Conditional independence property: weak union

Question

Let $(X,Y,W,Z)$ be disjoint sets of random variables each with finite space. Then prove that if $\Pr(X\mid W,Y \cup Z)=\Pr(X\mid W)$ then $\Pr(X\mid Y,Z \cup W) = \Pr(X\mid Z \cup W)$. This is sometimes referred to as weak union in conditional independence. i am having hard time to prove this. Can someone help me to prove this? Thanks

Answer 1

I assume that the question is about four random variables $X$, $Y$, $Z$, and $W$ with finite state space and that $\alpha\cup\beta$ is the joint random variable consisting of the pair $(\alpha,\beta)$.

The reason the equality is true is that the distribution of $X$ conditioned on $W$ and $Z$ is given by taking the expectation over $Y$ of the distribution of $X$ conditioned on $W$, $Y$, and $Z$. Since, by assumption, this latter distribution is independent of $Y$ and $Z$, taking the expectation over $Y$ does not change it. Therefore $$ {\Bbb P}(X\mid Z \cup W)={\Bbb P}(X\mid W), $$ and since ${\Bbb P}(X\mid W, Y \cup Z)$ and ${\Bbb P}(X\mid Y, Z \cup W)$ are two different ways of writing the same thing, which is the probability distribution of $X$ conditioned on $W$, $Y$, and $Z$, this proves the result.