Characterizations of risk aversion in cumulative prospect theory

Abstract

In this paper, we investigate the necessary and sufficient conditions for a decision maker to be monotone risk averse and left-monotone risk averse, respectively, in cumulative prospect theory (CPT). Our results show that the decision maker is more pessimistic than greedy if she is either monotone or left-monotone risk averse, which is similar to that of Chateauneuf et al. (Econ Theory 25(3):649–667, 2005) in the rank-dependent expected utility model. Detailed examples are presented to illustrate the main theorems. With this work, we make a progress in the characterizations of risk aversion in CPT, which is essential in understanding the features of CPT and its applications in finance and insurance.

Appendix

In the appendix, we first prove the relation (3.6), and then prove Theorems 3.1 and 4.1.

Proof of (3.6)

Denote $G_u^*:=\sup _{y\leqslant 0\leqslant x} \max \{RG_u^{+}(x),\,G_u^{+}(x),\,G_u^{-}(y)\}.$ To show the equation (3.6), it suffices to show that $G_{u}\leqslant G_u^*$. From the definition of $G_{u}$, for any $\varepsilon >0$, there exist $x_{1}<x_{2}\leqslant y_{1}<y_{2}$ such that

$$\begin{aligned} g_0:=\frac{(u(y_2)-u(y_1))/(y_2-y_1)}{(u(x_2)-u(x_1))/(x_2-x_1)}>G_u-\varepsilon . \end{aligned}$$

Next we consider the following five cases.

• (i) If $x_1\geqslant 0$, then $G_u^*\geqslant G_u^{+}(y_1)\geqslant g_0>G_u-\varepsilon$.

• (ii) If $x_{1}< 0 < x_{2}$, note that

  $$\begin{aligned} \frac{u(x_2)-u(x_1)}{x_2-x_1} = \lambda \frac{u(x_2)-u(0)}{x_2}+(1-\lambda ) \frac{u(0)-u(x_1)}{-x_1} \end{aligned}$$

  with $\lambda =x_2/(x_2-x_1)\in (0,1)$. Then at least one of ${(u(x_2)-u(0))}/{x_2}$ and ${(u(0)-u(x_1))}/{(-x_1)}$ is no larger than ${(u(x_2)-u(x_1))}/{(x_2-x_1)}$. If ${(u(x_2)-u(0))}/{x_2}\leqslant {(u(x_2)-u(x_1))}/{(x_2-x_1)}$, then

  $$\begin{aligned} G_u^*\geqslant G_u^+(y_1)\geqslant \frac{(u(y_2)-u(y_1))/(y_2-y_1)}{(u(x_2)-u(0))/x_2}\geqslant g_0>G_u-\varepsilon . \end{aligned}$$

  Otherwise, ${(u(0)-u(x_1))}/{(-x_1)}\leqslant {(u(x_2)-u(x_1))}/{(x_2-x_1)}$, then

  $$\begin{aligned} G_u^* \geqslant RG_u^+(y_1)\geqslant \frac{(u(y_2)-u(y_1))/(y_2-y_1)}{{(u(0)-u(x_1))}/{(-x_1)}}\geqslant g_0>G_u-\varepsilon . \end{aligned}$$

• (iii) If $x_{2}\leqslant 0\leqslant y_{1}$, then $G_u^*\geqslant RG^+(y_1)\geqslant g_0>G_u-\varepsilon$.

• (iv) If $y_{1}<0<y_{2}$, similar to (ii), one can show $G_u^*>G_u-\varepsilon$.

• (v) If $y_2\leqslant 0$, then $G_u^*\geqslant G_u^{-}(x_2)\geqslant g_0>G_u-\varepsilon$.

Combining the above five cases, we have that $G_u^*>G_u-\varepsilon$ for any $\varepsilon >0$, and hence, $G_u^*\geqslant G_u$. This completes the proof. $\square$

Next, we prove Theorems 3.1 and 4.1 in the following. We point out that the proofs are similar to Chateauneuf et al. [9] but much more complicated and technical. The proof of sufficiency involves two steps: first for random variables with finitely many values, and then using a continuity argument to extend sufficiency of the condition to random variables in general.

Proof of Theorem 3.1

Sufficiency.

