Fraction-Degree Reference Dependent Stochastic Dominance

Abstract

For addressing the Allis-type anomalies, a fractional degree reference dependent stochastic dominance rule is developed which is a generalization of the integer degree reference dependent stochastic dominance rules. This new rule can effectively explain why the risk comparison does not satisfy translational invariance and scaling invariance in some cases. The rule also has a good property that it is compatible with the endowment effect of risk. This rule can help risk-averse but not absolute risk-averse decision makers to compare risks relative to reference points. We present some tractable equivalent integral conditions for the fractional degree reference dependent stochastic dominance rule, as well as some practical applications for the rule in economics and finance.

Appendices

Appendices

1.1 Proof of Theorem 1

Proof

${\mathrm{[1]} \Rightarrow \mathrm{[2]}}$. Since $U_{\gamma }^{*}$ is invariant under translations, any $u\in U_{\gamma }^{*}$ can be approximated by a sequence of functions $\{u_{n}\in U_{\gamma }, n=1,2,\cdots \}$ as in the proof of Theorem 2.1 in Denuit and Müller (2002). From this result it follows that ${\mathrm{[1]}}$ implies ${\mathrm{[2]}}$.

${\mathrm{[2]} \Rightarrow \mathrm{[3]}}$. For a fixed $t\in \mathfrak {R}$, we define the consumption utility function u(x; t) with the following right derivative:

$$u'(x;t)=\left\{ \begin{array}{ll} \gamma , &{} x\le t~\text{ and }~ G(x)\le F(x), \\ 1, &{} x\le t~\text{ and }~ G(x)>F(x),\\ 0, &{} x>t. \end{array} \right.$$

Obviously, $u(x;t)\in U_{\gamma }^{*}$. Due to the integration by part, it holds that

$$\begin{aligned}&E\left[ v(Y,r,u(x;t))\right] -E\left[ v(X,r,u(x;t)\right] \\&~~~=\int _{-\infty }^{\infty }u'(x;t)\left( F(x)-G(x)\right) dx+\eta \lambda \int _{-\infty }^{r}u'(x;t)\left( F(x)-G(x)\right) dx\\&~~~~~~~+\eta \int _{r}^{\infty }u'(x;t)\left( F(x)-G(x)\right) dx\\&~~~=\left( 1+\eta \right) \left[ \int _{-\infty }^{\infty }u'(x;t)\left( F(x)-G(x)\right) f_{\eta , \lambda }(x;r)dx\right] \\&~~~=\left( 1+\eta \right) \left[ \int _{-\infty }^{t}\left[ \gamma \left( F(x)-G(x)\right) _{+}-\left( (G(x)-F(x)\right) _{+}\right] f_{\eta , \lambda }(x;r)dx\right] . \end{aligned}$$

Hence, for all $t\in \mathfrak {R}$, $\eta ^{*}>0$ and $\lambda ^{*}> 1$, $E\left[ v(Y,r,u(x;t))\right] \ge E\left[ v(X,r,u(x;t))\right]$ implies that

$$\int _{-\infty }^{t}\left[ \gamma \left( F(x)-G(x)\right) _{+}-\left( G(x)-F(x)\right) _{+}\right] f_{\eta ^{*}, \lambda ^{*}}(x;r)dx\ge 0.$$

${\mathrm{[3]} \Rightarrow \mathrm{[1]}}$. Let $u \in U_{\gamma }$. Without loss of generality we can assume

$$R: =\sup _{x\in \mathfrak {R}}u'(x)\in (0,\infty ).$$

For any fixed $n\ge 2$, define $\varepsilon _n=2^{-n}$ and K as the largest integer k for which

$$R(1-k\varepsilon _n)\ge \inf _{x\in \mathfrak {R}}u'(x),$$

and define a partition of a real line into intervals $[x_{k}, x_{k+1}]$ as follows: let $x_0=-\infty$, $x_{K+1}=\infty$ and

$$x_{k}=\sup \left\{ x:u'(x)\ge R(1-k\varepsilon _n)\right\} ,\qquad k=1,...,K .$$

Then we define

$$m_{k}=\sup \left\{ u'(x):x_{k-1}< x \le x_{k}\right\} =R\left[ 1-(k-1)\varepsilon _n\right] .$$

