# Conditional Expectation and Measurability

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?

• Sorry but how is the RHS $E(X\mathbf 1_S)$ problematic? – Did Jan 2 at 15:19
• 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 $Y$. – Michael Jan 2 at 19:29
• 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 $Y$. – Michael Jan 2 at 19:43