text:
\begin{document}

	\title{Conditional Independence Beyond Domain Separability \\
\Large{   Discussion of \cite{engelke2018graphical}}}
	\author[1]{Yuexia Zhang}
	\author[1,2]{Linbo Wang\thanks{E-mail address for correspondence: linbo.wang@utoronto.ca}}

	\affil[1]{Department of Computer and Mathematical Sciences, University of Toronto Scarborough}
	\affil[2]{Department of Statistical Sciences, University of Toronto}
	\clearpage \maketitle

	\section{A general definition of conditional independence}

We congratulate Engelke and Hitz on a thought-provoking paper on graphical models for extremes. A key contribution of the paper is the introduction of a novel definition of conditional independence for a multivariate Pareto distribution. Here, we outline a proposal  for independence and conditional independence of general random variables whose support is a general set $\Omega$ in $\mathbb{R}^d$. Our proposal includes the authors' definition of conditional independence, and the analogous definition of independence as special cases. By making our proposal independent of the context of extreme value theory, we  highlight the importance of the authors' contribution beyond this particular context.

\begin{definition}
    Suppose that   $\mathbf Y=(\mathbf Y_A,\mathbf Y_B, \mathbf Y_C)$ is a random vector with support $\Omega$ in $\mathbb{R}^d$, where  $A, B,C$ are disjoint sets whose union is $\{1,\ldots,d\}$.  Let $U\times V$ denote the Cartesian product of $U$ and $V$.
    \begin{itemize}
        \item[(a)] We say $\mathbf Y_A$ is \emph{conditionally outer  independent} of $\mathbf Y_C$ given $\mathbf Y_B$ if there exists a random vector $\mathbf W = (\mathbf W_A, \mathbf W_B, \mathbf W_C)$ with  support $L_A \times L_B \times L_C$ in $\mathbb{R}^d$ such that (i) $\Omega \subset L_A \times L_B \times  L_C$; (ii) $(\mathbf W \mid \mathbf W\in \Omega) \,{\buildrel d \over =}\, \mathbf Y$; (iii) $\mathbf W_A \ind \mathbf W_C\mid \mathbf W_B$. In this case, we write $\mathbf Y_A \ind_o \mathbf Y_C\mid \mathbf Y_B.$

        If $B = \emptyset$, we say $\mathbf Y_A$ is \emph{outer  independent} of $\mathbf Y_C$, denoted as $\mathbf Y_A \ind_o \mathbf Y_C$.
        \item[(b)] We say $\mathbf Y_A$ is \emph{conditionally inner independent} of $\mathbf Y_C$ given $\mathbf Y_B$ if for any $S_A \times S_B \times S_C \subset \Omega$ such that $S_k$ is a measurable subset of $\mathbb{R}^{dim(Y_k)}, k\in\{A,B,C\}$ and $P(\mathbf Y\in S_A \times S_B \times S_C)>0$, we have $\mathbf Y_A \ind \mathbf Y_C\mid (\mathbf Y_B, \mathbf Y \in S_A\times S_B \times S_C)$. In this case, we write $\mathbf Y_A \ind_i \mathbf Y_C\mid \mathbf Y_B.$

       If $B = \emptyset$, we say $\mathbf Y_A$ is \emph{inner independent} of $\mathbf Y_C$, denoted as $\mathbf Y_A \ind_i \mathbf Y_C$.
    \end{itemize}
\end{definition}

\begin{prop}
\label{prop:ind}
    Suppose $\Omega = [0,\infty)^d \setminus [0,1]^{d}$ as in EH's case. Then
   $$ \mathbf Y_A \ind_o \mathbf Y_C \mid \mathbf Y_B \quad \Leftrightarrow \quad \mathbf Y_A \ind_i \mathbf Y_C  \mid \mathbf Y_B \quad \Leftrightarrow \quad \mathbf Y_A \ind_e \mathbf Y_C \mid \mathbf Y_B.$$
   where $\ind_e$ denotes the notion of conditional independence introduced by \cite{engelke2018graphical}.
In particular, if $B = \emptyset$, then
   $$ \mathbf Y_A \ind_o \mathbf Y_C \quad \Leftrightarrow \quad \mathbf Y_A \ind_i \mathbf Y_C \quad \Leftrightarrow \quad \mathbf Y_A \ind_e \mathbf Y_C.$$
\end{prop}

\begin{remark}
 We do not place any distributional assumptions on $\mathbf Y$ in Proposition \ref{prop:ind}.
\end{remark}

 Engelke and Hitz showed that if $\mathbf Y$ is multivariate Pareto and admits a positive and continuous density, then $\mathbf Y_A \not\!\perp\!\!\!\perp_e \mathbf Y_C.$ This does not rule out the possibility of $\mathbf Y_A \ind_e \mathbf Y_C$ for general Pareto distributions.  For example, consider two independent standard Pareto distributions $X_1$ and $X_2$.

Following eqn. (6) in \cite{engelke2018graphical}, all the probability mass of $\mathbf Y$ lies on $(1,\infty)\times \{0\}$ and $\{0\}\times (1,\infty)$ so that it does not admit a density with respect to Lebesgue measure. Nevertheless  $Y_1 \ind_e Y_2$.
This observation is generalized in Proposition \ref{prop:eh}.

\begin{prop}
	\label{prop:eh}
Suppose $\Omega = \left[ ( [0,\infty)^{|A|} \setminus [0,1]^{|A|}) \times \{0\}^{|B|}  \times \{0\}^{|C|}\right] \cup \left[\{0\}^{|A|}  \times ([0,\infty)^{|B|} \setminus [0,1]^{|B|}) \times  \{0\}^{|C|} \right] \\ \cup \left[\{0\}^{|A|}  \times \{0\}^{|B|}  \times ([0,\infty)^{|C|} \setminus [0,1]^{|C|} ) \right],$ where $|A|$ is the cardinality of set $A$.  Then
	$$\mathbf Y_A \ind_o\mathbf Y_C \mid \mathbf Y_B \quad \Leftrightarrow \quad \mathbf Y_A \ind_i \mathbf Y_C  \mid \mathbf Y_B \quad \Leftrightarrow \quad \mathbf Y_A \ind_e \mathbf Y_C \mid \mathbf Y_B.$$
	In particular, if $B = \emptyset$, then
	$$\mathbf Y_A \ind_o \mathbf Y_C \quad \Leftrightarrow \quad \mathbf Y_A \ind_i \mathbf Y_C \quad \Leftrightarrow \quad \mathbf Y_A \ind_e \mathbf Y_C.$$
\end{prop}

\section{Proofs of propositions}

\begin{comment}
\subsection{Proof of Proposition 1 (a) in the case where  $B=\emptyset$}

\begin{proof}
We need to show
\begin{equation}
    \label{eqn:claim}
      Y_A \ind_o Y_C \Rightarrow Y_A \ind_i Y_C \Rightarrow Y_A \ind_e Y_C \Rightarrow Y_A \ind_o Y_C.
\end{equation}

The first two claims are straightforward. We now prove the third claim in \eqref{eqn:claim}.

Let $A_1 \times B_1 \subset S_1 = (0,1]\times (0,1], A_2 \times B_1 \subset S_2 = (1,\infty)\times (0,1], A_2 \times B_2 \subset S_3 = (1,\infty)\times (1,\infty),  A_1\times B_2 \subset S_4 = (0,1]\times (1,\infty)$. To minimize confusion of notation, we let $W_1 = W_A, W_2 = W_C.$
We let
$$
        {P(W_1 \in A_1, W_2\in B_1)} = \dfrac{P(W_1 \in A_1, W_2>1){P(W_1>1, W_2\in B_1)}}{P(W_1>1, W_2>1)}
$$
and
\begin{equation}
    \label{eqn:normal}
P(W_1\in A_i, W_2 \in B_j) = \lambda P (Y_1 \in A_i, Y_2 \in B_j), \quad (i,j)\neq (1,1),
\end{equation}
where $\lambda$ is a normalizing constant.
Simple calculation yields
$$
        \lambda =\dfrac{1}{1+ \dfrac{P(Y_1<1, Y_2>1) P(Y_1>1, Y_2<1)}{P(Y_1>1,Y_2>1)}}.
$$

It is easy to see that $(\mathbf W \mid \mathbf W \in L) \,{\buildrel d \over =}\, \mathbf Y.$
We now prove
\begin{equation}
    \label{eqn:ind}
    W_1 \ind W_2.
\end{equation}

Note that for $(i,j)= (2,1)$,
\begin{flalign*}
P(W_1 \in A_i, W_2 \in B_j) &= \lambda P (Y_1 \in A_i, Y_2 \in B_j) \\
&= \lambda P (Y_1 \in A_i, Y_2 \in B_j, Y_1 > 1) \\
&= \lambda  P (Y_2 \in B_j \mid Y_1 > 1, Y_1\in A_i) P(Y_1 > 1, Y_1\in A_i) \\
&= P (Y_2 \in B_j \mid Y_1 > 1) P(W_1 \in A_i) \quad (Y_1 \ind Y_2 \mid Y_1 > 1)
\end{flalign*}
So we have
\begin{flalign*}
    P(W_2\in B_j \mid W_1\in A_i) &= P (Y_2 \in B_j \mid Y_1 > 1)\\
    &= \dfrac{P (Y_2 \in B_j , Y_1 > 1)}{P(Y_1 > 1)} \\
    &= \dfrac{P (W_2 \in B_j , W_1 > 1)/\lambda}{P(W_1 > 1)/\lambda} \\
    &= \dfrac{P (W_2 \in B_j , W_1 > 1)}{P(W_1 > 1)} \\
    &= P(W_2\in B_j \mid W_1>1)
\end{flalign*}
Similarly we have
\begin{flalign}
P(W_2\in B_1 \mid W_1\in A_2) &= P(W_2\in B_1 \mid W_1>1) \label{eqn:3} \\
P(W_2\in B_2 \mid W_1\in A_2) &= P(W_2\in B_2 \mid W_1>1) \label{eqn:4}\\
P(W_1\in A_1 \mid W_2\in B_2) &= P(W_1\in A_1 \mid W_2>1)\label{eqn:5} \\
P(W_1\in A_2 \mid W_2\in B_2) &= P(W_1\in A_2 \mid W_2>1).\label{eqn:6}
\end{flalign}
In other words,
\begin{equation}
    \label{eqn:ind}
    W_1\ind W_2 \mid W_1 >1, W_1\ind W_2 \mid W_2>1.
\end{equation}

We now first show that  for any $A_1, B_1, A_2, B_2$, we have
\begin{equation}
\label{eqn:cons}
\dfrac{P(W_1 \in A_1, W_2\in B_1)}{P(W_1\in A_2, W_2\in B_1)} = \dfrac{P(W_1 \in A_1, W_2\in B_2)}{P(W_1\in A_2, W_2\in B_2)}.
\end{equation}
To see this, note that
\begin{flalign*}
& \ \ \dfrac{P(W_1 \in A_1, W_2\in B_2)}{P(W_1\in A_2, W_2\in B_2)}P(W_1\in A_2, W_2\in B_1)\\ &= \dfrac{P(W_1 \in A_1 \mid W_2\in B_2)}{P(W_1\in A_2 \mid W_2\in B_2)}P(W_1\in A_2, W_2\in B_1) \\
&= \dfrac{P(W_1 \in A_1 \mid W_2>1)}{P(W_1\in A_2 \mid W_2>1)} P(W_1\in A_2, W_2\in B_1) \quad \text{due to } \eqref{eqn:ind} \\
&= \dfrac{P(W_1\in A_2, W_2\in B_1)}{P(W_1\in A_2, W_2>1)}  P(W_1 \in A_1, W_2>1) \\
&= \dfrac{P(W_1>1, W_2\in B_1)}{P(W_1>1, W_2>1)}  P(W_1 \in A_1, W_2>1)  \quad \text{due to } \eqref{eqn:ind}.
\end{flalign*}
We then show that
\begin{equation}
    \label{eqn:7}
    P(W_2>1\mid W_1>1) = P(W_2>1\mid W_1<1)
\end{equation}
To see this, note that due to \eqref{eqn:cons},
$$
        \dfrac{P(W_1<1, W_2<1)}{P(W_1>1, W_2<1)}=\dfrac{P(W_1<1, W_2>1)}{P(W_1>1, W_2>1)}
$$
Then
\begin{flalign*}
\text{LHS of } \eqref{eqn:7} &= \dfrac{P(W_1>1, W_2>1)}{P(W_1>1)} \\
&= \dfrac{P(W_1>1, W_2>1)}{P(W_1>1,W_2>1) + P(W_1>1,W_2<1)}  \\
&= \dfrac{P(W_1<1, W_2>1)}{P(W_1<1,W_2>1) + P(W_1<1,W_2<1)}  \\
&= \dfrac{P(W_1<1, W_2>1)}{P(W_1<1)}  \\
&= P(W_2>1\mid  W_1<1) = \text{RHS of } \eqref{eqn:7}.
\end{flalign*}
Similarly we have
\begin{flalign*}
P(W_2>1\mid W_1>1) &= P(W_2>1\mid W_1<1)  \\
P(W_2<1\mid W_1>1) &= P(W_2<1\mid W_1<1) \\
P(W_1>1\mid W_2>1) &= P(W_1>1\mid W_2<1) \\
P(W_1<1\mid W_2>1) &= P(W_1<1\mid W_2<1).
\end{flalign*}

We then show that
\begin{equation}
    \label{eqn:9}
    P(W_2\in B_2\mid W_1>1) = P(W_2 \in B_2 \mid W_1<1)
\end{equation}
To see this, due to \eqref{eqn:cons},
\begin{flalign*}
\dfrac{P(W_1 <1, W_2\in B_1)}{P(W_1 >1, W_2\in B_1)} = \dfrac{P(W_1 <1, W_2>1)}{P(W_1 >1, W_2>1)}
\end{flalign*}
so that
$$
    \dfrac{P(W_2\in B_1\mid W_1<1)}{P(W_2\in B_1\mid W_1 > 1)} = \dfrac{P(W_2>1\mid W_1<1)}{P(W_2>1\mid W_1 > 1)} = 1
$$
where the last equality is due to \eqref{eqn:7}.
It follows directly that $P(W_2\in B_2\mid W_1 >1) = P(W_2\in B_2\mid W_1<1) = P(W_2\in B_2\mid W_1 \in A_2) = P(W_2\in B_2).$
Similarly we have
\begin{flalign*}
P(W_2\in B_2\mid W_1<1) &= P(W_2 \in B_2 \mid W_1 \in A_2) = P(W_2\in B_2) \\
P(W_2\in B_1\mid W_1<1) &= P(W_2 \in B_1 \mid W_1 \in A_2) = P(W_2\in B_1) \\
P(W_1\in A_2\mid W_2<1) &= P(W_1 \in A_2 \mid W_2 \in B_2) = P(W_1\in A_2) \\
P(W_1\in A_1\mid W_2<1) &= P(W_1 \in A_1 \mid W_2 \in B_2) = P(W_1\in A_1)
\end{flalign*}

We still need to show
\begin{flalign}
P(W_2 \in B_2 \mid W_1 \in A_1) &= P(W_2 \in B_2 \mid W_1 < 1) \label{eqn:l1}\\
P(W_2 \in B_1 \mid W_1 \in A_1) &= P(W_2 \in B_1 \mid W_1 < 1)\label{eqn:l2}\\
P(W_1 \in A_2 \mid W_2 \in B_1) &= P(W_1 \in A_2 \mid W_2 < 1)\label{eqn:l3} \\
P(W_1 \in A_1 \mid W_2 \in B_1) &= P(W_1 \in A_1 \mid W_2 < 1)\label{eqn:l4}
\end{flalign}
To show \eqref{eqn:l2}, note that
\begin{flalign*}
& \ \ P(W_2 \in B_1 \mid W_1 \in A_1) \\ &= \dfrac{P(W_2 \in B_1, W_1 \in A_1)}{P(W_1 \in A_1, W_2<1) + P(W_1 \in A_1, W_2>1)} \\
&= \dfrac{ \dfrac{P(W_1 \in A_1, W_2>1){P(W_1>1, W_2\in B_1)}}{P(W_1>1, W_2>1)} }{\dfrac{P(W_1 \in A_1, W_2>1){P(W_1>1, W_2<1)}}{P(W_1>1, W_2>1)} + P(W_1 \in A_1, W_2>1)} \\
&= \dfrac{ \dfrac{{P(W_1>1, W_2\in B_1)}}{P(W_1>1, W_2>1)} }{\dfrac{{P(W_1>1, W_2<1)}}{P(W_1>1, W_2>1)} + 1}
\end{flalign*}
does not depend on $A_1$.

To show \eqref{eqn:l1}, note that
\begin{flalign*}
    & \ \ P(W_2\in B_2\mid W_1\in A_1)    \\
    &= \dfrac{P(W_2\in B_2, W_1\in A_1)}{P(W_1 \in A_1)} \\
    &= \dfrac{ \lambda  P (Y_2\in B_2, Y_1 \in A_1, Y_2>1)}{P(W_1 \in A_1, W_2<1) + P(W_1 \in A_1, W_2>1)} \\
    &= \dfrac{ \lambda   P(Y_2\in B_2 \mid Y_2>1) P( Y_1 \in A_1, Y_2>1)}{ \dfrac{P(W_1 \in A_1, W_2>1){P(W_1>1, W_2<1)}}{P(W_1>1, W_2>1)} + P(W_1 \in A_1, W_2>1)} \\
    &= \dfrac{   P(Y_2\in B_2 \mid Y_2>1) P( W_1 \in A_1, W_2>1)}{ \dfrac{P(W_1 \in A_1, W_2>1){P(W_1>1, W_2<1)}}{P(W_1>1, W_2>1)} + P(W_1 \in A_1, W_2>1)} \\
      &= \dfrac{   P(Y_2\in B_2 \mid Y_2>1) }{ \dfrac{{P(W_1>1, W_2<1)}}{P(W_1>1, W_2>1)} + 1}
\end{flalign*}
does not depend on $A_1$.

\eqref{eqn:l3} and \eqref{eqn:l4} can be shown in a similar fashion. We have hence completed the proof.

\end{proof}
\end{comment}

\begin{comment}

\subsection{Proof of Proposition 1 (a)  for a general $B$}

We now prove for conditional independence, (3) implies (1).

\begin{proof}

Let $A_1 \times B_1\times C_1 \subset S_1, A_2 \times B_1 \times C_1 \subset S_2, A_2 \times B_2 \times C_1\subset S_3,  A_1\times B_2\times C_1 \subset S_4$. To minimize confusion of notation, we let $W_1 = W_A, W_2 = W_C, W_3 = W_B.$
Let
\begin{equation}
    \label{eqn:definition111}
        {P(W_1 \in A_1, W_2\in B_1, W_3\in C_1)} = \dfrac{P(W_1 \in A_1, W_2>1, W_3\in C_1){P(W_1>1, W_2\in B_1, W_3\in C_1)}}{P(W_1>1, W_2>1, W_3\in C_1)}
\end{equation}
and
\begin{equation}
    \label{eqn:normal}
P(W_1\in A_i, W_2 \in B_j, W_3 \in C_k) = \lambda P (Y_1 \in A_i, Y_2 \in B_j, Y_3 \in C_k), \quad (i,j,k)\neq (1,1,1),
\end{equation}
where $\lambda$ is a normalizing constant.
Simple calculation yields
$$
        \lambda =\dfrac{1}{1+ \dfrac{P(Y_1<1, Y_2>1, Y_3<1) P(Y_1>1, Y_2<1, Y_3<1)}{P(Y_1>1, Y_2>1, Y_3<1)}}.
$$

It is easy to see that $(\mathbf W \mid \mathbf W\in L) {\buildrel d \over =}\, \mathbf Y$.
We claim
$$
\label{eqn:imidmidd}
    W_1 \ind W_2\mid W_3.
$$
Note that for $(i,j,k)= (2,1,1)$,
\begin{flalign*}
& P(W_1 \in A_i, W_2 \in B_j, W_3 \in C_k)\\
&= \lambda P (Y_1 \in A_i, Y_2 \in B_j, Y_3 \in C_k) \\
&= \lambda P (Y_1 \in A_i, Y_2 \in B_j, Y_3 \in C_k, Y_1 > 1) \\
&= \lambda  P (Y_2 \in B_j \mid Y_1 > 1, Y_1\in A_i, Y_3 \in C_k) P(Y_1 > 1, Y_1\in A_i, Y_3 \in C_k) \\
&= P (Y_2 \in B_j \mid Y_1 > 1,Y_3 \in C_k) P(W_1 \in A_i,W_3 \in C_k) \quad ( Y_1 \ind Y_2 \mid Y_3, Y_1 > 1)
\end{flalign*}
So we have
\begin{flalign*}
    P(W_2\in B_j \mid W_1\in A_i,W_3 \in C_k) &= P (Y_2 \in B_j \mid Y_1 > 1,Y_3 \in C_k)\\
    &= \dfrac{P (Y_2 \in B_j , Y_1 > 1, Y_3 \in C_k)}{P(Y_1 > 1, Y_3 \in C_k)} \\
    &= \dfrac{P (W_2 \in B_j , W_1 > 1, W_3 \in C_k)/\lambda}{P(W_1 > 1, W_3 \in C_k)/\lambda} \\
    &= \dfrac{P (W_2 \in B_j , W_1 > 1,W_3 \in C_k)}{P(W_1 > 1, W_3 \in C_k)} \\
    &= P(W_2\in B_j \mid W_1>1,W_3 \in C_k)
\end{flalign*}
Similarly we have
\begin{flalign}
P(W_2\in B_1 \mid W_1\in A_2,W_3 \in C_1) &= P(W_2\in B_1 \mid W_1>1,W_3 \in C_1) \label{eqn:15} \\
P(W_2\in B_2 \mid W_1\in A_2,W_3 \in C_1) &= P(W_2\in B_2 \mid W_1>1,W_3 \in C_1) \label{eqn:16}\\
P(W_1\in A_1 \mid W_2\in B_2,W_3 \in C_1) &= P(W_1\in A_1 \mid W_2>1,W_3 \in C_1)\label{eqn:17} \\
P(W_1\in A_2 \mid W_2\in B_2,W_3 \in C_1) &= P(W_1\in A_2 \mid W_2>1,W_3 \in C_1).\label{eqn:18}\\
P(W_2\in B_1 \mid W_1\in A_2,W_3 \in C_2) &= P(W_2\in B_1 \mid W_1>1,W_3 \in C_2) \label{eqn:19} \\
P(W_2\in B_2 \mid W_1\in A_2,W_3 \in C_2) &= P(W_2\in B_2 \mid W_1>1,W_3 \in C_2) \label{eqn:20}\\
P(W_1\in A_1 \mid W_2\in B_2,W_3 \in C_2) &= P(W_1\in A_1 \mid W_2>1,W_3 \in C_2)\label{eqn:21} \\
P(W_1\in A_2 \mid W_2\in B_2,W_3 \in C_2) &= P(W_1\in A_2 \mid W_2>1,W_3 \in C_2).\label{eqn:22}
\end{flalign}

In addition, note that for $(i,j,k)= (2,1,2)$,
\begin{flalign*}
& P(W_1 \in A_i, W_2 \in B_j, W_3 \in C_k)\\
&= \lambda P (Y_1 \in A_i, Y_2 \in B_j, Y_3 \in C_k) \\
&= \lambda P (Y_1 \in A_i, Y_2 \in B_j, Y_3 \in C_k, Y_3 > 1) \\
&= \lambda  P (Y_2 \in B_j \mid Y_3 > 1, Y_1\in A_i, Y_3 \in C_k) P(Y_3 > 1, Y_1\in A_i, Y_3 \in C_k) \\
&= P (Y_2 \in B_j \mid Y_3 \in C_k) P(W_1 \in A_i,W_3 \in C_k) \quad (Y_1 \ind Y_2 \mid Y_3, Y_3 > 1 ).
\end{flalign*}

So we have
\begin{flalign*}
    P(W_2\in B_j \mid W_1\in A_i,W_3 \in C_k) &= P (Y_2 \in B_j \mid Y_3 \in C_k)\\
    &= \dfrac{P (Y_2 \in B_j , Y_3 \in C_k)}{P( Y_3 \in C_k)} \\
    &= \dfrac{P (W_2 \in B_j , W_3 \in C_k)/\lambda}{P( W_3 \in C_k)/\lambda} \\
    &= \dfrac{P (W_2 \in B_j , W_3 \in C_k)}{P( W_3 \in C_k)} \\
    &= P(W_2\in B_j \mid W_3 \in C_k)
\end{flalign*}
Similarly we have
\begin{flalign}
P(W_2\in B_1 \mid W_1\in A_2,W_3 \in C_2) &= P(W_2\in B_1 \mid W_3 \in C_2) \label{eqn:23} \\
P(W_2\in B_2 \mid W_1\in A_2,W_3 \in C_2) &= P(W_2\in B_2 \mid W_3 \in C_2) \label{eqn:24}\\
P(W_1\in A_1 \mid W_2\in B_2,W_3 \in C_2) &= P(W_1\in A_1 \mid W_3 \in C_2)\label{eqn:25} \\
P(W_1\in A_2 \mid W_2\in B_2,W_3 \in C_2) &= P(W_1\in A_2 \mid W_3 \in C_2).\label{eqn:26}\\
P(W_2\in B_1 \mid W_1\in A_1,W_3 \in C_2) &= P(W_2\in B_1 \mid W_3 \in C_2) \label{eqn:27} \\
P(W_2\in B_2 \mid W_1\in A_1,W_3 \in C_2) &= P(W_2\in B_2 \mid W_3 \in C_2) \label{eqn:28}\\
P(W_1\in A_1 \mid W_2\in B_1,W_3 \in C_2) &= P(W_1\in A_1 \mid W_3 \in C_2)\label{eqn:29} \\
P(W_1\in A_2 \mid W_2\in B_1,W_3 \in C_2) &= P(W_1\in A_2 \mid W_3 \in C_2).\label{eqn:30}
\end{flalign}

From \eqref{eqn:23}-\eqref{eqn:30}, we have $W_1\ind W_2\mid W_3 \in C_2 $ and $W_1\ind W_2\mid W_3>1 $.

