2.1. Definitions

The following definition of stochastic dominance of order $(1+\gamma)$ was given in [Reference Müller, Scarsini and Tsetlin25]. We first introduce the following notation. Let $\mathcal U$ be the set of all increasing functions on $\mathbb{R}$ . For $\gamma\in [0,1]$ , define

\begin{equation*} {\mathcal U}_\gamma =\{u \in {\mathcal U}\colon \text{u is differentiable}, \;0\le \gamma u'(y)\le u'(x)\ \text{for all}\ x\le y, \;x,y\in\mathbb{R} \}.\end{equation*}

Definition 2.1. ([Reference Müller, Scarsini and Tsetlin25].)Let X and Y be two random variables in $\mathbb{R}$ . We say that X is dominated by Y in stochastic dominance of order $(1+\gamma)$ , denoted by $X\preccurlyeq_{\mathrm{(1+\gamma)\hbox{-}SD}} Y$ , if

\begin{equation*} \mathbb{E} [u(X)] \le \mathbb{E} [u(Y)]\quad \text{for all}\ u\in {\mathcal U}_\gamma.\end{equation*}

The order $\preccurlyeq_{\mathrm{(1+\gamma)\hbox{-}SD}}$ cannot be defined for any $\gamma>1$ because $\mathcal U_\gamma$ is empty except for constant functions in this case. For $0\le \gamma_1<\gamma_2\le 1$ , $X\preccurlyeq_{\mathrm{(1+\gamma_1)\hbox{-}SD}} Y$ implies $X\preccurlyeq_{\mathrm{(1+\gamma_2)\hbox{-}SD}} Y$ since $\mathcal U_{\gamma_2} \subseteq \mathcal U_{\gamma_1}$ . This means that lower-degree stochastic dominance always implies higher-degree stochastic dominance. In Definition 2.1, the class $\mathcal U_\gamma$ of functions can be replaced by $\mathcal U_\gamma^\ast$ defined by

\begin{equation*} {\mathcal U}_\gamma^\ast = \biggl\{u\colon 0\le \gamma \frac{u(x_4)-u(x_3)}{x_4-x_3}\le \frac{u(x_2)-u(x_1)}{x_2-x_1}\ \text{for all}\ x_1<x_2\le x_3<x_4\biggr\}.\end{equation*}

It is obvious that $\preccurlyeq_{\mathrm{1\hbox{-}SD}}$ is equivalent to FSD while $\preccurlyeq_{\mathrm{2\hbox{-}SD}}$ is equivalent to SSD. The orders $\preccurlyeq_{\mathrm{1\hbox{-}SD}}$ and $\preccurlyeq_{\mathrm{2\hbox{-}SD}}$ are also denoted by $\preccurlyeq_{\mathrm{FSD}}$ and $\preccurlyeq_{\mathrm{SSD}}$ , respectively. Thus $(1+\gamma)$ -SD establishes an interpolation between FSD and SSD. For more properties of FSD, SSD, and other related stochastic orders, refer to [Reference Müller and Stoyan24] and [Reference Shaked and Shanthikumar26].

To investigate the properties of $(1+\gamma)$ -SD, we introduce the following $(1+\gamma)$ -pure stochastic dominance, denoted by $(1+\gamma)$ -PSD.

Definition 2.2. Let $X,Y\in L^1$ , and define

(2.1) \begin{equation} \gamma = \frac {\int_{-\infty}^\infty (G(x)-F(x))_+{\mathrm{d}} x }{\int_{-\infty}^\infty (G(x)-F(x))_-{\mathrm{d}} x}\end{equation}

with the convention that $0/0=0$ . X is said to be smaller than Y in the pure stochastic dominance of order $(1+\gamma)$ if $\gamma\in [0,1]$ and $X\preccurlyeq_{\mathrm{(1+\gamma)\hbox{-}SD}} Y$ . We denote this by $X\preccurlyeq_{\mathrm{(1+\gamma)-PSD}} Y$ .

In fact (2.1) can be replaced by

(2.2) \begin{equation} \int_0^1 (G^{-1}(\alpha)-F^{-1}(\alpha))_-{\mathrm{d}} \alpha = \gamma \int_0^1 (G^{-1}(\alpha)-F^{-1}(\alpha))_+ {\mathrm{d}} \alpha.\end{equation}

The motivation of the constraint condition (2.1) or (2.1) comes from Proposition 2.1, which gives a characterization of the order $\preccurlyeq_{\mathrm{(1+\gamma)\hbox{-}SD}}$ . For $\gamma=1$ , (2.1) is equivalent to $\mathbb{E}[X]=\mathbb{E}[Y]$ . Thus 2-PSD is exactly the concave order. Equation (2.1) appears to be similar to that used in [Reference Leshno and Levy14] to define $\epsilon$ -almost FSD as follows: Y dominates X by $\epsilon$ -almost FSD, denoted by $X\preccurlyeq_1^{{\mathrm{almost}}(\epsilon)} Y$ , if and only if

\begin{equation*} \frac {\int_{-\infty}^\infty (G(x)-F(x))_+{\mathrm{d}} x }{\int_{-\infty}^\infty (G(x)-F(x))_-{\mathrm{d}} x} \le \frac {\epsilon}{1-\epsilon} ,\end{equation*}

where $0<\epsilon<1/2$ . Therefore $X\preccurlyeq_{\mathrm{(1+\gamma)-PSD}} Y$ implies $X\preccurlyeq_1^{{\mathrm{almost}}(\epsilon)} Y$ with $\epsilon={\gamma}/{(1+\gamma)}$ . However, the converse is not true.

Hence $(1+\gamma)$ -PSD enables one to understand $(1+\gamma)$ -SD well. First, $(1+\gamma)$ -PSD can be used to characterize $(1+\gamma)$ -SD (see Theorem 3.2). Second, if Y dominates X in $(1+\gamma)$ -SD, and if there does not exist $Z\in L^1$ , not identically distributed with Y, such that Y dominates Z in the FSD and Z dominates X in $(1+\gamma)$ -SD, then Y dominates X in $(1+\gamma)$ -PSD (seeRemark 3.4). These two points are the motivation for us to introduce $(1+\gamma)$ -PSD.

It should be pointed out that the order $\preccurlyeq_{\mathrm{(1+\gamma)\hbox{-}PSD}}$ for $\gamma\in (0,1)$ is not a partial order because it does not possess transitivity, as illustrated by the following example.

Example 2.1. Let X, Y, and Z be three random variables with probability mass functions (PMFs) $\mathbb{P}(X=-1)=0.4$ , $\mathbb{P}(X=2)=0.1$ , $\mathbb{P}(X=3)=0.5$ , $\mathbb{P}(Y=0)=\mathbb{P}(Y=3)=1/2$ , and $\mathbb{P}(Z=2)=1$ . By Theorem 2.7 in [Reference Müller, Scarsini and Tsetlin25], it can be seen from the probability mass movement illustrated in Figure 1 that

\begin{equation*} X \preccurlyeq_{\mathrm{(1+1/2)\hbox{-}PSD}} Y \preccurlyeq_{\mathrm{(1+1/2)\hbox{-}PSD}} Z.\end{equation*}

However,

\begin{equation*} X \preccurlyeq_{\mathrm{(1+5/12)\hbox{-}PSD}} Z\quad \text{and}\quad X \not \preccurlyeq_{\mathrm{(1+1/2)\hbox{-}PSD}} Z.\end{equation*}

This means that the order $\preccurlyeq_{\mathrm{(1\,+\,1/2)\hbox{-}PSD}}$ does not possess transitivity.

Figure 1. Probability mass movement from Z to Y and then to X.

Let $X\sim F$ and $Y\sim G$ . For convenience, we will write $X\preccurlyeq_{\mathrm{order}} Y$ and $F\preccurlyeq_{\mathrm{order}} G$ interchangeably for any order relation $\preccurlyeq_{\mathrm{order}}$ .

2.2. Basic properties

In this subsection we list three basic properties of $(1+\gamma)$ -SD from [Reference Müller, Scarsini and Tsetlin25]. The first characterizes $(1+\gamma)$ -SD by using integral conditions, and the second and third are concerned with preservation properties of $(1+\gamma)$ -SD under transformations and under mixture, respectively.

