Bivariate almost stochastic dominance

Abstract

Univariate almost stochastic dominance has been widely studied and applied since its introduction by Leshno and Levy (Manag Sci 48:1074–1085, 2002). This paper extends this construction to the bivariate case by means of suitable two-attribute utility functions. After having confined correlation aversion and correlation loving to some acceptable levels, bivariate almost stochastic dominance rules are introduced for the preferences exhibiting confined correlation aversion and confined correlation loving. The impact of a change in risk in terms of bivariate almost stochastic dominance on optimal saving is analyzed as an application, as well as the effect of envy and altruism on income distributions. Finally, alternative definitions of bivariate almost stochastic dominance are discussed, as well as testing procedures for such dominance rules in financial problems.

Appendix: Proofs of the main results

1.1 A Proof of Theorem 1

Let us start with the “$\Rightarrow$” part. Integration by parts shows that

$$\begin{aligned} E[u(X_{1},X_{2})]&= u(b_{1},b_{2})-\int \limits _{a_{1}}^{b_{1}}u^{(1,0)}(x_{1},b_{2})\Pr [X_{1}\le x_{1}]\mathrm{d}x_{1} \nonumber \\&\quad -\int \limits _{a_{2}}^{b_{2}}u^{(0,1)}(b_{1},x_{2})\Pr [X_{2}\le x_{2}]\mathrm{d}x_{2} \nonumber \\&\quad +\int \limits _{a_{1}}^{b_{1}}\int \limits _{a_{2}}^{b_{2}}u^{(1,1)}(x_{1},x_{2})\Pr [X_{1}\le x_{1},X_{2}\le x_{2}]\mathrm{d}x_{1}\mathrm{d}x_{2}. \end{aligned}$$

(24)

Hence, the gain or loss in expected utility when switching from $(Y_{1},Y_{2})$ to $(X_{1},X_{2})$ can be written as

$$\begin{aligned}&E[u(X_{1},X_{2})]-E[u(Y_{1},Y_{2})] \\&\quad =\int \limits _{a_{1}}^{b_{1}}u^{(1,0)}(x_{1},b_{2})\Big (\Pr [Y_{1}\le x_{1}]-\Pr [X_{1}\le x_{1}]\Big )\mathrm{d}x_{1} \\&\qquad +\int \limits _{a_{2}}^{b_{2}}u^{(0,1)}(b_{1},x_{2})\Big (\Pr [Y_{2}\le x_{2}]-\Pr [X_{2}\le x_{2}]\Big )\mathrm{d}x_{2} \\&\qquad +\int \limits _{a_{1}}^{b_{1}}\int \limits _{a_{2}}^{b_{2}}\Big (\Pr [X_{1}\le x_{1},X_{2}\le x_{2}]-\Pr [Y_{1}\le x_{1},Y_{2}\le x_{2}]\Big ) u^{(1,1)}(x_{1},x_{2})\mathrm{d}x_{1}\mathrm{d}x_{2}. \end{aligned}$$

This expression can be traced back to Corollary 4 in Levy and Parouch (1974). The first two terms appearing in the expansion of $E[u(X_{1},X_{2})]-E[u(Y_{1},Y_{2})]$ are clearly negative as $u$ is non-decreasing and $X_{i}\preceq _{\text {fsd}}Y_{i}$ holds for $i=1,2$. Let us now show that the last one is also negative. To this end, consider $u$ in $\mathcal {U}_{\text {ca}}^{\varepsilon }$ such that $u^{(1,1)}\ne 0$ and denote

$$\begin{aligned} \gamma&= \inf \{-u^{(1,1)}\}>0 \\ \delta&= \sup \{-u^{(1,1)}\}>0. \end{aligned}$$

Note that (4) ensures that

$$\begin{aligned} \delta \le \gamma \left( \frac{1}{\varepsilon }-1\right) \Leftrightarrow \frac{\gamma }{\gamma +\delta }\ge \varepsilon . \end{aligned}$$

(25)