It follows that

$$\gamma (m_k-R\varepsilon _n)\le u'(x)\le m_k, \ \text{ for } x \in (x_{k-1}, x_k], \ k=1,\ldots , K+1.$$

This implies that for all $0\le k\le K$,

$$\begin{aligned}&\sum _{k=0}^{K}\int _{x_{k}}^{x_{k+1}}\left[ F(x)-G(x)\right] _{+}u'(x)f_{\eta ,\lambda }(x;r)dx\\&~~~\ge \gamma \sum _{k=0}^{K} \left( m_k-R\varepsilon _n\right) \int _{x_{k}}^{x_{k+1}}\left[ F(x)-G(x)\right] _{+}f_{\eta ,\lambda }(x;r)dx. \end{aligned}$$

and

$$\begin{aligned}&\sum _{k=0}^{K}\int _{x_{k}}^{x_{k+1}}\left[ G(x)-F(x)\right] _{+}u'(x)f_{\eta ,\lambda }(x;r)dx\\&~~~\le \sum _{k=0}^{K}m_k\int _{x_{k}}^{x_{k+1}}\left[ G(x)-F(x)\right] _{+}f_{\eta ,\lambda }(x;r)dx. \end{aligned}$$

Let

$$T_k=\int _{x_k}^{x_{k+1}}\left[ \gamma \left( F(x)-G(x)\right) _{+}-\left( G(x)-F(x)\right) _{+}\right] f_{\eta ,\lambda }(x;r)dx,$$

and

$$c_k=\int _{x_k}^{x_{k+1}}\left( F(x)-G(x)\right) _{+}f_{\eta ,\lambda }(x;r)dx.$$

Thus,

$$\begin{aligned} {E[v(Y,r,u)]-E[v(X,r,u)]}= & {} \sum _{k=0}^{K}\int _{x_{k}}^{x_{k+1}}\left[ F(x)-G(x)\right] _{+}u'(x)f_{\eta ,\lambda }(x;r)dx\\&~-\sum _{k=0}^{K}\int _{x_{k}}^{x_{k+1}}\left[ G(x)-F(x)\right] _{+}u'(x)f_{\eta ,\lambda }(x;r)dx\\\ge & {} \sum _{k=0}^{K}m_kT_k-\gamma R \varepsilon _n \sum _{k=0}^{K}c_k. \end{aligned}$$

Set

$$A(x,r)=\frac{f_{\eta ,\lambda }(x,r)}{f_{\eta ^{*},\lambda ^{*}}(x,r)}=\left\{ \begin{array}{ll} \frac{\frac{1+\eta \lambda }{1+\eta }}{\frac{1+\eta ^*\lambda ^{*}}{1+\eta ^{*}}}, &{} x\le r \\ 1, &{} x>r. \end{array} \right.$$

Note that for all $k=0, \ldots , K+1$,

$$\begin{aligned}&\sum _{i=0}^{k}T_i=\int _{-\infty }^{x_{k+1}}\left[ \gamma \left( F(x)-G(x)\right) _{+}-\left( G(x)-F(x)\right) _{+}\right] f_{\eta ^{*}, \lambda ^{*}}(x,r) A(x,r)dx \end{aligned}$$

and

$$\int _{-\infty }^{x_{k+1}}\left[ \gamma \left( F(x)-G(x)\right) _{+}-\left( G(x)-F(x)\right) _{+}\right] f_{\eta ^{*}, \lambda ^{*}}(x,r)dx\ge 0,$$

and A(x, r) is positive and noningcreasing, which implies that $\sum _{i=0}^{k}T_i \ge 0$. Furthermore, since $m_k$ is a decreasing non-negative sequences, $\sum _{i=0}^{k}m_iT_i\ge 0$. Therefore,

$$\begin{aligned} {E[v(Y,r,u)]-E[v(X,r,u)]}&\ge -\gamma R \varepsilon _n \int _{-\infty }^{\infty }\left( F(x)-G(x)\right) _{+}dx. \end{aligned}$$

Letting $n\rightarrow \infty$ yields part [1] holds. This completes the proof of the theorem.