Proposition 2.1. ([Reference Müller, Scarsini and Tsetlin25].)Let F and G be the distribution functions of X and Y, respectively. For $\gamma\in [0,1]$ , $X\preccurlyeq_{\mathrm{(1+\gamma)\hbox{-}{\rm SD}}} Y$ if and only if

(2.3) \begin{equation} \int_{-\infty}^t (G(x)-F(x))_+{\mathrm{d}} x \le \gamma \int_{-\infty}^t (G(x)-F(x))_-{\mathrm{d}} x\quad {for all}\ t\in\mathbb{R},\end{equation}

or, equivalently,

(2.4) \begin{equation} \int_0^p (G^{-1}(\alpha)-F^{-1}(\alpha))_-{\mathrm{d}} \alpha \le \gamma \int_0^p (G^{-1}(\alpha)-F^{-1}(\alpha))_+ {\mathrm{d}} \alpha\quad {for all}\ p\in (0,1].\end{equation}

In view ofProposition 2.1, it is seen that $X\preccurlyeq_{\mathrm{(1+\gamma)\hbox{-}PSD}} Y$ if and only if (2.1) and

\begin{equation*} \int^{\infty}_t (G(x)-F(x))_+{\mathrm{d}} x \ge \gamma \int^{\infty}_t (G(x)-F(x))_-{\mathrm{d}} x \quad \text{for all}\ t\in\mathbb{R}.\end{equation*}

Proposition 2.3. ([Reference Müller, Scarsini and Tsetlin25].)If random variables X, Y, and $\Theta$ satisfy

\begin{equation*} [X \mid \Theta=\theta]\preccurlyeq_{\mathrm{(1+\gamma)\hbox{-}{\rm SD}}} [Y\mid \Theta=\theta]\end{equation*}

for some $\gamma\in [0,1]$ and all $\theta$ in the support of $\Theta$ , then $X \preccurlyeq_{\mathrm{(1+\gamma)\hbox{-}{\rm SD}}} Y$ .

Immediate consequences of Proposition 2.3 are as follows. (i) Let $F_i$ and $G_i$ be the distribution functions of $X_i$ and $Y_i$ , respectively. For $\alpha\in (0,1)$ , assume that $Z_1\sim \alpha F_1 + (1-\alpha)F_2$ and $Z_2\sim \alpha G_1 + (1-\alpha)G_2$ . If $X_i\preccurlyeq_{\mathrm{(1+\gamma)\hbox{-}SD}} Y_i$ for $i=1, 2$ , then $Z_1 \preccurlyeq_{\mathrm{(1+\gamma)\hbox{-}SD}} Z_2$ . (ii) Let $X_1$ and $X_2$ be independent, and let $Y_1$ and $Y_2$ be independent. If $X_i\preccurlyeq_{\mathrm{(1+\gamma)\hbox{-}SD}} Y_i$ for $i=1, 2$ , then

(2.5) \begin{equation} X_1+X_2\preccurlyeq_{\mathrm{(1+\gamma)\hbox{-}SD}} Y_1+ Y_2.\end{equation}

3.1. Generating processes

For a better understanding of $(1+\gamma)$ -SD and $(1+\gamma)$ -PSD, we recall the definition of $\gamma$ -transfer, which is due to [Reference Müller, Scarsini and Tsetlin25].

Definition 3.1. ( $\gamma$ -transfer.) Let X and Y be two discrete random variables with PMFs f and g, respectively. We say that Y is obtained from X via a $\gamma$ -transfer if there exist $x_1<x_2<x_3<x_4$ and $\eta_1,\eta_2>0$ with $\eta_2(x_4-x_3)=\gamma \eta_1(x_2-x_1)$ such that

\begin{align*} g(x_1) & = f(x_1)-\eta_1,\\ g(x_2) & = f(x_2) +\eta_1,\\ g(x_3) & = f(x_3) +\eta_2,\\ g(x_4) & = f(x_4) -\eta_2,\\ g(z)&=f(z)\quad \text{for all other values z.}\end{align*}

In the definition of $\gamma$ -transfer, $\gamma$ is not necessarily restricted to be in [0,1], which can take any value in $\mathbb{R}_+=[0,\infty)$ . Further, $\gamma$ -spread is closely related to $\gamma$ -transfer: X is said to be obtained from Y by a $\gamma$ -spread if Y is obtained from X by a $\gamma$ -transfer. In a $\gamma$ -transfer, a mass of size $\eta_2$ is moved to the left from $x_4$ by $\Delta_2=x_4-x_3$ , while a mass of size $\eta_1$ is moved to the right from $x_1$ by $\Delta_1=x_2-x_1$ such that $\Delta_2\eta_2 =\gamma \Delta_1\eta_1$ . A $\gamma$ transfer increases the mean (i.e. $\mathbb{E} X\le \mathbb{E} Y$ ) for $\gamma\in [0,1]$ . In Example 2.1, Y is obtained from X by a $1/2$ -transfer, Z is obtained from Y by a $1/2$ -transfer, and Z is obtained from X by a $5/12$ -transfer.

In Definition 3.1, $\gamma$ -transfer can also be defined when $x_1< x_2=x_3< x_4$ . In this case the conditions $g(x_2)=f(x_2)+\eta_1$ and $g(x_3)=f(x_3)+\eta_2$ were replaced by $g(x_2)=f(x_2)+\eta_1 +\eta_2$ .

The following proposition states that $\gamma$ -transfers account for almost all mass transfers of $(1+\gamma)$ -SD. Specifically, part (ii) of Proposition 3.1 can be seen from the proof ofTheorem 2.8 of [Reference Müller, Scarsini and Tsetlin25]. For two random variables X and Y, we use $X \buildrel \mathrm{d} \over = Y$ to denote that X and Y have the same distribution, and let $\|X\|_\infty =\mathrm{ess\mbox{-}sup} (|X|)$ .

For $(1+\gamma)$ -PSD, we have the following result analogous to Proposition 3.1.

Proof. Without loss of generality, assume $\gamma\in (0,1]$ , and let F and G be the distribution functions of X and Y, respectively.

(i) The result can be obtained by modifying the proof of Theorem 2.7 in [Reference Müller, Scarsini and Tsetlin25]. We use the same notation as in [Reference Müller, Scarsini and Tsetlin25]. Define

\begin{equation*} A^+(p)=\int^p_0 (G^{-1}(\alpha)-F^{-1}(\alpha))_+ {\mathrm{d}} \alpha\quad \text{and}\quad A^-(p)=\int^p_0 (G^{-1}(\alpha)-F^{-1}(\alpha))_- {\mathrm{d}} \alpha.\end{equation*}

Let w(a) and v(a) be the smallest numbers satisfying

\begin{equation*} A^+(w(a))=\frac {a}{\gamma}\quad \text{and}\quad A^-(v(a))=a\quad \text{for}\ a\in [0, A^-(1)].\end{equation*}

In view of Proposition 2.1, we have $w(a)\le v(a)$ for all $a\in [0, A^-(1)]$ . For each $a\in [0, A^-(1)]$ , define

\begin{equation*} x_1(a)=F^{-1}(w(a)),\quad x_2(a)=G^{-1}(w(a)),\quad x_3(a)=G^{-1}(v(a)),\quad x_4(a)=F^{-1}(v(a)).\end{equation*}

Since X and Y both have finite outcomes, there exist $0=a_1<a_2<\cdots<a_k=A^-(1)$ such that the functions $x_1(a), \ldots, x_4(a)$ are constant on $(a_{i-1}, a_i]$ . Denote the corresponding values of these functions as $x_{\ell,i}=x_\ell(a)$ for $a\in (a_{i-1}, a_i]$ , $\ell=1, \ldots, 4$ . It was shown in [Reference Müller, Scarsini and Tsetlin25] that $x_{1,i}< x_{2,i}\le x_{3,i}<x_{4,i}$ , and that for each $i\in \{1, \ldots, k\}$ , the probability masses of F at points $x_{1,i}$ and $x_{4,i}$ , respectively, are moved to the points $x_{2,i}$ and $x_{3,i}$ of G by a $\gamma$ -transfer.

