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$X$ is a random variable. I wonder if there are some or no relations between

  • the subspace which $X - E(X|\mathcal G)$ is orthogonal to, which is the set of all random variables which are both $L^2$ and $\mathcal G$-measurable,

  • the set of all the random variables which generate the same $\mathcal G$,

and what they are.

I thought they might be the same, based on the relation between $E(X|Y)$ and $E(X|\sigma(Y))$. Then the second is also a vector subspace as the first, but it doesn't seem true to me. Thanks!

Tim
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    ${\rm E}(X\mid Y)\circ Y$ doesn't make sense since both ${\rm E}(X\mid Y)$ and $Y$ are defined on $\Omega$ with values in $\mathbb{R}$. And yes, ${\rm E}(X\mid Y)$ is, by definition, equal to ${\rm E}(X\mid \sigma(Y))$. – Stefan Hansen Mar 10 '14 at 17:07
  • @StefanHansen: Your comment was also part of my original confusion. Now, if I understand correctly, here $E(X|Y)$ is not defined on $\Omega$ but on the codomain of $Y$, (in the process going from $E(X|\sigma(Y))$ to $E(X|Y)$, we shift the domain $\Omega$ of $E(X|\sigma(Y))$ to the codomain of $Y$, by some factorization theorem such as Doob-Dinkyns theorem?) , see here http://en.wikipedia.org/wiki/Conditional_expectation#Conditioning_as_factorization – Tim Mar 10 '14 at 17:09
  • Either you define ${\rm E}(X\mid Y)$ to be ${\rm E}(X\mid \sigma(Y))$ or you define it to be $\psi(Y)$, where $\psi(y)={\rm E}(X\mid Y=y)$ (doesn't matter which one you choose, since it yields the same object). However, the article on wikipedia uses ${\rm E}(X\mid Y)$ to mean both which I strongly advise again. Maybe you can benefit from reading this post. – Stefan Hansen Mar 10 '14 at 17:21
  • I realized I made some error in my post. I changed to conditioning on sigma algebra instead of on random variable, so that we don't have to be bothered by which is the domain of $(X|Y)$. Do you have some thought about my posted question? Your reply in the link is a nice one. Thanks! – Tim Mar 10 '14 at 17:26

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