m}(x)$ is equivalent to

$$\begin{aligned} (-1)^{n-1}Eu^{(n)}(x+{\tilde{\epsilon }})\ge (-1)^{m-1}R_{n/m}(x)Eu^{(m)}(x+{\tilde{\epsilon }}). \end{aligned}$$

It is well known that the above inequality is true for all zero-mean ${\tilde{\epsilon }}$ if and only if it is true for all zero-mean binary ${\tilde{\epsilon }}$.¹⁴ Thus, the inequality is true for all zero-mean ${\tilde{\epsilon }}$ if and only if for all $x_2< x< x_1$,

$$\begin{aligned}&\left[ (-1)^{n-1}u^{(n)}(x_1)-(-1)^{m-1}R_{n/m}(x)u^{(m)} (x_1) \right] (x-x_2)\\&\quad + \left[ (-1)^{n-1}u^{(n)}(x_2)-(-1)^{m-1}R_{n/m} (x)u^{(m)}(x_2)\right] (x_1-x)\ge 0. \end{aligned}$$

This is equivalent to the condition that there exists a scaler $\kappa (x)$ such that

$$\begin{aligned}&(-1)^{m-1}u^{(m)}(x_1)\frac{R_{n/m}(x_1)-R_{n/m}(x)}{x_1-x}\ge \kappa (x)\\&\quad \ge (-1)^{m-1}u^{(m)}(x_2)\frac{R_{n/m}(x)-R_{n/m}(x_2)}{x-x_2}. \end{aligned}$$

The only candidate for $\kappa (x)$ is $u^{(m)}(x)R_{n/m}^\prime (x)$, and it follows that the above condition is equivalent to $(-1)^{m-1}\xi (w, x)\ge 0$, where $\xi (w, x)=u^{(m)}(x)(R_{n/m}(x)-R_{n/m}(w))- u^{(m)}(w)R_{n/m}^\prime (w)(x-w)$. This proves the first statement.

To prove the second statement, observe that $\xi (w, w)=0$ and

$$\begin{aligned} \frac{\partial \xi (w, x)}{\partial x}= & {} u^{(m+1)}(x)(R_{n/m}(x)-R_{n/m}(w)) \\&+u^{(m)}(x)R_{n/m}^\prime (x)- u^{(m)}(w)R_{n/m}^\prime (w), \end{aligned}$$

which implies $\frac{\partial \xi (w, x)}{\partial x}|_{x=w}=0$. It follows that to have $(-1)^{m-1}\xi (w, x)\ge 0$  $\forall w, x$, it is necessary that $(-1)^{m-1}\frac{\partial ^2\xi (w, x)}{\partial x^2}|_{x=w}\ge 0$  $\forall w$, which is equivalent to $R_{n/m}^{\prime \prime }(w)\ge 2R_{(m+1)/m}(w)R_{n/m}^\prime (w)$  $\forall w$.

To prove the third statement, let $f(x)=(-1)^{m-1}u^{(m)}(x)$ and $g(x)=(-1)^{n-1}u^{(n)}(x)$. Applying Lemma 4 in the last appendix, we immediately conclude that the third statement is true.

The proof of the fourth statement follows from the proof of Proposition 1 in Kimball (1993) and from the fact that $R_{(n+1)/n}$ is isomorphic to risk aversion as applied to $(-1)^{n-1}u^{(n-1)}$. $\square$