4
$\begingroup$

I am trying to solve this problem:

Let $X,Y$ be random variables on $(\Omega, \mathcal{F}, P)$ such that $E(Y|\mathcal{G}) = X$ and $E(Y^2|\mathcal{G}) = X^2$, where $\mathcal{G} \subset \mathcal{F}$ is a sigma-algebra. Prove that $X = Y$ almost surely.

What I have until now:

From the given property, it's easy to see that $\int_{G}XdP = \int_{G} YdP$, for any $G \in \mathcal{G}$. Also, $X$ is $\mathcal{G}$-measurable.

I know that if the integrals of two random variables, on the same probability space, over any element of the sigma-algebra in discussion agree, then they are equal almost surely. So, if I could prove that $Y$ is also $\mathcal{G}$-measurable I could apply this result. Or, in te other hand, if I could find a similar equality as above for any element $F \in \mathcal{F}$, then I think the problem is solved too. But I can't do it.

Also, I can't see how the information the $X^2$ can be helpful here. Any ideas? Thanks in advance for the support.

$\endgroup$
  • 1
    $\begingroup$ Hint: What is the variance of $Y$ over $\mathcal G$? $\mathsf{Var}(Y\mid\mathcal G) = \ldots$ $\endgroup$ – Graham Kemp May 8 '17 at 23:37
  • $\begingroup$ @GrahamKemp Oh yeah, I think I see it now. The conditional variance is going to be zero. That's cool, thanks! But, sorry, does that imply that $Y$ is $\mathcal{G}$-measurable? $\endgroup$ – Raul Guarini May 8 '17 at 23:40
5
$\begingroup$

Since $X$ is $\mathcal{G}$-measurable we have (by the product rule) that $$E(XY) = E(E(XY|\mathcal{G})) = E(XE(Y|\mathcal{G})) = E(X^2).$$ Similarly, $$E(Y^2) = E(E(Y^2|\mathcal{G})) = E(X^2).$$ Thus, $$E((X-Y)^2) = E(X^2)+E(Y^2) - 2E(XY) = 0.$$ Hence, $$P((X-Y)^2 = 0) = 1.$$ The last line follows from the non-negativity of $(X-Y)^2$.

$\endgroup$
  • $\begingroup$ That's beautiful, thanks a lot! $\endgroup$ – Raul Guarini May 9 '17 at 0:19
  • $\begingroup$ @RaulGuarini you got it...neat problem, thanks for the share $\endgroup$ – David May 9 '17 at 0:20

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.