Note that when $X\preccurlyeq_{\mathrm{(1+\gamma)\hbox{-}PSD}}Y$ , it follows from (2.2) that $A^{-}(1)=\gamma A^{+}(1)$ . Then $G^{-1}(p)\ge F^{-1}(p)$ for $p>v(a_k)$ , and $G^{-1}(p)\le F^{-1}(p)$ for $p>w(a_k)$ . This means that $F(x)\ge G(x)$ for $x>x_{4,k}$ and $F(x)\le G(x)$ for $x>x_{1,k}$ . Thus $F(x)=G(x)$ for all $x>x_{4,k}$ and the jumps of F and G occur in the points belonging to the set $\{x_{\ell,i}\colon i=1,\ldots, k; \ell=1, \ldots, 4\}$ . Therefore G can be obtained from F only by a sequence of k $\gamma$ -transfers.

(ii) First assume that F and G have finite crossings, that is, there exist $-\infty<x_0< x_1<\cdots<x_m<\infty$ such that either $F \le G $ or $F \ge G$ holds in $(x_{i-1},x_i)$ , $i=1,\ldots,m$ , where the supports of F and G are contained in $[x_0, x_m]$ . For $i=1,\ldots,m$ and $n\in\mathbb{N}$ , denote

\begin{equation*} x_{i,j} = \frac {1}{n} [(n-j) x_{i-1} + jx_{i} ], \quad j=0,\ldots,n. \end{equation*}

Define two random variables $X_n$ and $Y_n$ with distribution functions $F_n$ and $G_n$ , respectively, where

\begin{equation*} F_n (x) = \frac1{x_{i,j}-x_{i,j-1}}\int_{x_{i,j-1}}^{x_{i,j}} F(y)\,{\mathrm{d}} y,\quad G_n (x) = \frac1{x_{i,j}-x_{i,j-1}}\int_{x_{i,j-1}}^{x_{i,j}} G(y)\,{\mathrm{d}} y\end{equation*}

for $x\in [x_{i,j-1}, x_{i,j})$ . It is easy to see that $X_n$ and $Y_n$ both have finite outcomes. It can be verified that in each interval $[x_{i,j-1}, x_{i,j})$ , either $F_n \le G_n$ or $F_n\ge G_n$ , and the direction of the inequality is the same as $F \le G $ or $F \ge G$ on the same interval. Hence we have

\begin{equation*} \int_{x_{i,j-1}}^{x_{i,j}} (G_n(y)-F_n(y))_+{\mathrm{d}} y = \int_{x_{i,j-1}}^{x_{i,j}} (G(y)-F(y))_+{\mathrm{d}} y\end{equation*}

and

\begin{equation*} \int_{x_{i,j-1}}^{x_{i,j}} (G_n(y)-F_n(y))_-{\mathrm{d}} w = \int_{x_{i,j-1}}^{x_{i,j}} (G(y)-F(y))_-{\mathrm{d}} y.\end{equation*}

Then it follows from $F\preccurlyeq_{\mathrm{(1+\gamma)\hbox{-}PSD}} G$ that $F_n\preccurlyeq_{\mathrm{(1+\gamma)\hbox{-}PSD}} G_n$ . On the other hand, it is easy to see that $\max\{|X|, |Y|\} \le M\;:\!=\; x_m-x_0<\infty$ ,

\begin{equation*} |F^{-1}(\alpha)-F_n^{-1}(\alpha)| \le \frac1n \max\{x_{i,j}-x_{i,j-1}\} \le \frac {2M}{n},\quad \alpha\in (0,1),\end{equation*}

and

\begin{equation*} |G^{-1}(\alpha)-G_n^{-1}(\alpha)| \le \frac1n \max\{x_{i,j}-x_{i,j-1}\} \le \frac {2M}{n},\quad \alpha\in (0,1).\end{equation*}

Hence we can easily construct $X_n\sim F_n$ and $Y_n\sim G_n$ such that $|X_n-X|\le 2M/n$ and $|Y_n-Y|\le 2M/n$ .

Next, consider that X and Y have infinite crossings, that is, we have infinite intervals $\{(x_{i-1}, x_{i}),i\in I\}$ such that $G-F$ has the same sign in any one interval. Note that X and Y are both bounded. Then, for $n\in\mathbb{N}$ , the number of intervals with length larger than $1/n$ is finite. Then we can merge some of the remaining neighboring intervals to make the lengths smaller than $2/n$ and the number of intervals finite. Without loss of generality, assume that the transformed intervals are still denoted by $\{(x_{i-1}, x_{i}),i=1,\ldots,m\}$ . In each interval, either $G-F$ has the same sign or the length of the interval is less than $2/n$ . For the intervals where $G-F$ has the same sign, we use the same method as in the above case to define the values of $F_n$ and $G_n$ in the intervals. For the other intervals, take $(x_{i-1},x_i)$ as an example, where $G-F$ has different signs on $(x_{i-1},x_i)$ and $x_i-x_{i-1}< 2/n$ . Let $x^\ast\in (x_{i-1},x_i)$ such that $x^\ast-x_{i-1}$ is equal to the length of $A_i=\{x\in (x_{i-1},x_i)\colon F(x)\ge G(x) \}$ . Denote $A_i^{c} = (x_{i-1},x_i) \setminus A_i$ and define

\begin{equation*} F_n(x) = \frac1{x^\ast-x_{i-1}}\int_{A_i}F(y)\,{\mathrm{d}} y,\quad G_n (x) = \frac1{x^\ast-x_{i-1}}\int_{A_i} G(y)\,{\mathrm{d}} y\end{equation*}

for $x\in (x_{i-1},x^\ast)$ , and

\begin{equation*} F_n(x) = \frac1{x_i-x^\ast}\int_{A_i^c}F(y)\,{\mathrm{d}} y,\quad G_n (x) = \frac1{x_i-x^\ast}\int_{A_i^c} G(y)\,{\mathrm{d}} y\end{equation*}

for $x\in (x^\ast,x_i)$ . Then we have

\begin{equation*} \int_{x_{i-1}}^{x_i} (G_n(y)-F_n(y))_+{\mathrm{d}} w=\int_{x_{i-1}}^{x_i} (G(y)-F(y))_+{\mathrm{d}} y\end{equation*}

and

\begin{equation*} \int_{x_{i-1}}^{x_i} (G_n(y)-F_n(y))_-{\mathrm{d}} y =\int_{x_{i-1}}^{x_i} (G(y)-F(y))_-{\mathrm{d}} y.\end{equation*}

Then it can be checked that $F\preccurlyeq_{\mathrm{(1+\gamma)\hbox{-}PSD}} G$ implies $F_n\preccurlyeq_{\mathrm{(1+\gamma)\hbox{-}PSD}} G_n$ . The remaining proof is similar to the above case.

(iii) We modify the proof of Theorem 2.8 in [Reference Müller, Scarsini and Tsetlin25] for our purpose. For unbounded random variables X and Y, define

\begin{equation*} \psi(t)=\int^t_{-\infty} (G(x)-F(x))_+ {\mathrm{d}} x\quad \text{and}\quad \xi(t)=\gamma \int_{-\infty}^t (G(x)-F(x))_- {\mathrm{d}} x.\end{equation*}

We approximate X and Y by $X_n$ and $Y_n$ , respectively, as follows. Define

\begin{equation*} X_n=\begin{cases} x_n^\ast & \text{if $ X\le -n$,}\\ X & \text{if $-n<X\le n$,}\\ y_n^\ast & \text{if $ X>n$,} \end{cases}\end{equation*}

and

\begin{equation*} Y_n=\begin{cases} -n & \text{if $ Y\le -n$,}\\ Y & \text{if $ -n<Y\le n$,}\\ n & \text{if $ Y >n$,} \end{cases}\end{equation*}

where

\begin{equation*} x_n^\ast=-n - \frac {\xi(-n)}{\gamma F(-n)}\quad \text{and}\quad y_n^\ast= n+ \frac {\xi(n)-\psi(n) + \psi(-n)}{ {\overline F} (n)}.\end{equation*}

Since $\xi(t)\ge \psi(t)\ge 0$ for all t, it follows that $x_n^\ast\le -n$ and $y_n^\ast\ge n$ . Let $F_n$ and $G_n$ denote the distribution functions of $X_n$ and $Y_n$ , respectively, and define

\begin{equation*} \psi_n(t)=\int^t_{-\infty} (G_n(x)-F_n(x))_+ {\mathrm{d}} x\quad \text{and}\quad \xi_n(t)=\gamma \int_{-\infty}^t (G_n(x)-F_n(x))_- {\mathrm{d}} x.\end{equation*}

