Here Independence and conditional expectation is stated that $E(f(X)g(Y))=E(f(X))E(g(Y))$ iff $E(h(X)|Y) = E(h(X))$. Now I'm wondering if this generalizes to independence in the conditional sense, i.e. does it hold that $E(f(X)g(Y)|A)=E(f(X)|A)E(g(Y)|A)$ iff $E(h(X)|Y \cup A)=E(h(X)|A)$, where $Y \cup A$ is the $\sigma-$algebra generated by $Y \cup A$. I can't seem to find anything on the subject.
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1What is the notation $A\cup Y$ supposed to mean when $A$ is a class of events (or an event?) and $Y$ a random variable? – Did Oct 26 '14 at 08:36
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If the probability space is $(\Omega, \mathcal{F}, P)$ then $A$ is a sub $\sigma-$algebra of $\mathcal{F}$, $Y$ the $\sigma-$algebra generated by $Y$, that is ${Y^{-1}(B) ;:; B \in \mathcal{B}}$ and $Y \cup A$ the $\sigma-$algebra generated by $Y \cup A$ ie the smallest $\sigma-$algebra containing $Y \cup A$ – Anand Oct 26 '14 at 11:22
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Two faulty notations: "Y the σ−algebra generated by Y" and "Y∪A the σ−algebra generated by Y∪A". Please avoid these. – Did Oct 26 '14 at 12:35
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Well, I will just write $\sigma(Y)$ for the smallest $\sigma-$algebra generated by $Y$ and likewise $\sigma(\sigma(Y) \cup A)$. But this still doesn't answer my question, I've looked at it myself some more, I think I got the if part (from right to left) if I can prove $E(f(X)g(Y)|\sigma(\sigma(Y) \cup A))=E(g(Y)|\sigma(\sigma(Y) \cup A))E(f(X)|\sigma(\sigma(Y) \cup A))$, is this true, if so why? Still completely stuck on the only if part. – Anand Oct 26 '14 at 12:54
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A more basic question: To prove $\mathbb{E}(h(X) | \sigma(\sigma(Y) \cup A))=\mathbb{E}(h(X) |A))$ it has to be true that $\mathbb{E}(h(X) | \sigma(\sigma(Y) \cup A))$ is $A$-measurable, by definition it's $\sigma(\sigma(Y) \cup A)$-measurable, but then why is it $A$-measurable?
Thanx in advance.
Anand
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