• The discrete case. Noticing the relationship between the dispersive order and the mean-preserving out-stretch for random variables with finite outcomes, it suffices to show that $V_{u,h_1,h_2}$ preserves any one mean-preserving out-stretch. Let X and Y be two random variables with cumulative distribution functions supported on finite sets such that Y is a mean-preserving out-stretch of X with $(a,b,\alpha )$. Note that Y can be obtained from X through m mean-preserving out-stretches of X with $(a/m,b/m,\alpha )$. Hence, it suffices to show that $V_{u,h_1,h_2}$ preserves any one mean-preserving out-stretch $(a,b,\alpha )$ for $a,\,b>0$ small enough. Without loss of generality, let the probability mass functions of X and Y be given in Example 2.1 such that $x_i$ and $x_i-a$ have the same sign for $i=1,\ldots ,k$ and $x_j$ and $x_j+b$ have the same sign for $j=k,\ldots ,n$. Let $r\in \{1,\ldots ,n\}$ satisfy that $x_r\leqslant 0< x_{r+1}$. Considering this, three cases arise.

  • (a) $k>r$. Denote $\Delta V= V_{u,h_1,h_2}(X)-V_{u,h_1,h_2}(Y)$. It holds that

    $$\begin{aligned} \Delta V= & {} \sum _{i=1}^r (u(x_i) -u(x_{i}-a)) (h_2(p_i)-h_2(p_{i-1})) \\&+\, \sum _{i=r+1}^{k-1}(u(x_i) -u(x_{i}-a)) (h_1(1-p_{i-1})-h_1(1-p_i))\\&+ \,(u(x_k) -u(x_{k}-a)) (h_1(1-p_{k-1})-h_1(1-\alpha ))\\&- (u(x_k+b) -u(x_{k}))(h_1(1-\alpha )-h_1(1-p_k))\\&- \sum _{i=k+1}^n (u(x_i+b) -u(x_{i})) (h_1(1-p_{i-1})-h_1(1-p_i)). \end{aligned}$$

    Take $u(x_{i_0}) - u(x_{i_0}-a) = \min _{1\leqslant i\leqslant r}\{u(x_i) - u(x_{i}-a)\}$, $u(x_{i_1}) - u(x_{i_1}-a) = \min _{r+1\leqslant i\leqslant k}\{u(x_i) - u(x_{i}-a)\}$ and $u(x_{j_0}+b) - u(x_{j_0}) = \max _{k\leqslant j\leqslant n}\{u(x_j+b) - u(x_{j})\}$. It follows that

    $$\begin{aligned} \Delta V\geqslant & {} (u(x_{i_0}) - u(x_{i_0}-a)) h_2(p_r) +(u(x_{i_1}) - u(x_{i_1}-a)) (h_1(1-p_r)-h_1(1-\alpha )) \\&- (u(x_{j_0}+b) - u(x_{j_0}) ) h_1(1-\alpha )\\&{\mathop {=}\limits ^{\mathrm{sgn}}} \frac{u(x_{i_0}) - u(x_{i_0}-a)}{a}\frac{ h_2(p_r) }{\alpha } + \frac{u(x_{i_1}) - u(x_{i_1}-a)}{a} \frac{h_1(1-p_r)-h_1(1-\alpha )}{\alpha } \\&- \frac{u(x_{j_0}+b) - u(x_{j_0}) }{b} \frac{h_1(1-\alpha )}{1-\alpha }\\&{\mathop {=}\limits ^\mathrm{sgn}} \frac{(u(x_{i_0}) - u(x_{i_0}-a))/a}{(u(x_{j_0}+b) - u(x_{j_0}))/b} \cdot \frac{ h_2(p_r) / \alpha }{h_1(1-\alpha )/(1-\alpha )} \\&+ \frac{(u(x_{i_1}) - u(x_{i_1}-a))/a}{(u(x_{j_0}+b) - u(x_{j_0}))/b} \cdot \frac{(h_1(1-p_r)-h_1(1-\alpha ))/\alpha }{h_1(1-\alpha )/(1-\alpha )} - 1\\\geqslant & {} P_{p_r,1-\alpha }(h_1,h_2)/RG_u^{+}(x_{j_0}) + P_{p_r,1-\alpha }(h_1)/G_u^{+}(x_{j_0}) - 1, \end{aligned}$$

    where $x{\mathop {=}\limits ^\mathrm{sgn}} y$ represents that x and y have the same sign and the second step follows from $a\alpha =b(1-\alpha )$. From the second inequality in (3.9) we have that $\Delta V\geqslant 0$.

