Let $p(x,y)$ be the Joint density of the random variables $X,Y$ and $q(y) = \int p(x,y)dx$ and $p(x|y) = \frac{p(x,y)}{q(y)}$. Then

\begin{equation} \mathbb{E}(X|Y=y) = \int xp(x|y)dx. \end{equation} Now many books give this as a Definition for the conditional expectation. Can we also prove this?

The alternative definition is:

Let $(\Omega,\mathcal{F},\mathbb{P})$ be a probability space and $X$ a integrable random variable. Let $\mathcal{G}$ be a sub-sigma-algebra of $\mathcal{F}$. Then there exists a random variable $Y$ such that

\begin{equation} ~~Y~~\text{is}~~\mathcal{G}~\text{measurable}, \end{equation}

\begin{equation} ~~\mathbb{E}(|Y|) < \infty, \end{equation}

\begin{equation} \text{for every set G}~in ~\mathcal{G}~\text{we have}~ \end{equation}

\begin{equation} \int_{G}YdP = \int_{G}XdP, ~~~\forall G \in \mathcal{G} \end{equation}


closed as off-topic by Raskolnikov, JMP, Xander Henderson, Saad, jvdhooft May 24 '18 at 8:58

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    $\begingroup$ If you want to prove a definition, you'll have to provide an alternative definition from which the former derives as a special case or an equivalent. $\endgroup$ – Raskolnikov May 23 '18 at 23:26
  • $\begingroup$ I'm sorry...I forget it. I have edited the question $\endgroup$ – wayne May 24 '18 at 9:19