I know that $E[E(X\mid Y,Z)\mid Y]=E(X\mid Y)$.
What if $E[E(X\mid Y)\mid Y,Z]=?$ It has any meaning?
Conditional probability, $E[E(X\mid Y)\mid Y,Z]=?$
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$\begingroup$
probability
conditional-expectation
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0Use tower property of conditional expectation. – 2017-02-07
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0Thanks. The tower property can be applied to the first case. How do I use it in case of $E[E(X\mid Y)\mid Y,Z]$? – 2017-02-07
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0Note that $\sigma(Y)$ is a sub-sigma algebra of $\sigma(Y,Z)$ so by tower property $E(X|Y)=E[E(X|Y,Z)|Y]=E[E(X|Y)|Y,Z]$ – 2017-02-07
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0Thank you so much, Landon! – 2017-02-07
1 Answers
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$E(X|Y)$ is $Y$-measurable and thus $Y,Z$-measureable so can still be pulled out of the conditional expectation. So $E(E(X|Y)|Y,Z) = E(X|Y)$
It's like asking about $E(f(Y)|Y,Z)$. The $Z$ might seem to complicate things, but the fact that we're conditioning on $Y$ means we can treat the $f(Y)$ as constant and pull it out of the expectation, giving $f(Y)E(1|Y,Z) =f(Y).$