Then it can be checked that

\begin{equation*} \psi_n(t)=\begin{cases} 0 & \text{if $ t<-n$,}\\ \psi(t)-\psi(-n) & \text{if $ -n<t\le n$,}\\ \psi(n)-\psi(-n) + (t-n) {\overline F}(n) & \text{if $ n<t\le y_n^\ast$,}\\ \xi(n) & \text{if $ t>y_n^\ast$,} \end{cases}\end{equation*}

and

\begin{equation*} \xi_n(t)=\begin{cases} \xi(t) & \text{if $ -n\le t\le n$,}\\ \xi(n) & \text{if $ t>n$.} \end{cases}\end{equation*}

Thus $X_n$ and $Y_n$ are bounded, $\psi_n(t)\le \xi_n(t)$ for all $t\in\mathbb{R}$ , and $\psi_n(+\infty)=\xi(n)=\xi_n$ $(+\infty)$ . This means $X_n\preccurlyeq_{\mathrm{(1+\gamma)\hbox{-}PSD}} Y_n$ for all $n\in\mathbb{N}$ . On the other hand, $\psi(+\infty)=\xi$ $(+\infty)$ implies that $\mathbb{E} [X_n]\to \mathbb{E} [X]$ and $X_n\to X$ in distribution. Obviously, $\mathbb{E} [Y_n]\to \mathbb{E} [Y]$ and $Y_n\to Y$ in distribution. The desired result now follows from part (ii). This completes the proof.

3.2. Dual characterization

Let $\mathcal H$ denote the set of all probability perception functions h (also referred to as distortion functions in the actuarial literature), that is, $h\colon [0,1]\to [0,1]$ is increasing, satisfying $h(0)=0$ and $h(1)=1$ . For $\gamma \in [0,1]$ , define

\begin{equation*} \mathcal H_\gamma=\{h\in\mathcal H\colon h\ \text{is differentiable},\ 0\le \gamma h'(y)\le h'(x)\ \text{{for all}}\ 0\le x\le y\le 1\}\end{equation*}

and

\begin{equation*} \mathcal H_\gamma^\ast=\biggl\{h\in\mathcal H\colon 0\le \gamma \frac{h(p_4)-h(p_3)}{p_4-p_3}\le \frac{h(p_2)-h(p_1)}{p_2-p_1}\ \text{{for all}}\ 0\le p_1<p_2\le p_3<p_4\le1\biggr\},\end{equation*}

where h ^′(0) and h ^′(1) represent the right derivative at 0 and the left derivative at 1, respectively. Obviously, $\mathcal H_\gamma$ is the subset of $\mathcal H_\gamma^\ast$ containing all continuously differentiable $h\in\mathcal H_\gamma^\ast$ .

In the following theorem, we establish in the framework of Yaari’s dual theory that $(1+\gamma)$ -SD is equivalent to a common preference among all decision-makers with probability perception function $h\in \mathcal H_\gamma$ . This is a dual characterization of $(1+\gamma)$ -SD as the latter is originally defined via a common preference based on utility functions. For Yaari’s dual theory,see [Reference Yaari30].

Proof. Part (iii) is equivalent to (ii). It suffices to prove that (iii) $\Rightarrow$ (2.4) $\Rightarrow$ (ii).

To prove (iii) $\Rightarrow$ (2.4), for $p\in (0,1]$ , define a distortion function $h\in\mathcal H$ such that

\begin{equation*} h'(\alpha) = \begin{cases} \gamma & \text{if $ F^{-1}(\alpha)\le G^{-1}(\alpha)$ and $ \alpha\le p$,}\\ 1 & \text{if $ F^{-1}(\alpha)>G^{-1}(\alpha)$ and $\alpha\le p$,}\\ 0 & \text{if $ \alpha> p$.} \end{cases}\end{equation*}

It is easy to verify that $h\in\mathcal H_\gamma^\ast$ and hence (2.4) holds.

To prove the other direction (2.4) $\Rightarrow$ (ii), we use arguments similar to those in the proof of Theorem 2.4 of [Reference Müller, Scarsini and Tsetlin25]. For completeness, we give the details. Let $h\in \mathcal H_\gamma$ , i.e. h satisfies $0\le \gamma h'(y)\le h'(x)$ for all $0\le x\le y\le1$ . Then $R\;:\!=\; \sup_{v\in (0,1)}h'(v)\in (0,\infty)$ since $0\le h'(v)\le h'(1)/\gamma<\infty.$ For any fixed $n\ge 2$ , define $\epsilon_n=2^{-n}$ and K as the largest integer kfor which

\begin{equation*} R (1 - k \epsilon_n ) \ge \inf_{v\in (0,1)}h'(v), \end{equation*}

and define a partition of [0,1] into intervals $[v_k, v_{k+1}]$ as follows: $v_0=0$ , $v_{K+1}=1$ , and

\begin{equation*} v_k = \sup\{v\colon h'(v)\ge R (1 - k \epsilon_n) \},\quad k=1,\ldots,K.\end{equation*}

Then we define

\begin{equation*} m_k = \sup\{h'(v)\colon v_{k-1}\le v \le v_k\}= R (1 - (k-1) \epsilon_n). \end{equation*}

It follows that $\gamma m_{k+1} \le h'(v)\le m_k$ for $v\in [v_{k-1}, v_k]$ , i.e. $\gamma (m_k - R \epsilon_n) \le h'(v)\le m_k$ for all $x\in [v_{k-1}, v_k]$ and $k=1,\ldots,K+1$ . This implies that

\begin{align*}&\int_{v_{k-1}}^{v_k} (G^{-1}(\alpha)- F^{-1}(\alpha))\,{\mathrm{d}} h(\alpha)\\ & \quad = \int_{v_{k-1}}^{v_k} (G^{-1}(\alpha)- F^{-1}(\alpha))_+h'(\alpha) \,{\mathrm{d}} \alpha - \int_{v_{k-1}}^{v_k} (G^{-1}(\alpha)- F^{-1}(\alpha))_- h'(\alpha)\,{\mathrm{d}} \alpha\\ & \quad \ge \gamma (m_k - R \epsilon_n ) \int_{v_{k-1}}^{v_k}(G^{-1}(\alpha)- F^{-1}(\alpha))_+ {\mathrm{d}} \alpha - m_k \int_{v_{k-1}}^{v_k}(G^{-1}(\alpha)- F^{-1}(\alpha))_-{\mathrm{d}} \alpha\\ & \quad = m_k T_k -\epsilon_n c_k , \end{align*}

with

\begin{equation*} T_k = \gamma \int_{v_{k-1}}^{v_k}(G^{-1}(\alpha)- F^{-1}(\alpha))_+ {\mathrm{d}} \alpha -\int_{v_{k-1}}^{v_k}(G^{-1}(\alpha)- F^{-1}(\alpha))_- {\mathrm{d}} \alpha \end{equation*}

and

\begin{equation*} c_k = \gamma R\int_{v_{k-1}}^{v_k}(G^{-1}(\alpha)- F^{-1}(\alpha))_+ {\mathrm{d}} \alpha.\end{equation*}

Note that (2.4) implies $\sum_{i=1}^k T_i\ge 0$ for all $k=1,\ldots,K+1$ , which in turn implies $\sum_{k=1}^{K+1}m_k T_k \ge0$ for all decreasing non-negative sequences $m_k$ . Thus

\begin{align*} \int_0^1 (G^{-1}(\alpha)- F^{-1}(\alpha))\,{\mathrm{d}} h(\alpha)\ & \ge\ \sum_{k=1}^{K+1}(m_k T_k-\epsilon_n c_k)\\ & \ge\ -\epsilon_n\gamma R \int_0^1 (G^{-1}(\alpha)- F^{-1}(\alpha))_+ {\mathrm{d}} \alpha.\end{align*}

Letting $n\to\infty$ yields part (ii). This completes the proof of the theorem.

