Distortion representations of multivariate distributions

Abstract

The univariate distorted distributions were introduced in risk theory to represent changes (distortions) in the expected distributions of some risks. Later, they were also applied to represent distributions of order statistics, coherent systems, proportional hazard rate and proportional reversed hazard rate models, etc. In this paper we extend this concept to the multivariate setup. We show that, in some cases, they are a valid alternative to the copula representation, especially when the marginal distributions may not be easily handled. Several examples illustrate the applications of such representations in statistical modeling. They include the study of paired (dependent) ordered data, joint residual lifetimes, order statistics and coherent systems

Appendix

Proof of proposition 1

For every $i=1,\dots ,n$, $G_i$ is continuous and, hence, its range $Ran(G_i)$ contains the interval (0, 1). Let D be the distribution function of $(G_1(X_1),\dots ,G_n(X_n))$. Then it can be checked that D satisfies properties (i), (ii) and (iii) of Definition 1. Thus, $D\in {\mathcal {D}}_n$.

Moreover, for every $(x_1,\dots ,x_n)$ in ${\mathbb {R}}^n$, it follows that

$$\begin{aligned} F(x_1,\dots ,x_n)&=\Pr (X_1\le x_1,\dots ,X_n\le x_n)\\&=\Pr (G_1(X_1)\le G_1(x_1),\dots ,G_n(X_n)\le G_n(x_n))\\&=D(G_1(x_1),\dots ,G_n(x_n)), \end{aligned}$$

where in the second equality we use that $G_i$ is strictly increasing in the support of $X_i$. Hence is $F\equiv MDD(G_1,\dots ,G_n)$.$\square$

Proof of proposition 2

Since F is continuous, (1) holds for a unique copula C. Thus, for every $(x_1,\dots ,x_n)$ in ${\mathbb {R}}^n$, it follows that

$$\begin{aligned} D(G_1(x_1),\dots ,G_n(x_n))=C(F_1(x_1),\dots ,F_n(x_n)). \end{aligned}$$

For every $i=1,\dots ,n$, since $G_i$ is continuous, there exists $u_i\in (0,1)$ such that $x_i=G_i^{-1}(u_i)$, where $G_i^{-1}$ is quasi–inverse of G (see, e.g., Durante and Sempi 2016). Thus, it follows that

$$\begin{aligned} D(u_1,\dots ,u_n)=C(F_1(G_i^{-1}(u_1)),\dots ,F_n(G_i^{-1}(u_n))), \end{aligned}$$

which is the desired assertion. $\square$

Proof of proposition 3

Clearly, if (2) holds for some distribution functions $G_1,\dots ,G_n$ and $D\in {\mathcal {D}}_n$, then

$$\begin{aligned} \lim _{x_i\rightarrow -\infty }F(x_1,\dots ,x_n)=\lim _{x_i\rightarrow -\infty }D(G_1(x_1),\dots ,G_n(x_n)) =0 \end{aligned}$$

for $i=1,\dots ,n$ and

$$\begin{aligned} \lim _{\min (x_1,\dots ,x_n)\rightarrow +\infty }F(x_1,\dots ,x_n)&=\lim _{\min (x_1,\dots ,x_n)\rightarrow +\infty }D(G_1(x_1),\dots ,G_n(x_n))\\&=D(1,\dots ,1)=1 \end{aligned}$$

since $D \in {\mathcal {D}}_n$. Moreover, F is right-continuous in each variable since D is continuous and $G_1,\dots ,G_n$ are right-continuous.

Let us consider now $(x_1,\dots ,x_n)\in {\mathbb {R}}^n$ and $(y_1,\dots , y_n)\in {\mathbb {R}}^n$ such that $x_i\le y_i$ for $i=1,\dots ,n$. Then we define $u_i=G_i(x_i)$ and $v_i=G_i(y_i)$ for $i=1,\dots ,n$. As $G_i$ is a distribution function, we have $0\le u_i\le v_i\le 1$ for $i=1,\dots ,n$. Therefore

$$\begin{aligned} \triangle _{(x_1,\dots ,x_n)}^{(y_1,\dots ,y_n)} F=\triangle _{(u_1,\dots ,u_n)}^{(v_1,\dots ,v_n)} D\ge 0 \end{aligned}$$

since D satisfies property (iii) in Definition 1. Therefore, F is a proper multivariate distribution function. $\square$

Proof of Proposition 5

The joint distribution function of $(X_1,\dots ,X_m)$ can be written as

$$\begin{aligned} F_{1,\dots ,m}(x_1,\dots ,x_m)=F(x_1,\dots ,x_m,+\infty ,\dots ,+\infty ) \end{aligned}$$

for all $(x_1,\dots ,x_m)\in {\mathbb {R}}^m$. Then (5) is obtained from (2) taking into account that $G_i(+\infty )=1$ for any distribution function $G_i$ and $i=m+1,\dots ,n$. Finally, (5) implies $D_{1,\dots ,m}\in {\mathcal {D}}_m$. $\square$