In summary,
\begin{equation}
    \label{eqn:ind2}
    W_1\ind W_2 \mid W_3,W_1 >1, \quad W_1\ind W_2 \mid W_3, W_2>1, \quad W_1\ind W_2 \mid W_3, W_3>1.
\end{equation}

We now first show that  for any $A_1, B_1, C_1, A_2, B_2$, we have
\begin{equation}
\label{eqn:cons2}
\dfrac{P(W_1 \in A_1, W_2\in B_1\mid W_3\in C_1 )}{P(W_1\in A_2, W_2\in B_1\mid W_3 \in C_1)} = \dfrac{P(W_1 \in A_1, W_2\in B_2\mid W_3 \in C_1)}{P(W_1\in A_2, W_2\in B_2\mid W_3 \in C_1)},
\end{equation}

To prove \eqref{eqn:cons2}, note that
\begin{flalign*}
& \ \ \dfrac{P(W_1 \in A_1, W_2\in B_2\mid W_3\in C_1)}{P(W_1\in A_2, W_2\in B_2\mid W_3\in C_1)}P(W_1\in A_2, W_2\in B_1\mid W_3\in C_1)\\ &= \dfrac{P(W_1 \in A_1 \mid W_2\in B_2,W_3\in C_1)}{P(W_1\in A_2 \mid W_2\in B_2,W_3\in C_1)}P(W_1\in A_2, W_2\in B_1\mid W_3\in C_1) \\
&= \dfrac{P(W_1 \in A_1 \mid W_2>1,W_3\in C_1)}{P(W_1\in A_2 \mid W_2>1,W_3\in C_1)} P(W_1\in A_2, W_2\in B_1\mid W_3\in C_1) \quad \text{due to } \eqref{eqn:17},\eqref{eqn:18} \\
&= \dfrac{P(W_1\in A_2, W_2\in B_1 \mid W_3\in C_1)}{P(W_1\in A_2, W_2>1\mid W_3\in C_1)}  P(W_1 \in A_1, W_2>1\mid W_3\in C_1) \\
&= \dfrac{P(W_1>1, W_2\in B_1\mid W_3\in C_1)}{P(W_1>1, W_2>1\mid W_3\in C_1)}  P(W_1 \in A_1, W_2>1\mid W_3\in C_1)  \quad \text{due to } \eqref{eqn:15},\eqref{eqn:16}\\
&=P(W_1 \in A_1, W_2\in B_1\mid W_3\in C_1 )\quad \text{due to } \eqref{eqn:definition111}.
\end{flalign*}

We then show that
\begin{equation}
    \label{eqn:31}
    P(W_2>1\mid W_1>1,W_3\in C_1) = P(W_2>1\mid W_1<1,W_3\in C_1),
\end{equation}
To prove \eqref{eqn:31}, note that due to \eqref{eqn:cons2},
$$
        \dfrac{P(W_1<1, W_2<1\mid W_3\in C_1)}{P(W_1>1, W_2<1\mid W_3\in C_1)}=\dfrac{P(W_1<1, W_2>1\mid W_3\in C_1)}{P(W_1>1, W_2>1\mid W_3\in C_1)}
$$
Then
\begin{flalign*}
\text{LHS of } \eqref{eqn:31} &= \dfrac{P(W_1>1, W_2>1,W_3\in C_1)}{P(W_1>1,W_3\in C_1)} \\
&= \dfrac{P(W_1>1, W_2>1,W_3\in C_1)}{P(W_1>1,W_2>1,W_3\in C_1) + P(W_1>1,W_2<1,W_3\in C_1)}  \\
&= \dfrac{P(W_1<1, W_2>1\mid W_3\in C_1)}{P(W_1<1,W_2>1\mid W_3\in C_1) + P(W_1<1,W_2<1\mid W_3\in C_1)}  \\
&= \dfrac{P(W_1<1, W_2>1\mid W_3\in C_1)}{P(W_1<1\mid W_3\in C_1)}  \\
&= P(W_2>1\mid  W_1<1,W_3\in C_1) = \text{RHS of } \eqref{eqn:31}.
\end{flalign*}
Similarly we have
\begin{flalign*}
P(W_2>1\mid W_1>1,W_3\in C_1) &= P(W_2>1\mid W_1<1,W_3\in C_1)  \\
P(W_2<1\mid W_1>1,W_3\in C_1) &= P(W_2<1\mid W_1<1,W_3\in C_1) \\
P(W_1>1\mid W_2>1,W_3\in C_1) &= P(W_1>1\mid W_2<1,W_3\in C_1) \\
P(W_1<1\mid W_2>1,W_3\in C_1) &= P(W_1<1\mid W_2<1,W_3\in C_1).
\end{flalign*}

We then show that
\begin{equation}
    \label{eqn:32}
    P(W_2\in B_2\mid W_1>1,W_3\in C_1) = P(W_2 \in B_2 \mid W_1<1,W_3\in C_1)
\end{equation}
To see this, due to \eqref{eqn:cons2},
\begin{flalign*}
\dfrac{P(W_1 <1, W_2\in B_1\mid W_3\in C_1)}{P(W_1 >1, W_2\in B_1\mid W_3\in C_1)} = \dfrac{P(W_1 <1, W_2>1\mid W_3\in C_1)}{P(W_1 >1, W_2>1\mid W_3\in C_1)}
\end{flalign*}
so that
$$
    \dfrac{P(W_2\in B_1\mid W_1<1, W_3\in C_1)}{P(W_2\in B_1\mid W_1 > 1, W_3\in C_1)} = \dfrac{P(W_2>1\mid W_1<1, W_3\in C_1)}{P(W_2>1\mid W_1 > 1, W_3\in C_1)} = 1
$$
where the last equality is due to \eqref{eqn:31}. Due to \eqref{eqn:cons2},
it follows directly that $P(W_2\in B_2\mid W_1 >1, W_3\in C_1) = P(W_2\in B_2\mid W_1<1, W_3\in C_1) = P(W_2\in B_2\mid W_1 \in A_2, W_3\in C_1) = P(W_2\in B_2\mid W_3\in C_1).$
Similarly we have
\begin{flalign*}
P(W_2\in B_2\mid W_1<1, W_3\in C_1) &= P(W_2 \in B_2 \mid W_1 \in A_2, W_3\in C_1) = P(W_2\in B_2\mid  W_3\in C_1) \\
P(W_2\in B_1\mid W_1<1, W_3\in C_1) &= P(Y_2 \in B_1 \mid W_1 \in A_2, W_3\in C_1) = P(W_2\in B_1\mid  W_3\in C_1) \\
P(W_1\in A_2\mid W_2<1, W_3\in C_1) &= P(W_1 \in A_2 \mid W_2 \in B_2, W_3\in C_1) = P(W_1\in A_2\mid  W_3\in C_1) \\
P(W_1\in A_1\mid W_2<1, W_3\in C_1) &= P(W_1 \in A_1 \mid W_2 \in B_2, W_3\in C_1) = P(W_1\in A_1\mid  W_3\in C_1)
\end{flalign*}

We still need to show
\begin{flalign}
P(W_2 \in B_2 \mid W_1 \in A_1, W_3\in C_1) &= P(W_2 \in B_2 \mid W_1 < 1, W_3\in C_1) \label{eqn:33}\\
P(W_2 \in B_1 \mid W_1 \in A_1, W_3\in C_1) &= P(W_2 \in B_1 \mid W_1 < 1, W_3\in C_1)\label{eqn:34}\\
P(W_1 \in A_2 \mid W_2 \in B_1, W_3\in C_1) &= P(W_1 \in A_2 \mid W_2 < 1, W_3\in C_1)\label{eqn:35} \\
P(W_1 \in A_1 \mid W_2 \in B_1, W_3\in C_1) &= P(W_1 \in A_1 \mid W_2 < 1, W_3\in C_1)\label{eqn:36}
\end{flalign}
To show \eqref{eqn:34}, note that due to \eqref{eqn:definition111} and \eqref{eqn:cons2},
\begin{flalign*}
& \ \ P(W_2 \in B_1 \mid W_1 \in A_1, W_3\in C_1) \\ &= \dfrac{P(W_2 \in B_1, W_1 \in A_1, W_3\in C_1)}{P(W_1 \in A_1, W_2<1, W_3\in C_1) + P(W_1 \in A_1, W_2>1, W_3\in C_1)} \\
&= \dfrac{ \dfrac{P(W_1 \in A_1, W_2>1, W_3\in C_1){P(W_1>1, W_2\in B_1, W_3\in C_1)}}{P(W_1>1, W_2>1, W_3\in C_1)} }{\dfrac{P(W_1 \in A_1, W_2>1, W_3\in C_1){P(W_1>1, W_2<1, W_3\in C_1)}}{P(W_1>1,W_2>1, W_3\in C_1)} + P(W_1 \in A_1, W_2>1, W_3\in C_1)} \\
&= \dfrac{ \dfrac{{P(W_1>1, W_2\in B_1, W_3\in C_1)}}{P(W_1>1, W_2>1, W_3\in C_1)} }{\dfrac{{P(W_1>1, W_2<1, W_3\in C_1)}}{P(W_1>1, W_2>1, W_3\in C_1)} + 1}
\end{flalign*}
does not depend on $A_1$.

To show \eqref{eqn:33}, note that
\begin{flalign*}
    & \ \ P(W_2\in B_2\mid W_1\in A_1, W_3\in C_1)    \\
    &= \dfrac{P(W_2\in B_2, W_1\in A_1, W_3\in C_1)}{P(W_1 \in A_1, W_3\in C_1)} \\
    &= \dfrac{ \lambda  P (Y_2\in B_2, Y_1 \in A_1, Y_2>1, Y_3\in C_1)}{P(W_1 \in A_1, W_2<1, W_3\in C_1) + P(W_1 \in A_1, W_2>1, W_3\in C_1)} \\
    &= \dfrac{ \lambda   P(Y_2\in B_2 \mid Y_2>1, Y_3\in C_1) P( Y_1 \in A_1, Y_2>1, Y_3\in C_1)}{ \dfrac{P(W_1 \in A_1, W_2>1, W_3\in C_1){P(W_1>1, W_2<1, W_3\in C_1)}}{P(W_1>1, W_2>1, W_3\in C_1)} + P(W_1 \in A_1, W_2>1, W_3\in C_1)} \\
    &= \dfrac{   P(Y_2\in B_2 \mid Y_2>1, Y_3\in C_1) P( W_1 \in A_1, W_2>1, W_3\in C_1)}{ \dfrac{P(W_1 \in A_1, W_2>1, W_3\in C_1){P(W_1>1, W_2<1, W_3\in C_1)}}{P(W_1>1, W_2>1, W_3\in C_1)} + P(W_1 \in A_1, W_2>1, W_3\in C_1)} \\
      &= \dfrac{   P(Y_2\in B_2 \mid Y_2>1, Y_3\in C_1) }{ \dfrac{{P(W_1>1, W_2<1, W_3\in C_1)}}{P(W_1>1, W_2>1, W_3\in C_1)} + 1}
\end{flalign*}
does not depend on $A_1$.

\eqref{eqn:35} and \eqref{eqn:36} can be shown in a similar fashion. We have hence completed the proof.

\end{proof}
\end{comment}

\subsection{Proof of Proposition \ref{prop:ind} in the case where  $B=\emptyset$}

\begin{proof}
We need to show
\begin{equation}
    \label{eqn:claim1}
     \mathbf Y_A \ind_o \mathbf Y_C \Rightarrow \mathbf Y_A \ind_i \mathbf Y_C \Rightarrow \mathbf Y_A \ind_e \mathbf Y_C \Rightarrow \mathbf Y_A \ind_o \mathbf Y_C.
\end{equation}

The first two claims are straightforward. We now prove the third claim in \eqref{eqn:claim1}.

Let $H_k^1\subset L_k^1=[0,1]^{|k|}$, $H_k^2\subset L_k^2=[0,\infty)^{|k|}\setminus [0,1]^{|k|}$, $H_k=H_k^1\cup H_k^2$, $L_k=L_k^1\cup L_k^2$, $k\in \{A,C\}$. Obviously, $H_k^1\cap H_k^2=\emptyset$ and $L_k^1\cap L_k^2=\emptyset$. Besides, $\Omega = [0,\infty)^d \setminus [0,1]^{d} \subset L_A \times L_C$.

We let
\begin{equation}
    \label{eqn:definition1}
        {P(\mathbf W_A \in H_A^1, \mathbf W_C\in H_C^1)} = \dfrac{P(\mathbf W_A \in H_A^1, \mathbf W_C\in L_C^2){P(\mathbf W_A\in L_A^2, \mathbf W_C\in H_C^1)}}{P(\mathbf W_A\in L_A^2, \mathbf W_C\in L_C^2)},
\end{equation}
and
\begin{equation*}
P(\mathbf W_A\in H_A^i, \mathbf W_C\in H_C^j) = \lambda P (\mathbf Y_A \in H_A^i, \mathbf Y_C \in H_C^j), \quad (i,j)\neq (1,1),
\end{equation*}
where $\lambda$ is a normalizing constant.
Simple calculation yields
$$
        \lambda =\dfrac{1}{1+ \dfrac{P(\mathbf Y_A\in L_A^1, \mathbf Y_C \in L_C^2) P(\mathbf Y_A\in L_A^2, \mathbf Y_C\in L_C^1)}{P(\mathbf Y_A\in L_A^2,\mathbf Y_C\in L_C^2)}}.
$$

It is easy to see that $(\mathbf W \mid \mathbf W \in \Omega ) \,{\buildrel d \over =}\, \mathbf Y.$
We then prove
\begin{equation*}
    \mathbf W_A \ind \mathbf W_C.
\end{equation*}

If $k \in A$ and $\{\mathbf W_A \in H_A\}\cap \{\mathbf W_A: W_k>1\}=\emptyset$, then
\begin{equation}
\label{eqn:1}
P(\mathbf W_A\in H_A, \mathbf W_C\in H_C\mid W_k>1)=P(\mathbf W_A\in H_A \mid W_k>1)P(\mathbf W_C\in H_C \mid W_k>1)=0.
\end{equation}
If $k \in A$ and $\{\mathbf W_A \in H_A\}\cap \{\mathbf W_A: W_k>1\}\neq \emptyset$, then $\{\mathbf W_A \in H_A\}\cap \{\mathbf W_A: W_k>1\}=\{\mathbf W_A \in H_A^2, W_k>1\}$ and

\begin{flalign*}
&P(\mathbf W_A \in H_A,\mathbf W_C \in H_C, W_k>1) \\
=& \lambda P (\mathbf Y_A \in H_A,\mathbf Y_C \in H_C, Y_k>1) \\
=& \lambda  P (\mathbf Y_C\in H_C \mid  \mathbf Y_A\in H_A, Y_k > 1) P(\mathbf Y_A\in H_A, Y_k > 1) \\
=& P (\mathbf Y_C \in H_C \mid Y_k > 1) P(\mathbf W_A \in H_A, W_k>1) \quad (\mathbf Y_A \ind \mathbf Y_C \mid Y_k > 1).
\end{flalign*}
So we have
\begin{equation}
\begin{aligned}
 \label{eqn:2}
    P(\mathbf W_C\in H_C \mid \mathbf W_A\in H_A, W_k>1) &= P (\mathbf Y_C \in H_C \mid Y_k > 1)\\
    &= \dfrac{P (\mathbf Y_C \in H_C , Y_k > 1)}{P(Y_k > 1)} \\
    &= \dfrac{P (\mathbf W_C \in H_C , W_k > 1)/\lambda}{P(W_k > 1)/\lambda} \\
    &= \dfrac{P (\mathbf W_C \in H_C , W_k > 1)}{P(W_k > 1)} \\
    &= P(\mathbf W_C \in H_C \mid W_k>1).
\end{aligned}
\end{equation}
Thus, from \eqref{eqn:1} and \eqref{eqn:2}, we have for any $k \in A$, $\mathbf W_A \ind \mathbf W_C \mid W_k>1$. Similarly, we have for any $k \in C$, $\mathbf W_A \ind \mathbf W_C \mid W_k>1$. In summary,
\begin{equation}
\label{eqn:ind1}
   \forall k \in \{1,\ldots,d\}: \mathbf W_A \ind \mathbf W_C \mid W_k>1.
\end{equation}

We then show that  for any $H_A^1, H_C^1, H_A^2, H_C^2$, we have
\begin{equation}
\label{eqn:cons1}
\dfrac{P(\mathbf W_A \in H_A^1, \mathbf W_C\in H_C^1)}{P(\mathbf W_A\in H_A^2, \mathbf W_C\in H_C^1)} = \dfrac{P(\mathbf W_A \in H_A^1,\mathbf  W_C\in H_C^2)}{P(\mathbf W_A\in H_A^2, \mathbf W_C\in H_C^2)}.
\end{equation}

To prove \eqref{eqn:cons1}, note that
\begin{flalign*}
& \ \ \dfrac{P(\mathbf W_A \in H_A^1, \mathbf W_C\in H_C^2)}{P(\mathbf W_A\in H_A^2, \mathbf W_C\in H_C^2)}P(\mathbf W_A\in H_A^2, \mathbf W_C\in H_C^1)\\
&= \dfrac{P(\mathbf W_A \in H_A^1 \mid \mathbf W_C\in H_C^2)}{P(\mathbf W_A\in H_A^2 \mid \mathbf W_C\in H_C^2)}P(\mathbf W_A\in H_A^2, \mathbf W_C\in H_C^1) \\
&= \dfrac{P(\mathbf W_A \in H_A^1 \mid \mathbf W_C\in H_C^2, \mathbf W_C\in L_C^2)}{P(\mathbf W_A\in H_A^2 \mid \mathbf W_C\in H_C^2, \mathbf W_C\in L_C^2)}P(\mathbf W_A\in H_A^2, \mathbf W_C\in H_C^1) \\
&= \dfrac{P(\mathbf W_A \in H_A^1 \mid \mathbf W_C\in L_C^2)}{P(\mathbf W_A\in H_A^2 \mid \mathbf W_C\in L_C^2)} P(\mathbf W_A\in H_A^2, \mathbf W_C\in H_C^1) \quad \text{due to } \eqref{eqn:ind1} \\
&= \dfrac{P(\mathbf W_A\in H_A^2, \mathbf W_C\in H_C^1)}{P(\mathbf W_A\in H_A^2, \mathbf W_C\in L_C^2)}  P(\mathbf W_A \in H_A^1,\mathbf  W_C\in L_C^2) \\
&= \dfrac{P(\mathbf W_A\in H_A^2, \mathbf W_A\in L_A^2, \mathbf W_C\in H_C^1)}{P(\mathbf W_A\in H_A^2, \mathbf W_A\in L_A^2, \mathbf W_C\in L_C^2)}  P(\mathbf W_A \in H_A^1, \mathbf W_C\in L_C^2) \\
&= \dfrac{P(\mathbf W_A \in L_A^2, \mathbf W_C\in H_C^1)}{P(\mathbf W_A \in L_A^2, \mathbf W_C \in L_C^2)}  P(\mathbf W_A \in H_A^1, \mathbf W_C \in L_C^2) \quad \text{due to } \eqref{eqn:ind1}\\
&=P(\mathbf W_A \in H_A^1, \mathbf W_C\in H_C^1)\quad \text{due to } \eqref{eqn:definition1}.
\end{flalign*}

We then show that
\begin{equation}
    \label{eqn:3}
    P(\mathbf W_C \in L_C^2 \mid \mathbf W_A \in L_A^2) = P(\mathbf W_C \in L_C^2 \mid \mathbf W_A \in L_A^1).
\end{equation}
To prove \eqref{eqn:3}, note that due to \eqref{eqn:cons1},
$$
        \dfrac{P(\mathbf W_A \in L_A^1, \mathbf W_C \in L_C^1)}{P(\mathbf W_A \in L_A^2, \mathbf W_C \in L_C^1)}=\dfrac{P(\mathbf W_A \in L_A^1, \mathbf W_C \in L_C^2)}{P(\mathbf W_A \in L_A^2, \mathbf W_C \in L_C^2)}.
$$
Then
\begin{flalign*}
\text{LHS of } \eqref{eqn:3} &= \dfrac{P(\mathbf W_A \in L_A^2,\mathbf  W_C \in L_C^2)}{P(\mathbf W_A \in L_A^2)} \\
&= \dfrac{P(\mathbf W_A \in L_A^2, \mathbf W_C \in L_C^2)}{P(\mathbf W_A \in L_A^2,\mathbf W_C \in L_C^2) + P(\mathbf W_A \in L_A^2, \mathbf W_C \in L_C^1)}  \\
&= \dfrac{P(\mathbf W_A \in L_A^1, \mathbf W_C \in L_C^2)}{P(\mathbf W_A \in L_A^1,\mathbf W_C \in L_C^2) + P(\mathbf W_A \in L_A^1, \mathbf W_C \in L_C^1)}  \\
&= \dfrac{P(\mathbf W_A \in L_A^1, \mathbf W_C \in L_C^2)}{P(\mathbf W_A \in L_A^1)}  \\
&= P(\mathbf W_C \in L_C^2\mid  \mathbf W_A \in L_A^1) = \text{RHS of } \eqref{eqn:3}.
\end{flalign*}
Similarly we have
\begin{flalign*}
P(\mathbf W_C \in L_C^1\mid \mathbf W_A \in L_A^2) &= P(\mathbf W_C \in L_C^1\mid \mathbf W_A \in L_A^1).
\end{flalign*}

We then show that
\begin{equation*}
    P(\mathbf W_C\in H_C^2\mid \mathbf W_A \in L_A^2) = P(\mathbf W_C \in H_C^2 \mid\mathbf  W_A \in L_A^1).
\end{equation*}
To see this, due to \eqref{eqn:cons1},
\begin{flalign*}
\dfrac{P(\mathbf W_A \in L_A^1, \mathbf W_C\in H_C^1)}{P(\mathbf W_A \in L_A^2, \mathbf W_C\in H_C^1)} = \dfrac{P(\mathbf W_A \in L_A^1, \mathbf W_C \in L_C^2)}{P(\mathbf W_A \in L_A^2, \mathbf W_C \in L_C^2)}.
\end{flalign*}
Furthermore, due to \eqref{eqn:3},
$$
    \dfrac{P(\mathbf W_C\in H_C^1\mid \mathbf W_A \in L_A^1)}{P(\mathbf W_C\in H_C^1\mid \mathbf W_A \in L_A^2)} = \dfrac{P(\mathbf W_C \in L_C^2\mid \mathbf  W_A \in L_A^1)}{P(\mathbf W_C \in L_C^2\mid \mathbf W_A \in L_A^2)} = 1.
$$
Due to \eqref{eqn:ind1} and \eqref{eqn:cons1},
it follows directly that
\begin{multline}
\label{eqn:4}
P(\mathbf W_C\in H_C^2\mid \mathbf W_A \in L_A^1)
= P(\mathbf W_C\in H_C^2\mid \mathbf W_A \in L_A^2) \\
= P(\mathbf W_C\in H_C^2\mid \mathbf W_A \in H_A^2)
= P(\mathbf W_C\in H_C^2).
\end{multline}
Similarly we have
\begin{equation}
\label{eqn:5}
P(\mathbf W_C\in H_C^1\mid \mathbf W_A \in L_A^1)= P(\mathbf W_C\in H_C^1 \mid \mathbf W_A \in H_A^2) = P(\mathbf W_C\in H_C^1).
\end{equation}

We still need to show
\begin{flalign}
P(\mathbf W_C \in H_C^2 \mid \mathbf W_A \in H_A^1) &= P(\mathbf W_C \in H_C^2 \mid \mathbf W_A \in L_A^1), \label{eqn:6}\\
P(\mathbf W_C \in H_C^1 \mid \mathbf W_A \in H_A^1) &= P(\mathbf W_C \in H_C^1 \mid \mathbf W_A \in L_A^1).\label{eqn:7}
\end{flalign}

To show \eqref{eqn:6}, note that due to \eqref{eqn:ind1} and \eqref{eqn:cons1},

\begin{flalign*}
    & \ \ P(\mathbf W_C \in H_C^2\mid \mathbf W_A \in H_A^1)    \\
    &= \dfrac{P(\mathbf W_C \in H_C^2, \mathbf W_A \in H_A^1)}{P(\mathbf W_A \in H_A^1)} \\
    &= \dfrac{ \lambda  P (\mathbf Y_C \in H_C^2,\mathbf  Y_C\in L_C^2, \mathbf Y_A \in H_A^1)}{P(\mathbf W_A \in H_A^1, \mathbf W_C\in L_C^1) + P(\mathbf W_A \in H_A^1, \mathbf W_C\in L_C^2)} \\
    &= \dfrac{ \lambda   P(\mathbf Y_C \in H_C^2 \mid  \mathbf Y_C\in L_C^2) P(\mathbf Y_A \in H_A^1, \mathbf Y_C\in L_C^2)}{ \dfrac{P(\mathbf W_A \in H_A^1,  \mathbf W_C\in L_C^2){P(\mathbf W_A \in L_A^2, \mathbf W_C\in L_C^1)}}{P(\mathbf W_A \in L_A^2, \mathbf W_C\in L_C^2)} + P(\mathbf W_A \in H_A^1, \mathbf W_C\in L_C^2)} \\
    &= \dfrac{   P(\mathbf Y_C \in H_C^2 \mid \mathbf Y_C\in L_C^2) P(\mathbf W_A \in H_A^1, \mathbf W_C\in L_C^2)}{ \dfrac{P( \mathbf W_A \in H_A^1,  \mathbf W_C\in L_C^2){P(\mathbf W_A \in L_A^2,\mathbf  W_C\in L_C^1)}}{P(\mathbf W_A \in L_A^2, \mathbf W_C\in L_C^2)} + P(\mathbf W_A \in H_A^1, \mathbf W_C\in L_C^2)} \\
      &= \dfrac{   P(\mathbf Y_C \in H_C^2 \mid \mathbf  Y_C\in L_C^2) }{ \dfrac{{P(\mathbf W_A \in L_A^2, \mathbf W_C\in L_C^1)}}{P(\mathbf W_A \in L_A^2, \mathbf W_C\in L_C^2)} + 1}
\end{flalign*}
does not depend on $H_A^1$.