1.2 Proof of Proposition 1

Proof

Let $F_1(x)$ and $F_2(x)$ be two cdfs of Z and W, respectively. The proof is main to construct cdfs $F_1(x)$ and $F_2(x)$. When $x_1\le r\le x_2$, define $F_1(x)$ and $F_2(x)$ as

$$\begin{aligned} F_{1}(x)=\left\{ \begin{array}{ll} F(x), &{} x\ge x_2~\text{ or }~ x<x_1 \\ G(x)+\zeta _1^{*}, &{} x_1\le x< x_2, \end{array} \right. \end{aligned}$$

(19)

and

$$\begin{aligned} F_{2}(x)=\left\{ \begin{array}{ll} G(x), &{} x\ge x_2~\text{ or }~ x<x_1 \\ F(x)-\zeta _1^{*}, &{} x_1\le x< x_2, \end{array} \right. \end{aligned}$$

(20)

where $\zeta _1^{*}$ satisfies that

$$\begin{aligned} \gamma \zeta _1^{*}\left[ \frac{1+\eta ^{*}\lambda ^{*}}{1+\eta ^{*}}(r-x_1)+(x_2-r)\right] =\zeta _2(x_4-x_3) \end{aligned}$$

(21)

Obviously, $0\le \zeta _1^{*}\le \zeta _1$. Hence, $F_1(x)\le F(x)$ and $F_2(x)\ge G(x)$ for all $x\in \mathfrak {R}$, that is, $X\le _{1-SD}Z$ and $W\le _{1-SD} Y$. From (13), (19), we have that Z is simple spread of Z. Combing with (21), it holds that $Z \le _{(1+\gamma )-SD}^{r, \eta ^{*},\lambda ^{*}}Y$ but not $Z\le _{(1+\gamma )-SD}Y$. And since X is also a simple spread of W based on (13), (20), $X\le _{(1+\gamma )-SD}^{r, \eta ^{*},\lambda ^{*}}W$ but not $X\le _{(1+\gamma )-SD}W$ by (21).

When $x_2<r\le x_3$, we also define cdfs $F_1(x)$ as (19) and $F_2(x)$ as (20). But $\zeta _1^{*}$ satisfies that

$$\begin{aligned} \gamma \zeta _1^{*}(x_2-x_1)\frac{1+\eta ^{*}\lambda ^{*}}{1+\eta ^{*}}=\zeta _2(x_4-x_3) \end{aligned}$$

(22)

Similarly, we can obtain $X\le _{1-SD}Z$, $W\le _{1-SD}Y$, $Z\le _{(1+\gamma )-SD}^{r, \eta ^{*},\lambda ^{*}}Y$ but not $Z\le _{(1+\gamma )-SD}Y$ and $X\le _{(1+\gamma )-SD}^{r, \eta ^{*},\lambda ^{*}}W$ but not $X\le _{(1+\gamma )-SD}W$.

When $x_3<r\le x_4$, we define

$$\begin{aligned} F_1(x)=\left\{ \begin{array}{ll} F(x), &{} x<x_3~\text{ or }~x\ge x_4 \\ G(x)-\zeta _2^{*}, &{} x_3\le x< x_4 \end{array} \right. \end{aligned}$$

(23)

and

$$\begin{aligned} F_2(x)=\left\{ \begin{array}{ll} G(x), &{} x<x_3~\text{ or }~x\ge x_4 \\ F(x)-\zeta _2^{*}, &{} x_3\le x< x_4, \end{array} \right. \end{aligned}$$

(24)

where $\zeta _2^{*}$ satisfies that

$$\begin{aligned} \gamma \zeta _1(x_2-x_1)\frac{1+\eta ^{*}\lambda ^{*}}{1+\eta ^{*}}=\zeta _2^{*}\left[ (r-x_3)\frac{1+\eta ^{*}\lambda ^{*}}{1+\eta ^{*}}+(x_4-r)\right] \end{aligned}$$

(25)

Clearly, $\zeta _2^{*}\ge \zeta _2$. Hence, $F_1(x)\le F(x)$ and $F_2(x)\ge G(x)$ for all $x\in \mathfrak {R}$. That is, $X\le _{1-SD}Z$ and $W\le _{1-SD} Y$. Based on (13), (23), (25), we have $Z \le _{(1+\gamma )-SD}^{r, \eta ^{*},\lambda ^{*}}Y$ but not $Z\le _{(1+\gamma )-SD}Y$. And from (13), (24), (25), it holds that $X\le _{(1+\gamma )-SD}^{r, \eta ^{*},\lambda ^{*}}W$ but not $X\le _{(1+\gamma )-SD}W$. Combing these three cases, we complete the proof.

