Suppose that E(X|Y,Z,W)=E(X|Y)E(X| Y,Z,W) = E(X|Y), is it true that E(XZ|Y)=E(X|Y)E(Z|Y)E(XZ| Y) = E(X|Y)E(Z|Y)? [on hold]

Suppose that X,Y,Z,WX,Y,Z,W are all random variables.

If we have that E(X|Y,Z,W)=E(X|Y)E(X| Y,Z,W) = E(X|Y),

is it true that E(XZ|Y)=E(X|Y)E(Z|Y)E(XZ| Y) = E(X|Y)E(Z|Y) and that E(XW|Y)=E(X|Y)E(W|Y)E(XW| Y) = E(X|Y)E(W|Y)?

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1 Answer
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Yes
E(XZ∣Y)=E(E(XZ∣Y,Z,W)∣Y)=E(ZE(X∣Y,Z,W)∣Y)=E(ZE(X∣Y)∣Y)=E(X∣Y)E(Z∣Y),E(XZ\mid Y)=E(E(XZ\mid Y,Z,W)\mid Y)=E(ZE(X\mid Y,Z,W)\mid Y)=E(ZE(X\mid Y)\mid Y)=E(X\mid Y)E(Z\mid Y),

using the “tower” property, then “taking out what is known”. This is conditional independence.