283 lines
9.0 KiB
TeX
283 lines
9.0 KiB
TeX
\lecture{15}{2023-06-06}{}
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\subsection{Properties of Conditional Expectation}
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We want to derive some properties of conditional expectation.
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\begin{theorem}[Law of total expectation]
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\label{ceprop1}
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\yalabel{Law of Total Expectation}{Total Expectation}{totalexpectation}
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\[
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\bE[\bE[X | \cG ]] = \bE[X].
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\]
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\end{theorem}
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\begin{proof}
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Apply (b) from the definition for $G = \Omega \in \cG$.
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\end{proof}
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\begin{theorem}
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\label{ceprop2}
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If $X$ is $\cG$-measurable, then $X \overset{\text{a.s.}}{=}\bE[X | \cG]$.
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\end{theorem}
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\begin{proof}
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Suppose $\bP[X \neq Y] > 0$.
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Without loss of generality $\bP[X > Y] > 0$.
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Hence $\bP[ X > Y + \frac{1}{n}]> 0$ for some $n \in \N$.
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Let $A \coloneqq \{X > Y + \frac{1}{n}\}$.
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Then
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\[
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\int_A X \dif \bP \ge \frac{1}{n}\bP(A) + \int_A Y \dif \bP,
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\]
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contradicting property (b) from \yaref{conditionalexpectation}.
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\end{proof}
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\begin{example}
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Suppose $X \in L^1(\bP)$, $\cG \coloneqq \sigma(X)$.
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Then $X$ is measurable with respect to $\cG$.
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Hence $\bE[X | \cG] = X$.
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\end{example}
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\begin{theorem}[Linearity]
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\label{ceprop3}
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\yalabel{Linearity of Conditional Expectation}{Linearity}{celinearity}
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For all $a,b \in \R$
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we have
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\[
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\bE[a X_1 + bX_2 | \cG] = a \bE[X_1 | \cG] + b \bE[X_2|\cG].
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\]
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\end{theorem}
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\begin{proof}
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trivial\todo{add details}
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\end{proof}
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\begin{theorem}[Positivity]
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\label{ceprop4}
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\yalabel{Positivity of Conditional Expectation}{Positivity}{cpositivity}
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If $X \ge 0$, then $\bE[X | \cG] \ge 0$ a.s.
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\end{theorem}
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\begin{proof}
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Let $W $ be a version of $\bE[X | \cG]$.
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Suppose $\bP[ W < 0] > 0$.
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Then \[G \coloneqq \{W < -\frac{1}{n}\} \in \cG.\]
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For some $n \in \N$, we have $\bP[G] > 0$.
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However it follows that
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\[
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\int_G \bP[X | \cG] \dif \bP \le -\frac{1}{n} \bP[G] < 0 \le \int_G X \dif \bP.
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\]
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\end{proof}
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\begin{theorem}[Conditional monotone convergence theorem]
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\label{ceprop5}
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\yalabel{Conditional Monotone Converence Theorem}{MCT}{cmct}
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Let $X_n,X \in L^1(\Omega, \cF, \bP)$.
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Suppose $X_n \ge 0$ with $X_n \uparrow X$.
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Then $\bE[X_n|\cG] \uparrow \bE[X|\cG]$.
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\end{theorem}
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\begin{proof}
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Let $Z_n$ be a version of $\bE[X_n | Y]$.
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Since $X_n \ge 0$ and $X_n \uparrow$,
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by the \yaref{cpositivity},
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we have
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\[
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\bE[X_n | \cG] \overset{\text{a.s.}}{\ge } 0
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\]
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and
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\[
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\bE[X_n | \cG] \uparrow \text{a.s.}
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\]
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(consider $X_{n+1} - X_n$ ).
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Define $Z \coloneqq \limsup_{n \to \infty} Z_n$.
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Then $Z$ is $\cG$-measurable
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and $Z_n \uparrow Z$ a.s.
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Take some $G \in \cG$.
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We know by (b) % TODO REF
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that $\bE[Z_n \One_G] = \bE[X_n \One_G]$.
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The LHS increases to $\bE[Z \One_G]$ by the monotone
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convergence theorem.
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Again by MCT, $\bE[X_n \One_G]$ increases to
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$\bE[X \One_G]$.
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Hence $Z$ is a version of $\bE[X | \cG]$.
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\end{proof}
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\begin{theorem}[Conditional Fatou]
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\label{ceprop6}
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\yalabel{Conditional Fatou's Lemma}{Fatou}{cfatou}
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Let $X_n \in L^1(\Omega, \cF, \bP)$, $X_n \ge 0$.
