\lecture{16}{2023-06-13}{} % \subsection{Conditional expectation} \begin{refproof}{ceroleofindependence} Let $\cH$ be independent of $\sigma(\sigma(X), \cG)$. Then for all $H \in \cH$, we have that $\One_H$ and any random variable measurable with respect to either $\sigma(X)$ or $\cG$ must be independent. It suffices to consider the case of $X \ge 0$. Let $G \in \cG$ and $H \in \cH$. By assumption, $X \One_G$ and $\One_H$ are independent. Let $Z \coloneqq \bE[X | \cG]$. Then \begin{IEEEeqnarray*}{rCl} \underbrace{\bE[X;G \cap H]}_{\coloneqq \int_{G \cap H} X \dif \bP} &=& \bE[(X \One_G) \One_H]\\ &=& \bE[X \One_G] \bE[\One_H]\\ &=& \bE[Z \One_G] \bP(H)\\ &=& \bE[Z; G \cap H] \end{IEEEeqnarray*} The identity above means, that the measures $A \mapsto \bE[X; A]$ and $A \mapsto \bE[Z; A]$ agree on the $\sigma$-algebra $\sigma(\cG, \cH)$ for events of the form $G \cap H$. Since sets of this form generate $\sigma(\cG, \cH)$, these two measures must agree on $\sigma(\cG, \cH)$. The claim of the theorem follows by the uniqueness of conditional expectation. To deduce the second statement, choose $\cG = \{\emptyset, \Omega\}$. \end{refproof} \subsection{The Radon Nikodym Theorem} First, let us recall some basic facts: \begin{fact} Let $(\Omega, \cF, \mu)$ be a \vocab[Measure space!$\sigma$-finite]{$\sigma$-finite measure space}, i.e.~$\Omega$ can be decomposed into countably many subsets of finite measure. Let $f: \Omega \to [0, \infty)$ be measurable. Define $\nu(A) \coloneqq \int_A f \dif \mu$. Then $\nu$ is also a $\sigma$-finite measure on $(\Omega, \cF)$.\todo{Application of mct} Moreover, $\nu$ is finite iff $f$ is integrable. \end{fact} Note that in this setting, if $\mu(A) = 0$ it follows that $\nu(A) = 0$. The Radon Nikodym theorem is the converse of that: \begin{theorem}[Radon-Nikodym] \yalabel{Radon-Nikodym Theorem}{Radon-Nikodym}{radonnikodym} Let $\mu$ and $\nu$ be two $\sigma$-finite measures on $(\Omega, \cF)$. Suppose \[ \forall A \in \cF . ~ \mu(A) = 0 \implies \nu(A) = 0. \] Then \begin{enumerate}[(1)] \item there exists $Z: \Omega \to [0, \infty)$ measurable, such that \[\forall A \in \cF . ~ \nu(A) = \int_A Z \dif \mu.\] \item Such a $Z$ is unique up to equality a.e.~(w.r.t. $\mu$). \item $Z$ is integrable w.r.t.~$ \mu$ iff $\nu$ is a finite measure. \end{enumerate} Such a $Z$ is called the \vocab{Radon-Nikodym derivative}. \end{theorem} \begin{definition} Whenever the property $\forall A \in \cF, \mu(A) = 0 \implies \nu(A) = 0$ holds for two measures $\mu$ and $\nu$, we say that $\nu$ is \vocab{absolutely continuous} w.r.t.~$\mu$. This is written as $\nu \ll \mu$. \end{definition} \begin{definition}+ Two measures $\mu$ and $\nu$ on a measure space $(\Omega, \cF)$ are called \vocab{singular}, denoted $\mu \bot \nu$, iff there exists $A \in \cF$ such that \[ \mu(A) = \nu(A^c) = 0. \] \end{definition} With the \yaref{radonnikodym} we get a very short proof of the existence of conditional expectation: \begin{proof}[Second proof of \yaref{conditionalexpectation}] Let $(\Omega, \cF, \bP)$ as always, $X \in L^1(\bP)$ and $\cG \subseteq \cF$. It suffices to consider the case of $X \ge 0$. For all $G \in \cG$, define $\nu(G) \coloneqq \int_G X \dif \bP$. Obviously, $\nu \ll \bP$ on $\cG$. Then apply the \yaref{radonnikodym}. \end{proof} \begin{refproof}{radonnikodym} We will only sketch the proof. A full proof can be found in the official notes. \paragraph{Step 1: Uniqueness} \notes \paragraph{Step 2: Reduction to the finite measure case} \notes \paragraph{Step 3: Getting hold of $Z$} Assume now that $\mu$ and $\nu$ are two finite measures. Let \[ \cC \coloneqq \left\{f: \Omega \to [0,\infty] \middle| % \forall A \in \cF.