Total expectation under measure theory

Mathematics Asked by Jiexiong687691 on December 6, 2020

I want to show that $mathbb {E}(mathbb {E}(Xmid mathcal {F}))=mathbb {E}X$. My thought is that

begin{align*}mathbb {E}(mathbb {E}(Xmid mathcal {F}))=sum_{Omega_{i}} frac {mathbb {E}(X;Omega_{i})}{mathbb {P}(Omega_{i})}mathbb {P}(Omega_{i})=mathbb{E}X,

where $Omega_{i}’s$ are either finite or infinitely countable partition of $Omega$ into disjoint sets, and each one has positive probability. I wonder if it’s always true that there exists such $Omega_{i}’s$ such that $mathcal {F}=sigma(Omega_{1},Omega_{2},ldots)$ with the desired properties? Thanks.

One Answer

Not every sigma field is countably generated and not many sigma fields are generated by countable partitions.

$E(E(X|mathcal F) I_A)=EXI_A$ for all $A in mathcal F$ by the definition of $E(X|mathcal F)$. Just taking $A =Omega$ gives $E(E(X|mathcal F)) =EX$

Correct answer by Kavi Rama Murthy on December 6, 2020

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