311 lines
9.8 KiB
TeX
311 lines
9.8 KiB
TeX
\lecture{20}{2023-06-27}{}
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\begin{refproof}{ceismartingale}
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By the tower property (\autoref{cetower})
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it is clear that $(\bE[X | \cF_n])_n$
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is a martingale.
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First step:
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Assume that $X$ is bounded.
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Then, by \autoref{cjensen}, $|X_n| \le \bE[|X| | \cF_n]$,
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hence $\sup_{\substack{n \in \N \\ \omega \in \Omega}} | X_n(\omega)| < \infty$.
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Thus $(X_n)_n$ is a martingale in $L^{\infty} \subseteq L^2$.
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By the convergence theorem for martingales in $L^2$
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(\autoref{martingaleconvergencel2})
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there exists a random variable $Y$,
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such that $X_n \xrightarrow{L^2} Y$.
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Fix $m \in \N$ and $A \in \cF_m$.
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Then
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\begin{IEEEeqnarray*}{rCl}
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\int_A Y \dif \bP
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&=& \lim_{n \to \infty} \int_A X_n \dif \bP\\
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&=& \lim_{n \to \infty} \bE[X_n \One_A]\\
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&=& \lim_{n \to \infty} \bE[\bE[X | \cF_n] \One_A]\\
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&\overset{A \in \cF_n}{=}& \lim_{\substack{n \to \infty\\n \ge m}} \bE[X \One_A]\\
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\end{IEEEeqnarray*}
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Hence $\int_A Y \dif \bP = \int_A X \dif \bP$ for all $m \in \N, A \in \cF_m$.
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Since $\sigma(X) = \bigcup \cF_n$
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this holds for all $A \in \sigma(X)$.
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Hence $X = Y$ a.s., so $X_n \xrightarrow{L^2} X$.
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Since $(X_n)_n$ is uniformly bounded, this also means
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$X_n \xrightarrow{L^p} X$.
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Second step:
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Now let $X \in L^p$ be general and define
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\[
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X'(\omega) \coloneqq \begin{cases}
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X(\omega)& \text{ if } |X(\omega)| \le M,\\
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0&\text{ otherwise}
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\end{cases}
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\]
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for some $M > 0$.
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Then $X' \in L^\infty$ and
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\begin{IEEEeqnarray*}{rCl}
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\int | X - X'|^p \dif \bP &=& \int_{\{|X| > M\} } |X|^p \dif \bP \xrightarrow{M \to \infty} 0
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\end{IEEEeqnarray*}
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as $\bP$ is regular,
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i.e.~$\forall \epsilon > 0 . ~\exists k . ~
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\bP[|X|^p \in [-k,k]] \ge 1-\epsilon$.
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Take some $\epsilon > 0$ and $M$ large enough such that
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\[
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\int |X - X'| \dif \bP < \epsilon.
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\]
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Let $(X_n')_n$ be the martingale given by $(\bE[X' | \cF_n])_n$.
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Then $X_n' \xrightarrow{L^p} X'$ by the first step.
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It is
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\begin{IEEEeqnarray*}{rCl}
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\|X_n - X_n'\|_{L^p}^p
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&=& \bE[\bE[X - X' | \cF_n]^{p}]\\
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&\overset{\text{Jensen}}{\le}& \bE[\bE[(X - X')^p | \cF_n]]\\
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&=& \|X - X'\|_{L^p}^p\\
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&<& \epsilon.
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\end{IEEEeqnarray*}
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Hence
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\[
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\|X_n - X\|_{L^p} %
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\le \|X_n - X_n'\|_{L^p} + \|X_n' - X'\|_{L^p} + \|X - X'\|_{L^p} %
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\le 3 \epsilon.
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\]
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Thus $X_n \xrightarrow{L^p} X$.
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\end{refproof}
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For the proof of \autoref{martingaleisce},
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we need the following theorem, which we won't prove here:
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\begin{theorem}[Banach Alaoglu]
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\label{banachalaoglu}
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Let $X$ be a normed vector space and $X^\ast$ its
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continuous dual.
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Then the closed unit ball in $X^\ast$ is compact
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w.r.t.~the ${\text{weak}}^\ast$ topology.
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\end{theorem}
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\begin{fact}
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We have $L^p \cong (L^q)^\ast$ for $\frac{1}{p} + \frac{1}{q} = 1$
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via
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\begin{IEEEeqnarray*}{rCl}
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L^p &\longrightarrow & (L^q)^\ast \\
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f &\longmapsto & (g \mapsto \int g f \dif\bP)
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\end{IEEEeqnarray*}
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We also have $(L^1)^\ast \cong L^\infty$,
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however $ (L^\infty)^\ast \not\cong L^1$.
