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