s23-probability-theory/inputs/lecture_15.tex

\lecture{15}{2023-06-06}{}
\subsection{Properties of Conditional Expectation}

We want to derive some properties of conditional expectation.

\begin{theorem}[Law of total expectation]
    \label{ceprop1}
    \label{totalexpectation}
    \[
        \bE[\bE[X | \cG ]] = \bE[X].
    \]
\end{theorem}
\begin{proof}
    Apply (b) from the definition for $G = \Omega  \in \cG$.
\end{proof}
\begin{theorem}
    \label{ceprop2}
    If $X$ is $\cG$-measurable, then $X \overset{\text{a.s.}}{=}\bE[X | \cG]$.
\end{theorem}
\begin{proof}
    Suppose $\bP[X \neq Y] > 0$.
    Without loss of generality $\bP[X > Y] > 0$.
    Hence $\bP[ X > Y + \frac{1}{n}]> 0$ for some $n \in \N$.
    Let $A \coloneqq \{X > Y + \frac{1}{n}\}$.
    Then
    \[
    \int_A X \dif \bP \ge \frac{1}{n}\bP(A) + \int_A Y \dif \bP,
    \]
    contradicting property (b) from \autoref{conditionalexpectation}.
\end{proof}

\begin{example}
    Suppose $X \in L^1(\bP)$, $\cG \coloneqq  \sigma(X)$.
    Then $X$ is measurable with respect to $\cG$.
    Hence $\bE[X | \cG] = X$.
\end{example}

\begin{theorem}[Linearity]
    \label{ceprop3}
    \label{celinearity}
    For all $a,b \in \R$
    we have
    \[
        \bE[a X_1 + bX_2 | \cG] = a \bE[X_1 | \cG] + b \bE[X_2|\cG].
    \]
\end{theorem}
\begin{proof}
    trivial\todo{add details}
\end{proof}

\begin{theorem}[Positivity]
    \label{ceprop4}
    \label{cpositivity}
    If $X \ge  0$, then $\bE[X | \cG] \ge 0$ a.s.
\end{theorem}
\begin{proof}
    Let $W $ be a version of $\bE[X | \cG]$.
    Suppose $\bP[ W < 0] > 0$.
    Then \[G \coloneqq  \{W < -\frac{1}{n}\} \in \cG.\]
    For some $n \in \N$, we have $\bP[G] > 0$.
    However it follows that
    \[
        \int_G \bP[X | \cG] \dif \bP \le  -\frac{1}{n} \bP[G] < 0 \le  \int_G X \dif \bP.
    \]
\end{proof}
\begin{theorem}[Conditional monotone convergence theorem]
    \label{ceprop5}
    \label{mcmt}
    Let $X_n,X \in L^1(\Omega, \cF, \bP)$.
    Suppose $X_n \ge 0$ with $X_n \uparrow X$.
    Then $\bE[X_n|\cG] \uparrow \bE[X|\cG]$.
\end{theorem}
\begin{proof}
    Let $Z_n$ be a version of $\bE[X_n | Y]$.
    Since  $X_n \ge 0$ and $X_n \uparrow$,
    by \autoref{cpositivity},
    we have
    \[
        \bE[X_n | \cG] \overset{\text{a.s.}}{\ge } 0
    \]
    and
    \[
        \bE[X_n | \cG] \uparrow \text{a.s.}
    \]
    (consider $X_{n+1} - X_n$ ).

    Define $Z \coloneqq  \limsup_{n \to \infty} Z_n$.
    Then $Z$ is $\cG$-measurable
    and $Z_n \uparrow Z$ a.s.