Let $S^{c}$ denote the complement of $S$ in $[a_{1},b_{1}]\times [a_{2},b_{2}]$. Then,

$$\begin{aligned}&\int \limits _{a_{1}}^{b_{1}}\int \limits _{a_{2}}^{b_{2}}\Big (\Pr [X_{1}\le x_{1},X_{2}\le x_{2}]-\Pr [Y_{1}\le x_{1},Y_{2}\le x_{2}]\Big ) u^{(1,1)}(x_{1},x_{2})\mathrm{d}x_{1}\mathrm{d}x_{2} \\&\quad =\int \int \limits _{S}\Big (\Pr [Y_{1}\le x_{1},Y_{2}\le x_{2}]-\Pr [X_{1}\le x_{1},X_{2}\le x_{2}]\Big )\big (-u^{(1,1)}(x_{1},x_{2})\big )\mathrm{d}x_{1}\mathrm{d}_{2} \\&\qquad +\int \int \limits _{S^{c}}\Big (\Pr [Y_{1}\le x_{1},Y_{2}\le x_{2}]-\Pr [X_{1}\le x_{1},X_{2}\le x_{2}]\Big )\big (-u^{(1,1)}(x_{1},x_{2})\big ) \mathrm{d}x_{1}\mathrm{d}x_{2} \\&\quad \le \delta \int \int \limits _{S}\Big (\Pr [Y_{1}\le x_{1},Y_{2}\le x_{2}]-\Pr [X_{1}\le x_{1},X_{2}\le x_{2}]\Big )\mathrm{d}x_{1}\mathrm{d}x_{2} \\&\qquad +\,\gamma \int \int \limits _{S^{c}}\Big (\Pr [Y_{1}\le x_{1},Y_{2}\le x_{2}]-\Pr [X_{1}\le x_{1},X_{2}\le x_{2}]\Big )\mathrm{d}x_{1}\mathrm{d}x_{2} \\&\quad =-\gamma \int \limits _{a_{1}}^{b_{1}}\int \limits _{a_{2}}^{b_{2}}\Big |\Pr [X_{1}\le x_{1},X_{2}\le x_{2}]-\Pr [Y_{1}\le x_{1},Y_{2}\le x_{2}]\Big |\mathrm{d}x_{1}\mathrm{d}x_{2}\\&\qquad +\,(\gamma +\delta )\int \int \limits _{S}\Big (\Pr [Y_{1}\le x_{1},Y_{2}\le x_{2}]-\Pr [X_{1}\le x_{1},X_{2}\le x_{2}]\Big )\mathrm{d}x_{1}\mathrm{d}x_{2} \\&\quad \le 0\hbox { by (25)}, \end{aligned}$$

which ends the proof of the “$\Rightarrow$ ” part.

Let us now turn to the “$\Leftarrow$” part. We assume that $E[u(X_{1},X_{2})]\le E[u(Y_{1},Y_{2})]$ holds for all $u$ in $\mathcal {U}_{\text {ca}}^{\varepsilon }$ and we have to show that ( 5) holds. Let us proceed by contradiction and assume that

$$\begin{aligned}&\int \int \limits _{S}\Big (\Pr [Y_{1}\le x_{1},Y_{2}\le x_{2}]-\Pr [X_{1}\le x_{1},X_{2}\le x_{2}]\Big )\mathrm{d}x_{1}\mathrm{d}x_{2}\nonumber \\&\quad >\varepsilon \int \limits _{a_{1}}^{b_{1}}\int \limits _{a_{2}}^{b_{2}}\Big |\Pr [X_{1}\le x_{1},X_{2}\le x_{2}]-\Pr [Y_{1}\le x_{1},Y_{2}\le x_{2}]\Big | \mathrm{d}x_{1}\mathrm{d}x_{2}. \end{aligned}$$

(26)