To show \eqref{eqn:7}, note that due to \eqref{eqn:definition1} and \eqref{eqn:cons1},
\begin{flalign*}
& \ \ P(\mathbf W_C \in H_C^1 \mid \mathbf W_A \in H_A^1) \\ &= \dfrac{P(\mathbf W_A \in H_A^1,  \mathbf W_C \in H_C^1)}{P(\mathbf W_A \in H_A^1,  \mathbf W_C \in L_C^1) + P(\mathbf W_A \in H_A^1,\mathbf  W_C \in L_C^2)} \\
&= \dfrac{ \dfrac{P(\mathbf W_A \in H_A^1, \mathbf W_C \in L_C^2){P(\mathbf W_A \in L_A^2, \mathbf W_C \in H_C^1)}}{P(\mathbf W_A \in L_A^2,  \mathbf W_C \in L_C^2)} }{\dfrac{P(\mathbf W_A \in H_A^1, \mathbf W_C \in L_C^2){P(\mathbf W_A \in L_A^2, \mathbf  W_C \in L_C^1)}}{P(\mathbf W_A \in L_A^2,  \mathbf W_C \in L_C^2)} + P(\mathbf W_A \in H_A^1, \mathbf W_C \in L_C^2)} \\
&= \dfrac{ \dfrac{{P(\mathbf W_A \in L_A^2,  \mathbf W_C \in H_C^1)}}{P(\mathbf W_A \in L_A^2, \mathbf W_C \in L_C^2)} }{\dfrac{{P(\mathbf W_A \in L_A^2, \mathbf W_C \in L_C^1)}}{P(\mathbf W_A \in L_A^2, \mathbf W_C \in L_C^2)} + 1}
\end{flalign*}
does not depend on $H_A^1$.

From \eqref{eqn:4}, \eqref{eqn:5}, \eqref{eqn:6} and \eqref{eqn:7}, we have
\begin{equation*}
        P(\mathbf W_C \in H_C \mid \mathbf W_A \in H_A)=P(\mathbf W_C \in H_C).
\end{equation*}
Hence $\mathbf W_A \ind \mathbf W_C$.

\end{proof}

\subsection{Proof of Proposition \ref{prop:ind} for a general $B$}
\begin{proof}

We need to show
\begin{equation}
    \label{eqn:claim2}
      \mathbf Y_A \ind_o \mathbf Y_C \mid \mathbf Y_B \Rightarrow\mathbf  Y_A \ind_i \mathbf Y_C \mid \mathbf Y_B \Rightarrow \mathbf Y_A \ind_e \mathbf Y_C \mid \mathbf Y_B \Rightarrow \mathbf Y_A \ind_o \mathbf Y_C \mid\mathbf  Y_B.
\end{equation}

The first two claims are straightforward. We now prove the third claim in \eqref{eqn:claim2}.

Let $H_k^1\subset L_k^1=[0,1]^{|k|}$, $H_k^2\subset L_k^2=[0,\infty)^{|k|}\setminus [0,1]^{|k|}$, $H_k=H_k^1\cup H_k^2$, $L_k=L_k^1\cup L_k^2$, $k\in \{A,B,C\}$. Obviously, $H_k^1\cap H_k^2=\emptyset$ and $L_k^1\cap L_k^2=\emptyset$. Besides, $\Omega = [0,\infty)^d \setminus [0,1]^{d} \subset L_A \times L_B \times L_C$.

We let
\begin{multline}
\label{eqn:definition3}
        {P(\mathbf W_A \in H_A^1,\mathbf  W_B\in H_B^1, \mathbf W_C\in H_C^1)} \\
        = \dfrac{P(\mathbf W_A \in H_A^1, \mathbf W_B\in H_B^1, \mathbf W_C\in L_C^2){P(\mathbf W_A\in L_A^2, \mathbf W_B\in H_B^1, \mathbf W_C\in H_C^1)}}{P(\mathbf W_A\in L_A^2, \mathbf W_B\in H_B^1, \mathbf W_C\in L_C^2)},
\end{multline}
and for $(i, j, k)\neq (1, 1, 1)$,
\begin{equation*}
P(\mathbf W_A\in H_A^i, \mathbf W_B\in H_B^j, \mathbf W_C\in H_C^k) = \lambda P (\mathbf Y_A \in H_A^i, \mathbf Y_B\in H_B^j, \mathbf Y_C \in H_C^k),
\end{equation*}
where $\lambda$ is a normalizing constant.
Simple calculation yields
$$
        \lambda =\dfrac{1}{1+ \dfrac{P(\mathbf Y_A\in L_A^1,\mathbf  Y_B\in L_B^1, \mathbf Y_C \in L_C^2) P(\mathbf Y_A\in L_A^2, \mathbf Y_B\in L_B^1, \mathbf Y_C\in L_C^1)}{P(\mathbf Y_A\in L_A^2, \mathbf Y_B\in L_B^1, \mathbf Y_C\in L_C^2)}}.
$$

It is easy to see that $(\mathbf  W \mid  \mathbf  W \in \Omega ) \,{\buildrel d \over =}\, \mathbf  Y.$
We then prove
\begin{equation*}
   \mathbf  W_A \ind\mathbf  W_C \mid \mathbf W_B.
\end{equation*}

If $k \in A$ and $\{\mathbf W_A \in H_A\}\cap \{\mathbf W_A: W_k>1\}=\emptyset$, then
\begin{multline}
\label{eqn:37}
P(\mathbf W_A\in H_A, \mathbf W_C\in H_C\mid \mathbf W_B\in H_B, W_k>1)=P(\mathbf W_A\in H_A \mid \mathbf W_B\in H_B, W_k>1)\\
\times P(\mathbf W_C\in H_C \mid \mathbf W_B\in H_B, W_k>1)=0.
\end{multline}
If $k \in A$ and $\{\mathbf W_A \in H_A\}\cap \{\mathbf W_A: W_k>1\}\neq \emptyset$, then $\{\mathbf W_A \in H_A\}\cap \{\mathbf W_A: W_k>1\}=\{\mathbf W_A \in H_A^2, W_k>1\}$.
Note that
\begin{flalign*}
&P(\mathbf W_A \in H_A, \mathbf W_B \in H_B, \mathbf W_C \in H_C, W_k>1) \\
&= \lambda P (\mathbf Y_A \in H_A, \mathbf Y_B \in H_B, \mathbf Y_C \in H_C, Y_k>1) \\
&= \lambda  P (\mathbf Y_C\in H_C \mid \mathbf  Y_A\in H_A, \mathbf Y_B \in H_B, Y_k > 1) P( \mathbf Y_A\in H_A, \mathbf Y_B \in H_B, Y_k > 1) \\
&= P (\mathbf Y_C \in H_C \mid\mathbf  Y_B \in H_B, Y_k > 1) P(\mathbf W_A \in H_A, \mathbf W_B \in H_B, W_k>1) \quad (\mathbf Y_A \ind \mathbf Y_C \mid\mathbf  Y_B, Y_k > 1).
\end{flalign*}
So we have
\begin{equation}
\begin{aligned}
 \label{eqn:38}
    P(\mathbf W_C\in H_C \mid \mathbf W_A\in H_A, \mathbf W_B \in H_B, W_k>1) &= P (\mathbf Y_C \in H_C \mid\mathbf  Y_B \in H_B, Y_k > 1)\\
    &= \dfrac{P (\mathbf Y_C \in H_C, \mathbf Y_B \in H_B, Y_k > 1)}{P(\mathbf Y_B \in H_B, Y_k > 1)} \\
    &= \dfrac{P (\mathbf W_C \in H_C, \mathbf W_B \in H_B, W_k > 1)/\lambda}{P(\mathbf W_B \in H_B, W_k > 1)/\lambda} \\
    &= \dfrac{P (\mathbf W_C \in H_C, \mathbf W_B \in H_B, W_k > 1)}{P(\mathbf W_B \in H_B, W_k > 1)} \\
    &= P(\mathbf W_C \in H_C \mid \mathbf W_B \in H_B, W_k>1).
\end{aligned}
\end{equation}
Thus, from \eqref{eqn:37} and \eqref{eqn:38}, we have for any $k \in A$, $\mathbf W_A \ind \mathbf W_C \mid \mathbf W_B, W_k>1$.
Similarly, we have for any $k \in C$, $\mathbf W_A \ind \mathbf W_C \mid \mathbf W_B, W_k>1$.

If $k \in B$ and $\{\mathbf W_B \in H_B\}\cap \{\mathbf W_B: W_k>1\}=\emptyset$, then according to the definition in \citet[\S 7.4]{shorack2017probability},
\begin{equation}
    \label{eqn:39}
    P(\mathbf W_C \in H_C \mid \mathbf W_A \in H_A, \mathbf W_B \in H_B,  W_k>1)=P(\mathbf W_C \in H_C).
\end{equation}
If $k \in B$ and $\{\mathbf W_B \in H_B\}\cap \{\mathbf W_B: W_k>1\}\neq \emptyset$, then $\{\mathbf W_B\in H_B\}\cap \{\mathbf W_B: W_k>1\}=\{\mathbf W_B \in H_B^2, W_k>1\}$. Besides, similar to the proof of  \eqref{eqn:38}, we have
\begin{equation}
    \label{eqn:40}
    P(\mathbf W_C \in H_C \mid \mathbf W_A \in H_A, \mathbf W_B \in H_B,  W_k>1)=P(\mathbf W_C \in H_C \mid \mathbf W_B \in H_B,  W_k>1 ).
\end{equation}
Thus, from \eqref{eqn:39} and \eqref{eqn:40}, we have for any $k \in B$, $\mathbf W_A \ind \mathbf W_C \mid \mathbf W_B, W_k>1$.

In summary,
\begin{equation}
    \label{eqn:ind3}
   \forall k \in \{1,\ldots,d\}: \mathbf W_A\ind \mathbf W_C \mid \mathbf W_B, W_k>1.
\end{equation}

From  \eqref{eqn:ind3}, obviously, we have
\begin{equation}
   \label{eqn:41}
    P(\mathbf W_C \in H_C \mid \mathbf W_A \in H_A,\mathbf  W_B \in H_B^2)=P(\mathbf W_C \in H_C \mid \mathbf  W_B \in H_B^2).
\end{equation}

We then show that  for any $H_A^1, H_B^1, H_C^1, H_A^2, H_C^2$, we have
\begin{equation}
\label{eqn:cons3}
\dfrac{P(\mathbf W_A \in H_A^1, \mathbf W_C\in H_C^1\mid \mathbf W_B\in H_B^1 )}{P(\mathbf W_A\in H_A^2, \mathbf W_C\in H_C^1\mid \mathbf W_B \in H_B^1)} = \dfrac{P(\mathbf W_A \in H_A^1, \mathbf W_C\in H_C^2\mid \mathbf W_B \in H_B^1)}{P(\mathbf W_A\in H_A^2,\mathbf  W_C\in H_C^2\mid \mathbf W_B \in H_B^1)}.
\end{equation}

To prove \eqref{eqn:cons3}, note that
\begin{flalign*}
& \ \ \dfrac{P(\mathbf W_A \in H_A^1, \mathbf W_C\in H_C^2\mid \mathbf W_B \in H_B^1)}{P(\mathbf W_A\in H_A^2,\mathbf  W_C\in H_C^2\mid \mathbf W_B \in H_B^1)}P(\mathbf W_A\in H_A^2, \mathbf W_C\in H_C^1\mid \mathbf W_B\in H_B^1)\\ &= \dfrac{P(\mathbf W_A \in H_A^1 \mid \mathbf W_C\in H_C^2,\mathbf W_B\in H_B^1)}{P(\mathbf W_A\in H_A^2 \mid \mathbf W_C\in H_C^2,\mathbf W_B\in H_B^1)}P(\mathbf W_A\in H_A^2,\mathbf  W_C\in H_C^1\mid \mathbf W_B\in H_B^1) \\
&= \dfrac{P(\mathbf W_A \in H_A^1 \mid \mathbf W_C\in L_C^2, \mathbf W_B\in H_B^1)}{P(\mathbf W_A\in H_A^2 \mid \mathbf W_C\in L_C^2, \mathbf W_B\in H_B^1)} P(\mathbf W_A\in H_A^2, \mathbf W_C\in H_C^1\mid \mathbf W_B\in H_B^1) \quad \text{due to } \eqref{eqn:ind3} \\
&= \dfrac{P(\mathbf W_A\in H_A^2, \mathbf W_C\in H_C^1 \mid \mathbf W_B\in H_B^1)}{P(\mathbf W_A\in H_A^2, \mathbf W_C\in L_C^2 \mid \mathbf W_B\in H_B^1)}  P(\mathbf W_A \in H_A^1, \mathbf W_C\in L_C^2 \mid \mathbf W_B\in H_B^1) \\
&= \dfrac{P(\mathbf W_A \in L_A^2,\mathbf  W_C\in H_C^1\mid \mathbf W_B\in H_B^1)}{P(\mathbf W_A \in L_A^2, \mathbf W_C \in L_C^2 \mid \mathbf W_B\in H_B^1)}  P(\mathbf W_A \in H_A^1,\mathbf  W_C \in L_C^2\mid \mathbf W_B\in H_B^1) \quad \text{due to } \eqref{eqn:ind3}\\
&=P(\mathbf W_A \in H_A^1, \mathbf W_C\in H_C^1\mid \mathbf W_B\in H_B^1 )\quad \text{due to } \eqref{eqn:definition3}.
\end{flalign*}

We then show that
\begin{equation}
    \label{eqn:42}
    P(\mathbf W_C \in L_C^2 \mid \mathbf W_A \in L_A^2, \mathbf W_B\in H_B^1) = P(\mathbf W_C \in L_C^2 \mid \mathbf W_A \in L_A^1, \mathbf W_B\in H_B^1).
\end{equation}
To prove \eqref{eqn:42}, note that due to \eqref{eqn:cons3},
$$
        \dfrac{P(\mathbf W_A \in L_A^1, \mathbf W_C \in L_C^1\mid \mathbf W_B\in H_B^1)}{P(\mathbf W_A \in L_A^2, \mathbf W_C \in L_C^1\mid \mathbf W_B\in H_B^1)}=\dfrac{P(\mathbf W_A \in L_A^1, \mathbf W_C \in L_C^2\mid \mathbf W_B\in H_B^1)}{P(\mathbf W_A \in L_A^2, \mathbf W_C \in L_C^2\mid \mathbf W_B\in H_B^1)}.
$$
Then
\begin{flalign*}
\text{LHS of } \eqref{eqn:42} &= \dfrac{P(\mathbf W_A \in L_A^2, \mathbf W_C \in L_C^2, \mathbf W_B\in H_B^1)}{P(\mathbf W_A \in L_A^2,\mathbf W_B\in H_B^1)} \\
&= \dfrac{P(\mathbf W_A \in L_A^2, \mathbf W_C \in L_C^2,\mathbf W_B\in H_B^1)}{P(\mathbf W_A \in L_A^2,\mathbf W_C \in L_C^2,\mathbf W_B\in H_B^1) + P(\mathbf W_A \in L_A^2, \mathbf W_C \in L_C^1, \mathbf W_B\in H_B^1)}  \\
&= \dfrac{P(\mathbf W_A \in L_A^1, \mathbf W_C \in L_C^2\mid \mathbf W_B\in H_B^1)}{P(\mathbf W_A \in L_A^1,\mathbf W_C \in L_C^2\mid \mathbf W_B\in H_B^1) + P(\mathbf W_A \in L_A^1, \mathbf W_C \in L_C^1 \mid \mathbf W_B\in H_B^1)}  \\
&= \dfrac{P(\mathbf W_A \in L_A^1, \mathbf W_C \in L_C^2\mid \mathbf W_B\in H_B^1)}{P(\mathbf W_A \in L_A^1\mid \mathbf W_B\in H_B^1)}  \\
&= P(\mathbf W_C \in L_C^2\mid  \mathbf W_A \in L_A^1, \mathbf W_B\in H_B^1) = \text{RHS of } \eqref{eqn:42}.
\end{flalign*}
Similarly we have
\begin{flalign*}
P(\mathbf W_C \in L_C^1\mid \mathbf W_A \in L_A^2,\mathbf W_B\in H_B^1) &= P(\mathbf W_C \in L_C^1\mid \mathbf W_A \in L_A^1,\mathbf W_B\in H_B^1).
\end{flalign*}

We then show that
\begin{equation*}
    P(\mathbf W_C\in H_C^2\mid \mathbf W_A \in L_A^2,\mathbf W_B\in H_B^1) = P(\mathbf W_C \in H_C^2 \mid \mathbf W_A \in L_A^1,\mathbf W_B\in H_B^1).
\end{equation*}
To see this, due to \eqref{eqn:cons3},
\begin{flalign*}
\dfrac{P(\mathbf W_A \in L_A^1, \mathbf W_C\in H_C^1\mid \mathbf W_B\in H_B^1)}{P(\mathbf W_A \in L_A^2, \mathbf W_C\in H_C^1\mid \mathbf W_B\in H_B^1)} = \dfrac{P(\mathbf W_A \in L_A^1, \mathbf W_C \in L_C^2\mid \mathbf W_B\in H_B^1)}{P(\mathbf W_A \in L_A^2, \mathbf W_C \in L_C^2\mid \mathbf W_B\in H_B^1)}.
\end{flalign*}
Furthermore, due to \eqref{eqn:42},
$$
    \dfrac{P(\mathbf W_C\in H_C^1\mid\mathbf  W_A \in L_A^1, \mathbf W_B\in H_B^1)}{P(\mathbf W_C\in H_C^1\mid \mathbf W_A \in L_A^2,\mathbf  W_B\in H_B^1)} = \dfrac{P(\mathbf W_C \in L_C^2\mid \mathbf W_A \in L_A^1, \mathbf W_B\in H_B^1)}{P(\mathbf W_C \in L_C^2\mid \mathbf W_A \in L_A^2, \mathbf W_B\in H_B^1)} = 1.
$$
Due to \eqref{eqn:ind3} and \eqref{eqn:cons3},
it follows directly that
\begin{multline}
\label{eqn:43}
P(\mathbf W_C\in H_C^2\mid \mathbf W_A \in L_A^1, \mathbf W_B\in H_B^1) \\
= P(\mathbf W_C\in H_C^2\mid \mathbf W_A \in L_A^2, \mathbf W_B\in H_B^1) \\
= P(\mathbf W_C\in H_C^2\mid \mathbf W_A \in H_A^2, \mathbf W_B\in H_B^1) \\
= P(\mathbf W_C\in H_C^2\mid \mathbf W_B\in H_B^1).
\end{multline}
Similarly we have
\begin{multline}
\label{eqn:44}
P(\mathbf W_C\in H_C^1\mid \mathbf W_A \in L_A^1, \mathbf W_B\in H_B^1) \\= P(\mathbf W_C\in H_C^1 \mid \mathbf W_A \in H_A^2, \mathbf W_B\in H_B^1) = P(\mathbf W_C\in H_C^1\mid  \mathbf W_B\in H_B^1).
\end{multline}

We still need to show
\begin{flalign}
P(\mathbf W_C \in H_C^2 \mid \mathbf W_A \in H_A^1, \mathbf W_B\in H_B^1) &= P(\mathbf W_C \in H_C^2 \mid \mathbf W_A \in L_A^1, \mathbf W_B\in H_B^1), \label{eqn:45}\\
P(\mathbf W_C \in H_C^1 \mid \mathbf W_A \in H_A^1, \mathbf W_B\in H_B^1) &= P(\mathbf W_C \in H_C^1 \mid \mathbf W_A \in L_A^1, \mathbf W_B\in H_B^1).\label{eqn:46}
\end{flalign}

To show \eqref{eqn:45}, note that due to \eqref{eqn:ind3} and \eqref{eqn:cons3},
\begin{small}
\begin{flalign*}
    & \ \ P(\mathbf W_C \in H_C^2\mid \mathbf W_A \in H_A^1, \mathbf W_B\in H_B^1)    \\
    &= \dfrac{P(\mathbf W_C \in H_C^2, \mathbf W_A \in H_A^1, \mathbf W_B\in H_B^1)}{P(\mathbf W_A \in H_A^1, \mathbf W_B\in H_B^1)} \\
    &= \dfrac{ \lambda  P (\mathbf Y_C \in H_C^2, \mathbf Y_C\in L_C^2, \mathbf Y_A \in H_A^1,  \mathbf Y_B\in H_B^1)}{P(\mathbf W_A \in H_A^1, \mathbf W_B\in H_B^1, \mathbf W_C\in L_C^1) + P(\mathbf W_A \in H_A^1, \mathbf W_B\in H_B^1,\mathbf  W_C\in L_C^2)} \\
    &= \dfrac{ \lambda   P(\mathbf Y_C \in H_C^2 \mid  \mathbf Y_B\in H_B^1, \mathbf Y_C\in L_C^2) P(\mathbf Y_A \in H_A^1,  \mathbf Y_B\in H_B^1, \mathbf Y_C\in L_C^2)}{ \dfrac{P(\mathbf W_A \in H_A^1, \mathbf W_B\in H_B^1, \mathbf W_C\in L_C^2){P(\mathbf W_A \in L_A^2, \mathbf W_B\in H_B^1, \mathbf W_C\in L_C^1)}}{P(\mathbf W_A \in L_A^2, \mathbf W_B\in H_B^1, \mathbf W_C\in L_C^2)} + P(\mathbf W_A \in H_A^1, \mathbf W_B\in H_B^1, \mathbf W_C\in L_C^2)} \\
    &= \dfrac{   P(\mathbf Y_C \in H_C^2 \mid  \mathbf Y_B\in H_B^1,\mathbf  Y_C\in L_C^2) P(\mathbf W_A \in H_A^1,  \mathbf W_B\in H_B^1, \mathbf W_C\in L_C^2)}{ \dfrac{P( \mathbf W_A \in H_A^1,  \mathbf W_B\in H_B^1,\mathbf W_C\in L_C^2){P(\mathbf W_A \in L_A^2, \mathbf W_B\in H_B^1, \mathbf W_C\in L_C^1)}}{P(\mathbf W_A \in L_A^2,  \mathbf W_B\in H_B^1,\mathbf  W_C\in L_C^2)} + P(\mathbf W_A \in H_A^1, \mathbf W_B\in H_B^1, \mathbf W_C\in L_C^2)} \\
      &= \dfrac{   P(\mathbf Y_C \in H_C^2 \mid  \mathbf Y_B\in H_B^1, \mathbf Y_C\in L_C^2) }{ \dfrac{{P(\mathbf W_A \in L_A^2, \mathbf W_B\in H_B^1,\mathbf  W_C\in L_C^1)}}{P(\mathbf W_A \in L_A^2,\mathbf  W_B\in H_B^1, \mathbf W_C\in L_C^2)} + 1}
\end{flalign*}
\end{small}
does not depend on $H_A^1$.

To show \eqref{eqn:46}, note that due to \eqref{eqn:definition3} and \eqref{eqn:cons3},
\begin{small}
\begin{flalign*}
& \ \ P(\mathbf W_C \in H_C^1 \mid \mathbf W_A \in H_A^1, \mathbf W_B\in H_B^1) \\ &= \dfrac{P(\mathbf W_A \in H_A^1, \mathbf W_B\in H_B^1, \mathbf W_C \in H_C^1)}{P(\mathbf W_A \in H_A^1, \mathbf W_B\in H_B^1, \mathbf W_C \in L_C^1) + P(\mathbf W_A \in H_A^1, \mathbf W_B\in H_B^1, \mathbf W_C \in L_C^2)} \\
&= \dfrac{ \dfrac{P(\mathbf W_A \in H_A^1,\mathbf  W_B\in H_B^1, \mathbf W_C \in L_C^2){P(\mathbf W_A \in L_A^2,\mathbf  W_B\in H_B^1, \mathbf W_C \in H_C^1)}}{P(\mathbf W_A \in L_A^2, \mathbf W_B\in H_B^1, \mathbf W_C \in L_C^2)} }{\dfrac{P(\mathbf W_A \in H_A^1, \mathbf W_B\in H_B^1, \mathbf W_C \in L_C^2){P(\mathbf W_A \in L_A^2, \mathbf W_B\in H_B^1, \mathbf W_C \in L_C^1)}}{P(\mathbf W_A \in L_A^2,\mathbf  W_B\in H_B^1, \mathbf W_C \in L_C^2)} + P(\mathbf W_A \in H_A^1, \mathbf W_B\in H_B^1, \mathbf W_C \in L_C^2)} \\
&= \dfrac{ \dfrac{{P(\mathbf W_A \in L_A^2, \mathbf W_B\in H_B^1, \mathbf W_C \in H_C^1)}}{P(\mathbf W_A \in L_A^2, \mathbf W_B\in H_B^1, \mathbf W_C \in L_C^2)} }{\dfrac{{P(\mathbf W_A \in L_A^2, \mathbf W_B\in H_B^1, \mathbf W_C \in L_C^1)}}{P(\mathbf W_A \in L_A^2,\mathbf  W_B\in H_B^1, \mathbf W_C \in L_C^2)} + 1}
\end{flalign*}
\end{small}
does not depend on $H_A^1$.

From \eqref{eqn:43}, \eqref{eqn:44}, \eqref{eqn:45} and \eqref{eqn:46}, we have
\begin{equation*}
        P( \mathbf W_C \in H_C \mid \mathbf W_A \in H_A, \mathbf W_B \in H_B^1)=P(\mathbf W_C \in H_C \mid  \mathbf W_B \in H_B^1).
\end{equation*}
Besides, because of \eqref{eqn:41}, we can obtain that
\begin{equation*}
        P(\mathbf W_C \in H_C \mid \mathbf W_A \in H_A, \mathbf W_B \in H_B)=P(\mathbf W_C \in H_C \mid  \mathbf W_B \in H_B).
\end{equation*}

 We have hence completed the proof.

