AnswerBun.com

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,
end{align*}

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

Add your own answers!

Related Questions

Tangent Bundle of Product Manifold

1  Asked on January 5, 2022

 

Is this proof of propositional logic valid

2  Asked on January 5, 2022 by mauricio-vega

 

Solving equation $y^2-2ln(y)=x^2$

1  Asked on January 5, 2022 by majdgh

   

Possible Rule of thumb for sums of variances?

2  Asked on January 5, 2022 by portmadeleinecrumpet

   

Ask a Question

Get help from others!

© 2022 AnswerBun.com. All rights reserved. Sites we Love: PCI Database, MenuIva, UKBizDB, Menu Kuliner, Sharing RPP, SolveDir