~\int_A f \dif \mu \le \nu(A)\right\}. \] We have $\cC \neq \emptyset$ since $0 \in \cC$. The goal is to find a maximal function $Z$ in $\cC$. Obviously its integral will also be maximal. \begin{enumerate}[(a)] \item If $f,g \in \cC$, than $f \lor g$ (the pointwise maximum) s also in $\cC$. \item Suppose $\{f_n\}_{n \ge 1}$ is an increasing sequence in $\cC$. Let $f$ be the pointwise limit. Then $f \in \cC$. \item For all $f \in \cC$, we have \[ \int_\Omega f \dif \mu \le \nu(\Omega) < \infty. \] \end{enumerate} Define $\alpha \coloneqq \sup \{ \int f \dif \mu : f \in \cC\} \le \nu(\Omega) < \infty$. Let $f_n \in \cC, n\in \N$ be a sequence with $\int f_n \dif \mu \to \alpha$. Define $g_n \coloneqq \max \{f_1,\ldots,f_n\} \in \cC$. Applying (b), we get that the pointwise limit, $Z$, is an element of $\cC$. \paragraph{Step 4: Showing that our choice of $Z$ works} Define $\lambda(A) \coloneqq \nu(A) - \int_A Z \dif \mu \ge 0$. $\lambda$ is a measure. \begin{claim} $\lambda = 0$. \end{claim} \begin{subproof} Call $G \in \cF$ \emph{good} if the following hold: \begin{enumerate}[(i)] \item $\lambda(G) - \frac{1}{k}\mu(G) > 0$. \item $\forall B \subseteq G, B \in \cF. ~ \lambda(B) - \frac{1}{k}\mu(B) \ge 0$. \end{enumerate} Suppose we know that for all $A \in \cF, k \in \N$ we have $\lambda(A) \le \frac{1}{k} \mu(A)$. Then $\lambda(A) = 0$ since $\mu$ is finite. Assume the claim does not hold. Then there must be some $k \in \N$, $A \in \cF$ such that $\lambda(A) - \frac{1}{k} \mu(A) > 0$. Fix this $A$ and $k$. Then $A$ satisfies condition (i) of being good, but it need not satisfy (ii). The tricky part is to make $A$ smaller such that it also satisfies (ii).\notes \end{subproof} \end{refproof} \section{Martingales} \subsection{Definition} We have already worked with martingales, but we will define them rigorously now. \begin{definition}[Filtration] A \vocab{filtration} is a sequence $(\cF_n)$ of $\sigma$-algebras such that $\cF_n \subseteq \cF_{n+1}$ for all $n \ge 1$. \end{definition} Intuitively, we can think of a $\cF_n$ as the set of information we have gathered up to time $n$. Typically $\cF_n = \sigma(X_1, \ldots, X_n)$ for a sequence of random variables. \begin{definition} \label{def:martingale} Let $(\cF_n)$ be a filtration and $X_1,\ldots,X_n$ be random variables such that $X_i \in L^1(\bP)$. Then we say that $(X_n)_{n \ge 1}$ is an $(\cF_n)_n$-\vocab{martingale} if the following hold: \begin{itemize} \item $X_n$ is $\cF_n$-measurable for all $n$. ($X_n$ is \vocab[Sequence!adapted to a filtration]{adapted to the filtration} $\cF_n$ ). \item $\bE[X_{n+1} | \cF_n] \overset{\text{a.s.}}{=} X_n$ for all $n$. \end{itemize} $(X_n)_n$ is called a \vocab{submartingale}, if it is adapted to $\cF_n$ but \[ \bE[X_{n+1} | \cF_n] \overset{\text{a.s.}}{\ge} X_n. \] It is called a \vocab{supermartingale} if it is adapted but $\bE[X_{n+1} | \cF_n] \overset{\text{a.s.}}{\le} X_n$. \end{definition} \begin{corollary} \label{cor:convexmartingale} Suppose that $f: \R \to \R$ is a convex function such that $f(X_n) \in L^1(\bP)$. Suppose that $(X_n)_n$ is a martingale% \footnote{In this form it means, that there is some filtration, that we don't explicitly specify}. Then $(f(X_n))_n$ is a submartingale. Likewise, if $f$ is concave, then $((f(X_n))_n$ is a supermartingale. \end{corollary} \begin{proof} Apply \yaref{cjensen}. \end{proof} \begin{corollary} If $(X_n)_n$ is a martingale, then $\bE[X_n] = \bE[X_0]$. \end{corollary} \begin{example} The simple random walk: Let $\xi_1, \xi_2, ..$ iid, $\bP[\xi_i = 1] = \bP[\xi_i = -1] = \frac{1}{2}$, $X_n \coloneqq \xi_1 + \ldots + \xi_n$ and $\cF_n \coloneqq \sigma(\xi_1, \ldots, \xi_n) = \sigma(X_1, \ldots, X_n)$. Then $X_n$ is $\cF_n$-measurable. Showing that $(X_n)_n$ is a martingale is left as an exercise. \end{example} \begin{example} See exercise sheet 9. \todo{Copy} \end{example}