\end{proof}

\begin{comment}
\subsection{Proof of Proposition \ref{prop:eh} in the case where  $B=\emptyset$ and $|A|=|C|=1$ }

	We need to show
	\begin{equation}
	\label{eqn:claim31}
\text{the support of} \,\mathbf Y=(Y_1,Y_2)\,  \text{is}\, \Omega = \left( \{0\} \times (1,\infty) \right) \cup \left( (1,\infty) \times \{0\} \right)\Rightarrow Y_1 \ind_i Y_2,
	\end{equation}
		\begin{equation}
	\label{eqn:claim32}
	\text{the support of} \,\mathbf Y=(Y_1,Y_2)\,  \text{is}\, \Omega = \left( \{0\} \times (1,\infty) \right) \cup \left( (1,\infty) \times \{0\} \right)\Rightarrow Y_1 \ind_e Y_2,
	\end{equation}
			\begin{equation}
	\label{eqn:claim33}
	\text{the support of} \,\mathbf Y=(Y_1,Y_2)\,  \text{is}\, \Omega = \left( \{0\} \times (1,\infty) \right) \cup \left( (1,\infty) \times \{0\} \right)\Rightarrow Y_1 \ind_o Y_2.
	\end{equation}
Under \eqref{eqn:claim31}, \eqref{eqn:claim32} and  \eqref{eqn:claim33}, we can have if the support of $\mathbf Y=(Y_1,Y_2)$  is $\Omega = \left( \{0\} \times (1,\infty) \right) \cup \left( (1,\infty) \times \{0\} \right)$, then
\[
Y_1 \ind_o Y_2  \Leftrightarrow Y_1 \ind_i Y_2 \Leftrightarrow Y_1 \ind_e Y_2.
\]

First, we  show \eqref{eqn:claim31}. Since the support of $\mathbf Y=(Y_1,Y_2)$  is $\Omega = \left( \{0\} \times (1,\infty) \right) \cup \left( (1,\infty) \times \{0\} \right)$, then $S_1\times S_2 \subset \left( \{0\} \times (1,\infty) \right)$ or $S_1\times S_2 \subset \left( (1,\infty) \times \{0\} \right)$.
When $S_1\times S_2 \subset \left( \{0\} \times (1,\infty) \right)$, since $Y_1=0$ regardless of the value of $Y_2$, then $Y_1 \ind Y_2 \mid Y \in S_1 \times S_2$.
When $S_1\times S_2 \subset \left( (1,\infty) \times \{0\} \right)$, since $Y_2=0$ regardless of the value of $Y_1$, then $Y_1 \ind Y_2 \mid Y \in S_1 \times S_2$.
Thus, based on the definition of $Y_1 \ind_i Y_2$, we have $Y_1 \ind_i Y_2$.

Second, we show \eqref{eqn:claim32}. Since the support of $\mathbf Y=(Y_1,Y_2)$  is $\Omega = \left( \{0\} \times (1,\infty) \right) \cup \left( (1,\infty) \times \{0\} \right)$, then when $Y_1>1$, $\mathbf Y \in \left( (1,\infty) \times \{0\} \right)$. Since $Y_2=0$ regardless of the value of $Y_1$, then $Y_1 \ind Y_2 \mid Y_1>1$.  When $Y_2>1$, $\mathbf Y \in  \left( \{0\} \times (1,\infty) \right)$. Since $Y_1=0$ regardless of the value of $Y_2$, then $Y_1 \ind Y_2 \mid Y_2>1$. Thus, based on the definition of $Y_1 \ind_e Y_2$, we have $Y_1 \ind_e Y_2$.

Third, we show \eqref{eqn:claim33}.
Since the support of $\mathbf Y=(Y_1,Y_2)$  is $\Omega = \left( \{0\} \times (1,\infty) \right) \cup \left( (1,\infty) \times \{0\} \right)$, then
\[
P(Y_1>1, Y_2=0 \,\text{or}\, Y_1=0, Y_2>1)=P(Y_1>1, Y_2=0) + P(Y_1=0, Y_2>1)=1.
\]
Suppose $P(Y_1>1,Y_2=0)=\alpha $ $(0<\alpha<1)$,  then $P(Y_1=0,Y_2>1)=1-\alpha$.
Now we introduce two new random variables $W_1$ and $W_2$, assume that they have the distributions as follows:
\[
F(w_1)=
\begin{cases}
(1-p_1)w_1, \hspace{14em}   0 \leq  w_1 \leq 1,\\
(1-p_1)+P(Y_1\leq w_1 \mid Y_1>1, Y_2=0)p_1, \quad w_1>1,
\end{cases}
\]
and
\[
F(w_2)=
\begin{cases}
	(1-p_2)w_2, \hspace{14em}  0 \leq  w_2 \leq 1,\\
	(1-p_2)+P(Y_2\leq w_2 \mid Y_1=0, Y_2>1)p_2, \quad w_2>1,
\end{cases}
\]
where $p_1 \in (0,1)$ and $p_2 \in (0,1)$.
To make $W_1 \ind W_2$, let
\[
\begin{aligned}
&F(w_1,w_2)=F(w_1)F(w_2)=\\
&\begin{cases}
&(1-p_1)(1-p_2)w_1w_2, \hspace{26em} 0 \leq  w_1 \leq 1, 0 \leq  w_2 \leq 1,\\
&(1-p_1)w_1\{(1-p_2)+P(Y_2\leq w_2 \mid Y_1=0, Y_2>1)p_2\}, \hspace{12em}    0 \leq  w_1 \leq 1, w_2>1,\\
&\{(1-p_1)+P(Y_1\leq w_1 \mid Y_1>1, Y_2=0)p_1\}(1-p_2)w_2, \hspace{12em}  w_1>1, 0 \leq  w_2 \leq 1,\\
&\{(1-p_1)+P(Y_1\leq w_1 \mid Y_1>1, Y_2=0)p_1\}\{(1-p_2)+P(Y_2\leq w_2 \mid Y_1=0, Y_2>1)p_2\},  w_1>1, w_2>1.
\end{cases}
\end{aligned}
\]

  If the density function of $F(w_1,w_2)$ exists, then
\[
f(w_1,w_2)=
\begin{cases}
(1-p_1)(1-p_2), \hspace{11.5em} 0 \leq  w_1 \leq 1, 0 \leq  w_2 \leq 1,\\
(1-p_1)p_2\frac{\partial P(Y_1=0,Y_2 \leq w_2)}{\partial w_2}\frac{1}{1-\alpha},  \hspace{6em}    0 \leq  w_1 \leq 1, w_2>1,\\
p_1(1-p_2)\frac{\partial P(Y_1\leq w_1,Y_2 = 0)}{\partial w_1}\frac{1}{\alpha},  \hspace{6.9em}     w_1>1, 0 \leq w_2 \leq 1,\\
p_1p_2\frac{\partial P(Y_1\leq w_1,Y_2 = 0)}{\partial w_1}\frac{\partial P(Y_1=0,Y_2 \leq w_2)}{\partial w_2}\frac{1}{\alpha(1-\alpha)}, \quad w_1>1, w_2>1.
\end{cases}
\]
Since the support of $\mathbf Y = (Y_1,Y_2) $ is $\Omega = \left( \{0\} \times (1,\infty) \right) \cup \left( (1,\infty) \times \{0\} \right)$, then $\partial P(Y_1=0,Y_2 \leq w_2)/\partial w_2 \neq 0$ for $w_2 >1$ and $\partial P(Y_1\leq w_1,Y_2 = 0)/\partial w_1\neq 0$ for $w_2>1$. Thus, from $f(w_1,w_2)$, we have the support of $\mathbf W=(W_1,W_2)$ is $[0,\infty)^2$, which contains  $\Omega$.  Based on the definition of conditional density in \citet[\S 7.4]{shorack2017probability}, we have that the conditional density at $(w_1,0)$, $(w_1>1)$ and $(0,w_2)$, $(w_2>1)$ given that all the points are on the curve of $\Omega$ are
\[
\begin{aligned}
\frac{f(w_1,0)}{ \int_1^\infty f(w_1,0) {\rm d} w_1  + \int_1^\infty f(0, w_2) {\rm d} w_2  }=&\frac{ p_1(1-p_2)\frac{\partial P(Y_1\leq w_1,Y_2 = 0)}{\partial w_1}\frac{1}{\alpha}          }{ \frac{p_1(1-p_2)}{\alpha}P(Y_1>1,Y_2=0)+\frac{(1-p_1)p_2}{1-\alpha}P(Y_1=0,Y_2>1)}\\
=&\frac{p_1(1-p_2)\frac{\partial P(Y_1\leq w_1, Y_2=0 )}{\partial w_1}
}{\alpha\{p_1(1-p_2)+ (1-p_1)p_2 \}   },
\end{aligned}
\]
and
\[
\begin{aligned}
\frac{f(0,w_2)}{ \int_1^\infty f(w_1,0) {\rm d} w_1  + \int_1^\infty f(0, w_2) {\rm d} w_2  }=&\frac{(1-p_1)p_2\frac{\partial P(Y_1=0,Y_2 \leq w_2)}{\partial w_2}\frac{1}{1-\alpha}        }{ \frac{p_1(1-p_2)}{\alpha}P(Y_1>1,Y_2=0)+\frac{(1-p_1)p_2}{1-\alpha}P(Y_1=0,Y_2>1)}\\
=&\frac{(1-p_1)p_2\frac{\partial P(Y_1=0,Y_2\leq w_2)}{\partial w_2} }{(1-\alpha)\{p_1(1-p_2)+ (1-p_1)p_2 \}   }.
\end{aligned}
\]

Then we can get the value of $P(W_1\leq w_1, W_2\leq w_2\mid W_1>1, W_2=0 \, \text{or}\, W_1=0, W_2>1)$. When $0 \leq w_1 \leq 1$, $0 \leq w_2 \leq 1$,
\[
P(W_1\leq w_1, W_2\leq w_2\mid W_1>1, W_2=0 \, \text{or}\, W_1=0, W_2>1)=0=P(Y_1 \leq w_1, Y_2 \leq w_2).
\]
When $0 \leq w_1 \leq 1$, $w_2 > 1$,
\[
\begin{aligned}
P(W_1\leq w_1, W_2\leq w_2\mid W_1>1, W_2=0 \, \text{or}\, W_1=0, W_2>1)=&\frac{ \int _1^{w_2} f(0,t) {\rm d} t      }{\int _1^\infty f(w_1,0){\rm d} w_1+\int_1^\infty f(0,w_2){\rm d} w_2}\\
=&\frac{(1-p_1)p_2P(Y_1=0,Y_2 \leq w_2)  }{(1-\alpha)\{p_1(1-p_2)+(1-p_1)p_2  \}}\\
=&\frac{(1-p_1)p_2P(Y_1\leq w_1,Y_2 \leq w_2)  }{(1-\alpha)\{p_1(1-p_2)+(1-p_1)p_2  \}},
\end{aligned}
\]
where the last equation is due to the support of $\mathbf Y=(Y_1,Y_2)$  being $\Omega = \left( \{0\} \times (1,\infty) \right) \cup \left( (1,\infty) \times \{0\} \right)$.
When $w_1 > 1$, $0 \leq w_2 \leq 1$,
\[
\begin{aligned}
P(W_1\leq w_1, W_2\leq w_2\mid W_1>1, W_2=0 \, \text{or}\, W_1=0, W_2>1)=&\frac{ \int _1^{w_1} f(t,0) {\rm d} t      }{\int _1^\infty f(w_1,0){\rm d} w_1+\int_1^\infty f(0,w_2){\rm d} w_2}\\
=&\frac{p_1(1-p_2)P(Y_1 \leq w_1,Y_2 = 0)  }{\alpha\{p_1(1-p_2)+(1-p_1)p_2  \}}\\
=&\frac{p_1(1-p_2)P(Y_1\leq w_1,Y_2 \leq w_2)  }{\alpha\{p_1(1-p_2)+(1-p_1)p_2  \}}.
\end{aligned}
\]
When $w_1 > 1$, $w_2 > 1$, due to $P(Y_1 \leq w_1, Y_2 \leq w_2)=P(Y_1 \leq w_1, Y_2 = 0)+P(Y_1 = 0, Y_2 \leq w_2)$,
 \[
 \begin{aligned}
& P(W_1\leq w_1, W_2\leq w_2\mid W_1>1, W_2=0 \, \text{or}\, W_1=0, W_2>1)\\
 =&\frac{ \int _1^{w_1} f(t,0) {\rm d} t      }{\int _1^\infty f(w_1,0){\rm d} w_1+\int_1^\infty f(0,w_2){\rm d} w_2}+\frac{ \int _1^{w_2} f(0,t) {\rm d} t      }{\int _1^\infty f(w_1,0){\rm d} w_1+\int_1^\infty f(0,w_2){\rm d} w_2}\\
 =&\frac{p_1(1-p_2)P(Y_1 \leq w_1,Y_2 = 0)  }{\alpha\{p_1(1-p_2)+(1-p_1)p_2  \}}+\frac{(1-p_1)p_2P(Y_1=0,Y_2 \leq w_2)  }{(1-\alpha)\{p_1(1-p_2)+(1-p_1)p_2  \}}\\
 =&\frac{\{(1-\alpha)p_1(1-p_2)-\alpha (1-p_1)p_2\}P(Y_1 \leq w_1,Y_2 = 0)  }{\alpha(1-\alpha)\{p_1(1-p_2)+(1-p_1)p_2  \}}+\frac{(1-p_1)p_2P(Y_1\leq w_1,Y_2 \leq w_2)  }{(1-\alpha)\{p_1(1-p_2)+(1-p_1)p_2  \}}.
 \end{aligned}
 \]
In summary,
\begin{equation}
\label{eqn:48}
\begin{aligned}
&P(W_1\leq w_1, W_2\leq w_2\mid W_1>1, W_2=0 \, \text{or}\, W_1=0, W_2>1)=\\
&\begin{cases}
&0, \hspace{27.7em} 0 \leq  w_1 \leq 1, 0 \leq  w_2 \leq 1,\\
&\frac{ (1-p_1)p_2P(Y_1 \leq w_1, Y_2 \leq w_2) }{(1-\alpha)\{ p_1(1-p_2)+(1-p_1)p_2 \}   }, \hspace{18.2em} 0 \leq  w_1 \leq 1, w_2>1,\\
&\frac{ p_1(1-p_2)P(Y_1 \leq w_1, Y_2 \leq w_2) }{\alpha\{ p_1(1-p_2)+(1-p_1)p_2 \}   }, \hspace{18.7em}   w_1 >1, 0 \leq w_2 \leq 1,\\
& \frac{\{(1-\alpha)p_1(1-p_2) - \alpha(1-p_1)p_2 \} P(Y_1 \leq w_1, Y_2=0)}{\alpha(1-\alpha)\{ p_1(1-p_2)+(1-p_1)p_2  \} }+\frac{\{(1-p_1)p_2 P(Y_1 \leq w_1, Y_2\leq w_2)}{(1-\alpha)\{ p_1(1-p_2)+(1-p_1)p_2  \} }, \quad w_1>1,w_2>1.
\end{cases}
\end{aligned}
\end{equation}
From \eqref{eqn:48},
 $P(W_1\leq w_1, W_2\leq w_2\mid W_1>1, W_2=0 \, \text{or}\, W_1=0, W_2>1)=P(Y_1\leq w_1, Y_2 \leq w_2)$, if and only if
\[
\begin{cases}
\frac{(1-p_1)p_2}{(1-\alpha)\{p_1(1-p_2)+(1-p_1)p_2  \}  }=1,\\
\frac{p_1(1-p_2)}{\alpha\{p_1(1-p_2)+(1-p_1)p_2  \}  }=1,\\
\frac{\{(1-\alpha)p_1(1-p_2) - \alpha(1-p_1)p_2 \} }{\alpha(1-\alpha)\{ p_1(1-p_2)+(1-p_1)p_2  \} }=0,
\end{cases}
\]
or equivalently,
\begin{equation}
\label{eqn:49}
\alpha=\frac{p_1(1-p_2)}{p_1(1-p_2)+(1-p_1)p_2}.
\end{equation}
We have hence shown that under \eqref{eqn:49}, $(\mathbf W \mid \mathbf W\in \Omega) \,{\buildrel d \over =}\, \mathbf Y$ if the density function of $F(w_1,w_2)$ exists.

From $f(w_1,w_2)$, we can also get the conditional density at $(w_1,0)$, $(w_1>1)$ and $(0,w_2)$, $(w_2>1)$ given that all the points are on the curve of $((0,\infty)\times \{0\})\cup \{0\} \times (0,\infty)$. When $0 \leq w_1 \leq 1$,
\[
\begin{aligned}
&\frac{f(w_1,0)}{ \int_0^\infty f(w_1,0) {\rm d} w_1  + \int_0^\infty f(0, w_2) {\rm d} w_2  }\\
=&\frac{(1-p_1)(1-p_2)}{\int_0^1 f(w_1,0){\rm d}w_1+\int_1^\infty f(w_1,0){\rm d}w_1+\int_0^1 f(0,w_2){\rm d}w_2+\int_1^\infty f(0,w_2){\rm d} w_2}\\
=&\frac{(1-p_1)(1-p_2)}{(1-p_1)(1-p_2)+\frac{p_1(1-p_2)}{\alpha}P(Y_1>1,Y_2=0)+(1-p_1)(1-p_2)+\frac{(1-p_1)p_2}{1-\alpha}P(Y_1=0,Y_2>1)}\\
=&\frac{(1-p_1)(1-p_2)}{2-p_1-p_2}.
\end{aligned}
\]
When $w_1>1$,
\[
\begin{aligned}
&\frac{f(w_1,0)}{ \int_0^\infty f(w_1,0) {\rm d} w_1  + \int_0^\infty f(0, w_2) {\rm d} w_2}\\
=&\frac{p_1(1-p_2)\frac{\partial P(Y_1\leq w_1,Y_2=0)}{\partial w_1}\frac{1}{\alpha}       }{\int_0^1 f(w_1,0){\rm d}w_1+\int_1^\infty f(w_1,0){\rm d}w_1+\int_0^1 f(0,w_2){\rm d}w_2+\int_1^\infty f(0,w_2){\rm d} w_2}\\
=&\frac{p_1(1-p_2)\frac{\partial P(Y_1\leq w_1,Y_2=0)}{\partial w_1}\frac{1}{\alpha}  }{(1-p_1)(1-p_2)+\frac{p_1(1-p_2)}{\alpha}P(Y_1>1,Y_2=0)+(1-p_1)(1-p_2)+\frac{(1-p_1)p_2}{1-\alpha}P(Y_1=0,Y_2>1)}\\
=&\frac{p_1(1-p_2)\frac{\partial P(Y_1\leq w_1,Y_2=0)}{\partial w_1}\frac{1}{\alpha} }{2-p_1-p_2}.
\end{aligned}
\]
Thus, we have
\[
\frac{f(w_1,0)}{ \int_0^\infty f(w_1,0) {\rm d} w_1  + \int_0^\infty f(0, w_2) {\rm d} w_2}=
\begin{cases}
\frac{(1-p_1)(1-p_2)}{2-p_1-p_2}, \hspace{5.6em} 0 \leq w_1 \leq 1,\\
\frac{p_1(1-p_2)\frac{\partial P(Y_1\leq w_1,Y_2=0)}{\partial w_1}\frac{1}{\alpha} }{2-p_1-p_2}, \quad w_1 >1.
\end{cases}
\]

Similarly, we have
\[
\frac{f(0,w_2)}{ \int_0^\infty f(w_1,0) {\rm d} w_1  + \int_0^\infty f(0, w_2) {\rm d} w_2}=
\begin{cases}
\frac{(1-p_1)(1-p_2)}{2-p_1-p_2}, \hspace{6.5em} 0 \leq w_2 \leq 1,\\
\frac{(1-p_1)p_2\frac{\partial P(Y_1=0,Y_2 \leq w_2)}{\partial w_2}\frac{1}{1-\alpha} }{2-p_1-p_2}, \quad w_2 >1.
\end{cases}
\]

Then we can get the value of $P(W_1\leq w_1, W_2\leq w_2\mid W_1>0, W_2=0 \, \text{or}\, W_1=0, W_2>0)$. When $0 \leq w_1 \leq 1$, $0 \leq w_2 \leq 1$,
\[
\begin{aligned}
&P(W_1\leq w_1, W_2\leq w_2\mid W_1>0, W_2=0 \, \text{or}\, W_1=0, W_2>0)\\
=&\frac{\int_0^{w_1} f(t,0) {\rm d} t + \int_0^{w_2} f(0,t) {\rm d} t}{\int_0^\infty f(w_1,0) {\rm d} w_1 + \int_0^\infty f(0,w_2) {\rm d} w_2}\\
=&\frac{(1-p_1)(1-p_2)(w_1+w_2)}{2-p_1-p_2}.
\end{aligned}
\]
When $0 \leq w_1 \leq 1$, $w_2 > 1$,
\[
\begin{aligned}
&P(W_1\leq w_1, W_2\leq w_2\mid W_1>0, W_2=0 \, \text{or}\, W_1=0, W_2>0)\\
=&\frac{\int _0^{w_1} f(t,0) {\rm d} t  + \int _0^{1} f(0,t) {\rm d} t + \int _1^{w_2} f(0,t) {\rm d} t      }{\int _0^\infty f(w_1,0){\rm d} w_1+\int_0^\infty f(0,w_2){\rm d} w_2}\\
=&\frac{(1-p_1)(1-p_2)w_1 + (1-p_1)(1-p_2) +  (1-p_1)p_2P(Y_1=0,Y_2 \leq w_2)\frac{1}{1-\alpha} }{2-p_1-p_2}\\
=&\frac{(1-p_1)(1-p_2)(w_1+1) + \frac{(1-p_1)p_2}{1-\alpha}P(Y_1 \leq w_1,Y_2 \leq w_2)}{2-p_1-p_2}.
\end{aligned}
\]
When $w_1 > 1$, $0 \leq w_2 \leq 1$,
\[
\begin{aligned}
&P(W_1\leq w_1, W_2\leq w_2\mid W_1>0, W_2=0 \, \text{or}\, W_1=0, W_2>0)\\
=&\frac{\int _0^{1} f(t,0) {\rm d} t  + \int _1^{w_1} f(t,0) {\rm d} t + \int _0^{w_2} f(0,t) {\rm d} t}{\int _0^\infty f(w_1,0){\rm d} w_1+\int_0^\infty f(0,w_2){\rm d} w_2}\\
=&\frac{(1-p_1)(1-p_2) +p_1(1-p_2)P(Y_1 \leq w_1, Y_2 = 0)\frac{1}{\alpha} + (1-p_1)(1-p_2)w_2}{2-p_1-p_2}\\
=&\frac{(1-p_1)(1-p_2)(1+w_2)+\frac{p_1(1-p_2)}{\alpha}P(Y_1 \leq w_1, Y_2 \leq w_2)}{2-p_1-p_2}.
\end{aligned}
\]
When $w_1 > 1$, $w_2 > 1$, due to $P(Y_1 \leq w_1, Y_2 \leq w_2)=P(Y_1 \leq w_1, Y_2 = 0)+P(Y_1 = 0, Y_2 \leq w_2)$,
\[
\begin{aligned}
& P(W_1\leq w_1, W_2\leq w_2\mid W_1>0, W_2=0 \, \text{or}\, W_1=0, W_2>0)\\
=&\frac{ \int _0^1 f(t,0) {\rm d} t  + \int _1^{w_1} f(t,0) {\rm d} t + \int _0^1 f(0,t) {\rm d} t  + \int _1^{w_2} f(0,t) {\rm d} t }{\int _0^\infty f(w_1,0){\rm d} w_1+\int_0^\infty f(0,w_2){\rm d} w_2}\\
=&\frac{(1-p_1)(1-p_2) + p_1(1-p_2)P(Y_1 \leq w_1, Y_2 =0)\frac{1}{\alpha} +(1-p_1)(1-p_2) +(1-p_1)p_2P(Y_1=0,Y_2 \leq w_2)\frac{1}{1-\alpha}}{2-p_1-p_2}            \\
=&\frac{2(1-p_1)(1-p_2) + \frac{(1-\alpha)p_1(1-p_2)-\alpha(1-p_1)p_2}{\alpha(1-\alpha)}P(Y_1 \leq w_1, Y_2 =0)+\frac{(1-p_1)p_2}{1-\alpha}(Y_1\leq w_1,Y_2 \leq w_2)}{2-p_1-p_2}.
\end{aligned}
\]
Thus,
\[
\begin{aligned}
& P(W_1\leq w_1, W_2\leq w_2\mid W_1>0, W_2=0 \, \text{or}\, W_1=0, W_2>0)\\
&=\begin{cases}
&\frac{(1-p_1)(1-p_2)(w_1+w_2)}{2-p_1-p_2}, \hspace{22.4em} 0\leq w_1 \leq1, 0\leq w_2 \leq 1,\\
&\frac{(1-p_1)(1-p_2)(w_1+1) + \frac{(1-p_1)p_2}{1-\alpha}P(Y_1 \leq w_1,Y_2 \leq w_2)}{2-p_1-p_2},\hspace{13.1em} 0\leq w_1 \leq1,  w_2>1,\\
&\frac{(1-p_1)(1-p_2)(1+w_2)+\frac{p_1(1-p_2)}{\alpha}P(Y_1 \leq w_1, Y_2 \leq w_2)}{2-p_1-p_2},\hspace{13em} w_1>1, 0\leq w_2 \leq 1,\\
&\frac{2(1-p_1)(1-p_2) + \frac{(1-\alpha)p_1(1-p_2)-\alpha(1-p_1)p_2}{\alpha(1-\alpha)}P(Y_1 \leq w_1, Y_2 =0)+\frac{(1-p_1)p_2}{1-\alpha}(Y_1\leq w_1,Y_2 \leq w_2)}{2-p_1-p_2}, \quad w_1>1, w_2>1.
\end{cases}
\end{aligned}
\]
From the  formula of $P(W_1\leq w_1, W_2\leq w_2\mid W_1>0, W_2=0 \, \text{or}\, W_1=0, W_2>0)$, we can verify that $P(W_1\leq \infty, W_2\leq \infty \mid W_1>0, W_2=0 \, \text{or}\, W_1=0, W_2>0)=1$. Thus, we do not need to add any additional condition for $p_1, p_2$ and $\alpha$.

