3

The question is, Given a probability space $(\Omega, F, P)$, a sigma field $G \subset F$, random variables $X,Y$ and a measurable function $\phi$ such that $X \in G$, $Y$ is independent with $G$ and $E[|\phi(X,Y)|] < \infty$. Then,

Show that $E[\phi(X,Y)|G]= g(X)$, where $g(x) = E[\phi(x,Y)]$ for all $x$.

I proved this result when $X,Y$ are independent. What is direction to prove this if independence condition is not given?

User124356
  • 1,617
  • As $X$ is $G$-measurable (do you mean this with $X \in G$?) and $Y$ is $G$-independent, $X$ and $Y$ are independent.$\$Moreover: What do you mean by for all $x$? – martini Feb 06 '22 at 16:49
  • @martini yes $X$ is $G$- measurable means $X \in G$. – User124356 Feb 07 '22 at 00:44
  • Did you mean something like this? $$E[\phi(X,Y)|G]=E[\phi(x,Y)]|_{x=X}$$ – Brian Moehring Feb 07 '22 at 01:00
  • @BrianMoehring I corrected my question. – User124356 Feb 07 '22 at 01:38
  • Note that this is not necessarily true merely under the assumption $X,Y$ are independent. We need the stronger assumption that $X \in G$ and $Y$ is independent with $G$ (or something nearly as strong). The assumptions here are stronger than the assumption $X,Y$ are independent, however, as they imply $X,Y$ are independent. – Brian Moehring Feb 07 '22 at 07:31

1 Answers1

0

For functions of the product form $\phi(X,Y)=h(X)f(Y)$ the claim follows directly from $\mathbb E[f(Y)|{\cal G}]=\mathbb E[f(Y)]$ and $\mathbb E[h(X)|{\cal G}]= h(X)\,:$ $$ \mathbb E[h(X)f(Y)|{\cal G}]=h(X)E[f(Y)] $$ which is obviously equal to $E[h(x)f(Y)]|_{x=X}\,.$ To extend this to all measurable $\phi$ with $\mathbb E[\phi(X,X)]<+\infty$ I would look at the monotone class theorem.

Kurt G.
  • 14,198