Let us show that we can then construct a utility function $u$ in $\mathcal {U} _{\text {ca}}^{\varepsilon }$ such that $E[u(X_{1},X_{2})]>E[u(Y_{1},Y_{2})]$. Let $\gamma$ and $\delta$ be positive real numbers such that $\varepsilon =\frac{\gamma }{\gamma +\delta }$. Now consider a utility function $u$ such that $u^{(1,0)}(x_{1},b_{2})=0,\,u^{(0,1)}(b_{1},x_{2})=0$, $u^{(1,1)}=-\gamma$ on $S^{c}$ and $u^{(1,1)}=-\delta$ on $S$, that is, a utility $u$ proportional to $(x_{1}-b_{1})(x_{2}-b_{2})$ up to additive constants. We then see that

$$\begin{aligned}&E[u(X_{1},X_{2})]-E[u(Y_{1},Y_{2})] \\&\quad =\delta \int \int \limits _{S}\Big (\Pr [Y_{1}\le x_{1},Y_{2}\le x_{2}]-\Pr [X_{1}\le x_{1},X_{2}\le x_{2}]\Big )\mathrm{d}x_{1}\mathrm{d}x_{2} \\&\qquad +\gamma \int \int \limits _{S^{c}}\Big (\Pr [Y_{1}\le x_{1},Y_{2}\le x_{2}]-\Pr [X_{1}\le x_{1},X_{2}\le x_{2}]\Big )\mathrm{d}x_{1}\mathrm{d}x_{2} \\&\quad =-\gamma \int \limits _{a_{1}}^{b_{1}}\int \limits _{a_{2}}^{b_{2}}\Big |\Pr [X_{1}\le x_{1},X_{2}\le x_{2}]-\Pr [Y_{1}\le x_{1},Y_{2}\le x_{2}]\Big |\mathrm{d}x_{1}\mathrm{d}x_{2}\\&\qquad +(\gamma +\delta )\int \int \limits _{S}\Big (\Pr [Y_{1}\le x_{1},Y_{2}\le x_{2}]-\Pr [X_{1}\le x_{1},X_{2}\le x_{2}]\Big )\mathrm{d}x_{1}\mathrm{d}x_{2} \\&\quad >0, \end{aligned}$$

which ends the proof.

1.2 B Proof of Theorem 2

Let us start with the “$\Rightarrow$” part. By Corollary 1.6.12 in Denuit et al. (2005) applied to $X_{i}-a_{i}$ and $Y_{i}-a_{i}$, we have

$$\begin{aligned}&E[u(X_{1},X_{2})]-E[u(Y_{1},Y_{2})] \nonumber \\&\quad =\int \limits _{a_{1}}^{b_{1}}u^{(1,0)}(x_{1},a_{2})\Big (\Pr [X_{1}>x_{1}]-\Pr [Y_{1}>x_{1}]\Big )\mathrm{d}x_{1} \nonumber \\&\qquad +\int \limits _{a_{2}}^{b_{2}}u^{(0,1)}(a_{1},x_{2})\Big (\Pr [X_{2}>x_{2}]-\Pr [Y_{2}>x_{2}]\Big )\mathrm{d}x_{2} \nonumber \\&\qquad +\int \limits _{a_{1}}^{b_{1}}\int \limits _{a_{2}}^{b_{2}}\Big (\Pr [X_{1}>x_{1},X_{2}>x_{2}]-\Pr [Y_{1}>x_{1},Y_{2} >x_{2}]\Big ) u^{(1,1)}(x_{1},x_{2})\mathrm{d}x_{1}\mathrm{d}x_{2}. \end{aligned}$$

(27)

The first two terms appearing in the expansion of $E[u(X_{1},X_{2})]-E[u(Y_{1},Y_{2})]$ are negative based on our assumptions that $X_{i}\preceq _{\text {fsd}}Y_{i}$ holds for $i=1,2$. Let us now show that the last one is also negative. To this end, consider $u$ in $\mathcal {U} _{\text {cl}}^{\phi }$ such that $u^{(1,1)}\ne 0$ and denote

$$\begin{aligned} \underline{\theta }&= \inf \{u^{(1,1)}\}>0 \\ \bar{\theta }&= \sup \{u^{(1,1)}\}>0. \end{aligned}$$

