WeakUnion

Prove

{$ (X \bot Y,W | Z) \rightarrow (X \bot Y | Z,W) $}

Use Conditional Independence definition: {$ (A \bot B | C) \rightarrow p(A|B,C) = p(A|C) $}

on each of the two parts of the theorem

{$ (X \bot Y,W | Z) \rightarrow p(X|Y,W,Z) = p(X|Z) $}

{$ (X \bot Y | Z,W) \rightarrow P(X|Y,W,Z) = P(X|Z,W)$}

but {$ p(X|Z) = P(X|Z,W)$}

Why?

{$ p(X|Z)$} is the expectation over Y and W of {$ p(X|Y,W,Z) $}, and {$P(X|Z,W)$} is the expectation over Y of the same quantity. But since we assumed X is conditionally independent of Y and W, taking the expectation over Y or W or both does not change the value. Back to Lectures