1.3 Proof of Proposition 2

Proof

Without loss of generality, we assume $r_1< r_2$. Let X be a random variable with probability mass function $p_1(x)$. We take $x_1< x_2 \le x_3 <x_4$ with $x_2=r_1$ and $x_4=r_2$. Continue to take $\eta _1>0$, $\eta _2>0$, $0<\gamma \le 1$ such that $\gamma \eta _1\left( x_2-x_1\right) \frac{1+\eta ^{*}\lambda ^{*}}{1+\eta ^{*}}=\eta _2(x_4-x_3)$. Define random variable Y with probability mass function $p_2(x)$ as

$$\begin{aligned} p_2(x_1)= & {} p_1(x_1)-\eta _1,\\ p_2(x_2)= & {} p_1(x_2)+\eta _1,\\ p_2(x_3)= & {} p_1(x_3)+\eta _2,\\ p_2(x_4)= & {} p_1(x_4)-\eta _2,\\ p_2(x)= & {} p_1(x) \qquad \text{ for } \text{ all } \text{ other } \text{ values } \text{ x }. \end{aligned}$$

Let F(x) and G(x) be the cdfs of X and Y, respectively. Obviously, we have

$$F(x)-G(x)=\left\{ \begin{array}{cc} \eta _1, &{} x_1\le x<x_2 \\ -\eta _2, &{} x_3\le x<x_4 \\ 0, &{} \text{ otherwise } \end{array} \right.$$

Therefore, X is a simple spread of Y with a single crossing point $x_2$. And since $\gamma \eta _1\left( x_2-x_1\right) \frac{1+\eta ^{*}\lambda ^{*}}{1+\eta ^{*}}=\eta _2(x_4-x_3)$, we have

$$\gamma \int _{-\infty }^{x_2}(F(x)-G(x))f_{\eta ^{*}, \lambda ^{*}}(x,r_1)dx=\int _{x_2}^{\infty }(G(x)-F(x))f_{\eta ^{*}, \lambda ^{*}}(x,r_1)dx$$

Thus, $X\le _{(1+\gamma )-SD}^{r_1, \eta ^{*},\lambda ^{*}} Y$. And since

$$f_{\eta ^{*}, \lambda ^{*}}(x,r_1)< f_{\eta ^{*}, \lambda ^{*}}(x,r_2)~ \text{ on }~[x_3,x_4]$$

and

$$f_{\eta ^{*}, \lambda ^{*}}(x,r_1)= f_{\eta ^{*}, \lambda ^{*}}(x,r_2)~ \text{ on } \text{ otherwise },$$

we have

$$\begin{aligned} \gamma \int _{-\infty }^{x_2}(F(x)-G(x))f_{\eta ^{*}, \lambda ^{*}}(x,r_2)dx= & {} \gamma \int _{-\infty }^{x_2}(F(x)-G(x))f_{\eta ^{*}, \lambda ^{*}}(x,r_1)dx\\= & {} \int _{x_2}^{\infty }(G(x)-F(x))f_{\eta ^{*}, \lambda ^{*}}(x,r_1)dx\\\le & {} \int _{x_2}^{\infty }(G(x)-F(x))f_{\eta ^{*}, \lambda ^{*}}(x,r_2)dx. \end{aligned}$$

Thus, $X \nleqslant _{(1+\gamma )-SD}^{r_2,\eta ^{*},\lambda ^{*}} Y$. This completes the proof of proposition 2.

1.4 Proof of Theorem 2

To prove Theorem 2, we need the next lemma.

Lemma 1

Given cdfs F and G, let $F^{-1}(s)=\inf \{x:F(x)\ge s\}$ and $G^{-1}(s)=\inf \{x:G(x)\ge s\}$. $F\le _{(1+\gamma )-SD}^{r, \lambda ^{*},\eta ^{*}} G$ if and only if, for all $p\in [0,1]$,

$$\begin{aligned} \int _{0}^{p}H_{\gamma ,r}(G^{-1}(s))ds\ge \int _{0}^{p}H_{\gamma , r}(F^{-1}(s))ds \end{aligned}$$

(26)