  • (b) $r=k$. If $x_{r}=0$, then $\Delta V$ is expressed as the case in (a). If $x_r<0$, we similarly define $u(x_{i_0}) - u(x_{i_0}-a)=\min _{1\leqslant i\leqslant k}\{u(x_{i})-u(x_{i}-a) \}$ and $u(x_{j_0}+b) - u(x_{j_0})=\max _{k\leqslant j\leqslant n}\{u(x_{j}+b)-u(x_{j})\}$. Then it holds that

    $$\begin{aligned} \Delta V= & {} \sum _{i=1}^k (u(x_i) -u(x_{i}-a)) (h_2(p_i)-h_2(p_{i-1})) \\&+(u(x_{k})-u(x_{k}-a))(h_{2}(\alpha )-h_{2}(p_{k-1}))\\&-(u(x_{k}+b)-u(x_{k}))(h_{1}(1-\alpha )-h_{1}(1-p_{k})) \\&- \sum _{i=\ell +1}^n (u(x_i+b) -u(x_{i})) (h_1(1-p_{i-1})-h_1(1-p_i))\\\geqslant & {} (u(x_{i_0}) - u(x_{i_0}-a)) h_2(\alpha ) - (u(x_{j_0}+b) - u(x_{j_0}) ) h_1(1-\alpha )\\&{\mathop {=}\limits ^\mathrm{sgn}} \frac{\left( u(x_{i_{0}})-u(x_{i_{0}}-a)\right) /a}{(u(x_{j_{0}}+b)-u(x_{j_{0}}))/b} \cdot \frac{h_{2}(\alpha )/\alpha }{h_{1}(1-\alpha )/(1-\alpha )}-1\\\geqslant & {} P_{\alpha ,1-\alpha }(h_1,h_2)/RG_u^{+}(x_{j_0})-1\geqslant 0, \end{aligned}$$

    where the last inequality follows from the second inequality of (3.9) by noting that $P_{\alpha ,1-\alpha }(h_1)=0$.

  • (c) $k<r$. This is similar to (a). Note that in this case

    $$\begin{aligned} \Delta V =&\sum _{i=1}^{k-1}(u(x_{i})-u(x_{i}-a))(h_{2}(p_{i})-h_{2}(p_{i-1}))\\&+(u(x_{k})-u(x_{k}-a))(h_{2}(\alpha )-h_{2}(p_{k-1}))\\&-(u(x_{k}+b)-u(x_{k}))(h_{2}(p_{k})-h_{2}(\alpha ))\\&-\sum _{i=k+1}^{r}(u(x_{i}+b)-u(x_{i}))(h_{2}(p_{i})-h_{2}(p_{i-1}))\\&-\sum _{i=r+1}^{n}(u(x_{i}+b)-u(x_{i}))(h_{1}(1-p_{i-1})-h_{1}(1-p_{i})). \end{aligned}$$

    Similarly, denote $u(x_{i_{0}})-u(x_{i_{0}}-a)=\min _{1\leqslant i\leqslant k}\{u(x_{i})-u(x_{i}-a)\},$$u(x_{j_{0}}+b)-u(x_{j_{0}})=\max _{k\leqslant j\leqslant r}\{u(x_{j}+b)-u(x_{j})\},$ and $u(x_{j_{1}}+b)-u(x_{j_{1}})=\max _{r+1\leqslant j\leqslant n}\{u(x_{j}+b)-u(x_{j})\}.$ It follows from $a\alpha =b(1-\alpha )$ that