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\end{fact}
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\begin{refproof}{martingaleisce}
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Since $(X_n)_n$ is bounded in $L^p$, by \autoref{banachalaoglu},
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there exists $X \in L^p$ and a subsequence
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$(X_{n_k})_k$ such that for all $Y \in L^q$ ($\frac{1}{p} + \frac{1}{q} = 1$ )
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\[
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\int X_{n_k} Y \dif \bP \to \int XY \dif \bP
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\]
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(Note that this argument does not work for $p = 1$,
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because $(L^\infty)^\ast \not\cong L^1$).
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Let $A \in \cF_m$ for some fixed $m$
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and choose $Y = \One_A$.
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Then
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\begin{IEEEeqnarray*}{rCl}
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\int_A X \dif \bP
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&=& \lim_{k \to \infty} \int_A X_{n_k} \dif \bP\\
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&=& \lim_{k \to \infty} \bE[X_{n_k} \One_A]\\
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&\overset{\text{for }n_k \ge m}{=}& \bE[X_m \One_A].
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\end{IEEEeqnarray*}
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Hence $X_n = \bE[X | \cF_m]$ by the uniqueness of conditional expectation
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and by \autoref{ceismartingale},
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we get the convergence.
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\end{refproof}
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\subsection{Stopping Times}
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\begin{definition}[Stopping time]
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\label{def:stopping-time}
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A random variable $T: \Omega \to \N_0 \cup \{\infty\}$ on a filtered probability space $(\Omega, \cF, \{\cF_n\}_n, \bP)$ is called a \vocab{stopping time},
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if
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\[
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\{T \le n\} \in \cF_n
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\]
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for all $n \in \N$.
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Equivalently, $\{T = n\} \in \cF_n$ for all $n \in \N$.
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\end{definition}
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\begin{example}
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A constant random variable $T = c$ is a stopping time.
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\end{example}
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\begin{example}[Hitting times]
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For an adapted process $(X_n)_n$
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with values in $\R$ and $A \in \cB(\R)$, the \vocab{hitting time}
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\[
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T \coloneqq \inf \{n \in \N : X_n \in A\}
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\]
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is a stopping time,
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as
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\[
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\{T \le n \} = \bigcup_{k=1}^n \{X_k \in A\} \in \cF_n.
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\]
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However, the last exit time
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\[
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T \coloneqq \sup \{n \in \N : X_n \in A\}
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\]
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is not a stopping time.
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\end{example}
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\begin{example}
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Consider the simple random walk, i.e.
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$X_n$ i.i.d.~with $\bP[X_n = 1] = \bP[X_n = -1] = \frac{1}{2}$.
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Set $S_n \coloneqq \sum_{i=1}^{n} X_n$.
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Then
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\[
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T \coloneqq \inf \{n \in \N : S_n \ge A \lor S_n \le B\}
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\]
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is a stopping time.
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\end{example}
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\begin{fact}
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If $T_1, T_2$ are stopping times with respect to the same filtration,
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then
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\begin{itemize}
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\item $T_1 + T_2$,
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\item $\min \{T_1, T_2\}$ and
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\item $\max \{T_1, T_2\}$
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\end{itemize}
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are stopping times.
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\end{fact}
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\begin{warning}
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Note that $T_1 - T_2$ is not a stopping time.
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\end{warning}
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\begin{remark}
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There are two ways to look at the interaction between a stopping time $T$
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and a stochastic process $(X_n)_n$:
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\begin{itemize}
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\item The behaviour of $ X_n$ until $T$, i.e.
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\[
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X^T \coloneqq \left(X_{T \wedge n}\right)_{n \in \N}
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\]
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is called the \vocab{stopped process}.
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\item The value of $(X_n)_n)$ at time $T$,
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i.e.~looking at $X_T$.
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\end{itemize}
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\end{remark}
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\begin{example}
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If we look at a process
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\[ S_n = \sum_{i=1}^{n} X_i \]
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for some $(X_n)_n$,
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then
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\[ S^T = (\sum_{i=1}^{T \wedge n} X_i)_n \]
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and
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\[ S_T = \sum_{i=1}^{T} X_i. \]
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\end{example}
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\begin{theorem}
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If $(X_n)_n$ is a supermartingale and $T$ is a stopping time,
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then $X^T$ is also a supermartingale,
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and we have $\bE[X_{T \wedge n}] \le \bE[X_0]$ for all $n$.
