 # Let ( <mi mathvariant="normal">&#x03A3;<!-- Σ --> , F , P ) be a probability Brenden Tran 2022-06-14 Answered
Let $\left(\mathrm{\Sigma },F,P\right)$ be a probability space, let $X$ be an $F$-random variable, let $G$ be a sub-sigma algebra of $F$, let $\omega \in \mathrm{\Omega }$, let $A$ be the intersection of all the sets in $G$ that contain $\omega$, and suppose that $A\in F$. Then is it true that $E\left(X|G\right)\left(\omega \right)=E\left(X|A\right)$?
I just wrote this statement down based on my intuition of what conditional expectation means, and I want to verify if it’s actually true. Note that sigma algebras need not be closed under uncountable intersections, so $A$ need not be an element of $G$. But I assumed $A$ is at least an element of $F$ so that $E\left(X|A\right)$ is meaningful.
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No, when you condition on a sigma field, there is little connection to the basic definition of conditionings. The only result I know is when you condition on a discrete random variable. In other words, let $Z$ be an integrable random variable and let $X$ be a discrete random variable. Suppose all the values $X$ take with positive probability are $\left({x}_{k}\right).$ Then
$\mathbf{E}\left(Z\mid X\right)=\sum _{k}\mathbf{E}\left(Z\mid X={x}_{k}\right){\mathbf{1}}_{\left\{X={x}_{k}\right\}}$
and here $\mathbf{E}\left(Z\mid X={x}_{k}\right)$ is the usual conditional expectation over an event $A$ with positive probabiltiy $\mathbf{E}\left(Z\mid A\right)=\frac{1}{\mathbf{P}\left(A\right)}\underset{A}{\int }Zd\mathbf{P}.$ (Note that when $Z={\mathbf{1}}_{\left\{Y\in B\right\}}$ then you have
$\mathbf{E}\left(Z\mid A\right)=\frac{P\left(Y\in B,X={x}_{k}\right)}{P\left(X={x}_{k}\right)}=P\left(Y\in B\mid X={x}_{k}\right).\right)$