    $$\begin{aligned} \Delta V&\geqslant (u(x_{i_{0}})-u(x_{i_{0}}-a))h_{2}(\alpha )-(u(x_{j_{0}}+b)-u(x_{j_{0}}))\left( h_{2}(p_{r})-h_{2}(\alpha )\right) \\&\quad -(u(x_{j_{1}}+b)-u(x_{j_{1}}))h_{1}(1-p_{r})\\&\quad \overset{\mathrm {sgn}}{=}1-\frac{\left( u(x_{j_{0}}+b)-u(x_{j_{0}})\right) /b}{\left( u(x_{i_{0}})-u(x_{i_{0}}-a)\right) /a} \cdot \frac{\left( h_{2}(p_{r})-h_{2}(\alpha )\right) /(1-\alpha )}{h_{2}(\alpha )/\alpha }\\&\quad -\frac{(u(x_{j_{1}}+b)-u(x_{j_{1}}))/b}{\left( u(x_{i_{0}})-u(x_{i_{0}}-a)\right) /a}\cdot \frac{h_{1}(1-p_{r})/(1-\alpha )}{h_{2}(\alpha )/\alpha }\\&\geqslant 1- G_u^{-}(x_{i_0})P_{p_r,\alpha }(h_2) - RG_u^{-}(x_{i_0}) P_{p_r,\alpha }(h_2,h_1). \end{aligned}$$

    By the first inequality of (3.9), we have that $\Delta V\geqslant 0$. This ends the proof of sufficiency for the discrete case.

• The general case. Let X and Y be two general random variables such that $X\prec _{\mathrm{disp}} Y$ and $\mathbb {E}[X]=\mathbb {E}[Y]$. Then by Proposition 2.1, there exist two sequences of random variables $\{X_n\}$ and $\{Y_n\}$ each with finite support, such that (a) and (b) in Proposition 2.1 hold. By (a) and the proof of the discrete case, we have $V_{u,h_1,h_2}(X_n)\geqslant V_{u,h_1,h_2}(Y_n)$ for $n\in \mathbb {N}$. By (b), we have that $V_{u,h_1,h_2}(X_n)\leqslant V_{u,h_1,h_2}(X)$. Note that $Y-2^{-n}\leqslant Y_n$ for each $n\geqslant 1$ and u is continuous. It follows that $\lim _{n\rightarrow \infty } V_{u,h_1,h_2}(Y_n)=V_{u,h_1,h_2}(Y)$. Therefore, $V_{u,h_1,h_2}(X)\geqslant V_{u,h_1,h_2}(Y)$. This completes the proof of sufficiency for the general case.

Necessity. Similar to the proof of necessity in Theorem 1 of Chateauneuf et al. [9], we show it by contradiction. We only give a proof for the second inequality of (3.9), the proof of the first one is similar. Assume that the second inequality of (3.9) does not hold, that is, there exist ${\alpha },\,p\geqslant 0$ with ${\alpha }+p\leqslant 1$ and $y\geqslant 0$ such that

$$\begin{aligned} \frac{P_{p,{\alpha }}(h_1,h_2)}{RG_u^{+}(y)} + \frac{P_{p,{\alpha }}(h_1)}{G_u^{+}(y)}<1. \end{aligned}$$

Then there exist $x_1<x_2\leqslant 0\leqslant x_3< x_4\leqslant y< \min \{x_5, x_6\}$ such that

$$\begin{aligned}&\frac{(u(x_2) - u(x_1))/(x_2-x_1)}{(u(x_5) - u(y))/(x_5-y)} \cdot \frac{ h_2(p) / \alpha }{h_1(1-\alpha )/(1-\alpha )} \nonumber \\&\quad + \frac{(u(x_4) - u(x_3))/(x_4-x_3)}{(u(x_6) - u(y))/(x_6-y)}\cdot \frac{(h_1(1-p)-h_1(1-\alpha ))/\alpha }{h_1(1-\alpha )/(1-\alpha )} < 1. \end{aligned}$$

(6.1)

Without loss of generality, one can assume that $x_5=x_6$. This is due to that if $(u(x_6) - u(y))/(x_6-y) <(u(x_5) - u(y))/(x_5-y)$, the inequality of (6.1) still holds by replacing $(u(x_6) - u(y))/(x_6-y)$ by $(u(x_5) - u(y))/(x_5-y)$, and vice versa. Note that for a continuous function u, for any $z_1<z_2$, there exists $\delta _0>0$ such that

$$\begin{aligned} (0,\delta _0] \subset \left\{ z_2^*-z_1^*:\frac{u(z_2^*)-u(z_1^*)}{z_2^*-z_1^*} = \frac{u(z_2)-u(z_1)}{z_2-z_1},~ z_1\leqslant z_1^*<z_2^*\leqslant z_2\right\} . \end{aligned}$$

