172 lines
5.6 KiB
TeX
172 lines
5.6 KiB
TeX
\lecture{14}{2023-05-25}{Conditional expectation}
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\section{Conditional expectation}
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\subsection{Introduction}
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Consider a probability space $(\Omega, \cF, \bP)$
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and two events $A, B \in \cF$ with $\bP(B) > 0$.
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\begin{definition}
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The \vocab{conditional probability} of $A$ given $B$ is defined as
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\[
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\bP(A | B) \coloneqq \frac{\bP(A \cap B)}{\bP(B)}.
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\]
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\end{definition}
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Suppose we have two random variables $X$ and $Y$ on $\Omega$,
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such that $X$ takes distinct values $x_1, x_2,\ldots, x_{m}$
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and $Y$ takes distinct values $y_1,\ldots, y_n$.
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Then for this case, define the \vocab{conditional expectation}
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of $X$ given $Y = y_j$ as
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\[
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\bE[X | Y = y_j] \coloneqq \sum_{i=1}^m x_i \bP[X=x_i | Y = y_j].
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\]
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The random variable $Z = \bE[X | Y]$
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is defined as follows:
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If $Y(\omega) = y_j$ then
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\[
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Z(\omega) \coloneqq \underbrace{\bE[X | Y = y_j]}_{\text{\reflectbox{$\coloneqq$}} z_j}.
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\]
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Note that $\Omega_j \coloneqq \{\omega : Y(\omega) = y_j\}$
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defines a partition of $\Omega$ and on each $\Omega_j$
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(``the $j^{\text{th}}$ $Y$-atom'')
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$ Z$ is constant.
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Let $\cG \coloneqq \sigma(Y)$.
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Then $Z$ is measurable with respect to $\cG$.
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Furthermore
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\begin{IEEEeqnarray*}{rCl}
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\int_{\{Y = y_j\} } Z \dif \bP &=& z_j \int_{\{Y = y_j\}} \dif \bP\\
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&=& z_j \bP[Y=y_j]\\
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&=&\sum_{i=1}^m x_i \bP[X = x_i | Y = y_j] \bP[Y = y_j]\\
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&=&\sum_{i=1}^m x_i \bP[X = x_i, Y = y_j]\\
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&=& \int_{\{Y = y_j\}} X \dif \bP.
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\end{IEEEeqnarray*}
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Hence
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\[
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\int_{G} Z \dif \bP = \int_{G} X \dif \bP
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\]
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for all $G \in \cG$.
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We now want to generalize this to arbitrary random variables.
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\begin{theorem}
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\label{conditionalexpectation}
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Let $(\Omega, \cF, \bP)$ be a probability space, $X \in L^1(\bP)$
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and $\cG \subseteq \cF$ a sub-$\sigma$-algebra.
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Then there exists a random variable $Z$
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such that
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\begin{enumerate}[(a)]
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\item $Z$ is $\cG$-measurable and $Z \in L^1(\bP)$,
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\item $\int_G Z \dif \bP = \int_G X \dif \bP$
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for all $G \in \cG$.
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\end{enumerate}
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Such a $Z$ is unique up to sets of measure $0$ and is
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called the \vocab{conditional expectation} of $X$ given
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the $\sigma$-algebra $\cG$ and written
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$Z = \bE[X | \cG]$.
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\end{theorem}
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\begin{remark}
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Suppose $\cG = \{\emptyset, \Omega\}$,
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then
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\[
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\bE[X | \cG] = (\omega \mapsto \bE[X])
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\]
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is a constant random variable.
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\end{remark}
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\begin{definition}[Conditional probability]
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Let $A \subseteq \Omega$ be an event and $\cG \subseteq \cF$
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a sub-$\sigma$-algebra.
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We define the \vocab{conditional probability} of $A$ given $\cG$ by
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\[
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\bP[A | \cG] \coloneqq \bE[\One_A | \cG].