    Take some $G \in \cG$.
    We know by (b) % TODO REF
    that $\bE[Z_n \One_G] = \bE[X_n \One_G]$.
    The LHS increases to  $\bE[Z \One_G]$ by the monotone
    convergence theorem.
    Again by MCT, $\bE[X_n \One_G]$ increases to
    $\bE[X \One_G]$.
    Hence $Z$ is a version of $\bE[X | \cG]$.
\end{proof}

\begin{theorem}[Conditional Fatou]
    \label{ceprop6}
    \label{cfatou}
    Let $X_n \in L^1(\Omega, \cF, \bP)$, $X_n \ge 0$.
    Then
    \[
        \bE[ \liminf_{n \to \infty} X_n | \cG] \le  \liminf_{n \to \infty} \bE[X_n | \cG].
    \]
\end{theorem}
\begin{proof}
    \notes
\end{proof}
\begin{theorem}[Conditional dominated convergence theorem]
    \label{ceprop7}
    \label{cdct}
    Let $X_n,X \in L^1(\Omega, \cF, \bP)$.
    Suppose $|X_n(\omega)| < X(\omega)$ a.e.~
    and  $\int |X| \dif \bP < \infty$.
    Then $X_n(\omega) \to  X\left( \omega \right) \implies \bE[ X_n | \cG] \to \bE[X | \cG]$.

\end{theorem}
\begin{proof}
    \notes
\end{proof}

Recall
\begin{fact}[Jensen's inequality]
    If $c : \R \to  \R$ is convex and $\bE[|c \circ X|] < \infty$,
    then $\bE[c \circ X] \overset{\text{a.s.}}{\ge} c(\bE[X])$.
\end{fact}

For conditional expectation, we have
\begin{theorem}[Conditional Jensen's inequality]
    \label{ceprop8}
    \label{cjensen}
    Let $X \in L^1(\Omega, \cF, \bP)$.
    If $c : \R \to  \R$ is convex and $\bE[|c \circ X|] < \infty$,
    then $\bE[c \circ X | \cG] \ge c(\bE[X | \cG])$ a.s.
\end{theorem}
\begin{fact}
    \label{convapprox}
    If $c$ is convex, then there are two sequences of real numbers
    $a_n, b_n \in \R$
    such that
    \[
    c(x) = \sup_n(a_n x + b_n).
    \]
\end{fact}
\begin{refproof}{cjensen}
    By \autoref{convapprox}, $c(x) \ge  a_n X + b_n$
    for all $n$.
    Hence
    \[
        \bE[c(X) | \cG] \ge a_n \bE[X | \cG] + \bE[b_n | \cG]
        = a_n \bE[X | \cG] + b_n \text{ a.s.}
    \]
    for all $n$.
    Using that  a countable union of sets o f measure zero has measure zero,
    we conclude that a.s~this happens simultaneously for all $n$.
    Hence
    \[
        \bE[c(X) | \cG] \ge  \sup_n (a_n \bE[X | \cG] + b_n) \overset{\text{\autoref{convapprox}}}{=} c(\bE(X | \cG)).
    \]
\end{refproof}

Recall
\begin{fact}[Hölder's inequality]
    Let $p,q \ge  1$ such that $\frac{1}{p} + \frac{1}{q} = 1$.
    Suppose $X \in L^p(\bP)$ and $Y \in L^q(\bP)$.
    Then
    \[
        \bE(X Y) \le  \underbrace{\bE(|X|^p)^{\frac{1}{p}}}_{\text{\reflectbox{$\coloneqq$}} \|X\|_{L^p}}  \bE(|Y|^q)^{\frac{1}{q}}.
    \]
\end{fact}

\begin{theorem}[Conditional Hölder's inequality]
    \label{ceprop9}
    \label{choelder}
    Let $p,q \ge  1$ such that $\frac{1}{p} + \frac{1}{q} = 1$.
    Suppose $X \in L^p(\bP)$ and $Y \in L^q(\bP)$.
    Then
    \[
        \bE(X Y | \cG) \le \bE(|X|^p | \cG)^{\frac{1}{p}} \bE(|Y|^q | \cG)^{\frac{1}{q}}.
    \]
\end{theorem}
\begin{proof}
    \todo{Exercise}
\end{proof}