Remark 3.1. To get a better understanding of the dual characterization of $(1+\gamma)$ -SD, we introduce the following index $Q_f$ of a probability perception function h:

\begin{equation*} Q_h =\sup_{0\le p_1<p_2\le p_3<p_4\le 1} \frac {(h(p_4)-h(p_3))/(p_4-p_3)}{(h(p_1)-h(p_1))/(p_2-p_1)}.\end{equation*}

Here we use the convention that $a/0=+\infty$ for any real number $a>0$ and $0/0=0$ . As mentioned in [Reference Chateauneuf, Cohen and Meilijson5], $Q_h$ is an index of non-concavity of $h\in\mathcal H$ , $Q_h\ge 1$ , and $Q_h=1$ corresponds exactly to concavity. Thus $1/Q_h$ can be regarded as an index of the greediness of a decision-maker with probability perception function h (in short, a decision-maker h). That is, for $h_1, h_2\in \mathcal H$ , $Q_{h_1}<Q_{h_2}$ means that $h_1$ is more greedy than $h_2$ . Therefore, for $\gamma\in [0,1]$ ,

\begin{equation*} \mathcal H_\gamma^\ast=\biggl\{h\in\mathcal H\colon Q_h \le \frac {1}{\gamma} \biggr\}\end{equation*}

denotes the set of decision-makers with index of greediness larger than or equal to $\gamma$ . On the other hand, ${\mathcal U}_\gamma^\ast$ or ${\mathcal U}_\gamma^\ast$ has a similar interpretation in terms of risk aversion.

The order $X\preccurlyeq_{\mathrm{(1+\gamma)\hbox{-}SD}} Y$ is defined in Definition 2.1 by comparing expected utilities $\mathbb{E} [u(X)]$ and $\mathbb{E} [u(Y)]$ for all utility functions $u\in {\mathcal U}_\gamma$ , while the dual characterization in Theorem 3.1 compares the expected values $\mathbb{E} [X_h]$ and $\mathbb{E} [Y_h]$ of random variables $X_h$ and $Y_h$ for all probability perception functions $h\in \mathcal H_\gamma$ , where $X_h$ and $Y_h$ have the distorted distribution functions h(F(x)) and h(G(x).

3.3. Separation theorem

We establish a separation theorem similar to the classic separation theorem for 2-SD. That is, a $(1+\gamma)$ -SD can be separated by an FSD and a ( $1+\gamma$ )-PSD.

Theorem 3.2. For $X,Y\in L^1$ , $X\preccurlyeq_{\mathrm{(1+\gamma)\hbox{-}{\rm SD}}} Y$ if and only if there exist $Z_1, Z_2\in L^1$ such that

(3.1) \begin{equation} X\preccurlyeq_{\mathrm{(1+\gamma)\hbox{-}{\rm PSD}}} Z_1 \preccurlyeq_{\mathrm{FSD}} Y\end{equation}

and

(3.2) \begin{equation} X\preccurlyeq_{\mathrm{FSD}} Z_2 \preccurlyeq_{\mathrm{(1+\gamma)\hbox{-}{\rm PSD}}} Y.\end{equation}

Proof. The sufficiency is trivial. It requires us to prove the necessity. To this end, let F and G denote the distribution functions of X and Y, respectively, and assume that $X\preccurlyeq_{\mathrm{(1+\gamma)\hbox{-}SD}} Y$ but $X\not\preccurlyeq_{\mathrm{(1+\gamma)\hbox{-}PSD}} Y$ , i.e. (2.1) does not hold. By Proposition 2.1, we have

\begin{equation*} \Delta\;:\!=\; \gamma \int_{-\infty}^\infty (G(x)-F(x))_-{\mathrm{d}} x-\int_{-\infty}^\infty (G(x)-F(x))_+{\mathrm{d}} x >0.\end{equation*}

For $t\in \overline{\mathbb{R}}=[-\infty,\infty]$ , define

(3.3) \begin{equation} \delta_t(x) = (G(x)-F(x))_-1_{\{x\ge t\}},\end{equation}

where $\delta_{-\infty}(x)\equiv (G(x)-F(x))_-$ and $\delta_{\infty}(x)\equiv0$ . Note that $\delta_t(x)$ is decreasing in $t\in\overline{\mathbb{R}}$ for each fixed x, $\int_{-\infty}^\infty \delta_t(x)\,{\mathrm{d}} x$ is continuous in $t\in\overline{\mathbb{R}}$ , and

\begin{equation*} \gamma \int_{-\infty}^\infty \delta_{\infty}(x)\,{\mathrm{d}} x=0<\Delta \le \gamma \int_{-\infty}^\infty \delta_{-\infty}(x)\,{\mathrm{d}} x.\end{equation*}

Then there exists $t_0\in\overline{\mathbb{R}}$ such that

(3.4) \begin{equation} \gamma \int_{-\infty}^\infty \delta_{t_0}(x) \,{\mathrm{d}} x = \Delta.\end{equation}

We define

\begin{equation*} H_1(x) = G(x) + \delta_{t_0}(x),\quad x\in\mathbb{R},\end{equation*}

which is an increasing and right-continuous function. From (3.3), we have that

\begin{equation*} G(x)\le H_1(x)\le G(x)+(G(x)-F(x))_- = F(x) \vee G(x),\quad x\in\mathbb{R},\end{equation*}

and hence $H_1$ is a distribution function on $\mathbb{R}$ such that $H_1\preccurlyeq_{\mathrm{FSD}}G$ . From (3.4), one can verify that

(3.5) \begin{equation} \int_{-\infty}^\infty (H_1(x)-F(x))_+{\mathrm{d}} x = \gamma \int_{-\infty}^\infty (H_1(x)-F(x))_-{\mathrm{d}} x\end{equation}

and

(3.6) \begin{equation} \int_{-\infty}^t (H_1(x)-F(x))_+{\mathrm{d}} x \le \gamma \int_{-\infty}^t (H_1(x)-F(x))_-{\mathrm{d}} x\quad \text{{for all}}\ t\in\mathbb{R}.\end{equation}

In fact, (3.5) follows from the fact that $(H_1(x)-F(x))_+=(G(x)-F(x))_+$ and $(H_1(x)-F(x))_-=(G(x)-F(x))_- - \delta_{t_0}(x)$ for all $x\in\mathbb{R}$ . Equation (3.6) follows from the facts that $H_1(x)=G(x)$ for $x<t_0$ and $H_1(x)=G(x)\vee F(x)$ for $x \ge t_0$ , which implies the inequality

\begin{equation*} \int_t^{\infty} (H_1(x)-F(x))_+{\mathrm{d}} x \ge \gamma \int^{\infty}_t (H_1(x)-F(x))_-{\mathrm{d}} x\quad \text{{for all}}\ t\in\mathbb{R}.\end{equation*}

This implies that $F\preccurlyeq_{\mathrm{(1+\gamma)\hbox{-}SD}} H_1$ . Then (3.1) follows by taking $Z_1$ as a random variable having distribution function $H_1$ .

A similar argument to the above can be applied to obtain (3.2) by choosing $H_2(x)= F(x)- \eta_{t_1}(x)$ , where

(3.7) \begin{equation} \eta_t(x)=(F(x)-G(x))_+ 1_{\{x< t\}}\end{equation}

and $t_1\in\overline{\mathbb{R}}$ such that

(3.8) \begin{equation} \gamma \int_{-\infty}^\infty \eta_{t_1}(x) \,{\mathrm{d}} x = \Delta.\end{equation}

This completes the proof of the theorem.

Remark 3.2. For $\gamma=1$ , 2-PSD is the concave order. The separation result in Theorem 3.2 reduces to the separation theorem for the SSD: $X\preccurlyeq_{\rm SSD} Y$ if and only if there exists a random variable Z such that

\begin{equation*} X\preccurlyeq_{\mathrm{cv}} Z \preccurlyeq_{\mathrm{FSD}} Y\quad \text{or}\quad X\preccurlyeq_{\mathrm{FSD}} Z \preccurlyeq_{\mathrm{cv}} Y.\end{equation*}

This is a well-known result; see parts (c) and (d) of Theorem 4.A.6 in [Reference Shaked and Shanthikumar26]. There are several proofs of the above separation theorem for the SSD in the literature, for example [Reference Makowski19] and [22]. For $\gamma=1$ , the proof of Theorem 3.2 is new and differs from those in the literature.

1. (i) If $X \preccurlyeq_{\mathrm{(1+\gamma)\hbox{-}SD}} Y$ , and if there does not exist $Z\in L^1$ such that $Z\not\stackrel {\rm d}{=} Y$ , $Z\preccurlyeq_{\mathrm{FSD}} Y$ and $ X \preccurlyeq_{\mathrm{(1+\gamma)\hbox{-}SD}} Z$ , then $X \preccurlyeq_{\mathrm{(1+\gamma)\hbox{-}PSD}} Y$ .