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\]
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\end{definition}
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\subsection{Existence of conditional probability}
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We will give two different proves of \autoref{conditionalexpectation}.
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The first one will use orthogonal projections.
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The second will use the Radon-Nikodym theorem.
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We'll first do the easy proof, derive some properties
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and then do the harder proof.
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\begin{lemma}
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\label{orthproj}
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Suppose $H$ is a \vocab{Hilbert space},
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i.e.~$H$ is a vector space with an inner product $\langle \cdot, \cdot \rangle_H$ which defines a norm by $\|x\|_H^2 = \langle x, x\rangle_H$
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making $H$ a complete metric space.
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For any $x \in H$ and $K \subseteq H$ closed,
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there exists a unique $z \in K$ such that the following equivalent conditions hold:
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\begin{enumerate}[(a)]
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\item $\forall y \in K : \langle x-z, y\rangle_H = 0$,
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\item $\forall y \in K: \|z-x\|_H \le \|z-x\|_H$.
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\end{enumerate}
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\end{lemma}
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\begin{proof}
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\notes
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\end{proof}
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\begin{refproof}{conditionalexpectation}
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Almost sure uniqueness of $Z$:
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Suppose $X \in L^1$ and $Z$ and $Z'$ satisfy (a) and (b).
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We need to show that $\bP[Z \neq Z'] = 0$.
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By (a), we have $Z, Z' \in L^1(\Omega, \cG, \bP)$.
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By (b), $\bE[(Z - Z') \One_G] = 0$ for all $G \in \cG$.
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Assume that $\bP[Z > Z'] > 0$.
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Since $\{Z > Z' + \frac{1}{n}\} \uparrow \{Z > Z'\}$,
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we see that $\bP[Z > Z' + \frac{1}{n}] > 0$ for some $n$.
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However $\{Z > Z' + \frac{1}{n}\} \in \cG$,
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which is a contradiction, since
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\[
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\bE[(Z - Z') \One_{Z - Z' > \frac{1}{n}}] \ge \frac{1}{n} \bP[ Z - Z' > \frac{1}{n}] > 0.
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\]
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\bigskip
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Existence of $\bE(X | \cG)$ for $X \in L^2$:
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Let $H = L^2(\Omega, \cF, \bP)$
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and $K = L^2(\Omega, \cG, \bP)$.
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$K$ is closed, since a pointwise limit of $\cG$-measurable
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functions is $\cG$ measurable (if it exists).
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By \autoref{orthproj},
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there exists $z \in K$ such that
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\[\bE[(X - Z)^2] = \inf \{ \bE[(X- W)^2] ~|~ W \in L^2(\cG)\}\]
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and
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\begin{equation}
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\forall Y \in L^2(\cG) : \langle X - Z, Y\rangle = 0.
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\label{lec13_boxcond}
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\end{equation}
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Now, if $G \in \cG$, then $Y \coloneqq \One_G \in L^2(\cG)$
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and by \eqref{lec13_boxcond} $\bE[Z \One_G] = \bE[X \One_G]$.
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\bigskip
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Existence of $\bE(X | \cG)$ for $X \in L^1$ :
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Let $X = X^+ - X^-$.
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It suffices to show (a) and (b) for $X^+$.
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Choose bounded random variables $X_n \ge 0$ such that $X_n \uparrow X$.
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Since each $X_n \in L^2$, we can choose a version $Z_n$ of $\bE(X_n | \cG)$.
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\begin{claim}
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$0 \overset{\text{a.s.}}{\le} Z_n \uparrow$.
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\end{claim}
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\begin{subproof}
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\notes
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\end{subproof}
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Define $Z(\omega) \coloneqq \limsup_{n \to \infty} Z_n(\omega)$.
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Then $Z$ is $\cG$-measurable and since $Z_n \uparrow Z$,
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by MCT, $\bE(Z \One_G) = \bE(X \One_G)$ for all $G \in \cG$.
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\end{refproof}
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