Note that (4) ensures that

$$\begin{aligned} \bar{\theta }\le \underline{\theta }\left( \frac{1}{\phi }-1\right) \Leftrightarrow \frac{\underline{\theta }}{\bar{\theta }+\underline{\theta }} \ge \phi . \end{aligned}$$

(28)

Let $\hat{S}^c$ denote the complement of $\hat{S}$ in $[a_1,b_1]\times [a_2,b_2]$. Then,

$$\begin{aligned}&\int \limits _{a_{1}}^{b_{1}}\int \limits _{a_{2}}^{b_{2}}\Big (\Pr [X_{1}>x_{1},X_{2}>x_{2}]-\Pr [Y_{1}>x_{1},Y_{2}>x_{2}]\Big ) u^{(1,1)}(x_{1},x_{2})\mathrm{d}x_{1}\mathrm{d}x_{2} \\&\quad =\int \int \limits _{\hat{S}}\Big (\Pr [X_{1}>x_{1},X_{2}>x_{2}]-\Pr [Y_{1}>x_{1},Y_{2}>x_{2}]\Big )u^{(1,1)}(x_{1},x_{2})\mathrm{d}x_{1}\mathrm{d}x_{2} \\&\qquad +\int \int \limits _{\hat{S}^{c}}\Big (\Pr [X_{1}>x_{1},X_{2}>x_{2}]-\Pr [Y_{1}>x_{1},Y_{2}>x_{2}]\Big )u^{(1,1)}(x_{1},x_{2})\mathrm{d}x_{1}\mathrm{d}x_{2} \\&\quad \le \bar{\theta }\int \int \limits _{\hat{S}}\Big (\Pr [X_{1}>x_{1},X_{2}>x_{2}]-\Pr [Y_{1}>x_{1},Y_{2}>x_{2}]\Big )\mathrm{d}x_{1}\mathrm{d}_{2} \\&\qquad +\,\underline{\theta }\int \int \limits _{\hat{S}^{c}}\Big (\Pr [X_{1}>x_{1},X_{2}>x_{2}]-\Pr [Y_{1}>x_{1},Y_{2}>x_{2}]\Big )\mathrm{d}x_{1}\mathrm{d}x_{2} \\&\quad =-\underline{\theta }\int \limits _{a_{1}}^{b_{1}}\int \limits _{a_{2}}^{b_{2}}\Big |\Pr [X_{1}>x_{1},X_{2}>x_{2}]-\Pr [Y_{1}>x_{1},Y_{2}>x_{2}]\Big |\mathrm{d}x_{1}\mathrm{d}x_{2} \\&\qquad +\,(\bar{\theta }+\underline{\theta })\int \int \limits _{\hat{S}}\Big (\Pr [X_{1}>x_{1},X_{2}>x_{2}]-\Pr [Y_{1}>x_{1},Y_{2}>x_{2}]\Big )\mathrm{d}x_{1}\mathrm{d}x_{2} \\&\quad \le 0\hbox { by (25)}, \end{aligned}$$

which ends the proof of the “$\Rightarrow$ ” part.

With Eq. (27), the proof for the “$\Leftarrow$” part is similar to the one for “$\Leftarrow$” part in the proof for Theorem 1. Thus, it is omitted.

1.3 C Proof of Property 1

Let us first consider the implication with $\preceq _{\text {ca} }^{\varepsilon }$. Under (10) integration by parts gives

$$\begin{aligned}&E[X_{1}X_{2}]-E[Y_{1}Y_{2}]\\&\quad =\int \limits _{a_{1}}^{b_{1}}\int \limits _{a_{2}}^{b_{2}}\Big (\Pr [X_{1}\le x_{1},X_{2}\le x_{2}]-\Pr [Y_{1}\le x_{1},Y_{2}\le x_{2}]\Big ) \mathrm{d}x_{1}\mathrm{d}x_{2}. \end{aligned}$$