\begin{theorem}[Tower property]
    \label{ceprop10}
    \label{cetower}
    Suppose $\cF \supset \cG \supset \cH$ are sub-$\sigma$-algebras.
    Then
    \[
        \bE\left[\bE[X | \cG] \mid \cH\right] \overset{\text{a.s.}}{=} \bE[X | \cH].
    \]
\end{theorem}
\begin{proof}
    By definition, $\bE[\bE[X | \cG] | \cH]$ is  $\cH$-measurable.
    For any $H \in \cH$, we have
    \begin{IEEEeqnarray*}{rCl}
        \int_H \bE[\bE[X | \cG] | \cH] \dif \bP
        &=& \int_{H} \bE[X | \cG] \dif \bP\\
        &=& \int_H X \dif \bP.
    \end{IEEEeqnarray*}
    Hence $\bE[\bE[X | \cG] | \cH] \overset{\text{a.s.}}{=} \bE[X | \cH]$.
\end{proof}

\begin{theorem}[Taking out what is known]
    \label{ceprop11}
    \label{takingoutwhatisknown}

    If $Y$ is $\cG$-measurable and bounded, then
    \[
        \bE[YX| \cG] \overset{\text{a.s.}}{=} Y \bE[X | \cG].
    \]
\end{theorem}
\begin{proof}
    Assume w.l.o.g.~$X \ge 0$.
Assume $Y = \One_B$, then $Y$ simple, then take the limit (using that $Y$ is bounded).
    \todo{Exercise}
\end{proof}

\begin{definition}
    Let $\cG$ and $\cH$ be $\sigma$-algebras.
    We call $\cG$ and $\cH$ \vocab[$\sigma$-algebra!independent]{independent},
    if $\bP(G \cap H) = \bP(G) \bP(H)$
    for all events $G \in \cG$, $H \in \cH$.
\end{definition}

\begin{theorem}[Role of independence]
    \label{ceprop12}
    \label{ceroleofindependence}
    Let $X$ be a random variable,
    and let $\cG, \cH$ be $\sigma$-algebras.

    If $\cH$ is independent of $\sigma\left( \sigma(X), \cG \right)$,
    then
    \[
    \bE[X | \sigma(\cG, \cH)] \overset{\text{a.s.}}{=} \bE[X | \cG].
    \]

    In particular, if $X$ is independent of $\cG$,
    then
    \[
    \bE[X | \cG] \overset{\text{a.s.}}{=} \bE[X].
    \]
\end{theorem}

\begin{example}[Martingale property of the simple random walk]
    Suppose $X_1,X_2,\ldots$ are i.i.d.~with $\bP[X_i = 1] = \bP[X_i = -1] = \frac{1}{2}$.
    Let $S_n \coloneqq  \sum_{i=1}^n X_i$ be the \vocab{simple random walk}.
    Let $\cF$ denote the $\sigma$-algebra on the product space.
    Define $\cF_n \coloneqq  \sigma(X_1,\ldots)$.
    Intuitively, $\cF_n$ contains all the information gathered until time  $n$.
    We have $\cF_1 \subset \cF_2 \subset \cF_3 \subset \ldots$

    For $\bE[S_{n+1} | \cF_n]$ we obtain
    \begin{IEEEeqnarray*}{rCl}
        \bE[S_{n+1} | \cF_n] &\overset{\text{\autoref{celinearity}}}{=}&
        \bE[S_n | \cF_n] + \bE[X_{n+1} | \cF_n]\\
                             &\overset{\text{a.s.}}{=}& S_n + \bE[X_{n+1} | \cF_n]\\
                             &\overset{\text{\autoref{ceprop12}}}{=}& S_{n} + \bE[X_n]\\
                             &=& S_n.
    \end{IEEEeqnarray*}
\end{example}