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Then
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\[
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\bE[ \liminf_{n \to \infty} X_n | \cG] \le \liminf_{n \to \infty} \bE[X_n | \cG].
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\]
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\end{theorem}
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\begin{proof}
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\notes
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\end{proof}
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\begin{theorem}[Conditional dominated convergence theorem]
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\label{ceprop7}
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\yalabel{Conditional Dominated Convergence Theorem}{DCT}{cdct}
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Let $X_n,Y \in L^1(\Omega, \cF, \bP)$.
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Suppose that $\sup_n |X_n(\omega)| < Y(\omega)$ a.e.~
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and that $X_n$ converges to a pointwise limit $X$.
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Then $\bE[ X_n | \cG] \to \bE[X | \cG]$ a.e.
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\end{theorem}
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\begin{proof}
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\notes
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\end{proof}
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Recall
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\begin{fact}[Jensen's inequality]
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\yalabel{Jensen's Inequality}{Jensen}{jensen}
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If $c : \R \to \R$ is convex and $\bE[|c \circ X|] < \infty$,
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then $\bE[c \circ X] \overset{\text{a.s.}}{\ge} c(\bE[X])$.
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\end{fact}
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For conditional expectation, we have
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\begin{theorem}[Conditional Jensen's inequality]
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\label{ceprop8}
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\yalabel{Jensen's Inequality}{Jensen}{cjensen}
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Let $X \in L^1(\Omega, \cF, \bP)$.
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If $c : \R \to \R$ is convex and $\bE[|c \circ X|] < \infty$,
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then $\bE[c \circ X | \cG] \ge c(\bE[X | \cG])$ a.s.
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\end{theorem}
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\begin{fact}
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\label{convapprox}
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If $c$ is convex, then there are two sequences of real numbers
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$a_n, b_n \in \R$
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such that
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\[
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c(x) = \sup_n(a_n x + b_n).
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\]
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\end{fact}
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\begin{refproof}{cjensen}
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By \yaref{convapprox}, $c(x) \ge a_n X + b_n$
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for all $n$.
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Hence
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\[
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\bE[c(X) | \cG] \ge a_n \bE[X | \cG] + \bE[b_n | \cG]
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= a_n \bE[X | \cG] + b_n \text{ a.s.}
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\]
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for all $n$.
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Using that a countable union of sets o f measure zero has measure zero,
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we conclude that a.s~this happens simultaneously for all $n$.
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Hence
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\[
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\bE[c(X) | \cG] \ge \sup_n (a_n \bE[X | \cG] + b_n) \overset{\yaref{convapprox}}{=} c(\bE(X | \cG)).
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\]
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\end{refproof}
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Recall
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\begin{fact}[Hölder's inequality]
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\yalabel{Hölder's Inequality}{Hölder}{thm:hoelder}
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Let $p,q \ge 1$ such that $\frac{1}{p} + \frac{1}{q} = 1$.
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Suppose $X \in L^p(\bP)$ and $Y \in L^q(\bP)$.
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Then
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\[
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\bE(X Y) \le \underbrace{\bE(|X|^p)^{\frac{1}{p}}}_{\text{\reflectbox{$\coloneqq$}} \|X\|_{L^p}} \bE(|Y|^q)^{\frac{1}{q}}.
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\]
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\end{fact}
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\begin{theorem}[Conditional Hölder's inequality]
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\label{ceprop9}
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\yalabel{Hölder's Inequality}{Hölder}{choelder}
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Let $p,q \ge 1$ such that $\frac{1}{p} + \frac{1}{q} = 1$.
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Suppose $X \in L^p(\bP)$ and $Y \in L^q(\bP)$.
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Then
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\[
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\bE(X Y | \cG) \le \bE(|X|^p | \cG)^{\frac{1}{p}} \bE(|Y|^q | \cG)^{\frac{1}{q}}.
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\]
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\end{theorem}
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% TODO
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% \begin{proof}
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% Take some $G \in \cG$.
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% We first consider the case of $|X(\omega)|^p, |Y(\omega)|^p > 0$
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% a.s.~for $\omega \in G$.