Now,

$$\begin{aligned} 0&\le \int \int \limits _{S}\Big (\Pr [Y_{1}\le x_{1},Y_{2}\le x_{2}]-\Pr [X_{1}\le x_{1},X_{2}\le x_{2}]\Big )\mathrm{d}x_{1}\mathrm{d}x_{2} \\&\le \varepsilon \left( \int \int \limits _{S^{c}}\Big (\Pr [X_{1}\le x_{1},X_{2}\le x_{2}]-\Pr [Y_{1}\le x_{1},Y_{2}\le x_{2}]\Big ) \mathrm{d}x_{1}\mathrm{d}x_{2}\right. \\&\quad \left. -\int \int \limits _{S}\Big (\Pr [X_{1}\le x_{1},X_{2}\le x_{2}]-\Pr [Y_{1}\le x_{1},Y_{2}\le x_{2}]\Big )\mathrm{d}x_{1}\mathrm{d}x_{2}\right) \end{aligned}$$

so that

$$\begin{aligned} 0&\le \int \int \limits _{S^{c}}\Big (\Pr [X_{1}\le x_{1},X_{2}\le x_{2}]-\Pr [Y_{1}\le x_{1},Y_{2}\le x_{2}]\Big )\mathrm{d}x_{1}\mathrm{d}x_{2} \\&\quad +\frac{1-\varepsilon }{\varepsilon }\int \int \limits _{S}\Big (\Pr [X_{1}\le x_{1},X_{2}\le x_{2}]-\Pr [Y_{1}\le x_{1},Y_{2}\le x_{2}]\Big ) \mathrm{d}x_{1}\mathrm{d}x_{2}. \end{aligned}$$

As $\varepsilon <0.5\Rightarrow \frac{1-\varepsilon }{\varepsilon }>1$ and $\int \int _{S}\Big (\Pr [X_{1}\le x_{1},X_{2}\le x_{2}]-\Pr [Y_{1}\le x_{1},Y_{2}\le x_{2}]\Big )\mathrm{d}x_{1}\mathrm{d}x_{2}\le 0$ we have

$$\begin{aligned} 0&\le \int \int \limits _{S^{c}}\Big (\Pr [X_{1}\le x_{1},X_{2}\le x_{2}]-\Pr [Y_{1}\le x_{1},Y_{2}\le x_{2}]\Big )\mathrm{d}x_{1}\mathrm{d}x_{2} \\&\quad +\int \int \limits _{S}\Big (\Pr [X_{1}\le x_{1},X_{2}\le x_{2}]-\Pr [Y_{1}\le x_{1},Y_{2}\le x_{2}]\Big )\mathrm{d}x_{1}\mathrm{d}x_{2} \\&= E[X_{1}X_{2}]-E[Y_{1}Y_{2}]. \end{aligned}$$

This ends the proof since

$$\begin{aligned} Cov[X_{1},X_{2}]-Cov[Y_{1},Y_{2}]=E[X_{1}X_{2}]-E[Y_{1}Y_{2}] \end{aligned}$$

under (10).

Let us now turn to the implication with $\preceq _{\text {cl}}^{\phi }$. Under (10), integration by parts gives

$$\begin{aligned}&E[X_{1}X_{2}]-E[Y_{1}Y_{2}]\\&\quad =\int \limits _{a_{1}}^{b_{1}}\int \limits _{a_{2}}^{b_{2}}\Big (\Pr [X_{1}>x_{1},X_{2}>x_{2}]-\Pr [Y_{1}>x_{1},Y_{2}>x_{2}]\Big )\mathrm{d}x_{1}\mathrm{d}x_{2}. \end{aligned}$$

Now,

$$\begin{aligned} 0&\le \int \int \limits _{\hat{S}}\Big (\Pr [X_{1}>x_{1},X_{2}>x_{2}]-\Pr [Y_{1}>x_{1},Y_{2}>x_{2}]\Big )\mathrm{d}x_{1}\mathrm{d}x_{2} \\&\le \phi \left( -\int \int \limits _{\hat{S}^{c}}\Big (\Pr [X_{1}>x_{1},X_{2}>x_{2}]-\Pr [Y_{1}>x_{1},Y_{2}>x_{2}]\Big ) \mathrm{d}x_{1}\mathrm{d}x_{2}\right. \\&\quad \left. +\int \int \limits _{\hat{S}}\Big (\Pr [X_{1}>x_{1},X_{2}>x_{2}]-\Pr [Y_{1}>x_{1},Y_{2}>x_{2}]\Big ) \mathrm{d}x_{1}\mathrm{d}_{2}\right) \end{aligned}$$