Proof of Proposition 7

The conditional PDF of $(X_2|X_1=x_1)$ can be written as

$$\begin{aligned} f_{2|1}(x_2|x_1)=\frac{f(x_1,x_2)}{f_1(x_1)} \end{aligned}$$

for all $x_1,x_2$ such that $f_1(x_1)>0$. Then by using (7) and the fact that

$$\begin{aligned} f_1(x_1)=g_1(x_1)D'_1(G_1(x_1))>0, \end{aligned}$$

where $D_1(u):=D(u,1)$ and $D'_1(u)=\partial _1 D(u,1)$, we obtain

$$\begin{aligned} f_{2|1}(x_2|x_1) =g_2(x_2)\frac{\partial _{1,2}D(G_1(x_1),G_2(x_2))}{\partial _1D(G_1(x_1),1)}. \end{aligned}$$

Thus, the conditional distribution function can be obtained as

$$\begin{aligned} F_{2|1}(x_2|x_1)=\int _{-\infty }^{x_2} f_{2|1}(z|x_1) dz= \int _{-\infty }^{x_2} g_2(z)\ \frac{\partial _{1,2}D(G_1(x_1),G_2(z))}{\partial _1D(G_1(x_1),1)}dz. \end{aligned}$$

Now, if we assume $\lim _{v\rightarrow 0^+}\partial _1D(G_1(x_1),v)=0$, then

$$\begin{aligned} F_{2|1}(x_2|x_1)= \left[ \frac{ \partial _{1}D(G_1(x_1),G_2(z))}{\partial _1D(G_1(x_1),1)}\right] _{z=-\infty }^{x_2}=\frac{\partial _{1}D(G_1(x_1),G_2(x_2))}{\partial _1D(G_1(x_1),1)}. \end{aligned}$$

Hence, (8) holds. $\square$

Proof of Proposition10

First we note that $\Pr (X_i>t)\ge \Pr (X_1>t,\dots ,X_n>t)>0$ for $i=1,\dots ,n$. So we can consider the survival functions $\bar{F}_{1,t},\dots ,\bar{F}_{n,t}$ of the marginal residual lifetimes at time t. Then we note that ${\bar{F}}_t$ can be written as

$$\begin{aligned} {\bar{F}}_t(x_1,\dots ,x_n)&=\Pr (X_1-t>x_1,\dots ,X_n-t>x_n|X_1>t,\dots ,X_n>t)\\&=\frac{\Pr (X_1>t+x_1,\dots ,X_n>t+x_n)}{\Pr (X_1>t,\dots ,X_n>t)}\\&=\frac{ {\bar{F}}(t+x_1,\dots ,t+x_n)}{ {\bar{F}}(t,\dots ,t)} \end{aligned}$$

for $x_1,\dots ,x_n\ge 0$. Now we use the following copula representation for ${\bar{F}}$ (obtained from Sklar’s theorem) ${\bar{F}}(x_1,\dots ,x_n)={\hat{C}}({\bar{F}}_1(x_1),\dots ,{\bar{F}}_n(x_n)),$ where ${\hat{C}}$ is a continuous survival copula of ${\bar{F}}$. Hence

$$\begin{aligned} {\bar{F}}_t(x_1,\dots ,x_n)&=\frac{ {\bar{F}}(t+x_1,\dots ,t+x_n)}{ {\bar{F}}(t,\dots ,t)}\\&=\frac{{\hat{C}}({\bar{F}}_1(t+x_1),\dots ,{\bar{F}}_n(t+x_n))}{{\hat{C}}({\bar{F}}_1(t),\dots ,{\bar{F}}_n(t))}\\&=\frac{{\hat{C}}({\bar{F}}_1(t){\bar{F}}_{1,t}(x_1),\dots ,{\bar{F}}_n(t) {\bar{F}}_{n,t}(x_n))}{{\hat{C}}({\bar{F}}_1(t),\dots ,{\bar{F}}_n(t))}\\&={\hat{D}}_t(\bar{F}_{1,t}(x_1),\dots ,\bar{F}_{n,t}(x_n)) \end{aligned}$$

and (10) holds for the function $D_t$ in (11). Hence ${\hat{D}}_t\in {\mathcal {D}}_n$. $\square$

Survival function of ${\mathbf {X}}^{(n)}_t$ given by (12) We aim at calculating the survival function of

$$\begin{aligned} {\mathbf {X}}^{(n,\le )}_t:=(X_1-t,\dots ,X_n-t|X_1>t,\dots ,X_{n-1}>t,X_n\le t) \end{aligned}$$