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% Then
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% \begin{IEEEeqnarray*}{rCl}
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% \int_G\frac{\bE[|XY| ~ |\cG]}%
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% {\bE[\bE[|X|^p | \cG]^{\frac{1}{p}} \bE[|Y|^p| \cG]^{\frac{1}{q}}}
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% \dif \bP
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% &=& \int_G \frac{|X|}{\bE[|X|^p]^{\frac{1}{p}}} \frac{|Y|}{\bE[|Y|^q]^{\frac{1}{q}}}
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% \dif \bP\\
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% &\le& \left(\int_G \frac{|X|^p}{\bE[|X|^p]} \dif \bP\right)^p
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% \left(\int_G \frac{|Y|^q}{\bE[|Y|^q]} \dif \bP\right)^q\\
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% &=& \bE[\One_G]
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% \end{IEEEeqnarray*}
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% \end{proof}
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\begin{theorem}[Tower property]
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\label{ceprop10}
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\yalabel{Tower Property}{Tower}{cetower}
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Suppose $\cF \supset \cG \supset \cH$ are sub-$\sigma$-algebras.
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Then
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\[
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\bE\left[\bE[X | \cG] \mid \cH\right] \overset{\text{a.s.}}{=} \bE[X | \cH].
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\]
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\end{theorem}
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\begin{proof}
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By definition, $\bE[\bE[X | \cG] | \cH]$ is $\cH$-measurable.
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For any $H \in \cH$, we have
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\begin{IEEEeqnarray*}{rCl}
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\int_H \bE[\bE[X | \cG] | \cH] \dif \bP
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&=& \int_{H} \bE[X | \cG] \dif \bP\\
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&=& \int_H X \dif \bP.
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\end{IEEEeqnarray*}
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Hence $\bE[\bE[X | \cG] | \cH] \overset{\text{a.s.}}{=} \bE[X | \cH]$.
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\end{proof}
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\begin{theorem}[Taking out what is known]
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\label{ceprop11}
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\label{takingoutwhatisknown}
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If $Y$ is $\cG$-measurable and bounded, then
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\[
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\bE[YX| \cG] \overset{\text{a.s.}}{=} Y \bE[X | \cG].
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\]
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\end{theorem}
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\begin{proof}
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Assume w.l.o.g.~$X \ge 0$.
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Assume $Y = \One_B$, then $Y$ simple, then take the limit (using that $Y$ is bounded).
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\todo{Exercise}
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\end{proof}
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\begin{definition}
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Let $\cG$ and $\cH$ be $\sigma$-algebras.
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We call $\cG$ and $\cH$ \vocab[$\sigma$-algebra!independent]{independent},
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if $\bP(G \cap H) = \bP(G) \bP(H)$
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for all events $G \in \cG$, $H \in \cH$.
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\end{definition}
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\begin{theorem}[Role of independence]
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\label{ceprop12}
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\yalabel{Role of Independence}{Independence}{ceroleofindependence}
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Let $X$ be a random variable,
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and let $\cG, \cH$ be $\sigma$-algebras.
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If $\cH$ is independent of $\sigma\left( \sigma(X), \cG \right)$,
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then
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\[
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\bE[X | \sigma(\cG, \cH)] \overset{\text{a.s.}}{=} \bE[X | \cG].
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\]
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In particular, if $X$ is independent of $\cG$,
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then
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\[
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\bE[X | \cG] \overset{\text{a.s.}}{=} \bE[X].
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\]
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\end{theorem}
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\begin{example}[Martingale property of the simple random walk]
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Suppose $X_1,X_2,\ldots$ are i.i.d.~with $\bP[X_i = 1] = \bP[X_i = -1] = \frac{1}{2}$.
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Let $S_n \coloneqq \sum_{i=1}^n X_i$ be the \vocab{simple random walk}.
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Let $\cF$ denote the $\sigma$-algebra on the product space.
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Define $\cF_n \coloneqq \sigma(X_1,\ldots)$.
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Intuitively, $\cF_n$ contains all the information gathered until time $n$.
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We have $\cF_1 \subset \cF_2 \subset \cF_3 \subset \ldots$
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For $\bE[S_{n+1} | \cF_n]$ we obtain
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\begin{IEEEeqnarray*}{rCl}
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\bE[S_{n+1} | \cF_n] &\overset{\yaref{celinearity}}{=}&
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\bE[S_n | \cF_n] + \bE[X_{n+1} | \cF_n]\\
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&\overset{\text{a.s.}}{=}& S_n + \bE[X_{n+1} | \cF_n]\\
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&\overset{\yaref{ceroleofindependence}}{=}& S_{n} + \bE[X_n]\\
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&=& S_n.
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\end{IEEEeqnarray*}
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\end{example}
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