From b7e2d7bd3c8a65083eff939208b7acade009f233 Mon Sep 17 00:00:00 2001 From: Josia Pietsch Date: Mon, 17 Jul 2023 15:18:26 +0200 Subject: [PATCH] improved notes on lecture 17 --- inputs/lecture_17.tex | 128 ++++++++++++++++++++++++++++-------------- 1 file changed, 86 insertions(+), 42 deletions(-) diff --git a/inputs/lecture_17.tex b/inputs/lecture_17.tex index f2ee73e..028d331 100644 --- a/inputs/lecture_17.tex +++ b/inputs/lecture_17.tex @@ -8,6 +8,12 @@ variables $(X_t)_{t \in T}$ for some index set $T$. In this lecture we will consider the case $T = \N$. \end{definition} +\begin{definition}[Previsible process] + Consider a filtration $(\cF_n)_{n \ge 0}$. + A stochastic process $(C_n)_{n \ge 1}$ + is called \vocab[Stochastic process!previsible]{previsible}, + iff $C_n$ is $\cF_{n-1}$-measurable. +\end{definition} \begin{goal} What about a ``gambling strategy''? @@ -19,30 +25,52 @@ Suppose that there is another stochastic process $(C_n)_{n \ge 1}$ such that $C_n$ is determined by the information gathered up until time $n$, - i.e.~$C_n$ is measurable with respect to $\cF_{n-1}$. - If such process $C_n$ exists, we say that $ C_n$ is - \vocab[Stochastic process!previsible]{previsible}. + i.e.~$C_n$ is previsible. Think of $C_n$ as our strategy of playing the game. Then $C_n(X_n - X_{n-1})$ defines the win in the $n$-th game, while - \[ - Y_n \coloneqq \sum_{j=1}^n C_j(X_j - X_{j-1}) - \] + \begin{equation} + Y_n \coloneqq \sum_{j=1}^n C_j(X_j - X_{j-1}) + \label{eqn:cumulative-win-process} + \end{equation} defines the cumulative win process. \end{goal} \begin{lemma} + \label{lem:gambling-strategy} If $(C_n)_{n \ge 1}$ is previsible and $(X_n)_{n \ge 0}$ - is a (sub/super-) martingale + is a martingale and there exists a constant $K_n$ such that $|C_n(\omega)| \le K_n$. - Then $(Y_n)_{n \ge 1}$ is also a (sub/super-) martingale. + Then $(Y_n)_{n \ge 1}$ defined in \eqref{eqn:cumulative-win-process} + is also a martingale. \end{lemma} -\begin{proof} - Exercise. \todo{Copy Exercise 10.4} -\end{proof} \begin{remark} The assumption of $K_n$ being constant can be weakened to - $C_n \in L^p(\bP)$, $X_n \in ^q(\bP)$ with $\frac{1}{p} + \frac{1}{q} = 1$. + $C_n \in L^p(\bP)$, $X_n \in L^q(\bP)$ with $\frac{1}{p} + \frac{1}{q} = 1$. + + If $C_n \ge 0$ the assumption of $(X_n)_{n\ge 0}$ + being a martingale can be weakened to it being + a sub-/supermartingale. + Then $(Y_n)_{n \ge 1}$ is a sub-/supermartingale as well. \end{remark} +\begin{refproof}{lem:gambling-strategy} + It is clear that $Y_n$ is $\cF_n$-measurable. + Suppose that $C_n \in L^p(\bP)$ and $X_n \in L^{q}(\bP)$ + for all $n$. + We have + \begin{IEEEeqnarray*}{rCl} + \|Y_n\|_{L^1} + &\le& \sum_{i=1}^{n} \|C_i(X_i - X_{i-1})\|_{L^1}\\ + &\overset{\text{Hölder}}{\le}& \sum_{i=1}^{n} \|C_i\|_{L^p} \|(X_i - X_{i-1})\|_{L^q} \\ + &<&\infty + \end{IEEEeqnarray*} + and + \begin{IEEEeqnarray*}{rCl} + \bE[Y_{n+1} - Y_n | \cF_n] + &=& \bE[C_{n+1} (X_{n+1} - X_n) | \cF_n]\\ + &=& C_{n+1} (\bE[X_{n+1} | \cF_n] - X_n)\\ + &=& 0. + \end{IEEEeqnarray*} +\end{refproof} Suppose we have $(X_n)$ adapted, $X_n \in L^1(\bP)$, $(C_n)_{n \ge 1}$ previsible. @@ -51,21 +79,29 @@ Pick two real numbers $a < b$. Wait until $X_n \le a$, then start playing. Stop playing when $X_n \ge b$. I.e.