2. (ii) If $X \preccurlyeq_{\mathrm{(1+\gamma)\hbox{-}SD}} Y$ , and if there does not exist $Z\in L^1$ such that $Z\not\stackrel {\rm d}{=} X$ , $X\preccurlyeq_{\mathrm{FSD}} Z$ and $Z \preccurlyeq_{\mathrm{(1+\gamma)\hbox{-}SD}} Y$ , then $X \preccurlyeq_{\mathrm{(1+\gamma)\hbox{-}PSD}} Y$ .

3.4. Strassen’s representation

A famous result of Strassen [Reference Strassen28] states that $X\preccurlyeq_{\mathrm{SSD}} Y$ if and only if there exist random variables $\widehat X$ and $\widehat Y$ defined on a common probability space with the same distributions as X and Y such that $\mathbb{E} [\widehat X \mid \widehat Y ] \le \widehat Y$ , a.s. Müller and Rüschendorf [Reference Müller and Rüschendorf23] presented an elementary and constructive proof of this result on the real line. For more details on Strassen’s theorem and extensions, see [Reference Armbruster1], [Reference Elton and Hill9], [Reference Elton and Hill10], [Reference Lindvall18], and references therein.

For $(1+\gamma)$ -SD, we have the following partial Strassen’s representation.

Theorem 3.3. Let X and Y be two random variables. If there exist $\widehat X$ and $\widehat Y$ on the same probability space such that $\widehat X \buildrel \mathrm{d} \over = X$ , $\widehat Y \buildrel \mathrm{d} \over = Y$ , and

(3.9) \begin{equation} \mathbb{E} [ (\widehat Y-\widehat X )_- \mid \widehat Y ] \le \gamma\,\mathbb{E} [ (\widehat Y-\widehat X )_+ \mid \widehat Y]\quad a.s.\end{equation}

for some $\gamma\in [0,1]$ , then $X \preccurlyeq_{\mathrm{(1+\gamma)\hbox{-}{\rm SD}}} Y$ .

Proof. First we assert that, for any random variable Z,

\begin{equation*} \mathbb{E} [Z_+]\le \gamma\, \mathbb{E}[Z_-] \Longrightarrow Z\preccurlyeq_{\mathrm{(1+\gamma)\hbox{-}SD}} 0,\end{equation*}

which can be seen by verifying (2.3). Then it follows from (2.5) that for any $y\in\mathbb{R}$ , $Z+y\preccurlyeq_{\mathrm{(1+\gamma)\hbox{-}SD}} y$ . Let $\mathrm{supp}(G)$ denote the support of the distribution function of Y. Note that (3.9) implies that, for almost all $y\in \mathrm{supp}(G)$ ,

\begin{equation*} \mathbb{E} [ (\widehat X-y )_+ \mid \widehat Y= y ] \le \gamma\,\mathbb{E} [ (\widehat X-y )_- \mid \widehat Y= y ],\end{equation*}

and hence $ [\widehat X\mid \widehat Y=y ] \preccurlyeq_{\mathrm{(1+\gamma)\hbox{-}SD}} y$ . Then, for any $\phi\in\mathcal U_\gamma^\ast$ , we have $\mathbb{E} [\phi(\widehat X)\mid \widehat Y=y ] \le \phi(y)$ for almost all $y\in \mathrm{supp}(G)$ . Hence

\begin{equation*} \mathbb{E}[\phi(X)] = \mathbb{E} [\phi(\widehat X) ]=\mathbb{E} \{\mathbb{E} [\phi(\widehat X)\mid \widehat Y ] \} \le \mathbb{E} [\phi(\widehat Y) ]=\mathbb{E} [\phi(Y)].\end{equation*}

We thus complete the proof.

For $\gamma=1$ , (3.9) reduces to $\mathbb{E} [\widehat X \mid \widehat Y ] \le \widehat Y$ a.s. In Theorem 3.3, (3.9) is a sufficient condition for $(1+\gamma)$ -SD. However, it is not a necessary condition, as illustrated by the following counterexample.

Example 3.1. Let X and Y be two binary random variables with PMFs

\begin{equation*} \mathbb{P}(X=0)=\mathbb{P}(X=4)=1/2\quad \text{and}\quad \mathbb{P}(Y=2)=\mathbb{P}(Y=3)=1/2.\end{equation*}

Then Y is a $1/2$ -transfer of X and hence $X\preccurlyeq_{\mathrm{(1+1/2)\hbox{-}SD}} Y$ . Assume that there exist $\widehat X$ and $\widehat Y$ on the same probability space such that $\widehat X \buildrel \mathrm{d} \over = X$ , $\widehat Y \buildrel \mathrm{d} \over = Y$ , and (3.9) holds with $\gamma=1/2$ . Denote $b=\mathbb{P}(\widehat X=0\mid \widehat Y=3)$ . From (3.9) it follows that $b\ge 2$ . However, $b\in [0,1]$ . This is a contradiction. Therefore (3.9) is not necessary for $(1+\gamma)$ -SD.

To state the next proposition, we recall the definition ofcomonotonicity. A random vector $(X_1,\ldots$ , $X_n)$ is said to be comonotonic if there exist non-decreasing functions $g_i\ (i=1,\ldots,n)$ , and a random variable W such that $(X_1,\ldots,X_n) \buildrel \mathrm{d} \over = (g_1(W),\ldots,g_n(W))$ . For more on comonotonicity, see [Reference Deelstra, Dhaene and Vanmaele7], [Reference Dhaene, Denuit, Goovaerts, Kaas and Vyncke8], and references therein.

Proposition 3.3. Let F and G be two distribution functions. If G is continuous on $\mathbb{R}$ , then $F\preccurlyeq_{\mathrm{(1+\gamma)\hbox{-}{\rm SD}}} G$ for $\gamma\in[0,1]$ if and only if there exist $X\sim F$ and $Y\sim G$ on the same probability space such that they are comonotonic and

(3.10) \begin{equation} \mathbb{E}[(Y-X)_- \mid Y\le y] \le \gamma\,\mathbb{E}[(Y-X)_+ \mid Y\le y]\quad for all \ y\in\mathbb{R}.\end{equation}

Proof. To show sufficiency, let U be a random variable uniformly distributed on (0, 1). Then we have $(X,Y) \buildrel \mathrm{d} \over = (F^{-1}(U),G^{-1}(U))$ and hence, for $y\in\mathbb{R}$ ,

\begin{align*} \mathbb{E}[(Y-X)_- \mid Y\le y] & = \mathbb{E} [ (G^{-1}(U)-F^{-1}(U) )_- \mid G^{-1}(U)\le y ]\\ & = \mathbb{E} [ (G^{-1}(U)-F^{-1}(U) )_- \mid U\le G(y) ]\\ & = \frac 1{G(y)} \int_0^{G(y)} (G^{-1}(\alpha)-F^{-1}(\alpha) )_- {\mathrm{d}} \alpha.\end{align*}

Similarly,

\begin{equation*} \mathbb{E}[(Y-X)_+ \mid Y\le y] = \frac1{G(y)}\int_0^{G(y)}(G^{-1}(\alpha)-F^{-1}(\alpha))_+ {\mathrm{d}} \alpha.\end{equation*}

Since G is continuous, then for any $p\in (0,1)$ there exists $y\in\mathbb{R}$ such that $G(y)=p$ . It follows from (3.10) that (2.4) holds for all $p\in (0,1)$ . It is obvious to check that (2.4) holds for $p=1$ by the continuity of the two functions of (2.4). Therefore we have $F\preccurlyeq_{\mathrm{(1+\gamma)\hbox{-}SD}} G$ .

Necessity follows immediately by taking $(X,Y)\;:\!=\; (F^{-1}(U),G^{-1}(U))$ with U uniformly distributed on (0,1). This completes the proof.

