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Let us say we have two real valued random variables $X,Y$ on a probability space $(\Omega, \mathcal{F},\mathbb{P})$. Denote by $\mathcal{F}_X, \mathcal{F}_Y$ the sub-sigma algebras of $\mathcal{F}$ generated by $X,Y$ (endowing $\mathbb{R}$ with the Borel $\sigma$ algebra).

The conditional expectation of $X$ given $Y$, denoted by $\mathbb{E}(X|Y)$ is the (unique up to measure $0$) random variable satisfying: $$ \int_S \mathbb{E}(X|Y)(\omega) \mathrm{d}\mathbb{P}(\omega) = \int_S X(\omega) \mathrm{d}\mathbb{P}(\omega) $$ for any $S\in \mathcal{F}_Y$. Now, it seems to me we are trying to generalize the idea of restricting our probability measure on $X$ (which is $\mathcal{F}_X$ measurable, by definition), to sets generated by preimages of $Y$ (hence we consider $S \in \mathcal{F}_Y$). Thus, in an intuitive sense, we may take probability weighted averages of of $X$ over events that depend on $Y$. However, the problem (I think???) is that we cannot really integrate $X$ over sets in $\mathcal{F}_Y$, and hence conditional expectation solves this for us. But how then would the right hand side make sense in this expression?

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    $\begingroup$ Sorry but how is the RHS $E(X\mathbf 1_S)$ problematic? $\endgroup$ – Did Jan 2 at 15:19
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    $\begingroup$ The random variable $X$ itself will satisfy that integral property. So your definition needs another condition: You also want the random variable $E[X|Y]$ to be $\sigma(Y)$-measurable. In other words, it should be a pure function of $Y$. $\endgroup$ – Michael Jan 2 at 19:29
  • $\begingroup$ On your question: You can define those integrals for any sets $S\in \mathcal{F}$. The fact that you only need those integrals to hold over sets $S \in \mathcal{F}_Y$ can be viewed as a feature and allows you to view $E[X|Y]$ as a random variable on a smaller $\sigma$-algebra $\mathcal{F}_Y$. In other words, we can calculate the integral on the left only using knowledge of the distribution of $Y$. If we tried to integrate over a set $S$ that is not in $\mathcal{F}_Y$, we would need to know more than just the distribution of $Y$. $\endgroup$ – Michael Jan 2 at 19:43

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