Proof

We can use the similar proof of Müller et al. (2017). Define the function

$$t\rightarrow h(x):=\int _{-\infty }^{x}\left[ F(t)-G(t)\right] dH_{\gamma ,r}(t).$$

Then, $F\le _{(1+\gamma )-SD}^{r, \lambda ^{*},\eta ^{*}} G$ if and only if for all x, $h(x)\ge 0$. Since $H_{\gamma ,r}^{'}(t)\ge 0$, the function h assumes local minima in the points $b_k$ where the distribution functions F and G cross, going from $F\le G$ left of $b_k$ to $F>G$ right of $b_k$. If we define $p_k:=G(b_k)$ and

$$\tilde{h}(p):=\int _{0}^{p}\left[ H_{\gamma ,r}(G^{-1}(s))-H_{\gamma ,r}(F^{-1}(s))\right] ds,$$

then we have $h(b_k)=\tilde{h}(p_k)$ and the function $\tilde{h}$ assumes its local minima in the points $p_k$. Therefore, $h\ge 0$ if and only if $\tilde{h}\ge 0$.

Proof

Based on the idea of Müller et al. (2017), define

$$A_{1}(p):=\int _{0}^{p}\left[ H_{\gamma , r}(G^{-1}(s))-H_{\gamma , r}(F^{-1}(s))\right] _{+}ds$$

and

$$A_2(p):=\int _{0}^{p}\left[ H_{\gamma , r}(F^{-1}(s))-H_{\gamma , r}(G^{-1}(s))\right] _{+}ds.$$

Without the loss of generality, we assume that $A_{2}(1)>0$. Let $\alpha (a)$ and $\beta (a)$ be the smallest probabilities that solve

$$A_{1}(\alpha (a))=a ~~~ \text{ and }~~~ A_{2}(\beta (a))=a,~~~0<a< A_{2}(1).$$

It follows from Lemma 1 that $A_{1}(p)\ge A_{2}(p)$. Hence, $\alpha (a)\le \beta (a)$ for all $0<a<A_{2}(1)$. We set

$$x_1(a):= F^{-1}(\alpha (a)),~ x_2(a):=G^{-1}(\alpha (a))$$

and

$$x_3(a):= G^{-1}(\beta (a)),~x_4(a):=F^{-1}(\beta (a)).$$

Since X and Y assume only finitely many values, there is a sequence $0=a_1<\cdots <a_k\le A_{2}(1)$ such that $a\mapsto x_1(a), \cdots , x_4(a)$ are constant on $(a_{i-1}, a_i)$. Denote the corresponding values of these functions as

$$x_{l,i}=x_l(a)~~~~~ \text{ for }~ a\in (a_{i-1}, a_i),~l=1,\ldots , 4.$$

Moreover, for $i=1,\ldots ,k$, at the points $x_{1,i}$ and $x_{4,i}$ the function F has jumps of sizes at least $\zeta _{1i}$ and $\zeta _{2i}$, and at the corresponding points $x_{2,i}$ and $x_{4,i}$ the function G has jumps of sizes at least $\zeta _{1i}$ and $\zeta _{2i}$, where $\zeta _{1i}$ and $\zeta _{2i}$ are given by the equation

$$\zeta _{1i}\left( H_{\gamma ,r}(x_{2,i})-H_{\gamma ,r}(x_{1,i})\right) =\zeta _{2i}\left( H_{\gamma ,r}(x_{4,i})-H_{\gamma ,r}(x_{3,i})\right) =a_i-a_{i-1}~$$

For $x>x_{4,k}$, we have $F(x)>G(x)$. Thus, G is obtained from F by a sequence of k $(\gamma ,r)$-transfers described by the corresponding x’s and $\zeta$’s above, plus a finite number of increasing transfers moving the mass from F to G right of $x_{4,k}$. This completes the proof.

1.5 Proof of Theorem 3

Proof

Note that for random variables $X_n$, X with distribution functions $F_n$, F the convergence $X_n\Rightarrow X$ mentioned in the theorem holds if and only if

$$\int _{-\infty }^{\infty }|F_{n}(x)-F(x)|dx\rightarrow 0,$$

and since $0<H_{\gamma ,r}^{'}\le \frac{1+\eta ^{*}\lambda ^{*}}{1+\eta ^{*}}$, it holds that

$$\int _{-\infty }^{\infty }|F_n(x)-F(x)|dH_{\gamma ,r}(x)\rightarrow 0.$$

This implies that, for any $t\in \mathfrak {R}$,

$$\int _{-\infty }^{t}\left( F_n(x)-F(x)\right) dH_{\gamma ,r}(x)\rightarrow 0.$$

The if-part thus follows from (14).