To this end, consider that

$$\begin{aligned} {\bar{F}}^{(n,\le )}_t(x_1&,\dots ,x_{n-1})\\&=\Pr (X_1-t>x_1,\dots ,X_{n-1}-t>x_{n-1}|X_1>t,\dots ,X_{n-1}>t,X_n\le t)\\&=\frac{\Pr (X_1>t+x_1,\dots ,X_{n-1}>t+x_{n-1},X_n\le x_n)}{\Pr (X_1>t,\dots ,X_{n-1}>t,X_n\le t)}\\&=\frac{\Pr (X_1>t+x_1,\dots ,X_{n-1}>t+x_{n-1})}{\Pr (X_1>t,\dots ,X_{n-1}>t,X_n\le t)}\\ {}&\quad -\frac{\Pr (X_1>t+x_1,\dots ,X_{n-1}>t+x_{n-1},X_n>t)}{\Pr (X_1>t,\dots ,X_{n-1}>t,X_n\le t)}\\&=\frac{ {\bar{F}}(t+x_1,\dots ,t+x_{n-1},0)- {\bar{F}}(t+x_1,\dots ,t+x_{n-1},t)}{\Pr (X_1>t,\dots ,X_{n-1}>t,X_n\le t)}\\&=\frac{ {\bar{F}}(t+x_1,\dots ,t+x_{n-1},0)- {\bar{F}}(t+x_1,\dots ,t+x_{n-1},t)}{{\bar{F}}(t,\dots ,t,0)- {\bar{F}}(t,\dots ,t)}\\&=\frac{ {\hat{C}}({\bar{F}}_1(t+x_1),\dots ,{\bar{F}}_{n-1}(t+x_{n-1}), 1)}{{\hat{C}}({\bar{F}}_1(t),\dots ,{\bar{F}}_{n-1}(t),1)- {\hat{C}}({\bar{F}}_1(t),\dots ,{\bar{F}}_{n}(t))} \\ {}&\quad -\frac{ {\hat{C}}({\bar{F}}_1(t+x_1),\dots ,{\bar{F}}_{n-1}(t+x_{n-1}), {\bar{F}}_n(t))}{{\hat{C}}({\bar{F}}_1(t),\dots ,{\bar{F}}_{n-1}(t),1)- {\hat{C}}({\bar{F}}_1(t),\dots ,{\bar{F}}_{n}(t))} \end{aligned}$$

for $x_1,\dots ,x_{n-1}\ge 0$. Hence it can be written as

$$\begin{aligned} {\bar{F}}^{(n,\le )}_t(x_1,\dots ,x_{n-1})=D^{(n,\le )}_t(\bar{F}_{1,t}(x_1), \dots ,\bar{F}_{n-1,t}(x_{n-1})), \end{aligned}$$

where $\bar{F}_{1,t},\dots ,\bar{F}_{n-1,t}$ are the survival functions of the univariate residual lifetimes,

$$\begin{aligned} D^{(n,\le )}_t({\mathbf {u}})=\frac{ {\hat{C}}({\bar{F}}_1(t)u_1,\dots ,{\bar{F}}_{n-1}(t)u_{n-1}, 1)- {\hat{C}}({\bar{F}}_1(t)u_1,\dots ,{\bar{F}}_{n-1}(t) u_{n-1}, {\bar{F}}_n(t))}{{\hat{C}}({\bar{F}}_1(t),\dots ,{\bar{F}}_{n-1}(t),1)- {\hat{C}}({\bar{F}}_1(t),\dots ,{\bar{F}}_{n}(t))} \end{aligned}$$

for ${\mathbf {u}}=(u_1,\dots ,u_{n})\in [0,1]^n$ and $D^{(n,\le )}_t\in {\mathcal {D}}_{n-1}$. $\square$

Proof of Proposition 11

The distortion function for $n=3$ can be obtained as follows for $0\le u_1\le u_2 \le u_3\le 1$. If we assume $F(x)=x$ for $x\in [0,1]$, then

$$\begin{aligned} D(u_1,u_2,u_3)&=\Pr (X_{1:3}\le u_1,X_{2:3}\le u_2,X_{3:3}\le u_3)\\&=\Pr ( (A_1\cup A_2\cup A_3)\cap (A_{1,2}\cup A_{1,3}\cup A_{2,3})\cap A_{1,2,3}), \end{aligned}$$

where $A_i=\{ X_i\le u_1\}$, $A_{i,j}=\{ X_i\le u_2\}\cap \{ X_j \le u_2\}$, and $A_{1,2,3}=\{ X_1\le u_3\}\cap \{ X_2 \le u_3\}\cap \{ X_3 \le u_3\}$ for $i,j\in \{1,2,3\}$. Hence