It then follows that there exist $x_1^*<x_2^*\leqslant 0\leqslant x_3^*< x_4^*\leqslant y^*< x_5^*$ with $x_2^*-x_1^*=x_4^*-x_3^*$ and $(x_5^*-y^*)(1-\alpha )=(x_2^*-x_1^*)\alpha$ such that

$$\begin{aligned}&\frac{(u(x_2^*) - u(x_1^*))/(x_2^*-x_1^*)}{(u(x_5^*) - u(y^*))/(x_5^*-y^*)}\cdot \frac{ h_2(p) / \alpha }{h_1(1-\alpha )/(1-\alpha )} \nonumber \\&\quad + \frac{(u(x_4^*) - u(x_3^*))/(x_4^*-x_3^*)}{(u(x_5^*) - u(y^*))/(x_5^*-y^*)} \cdot \frac{(h_1(1-p)-h_1(1-\alpha ))/\alpha }{h_1(1-\alpha )/(1-\alpha )} < 1. \end{aligned}$$

(6.2)

We define two random variables X and Y such that

Then it can be verified that Y is a mean-preserving out-stretch $(x_2^*-x_1^*,x_5^*-y,\alpha )$ of X by Example 2.1, but $V_{u,h_1,h_2}(X) < V_{u,h_1,h_2}(Y)$ by (6.2), which yields a contradiction with that $V_{u,h_1,h_2}$ is monotone risk averse. This ends the proof. $\square$

Proof of Theorem 4.1

Sufficiency.

• The discrete case. Let X and Y be two random variables with finite outcomes such that $X\prec _{\mathrm{lir}}Y$. By Proposition 2.2, without loss of generality, we assume that X and Y are defined in Example 2.2, and, moreover, the special mean-preserving left-stretch between X and Y satisfies that $s=1$ and $a>0$ is small enough such that $x_k-a$ and $x_k$ have the same sign. Suppose that r is an integer such that $x_{r}\leqslant 0<x_{r+1}$.

  • (a) $r\leqslant k$ and $x_r< 0$. Denote $\Delta V= V(X)-V(Y)$. It holds that

    $$\begin{aligned} \Delta V= & {} \sum _{i=1}^r (u(x_i) -u(x_{i}-a)) (h_2(p_i)-h_2(p_{i-1})) \\&+ \sum _{i=r+1}^{k-1}(u(x_i) -u(x_{i}-a)) (h_1(1-p_{i-1})-h_1(1-p_i))\\&+\ (u(x_k)-u(x_k-a))(h_1(1-p_{k-1})-h_1(1-\alpha ))\\&-\ (u(x_{k+1})-u(x_k))(h_1(1-\alpha )-h_1(1-p_k)). \end{aligned}$$

    Let $u(x_{i_0}) - u(x_{i_0}-a) = \min _{1\leqslant i\leqslant r}(u(x_i) - u(x_{i}-a))$ and $u(x_{j_{0}})-u(x_{j_{0}}-a)=\min _{r+1\leqslant j\leqslant k}(u(x_{j})-u(x_{j}-a))$. It follows that