If the density function of $F(w_1,w_2)$ does not exist, since
$P(W_1\leq w_1, W_2\leq w_2\mid W_1>1, W_2=0 \, \text{or}\, W_1=0, W_2>1)$ can be defined by
\begin{multline*}
P(W_1\leq w_1, W_2\leq w_2\mid W_1>1, W_2=0 \, \text{or}\, W_1=0, W_2>1)P( W_1>1, W_2=0 \, \text{or}\, W_1=0, W_2>1)\\
=P(W_1\leq w_1, W_2\leq w_2, W_1>1, W_2=0 \, \text{or}\, W_1=0, W_2>1),
\end{multline*}
and $P( W_1>1, W_2=0 \, \text{or}\, W_1=0, W_2>1)=P(W_1\leq w_1, W_2\leq w_2, W_1>1, W_2=0 \, \text{or}\, W_1=0, W_2>1)=0$, then
$P(W_1\leq w_1, W_2\leq w_2\mid W_1>1, W_2=0 \, \text{or}\, W_1=0, W_2>1)$ can take any value. If we take
$P(W_1\leq w_1, W_2\leq w_2\mid W_1>1, W_2=0 \, \text{or}\, W_1=0, W_2>1)=P(Y_1\leq w_1, Y_2 \leq w_2)$, then we have $(\mathbf W \mid \mathbf W\in \Omega) \,{\buildrel d \over =}\, \mathbf Y$.

\end{comment}

\subsection{Proof of Proposition \ref{prop:eh} in the case where  $B=\emptyset$ }

\begin{proof}
		Use the same notation as before, denote $[0,1]^{|k|}$ as $L_k^1$, $ [0,\infty)^{|k|} \setminus [0,1]^{|k|} $ as $L_k^2$ for $k \in \{A,C\}$. Besides, denote $\{0\}^{|k|}$ as $\bm{0}_k$.

		We need to show
	\begin{equation}
	\label{eqn:claim41}
	\text{the support of} \,\mathbf Y=(\mathbf Y_A,\mathbf Y_C)\,  \text{is}\, \Omega =\left( L_A^2 \times \bm{0}_{C}\right) \cup  \left( \bm{0}_{A}\times L_C^2  \right)\Rightarrow \mathbf Y_A \ind_i \mathbf Y_C,
	\end{equation}
	\begin{equation}
	\label{eqn:claim42}
	\text{the support of} \,\mathbf Y=(\mathbf Y_A,\mathbf Y_C)\,  \text{is}\, \Omega =\left( L_A^2 \times \bm{0}_{C}\right) \cup  \left( \bm{0}_{A}\times L_C^2  \right)\Rightarrow \mathbf Y_A \ind_e \mathbf Y_C,
	\end{equation}
	\begin{equation}
	\label{eqn:claim43}
	\text{the support of} \,\mathbf Y=(\mathbf Y_A,\mathbf Y_C)\,  \text{is}\, \Omega =\left( L_A^2 \times \bm{0}_{C}\right) \cup  \left( \bm{0}_{A}\times L_C^2  \right)\Rightarrow \mathbf Y_A \ind_o \mathbf Y_C.
	\end{equation}
	Under \eqref{eqn:claim41}, \eqref{eqn:claim42} and  \eqref{eqn:claim43}, we can obtain that if the support of $\mathbf Y=(\mathbf Y_A,\mathbf Y_C)$  is $\Omega =\left( L_A^2 \times \bm{0}_{C}\right) \cup  \left( \bm{0}_{A}\times L_C^2  \right)$, then
	\[
	\mathbf Y_A \ind_o \mathbf Y_C  \Leftrightarrow \mathbf Y_A \ind_i \mathbf Y_C  \Leftrightarrow \mathbf Y_A \ind_e \mathbf Y_C .
	\]

	First, we  show \eqref{eqn:claim41}. Since the support of $\mathbf Y=(\mathbf Y_A,\mathbf Y_C)$  is $\Omega =\left( L_A^2 \times \bm{0}_{C}\right) \cup  \left( \bm{0}_{A}\times L_C^2  \right)$, then $S_A\times S_C \subset \left( L_A^2 \times \bm{0}_{C}\right) $ or $S_A\times S_C \subset \left( \bm{0}_{A}\times L_C^2  \right)$.
	When $S_A\times S_C \subset \left( L_A^2 \times \bm{0}_{C}\right) $, since $\mathbf Y_C=\bm{0}_{C}$ regardless of the value of $\mathbf Y_A$, then $\mathbf Y_A \ind \mathbf Y_C \mid \left(\mathbf Y \in S_A \times S_C\right)$.
	When $S_A\times S_C \subset  \left( \bm{0}_{A}\times L_C^2  \right)$, since $\mathbf Y_A=\bm{0}_{A}$ regardless of the value of $\mathbf Y_C$, then $\mathbf Y_A \ind \mathbf Y_C \mid \left(\mathbf Y \in S_A \times S_C\right)$.
	Thus, based on the definition of $\mathbf Y_A \ind_i \mathbf Y_C$, we have $\mathbf Y_A \ind_i \mathbf Y_C$.

	Second, we show \eqref{eqn:claim42}. Since the support of $\mathbf Y=(\mathbf Y_A,\mathbf Y_C)$  is $\Omega =\left( L_A^2 \times \bm{0}_{C}\right) \cup  \left( \bm{0}_{A}\times L_C^2  \right)$, then when $Y_k>1$ for some $k \in A$, we have $\mathbf Y \in\left( L_A^2 \times \bm{0}_{C}\right)$. Since $\mathbf Y_C=\bm{0}_{C}$ regardless of the value of $\mathbf Y_A$, then $\mathbf Y_A \ind \mathbf Y_C \mid Y_k>1$.  When $Y_k>1$ for some $k\in C$, we have $\mathbf  Y \in \left( \bm{0}_{A}\times L_C^2  \right)$. Since $\mathbf Y_A=\bm{0}_{A}$ regardless of the value of $\mathbf Y_C$, then $\mathbf Y_A \ind \mathbf Y_C \mid Y_k>1$. Thus, based on the definition of $\mathbf Y_A \ind_e \mathbf Y_C$, we have $\mathbf Y_A \ind_e \mathbf Y_C$.

	Third, we show \eqref{eqn:claim43}.  Since the support of $\mathbf{Y}=(\mathbf{Y}_A,\mathbf{Y}_C)$  is $\Omega =\left( L_A^2 \times \bm{0}_{C}\right) \cup  \left( \bm{0}_{A}\times L_C^2  \right)$, then
    \begin{multline*}
	P(\mathbf{Y}_A \in  L_A^2, \mathbf{Y}_C=\bm{0}_C\,\text{or}\, \mathbf{Y}_A=\bm{0}_A, \mathbf{Y}_C \in L_C^2  )\\
	=P(\mathbf{Y}_A \in  L_A^2, \mathbf{Y}_C=\bm{0}_C) + P(\mathbf{Y}_A=\bm{0}_A, \mathbf{Y}_C \in L_C^2)=1.
	\end{multline*}
	Suppose $P(\mathbf{Y}_A \in  L_A^2, \mathbf{Y}_C=\bm{0}_C)=\alpha$ $(0<\alpha<1)$, then $P(\mathbf{Y}_A=\bm{0}_A, \mathbf{Y}_C \in L_C^2)=1-\alpha$. Now we introduce two new random vectors $\mathbf{W}_A$ and $\mathbf W_C$, assume that they have the distributions as follows:
	\[
F_{\mathbf{W}_A}(\mathbf w_A)=
\begin{cases}
	(1-p_A)P(\mathbf U_A\leq \mathbf w_A),& \mathbf w_A \in  L_A^1,\\
	(1-p_A)+p_AP(\mathbf Y_A \leq \mathbf w_A \mid \mathbf Y_A \in L_A^2, \mathbf Y_C=\bm{0}_C), & \mathbf w_A \in L_A^2,
	\end{cases}
	\]
	and
	\[
F_{\mathbf W_C}(\mathbf w_C)=
\begin{cases}
(1-p_C)P(\mathbf U_C \leq \mathbf w_C), &  \mathbf w_C \in L_C^1,\\
(1-p_C)+p_C P(\mathbf Y_C \leq \mathbf w_C \mid \mathbf Y_A=\bm 0_A, \mathbf Y_C \in L_C^2), & \mathbf w_C \in L_C^2,
\end{cases}
	\]
	where $p_A \in (0,1)$, $p_C \in (0,1)$, $\mathbf U_A$ is a random vector which follows the uniform distribution in $L_A^1$, and $\mathbf U_C$ is a random vector which follows the uniform distribution in $L_C^1$.

	To make $\mathbf W_A \ind \mathbf W_C$, let $F(\mathbf w_A,\mathbf w_C)=F_{\mathbf W_A}(\mathbf w_A)F_{\mathbf W_C}(\mathbf w_C)$. Thus,
	if $\mathbf w_A \in L_A^1, \mathbf w_C \in L_C^1$,
	\[
	 F(\mathbf w_A,\mathbf w_C)= (1-p_A)(1-p_C)P(\mathbf U_A\leq \mathbf w_A)P(\mathbf U_C \leq \mathbf w_C).
	\]
	If $\mathbf w_A \in L_A^1, \mathbf w_C \in L_C^2$,
	\[
	 F(\mathbf w_A, \mathbf w_C)=(1-p_A)P(\mathbf U_A \leq \mathbf w_A)\big\{(1-p_C)+p_C P(\mathbf Y_C \leq \mathbf w_C \mid \mathbf Y_A=\bm 0_A, \mathbf Y_C \in L_C^2)\big\}.
	\]
	If $\mathbf w_A \in L_A^2, \mathbf w_C \in L_C^1$,
	\[
	F(\mathbf w_A,\mathbf w_C)=\big\{(1-p_A)+p_A P(\mathbf Y_A \leq \mathbf w_A \mid \mathbf Y_A \in L_A^2, \mathbf Y_C=\bm 0_C)\big\}(1-p_C)P(\mathbf U_C \leq \mathbf w_C).
	\]
	If $\mathbf w_A \in L_A^2, \mathbf w_C \in L_C^2$, then
	\begin{multline*}
	F(\mathbf w_A,\mathbf w_C)=\big\{(1-p_A)+p_A P(\mathbf Y_A \leq \mathbf  w_A \mid \mathbf  Y_A \in L_A^2, \mathbf Y_C=\bm 0_C)\big\}\\
	\times \big\{(1-p_C)+p_C P(\mathbf Y_C \leq \mathbf w_C \mid \mathbf Y_A=\bm 0_A, \mathbf Y_C \in L_C^2)\big\}.
	\end{multline*}

If the density function of $F(\mathbf w_A,\mathbf w_C)$ exists, then
\[
f(\mathbf w_A,\mathbf w_C)=
\begin{cases}
(1-p_A)(1-p_C), & \mathbf w_A \in L_A^1, \mathbf w_C \in L_C^1,\\
\frac{(1-p_A)p_C}{1-\alpha}\frac{\partial P(\mathbf Y_A=\bm 0_A,\mathbf Y_C \leq \mathbf w_C)}{\partial \mathbf w_C},  &   \mathbf w_A \in L_A^1, \mathbf w_C \in L_C^2,\\
\frac{p_A(1-p_C)}{\alpha}\frac{\partial P(\mathbf Y_A\leq \mathbf w_A,\mathbf Y_C = \bm 0_C)}{\partial \mathbf w_A},  &    \mathbf w_A \in L_A^2, \mathbf w_C \in L_C^1,\\
\frac{p_Ap_C}{\alpha(1-\alpha)}\frac{\partial P(\mathbf Y_A\leq \mathbf w_A,\mathbf Y_C = \bm 0_C)}{\partial \mathbf w_A}\frac{\partial P(\mathbf Y_A=\bm 0_A,\mathbf Y_C \leq \mathbf w_C)}{\partial \mathbf w_C}, & \mathbf w_A \in L_A^2, \mathbf w_C \in L_C^2.
\end{cases}
\]

Since the support of $\mathbf Y = (\mathbf Y_A,\mathbf Y_C) $ is $\Omega =\left( L_A^2 \times \bm{0}_{C}\right) \cup  \left( \bm{0}_{A}\times L_C^2  \right)$, then $\partial P(\mathbf Y_A=\bm 0_A,\mathbf Y_C \leq \mathbf w_C)/\partial \mathbf w_C \neq 0$ for $\mathbf w_C  \in L_C^2$ and $\partial P(\mathbf Y_A\leq \mathbf w_A,\mathbf Y_C = \bm 0_C)/\partial \mathbf w_A\neq 0$ for $\mathbf w_A \in L_A^2$. Thus, from $f(\mathbf w_1,\mathbf w_2)$, we have the support of $\mathbf W=(\mathbf W_A,\mathbf W_C)$ is $[0,\infty)^{|A|+|C|}$, which contains  $\Omega$.  Based on the definition of conditional density in \citet[\S 7.4]{shorack2017probability}, we have that the conditional density at $(\mathbf w_A,\bm 0_C)$, $(\mathbf w_A \in L_A^2)$ and $(\bm 0_A,\mathbf w_C)$, $(\mathbf w_C \in L_C^2)$ given that all the points are on the curve of $\Omega$ are
\[
\begin{aligned}
&\frac{f(\mathbf w_A,\bm 0_C)}{ \int_{L_A^2} f(\mathbf w_A,\bm 0_C) {\rm d} \mathbf w_A  + \int_{L_C^2} f(\bm 0_A, \mathbf w_C) {\rm d} \mathbf w_C  }\\
=&\frac{ \frac{p_A(1-p_C)}{\alpha}\frac{\partial P(\mathbf Y_A\leq \mathbf w_A,\mathbf Y_C = \bm 0_C)}{\partial \mathbf w_A}}{ \frac{p_A(1-p_C)}{\alpha}P(\mathbf Y_A \in L_A^2,\mathbf Y_C=\bm 0_C)+\frac{(1-p_A)p_C}{1-\alpha}P(\mathbf Y_A=\bm 0_A,\mathbf Y_C\in L_C^2)}\\
=&\frac{p_A(1-p_C)\frac{\partial P(\mathbf Y_A\leq \mathbf w_A, \mathbf Y_C=\bm 0_C )}{\partial \mathbf w_A}
}{\alpha\{p_A(1-p_C)+ (1-p_A)p_C \}   },
\end{aligned}
\]
and
\[
\begin{aligned}
&\frac{f(\bm 0_A,\mathbf w_C)}{ \int_{L_A^2} f(\mathbf w_A,\bm 0_C) {\rm d} \mathbf w_A  + \int_{L_C^2} f(\bm 0_A, \mathbf w_C) {\rm d} \mathbf w_C  }\\
=&\frac{\frac{(1-p_A)p_C}{1-\alpha}\frac{\partial P(\mathbf Y_A=\bm 0_A,\mathbf Y_C \leq \mathbf w_C)}{\partial \mathbf w_C}}{ \frac{p_A(1-p_C)}{\alpha}P(\mathbf Y_A\in L_A^2,\mathbf Y_C=\bm 0_C)+\frac{(1-p_A)p_C}{1-\alpha}P(\mathbf Y_A=\bm 0_A,\mathbf Y_C \in L_C^2)}\\
=&\frac{(1-p_A)p_C\frac{\partial P(\mathbf Y_A=\bm 0_A,\mathbf Y_C\leq \mathbf w_C)}{\partial \mathbf w_C} }{(1-\alpha)\{p_A(1-p_C)+ (1-p_A)p_C\}   }.
\end{aligned}
\]

Then we can get the value of $P(\mathbf W_A\leq \mathbf w_A, \mathbf W_C\leq \mathbf w_C\mid \mathbf W_A\in L_A^2, \mathbf  W_C=\bm 0_C \, \text{or}\, \mathbf W_A=\bm 0_A, \mathbf W_C\in L_C^2)$. When  $\mathbf w_A \in L_A^1, \mathbf w_C \in L_C^1$,
\[
P(\mathbf W_A\leq \mathbf w_A, \mathbf W_C\leq \mathbf w_C\mid \mathbf W_A\in L_A^2, \mathbf  W_C=\bm 0_C \, \text{or}\, \mathbf W_A=\bm 0_A, \mathbf W_C\in L_C^2)=0=P(\mathbf Y_A \leq \mathbf w_A, \mathbf Y_C \leq \mathbf w_C).
\]
When $\mathbf w_A \in L_A^1, \mathbf w_C \in L_C^2$,
\[
\begin{aligned}
&P(\mathbf W_A\leq \mathbf w_A, \mathbf W_C\leq \mathbf w_C\mid \mathbf W_A\in L_A^2, \mathbf  W_C=\bm 0_C \, \text{or}\, \mathbf W_A=\bm 0_A, \mathbf W_C\in L_C^2)\\
=&\frac{ \int _{[\bm{0}_C,\mathbf w_C]\setminus[0,1]^{|C|} } f(\bm 0_A,\mathbf t) {\rm d} \mathbf t  }{\int _{L_A^2} f(\mathbf w_A,\bm 0_C){\rm d} \mathbf w_A+\int_{L_C^2} f(\bm 0_A,\mathbf w_C){\rm d} \mathbf w_C}\\
=&\frac{(1-p_A)p_CP(\mathbf Y_A=\bm 0_A,\mathbf Y_C \leq \mathbf w_C)  }{(1-\alpha)\{p_A(1-p_C)+(1-p_A)p_C  \}}\\
=&\frac{(1-p_A)p_CP(\mathbf Y_A\leq \mathbf w_A,\mathbf Y_C \leq \mathbf w_C)  }{(1-\alpha)\{p_A(1-p_C)+(1-p_A)p_C  \}},
\end{aligned}
\]
where the last equation is due to the support of $\mathbf{Y}=(\mathbf{Y}_A,\mathbf{Y}_C)$  being $\Omega =\left( L_A^2 \times \bm{0}_{C}\right) \cup  \left( \bm{0}_{A}\times L_C^2  \right)$.
When $\mathbf w_A \in L_A^2, \mathbf w_C \in L_C^1$,
\[
\begin{aligned}
&P(\mathbf W_A\leq \mathbf w_A, \mathbf W_C\leq \mathbf w_C\mid \mathbf W_A\in L_A^2, \mathbf  W_C=\bm 0_C \, \text{or}\, \mathbf W_A=\bm 0_A, \mathbf W_C\in L_C^2)\\
=&\frac{ \int _{[\bm{0}_A,\mathbf w_A]\setminus[0,1]^{|A|} }  f(\mathbf t,\bm 0_C) {\rm d} \mathbf t      }{\int _{L_A^2} f(\mathbf w_A,\bm 0_C){\rm d} \mathbf w_A+\int_{L_C^2} f(\bm 0_A,\mathbf w_C){\rm d} \mathbf w_C}\\
=&\frac{p_A(1-p_C)P(\mathbf Y_A \leq \mathbf w_A,\mathbf Y_C = \bm 0_C)  }{\alpha\{p_A(1-p_C)+(1-p_A)p_C \}}\\
=&\frac{p_A(1-p_C)P(\mathbf Y_A\leq \mathbf w_A,\mathbf Y_C \leq \mathbf w_C)  }{\alpha\{p_A(1-p_C)+(1-p_A)p_C\}}.
\end{aligned}
\]
When $\mathbf w_A \in L_A^2, \mathbf w_C \in L_C^2$, due to $P(\mathbf Y_A \leq \mathbf w_A, \mathbf Y_C \leq \mathbf w_C)=P(\mathbf Y_A \leq \mathbf w_A, \mathbf Y_C = \bm 0_C)+P(\mathbf Y_A = \bm 0_A, \mathbf Y_C \leq \mathbf w_C)$,
\[
\begin{aligned}
&P(\mathbf W_A\leq \mathbf w_A, \mathbf W_C\leq \mathbf w_C\mid \mathbf W_A\in L_A^2, \mathbf  W_C=\bm 0_C \, \text{or}\, \mathbf W_A=\bm 0_A, \mathbf W_C\in L_C^2)\\
=&\frac{\int _{[\bm{0}_A,\mathbf w_A]\setminus[0,1]^{|A|} }  f(\mathbf t,\bm 0_C) {\rm d} \mathbf t }{\int _{L_A^2} f(\mathbf w_A,\bm 0_C){\rm d} \mathbf w_A+\int_{L_C^2} f(\bm 0_A,\mathbf w_C){\rm d} \mathbf w_C}+\frac{\int _{[\bm{0}_C,\mathbf w_C]\setminus[0,1]^{|C|} } f(\bm 0_A,\mathbf t) {\rm d} \mathbf t  }{\int _{L_A^2} f(\mathbf w_A,\bm 0_C){\rm d} \mathbf w_A+\int_{L_C^2} f(\bm 0_A,\mathbf w_C){\rm d} \mathbf w_C}\\
=&\frac{p_A(1-p_C)P(\mathbf Y_A \leq \mathbf w_A,\mathbf Y_C = \bm 0_C)  }{\alpha\{p_A(1-p_C)+(1-p_A)p_C \}}+\frac{(1-p_A)p_CP(\mathbf Y_A=\bm 0_A,\mathbf Y_C \leq \mathbf w_C)  }{(1-\alpha)\{p_A(1-p_C)+(1-p_A)p_C  \}}\\
=&\frac{\{(1-\alpha)p_A(1-p_C)-\alpha (1-p_A)p_C\}P(\mathbf Y_A\leq \mathbf w_A,\mathbf Y_C = \bm 0_C)  }{\alpha(1-\alpha)\{p_A(1-p_C)+(1-p_A)p_C  \}}+\frac{(1-p_A)p_CP(\mathbf Y_A\leq \mathbf w_A,\mathbf Y_C \leq \mathbf w_C)  }{(1-\alpha)\{p_A(1-p_C)+(1-p_A)p_C  \}}.
\end{aligned}
\]

In summary,
\begin{equation}
\label{eqn:50}
\begin{aligned}
&P(\mathbf W_A\leq \mathbf w_A, \mathbf W_C\leq \mathbf w_C\mid \mathbf W_A\in L_A^2, \mathbf  W_C=\bm 0_C \, \text{or}\, \mathbf W_A=\bm 0_A, \mathbf W_C\in L_C^2)=\\
&\begin{cases}
0, &\mathbf w_A \in L_A^1, \mathbf w_C \in L_C^1,\\
\frac{(1-p_A)p_CP(\mathbf Y_A\leq \mathbf w_A,\mathbf Y_C \leq \mathbf w_C)  }{(1-\alpha)\{p_A(1-p_C)+(1-p_A)p_C\}}, & \mathbf w_A \in L_A^1, \mathbf w_C \in L_C^2,\\
\frac{p_A(1-p_C)P(\mathbf Y_A\leq \mathbf w_A,\mathbf Y_C \leq \mathbf w_C)  }{\alpha\{p_A(1-p_C)+(1-p_A)p_C\}}, & \mathbf w_A \in L_A^2, \mathbf w_C \in L_C^1,\\
 \frac{\{(1-\alpha)p_A(1-p_C)-\alpha (1-p_A)p_C\}P(\mathbf Y_A\leq \mathbf w_A,\mathbf Y_C = \bm 0_C)  }{\alpha(1-\alpha)\{p_A(1-p_C)+(1-p_A)p_C  \}}+\frac{(1-p_A)p_CP(\mathbf Y_A\leq \mathbf w_A,\mathbf Y_C \leq \mathbf w_C)  }{(1-\alpha)\{p_A(1-p_C)+(1-p_A)p_C  \}},& \mathbf w_A \in L_A^2, \mathbf w_C \in L_C^2.
\end{cases}
\end{aligned}
\end{equation}
From \eqref{eqn:50},  $P(\mathbf W_A\leq \mathbf w_A, \mathbf W_C\leq \mathbf w_C\mid \mathbf W_A\in L_A^2, \mathbf  W_C=\bm 0_C \, \text{or}\, \mathbf W_A=\bm 0_A, \mathbf W_C\in L_C^2)=P(\mathbf Y_A\leq \mathbf w_A,\mathbf Y_C \leq \mathbf w_C)$,  if and only if
\[
\begin{cases}
\frac{(1-p_A)p_C}{(1-\alpha)\{p_A(1-p_C)+(1-p_A)p_C  \}  }=1,\\
\frac{p_A(1-p_C)}{\alpha\{p_A(1-p_C)+(1-p_A)p_C  \}  }=1,\\
\frac{\{(1-\alpha)p_A(1-p_C) - \alpha(1-p_A)p_C \} }{\alpha(1-\alpha)\{ p_A(1-p_C)+(1-p_A)p_C  \} }=0,
\end{cases}
\]
or equivalently,
\begin{equation}
\label{eqn:51}
\alpha=\frac{p_A(1-p_C)}{p_A(1-p_C)+(1-p_A)p_C}.
\end{equation}
We have hence shown that under condition \eqref{eqn:51}, $(\mathbf W \mid \mathbf W\in \Omega) \,{\buildrel d \over =}\, \mathbf Y$ if the density function of $F(\mathbf w_A,\mathbf w_C)$ exists.

If the density function of  $F(\mathbf w_A,\mathbf w_C)$  does not exist, from the definition of conditional probability in \citet[\S 7.4]{shorack2017probability}, we know that \eqref{eqn:50} still satisfies the condition for conditional probability. Thus, we can define $P(\mathbf W_A\leq \mathbf w_A, \mathbf W_C\leq \mathbf w_C\mid \mathbf W_A\in L_A^2, \mathbf  W_C=\bm 0_C \, \text{or}\, \mathbf W_A=\bm 0_A, \mathbf W_C\in L_C^2)$ as that in \eqref{eqn:50}. Under condition \eqref{eqn:51}, we also have $P(\mathbf W_A\leq \mathbf w_A, \mathbf W_C\leq \mathbf w_C\mid \mathbf W_A\in L_A^2, \mathbf  W_C=\bm 0_C \, \text{or}\, \mathbf W_A=\bm 0_A, \mathbf W_C\in L_C^2)=P(\mathbf Y_A\leq \mathbf w_A,\mathbf Y_C \leq \mathbf w_C)$. Then $(\mathbf W \mid \mathbf W\in \Omega) \,{\buildrel d \over =}\, \mathbf Y$.