For the only-if-part, if X, Y are are bounded, then the proof is similar to Müller et al. (2017). We can define for any $n\in \mathbb {N}$

$$X_n=\frac{i}{n},~\text{ if }~ \frac{i}{n}\le X<\frac{i+1}{n}, ~i\in \mathbb {Z}$$

and

$$Y_n=\frac{i+1}{n},~\text{ if }~ \frac{i}{n}\le Y<\frac{i+1}{n},~i \in \mathbb {Z}.$$

Then $X_n$ and $Y_n$ have finite support with $X_n\le _{1-SD} X$ and $Y\le _{1-SD} Y_n$. Therefore,

$$X_n\le _{(1+\gamma )-SD}^{r, \lambda ^{*},\eta ^{*}}X\le _{(1+\gamma )-SD}^{r, \lambda ^{*},\eta ^{*}}Y \le _{(1+\gamma )-SD}^{r, \lambda ^{*},\eta ^{*}} Y_n$$

and $X_n\Rightarrow X$ and $Y_n\Rightarrow Y$.

If X and Y are unbounded, then we define

$$\begin{aligned} X_n:=\left\{ \begin{array}{ll} x_n^{*}, &{} \text{ if }~X<-n, \\ X, &{} \text{ if }~-n\le X\le n, \\ n, &{} \text{ if }~n<X, \end{array} \right. \end{aligned}$$

(27)

and

$$\begin{aligned} Y_n:=\left\{ \begin{array}{ll} -n, &{}~ \text{ if }~Y<-n, \\ Y, &{}~ \text{ if }~-n\le Y\le n, \\ n, &{}~ \text{ if }~n<Y, \end{array} \right. \end{aligned}$$

(28)

where $x_n^{*}$ satisfies that

$$H_{\gamma ,r}(x_n^{*})=H_{\gamma ,r}(-n)-\frac{\int _{-\infty }^{-n}F(x)dH_{\gamma ,r}(x)}{P(X<-n)}.$$

An easy calculation for the corresponding distribution functions $F_n$, $G_n$ shows that

$$\begin{aligned}&\int _{-\infty }^{t}\left( F_n(x)-G_n(x)\right) dH_{\gamma ,r}(x)\\&~~~~=\left\{ \begin{array}{ll} 0, &{} t\le x_n^{*}, \\ P(X<-n)\left( H_{\gamma ,r}(t)-H_{\gamma ,r}(x_n^{*})\right) , &{} x_n^{*}< t\le -n, \\ \int _{-\infty }^{t}\left( F(x)-G(x)\right) dH_{\gamma ,r}(x)+\int _{-\infty }^{-n}G(x)dH_{\gamma ,r}(x), &{} -n<t\le n,\\ 0,&{}t>n. \end{array} \right. \end{aligned}$$

Thus $X\le _{(1+\gamma )-SD}^{r, \lambda ^{*},\eta ^{*}}Y$, that is,

$$\int _{-\infty }^{t}\left( F(x)-G(x)\right) dH_{\gamma ,r}(x)\ge 0, \text{ for } \text{ all }~~t\in \mathbb {R}$$

implies that

$$\int _{-\infty }^{t}\left( F_n(x)-G_n(x)\right) dH_{\gamma ,r}(x)\ge 0 ~~\text{ for } \text{ all }~~t\in \mathbb {R}.$$

Then, it follows that $X\le _{(1+\gamma )-SD}^{r, \lambda ^{*},\eta ^{*}}Y$ implies that $X_n\le _{(1+\gamma )-SD}^{r, \lambda ^{*},\eta ^{*}}Y_n$, and obviously $X_n$, $Y_n$ are bounded and $X_n\Rightarrow X$, and $Y_n\Rightarrow Y$.

For each fixed n we can approximate $X_n$ and $Y_n$ by sequences $\{X_{nn}\}$ and $\{Y_{nn}\}$ as in (27) and (28). Then, the sequences $\{X_{nn}\}$ and $\{Y_{nn}\}$ fulfill the conditions of the theorem. This completes the proof.