    $$\begin{aligned}&\Delta V \geqslant (u(x_{i_0}) - u(x_{i_0}-a)) h_2(p_r) + (u(x_{j_0})\\&\quad - u(x_{j_0}-a))( h_1(1-p_r)-h_1(1-\alpha )) \\&\quad - (u(x_{k+1}) - u(x_{k}) ) (h_1(1-\alpha )-h_1(1-p_k))\\&\quad {\mathop {=}\limits ^\mathrm{sgn}} \frac{u(x_{i_0}) - u(x_{i_0}-a)}{a} \cdot \frac{h_2(p_r)}{\alpha } + \frac{u(x_{j_0}) - u(x_{j_0}-a)}{a} \\&\quad \cdot \frac{ h_1(1-p_r)-h_1(1-\alpha )}{\alpha }\\&\quad - \frac{u(x_{k+1}) - u(x_{k})}{x_{k+1}-x_k} \cdot \frac{h_1(1-\alpha )-h_1(1-p_k)}{p_k-\alpha }\\&\quad {\mathop {=}\limits ^\mathrm{sgn}} \frac{(u(x_{i_0}) - u(x_{i_0}-a))/a}{(u(x_{k+1}) - u(x_{k}))/(x_{k+1}-x_{k})} \cdot \frac{ h_2(p_r) / \alpha }{(h_1(1-\alpha )-h_1(1-p_k))/(p_k-\alpha )}\\&\quad + \frac{(u(x_{j_0}) - u(x_{j_0}-a))/a}{(u(x_{k+1}) - u(x_{k}))/(x_{k+1}-x_{k})} \cdot \frac{(h_1(1-p_r)-h_1(1-\alpha ))/\alpha }{(h_1(1-\alpha )-h_1(1-p_k))/(p_k-\alpha )} - 1\\&\quad \geqslant P_{p_r,1-\alpha ,1-p_k}^L(h_1,h_2)/RG_u^+(x_k) + P_{1-p_r,1-\alpha ,1-p_k}^L(h_1)/G^{+}_u(x_k) - 1\geqslant 0, \end{aligned}$$

    where the second step follows from $a\alpha =(x_{k+1}-x_k)(p_k-\alpha )$ in (2.4) and the last step is from the first inequality of (4.4).

  • (b) $r=k$ and $x_r=0$. Similarly, we define $u(x_{i_0}) - u(x_{i_0}-a)= \min _{1\leqslant i\leqslant k}(u(x_i) - u(x_{i}-a))$. Then it holds that

    $$\begin{aligned} \Delta V= & {} \sum _{i=1}^{k-1} (u(x_i) -u(x_{i}-a)) (h_2(p_i)-h_2(p_{i-1}))+(u(x_k) \\&-u(x_{k}-a)) (h_2(\alpha )-h_2(p_{k-1})) - (u(x_{k+1}) -u(x_{k})) (h_1(1-\alpha )-h_1(1-p_{k}))\\\geqslant & {} (u(x_{i_0}) - u(x_{i_0}-a)) h_2(\alpha ) - (u(x_{k+1}) -u(x_{k})) (h_1(1-\alpha )-h_1(1-p_{k}))\\&{\mathop {=}\limits ^\mathrm{sgn}} \frac{u(x_{i_0}) - u(x_{i_0}-a)}{a} \cdot \frac{ h_2(\alpha )}{\alpha } - \frac{u(x_{k+1}) -u(x_{k})}{x_{k+1}-x_{k}} \cdot \frac{h_1(1-\alpha )-h_1(1-p_{k})}{p_k-\alpha }\\&{\mathop {=}\limits ^\mathrm{sgn}} \frac{(u(x_{i_0}) - u(x_{i_0}-a))/a}{(u(x_{k+1}) - u(x_{k}))/(x_{k+1}-x_{k})} \cdot \frac{ h_2(\alpha ) / \alpha }{(h_1(1-\alpha )-h_1(1-p_k))/(p_k-\alpha )} -1 \\\geqslant & {} P_{\alpha ,1-\alpha ,1-p_k}^L(h_1,h_2)/RG_u^+(x_k)-1\geqslant 0, \end{aligned}$$

    where the third step follows from $a\alpha =(x_{k+1}-x_k)(p_k-\alpha )$ and the last step is from the first inequality of (4.4) by noting that $P_{1-\alpha ,1-\alpha ,1-p_k}^L(h_1)=0$.

  • (c) $r > k$. Similarly, we define $u(x_{i_0}) - u(x_{i_0}-a)= \min _{1\leqslant i\leqslant k}(u(x_i) - u(x_{i}-a))$. Note that in this case

