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The conditional probability of a continuous random variable $X$ given $Y=y$ is defined as

$1) \ E[X|Y=y] = \displaystyle \int x \frac{f_{X,Y}(x,y)}{f_{Y}(y)}dx $

My question is the following:

Does a definition similar to $(1)$ exist for $E[X|\mathcal{H}]$?

Where $X: \Omega \to \mathbb{R}^{n}$ is a random variable such that $E[|X|] < \infty$, and $\mathcal{H} \subset \mathcal{F}$ is a $\sigma$-algebra

I'm aware from wiki and other sources that by definition $E[X|\mathcal{H}]$ is the (a.s. unique) function from $\Omega$ to $ \mathbb{R}^{n}$ satisfying the following:

$a)$ $E[X|\mathcal{H}]$ is $\mathcal{H}$ - measurable

$b) $ $\int_{H}E[X|\mathcal{H}]dP = \int_{H}X dP$ for all $H \in \mathcal{H}$.

However I am still struggling to construct an intuitive understanding of what $E[X|\mathcal{H}]$ means in comparison to $(1)$.

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seraphimk
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    The sound option in your situation is to forget (1) altogether, which is a very weak way of defining $E(X\mid Y=y)$, and to turn altogether to the sensible definition, valid even when no density exists, which says that $E(X\mid Y=y)=u(y)$ where $E(X\mid Y)=u(Y)$ and $u$ is any measurable function satisfying the two necessary axioms for $u(Y)$ to be $E(X\mid Y)$, which are simply the transcriptions of (a) and (b) to the case $\mathcal H=\sigma(Y)$. – Did Apr 18 '18 at 13:53
  • @Did so in the case of $E[X|Y=y]$ in terms of sigma algebras would the equivalent be $E[X|Y=y]=E[X|h]$ where $h \in \sigma(Y)$ – seraphimk Apr 18 '18 at 19:19
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    ?? Not. At. All. – Did Apr 18 '18 at 21:55
  • @Did Okay. Take 2. $\sigma(Y)$ is the sigma algebra generated by the random variable $Y$. A member of $\sigma(Y)$ is a measurable subset of possible outcomes. So given an event $H \in \sigma(Y)$, the conditional expectation $E[X|H]$ (i.e. the expectation of $X$ given the event $H$) can be found by evaluating $E[X|\sigma(Y)]$ at $H$ (i.e. the definition $(b)$) correct? So in that sense does evaluating $E[X|\sigma(Y)$ at $H$ give $E[X|Y=y]$? – seraphimk Apr 19 '18 at 14:26
  • "correct?" Well, not at all. "Evaluating $E(X\mid Y)$ at $H$" has no sense since $E(X\mid Y)$ is not, in general, constant on $H$. Again, please refer to the definition that $E(X\mid Y)=u(Y)$ where $u$ is such that $E(Xv(Y))=E(u(Y)v(Y))$ for every bounded measurable function $v$. – Did Apr 19 '18 at 15:56
  • @Did Thanks for the replies, though I am still a bit confused. I have however found a similar question which sort of covered some the issues I was having regarding the intuition behind $E[X|\mathcal{H}]$ (https://math.stackexchange.com/questions/23600/intuition-behind-conditional-expectation?rq=1 ) I also found the following Appendix which has also been useful (http://www.math.uwaterloo.ca/~dlmcleis/book/appendixa2.pdf) – seraphimk Apr 19 '18 at 18:05

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