so that

$$\begin{aligned} 0&\ge \int \int \limits _{\hat{S}^{c}}\Big (\Pr [X_{1}>x_{1},X_{2}>x_{2}]-\Pr [Y_{1}>x_{1},Y_{2}>x_{2}]\Big )\mathrm{d}x_{1}\mathrm{d}x_{2} \\&\quad +\frac{1-\phi }{\phi }\int \int \limits _{\hat{S}}\Big (\Pr [X_{1}>x_{1},X_{2}>x_{2}]-\Pr [Y_{1}>x_{1},Y_{2}>x_{2}]\Big )\mathrm{d}x_{1}\mathrm{d}x_{2}. \end{aligned}$$

As $\phi <0.5\Rightarrow \frac{1-\phi }{\phi }>1$ and $\int \int _{\hat{S}} \Big (\Pr [X_{1}>x_{1},X_{2}>x_{2}]-\Pr [Y_{1}>x_{1},Y_{2}>x_{2}]\Big ) \mathrm{d}x_{1}\mathrm{d}x_{2}\ge 0$ we have

$$\begin{aligned} 0&\ge \int \int \limits _{\hat{S}^{c}}\Big (\Pr [X_{1}>x_{1},X_{2}>x_{2}]-\Pr [Y_{1}>x_{1},Y_{2}>x_{2}]\Big )\mathrm{d}x_{1}\mathrm{d}x_{2} \\&\quad +\int \int \limits _{\hat{S}}\Big (\Pr [X_{1}>x_{1},X_{2}>x_{2}]-\Pr [Y_{1}>x_{1},Y_{2}>x_{2}]\Big )\mathrm{d}x_{1}\mathrm{d}x_{2} \\&= E[X_{1}X_{2}]-E[Y_{1}Y_{2}]. \end{aligned}$$

This ends the proof.

1.4 D Proof of Property 2

Consider $v\in \mathcal {U}_{\text {ssd}}^{\varepsilon }$. The announced result is valid if we can prove that the bivariate utility function $u$ defined as

$$\begin{aligned} u(x_{1},x_{2})=v(w_{1}x_{1}+w_{2}x_{2}) \end{aligned}$$

belongs to $\mathcal {U}_{\text {ca}}^{\varepsilon }$. To this end, notice that

$$\begin{aligned} u^{(1,1)}(x_{1},x_{2})=w_{1}w_{2}v^{(2)}(w_{1}x_{1}+w_{2}x_{2}) \end{aligned}$$

so that

$$\begin{aligned} -u^{(1,1)}(x_{1},x_{2})&= -w_{1}w_{2}v^{(2)}(w_{1}x_{1}+w_{2}x_{2}) \\&\le w_{1}w_{2}\inf \{-v^{(2)}(w_{1}x_{1}+w_{2}x_{2})\}\left( \frac{1}{ \varepsilon }-1\right) \\&= \inf \{-u^{(1,1)}\}\left( \frac{1}{\varepsilon }-1\right) . \end{aligned}$$

This ends the proof.

1.5 E Proof of Theorem 3

Notice that

$$\begin{aligned} \int \limits _{x}^{b}\Pr [X>t]\mathrm{d}t=E[(X-x)_{+}] \end{aligned}$$

where $\xi _{+}$ denotes the positive part of the real $\xi$ (equal to 0 if $\xi$ is negative and to $\xi$ otherwise). According to Corollary 1.6.10 in Denuit et al. (2005) applied to $X-a$ and $Y-a$, we have

$$\begin{aligned}&E[v(X)]-E[v(Y)] \\&\quad = v^{\prime }(a)\big (E[X]-E[Y]\big )+\int \limits _{a}^{b}v^{(2)}(x)\int \limits _{x}^{b}\Big ( \Pr [X>t]-\Pr [Y>t]\Big )\mathrm{d}t\mathrm{d}x. \end{aligned}$$