    $$\begin{aligned} \Delta V= & {} \sum _{i=1}^{k-1} (u(x_i) -u(x_{i}-a)) (h_2(p_i)-h_2(p_{i-1}))+(u(x_k)\\&-u(x_{k}-a)) (h_2(\alpha )-h_2(p_{k-1})) \\&- (u(x_{k+1}) -u(x_{k})) (h_2(p_{k})-h_2(\alpha ))\\&{\mathop {=}\limits ^{\mathrm{sgn}}} \frac{u(x_{i_0}) -u(x_{i_0}-a)}{a} \frac{h_2(\alpha )}{\alpha } - \frac{u(x_{k+1}) -u(x_{k})}{x_{k+1}-x_k} \frac{h_2(p_{k})-h_2(\alpha )}{p_{k}-\alpha }\\&{\mathop {=}\limits ^{\mathrm{sgn}}} \frac{(u(x_{i_0}) - u(x_{i_0}-a))/a}{(u(x_{k+1}) - u(x_{k}))/(x_{k+1}-x_{k})} \cdot \frac{ h_2(\alpha ) / \alpha }{(h_2(p_{k})-h_2(\alpha ))/(p_{k}-\alpha )} -1 \\\geqslant & {} \hat{P}_{h_2^*} / {G^-_u(x_{i_0})}-1 \geqslant 0, \end{aligned}$$

    where the last step follows from the second inequality of (4.4). This completes the proof of sufficiency for the discrete case.

• The general case. Let X and Y be two general random variables such that $X\prec _{\mathrm{lir}} Y$ and $\mathbb {E}[X]=\mathbb {E}[Y]$. Then by Proposition 2.3, there exist two sequences of random variables $\{X_n\}$ and $\{Y_n\}$ such that (a) and (b) in Proposition 2.3 hold. By (a) and the proof of the discrete case, we have $V_{u,h_1,h_2}(X_n)\geqslant V_{u,h_1,h_2}(Y_n)$ for $n\in \mathbb {N}$. By (b) and Proposition 3.1 of Mao and Hu [19], one can similarly prove that $\lim _{n\rightarrow \infty } V_{u,h_1,h_2}(X_n)=V_{u,h_1,h_2}(X)$ and $\lim _{n\rightarrow \infty } V_{u,h_1,h_2}(Y_n)=V_{u,h_1,h_2}(Y)$. Therefore, $V_{u,h_1,h_2}(X)\geqslant V_{u,h_1,h_2}(Y)$. This completes the proof of sufficiency for the general case.

Necessity. We show the necessity by contradiction similarly to that in the proof of Theorem 3.1. Assume that the first inequality of (4.4) does not hold, that is, there exist $0\leqslant p\leqslant \alpha \leqslant q\leqslant 1$ and $y\geqslant 0$ such that

$$\begin{aligned} \frac{P_{p,1-\alpha ,1-q}^{L}(h_{1},h_{2})}{RG_{u}^{+}(y)}+\frac{ P_{1-p,1-\alpha ,1-q}^{L}(h_{1})}{G_{u}^{+}(y)}<1. \end{aligned}$$

Then there exist $x_{1}<x_{2}\leqslant 0\leqslant x_{3}<x_{4}\leqslant y<\min \{x_{5},x_{6}\}$ such that

$$\begin{aligned}&\frac{\left( u(x_{2})-u(x_{1})\right) /(x_{2}-x_{1})}{\left( u(x_{5})-u(y)\right) /(x_{5}-y)}\cdot \frac{h_{2}(p)/\alpha }{\left( h_{1}(1-\alpha )-h_{1}(1-q)\right) /(q-\alpha )}\\&\quad + \frac{\left( u(x_{4})-u(x_{3})\right) /(x_{4}-x_{3})}{\left( u(x_{6})-u(y)\right) /(x_{6}-y)}\cdot \frac{\left( h_{1}(1-p)-h_{1}(1-\alpha )\right) /\alpha }{\left( h_{1}(1-\alpha )-h_{1}(1-q)\right) /(q-\alpha )}<1. \end{aligned}$$