In summary, based on the definition of $\mathbf Y_A \ind_o \mathbf Y_C$, we have $\mathbf Y_A \ind_o \mathbf Y_C$.

\end{proof}

\begin{comment}
\subsection{Proof of Proposition \ref{prop:eh} in the case where $|A|=|B|=|C|=1$}

\begin{proof}
	We need to show
	\begin{multline}
	\label{eqn:claim51}
	\text{the support of} \,\mathbf Y=(Y_1,Y_2,Y_3)\,  \text{is}\, \Omega = \left[(1,\infty) \times \{0\} \times \{0\} \right] \cup \left[\{0\} \times (1,\infty) \times \{0\} \right]\cup \left[\{0\} \times \{0\} \times (1,\infty)\right]\\
	\Rightarrow Y_1 \ind_i Y_2 \mid Y_3,
	\end{multline}
	\begin{multline}
	\label{eqn:claim52}
		\text{the support of} \,\mathbf Y=(Y_1,Y_2,Y_3)\,  \text{is}\, \Omega =  \left[(1,\infty) \times \{0\} \times \{0\} \right] \cup \left[\{0\} \times (1,\infty) \times \{0\} \right]\cup \left[\{0\} \times \{0\} \times (1,\infty)\right]\\
		\Rightarrow Y_1 \ind_e Y_2 \mid Y_3,
	\end{multline}
	\begin{multline}
	\label{eqn:claim53}
		\text{the support of} \,\mathbf Y=(Y_1,Y_2,Y_3)\,  \text{is}\, \Omega =  \left[(1,\infty) \times \{0\} \times \{0\} \right] \cup \left[\{0\} \times (1,\infty) \times \{0\} \right]\cup \left[\{0\} \times \{0\} \times (1,\infty)\right]\\
		\Rightarrow Y_1 \ind_o Y_2 \mid Y_3.
	\end{multline}
	Under \eqref{eqn:claim51}, \eqref{eqn:claim52} and  \eqref{eqn:claim53}, we can have if the support of $\mathbf Y=(Y_1,Y_2,Y_3)$  is $\Omega =  \left[(1,\infty) \times \{0\} \times \{0\} \right] \cup \left[\{0\} \times (1,\infty) \times \{0\} \right]\cup \left[\{0\} \times \{0\} \times (1,\infty)\right]$, then
	\[
	Y_1 \ind_o Y_2 \mid Y_3 \Leftrightarrow Y_1 \ind_i Y_2 \mid Y_3 \Leftrightarrow Y_1 \ind_e Y_2 \mid Y_3.
	\]

	First, we  show \eqref{eqn:claim51}. Since the support of $\mathbf Y=(Y_1,Y_2,Y_3)$  is $\Omega = \left[(1,\infty) \times \{0\} \times \{0\} \right] \cup \left[\{0\} \times (1,\infty) \times \{0\} \right]\cup \left[\{0\} \times \{0\} \times (1,\infty)\right]$, then $S_1\times S_2\times S_3 \subset \left[\{0\} \times \{0\} \times (1,\infty) \right]$, $ \left[\{0\} \times (1,\infty) \times \{0\} \right]$ or $ \left[ \{0\} \times \{0\} \times (1,\infty)\right]$.
	When $S_1\times S_2 \times S_3 \subset \left[ (1,\infty) \times \{0\} \times \{0\} \right]$, $Y_3=0$. Given $Y_3=0$, since $Y_2=0$ regardless of the value of $Y_1$, then $Y_1 \ind Y_2 \mid (Y_3, \mathbf Y \in S_1 \times S_2 \times S_3)$.
	When $S_1\times S_2\times S_3 \subset \left[\{0\} \times (1,\infty) \times \{0\} \right]$, $Y_3=0$. Given $Y_3=0$, since $Y_1=0$ regardless of the value of $Y_2$, then $Y_1 \ind Y_2 \mid (Y_3, \mathbf Y \in S_1 \times S_2 \times S_3)$. When $S_1\times S_2\times S_3 \subset \left[\{0\} \times \{0\}\times (1,\infty)  \right]$, $Y_3\in (1,\infty)$. Given $Y_3=y_3$, since $Y_1=Y_2=0$, then $Y_1 \ind Y_2 \mid (Y_3, \mathbf Y \in S_1 \times S_2 \times S_3)$.
	Thus, based on the definition of $Y_1 \ind_i Y_2 \mid Y_3$, we have $Y_1 \ind_i Y_2 \mid Y_3$.

	Second, we show \eqref{eqn:claim52}. Since the support of $\mathbf Y=(Y_1,Y_2,Y_3)$  is $\Omega = \left[(1,\infty) \times \{0\} \times \{0\} \right] \cup \left[\{0\} \times (1,\infty) \times \{0\} \right]\cup \left[\{0\} \times \{0\} \times (1,\infty)\right]$, then when $Y_1>1$, $\mathbf Y \in \left[ (1,\infty) \times \{0\} \right]$. Since $Y_2=0$ regardless of the value of $Y_1$, then $Y_1 \ind Y_2 \mid Y_1>1$.  When $Y_2>1$, $\mathbf Y \in  \left[ \{0\} \times (1,\infty) \right]$. Since $Y_1=0$ regardless of the value of $Y_2$, then $Y_1 \ind Y_2 \mid Y_2>1$. Thus, based on the definition of $Y_1 \ind_e Y_2$, we have $Y_1 \ind_e Y_2$.

	Third, we show \eqref{eqn:claim53}. Since the support of $\mathbf Y=(Y_1,Y_2,Y_3)$  is $\Omega = \left[(1,\infty) \times \{0\} \times \{0\} \right] \cup \left[\{0\} \times (1,\infty) \times \{0\} \right]\cup \left[\{0\} \times \{0\} \times (1,\infty)\right]$, then
\begin{multline*}
P(Y_1>1, Y_2=0, Y_3=0 \,\text{or}\, Y_1=0, Y_2>1, Y_3=0\,\text{or}\,Y_1=0, Y_2=0, Y_3>1)\\=P(Y_1>1, Y_2=0, Y_3=0) + P(Y_1=0, Y_2>1, Y_3=0)+ P(Y_1=0, Y_2=0, Y_3>1)=1.
\end{multline*}
Suppose that $P(Y_1>1,Y_2=0, Y_3=0)=\alpha_1$ and $P(Y_1=0,Y_2>1, Y_3=0)=\alpha_2$, then $P(Y_1=0, Y_2=0, Y_3>1)=1-\alpha_1-\alpha_2$, where $\alpha_1$ and $\alpha_2$ satisfy the conditions of $\alpha_1>0, \alpha_2>0$ and $0<\alpha_1+\alpha_2<1$.
	Now we introduce three new random variables $W_1$, $W_2$ and $W_3$, assume that they have the distributions as follows:
	\[
	F(w_1)=
	\begin{cases}
	(1-p_1)w_1, &  0 \leq  w_1 \leq 1,\\
	(1-p_1)+P(Y_1\leq w_1 \mid Y_1>1, Y_2=0, Y_3=0)p_1, & w_1>1,
	\end{cases}
	\]
	\[
	F(w_2)=
	\begin{cases}
	(1-p_2)w_2, & 0 \leq  w_2 \leq 1,\\
	(1-p_2)+P(Y_2\leq w_2 \mid Y_1=0, Y_2>1, Y_3=0)p_2, & w_2>1,
	\end{cases}
	\]
	and
	\[
		F(w_3)=
	\begin{cases}
		(1-p_3)w_3, & 0 \leq  w_3 \leq 1,\\
		(1-p_3)+P(Y_3\leq w_3 \mid Y_1=0, Y_2=0, Y_3>1)p_3, & w_3>1,
	\end{cases}
	\]
	where $p_1 \in (0,1)$, $p_2 \in (0,1)$ and $p_3 \in (0,1)$.
	To make $W_1,W_2,W_3$ independent, let $F(w_1,w_2,w_3)=F(w_1)F(w_2)F(w_3)$. Then,
	\[
	\begin{aligned}
	&F(w_1,w_2,w_3)=\\
	&\begin{cases}
&	(1-p_1)(1-p_2)(1-p_3)w_1w_2w_3,\hspace{6cm} 0 \leq w_1 \leq 1, 0\leq w_2 \leq 1, 0 \leq w_3 \leq 1,\\
&	(1-p_1)(1-p_2)w_1w_2\{(1-p_3)+P(Y_3\leq w_3 \mid Y_1=0, Y_2=0, Y_3>1)p_3\},0 \leq w_1 \leq 1, 0\leq w_2\leq 1, w_3>1,\\
&	(1-p_1)(1-p_3)w_1w_3\{(1-p_2)+P(Y_2\leq w_2 \mid Y_1=0, Y_2>1, Y_3=0)p_2\}, 0 \leq w_1 \leq 1, w_2>1, 0 \leq w_3 \leq 1,\\
& (1-p_1)w_1\{(1-p_2)+P(Y_2\leq w_2 \mid Y_1=0, Y_2>1, Y_3=0)p_2\}\\
& \hspace{2cm}\times \{(1-p_3)+P(Y_3\leq w_3 \mid Y_1=0, Y_2=0, Y_3>1)p_3\},\hspace{2cm} 0 \leq w_1 \leq 1, w_2>1, w_3 > 1,\\
& \{(1-p_1)+P(Y_1\leq w_1 \mid Y_1>1, Y_2=0, Y_3=0)p_1\}(1-p_2)(1-p_3)w_2w_3, \hspace{0.1cm}w_1 > 1,0\leq w_2\leq 1, 0\leq w_3 \leq 1,\\
& \{(1-p_1)+P(Y_1\leq w_1 \mid Y_1>1, Y_2=0, Y_3=0)p_1\}(1-p_2)w_2\\
&\hspace{2cm}\times\{(1-p_3)+P(Y_3\leq w_3 \mid Y_1=0, Y_2=0, Y_3>1)p_3\},\hspace{2cm} w_1>1, 0\leq w_2 \leq 1, w_3>1,\\
& \{(1-p_1)+P(Y_1\leq w_1 \mid Y_1>1, Y_2=0, Y_3=0)p_1\}\\
&\hspace{2cm}\times \{(1-p_2)+P(Y_2\leq w_2 \mid Y_1=0, Y_2>1, Y_3=0)p_2\}(1-p_3)w_3, \hspace{0.2cm}w_1>1,w_2>1,0\leq w_3\leq 1,\\
&\{(1-p_1)+P(Y_1\leq w_1 \mid Y_1>1, Y_2=0, Y_3=0)p_1\}\{(1-p_2)+P(Y_2\leq w_2 \mid Y_1=0, Y_2>1, Y_3=0)p_2\}\\
&\hspace{2cm}\times\{(1-p_3)+P(Y_3\leq w_3 \mid Y_1=0, Y_2=0, Y_3>1)p_3\},  \hspace{2.6cm} w_1>1, w_2>1, w_3 > 1.
	\end{cases}
	\end{aligned}
	\]
	Furthermore, we have $W_1 \ind W_2 \mid W_3$.

  If the density function of $F(w_1,w_2,w_3)$ exists, then
\[
\begin{aligned}
&f(w_1,w_2,w_3)=\\
&\begin{cases}
&(1-p_1)(1-p_2)(1-p_3), \hspace{15.1em} 0 \leq  w_1 \leq 1, 0 \leq  w_2 \leq 1, 0 \leq  w_3 \leq 1,\\
&\frac{(1-p_1)(1-p_2)p_3}{1-\alpha_1-\alpha_2}\frac{\partial P(Y_1=0,Y_2=0,Y_3 \leq w_3)}{\partial w_3},  \hspace{13.2em}    0 \leq  w_1 \leq 1, 0 \leq  w_2 \leq 1, w_3>1,\\
&\frac{(1-p_1)p_2(1-p_3)}{\alpha_2}\frac{\partial P(Y_1=0,Y_2 \leq w_2,Y_3=0)}{\partial w_2},  \hspace{13.2em}    0\leq w_1\leq 1, w_2>1, 0 \leq w_3 \leq 1,\\
&\frac{(1-p_1)p_2p_3}{\alpha_2(1-\alpha_1-\alpha_2)}\frac{\partial P(Y_1=0,Y_2\leq w_2,Y_3 = 0)}{\partial w_2}\frac{\partial P(Y_1=0,Y_2=0,Y_3\leq w_3)}{\partial w_3},  \hspace{7.3em} 0\leq w_1\leq 1, w_2>1,w_3>1,\\
&\frac{p_1(1-p_2)(1-p_3)}{\alpha_1}\frac{\partial P(Y_1\leq w_1,Y_2=0,Y_3 = 0)}{\partial w_2}, \hspace{13.1em} w_1>1, 0\leq w_2\leq 1,0\leq w_3\leq 1,\\
&\frac{p_1(1-p_2)p_3}{\alpha_1(1-\alpha_1-\alpha_2)}\frac{\partial P(Y_1\leq w_1,Y_2=0,Y_3 = 0)}{\partial w_1}\frac{\partial P(Y_1=0,Y_2=0,Y_3\leq w_3)}{\partial w_3}, \hspace{7.3em} w_1>1, 0\leq w_2\leq 1, w_3> 1,\\
&\frac{p_1p_2(1-p_3)}{\alpha_1\alpha_2}\frac{\partial P(Y_1\leq w_1,Y_2=0,Y_3 = 0)}{\partial w_1}\frac{\partial P(Y_1=0,Y_2\leq w_2,Y_3=0)}{\partial w_2}, \hspace{8.2em} w_1>1, w_2>1,0\leq w_3\leq 1,\\
&\frac{p_1p_2p_3}{\alpha_1\alpha_2(1-\alpha_1-\alpha_2)}\frac{\partial P(Y_1\leq w_1,Y_2=0,Y_3 = 0)}{\partial w_1}\frac{\partial P(Y_1=0,Y_2\leq w_2,Y_3=0)}{\partial w_2}\frac{\partial P(Y_1=0,Y_2=0,Y_3\leq w_3)}{\partial w_3},  w_1>1, w_2>1, w_3>1.
\end{cases}
\end{aligned}
\]
Since the support of $\mathbf Y = (Y_1,Y_2,Y_3) $ is $\Omega = \left[(1,\infty) \times \{0\} \times \{0\} \right] \cup \left[\{0\} \times (1,\infty) \times \{0\} \right]\cup \left[\{0\} \times \{0\} \times (1,\infty)\right]$, then $\partial P(Y_1=0,Y_2=0,Y_3 \leq w_3)/\partial w_3 \neq 0$ for $w_3 >1$, $\partial P(Y_1=0,Y_2\leq w_2,Y_3 = 0)/\partial w_2 \neq 0$ for $w_2 >1$ and $\partial P(Y_1\leq w_1,Y_2 = 0,Y_3=0)/\partial w_1\neq 0$ for $w_1>1$. Thus, from $f(w_1,w_2,w_3)$, we have the support of $\mathbf W=(W_1,W_2,W_3)$ is $[0,\infty)^3$, which contains  $\Omega$.  Based on the definition of conditional density in \citet[\S 7.4]{shorack2017probability}, we have that the conditional densities at $(w_1,0,0)$, $(w_1>1)$, $(0,w_2,0)$, $(w_2>1)$, $(0,0,w_3)$, $(w_3>1)$, given that all the points are on the curve of $\Omega$ are
\[
\begin{aligned}
&\frac{f(w_1,0,0)}{ \int_1^\infty f(w_1,0,0) {\rm d} w_1  + \int_1^\infty f(0, w_2, 0) {\rm d} w_2 + \int_1^\infty f(0, 0, w_3) {\rm d} w_3}\\
=&\frac{p_1(1-p_2)(1-p_3)\frac{\partial P(Y_1\leq w_1, Y_2=0, Y_3=0 )}{\partial w_1}
}{\alpha_1\{p_1(1-p_2)(1-p_3)+ (1-p_1)p_2(1-p_3)+(1-p_1)(1-p_2)p_3 \}   },
\end{aligned}
\]
\[
\begin{aligned}
&\frac{f(0,w_2,0)}{ \int_1^\infty f(w_1,0,0) {\rm d} w_1  + \int_1^\infty f(0, w_2,0) {\rm d} w_2 + \int_1^\infty f(0, 0, w_3) {\rm d} w_3 }\\
=&\frac{(1-p_1)p_2(1-p_3)\frac{\partial P(Y_1=0,Y_2\leq w_2,Y_3=0)}{\partial w_2} }{\alpha_2\{p_1(1-p_2)(1-p_3)+ (1-p_1)p_2(1-p_3)+(1-p_1)(1-p_2)p_3 \}   },
\end{aligned}
\]
and
\[
\begin{aligned}
&\frac{f(0,0,w_3)}{ \int_1^\infty f(w_1,0,0) {\rm d} w_1  + \int_1^\infty f(0, w_2,0) {\rm d} w_2 + \int_1^\infty f(0, 0, w_3) {\rm d} w_3 }\\
=&\frac{(1-p_1)(1-p_2)p_3\frac{\partial P(Y_1=0,Y_2=0,Y_3\leq w_3)}{\partial w_3} }{(1-\alpha_1-\alpha_2)\{p_1(1-p_2)(1-p_3)+ (1-p_1)p_2(1-p_3)+(1-p_1)(1-p_2)p_3 \}   }.
\end{aligned}
\]
Then we can get the value of $P(W_1\leq w_1, W_2\leq w_2, W_3\leq w_3\mid W_1>1, W_2=0, W_3=0 \, \text{or}\, W_1=0, W_2>1, W_3=0\, \text{or}\, W_1=0, W_2=0, W_3>1)=P(W_1\leq w_1, W_2\leq w_2, W_3\leq w_3\mid \mathbf W\in \Omega)$. When $0 \leq w_1 \leq 1$, $0 \leq w_2 \leq 1$, $0 \leq w_3 \leq 1$,
\[
P(W_1\leq w_1, W_2\leq w_2, W_3\leq w_3\mid \mathbf W\in \Omega)
=0=P(Y_1 \leq w_1, Y_2 \leq w_2, Y_3 \leq w_3).
\]
When $0 \leq w_1 \leq 1$, $0 \leq w_2 \leq 1$, $w_3 > 1$,
\[
\begin{aligned}
  &P(W_1\leq w_1, W_2\leq w_2, W_3\leq w_3\mid \mathbf W\in \Omega)\\
  =&\frac{\int_1^{w_3} f(0,0,t){\rm d} t}{ \int_1^\infty f(w_1,0,0) {\rm d} w_1  + \int_1^\infty f(0, w_2,0) {\rm d} w_2 + \int_1^\infty f(0, 0, w_3) {\rm d} w_3 }\\
    =&\frac{(1-p_1)(1-p_2)p_3P(Y_1=0,Y_2=0,Y_3\leq w_3)}{(1-\alpha_1-\alpha_2)\{p_1(1-p_2)(1-p_3)+ (1-p_1)p_2(1-p_3)+(1-p_1)(1-p_2)p_3 \}}\\
        =&\frac{(1-p_1)(1-p_2)p_3P(Y_1\leq w_1,Y_2\leq w_2,Y_3\leq w_3)}{(1-\alpha_1-\alpha_2)\{p_1(1-p_2)(1-p_3)+ (1-p_1)p_2(1-p_3)+(1-p_1)(1-p_2)p_3 \}}.
\end{aligned}
\]
When $0 \leq w_1 \leq 1$, $w_2>1$, $0\leq w_3 \leq 1$,
\[
\begin{aligned}
  &P(W_1\leq w_1, W_2\leq w_2, W_3\leq w_3\mid \mathbf W\in \Omega)\\
  =&\frac{\int_1^{w_2} f(0,t,0){\rm d} t}{ \int_1^\infty f(w_1,0,0) {\rm d} w_1  + \int_1^\infty f(0, w_2,0) {\rm d} w_2 + \int_1^\infty f(0, 0, w_3) {\rm d} w_3 }\\
    =&\frac{(1-p_1)p_2(1-p_3)P(Y_1=0,Y_2\leq w_2,Y_3=0)}{\alpha_2\{p_1(1-p_2)(1-p_3)+ (1-p_1)p_2(1-p_3)+(1-p_1)(1-p_2)p_3 \}}\\
        =&\frac{(1-p_1)p_2(1-p_3)P(Y_1\leq w_1,Y_2\leq w_2,Y_3\leq w_3)}{\alpha_2\{p_1(1-p_2)(1-p_3)+ (1-p_1)p_2(1-p_3)+(1-p_1)(1-p_2)p_3 \}}.
\end{aligned}
\]
When $0 \leq w_1 \leq 1$, $w_2>1$, $w_3>1$,	due to $P(Y_1\leq w_1,Y_2\leq w_2,Y_3\leq w_3)=P(Y_1=0,Y_2\leq w_2,Y_3=0)+P(Y_1=0,Y_2=0,Y_3\leq w_3)$,
\[
\begin{aligned}
  &P(W_1\leq w_1, W_2\leq w_2, W_3\leq w_3\mid \mathbf W\in \Omega)\\
  =&\frac{\int_1^{w_2} f(0,t,0){\rm d} t+\int_1^{w_3} f(0,0,t){\rm d} t}{ \int_1^\infty f(w_1,0,0) {\rm d} w_1  + \int_1^\infty f(0, w_2,0) {\rm d} w_2 + \int_1^\infty f(0, 0, w_3) {\rm d} w_3 }\\
        =&\frac{\{(1-p_1)p_2(1-p_3)(1-\alpha_1-\alpha_2)-(1-p_1)(1-p_2)p_3\alpha_2\}P(Y_1=0,Y_2\leq w_2,Y_3=0)}{\alpha_2(1-\alpha_1-\alpha_2)\{p_1(1-p_2)(1-p_3)+ (1-p_1)p_2(1-p_3)+(1-p_1)(1-p_2)p_3 \}}\\
       &+ \frac{(1-p_1)(1-p_2)p_3P(Y_1\leq w_1,Y_2\leq w_2,Y_3\leq w_3)}{(1-\alpha_1-\alpha_2)\{p_1(1-p_2)(1-p_3)+ (1-p_1)p_2(1-p_3)+(1-p_1)(1-p_2)p_3 \}}.
\end{aligned}
\]
When $ w_1>1$, $0\leq w_2\leq 1$, $0\leq w_3 \leq 1$,
\[
\begin{aligned}
  &P(W_1\leq w_1, W_2\leq w_2, W_3\leq w_3\mid \mathbf W\in \Omega)\\
  =&\frac{\int_1^{w_1} f(t,0,0){\rm d} t}{ \int_1^\infty f(w_1,0,0) {\rm d} w_1  + \int_1^\infty f(0, w_2,0) {\rm d} w_2 + \int_1^\infty f(0, 0, w_3) {\rm d} w_3 }\\
    =&\frac{p_1(1-p_2)(1-p_3)P(Y_1\leq w_1,Y_2=0,Y_3=0)}{\alpha_1\{p_1(1-p_2)(1-p_3)+ (1-p_1)p_2(1-p_3)+(1-p_1)(1-p_2)p_3 \}}\\
        =&\frac{p_1(1-p_2)(1-p_3)P(Y_1\leq w_1,Y_2\leq w_2,Y_3\leq w_3)}{\alpha_1\{p_1(1-p_2)(1-p_3)+ (1-p_1)p_2(1-p_3)+(1-p_1)(1-p_2)p_3 \}}.
\end{aligned}
\]
When $ w_1 > 1$, $0\leq w_2\leq 1$, $w_3>1$,	due to $P(Y_1\leq w_1,Y_2\leq w_2,Y_3\leq w_3)=P(Y_1\leq w_1,Y_2=0,Y_3=0)+P(Y_1=0,Y_2=0,Y_3\leq w_3)$,
\[
\begin{aligned}
  &P(W_1\leq w_1, W_2\leq w_2, W_3\leq w_3\mid \mathbf W\in \Omega)\\
  =&\frac{\int_1^{w_1} f(t,0,0){\rm d} t+\int_1^{w_3} f(0,0,t){\rm d} t}{ \int_1^\infty f(w_1,0,0) {\rm d} w_1  + \int_1^\infty f(0, w_2,0) {\rm d} w_2 + \int_1^\infty f(0, 0, w_3) {\rm d} w_3 }\\
        =&\frac{\{p_1(1-p_2)(1-p_3)(1-\alpha_1-\alpha_2)-(1-p_1)(1-p_2)p_3\alpha_1\}P(Y_1\leq w_1,Y_2=0,Y_3=0)}{\alpha_1(1-\alpha_1-\alpha_2)\{p_1(1-p_2)(1-p_3)+ (1-p_1)p_2(1-p_3)+(1-p_1)(1-p_2)p_3 \}}\\
       &+ \frac{(1-p_1)(1-p_2)p_3P(Y_1\leq w_1,Y_2\leq w_2,Y_3\leq w_3)}{(1-\alpha_1-\alpha_2)\{p_1(1-p_2)(1-p_3)+ (1-p_1)p_2(1-p_3)+(1-p_1)(1-p_2)p_3 \}}.
\end{aligned}
\]
When $ w_1 > 1$, $w_2>1$, $0\leq w_3\leq 1$,	due to $P(Y_1\leq w_1,Y_2\leq w_2,Y_3\leq w_3)=P(Y_1\leq w_1,Y_2=0,Y_3=0)+P(Y_1=0,Y_2\leq w_2,Y_3=0)$,
\[
\begin{aligned}
  &P(W_1\leq w_1, W_2\leq w_2, W_3\leq w_3\mid \mathbf W\in \Omega)\\
  =&\frac{\int_1^{w_1} f(t,0,0){\rm d} t+\int_1^{w_2} f(0,t,0){\rm d} t}{ \int_1^\infty f(w_1,0,0) {\rm d} w_1  + \int_1^\infty f(0, w_2,0) {\rm d} w_2 + \int_1^\infty f(0, 0, w_3) {\rm d} w_3 }\\
        =&\frac{\{p_1(1-p_2)(1-p_3)\alpha_2-(1-p_1)p_2(1-p_3)\alpha_1\}P(Y_1\leq w_1,Y_2=0,Y_3=0)}{\alpha_1\alpha_2\{p_1(1-p_2)(1-p_3)+ (1-p_1)p_2(1-p_3)+(1-p_1)(1-p_2)p_3 \}}\\
       &+ \frac{(1-p_1)p_2(1-p_3)P(Y_1\leq w_1,Y_2\leq w_2,Y_3\leq w_3)}{\alpha_2\{p_1(1-p_2)(1-p_3)+ (1-p_1)p_2(1-p_3)+(1-p_1)(1-p_2)p_3 \}}.
\end{aligned}
\]
When $ w_1 > 1$, $w_2>1$, $w_3>1$,	due to $P(Y_1\leq w_1,Y_2\leq w_2,Y_3\leq w_3)=P(Y_1\leq w_1,Y_2=0,Y_3=0)+P(Y_1=0,Y_2\leq w_2,Y_3=0)+P(Y_1=0,Y_2=0,Y_3\leq w_3)$,
\begin{multline*}
  P(W_1\leq w_1, W_2\leq w_2, W_3\leq w_3\mid \mathbf W\in \Omega)\\
  =\frac{\int_1^{w_1} f(t,0,0){\rm d} t+\int_1^{w_2} f(0,t,0){\rm d} t+\int_1^{w_3} f(0,0,t){\rm d} t}{ \int_1^\infty f(w_1,0,0) {\rm d} w_1  + \int_1^\infty f(0, w_2,0) {\rm d} w_2 + \int_1^\infty f(0, 0, w_3) {\rm d} w_3 }\\
        =\Bigg[\frac{\{p_1(1-p_2)(1-p_3)(1-\alpha_1-\alpha_2)-(1-p_1)(1-p_2)p_3\alpha_1\}P(Y_1\leq w_1,Y_2=0,Y_3=0)}{\alpha_1(1-\alpha_1-\alpha_2)}\\
      + \frac{\{(1-p_1)p_2(1-p_3)(1-\alpha_1-\alpha_2)-(1-p_1)(1-p_2)p_3\alpha_2\}P(Y_1=0,Y_2\leq w_2,Y_3=0)}{\alpha_2(1-\alpha_1-\alpha_2)}\\
      +\frac{(1-p_1)(1-p_2)p_3P(Y_1\leq w_1,Y_2\leq w_2,Y_3\leq w_3)}{1-\alpha_1-\alpha_2}\Bigg]\\
      \times \frac{1}{p_1(1-p_2)(1-p_3)+ (1-p_1)p_2(1-p_3)+(1-p_1)(1-p_2)p_3}\\.
\end{multline*}
By comparison, we can find that $P(W_1\leq w_1, W_2\leq w_2, W_3\leq w_3\mid \mathbf W\in \Omega)=P(Y_1\leq w_1,Y_2\leq w_2,Y_3\leq w_3)$ if and only if
\begin{equation}
\label{eqn:conditon2}
\begin{cases}
\alpha_1=\frac{p_1(1-p_2)(1-p_3)}{p_1(1-p_2)(1-p_3)+ (1-p_1)p_2(1-p_3)+(1-p_1)(1-p_2)p_3},\\
\alpha_2=\frac{(1-p_1)p_2(1-p_3)}{p_1(1-p_2)(1-p_3)+ (1-p_1)p_2(1-p_3)+(1-p_1)(1-p_2)p_3}.
\end{cases}
\end{equation}
We have hence shown that \eqref{eqn:conditon2}, $(\mathbf W \mid \mathbf W\in \Omega) \,{\buildrel d \over =}\, \mathbf Y$ if the density function of $F(w_1,w_2,w_3)$ exists.