$$\begin{aligned} D(u_1,u_2,u_3)&=\Pr ( B_1\cup \dots \cup B_9), \end{aligned}$$

(21)

where $B_1=A_1\cap A_{1,2}\cap A_{1,2,3}$, $B_2=A_2\cap A_{1,2}\cap A_{1,2,3}$, $B_3=A_3\cap A_{1,2}\cap A_{1,2,3}$, $B_4=A_1\cap A_{1,3}\cap A_{1,2,3}$, $B_5=A_2\cap A_{1,3}\cap A_{1,2,3}$, $B_6=A_3\cap A_{1,3}\cap A_{1,2,3}$, $B_7=A_1\cap A_{2,3}\cap A_{1,2,3}$, $B_8=A_2\cap A_{2,3}\cap A_{1,2,3}$, and $B_9=A_3\cap A_{2,3}\cap A_{1,2,3}$. Hence, the formula for D is obtained by applying the inclusion-exclusion formula to (21) taking into account that all these probabilities can be computed from C. For example

$$\begin{aligned} \Pr (B_1)=\Pr (A_1\cap A_{1,2}\cap A_{1,2,3})=\Pr (X_1\le u_1,X_2\le u_2,X_3\le u_3)=C(u_1,u_2,u_3) \end{aligned}$$

and

$$\begin{aligned} \Pr (B_1\cap B_2)&=\Pr (A_1\cap A_2\cap A_{1,2}\cap A_{1,2,3})\\&=\Pr (X_1\le u_1,X_2\le u_1,X_3\le u_3)\\&=C(u_1,u_1,u_3) \end{aligned}$$

$0\le u_1\le u_2 \le u_3\le 1$. The other probabilities can be obtained in a similar way. Clearly, this procedure can also be applied to the n dimensional case (but the expression for D gets really involved).

Proof of Proposition 12

Let ${\mathcal {C}}_1,\dots , {\mathcal {C}}_s$ and ${\mathcal {C}}^*_1,\dots , {\mathcal {C}}^*_{s^*}$ be the minimal cut sets of T and $T^*$, respectively. Hence the joint distribution $F(x,y)=\Pr (T\le x, T^*\le y)$ can be written as

$$\begin{aligned} F(x,y)&=\Pr (\min _{i=1,\dots ,s}\max _{k\in {\mathcal {C}}_i}X_k \le x, \min _{j=1,\dots ,s^*}\max _{k\in {\mathcal {C}}^*_j}X_k \le y)\\&=\Pr ((A_1\cup \dots \cup A_s)\cap (A^*_1\cup \dots \cup A^*_{s^*}))\\&=\Pr \left( \cup _{i=1}^s\cup _{j=1}^{s^*} B_{i,j} \right) , \end{aligned}$$

where $A_i:=\{\max _{k\in {\mathcal {C}}_i}X_k \le x\}$, $A^*_j:=\{\max _{k\in {\mathcal {C}}^*_j}X_k \le y\}$ and $B_{i,j}:=A_i\cap A^*_j$. Now, we can apply the inclusion-exclusion formula to the union of the sets $B_{i,j}$. Moreover we note that, if $x\le y$, then

$$\begin{aligned} \Pr (B_{i,j})&=\Pr \left( \max _{k\in {\mathcal {C}}_i} X_k \le x, \max _{k\in {\mathcal {C}}^*_j}X_k \le y \right) \\&=\Pr \left( \max _{k\in {\mathcal {C}}_i}X_k \le x, \max _{k\in {\mathcal {C}}^*_j-{\mathcal {C}}_i}X_k \le y \right) =C_{{\mathcal {C}}_i,{\mathcal {C}}^*_j}(F(x),F(y)), \end{aligned}$$

where ${\mathcal {C}}^*_j-{\mathcal {C}}_i ={\mathcal {C}}^*_j\cap \mathcal {{\bar{C}}}_i$ (${\bar{A}}$ is the complementary set of the set A), $C_{{\mathcal {C}}_i,{\mathcal {C}}^*_j}(u,v):=C(u_1,\dots ,u_n)$, $u_k=F(x)$ if $k\in {\mathcal {C}}_i$, $u_k=F(y)$ if $k\in {\mathcal {C}}^*_j-{\mathcal {C}}_i$, and $u_k=1$ if $k\notin {\mathcal {C}}_i\cup {\mathcal {C}}^*_j$. Similar expressions can be obtained for the other probabilities in the inclusion-exclusion formula as $\Pr (B_{i,j}\cap B_{\ell ,r})$, $\dots$ and for $x>y$. Hence, we obtain (19). $\square$