By the same arguments in the proof of Theorem 3.1, there exist $x_{1}^{*}<x_{2}^{*}\leqslant 0\leqslant x_{3}^{*}<x_{4}^{*}\leqslant y^{*}\leqslant x_{5}^{*}$ with $x_{2}^{*}-x_{1}^{*}=x_{4}^{*}-x_{3}^{*}$ and $\left( x_{5}^{*}-y^{*}\right) (q-\alpha )=(x_{2}^{*}-x_{1}^{*})\alpha$ such that

$$\begin{aligned}&\frac{\left( u(x_{2}^{*})-u(x_{1}^{*})\right) /(x_{2}^{*}-x_{1}^{*})}{\left( u(x_{5}^{*})-u(y^{*})\right) /(x_{5}^{*}-y^{*})}\cdot \frac{h_{2}(p)/\alpha }{\left( h_{1}(1-\alpha )-h_{1}(1-q)\right) /(q-\alpha )} \nonumber \\&\quad +\frac{\left( u(x_{4}^{*})-u(x_{3}^{*})\right) /(x_{4}^{*}-x_{3}^{*})}{\left( u(x_{5}^{*})-u(y^{*})\right) /(x_{5}^{*}-y^{*})}\cdot \frac{\left( h_{1}(1-p)-h_{1}(1-\alpha )\right) /\alpha }{ \left( h_{1}(1-\alpha )-h_{1}(1-q)\right) /(q-\alpha )}<1. \end{aligned}$$

(6.3)

We define two random variables X and Y such that for $p\leqslant \alpha \leqslant q$

where $a=x_{2}^{*}-x_{1}^{*}$. Then it can be verified that Y is a mean-preserving left-stretch of X by Example 2.2. However,

$$\begin{aligned}&V_{u,h_{1},h_{2}}(X)-V_{u,h_{1},h_{2}}(Y) \nonumber \\&\quad =\left( u(x_{2}^{*})-u(x_{1}^{*})\right) h_{2}(p)+\left( u(x_{4}^{*})-u(x_{3}^{*})\right) \left( h_{1}(1-p)-h_{1}(1-\alpha )\right) \nonumber \\&\quad -\left( u(x_{5}^{*})-u(y^{*})\right) \left( h_{1}(1-\alpha )-h_{1}(1-q)\right) <0, \end{aligned}$$

(6.4)

where the inequality is due to relation (6.3). This yields a contradiction with that $V_{u,h_{1},h_{2}}$ is left-monotone risk averse. Next, we turn to the second inequality of (4.4). Suppose it does not hold, that is, there exists $x\leqslant 0$ such that

$$\begin{aligned} \frac{\hat{P}_{{h^*_{2}}}}{G_{u}^{-}(x)}<1. \end{aligned}$$

Then there exist $x_{1}<x\leqslant x_{2}\leqslant x_{3}\leqslant 0$ such that

$$\begin{aligned} \frac{\left( u(x)-u(x_{1})\right) /(x-x_{1})}{\left( u(x_{3})-u(x_{2})\right) /(x_{3}-x_{2})}\cdot \frac{h_{2}(\alpha )/\alpha }{ (h_{2}(p)-h_{2}(\alpha ))/(p-\alpha )}<1. \end{aligned}$$

By the same arguments in the proof of Theorem 3.1, there exist $x_{1}^{*}<x^{*}\leqslant x_{2}^{*}\leqslant x_{3}^{*}\leqslant 0$ with $\left( x_{3}^{*}-x_{2}^{*}\right) (p-\alpha )=(x^{*}-x_{1}^{*})\alpha$ such that

$$\begin{aligned} \frac{\left( u(x^{*})-u(x_{1}^{*})\right) /(x^{*}-x_{1}^{*}) }{\left( u(x_{3}^{*})-u(x_{2}^{*})\right) /(x_{3}^{*}-x_{2}^{*})}\cdot \frac{h_{2}(\alpha )/\alpha }{(h_{2}(p)-h_{2}(\alpha ))/(p-\alpha )}<1. \end{aligned}$$

(6.5)

We define two random variables X and Y such that for $p\geqslant \alpha$

where $a=x^{*}-x_{1}^{*}$. Then it can be verified that Y is a mean-preserving left-stretch of X by Example 2.2. However,

$$\begin{aligned} V_{u,h_{1},h_{2}}(X)-V_{u,h_{1},h_{2}}(Y)=\left( u(x^{*})-u(x_{1}^{*})\right) h_{2}(\alpha )-\left( u(x_{3}^{*})-u(x_{2}^{*})\right) (h_{2}(p)-h_{2}(\alpha ))<0 \end{aligned}$$

due to (6.5), which again yields a contradiction. This completes the proof of the necessity. $\square$