~define -\begin{itemize} - \item $C_1 \coloneqq 0$, - \item $C_n \coloneqq \One_{\{C_{n-1} = 1\}} \cdot \One_{\{X_{n-1} \le b\}} + \One_{\{C_{n-1} = 0\} } \One_{\{X_{n-1} < a\}}$. -\end{itemize} + +\begin{equation} + \begin{aligned} + C_1 &\coloneqq 0,\\ + C_n &\coloneqq + \One_{\{C_{n-1} = 1\}} \cdot \One_{\{X_{n-1} \le b\}} + + \One_{\{C_{n-1} = 0\} } \One_{\{X_{n-1} < a\}}. + \end{aligned} + \label{eqn:upcrossing-strategy} +\end{equation} \begin{definition} Fix $N \in \N$ and let -\[U_n^X([a,b]) \coloneqq \# \{\text{Upcrossings of $[a,b]$ made by $n \mapsto X_n(\omega)$ by time $n$}\},\] -i.e.~$U_n([a,b])(\omega)$ is the largest $k \in \N_0$ such that we can find a +\[U_N^X([a,b]) \coloneqq \# \{\text{Upcrossings of $[a,b]$ made by $n \mapsto X_n(\omega)$ by time $N$}\},\] +i.e.~$U_N([a,b])(\omega)$ is the largest $k \in \N_0$ such that we can find a sequence $0 \le s_1 < t_1 < s_2 < t_2 < \ldots < s_k < t_k \le N$ such that $X_{s_j}(\omega) < a$ and $X_{t_j}(\omega) > b$ for all $1 \le j \le k$. \end{definition} Clearly $U_N^X([a,b]) \uparrow$ as $N$ increases. -It follows that the monotonic limit $U_\infty([a,b]) \coloneqq \lim_{N \to \infty} U_N([a,b])$ exists pointwise. +It follows that the monotonic limit +\[U_\infty([a,b]) \coloneqq \lim_{N \to \infty} U_N([a,b])\] +exists pointwise. \begin{lemma} % Lemma 1 \label{lec17l1} \[ @@ -74,31 +110,37 @@ It follows that the monotonic limit $U_\infty([a,b]) \coloneqq \lim_{N \to \inf \{\omega: U^{Z}_\infty([a,b])(\omega) = \infty\} \] for every sequence of measurable functions $(Z_n)_{n \ge 1}$. - \end{lemma} \begin{lemma} % 2 \label{lec17l2} - Let $Y_n(\omega) \coloneqq \sum_{j=1}^n C_j(X_j - X_{j-1})$. + Let $Y_n(\omega) \coloneqq \sum_{j=1}^n C_j(X_j - X_{j-1})$, + where $C_n $ is defined as in \eqref{eqn:upcrossing-strategy} Then \[Y_N \ge (b -a) U_N([a,b]) - (X_N - a)^{-}.\] \end{lemma} \begin{proof} Every upcrossing of $[a,b]$ increases the value of $Y$ by $(b-a)$, - while the last intverval of play $(X_n -a)^{-}$ overemphasizes the loss. + while the last interval of play $(X_n -a)^{-}$ overemphasizes the loss. \end{proof} \begin{lemma} %3 \label{lec17l3} Suppose $(X_n)_n$ is a supermartingale. Then in the above setup - \[(b-a) \bE[U_N([a,b])] \le \bE[(X_n - a)^-].\] + \[(b-a) \bE[U_N([a,b])] \le \bE[(X_N - a)^-].\] \end{lemma} \begin{proof} - This is obvious from \autoref{lec17l2} - and the supermartingale property. + Since $C_n \ge 0$, + by \autoref{lem:gambling-strategy} we have that $Y_n$ is a supermartingale. + Hence $\bE[Y_N] \le \bE[Y_1] = 0$. + From \autoref{lec17l2} it follows that + \[ + (b-a) \bE[U_N([a,b])] \le \bE[Y_n] + \bE[(X_N-a)^-] \le \bE[(X_N-a)^-]. + \] + \end{proof} \begin{corollary} Let $(X_n)_n$ be a - \vocab[Supermartingale!bounded]{supermartingale bounded in $L^1(\bP)$}, + \vocab[Supermartingale!bounded in $L^1$]{supermartingale bounded in $L^1(\bP)$}, i.e.