3.5. Bivariate characterization

To state the bivariate characterization for $(1+\gamma)$ -SD, we introduce the following class of bivariate functions:

(3.11) \begin{equation} \mathcal G_{\gamma} = \{\phi\colon \mathbb{R}^2\to\mathbb{R}\mid x \mapsto \phi(x,y)-\phi(y,x)\in \mathcal U_\gamma^\ast\ \text{for each y}\}.\end{equation}

Proposition 3.4. Let X and Y be two independent random variables. Then $X\preccurlyeq_{\mathrm{(1+\gamma)\hbox{-}{\rm SD}}} Y$ if and only if

\begin{equation*} \mathbb{E}[\phi(X,Y)] \le \mathbb{E}[\phi(Y,X)]\quad for all\ \phi\in \mathcal G_{\gamma}.\end{equation*}

Proof. The sufficiency is trivial by noting that, for any $u\in \mathcal U_\gamma^\ast$ , the bivariate function $\phi(x,y)\;:\!=\; u(x)$ belongs to the set $\mathcal G_\gamma$ . To see the necessity, for any $\phi\in\mathcal G_\gamma$ , define

\begin{equation*} u(x) \;:\!=\; \mathbb{E}[\phi(x,Y)] -\mathbb{E}[\phi(Y,x)],\quad x\in\mathbb{R}.\end{equation*}

It can be easily verified that $u\in \mathcal U_\gamma^\ast$ and thus

\begin{equation*}\mathbb{E}[\phi(X,Y)] -\mathbb{E}[\phi(Y,X)] = \mathbb{E}[u(X)] \le \mathbb{E}[u(Y)] =0.\end{equation*}

Necessity then follows, and hence we complete the proof.

It is worth noting that the above result still holds true if all $\mathcal U_\gamma^\ast$ are replaced by $\mathcal U_\gamma$ . For $\gamma=1$ , the equivalence characterization was implicitly given in [Reference Shanthikumar and Yao27]; see also Theorem 4.A.7 of [Reference Shaked and Shanthikumar26]. An application of Proposition 3.4 is given in Example 5.6.

4.1. Closure under p-quantile truncation

Proposition 4.1. Let X and Y be two continuous random variables with respective distribution functions F and G. If $X \preccurlyeq_{\mathrm{(1+\gamma)\hbox{-}{\rm SD}}} Y$ , then

(4.1) \begin{equation} [X \mid X \le F^{-1}(p) ] \preccurlyeq_{\mathrm{(1+\gamma)\hbox{-}{\rm SD}}} [Y \mid Y \le G^{-1}(p) ],\quad p\in (0,1).\end{equation}

Proof. Let $\phi\in\mathcal{U}_\gamma$ , and suppose that $X\preccurlyeq_{\mathrm{(1+\gamma)\hbox{-}SD}} Y$ . From Proposition 2.2 it follows that $\phi(X)\preccurlyeq_{\mathrm{SSD}} \phi(Y)$ . Since $F_{\phi(X)}^{-1}(\alpha) =\phi(F^{-1}(\alpha))$ for each $\alpha$ , by Theorems 4.A.1 and 4.A.3 in [Reference Shaked and Shanthikumar26], we have

\begin{equation*} \int^p_0 \phi(F^{-1}(\alpha)) \,{\mathrm{d}} \alpha \le \int^p_0 \phi(G^{-1}(\alpha)) \,{\mathrm{d}} \alpha, \quad p\in (0,1),\end{equation*}

or, equivalently, $\mathbb{E} [\phi(X) \mid X\le F^{-1}(p) ] \le \mathbb{E} [\phi(Y) \mid Y \le G^{-1}(p) ]$ since F and G are continuous. This means that (4.1) holds.

From the proof of Proposition 4.1, we conclude that if X and Y are general random variables (not necessarily continuous), then $F \preccurlyeq_{\mathrm{(1+\gamma)\hbox{-}SD}} G$ implies that $ F^{-1}(U) 1_{\{U\le p\}}$ $\preccurlyeq_{\mathrm{(1+\gamma)\hbox{-}SD}} G^{-1}(U) 1_{\{U\le p\}}$ for $p\in (0,1)$ , where U is a random variable uniformly distributed on (0,1).

Remark 4.1. When $\gamma=1$ , the result of Proposition 4.1 was implicitly given by Theorem 4.A.42 of [Reference Shaked and Shanthikumar26] without the constraint of continuity. However, we point out that the condition of continuity is necessary. To see it, we give a counter-example. Define two random variables X and Y with PMFs given by $\mathbb{P}(X=0)=0.625$ , $\mathbb{P}(X=4)=0.375$ and $\mathbb{P}(Y=1)=0.7$ , $\mathbb{P}(Y=2)=0.1$ , $\mathbb{P}(Y=3)=0.2$ . It is easy to verify that $\mathbb{E}[X]=1.5=\mathbb{E}[Y]$ , and hence $X\preccurlyeq_{\mathrm{SSD}} Y$ . Let F and G denote the distribution functions of X and Y, respectively. Note that, for $p=0.7$ ,

\begin{equation*} [X\mid X\le F^{-1}(p)]=[X\mid X\le 4]=X\end{equation*}

and

\begin{equation*} [Y\mid Y\le G^{-1}(p)]=[Y\mid Y\le 1]=1.\end{equation*}

Thus $[X\mid X\le F^{-1}(p)] \not \prec_{\mathrm{SSD}} [Y\mid Y\le G^{-1}(p)]$ .

4.2. Closure under comonotonic sums

Equation (2.5) states that $(1+\gamma)$ -SD is closed under independent sums. With Theorem 3.1 we can prove that $(1+\gamma)$ -SD is closed under comonotonic sums.

Proof. Let $F_i$ and F denote the distribution functions of $X_i$ and $X_1+X_2$ , respectively. Similarly, let $G_i$ and G denote the distribution functions of $Y_i$ and $Y_1+Y_2$ , respectively. Since $X_1$ and $X_2$ are comonotonic, it follows from [Reference Dhaene, Denuit, Goovaerts, Kaas and Vyncke8] that $F^{-1}(\alpha) = F_1^{-1}(\alpha) + F_2^{-1}(\alpha)$ for all $\alpha\in (0,1)$ . Similarly, $G^{-1}(\alpha) = G_1^{-1}(\alpha) + G_2^{-1}(\alpha)$ for all $\alpha\in (0,1)$ . By Theorem 3.1 (iii), $X_i\preccurlyeq_{\mathrm{(1+\gamma)\hbox{-}SD}} Y_i$ implies that

\begin{equation*} \int_0^1 F_i^{-1}(\alpha)\,{\mathrm{d}} h(\alpha) \le \int_0^1 G_i^{-1}(\alpha)\,{\mathrm{d}} h(\alpha)\quad \text{for all}\ h\in \mathcal H_\gamma^\ast, \ i=1, 2.\end{equation*}

Thus

\begin{equation*} \int_0^1 F^{-1}(\alpha)\,{\mathrm{d}} h(\alpha) \le \int_0^1 G^{-1}(\alpha)\,{\mathrm{d}} h(\alpha)\quad \text{for all}\ h\in \mathcal H_\gamma^\ast,\end{equation*}

implying $X_1+X_2\preccurlyeq_{\mathrm{(1+\gamma)\hbox{-}SD}} Y_1+Y_2$ by applying Theorem 3.1 (iii) again. This completes the proof of the proposition.

4.3. Closure under minima

We first present a general result concerning the preservation of $(1+\gamma)$ -SD under increasing and concave transforms.

Proposition 4.3. Let $X_1, \ldots, X_n$ be a set of independent random variables, and let $Y_1, \ldots, Y_n$ be another set of independent random variables. If $X_i \preccurlyeq_{\mathrm{(1+\gamma)\hbox{-}{\rm SD}}} Y_i$ for $i=1, \ldots, n$ and $\gamma\in [0,1]$ , then

(4.2) \begin{equation} g(X_1, \ldots, X_n) \preccurlyeq_{\mathrm{(1+\gamma)\hbox{-}{\rm SD}}} g(Y_1, \ldots, Y_n)\end{equation}

for every increasing and component-wise concave function g.

