# Conditional Expectations

To define an expectation conditional on an event in a probability space is essentially no more than defining a conditional probability, then constructing the expectation as an integral with respect to this measure. At an abstract level, it is often more useful to define an expectation conditional on a sigma-algebra. Informally, we want to define the expectation conditional on every event in a sigma-algebra simultaneously. Most importantly, the result is measurable with respect to this sigma-algebra, so is a random variable itself under suitable conditions.

We want to proof that such a construction exists. We take $X$ a $\mathcal{F}$-measurable random variable, and consider a sub-sigma-algebra $\mathcal{G}\subset\mathcal{F}$. We want to define $Y=\mathbb{E}[X|\mathcal{G}]$, which is integrable, $\mathcal{G}$-measurable, and such that $\mathbb{E}X1_A=\mathbb{E}Y1_A$, for all $A\in\mathcal{G}$.

We also want to show that this conditional expectation is, up to null events, unique. When $\mathcal{G}=\sigma(B_i,i\in I)$ is generated by a countable collection of disjoint events, then we can define $Y:=\sum_{i\in I}\mathbb{E}[X|B_i]1_{B_i},$

and verify that this satisfies the conditions.

We proceed in the general case. Uniqueness is easy. Suppose have $Y,Y'$ satisfying the conditions. Then the event $A=\{Y>Y'\}\in\mathcal{G}$. Substituting into the definition gives: $\mathbb{E}[(Y-Y')1_A]=\mathbb{E}[X1_A]-\mathbb{E}[X1_A]=0$

But $(Y-Y')1_A\geq 0$ and so we conclude that $\mathbb{P}(A)=0$, and $Y'\geq Y$ almost surely. Of course the reverse argument applies also, and so $Y=Y'$ a.s.

For existence, we exploit a property of Hilbert spaces. We initially assume $X\in L^2$. We can decompose the host space as $L^2(\mathcal{F})=L^2(\mathcal{G})+L^2(\mathcal{G})^{\perp}$ and $X=Y+Z$

in this orthogonal projection. The operator on this space is $\langle X,Y\rangle:=\mathbb{E}XY$, and so $1_A\in L^2(\mathcal{G})\Rightarrow \mathbb{E}[Z1_A]=0$,

From this, we conclude that $Y$ is suitable. For what follows, observe that $\{Y<0\}$ is $\mathcal{G}$-measurable, and so by a similar argument to before, $X\geq 0$ a.s. implies $Y\geq 0$ a.s.

For general $X\geq 0$, set $X_n:=X\wedge n\uparrow X$, and $Y_n:=\mathbb{E}[X_n|\mathcal{G}]$. By our previous observation, everything relevant is almost surely non-negative, and we can apply monotone convergence to both sides of the relation $\mathbb{E}[X_n1_A]=\mathbb{E}[Y_n1_A]$

to obtain the definition, and take $A=\Omega$ to check integrability. Separating general $X$ into positive and negative parts gives the result for generally supported random variables.