~$\sup_n \bE[|X_n|] < \infty$. Then $(b-a) \bE(U_\infty) \le |a| + \sup_n \bE(|X_n|)$. In particular, $\bP[U_\infty = \infty] = 0$. @@ -110,12 +152,21 @@ It follows that the monotonic limit $U_\infty([a,b]) \coloneqq \lim_{N \to \inf Since $U_N(\cdot) \ge 0$ and $U_N(\cdot ) \uparrow U_\infty(\cdot )$, by the monotone convergence theorem \[ - \bE(U_n([a,b])] \uparrow \bE[U_\infty([a,b])]. + \bE(U_N([a,b])] \uparrow \bE[U_\infty([a,b])]. \] \end{proof} -Assume now, that our process $(X_n)_{n \ge 1}$ is a supermartingale +Let us now consider the case that our process $(X_n)_{n \ge 1}$ is a supermartingale bounded in $L^1(\bP)$. + +\begin{theorem}[Doob's martingale convergence theorem] + \label{doobmartingaleconvergence} + \label{doob} + Any supermartingale bounded in $L^1$ converges almost surely to a + random variable, which is almost surely finite. + In particular, any non-negative supermartingale converges a.s.~to a finite random variable. +\end{theorem} +\begin{refproof}{doobmartingaleconvergence} Let \[ \Lambda \coloneqq \{\omega | X_n(\omega) \text{ does not converge to anything in $[-\infty,\infty]$}\}. @@ -127,10 +178,11 @@ We have &=& \bigcup_{a,b \in \Q} \underbrace{\{\omega | \liminf_N X_N(\omega) < a < b < \limsup_N X_N(\omega)\}}_{\Lambda_{a,b}} \\ \end{IEEEeqnarray*} -We have $\Lambda_{a,b} \subseteq \{\omega : U_{\infty}([a,b])(\omega) = \infty\} $ by lemma 1. % TODO REF -By lemma 3 % TODO REF -we have $\bP(\Lambda_{a,b}) = 0$, hence $\bP(\Lambda) = 0$. -Hence there exists a random variable $X_\infty$ such that $X_n \xrightarrow{a.s.} X_\infty$. +We have $\Lambda_{a,b} \subseteq \{\omega : U_{\infty}([a,b])(\omega) = \infty\}$ +by \autoref{lec17l1}. +By \autoref{lec17l3} we have $\bP(\Lambda_{a,b}) = 0$, +hence $\bP(\Lambda) = 0$. +Thus there exists a random variable $X_\infty$ such that $X_n \xrightarrow{a.s.} X_\infty$. \begin{claim} $\bP[X_\infty \in \{\pm \infty\}] = 0$. @@ -146,14 +198,7 @@ Hence there exists a random variable $X_\infty$ such that $X_n \xrightarrow{a.s. \end{IEEEeqnarray*} \end{subproof} -We have thus shown -\begin{theorem}[Doob's martingale convergence theorem] - \label{doobmartingaleconvergence} - \label{doob} - Any supermartingale bounded in $L^1$ converges almost surely to a - random variable, which is almost surely finite. - In particular, any non-negative supermartingale converges a.s.~to a finite random variable. -\end{theorem} + The second part follows from \begin{claim} Any non-negative supermartingale is bounded in $L^1$. @@ -163,5 +208,4 @@ The second part follows from Since the supermartingale is non-negative, we have $\bE[|X_n|] = \bE[X_n]$ and since it is a supermartingale $\bE[X_n] \le \bE[X_0]$. \end{subproof} - -\todo{rearrange proof} +\end{refproof}