If the density function of  $F(w_1,w_2,w_3)$ does not exist, from the definition of conditional probability in \citet[\S 7.4]{shorack2017probability}, we know that the above formulas for $P(W_1\leq w_1, W_2\leq w_2, W_3\leq w_3\mid \mathbf W\in \Omega)$ still satisfy the condition for conditional probability. Thus, we still can define $P(W_1\leq w_1, W_2\leq w_2, W_3\leq w_3\mid \mathbf W\in \Omega)$ in the same way as before. Under condition \eqref{eqn:conditon2}, we also have $P(W_1\leq w_1, W_2\leq w_2, W_3\leq w_3\mid \mathbf W\in \Omega)=P(Y_1\leq w_1,Y_2\leq w_2,Y_3\leq w_3)$. Then $(\mathbf W \mid \mathbf W\in \Omega) \,{\buildrel d \over =}\, \mathbf Y$.

	In summary, based on the definition of $Y_1 \ind_o Y_2\mid Y_3$, we have $Y_1 \ind_o Y_2\mid Y_3$.
\end{proof}
\end{comment}

\subsection{Proof of Proposition \ref{prop:eh} for a general $B$}

\begin{proof}
	We need to show
	\begin{multline}
	\label{eqn:claim61}
	\text{the support of} \,\mathbf Y=(\mathbf Y_A,\mathbf Y_B,\mathbf Y_C)\,  \text{is}\, \Omega =\left( L_A^2 \times \bm{0}_{B}\times \bm{0}_{C} \right) \cup  \left( \bm{0}_{A}\times L_B^2\times  \bm{0}_{C} \right) \cup  \left( \bm{0}_{A}\times \bm{0}_{B}\times L_C^2  \right)\\
	\Rightarrow \mathbf Y_A \ind_i \mathbf Y_C \mid  \mathbf Y_B,
	\end{multline}
	\begin{multline}
	\label{eqn:claim62}
	\text{the support of} \,\mathbf Y=(\mathbf Y_A,\mathbf Y_B,\mathbf Y_C)\,  \text{is}\, \Omega =\left( L_A^2 \times \bm{0}_{B}\times \bm{0}_{C} \right) \cup  \left( \bm{0}_{A}\times L_B^2\times  \bm{0}_{C} \right) \cup  \left( \bm{0}_{A}\times \bm{0}_{B}\times L_C^2  \right)\\
\Rightarrow \mathbf Y_A \ind_e \mathbf Y_C \mid  \mathbf Y_B,
	\end{multline}
	\begin{multline}
	\label{eqn:claim63}
	\text{the support of} \,\mathbf Y=(\mathbf Y_A,\mathbf Y_B,\mathbf Y_C)\,  \text{is}\, \Omega =\left( L_A^2 \times \bm{0}_{B}\times \bm{0}_{C} \right) \cup  \left( \bm{0}_{A}\times L_B^2\times  \bm{0}_{C} \right) \cup  \left( \bm{0}_{A}\times \bm{0}_{B}\times L_C^2  \right)\\
\Rightarrow \mathbf Y_A \ind_o \mathbf Y_C \mid  \mathbf Y_B.
	\end{multline}
	Under \eqref{eqn:claim61}, \eqref{eqn:claim62} and  \eqref{eqn:claim63}, we can have if the support of $\mathbf Y=(\mathbf Y_A,\mathbf Y_B,\mathbf Y_C)$  is $\Omega =\left( L_A^2 \times \bm{0}_{B}\times \bm{0}_{C} \right) \cup  \left( \bm{0}_{A}\times L_B^2\times  \bm{0}_{C} \right) \cup  \left( \bm{0}_{A}\times \bm{0}_{B}\times L_C^2  \right)$, then
	\[
	\mathbf Y_A \ind_o \mathbf Y_C \mid  \mathbf Y_B \Leftrightarrow \mathbf Y_A \ind_i \mathbf Y_C \mid  \mathbf Y_B\Leftrightarrow \mathbf Y_A \ind_e \mathbf Y_C \mid  \mathbf Y_B.
	\]

	First, we  show \eqref{eqn:claim61}. Since the support of $\mathbf Y=(\mathbf Y_A,\mathbf Y_B,\mathbf Y_C)$  is $\Omega =\left( L_A^2 \times \bm{0}_{B}\times \bm{0}_{C} \right) \cup  \left( \bm{0}_{A}\times L_B^2\times  \bm{0}_{C} \right) \\
	\cup  \left( \bm{0}_{A}\times \bm{0}_{B}\times L_C^2  \right)$, then $S_A\times S_B\times S_C \subset \left( L_A^2 \times \bm{0}_{B}\times \bm{0}_{C} \right)$, $\left( \bm{0}_{A}\times L_B^2\times  \bm{0}_{C} \right)$ or $\left( \bm{0}_{A}\times \bm{0}_{B}\times L_C^2  \right)$.
	When $S_A\times S_B \times S_C \subset\left( L_A^2 \times \bm{0}_{B}\times \bm{0}_{C} \right) $, $\mathbf Y_B=\bm{0}_{B}$. Given $\mathbf Y_B=\bm{0}_{B}$, since $\mathbf Y_C=\bm{0}_{C}$ regardless of the value of $\mathbf Y_A$, then $\mathbf Y_A \ind \mathbf Y_C \mid (\mathbf Y_B, \mathbf Y \in S_A \times S_B \times S_C)$.
	When $S_A\times S_B\times S_C \subset\left( \bm{0}_{A}\times L_B^2\times  \bm{0}_{C} \right)$, $\mathbf Y_B \in L_B^2$. Given $\mathbf Y_B=\mathbf y_B$, since $\mathbf Y_A=\bm{0}_{A}$ and  $\mathbf Y_C=\bm{0}_{C}$,  then $\mathbf Y_A \ind \mathbf Y_C \mid (\mathbf Y_B, \mathbf Y \in S_A \times S_B \times S_C)$. When $S_A\times S_B\times S_C \subset \left( \bm{0}_{A}\times \bm{0}_{B}\times L_C^2  \right)$, $\mathbf Y_B=\bm{0}_{B}$. Given $\mathbf Y_B=\bm{0}_{B}$, since $\mathbf Y_A=\bm{0}_{A}$ regardless of the value of $\mathbf Y_C$, then $\mathbf Y_A \ind \mathbf Y_C \mid (\mathbf Y_B, \mathbf Y \in S_A \times S_B \times S_C)$.
	Thus, based on the definition of $\mathbf Y_A \ind_i \mathbf Y_C \mid  \mathbf Y_B$, we have $\mathbf Y_A \ind_i \mathbf Y_C \mid  \mathbf Y_B$.

	Second, we show \eqref{eqn:claim62}. Since the support of $\mathbf Y=(\mathbf Y_A,\mathbf Y_B,\mathbf Y_C)$  is $\Omega =\left( L_A^2 \times \bm{0}_{B}\times \bm{0}_{C} \right) \cup  \left( \bm{0}_{A}\times L_B^2\times  \bm{0}_{C} \right)\\ \cup  \left( \bm{0}_{A}\times \bm{0}_{B}\times L_C^2  \right)$, then when $Y_k>1$ for some $k\in A$, we have $\mathbf Y \in \left( L_A^2 \times \bm{0}_{B}\times \bm{0}_{C} \right)$ and $\mathbf{Y}_B=\bm{0}_{B}$. Given $\mathbf Y_B=\bm{0}_{B}$ and $Y_k>1$, since $\mathbf{Y}_C=\bm{0}_{C}$ regardless of the value of $\mathbf Y_A$, then $\mathbf Y_A \ind \mathbf Y_C \mid  (\mathbf Y_B, Y_k>1)$.  When $Y_k>1$ for some $k\in B$, we have $\mathbf Y \in   \left( \bm{0}_{A}\times L_B^2\times  \bm{0}_{C} \right)$ and $\mathbf Y_B \in  L_B^2$. Given $\mathbf Y_B=\mathbf y_B$ and $Y_k>1$, since $\mathbf Y_A=\bm{0}_{A}$ and  $\mathbf Y_C=\bm{0}_{C}$, then $\mathbf Y_A \ind \mathbf Y_C \mid  (\mathbf Y_B, Y_k>1)$. When $Y_k>1$ for some $k\in C$, we have $\mathbf Y \in    \left( \bm{0}_{A}\times \bm{0}_{B}\times L_C^2  \right)$ and $\mathbf Y_B =\bm{0}_{B}$. Given $\mathbf Y_B=\bm{0}_{B}$ and $Y_k>1$, since $\mathbf Y_A=\bm{0}_{A}$ regardless of the value of $\mathbf Y_C$, then $\mathbf Y_A \ind \mathbf Y_C \mid  (\mathbf Y_B, Y_k>1)$.
	Thus, based on the definition of $\mathbf Y_A \ind_e \mathbf Y_C \mid  \mathbf Y_B$, we have $\mathbf Y_A \ind_e \mathbf Y_C \mid  \mathbf Y_B$.

	Third, we show \eqref{eqn:claim63}.
	Since the support of $\mathbf Y=(\mathbf Y_A,\mathbf Y_B,\mathbf Y_C)$  is $\Omega =\left( L_A^2 \times \bm{0}_{B}\times \bm{0}_{C} \right) \cup  \left( \bm{0}_{A}\times L_B^2\times  \bm{0}_{C} \right)\\ \cup  \left( \bm{0}_{A}\times \bm{0}_{B}\times L_C^2  \right)$, then
\begin{multline*}
P(\mathbf Y_A\in L_A^2,\mathbf Y_B=\bm{0}_{B},\mathbf Y_C=\bm{0}_{C} \,\text{or}\, \mathbf Y_A=\bm{0}_{A}, \mathbf Y_B\in L_B^2, \mathbf Y_C=\bm{0}_{C}\,\text{or}\,\mathbf Y_A=\bm{0}_{A}, \mathbf Y_B=\bm{0}_{B}, \mathbf Y_C\in L_C^2)\\=P(\mathbf Y_A\in L_A^2,\mathbf Y_B=\bm{0}_{B},\mathbf Y_C=\bm{0}_{C}) + P(\mathbf Y_A=\bm{0}_{A}, \mathbf Y_B\in L_B^2, \mathbf Y_C=\bm{0}_{C})\\
+ P(\mathbf Y_A=\bm{0}_{A}, \mathbf Y_B=\bm{0}_{B}, \mathbf Y_C\in L_C^2)=1.
\end{multline*}
Suppose that $P(\mathbf Y_A\in L_A^2,\mathbf Y_B=\bm{0}_{B},\mathbf Y_C=\bm{0}_{C})=\alpha_1$ and $P(\mathbf Y_A=\bm{0}_{A}, \mathbf Y_B\in L_B^2, \mathbf Y_C=\bm{0}_{C})=\alpha_2$, then $P(\mathbf Y_A=\bm{0}_{A}, \mathbf Y_B=\bm{0}_{B}, \mathbf Y_C\in L_C^2)=1-\alpha_1-\alpha_2$, where $\alpha_1$ and $\alpha_2$ satisfy the conditions of $\alpha_1>0, \alpha_2>0$ and $0<\alpha_1+\alpha_2<1$.
	Now we introduce three new random vectors $\mathbf W_A$, $\mathbf W_B$ and $\mathbf W_C$, assume that they have the distributions as follows:
	\[
F_{\mathbf{W}_A}(\mathbf w_A)=
\begin{cases}
(1-p_A)P(\mathbf U_A\leq \mathbf w_A),& \mathbf w_A \in  L_A^1,\\
(1-p_A)+p_AP(\mathbf Y_A \leq \mathbf w_A \mid \mathbf Y_A \in L_A^2, \mathbf Y_B=\bm{0}_B, \mathbf Y_C=\bm{0}_C), & \mathbf w_A \in L_A^2,
\end{cases}
\]
\[
F_{\mathbf W_B}(\mathbf w_B)=
\begin{cases}
(1-p_B)P(\mathbf U_B \leq \mathbf w_B), &  \mathbf w_B \in L_B^1,\\
(1-p_B)+p_B P(\mathbf Y_B \leq \mathbf w_B \mid \mathbf Y_A=\bm{0}_A, \mathbf Y_B \in L_B^2, \mathbf Y_C=\bm{0}_C), & \mathbf w_B \in L_B^2,
\end{cases}
\]
and
\[
F_{\mathbf W_C}(\mathbf w_C)=
\begin{cases}
(1-p_C)P(\mathbf U_C \leq \mathbf w_C), &  \mathbf w_C \in L_C^1,\\
(1-p_C)+p_C P(\mathbf Y_C \leq \mathbf w_C \mid \mathbf Y_A=\bm{0}_A, \mathbf Y_B=\bm{0}_B,\mathbf Y_C \in L_C^2), & \mathbf w_C \in L_C^2,
\end{cases}
\]
where $p_A \in (0,1)$, $p_B \in (0,1)$ $p_C \in (0,1)$, and  $\mathbf U_A$, $\mathbf U_B$, $\mathbf U_C$ are random vectors which follow the uniform distributions in $L_A^1$,  $L_B^1$ and  $L_C^1$, respectively.

To make $\mathbf W_A,\mathbf W_B,\mathbf W_C$ independent, let $F(\mathbf w_A,\mathbf w_B,\mathbf w_C)=F_{\mathbf W_A}(\mathbf w_A)F_{\mathbf W_B}(\mathbf w_B)F_{\mathbf W_C}(\mathbf w_C)$.  Furthermore, we have $\mathbf W_A \ind \mathbf W_C \mid \mathbf W_B$. Besides,
	if $\mathbf w_A \in L_A^1, \mathbf w_B \in L_B^1, \mathbf w_C \in L_C^1$,
	\[
	 F(\mathbf w_A,\mathbf w_B,\mathbf w_C)= (1-p_A)(1-p_B)(1-p_C)P(\mathbf U_A\leq \mathbf w_A)P(\mathbf U_B\leq \mathbf w_B)P(\mathbf U_C \leq \mathbf w_C).
	\]
	If $\mathbf w_A \in L_A^1, \mathbf w_B \in L_B^1, \mathbf w_C \in L_C^2$,
	\begin{multline*}
	 F(\mathbf w_A, \mathbf w_B, \mathbf w_C)=(1-p_A)(1-p_B)P(\mathbf U_A \leq \mathbf w_A)P(\mathbf U_B \leq \mathbf w_B)\\
	 \times \big\{(1-p_C)+p_C P(\mathbf Y_C \leq \mathbf w_C \mid \mathbf Y_A=\bm 0_A, \mathbf Y_B=\bm 0_B, \mathbf Y_C \in L_C^2)\big\}.
	\end{multline*}
	If $\mathbf w_A \in L_A^1, \mathbf w_B \in L_B^2, \mathbf w_C \in L_C^1$,
	\begin{multline*}
	 F(\mathbf w_A, \mathbf w_B, \mathbf w_C)=(1-p_A)(1-p_C)P(\mathbf U_A \leq \mathbf w_A)P(\mathbf U_C \leq \mathbf w_C)\\
	 \times \big\{(1-p_B)+p_B P(\mathbf Y_B \leq \mathbf w_B \mid \mathbf Y_A=\bm 0_A,  \mathbf Y_B \in L_B^2, \mathbf Y_C=\bm 0_C)\big\}.
	\end{multline*}
	If $\mathbf w_A \in L_A^1, \mathbf w_B \in L_B^2, \mathbf w_C \in L_C^2$,
	\begin{multline*}
	 F(\mathbf w_A, \mathbf w_B, \mathbf w_C)=(1-p_A)P(\mathbf U_A \leq \mathbf w_A) \big\{(1-p_B)+p_B P(\mathbf Y_B \leq \mathbf w_B \mid \mathbf Y_A=\bm 0_A,  \mathbf Y_B \in L_B^2, \mathbf Y_C=\bm 0_C)\big\}\\
	 \times \big\{(1-p_C)+p_C P(\mathbf Y_C \leq \mathbf w_C \mid \mathbf Y_A=\bm 0_A, \mathbf Y_B=\bm 0_B, \mathbf Y_C \in L_C^2)\big\}.
	\end{multline*}
	If $\mathbf w_A \in L_A^2, \mathbf w_B \in L_B^1, \mathbf w_C \in L_C^1$,
	\begin{multline*}
	 F(\mathbf w_A, \mathbf w_B, \mathbf w_C)=(1-p_B)(1-p_C)P(\mathbf U_B \leq \mathbf w_B)P(\mathbf U_C \leq \mathbf w_C)\\
	 \times \big\{(1-p_A)+p_AP(\mathbf Y_A \leq \mathbf w_A \mid \mathbf Y_A \in L_A^2, \mathbf Y_B=\bm{0}_B, \mathbf Y_C=\bm{0}_C)\big\}.
	\end{multline*}
		If $\mathbf w_A \in L_A^2, \mathbf w_B \in L_B^1, \mathbf w_C \in L_C^2$,
	\begin{multline*}
	 F(\mathbf w_A, \mathbf w_B, \mathbf w_C)=(1-p_B)P(\mathbf U_B \leq \mathbf w_B) \big\{(1-p_A)+p_AP(\mathbf Y_A \leq \mathbf w_A \mid \mathbf Y_A \in L_A^2, \mathbf Y_B=\bm{0}_B, \mathbf Y_C=\bm{0}_C)\big\}\\
	 \times \big\{(1-p_C)+p_C P(\mathbf Y_C \leq \mathbf w_C \mid \mathbf Y_A=\bm 0_A, \mathbf Y_B=\bm 0_B, \mathbf Y_C \in L_C^2)\big\}.
	\end{multline*}
	If $\mathbf w_A \in L_A^2, \mathbf w_B \in L_B^2, \mathbf w_C \in L_C^1$,
	\begin{multline*}
	 F(\mathbf w_A, \mathbf w_B, \mathbf w_C)=(1-p_C)P(\mathbf U_C \leq \mathbf w_C) \big\{(1-p_A)+p_AP(\mathbf Y_A \leq \mathbf w_A \mid \mathbf Y_A \in L_A^2, \mathbf Y_B=\bm{0}_B, \mathbf Y_C=\bm{0}_C)\big\}\\
	 \times  \big\{(1-p_B)+p_B P(\mathbf Y_B \leq \mathbf w_B \mid \mathbf Y_A=\bm 0_A,  \mathbf Y_B \in L_B^2, \mathbf Y_C=\bm 0_C)\big\}.
	\end{multline*}
		If $\mathbf w_A \in L_A^2, \mathbf w_B \in L_B^2, \mathbf w_C \in L_C^2$,
	\begin{multline*}
	 F(\mathbf w_A, \mathbf w_B, \mathbf w_C)= \big\{(1-p_A)+p_AP(\mathbf Y_A \leq \mathbf w_A \mid \mathbf Y_A \in L_A^2, \mathbf Y_B=\bm{0}_B, \mathbf Y_C=\bm{0}_C)\big\}\\
	 \times  \big\{(1-p_B)+p_B P(\mathbf Y_B \leq \mathbf w_B \mid \mathbf Y_A=\bm 0_A,  \mathbf Y_B \in L_B^2, \mathbf Y_C=\bm 0_C)\big\}\\
	 \times \big\{(1-p_C)+p_C P(\mathbf Y_C \leq \mathbf w_C \mid \mathbf Y_A=\bm 0_A, \mathbf Y_B=\bm 0_B, \mathbf Y_C \in L_C^2)\big\}.
	\end{multline*}

