Conditional Expectation

Definition

Conditional Expectation of a Random Variable

$E(X|Y=y)={\int }_{-\infty }^{\infty }x{f}_{X|Y}(x|y)dx$

Conditional Expectation of a Function

$E[g(x)|Y=y]={\int }_{-\infty }^{\infty }g(x)f_{X|Y}(x|y)dx$

Conditional Expectation with respect to a Sub-Sigma-Algebra

Consider a Probability Space $(\Omega ,\mathcal{F},P)$, a Random Variable $X:\Omega \to \mathbb{R}$, and a sub-Sigma-Algebra $\mathcal{G}\subset \mathcal{F}$. A conditional expectation of $X$ given $\mathcal{G}$, denoted as $E(X|\mathcal{G}):\Omega \to \mathbb{R}$ is a function which satisfies: $\forall G\in \mathcal{G},{\int }_{G}E(X|\mathcal{G})dP={\int }_{G}XdP$

Existence of Conditional Distribution

Define a measure ${\mu }^X(F)={\int }_{F}XdP$ and its restriction ${\mu }^X{|}_{\mathcal{G}}:={\mu }^X\circ g$ where $g:\mathcal{G}\to \mathcal{F},g(x)=x$, and $P{|}_{\mathcal{G}}:=P\circ g$. Then, ${\mu }^X{|}_{\mathcal{G}}\ll P{|}_{\mathcal{G}}$ (${\mu }^X{|}_{\mathcal{G}}$ is absolute continuous with respect to $P{|}_{\mathcal{G}}$) and there exists a Radon–Nikodym Theorem $\frac{d{\mu }^X{|}_{\mathcal{G}}}{dP{|}_{\mathcal{G}}}=:E(X|\mathcal{G})$ satisfying $\forall G\in \mathcal{G},{\mu }^X{|}_{\mathcal{G}}(G)={\int }_{G}E(X|\mathcal{G})d{P}_{\mathcal{G}}={\int }_{G}XdP={\mu }^X(G)$ and it is called the conditional expectation of $X$ given $\mathcal{G}$.

Conditional Expectation with respect to a Random Variable

Suppose a Probability Space $(\Omega ,\mathcal{F},P)$, and random variables $X:\Omega \to \mathbb{R}$ and $Y:\Omega \to \mathbb{R}$. The conditional expectation of $X$ given $Y$, denoted as $E(X|Y):\Omega \to \mathbb{R}$ is defined as $E(X|\sigma (Y))=E(X|\{Y^{-1}(B)|B\in \mathcal{R}\}\subset \mathcal{F})$ where $\sigma (Y)$ is the sigma-field generated by random variable $Y$

Facts

If $X$ is $\mathcal{G}$-measurable, then $E(XY|\mathcal{G})=XE(Y|\mathcal{G})$, where $\mathcal{G}\subset \mathcal{F}$ is sub-Sigma-Field

Since $X$ is $\mathcal{G}$-measurable, $\exists$ $\mathcal{G}$-measurable function $h$ s.t. $X=h(\mathcal{G})$ By the property of conditional expectation, $E(XY|\mathcal{G})=E(h(\mathcal{G})Y|\mathcal{G})=h(\mathcal{G})E(Y|G)=XE(Y|\mathcal{G})$.