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If $(X_n)_n$ is a martingale, then so is $X^T$
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and $\bE[X_{T \wedge n}] = \bE[X_0]$.
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\end{theorem}
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\begin{proof}
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First, we need to show that $X^T$ is adapted.
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This is clear since
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\begin{IEEEeqnarray*}{rCl}
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X^T_n &=& X_T \One_{T < n} + X_n \One_{T \ge n}\\
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&=& \sum_{k=1}^{n-1} X_k \One_{T = k} + X_n \One_{T \ge n}.
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\end{IEEEeqnarray*}
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It is also clear that $X^T_n$ is integrable since
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\[
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\bE[|X^T_n|] \le \sum_{k=1}^{n} \bE[|X_k|] < \infty.
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\]
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We have
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\begin{IEEEeqnarray*}{rCl}
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&&\bE[X^T_n - X^T_{n-1} | \cF_{n-1}]\\
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&=& \bE\left[X_n \One_{\{T \ge n\}} + \sum_{k=1}^{n-1} X_k \One_{\{ T = k\} }
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- X_{n-1}(\One_{T \ge n} + \One_{\{T = n-1\}})\right.\\
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&&\left.+ \sum_{k=1}^{n-2} X_k \One_{\{T = k\} } \middle| \cF_{n-1}\right]\\
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&=& \bE[(X_n - X_{n-1}) \One_{\{ T \ge n\} } | \cF_{n-1}]\\
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&=& \One_{\{ T \ge n\}} (\bE[X_n | \cF_{n-1}] - X_{n-1})
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\begin{cases}
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\le 0\\
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= 0 \text{ if $(X_n)_n$ is a martingale}.
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\end{cases}
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\end{IEEEeqnarray*}
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\end{proof}
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\begin{remark}
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\label{roptionalstoppingi}
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We now want a similar statement for $X_T$.
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In the case that $T \le M$ is bounded,
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we get from the above that
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\[
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\bE[X_T] \overset{n \ge M}{=} \bE[X^T_n] \begin{cases}
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\le \bE[X_0] & \text{ supermartingale},\\
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= \bE[X_0] & \text{ martingale}.
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\end{cases}
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\]
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However if $T$ is not bounded, this does not hold in general.
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\end{remark}
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\begin{example}
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Let $(S_n)_n$ be the simple random walk
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and take $T \coloneqq \inf \{n : S_n = 1\}$.
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Then $\bP[T < \infty] = 1$, but
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\[
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1 = \bE[S_T] \neq \bE[S_0] = 0.
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\]
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\end{example}
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\begin{theorem}[Optional Stopping]
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\label{optionalstopping}
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Let $(X_n)_n$ be a supermartingale
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and let $T$ be a stopping time
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taking values in $\N$.
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If one of the following holds
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\begin{enumerate}[(i)]
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\item $T \le M$ is bounded,
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\item $(X_n)_n$ is uniformly bounded
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and $T < \infty$ a.s.,
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\item $\bE[T] < \infty$
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and $|X_n(\omega) - X_{n-1}(\omega)| \le K$
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for all $n \in \N, \omega \in \Omega$ and
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some $K > 0$,
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\end{enumerate}
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then $\bE[X_T] \le \bE[X_0]$.
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If $(X_n)_n$ even is a martingale, then
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under the same conditions
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$\bE[X_T] = \bE[X_0]$.
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\end{theorem}
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\begin{proof}
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(i) was already done in \autoref{roptionalstoppingi}.
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(ii): Since $(X_n)_n$ is bounded, we get that
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\begin{IEEEeqnarray*}{rCl}
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\bE[|X_T - X_0|] &\overset{\text{dominated convergence}}{=}& \lim_{n \to \infty} \bE[|X_{T \wedge n} - X_0|]\\
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&\overset{\text{part (i)}}{\le}& 0.
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\end{IEEEeqnarray*}
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(iii): It is
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\begin{IEEEeqnarray*}{rCl}
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|X_{T \wedge n}- X_0| &\le& | \sum_{k=1}^{T \wedge n} X_k - X_{k-1}|\\
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&\le & (T \wedge n) \cdot K\\
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&\le & T \cdot K < \infty.
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\end{IEEEeqnarray*}
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Hence, we can apply dominated convergence and obtain
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\begin{IEEEeqnarray*}{rCl}
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\bE[X_T - X_0] &=& \lim_{n \to \infty} \bE[X_{T \wedge n} - X_0].
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\end{IEEEeqnarray*}
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Thus, we can apply (ii).
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The statement about martingales follows from
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applying this to $(X_n)_n$ and $(-X_n)_n$,
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which are both supermartingales.
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\end{proof}
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