Conditional Expectation

Theorem

Let $(\Omega ,\Sigma ,\mathbb{P})$ be a probability space and ${\Sigma }_0\subset \Sigma$ a sigma-algebra. There exists a unique linear continuous map $T:L^2(\Omega ,\Sigma ,\mathbb{P})\to L^2(\Omega ,{\Sigma }_0,\mathbb{P})$, such that for any $X\in L^2(\Omega ,\Sigma ,\mathbb{P})$ and $A\in {\Sigma }_0$, ${\int }_{A}Xd\mathbb{P}={\int }_{A}T(X)d\mathbb{P}.$

Conditional Expectation

The unique map $T$ satisfying previous theorem is called the conditional expectation with respect to the sigma-algebra ${\Sigma }_0$. For $X\in L^2(\Omega ,\Sigma ,\mathbb{P})$, the conditional expectation of $X$ given ${\Sigma }_0$ will be denoted by $E(X\mid {\Sigma }_0):=T(X)$.

Example

Conditioning Over a Random Variable Let $X_0:\Omega \to \mathbb{R}$ be a random variable and $\mathcal{B}$ denote the Borel sigma algebra of $\mathbb{R}$. For $X\in L^2(\Omega ,\Sigma ,\mathbb{P})$, define $E[X\mid X_0]:=E(X\mid \sigma (X_0)),$ where $\sigma (X_0):=\{X_0^{-1}(B):B\in \mathcal{B}\}$ is the sigma-algebra generated by $X_0$. $E(\cdot \mid X_0)$ is referred to as the conditional expectation given the random variable $X_0$.

Proposition

Let $(\Omega ,\Sigma ,\mathbb{P})$ be a probability space and ${\Sigma }_0\subset \Sigma$ a sub-sigma-algebra. The following properties hold for any $X$ and $Y$ in $L^2(\Omega ,\Sigma ,\mathbb{P})$. (i) Linearity: For any $a,b\in \mathbb{R}$, $E(aX+bY\mid {\Sigma }_0)=aE(X\mid {\Sigma }_0)+bE(Y\mid {\Sigma }_0)$ almost surely. (ii) Monotonicity: Suppose that $X\le Y$ almost surely. Then, $E(X\mid {\Sigma }_0)\le E(Y\mid {\Sigma }_0)$ almost surely. (iii) The Tower Property: Let ${\Sigma }_1$ be a sub-sigma-algebra of ${\Sigma }_0$. Then, $E(E(X\mid {\Sigma }_0)\mid {\Sigma }_1)=E(X\mid {\Sigma }_1)=E(E(X\mid {\Sigma }_1)\mid {\Sigma }_0).(1.5)$ (iv) Jensen’s Inequality For any convex $\varphi :\mathbb{R}\to \mathbb{R}$, $\varphi (E(X\mid {\Sigma }_0))\le E(\varphi (X)\mid {\Sigma }_0).(1.6)$ (v) If ${\Sigma }_0$ is the trivial sigma-algebra, ${\Sigma }_0=\{\varnothing ,\Omega \}$, then $E(X\mid {\Sigma }_0)=E(X)$. (vi)Law of Total Expectation: $E(E(X\mid Y))=E(X).(1.7)$ (vii)Taking Out What Is Known: If $Y$ is ${\Sigma }_0$-measurable, then $E(Y\cdot X\mid {\Sigma }_0)=Y\cdot E(X\mid {\Sigma }_0)$. (viii) Extension to $L^1$: We have that $\mathbb{E}[E(X\mid {\Sigma }_0)]\le \mathbb{E}|X|.$ Consequently, $E(\cdot \mid {\Sigma }_0)$ extends to a bounded linear map from $L^1(\Omega ,\Sigma ,\mathbb{P})$ to $L^1(\Omega ,{\Sigma }_0,\mathbb{P})$.