I have a relationship to prove for the random variables $x,y,z$: -
$(x\perp z | y) \hspace{5mm} and \hspace{5mm} (x \perp y | z) \Rightarrow (x \perp z,y)$
We are able to rewrite the terms on LHS using conditional independence relation. However, we are facing difficulty interpreting the term on the RHS i.e. $(x \perp z,y)$. I mean how to write this mathematically ?
We tried to use $(x \perp z,y) = (x \perp z)(x \perp y )$ but we aren't sure if it is correct ?
Any help ? We do not need the answer to the above proving question; we will do it our self; just want to know how to decompose the term in the RHS
$(x\perp y,z)$ exactly when for all subsets ($A,B,C$) of the supports of the respective random variables we have: $$\mathsf P(x\in A, y\in B,z\in C)=\mathsf P(x\in A)\cdot\mathsf P(y\in B,z\in C)$$
Similarly $(x\perp y\mid z)$ exactly when $$\mathsf P(x\in A,y\in B\mid z\in C)=\mathsf P(x\in A\mid z\in C)\cdot \mathsf P(y\in B\mid z\in C)$$
Likewise $(x\perp z\mid y)$ exactly when $$\mathsf P(x\in A,z\in C\mid y\in B)=\mathsf P(x\in A\mid y\in B)\cdot \mathsf P(z\in C\mid y\in B)$$