1.6 Proof of Theorem 4

To prove Theorem 4, we need Lemma 2.

Lemma 2

Define

$$\begin{aligned} A(x;H)=\frac{f_{\eta , \lambda }(x; H)}{f_{\eta ^{*},\lambda ^{*}}(x; H)} \end{aligned}$$

(29)

Then, A(x; H) is nonegative and nonincreasing.

Proof

 Obviously, A(x; H) is greater than zero. Since

$$A(x; H)=\frac{\eta (\lambda -1)(1+\eta ^{*})}{\eta ^{*}(\lambda ^{*}-1)(1+\eta )}+ \frac{\frac{1+\eta \lambda }{1+\eta }-\frac{(1+\eta ^{*}\lambda ^{*})\eta (\lambda -1)}{(1+\eta )\eta ^{*}(\lambda ^{*}-1)}}{f_{\eta ^{*},\lambda ^{*}}(x;H)}$$

and cdf H(x) is nondecreasing in x, to prove A(x; H) is nonincreasing is just to prove

$$\frac{1+\eta \lambda }{1+\eta }-\frac{(1+\eta ^{*}\lambda ^{*})\eta (\lambda -1)}{(1+\eta )\eta ^{*}(\lambda ^{*}-1)}$$

is less than zero.

$$\begin{aligned}&\frac{1+\eta \lambda }{1+\eta }-\frac{(1+\eta ^{*}\lambda ^{*})\eta (\lambda -1)}{(1+\eta )\eta ^{*}(\lambda ^{*}-1)}\\&~~~~~~=\frac{1}{\eta ^{*}(1+\eta )(\lambda ^{*}-1)}\left[ \eta ^{*}(\lambda ^{*}-1)-\eta (\lambda -1)-\eta \eta ^{*}(\lambda -\lambda ^{*})\right] . \end{aligned}$$

From $\eta \ge \eta ^{*}\ge 0$, $\lambda \ge \lambda ^{*}\ge 1$, we have

$$\frac{1+\eta \lambda }{1+\eta }-\frac{(1+\eta ^{*}\lambda ^{*})\eta (\lambda -1)}{(1+\eta )\eta ^{*}(\lambda ^{*}-1)} \le 0.$$

This completes the proof of Lemma 2.

Proof

 Let $X\vee R = \max \left\{ X, R\right\}$ and $X\wedge R = \min \left\{ X,R\right\}$ with cdfs $F_{X\vee R}(x)$ and $F_{X\wedge R}(x)$, respectively. Then,

$$\begin{aligned}&E\left[ v(X; R,u)\right] \\&~~~~= E\left[ u(X)+\eta u(X\vee R)+\eta \lambda u(X\wedge R)-\eta (1+\lambda )u(R)\right] \nonumber \\&~~~~=\int _{-\infty }^{\infty }u(x)dF(x)+\eta \int _{-\infty }^{\infty }u(x)dF_{X\vee R}(x)+ \eta \lambda \int _{-\infty }^{\infty }u(x)dF_{X\wedge R}(x)\nonumber \\&~~~~~~~-\eta (1+\lambda )\int _{-\infty }^{\infty }u(x)dH(x)\nonumber \\&~~~~=\int _{-\infty }^{\infty }u(x)dF(x)+\eta \int _{-\infty }^{\infty }u(x)d\left[ F(x)H(x)\right] -\eta (1+\lambda )\int _{-\infty }^{\infty }u(x)dH(x)\nonumber \\&~~~~~~~+\eta \lambda \int _{-\infty }^{\infty }u(x)d\left[ F(x)+H(x)-H(x)F(x)\right] \nonumber \\&~~~~=\int _{-\infty }^{\infty }u(x)d\left[ F(x)\left( 1+\eta \lambda -\eta (\lambda -1)H(x)\right) \right] -\eta \int _{-\infty }^{\infty }u(x)dH(x)\nonumber \\&~~~~ =\int _{-\infty }^{\infty }u(x)d\left[ F(x)(1+\eta )f_{\eta ,\lambda }(x, H)\right] -\eta \int _{-\infty }^{\infty }u(x)dH(x). \end{aligned}$$

Thus,

$$E\left[ v(X; R,u)\right] -E\left[ v(Y; R,u)\right] =(1+\eta )\int _{-\infty }^{\infty }u(x)d\left[ \left( F(x)-G(x)\right) f_{\eta ,\lambda }(x, H)\right]$$

Combing with Lemma 2, the equivalent integral condition of

$$E\left[ v(X; R, H)\right] \le E\left[ v(Y; R, H)\right]$$

for all $u\in U_{\gamma }$, and $\eta \ge \eta ^{*}, \lambda \ge \lambda ^{*}$ follows in a similar manner of the proof of Theorem 1.