Proof. Without loss of generality, assume that all random variables $X_i$ and $Y_i$ are independent. The proof is by induction on n. For $n=1$ , the result is just Proposition 2.2. Assume that (4.2) holds true for $n=m-1\ge 1$ . Let $g\colon \mathbb{R}^m\to\mathbb{R}$ be an increasing and component-wise concave function, and let $u\in \mathcal{U}_\gamma$ . Then $u(g(x_1,\ldots, x_m))\in\mathcal {U}_\gamma$ with respect to $x_j$ with other $x_i$ fixed, and hence

\begin{align*} \mathbb{E} [u(g(X_1, X_2,\ldots, X_m))\mid X_1=x] & = \mathbb{E} [u(g(x, X_2,\ldots, X_m))]\\ &\le \mathbb{E} [u(g(x, Y_2,\ldots, Y_m))]\\ &= \mathbb{E} [u(g(X_1, Y_2,\ldots, Y_m))\mid X_1=x],\end{align*}

where the equality follows from the independence of all random variables, and the inequality follows from the induction assumption. Thus

\begin{equation*} \mathbb{E} [u(g(X_1, X_2,\ldots, X_m))] \le \mathbb{E} [u(g(X_1, Y_2,\ldots, Y_m))].\end{equation*}

Similarly, we have

\begin{equation*} \mathbb{E} [u(g(X_1, Y_2,\ldots, Y_m))] \le \mathbb{E} [u(g(Y_1, Y_2,\ldots, Y_m))].\end{equation*}

This proves the desired result.

From Proposition 4.3 we obtain the following corollary by observing that $\min\{x_1, \ldots, x_n\}$ is an increasing and component-wise concave function.

Corollary 4.1. Let $X_1, \ldots, X_n$ be a set of independent random variables, and let $Y_1, \ldots, Y_n$ be another set of independent random variables. If $X_i \preccurlyeq_{\mathrm{(1+\gamma)\hbox{-}{\rm SD}}} Y_i$ for $i=1, \ldots, n$ and $\gamma\in [0,1]$ , then

\begin{equation*} \min\{X_1, X_2, \ldots, X_n\} \preccurlyeq_{\mathrm{(1+\gamma)\hbox{-}{\rm SD}}} \min\{Y_1, Y_2, \ldots, Y_n\}.\end{equation*}

For $\gamma =1$ , Corollary 4.1 for SSD was implicitly given in [Reference Li, Li and Jing17]; see, for example, the paragraph after Corollary 4.A.16 in [Reference Shaked and Shanthikumar26].

4.4. Closure under distortion

Under suitable conditions, $(1+\gamma)$ -SD is preserved under a distortion transformation on the space of distribution functions.

Proof. Denote $F_h=h(F)$ and $G_h=h(G)$ . Then their left inverse functions are $F_h^{-1}(p)=F^{-1}(h^{-1}(p))$ and $G_h^{-1}(p)=G^{-1}(h^{-1}(p))$ , where all the inversion functions are left inverse. Then, for any $\phi\in \mathcal H_{\gamma/\beta}^\ast$ , we have

\begin{equation*} \int_0^1 F_h^{-1}(\alpha) \,{\mathrm{d}} \phi(\alpha)= \int_0^1 F^{-1}(h^{-1}(\alpha)) \,{\mathrm{d}} \phi(\alpha) = \int_0^1 F^{-1}(\alpha) \,{\mathrm{d}} \phi(h(\alpha)).\end{equation*}

Similarly,

\begin{equation*} \int_0^1 G_h^{-1}(\alpha) \,{\mathrm{d}} \phi(\alpha) = \int_0^1 G^{-1}(\alpha) \,{\mathrm{d}} \phi(h(\alpha)).\end{equation*}

Note that for any $\phi\in \mathcal H_{\gamma/\beta}^\ast$ it can be verified that $\phi(h)\in \mathcal H_{\gamma}^\ast$ . To see this, for any $0\le p_1<p_2\le p_3<p_4\le 1$ we have

\begin{align*} \gamma \frac{\phi(h(p_4))-\phi(h(p_3))}{p_4-p_3} & = \frac{\gamma}{\beta} \frac{\phi(h(p_4))-\phi(h(p_3))}{h(p_4)-h(p_3)} \cdot \beta \frac{h(p_4)-h(p_3)}{p_4-p_3} \\ &\le \frac{\phi(h(p_2))-\phi(h(p_1))}{h(p_2)-h(p_1)} \cdot \frac{h(p_2)-h(p_1)}{p_2-p_1} \\ & = \frac{\phi(h(p_2))-\phi(h(p_1))}{p_2-p_1},\end{align*}

where the inequality follows from the fact that $\phi\in \mathcal H_{\gamma/\beta}^\ast$ , $h\in \mathcal H_\beta^\ast$ , and h is increasing. Then the desired result follows immediately from Theorem 3.1 (iii).

For SSD ( $\gamma=\beta=1$ ), Proposition 4.4 was implicitly given in Theorem 4.2 of [Reference Tsukahara29], which was proved by using the fact that any concave $h\in \mathcal{H}$ can be approximated by a sequence of piecewise linear concave distortion functions of the form $h_\alpha(t)=\min\{t/\alpha, 1\}$ , $0<\alpha\le 1$ .

4.5. Equivalence characterization

In the expected utility theory, a decision-maker is risk-averse if she has an increasing and concave utility function. The next proposition states that, for two risks X and Y satisfying $X\preccurlyeq_{\mathrm{(1+\gamma)\hbox{-}SD}} Y$ , if a risk-averse decision-maker is indifferent between X and Y, then X and Y are identically distributed.

Proof. By Proposition 4.3, it suffices to show the case when $\phi(x)=x$ for $x\in\mathbb{R}$ , i.e. $\mathbb{E}[X]=\mathbb{E}[Y]$ . Let F and G denote the distribution functions of X and Y, respectively. By Proposition 2.1, we have that $X\preccurlyeq_{\mathrm{(1+\gamma)\hbox{-}{\rm SD}}} Y$ if and only if (2.3) holds, that is,

(4.3) \begin{equation} (1-\gamma) \int_{-\infty}^t(G(x)-F(x))_+{\mathrm{d}} x \le \gamma \int_{-\infty}^t(F(x)-G(x))\,{\mathrm{d}} x \quad \text{for all}\ t\in\mathbb{R}.\end{equation}

Note that

\begin{equation*} \mathbb{E}[Y] -\mathbb{E}[X] = \int_{-\infty}^\infty (F(x)-G(x)) \,{\mathrm{d}} x.\end{equation*}

Then, taking $t\to\infty$ in (4.3) yields

\begin{equation*} (1-\gamma) \int_{-\infty}^\infty(G(x)-F(x))_+{\mathrm{d}} x \le \gamma(\mathbb{E}[Y]-\mathbb{E}[X])=0,\end{equation*}

which implies that $(G(x)-F(x))_+=0$ for all $x\in\mathbb{R}$ , i.e. $G(x)\le F(x)$ for all $x\in\mathbb{R}$ . Thus we have $X \preccurlyeq_{\mathrm{st}}Y$ . By $\mathbb{E}[X]=\mathbb{E}[Y]$ , it follows from Theorem 1.A.8 of [Reference Shaked and Shanthikumar26] that $X \buildrel \mathrm{d} \over = Y$ . This completes the proof.

In the literature, several authors have investigated conditions under which ordered random variables are equal in distribution; see, for example, [Reference Bhattacharjee3], [Reference Bhattacharjee and Bhattacharya4], [Reference Cheung, Dhaene, Kukush and Linders6], and [Reference Li and Zhu16].

An immediate consequence of Proposition 4.5 is the following corollary.

Corollary 4.2. Let $X_1,X_2, \ldots, X_n$ and $Y_1, Y_2, \ldots, Y_n$ be two collections of independent and identically distributed random variables. If $X_1\preccurlyeq_{\mathrm{(1+\gamma)\hbox{-}{\rm SD}}} Y_1$ and

\begin{equation*} \mathbb{E}\Bigl[\min_{1\le i\le n} X_i\Bigr] = \mathbb{E}\Bigl[\min_{1\le i\le n} Y_i\Bigr], \end{equation*}

then $X_1 \buildrel \mathrm{d} \over = Y_1$ .

Proof. From Corollary 4.1 it follows that $\min_{1\le i\le n} X_i\}\preccurlyeq_{\mathrm{(1+\gamma)\hbox{-}SD}} \min_{1\le i\le n} Y_i$ . By Proposition 4.5, $\mathbb{E}[\min_{1\le i\le n} X_i] = \mathbb{E}[\min_{1\le i\le n} Y_i]$ implies that $\min_{1\le i\le n} X_i \buildrel \mathrm{d} \over = \min_{1\le i\le n} Y_i$ . Therefore, by the relation between the survival functions of $X_1$ and $\min_{1\le i\le n} X_i$ , we have $X_1 \buildrel \mathrm{d} \over = Y_1$ .