	If the density function of $F(\mathbf w_A,\mathbf w_B,\mathbf w_C)$ exists, then by calculation
	\[
\begin{aligned}
&f(\mathbf w_A,\mathbf w_B,\mathbf w_C)=\\
&\begin{cases}
&(1-p_A)(1-p_B)(1-p_C), \hspace{17.3em} \mathbf w_A \in L_A^1,\mathbf w_B \in L_B^1, \mathbf w_C \in L_C^1,\\
&\frac{(1-p_A)(1-p_B)p_C}{1-\alpha_1-\alpha_2}\frac{\partial P(\mathbf Y_A=\bm 0_A,\mathbf Y_B=\bm 0_B,\mathbf Y_C \leq \mathbf w_C)}{\partial \mathbf w_C},  \hspace{10.6em}    \mathbf w_A \in L_A^1,\mathbf w_B \in L_B^1, \mathbf w_C \in L_C^2,\\
&\frac{(1-p_A)p_B(1-p_C)}{\alpha_2}\frac{\partial P(\mathbf Y_A=\bm 0_A,\mathbf Y_B \leq \mathbf w_B,\mathbf Y_C=\bm 0_C)}{\partial \mathbf w_B},  \hspace{10.7em}    \mathbf w_A \in L_A^1,\mathbf w_B \in L_B^2, \mathbf w_C \in L_C^1,\\
&\frac{(1-p_A)p_B p_C}{\alpha_2(1-\alpha_1-\alpha_2)}\frac{\partial P(\mathbf Y_A=\bm 0_A,\mathbf Y_B\leq \mathbf w_B,\mathbf Y_C=\bm 0_C)}{\partial \mathbf w_B}\frac{\partial P(\mathbf Y_A=\bm 0_A,\mathbf Y_B=\bm 0_B,\mathbf Y_C\leq \mathbf w_C)}{\partial \mathbf w_C},  \hspace{0.7em} \mathbf w_A \in L_A^1,\mathbf w_B \in L_B^2, \mathbf w_C \in L_C^2,\\
&\frac{p_A(1-p_B)(1-p_C)}{\alpha_1}\frac{\partial P(\mathbf Y_A\leq \mathbf w_A,\mathbf Y_B=\bm 0_B,\mathbf Y_C=\bm 0_C)}{\partial \mathbf w_A}, \hspace{10.7em} \mathbf w_A \in L_A^2,\mathbf w_B \in L_B^1, \mathbf w_C \in L_C^1,\\
&\frac{p_A(1-p_B)p_C}{\alpha_1(1-\alpha_1-\alpha_2)}\frac{\partial P(\mathbf Y_A\leq \mathbf w_A,\mathbf Y_B=\bm 0_B,\mathbf Y_C=\bm 0_C)}{\partial \mathbf w_A}\frac{\partial P(\mathbf Y_A=\bm 0_A,\mathbf Y_B=\bm 0_B,\mathbf Y_C\leq \mathbf w_C)}{\partial \mathbf w_C}, \hspace{0.7em} \mathbf w_A \in L_A^2,\mathbf w_B \in L_B^1, \mathbf w_C \in L_C^2,\\
&\frac{p_A p_B(1-p_C)}{\alpha_1\alpha_2}\frac{\partial P(\mathbf Y_A\leq \mathbf w_A,\mathbf Y_B=\bm 0_B,\mathbf Y_C=\bm 0_C)}{\partial \mathbf w_A}\frac{\partial P(\mathbf Y_A=\bm 0_A,\mathbf Y_B\leq \mathbf w_B,\mathbf Y_C=\bm 0_C)}{\partial \mathbf w_B}, \hspace{1em} \mathbf w_A \in L_A^2,\mathbf w_B \in L_B^2, \mathbf w_C \in L_C^1,\\
&\frac{p_A p_B p_C}{\alpha_1\alpha_2(1-\alpha_1-\alpha_2)}\frac{\partial P(\mathbf Y_A\leq \mathbf w_A,\mathbf Y_B=\bm 0_B,\mathbf Y_C=\bm 0_C)}{\partial \mathbf w_A}\\
&\hspace{2em}\times \frac{\partial P(\mathbf Y_A=\bm 0_A,\mathbf Y_B\leq \mathbf w_B,\mathbf Y_C=\bm 0_C)}{\partial \mathbf w_B} \frac{\partial P(\mathbf Y_A=\bm 0_A,\mathbf Y_B=\bm 0_B,\mathbf Y_C\leq \mathbf w_C)}{\partial \mathbf w_C},\hspace{3em}\mathbf w_A \in L_A^2,\mathbf w_B \in L_B^2, \mathbf w_C \in L_C^2.
\end{cases}
\end{aligned}
\]
Since the support of $\mathbf Y=(\mathbf Y_A,\mathbf Y_B,\mathbf Y_C)$  is $\Omega =\left( L_A^2 \times \bm{0}_{B}\times \bm{0}_{C} \right) \cup  \left( \bm{0}_{A}\times L_B^2\times  \bm{0}_{C} \right) \cup  \left( \bm{0}_{A}\times \bm{0}_{B}\times L_C^2  \right)$, then $\partial P(\mathbf Y_A=\bm 0_A,\mathbf Y_B=\bm 0_B,\mathbf Y_C \leq \mathbf w_C)/\partial \mathbf w_C \neq 0$ for $\mathbf w_C \in L_C^2$, $\partial P(\mathbf Y_A=\bm 0_A,\mathbf Y_B \leq \mathbf w_B,\mathbf Y_C=\bm 0_C)/\partial \mathbf w_B \neq 0$ for $\mathbf w_B \in L_B^2$ and $\partial P(\mathbf Y_A\leq \mathbf w_A,\mathbf Y_B=\bm 0_B,\mathbf Y_C=\bm 0_C)/\partial \mathbf w_A\neq 0$ for $\mathbf w_A\in L_A^2$. Thus, from $f(\mathbf w_A,\mathbf w_B,\mathbf w_C)$, we
can obtain that the support of $\mathbf W=(\mathbf W_A,\mathbf W_B,\mathbf W_C)$ is $[0,\infty)^{|A|+|B|+|C|}$, which contains  $\Omega$.  Based on the definition of conditional density in \citet[\S 7.4]{shorack2017probability}, if  $\mathbf w_A
\in L_A^2$, $\mathbf w_B \in L_B^2$ and  $\mathbf w_C \in L_C^2$, then we have that the conditional densities at $(\mathbf w_A,\bm 0_B,\bm 0_C)$, $(\bm 0_A,\mathbf w_B,\bm 0_C)$ and  $(\bm 0_A,\bm 0_B,\mathbf w_C)$,  given that all the points are on the curve of $\Omega$ are
\[
\begin{aligned}
&\frac{f(\mathbf w_A,\bm 0_B,\bm 0_C)}{\int_{L_A^2} f(\mathbf w_A,\bm 0_B,\bm 0_C) {\rm d} \mathbf w_A  + \int_{L_B^2} f(\bm 0_A, \mathbf w_B, \bm 0_C) {\rm d} \mathbf w_B + \int_{L_C^2} f(\bm 0_A, \bm 0_B, \mathbf w_C) {\rm d} \mathbf w_C}\\
=&\frac{p_A(1-p_B)(1-p_C)\frac{\partial P(\mathbf Y_A\leq \mathbf w_A, \mathbf Y_B=\bm 0_B, \mathbf Y_C=\bm 0_C )}{\partial \mathbf w_A}
}{\alpha_1\{p_A(1-p_B)(1-p_C)+ (1-p_A)p_B(1-p_C)+(1-p_A)(1-p_B)p_C \}   },
\end{aligned}
\]
\[
\begin{aligned}
&\frac{f(\bm 0_A,\mathbf w_B,\bm 0_C)}{ \int_{L_A^2} f(\mathbf w_A,\bm 0_B,\bm 0_C) {\rm d} \mathbf w_A  + \int_{L_B^2} f(\bm 0_A, \mathbf w_B,\bm 0_C) {\rm d} \mathbf w_B + \int_{L_C^2} f(\bm 0_A, \bm 0_B, \mathbf w_C) {\rm d} \mathbf w_C}\\
=&\frac{(1-p_A)p_B(1-p_C)\frac{\partial P(\mathbf Y_A=\bm 0_A,\mathbf Y_B\leq \mathbf w_B,\mathbf Y_C=\bm 0_C)}{\partial \mathbf w_B} }{\alpha_2\{p_A(1-p_B)(1-p_C)+ (1-p_A)p_B(1-p_C)+(1-p_A)(1-p_B)p_C \}   },
\end{aligned}
\]
and
\[
\begin{aligned}
&\frac{f(\bm 0_A,\bm 0_B,\mathbf w_C)}{ \int_{L_A^2} f(\mathbf w_A,\bm 0_B,\bm 0_C) {\rm d} \mathbf w_A  + \int_{L_B^2} f(\bm 0_A, \mathbf w_B,\bm 0_C) {\rm d} \mathbf w_B+ \int_{L_C^2} f(\bm 0_A, \bm 0_B, \mathbf w_C) {\rm d} \mathbf w_C }\\
=&\frac{(1-p_A)(1-p_B)p_C\frac{\partial P(\mathbf Y_A=\bm 0_A,\mathbf Y_B=\bm 0_B,\mathbf Y_C\leq \mathbf w_C)}{\partial \mathbf w_C} }{(1-\alpha_1-\alpha_2)\{p_A(1-p_B)(1-p_C)+ (1-p_A)p_B(1-p_C)+(1-p_A)(1-p_B)p_C \}   }.
\end{aligned}
\]
Then we can get the value of $P(\mathbf W_A\leq \mathbf w_A, \mathbf W_B\leq \mathbf w_B, \mathbf W_C\leq \mathbf w_C\mid \mathbf W_A\in L_A^2,\mathbf W_B=\bm{0}_{B},\mathbf W_C=\bm{0}_{C} \,\text{or}\, \mathbf W_A=\bm{0}_{A}, \mathbf W_B\in L_B^2, \mathbf W_C=\bm{0}_{C}\,\text{or}\,\mathbf W_A=\bm{0}_{A}, \mathbf W_B=\bm{0}_{B}, \mathbf W_C\in L_C^2)=P(\mathbf W_A\leq \mathbf w_A, \mathbf W_B\leq \mathbf w_B, \mathbf W_C\leq \mathbf w_C\mid \mathbf W\in \Omega)$. When $\mathbf w_A \in L_A^1,\mathbf w_B \in L_B^1, \mathbf w_C \in L_C^1$,
\[
P(\mathbf W_A\leq \mathbf w_A, \mathbf W_B\leq \mathbf w_B, \mathbf W_C\leq \mathbf w_C\mid \mathbf W\in \Omega)
=0=P(\mathbf Y_A\leq \mathbf w_A, \mathbf Y_B\leq \mathbf w_B, \mathbf Y_C\leq \mathbf w_C).
\]
When $\mathbf w_A \in L_A^1,\mathbf w_B \in L_B^1, \mathbf w_C \in L_C^2$,
\[
\begin{aligned}
  &P(\mathbf W_A\leq \mathbf w_A, \mathbf W_B\leq \mathbf w_B, \mathbf W_C\leq \mathbf w_C\mid \mathbf W\in \Omega)\\
  =&\frac{\int _{[\bm{0}_C,\mathbf w_C]\setminus[0,1]^{|C|} } f(\bm 0_A,\bm 0_B,\mathbf t){\rm d} \mathbf t}{ \int_{L_A^2} f(\mathbf w_A,\bm 0_B,\bm 0_C) {\rm d} \mathbf w_A  + \int_{L_B^2} f(\bm 0_A, \mathbf w_B,\bm 0_C) {\rm d} \mathbf w_B + \int_{L_C^2} f(\bm 0_A, \bm 0_B, \mathbf w_C) {\rm d} \mathbf w_C }\\
    =&\frac{(1-p_A)(1-p_B)p_C P(\mathbf Y_A=\bm 0_A,\mathbf Y_B=\bm 0_B,\mathbf Y_C\leq \mathbf w_C)}{(1-\alpha_1-\alpha_2)\{p_A(1-p_B)(1-p_C)+ (1-p_A)p_B(1-p_C)+(1-p_A)(1-p_B)p_C \}}\\
        =&\frac{(1-p_A)(1-p_B)p_C P(\mathbf Y_A\leq \mathbf w_A,\mathbf Y_B\leq \mathbf w_B,\mathbf Y_C\leq \mathbf w_C)}{(1-\alpha_1-\alpha_2)\{p_A(1-p_B)(1-p_C)+ (1-p_A)p_B(1-p_C)+(1-p_A)(1-p_B)p_C \}}.
\end{aligned}
\]
When $\mathbf w_A \in L_A^1,\mathbf w_B \in L_B^2, \mathbf w_C \in L_C^1$,
\[
\begin{aligned}
  &P(\mathbf W_A\leq \mathbf w_A, \mathbf W_B\leq \mathbf w_B, \mathbf W_C\leq \mathbf w_C\mid \mathbf W\in \Omega)\\
  =&\frac{\int _{[\bm{0}_B,\mathbf w_B]\setminus[0,1]^{|B|} } f(\bm{0}_A,\mathbf t,\bm{0}_C){\rm d} \mathbf t}{ \int_{L_A^2} f(\mathbf w_A,\bm{0}_B,\bm{0}_C) {\rm d} \mathbf w_A  + \int_{L_B^2} f(\bm{0}_A, \mathbf w_B,\bm{0}_C) {\rm d} \mathbf w_B + \int_{L_C^2} f(\bm{0}_A, \bm{0}_B, \mathbf w_C) {\rm d} \mathbf w_C }\\
    =&\frac{(1-p_A)p_B(1-p_C)P(\mathbf Y_A= \bm{0}_A,\mathbf Y_B\leq \mathbf w_B,\mathbf Y_C=\bm{0}_C)}{\alpha_2\{p_A(1-p_B)(1-p_C)+ (1-p_A)p_B(1-p_C)+(1-p_A)(1-p_B)p_C \}}\\
        =&\frac{(1-p_A)p_B(1-p_C)P(\mathbf Y_A\leq \mathbf w_A,\mathbf Y_B\leq \mathbf w_B,\mathbf Y_C\leq \mathbf w_C)}{\alpha_2\{p_A(1-p_B)(1-p_C)+ (1-p_A)p_B(1-p_C)+(1-p_A)(1-p_B)p_C \}}.
\end{aligned}
\]
When $\mathbf w_A \in L_A^1,\mathbf w_B \in L_B^2, \mathbf w_C \in L_C^2$,	due to $P(\mathbf Y_A\leq \mathbf w_A,\mathbf Y_B\leq \mathbf w_B,\mathbf Y_C\leq \mathbf w_C)=P(\mathbf Y_A=\bm 0_A,\mathbf Y_B\leq \mathbf w_B,\mathbf Y_C=\bm 0_C)+P(\mathbf Y_A=\bm 0_A,\mathbf Y_B=\bm 0_B,\mathbf Y_C\leq \mathbf w_C)$,
\[
\begin{aligned}
  &P(\mathbf W_A\leq \mathbf w_A, \mathbf W_B\leq \mathbf w_B, \mathbf W_C\leq \mathbf w_C\mid \mathbf W\in \Omega)\\
  =&\frac{\int _{[\bm{0}_B,\mathbf w_B]\setminus[0,1]^{|B|} } f(\bm{0}_A,\mathbf t,\bm{0}_C){\rm d} \mathbf t+\int _{[\bm{0}_C,\mathbf w_C]\setminus[0,1]^{|C|} } f(\bm{0}_A,\bm{0}_B,\mathbf t){\rm d} \mathbf t}{  \int_{L_A^2} f(\mathbf w_A,\bm{0}_B,\bm{0}_C) {\rm d} \mathbf w_A  + \int_{L_B^2} f(\bm{0}_A, \mathbf w_B,\bm{0}_C) {\rm d} \mathbf w_B + \int_{L_C^2} f(\bm{0}_A, \bm{0}_B, \mathbf w_C) {\rm d} \mathbf w_C }\\
        =&\frac{\{(1-p_A)p_B(1-p_C)(1-\alpha_1-\alpha_2)-(1-p_A)(1-p_B)p_C\alpha_2\}P(\mathbf Y_A=\bm{0}_A,\mathbf Y_B\leq \mathbf w_B,\mathbf Y_C=\bm{0}_C)}{\alpha_2(1-\alpha_1-\alpha_2)\{p_A(1-p_B)(1-p_C)+ (1-p_A)p_B(1-p_C)+(1-p_A)(1-p_B)p_C \}}\\
       &+ \frac{(1-p_A)(1-p_B)p_C P(\mathbf Y_A\leq \mathbf w_A,\mathbf Y_B\leq \mathbf w_B,\mathbf Y_C\leq \mathbf w_C)}{(1-\alpha_1-\alpha_2)\{p_A(1-p_B)(1-p_C)+ (1-p_A)p_B(1-p_C)+(1-p_A)(1-p_B)p_C \}}.
\end{aligned}
\]
When $\mathbf w_A \in L_A^2,\mathbf w_B \in L_B^1, \mathbf w_C \in L_C^1$,
\[
\begin{aligned}
  &P(\mathbf W_A\leq \mathbf w_A, \mathbf W_B\leq \mathbf w_B, \mathbf W_C\leq \mathbf w_C\mid \mathbf W\in \Omega)\\
  =&\frac{\int _{[\bm{0}_A,\mathbf w_A]\setminus[0,1]^{|A|} } f(\mathbf t,\bm{0}_B,\bm{0}_C){\rm d} \mathbf t}{ \int_{L_A^2} f(\mathbf w_A,\bm{0}_B,\bm{0}_C) {\rm d} \mathbf w_A  + \int_{L_B^2} f(\bm{0}_A, \mathbf w_B,\bm{0}_C) {\rm d} \mathbf w_B + \int_{L_C^2} f(\bm{0}_A, \bm{0}_B, \mathbf w_C) {\rm d} \mathbf w_C }\\
    =&\frac{p_A(1-p_B)(1-p_C)P(\mathbf Y_A\leq \mathbf w_A,\mathbf Y_B=\bm{0}_B,\mathbf Y_C=\bm{0}_C)}{\alpha_1\{p_A(1-p_B)(1-p_C)+ (1-p_A)p_B(1-p_C)+(1-p_A)(1-p_B)p_C \}}\\
        =&\frac{p_A(1-p_B)(1-p_C)P(\mathbf Y_A\leq \mathbf w_A,\mathbf Y_B\leq \mathbf w_B,\mathbf Y_C\leq \mathbf w_C)}{\alpha_1\{p_A(1-p_B)(1-p_C)+ (1-p_A)p_B(1-p_C)+(1-p_A)(1-p_B)p_C \}}.
\end{aligned}
\]
When $\mathbf w_A \in L_A^2,\mathbf w_B \in L_B^1, \mathbf w_C \in L_C^2$,	due to $P(\mathbf Y_A\leq \mathbf w_A,\mathbf Y_B\leq \mathbf w_B,\mathbf Y_C\leq \mathbf w_C)=P(\mathbf Y_A\leq \mathbf w_A,\mathbf Y_B=\bm 0_B,\mathbf Y_C=\bm 0_C)+P(\mathbf Y_A=\bm 0_A,\mathbf Y_B=\bm 0_B,\mathbf Y_C\leq \mathbf w_C)$,
\[
\begin{aligned}
  &P(\mathbf W_A\leq \mathbf w_A, \mathbf W_B\leq \mathbf w_B, \mathbf W_C\leq \mathbf w_C\mid \mathbf W\in \Omega)\\
  =&\frac{\int _{[\bm{0}_A,\mathbf w_A]\setminus[0,1]^{|A|} } f(\mathbf t,\bm{0}_B,\bm{0}_C){\rm d} \mathbf t+\int _{[\bm{0}_C,\mathbf w_C]\setminus[0,1]^{|C|} } f(\bm{0}_A,\bm{0}_B,\mathbf t){\rm d} \mathbf t}{  \int_{L_A^2} f(\mathbf w_A,\bm{0}_B,\bm{0}_C) {\rm d} \mathbf w_A  + \int_{L_B^2} f(\bm{0}_A, \mathbf w_B,\bm{0}_C) {\rm d} \mathbf w_B + \int_{L_C^2} f(\bm{0}_A, \bm{0}_B, \mathbf w_C) {\rm d} \mathbf w_C }\\
        =&\frac{\{p_A(1-p_B)(1-p_C)(1-\alpha_1-\alpha_2)-(1-p_A)(1-p_B)p_C\alpha_1\}P(\mathbf Y_A\leq \mathbf w_A,\mathbf Y_B=\bm{0}_B,\mathbf Y_C=\bm{0}_C)}{\alpha_1(1-\alpha_1-\alpha_2)\{p_A(1-p_B)(1-p_C)+ (1-p_A)p_B(1-p_C)+(1-p_A)(1-p_B)p_C \}}\\
       &+ \frac{(1-p_A)(1-p_B)p_C P(\mathbf Y_A\leq \mathbf w_A,\mathbf Y_B\leq \mathbf w_B,\mathbf Y_C\leq \mathbf w_C)}{(1-\alpha_1-\alpha_2)\{p_A(1-p_B)(1-p_C)+ (1-p_A)p_B(1-p_C)+(1-p_A)(1-p_B)p_C \}}.
\end{aligned}
\]
When $\mathbf w_A \in L_A^2,\mathbf w_B \in L_B^2, \mathbf w_C \in L_C^1$,	due to $P(\mathbf Y_A\leq \mathbf w_A,\mathbf Y_B\leq \mathbf w_B,\mathbf Y_C\leq \mathbf w_C)=P(\mathbf Y_A\leq \mathbf w_A,\mathbf Y_B=\bm 0_B,\mathbf Y_C=\bm 0_C)+P(\mathbf Y_A=\bm 0_A,\mathbf Y_B\leq \mathbf w_B,\mathbf Y_C=\bm 0_C)$,
\[
\begin{aligned}
  &P(\mathbf W_A\leq \mathbf w_A, \mathbf W_B\leq \mathbf w_B, \mathbf W_C\leq \mathbf w_C\mid \mathbf W\in \Omega)\\
  =&\frac{\int _{[\bm{0}_A,\mathbf w_A]\setminus[0,1]^{|A|} } f(\mathbf t,\bm{0}_B,\bm{0}_C){\rm d} \mathbf t+\int _{[\bm{0}_B,\mathbf w_B]\setminus[0,1]^{|B|} } f(\bm{0}_A,\mathbf t,\bm{0}_C){\rm d}\mathbf t}{  \int_{L_A^2} f(\mathbf w_A,\bm{0}_B,\bm{0}_C) {\rm d} \mathbf w_A  + \int_{L_B^2} f(\bm{0}_A, \mathbf w_B,\bm{0}_C) {\rm d} \mathbf w_B + \int_{L_C^2} f(\bm{0}_A, \bm{0}_B, \mathbf w_C) {\rm d} \mathbf w_C }\\
        =&\frac{\{p_A(1-p_B)(1-p_C)\alpha_2-(1-p_A)p_B(1-p_C)\alpha_1\}P(\mathbf Y_A\leq \mathbf w_A,\mathbf Y_B=\bm{0}_B,\mathbf Y_C=\bm{0}_C)}{\alpha_1\alpha_2\{p_A(1-p_B)(1-p_C)+ (1-p_A)p_B(1-p_C)+(1-p_A)(1-p_B)p_C \}}\\
       &+ \frac{(1-p_A)p_B(1-p_C)P(\mathbf Y_A\leq \mathbf w_A,\mathbf Y_B\leq \mathbf w_B,\mathbf Y_C\leq \mathbf w_C)}{\alpha_2\{p_A(1-p_B)(1-p_C)+ (1-p_A)p_B(1-p_C)+(1-p_A)(1-p_B)p_C \}}.
\end{aligned}
\]
When $\mathbf w_A \in L_A^2,\mathbf w_B \in L_B^2, \mathbf w_C \in L_C^2$,	due to $P(\mathbf Y_A\leq \mathbf w_A,\mathbf Y_B\leq \mathbf w_B,\mathbf Y_C\leq \mathbf w_C)=P(\mathbf Y_A\leq \mathbf w_A,\mathbf Y_B=\bm 0_B,\mathbf Y_C=\bm 0_C)+P(\mathbf Y_A=\bm 0_A,\mathbf Y_B\leq \mathbf w_B,\mathbf Y_C=\bm 0_C)+P(\mathbf Y_A=\bm 0_A,\mathbf Y_B=\bm 0_B,\mathbf Y_C\leq \mathbf w_C)$,
\begin{multline*}
  P(\mathbf W_A\leq \mathbf w_A, \mathbf W_B\leq \mathbf w_B, \mathbf W_C\leq \mathbf w_C\mid \mathbf W\in \Omega)\\
  =\frac{\int _{[\bm{0}_A,\mathbf w_A]\setminus[0,1]^{|A|} } f(\mathbf t,\bm{0}_B,\bm{0}_C){\rm d} \mathbf t+\int _{[\bm{0}_B,\mathbf w_B]\setminus[0,1]^{|B|} } f(\bm{0}_A,\mathbf t,\bm{0}_C){\rm d} \mathbf t+\int _{[\bm{0}_C,\mathbf w_C]\setminus[0,1]^{|C|} } f(\bm{0}_A,\bm{0}_B,\mathbf t){\rm d} \mathbf t}{  \int_{L_A^2} f(\mathbf w_A,\bm{0}_B,\bm{0}_C) {\rm d} \mathbf w_A  + \int_{L_B^2} f(\bm{0}_A, \mathbf w_B,\bm{0}_C) {\rm d} \mathbf w_B + \int_{L_C^2} f(\bm{0}_A, \bm{0}_B, \mathbf w_C) {\rm d} \mathbf w_C }\\
        =\Bigg[\frac{\{p_A(1-p_B)(1-p_C)(1-\alpha_1-\alpha_2)-(1-p_A)(1-p_B)p_C\alpha_1\}P(\mathbf Y_A\leq \mathbf w_A,\mathbf Y_B=\bm{0}_B,\mathbf Y_C=\bm{0}_C)}{\alpha_1(1-\alpha_1-\alpha_2)}\\
      + \frac{\{(1-p_A)p_B(1-p_C)(1-\alpha_1-\alpha_2)-(1-p_A)(1-p_B)p_C\alpha_2\}P(\mathbf Y_A=\bm{0}_A,\mathbf Y_B\leq \mathbf w_B,\mathbf Y_C=\bm{0}_C)}{\alpha_2(1-\alpha_1-\alpha_2)}\\
      +\frac{(1-p_A)(1-p_B)p_C P(\mathbf Y_A\leq \mathbf w_A,\mathbf Y_B\leq \mathbf w_B,\mathbf Y_C\leq \mathbf w_C)}{1-\alpha_1-\alpha_2}\Bigg]\\
      \times \frac{1}{p_A(1-p_B)(1-p_C)+ (1-p_A)p_B(1-p_C)+(1-p_A)(1-p_B)p_C}.
\end{multline*}
By comparison, we can find that $P(\mathbf W_A\leq \mathbf w_A, \mathbf W_B\leq \mathbf w_B, \mathbf W_C\leq \mathbf w_C\mid \mathbf W\in \Omega)=P(\mathbf Y_A\leq \mathbf w_A, \mathbf Y_B\leq \mathbf w_B, \mathbf Y_C\leq \mathbf w_C)$ if and only if
\begin{equation}
\label{eqn:conditon2}
\begin{cases}
\alpha_1=\frac{p_A(1-p_B)(1-p_C)}{p_A(1-p_B)(1-p_C)+ (1-p_A)p_B(1-p_C)+(1-p_A)(1-p_B)p_C},\\
\alpha_2=\frac{(1-p_A)p_B(1-p_C)}{p_A(1-p_B)(1-p_C)+ (1-p_A)p_B(1-p_C)+(1-p_A)(1-p_B)p_C}.
\end{cases}
\end{equation}
We have hence shown that under condition \eqref{eqn:conditon2}, $(\mathbf W \mid \mathbf W\in \Omega) \,{\buildrel d \over =}\, \mathbf Y$ if the density function of $F(\mathbf w_A,\mathbf w_B,\mathbf w_C)$ exists.

If the density function of  $F(\mathbf w_A,\mathbf w_B,\mathbf w_C)$ does not exist, from the definition of conditional probability in \citet[\S 7.4]{shorack2017probability}, we know that the above formulas for $P(\mathbf W_A\leq \mathbf w_A, \mathbf W_B\leq \mathbf w_B, \mathbf W_C\leq \mathbf w_C\mid \mathbf W\in \Omega)$ still satisfy the condition for conditional probability. Thus, we still can define $P(\mathbf W_A\leq \mathbf w_A, \mathbf W_B\leq \mathbf w_B, \mathbf W_C\leq \mathbf w_C\mid \mathbf W\in \Omega)$ in the same way as before. Under condition \eqref{eqn:conditon2}, we also have $P(\mathbf W_A\leq \mathbf w_A, \mathbf W_B\leq \mathbf w_B, \mathbf W_C\leq \mathbf w_C\mid \mathbf W\in \Omega)=P(\mathbf Y_A\leq \mathbf w_A,\mathbf Y_B\leq \mathbf w_B,\mathbf Y_C\leq \mathbf w_C)$. Then $(\mathbf W \mid \mathbf W\in \Omega) \,{\buildrel d \over =}\, \mathbf Y$.

	In summary, according to the definition of $\mathbf Y_A \ind_o \mathbf Y_C \mid  \mathbf Y_B$, we have $\mathbf Y_A \ind_o \mathbf Y_C \mid  \mathbf Y_B$.
\end{proof}

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\end{document}