1.7 Proof of Proposition 3

Proof

Since $X_1$ is a simple spread of $Y_1$ with crossing point $x_0$ and

$$\begin{aligned}&\gamma _0\int _{-\infty }^{x_0}\left[ F_1(x)-G_1(x)\right] f_{\eta ^{*},\lambda ^{*}}(x,G_1)dx\\&~~~~~~~~=\int _{x_0}^{\infty }\left[ G_1(x)-F_1(x)\right] f_{\eta ^{*},\lambda ^{*}}(x,G_1)dx \end{aligned}$$

it holds that $X_1\le _{(1+\gamma _0)-SD}^{Y_1,\eta ^{*},\lambda ^{*}}Y_1$. But we have

$$\begin{aligned}&\int _{x_0}^{\infty }\left[ G_2(x)-F_2(x)\right] f_{\eta ^{*},\lambda ^{*}}(x,G_2)dx\\&~~~~~~=k\int _{x_0}^{\infty }\left[ G_1(x)-F_1(x)\right] f_{\eta ^{*},\lambda ^{*}}(x,G_1)dx\\&~~~~~=k\gamma _0\int _{-\infty }^{x_0}\left[ F_1(x)-G_1(x)\right] f_{\eta ^{*},\lambda ^{*}}(x,G_1)dx\\&~~~~~~\ge \gamma _0\int _{-\infty }^{x_0}\left[ F_2(x)-G_2(x)\right] f_{\eta ^{*},\lambda ^{*}}(x,G_2)dx. \end{aligned}$$

That is, $X_2\nleq _{(1+\gamma _0)-SD}^{Y_2,\eta ^{*},\lambda ^{*}}Y_2$. This completes the proof.

1.8 Proof of Proposition 4

Proof

Since X is a simple spread of Y with crossing point $x_0$ and $\gamma _1\le 1$, we have $X\le _{(1+\gamma _1)-SD}^{R_1\eta ^{*},\lambda ^{*}} Y$. For case [1], it holds that $H_2(x)\ge H_1(x)$ on $[-\infty ,x_0)$ and $H_2(x)\le H_1(x)$ on $[x_0,\infty )$. Thus, we have

$$\begin{aligned}&\gamma _1\int _{-\infty }^{x_0}\left[ F(x)-G(x)\right] f_{\eta ^{*},\lambda ^{*}}(x,H_2)dx\nonumber \\&~~~~~~\le \gamma _1\int _{-\infty }^{x_0}\left[ F(x)-G(x)\right] f_{\eta ^{*},\lambda ^{*}}(x,H_1)dx \nonumber \\&~~~~~~=\int _{x_0}^{\infty }\left[ G(x)-F(x)\right] f_{\eta ^{*},\lambda ^{*}}(x,H_1)dx \\&~~~~~~\le \int _{x_0}^{\infty }\left[ G(x)-F(x)\right] f_{\eta ^{*},\lambda ^{*}}(x,H_2)dx.\nonumber \end{aligned}$$

(30)

That is, $X\nleq _{(1+\gamma _1)-SD}^{R_2, \eta ^{*},\lambda ^{*}} Y$. For case [2], it holds that $H_2(x)\le H_1(x)$ on $[-\infty ,x_0)$ and $H_2(x)\ge H_1(x)$ on $[x_0,\infty )$. Thus, in contrast to (30), we have

$$\gamma _1\int _{-\infty }^{x_0}\left[ F(x)-G(x)\right] f_{\eta ^{*},\lambda ^{*}}(x,H_2)dx\ge \int _{x_0}^{\infty }\left[ G(x)-F(x)\right] f_{\eta ^{*},\lambda ^{*}}(x,H_2)dx.$$

That is, $X\le _{(1+\gamma _1)-SD}^{R_2, \eta ^{*},\lambda ^{*}} Y$. This completes the proof.