The first term is negative based on our assumptions. Let us now show that the second term of the above equation is also negative. Consider $v$ in $\mathcal {U}_{\text {rl}}^{\phi }$ such that $v^{(2)}\ne 0$ and denote

$$\begin{aligned} \underline{\lambda }&= \inf \{v^{(2)}\}>0 \\ \bar{\lambda }&= \sup \{v^{(2)}\}>0. \end{aligned}$$

Note that (14) ensures that

$$\begin{aligned} \bar{\lambda }\le \underline{\lambda }\left( \frac{1}{\phi }-1\right) \Leftrightarrow \frac{\underline{\lambda }}{\bar{\lambda }+\underline{\lambda }}\ge \phi . \end{aligned}$$

(29)

Then,

$$\begin{aligned}&\int \limits _{a}^{b}v^{(2)}(x)\int \limits _{x}^{b}\Big (\Pr [X>t]-\Pr [Y>t]\Big )\mathrm{d}t\mathrm{d}x \\&\quad =\int \limits _{\Omega }v^{(2)}(x)\int \limits _{x}^{b}\Big (\Pr [X>t]-\Pr [Y>t]\Big ) \mathrm{d}t\mathrm{d}x\\&\qquad +\int \limits _{\Omega ^{c}}v^{(2)}(x)\int \limits _{x}^{b}\Big (\Pr [X>t]-\Pr [Y>t]\Big ) \mathrm{d}t\mathrm{d}x \\&\quad \le \bar{\lambda }\int \limits _{\Omega }\int \limits _{x}^{b}\Big (\Pr [X>t]-\Pr [Y>t]\Big ) \mathrm{d}t\mathrm{d}x+\underline{\lambda }\int \limits _{\Omega ^{c}}\int \limits _{x}^{b}\Big (\Pr [X>t] -\Pr [Y>t]\Big )\mathrm{d}t\mathrm{d}x \\&\quad =\left( \underline{\lambda }+\bar{\lambda }\right) \int \limits _{\Omega }\int \limits _{x}^{b}\Big (\Pr [X>t]-\Pr [Y>t]\Big )\mathrm{d}t\mathrm{d}x \\&\qquad -\bar{\lambda }\int \limits _{a}^{b}\Big |\int \limits _{x}^{b}\Big (\Pr [X>t]-\Pr [Y>t]\Big )\mathrm{d}t \Big |\mathrm{d}x \\&\quad \le 0\hbox { by (14)}, \end{aligned}$$

where $\Omega ^{c}$ denotes the complement of $\Omega$ in $[a,b]$. It ends the proof of the “$\Rightarrow$” part.

With Eq. (27), the “$\Leftarrow$ ” part is similar to the “$\Leftarrow$ ” part in the proof for Theorem 1. Thus, it is omitted.

1.6 F Proof of Property 3

Consider $v\in \mathcal {U}_{\text {cl}}^{\phi }$. The announced result is valid if we can prove that the bivariate utility function $u$ defined as

$$\begin{aligned} u(x_{1},x_{2})=v(w_{1}x_{1}+w_{2}x_{2}) \end{aligned}$$

belongs to $\mathcal {U}_{\text {cl}}^{\phi }$. To this end, notice that

$$\begin{aligned} u^{(1,1)}(x_{1},x_{2})=w_{1}w_{2}v^{(2)}(w_{1}x_{1}+w_{2}x_{2}) \end{aligned}$$

so that

$$\begin{aligned} u^{(1,1)}(x_{1},x_{2})&= w_{1}w_{2}v^{(2)}(w_{1}x_{1}+w_{2}x_{2}) \\&\le w_{1}w_{2}\inf \{v^{(2)}(w_{1}x_{1}+w_{2}x_{2})\}\left( \frac{1}{\phi }-1\right) \\&= \inf \{u^{(1,1)}\}\left( \frac{1}{\phi }-1\right) . \end{aligned}$$

This ends the proof.