From 5304f38fbbe1b6abe8ba0d4b3613996489c23f73 Mon Sep 17 00:00:00 2001 From: Josia Pietsch Date: Thu, 15 Jun 2023 17:51:32 +0200 Subject: [PATCH] lecture 17 --- inputs/lecture_16.tex | 10 +-- inputs/lecture_17.tex | 166 +++++++++++++++++++++++++++++++++++++++++ probability_theory.tex | 1 + 3 files changed, 172 insertions(+), 5 deletions(-) create mode 100644 inputs/lecture_17.tex diff --git a/inputs/lecture_16.tex b/inputs/lecture_16.tex index 50a2ef5..e78e929 100644 --- a/inputs/lecture_16.tex +++ b/inputs/lecture_16.tex @@ -221,8 +221,7 @@ Typically $\cF_n = \sigma(X_1, \ldots, X_n)$ for a sequence of random variables. \begin{example} - \begin{itemize} - \item The simple random walk: + The simple random walk: Let $\xi_1, \xi_2, ..$ iid, $\bP[\xi_i = 1] = \bP[\xi_i = -1] = \frac{1}{2}$, @@ -231,7 +230,8 @@ Typically $\cF_n = \sigma(X_1, \ldots, X_n)$ for a sequence of random variables. Then $X_n$ is $\cF_n$-measurable. Showing that $(X_n)_n$ is a martingale is left as an exercise. - \item See exercise sheet 9. - \item The branching process (next lecture). - \end{itemize} +\end{example} +\begin{example} + See exercise sheet 9. + \todo{Copy} \end{example} diff --git a/inputs/lecture_17.tex b/inputs/lecture_17.tex new file mode 100644 index 0000000..0bda9c9 --- /dev/null +++ b/inputs/lecture_17.tex @@ -0,0 +1,166 @@ +\lecture{17}{2023-06-15}{} + +\begin{definition}[Stochastic process] + % TODO +\end{definition} + +\begin{goal} + What about a ``gambling strategy''? + + Consider a stochastic process $(X_n)_{n \in \N}$. + + Note that the increments $X_{n+1} - X_n$ can be thought of as the win + or loss per round of a game. + 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}. + 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 + \[ + Y_n \coloneqq \sum_{j=1}^n C_j(X_j - X_{j-1}) + \] + defines the cumulative win process. +\end{goal} +\begin{lemma} + If $(C_n)_{n \ge 1}$ is previsible and $(X_n)_{n \ge 0}$ + is a (sub/super-) 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. +\end{lemma} +\begin{proof} + Exercise. \todo{Copy} +\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 $\frac{1}{p} + \frac{1}{q} = 1$. +\end{remark} + +Suppose we have $(X_n)$ adapted, $X_n \in L^1(\bP)$, +$(C_n)_{n \ge 1}$ previsible. +We play according to the following principle: +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{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 +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. +\begin{lemma} % Lemma 1 + \[ + \{\omega | \liminf_{N \to \infty} Z_N(\omega) < a < b < + \limsup_{N \to \infty} Z_N(\omega)\} \subseteq + \{\omega: U^{Z}_\infty([a,b])(\omega) = \infty\} + \] + for every sequence of measurable functions $(Z_n)_{n \ge 1}$. + +\end{lemma} +\begin{lemma} % 2 + Let $Y_n(\omega) \coloneqq \sum_{j=1}^n C_j(X_j - X_{j-1})$. + 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. +\end{proof} + +\begin{lemma} %3 + Suppose $(X_n)_n$ is a supermartingale. + Then in the above setup, $(b-a) \bE[U_N([a,b])] \le \bE[(X_n - a)^-]$. +\end{lemma} +\begin{proof} + Obvious from lemma 2 % TODO REF + and the supermartingale property. +\end{proof} +\begin{corollary} + Let $(X_n)_n$ be a \vocab[Supermartingale!bounded]{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, $\bP[U_\infty = \infty] = 0$. +\end{corollary} +\begin{proof} + By lemma 3 % TODO REF + we have that + \[(b-a) \bE[U_N([a,b])] \le \bE[ | X_N| ] + |a| \le \sup_n \bE[|X_n|] + |a|.\] + Since $U_N(\cdot) \ge 0$ and $U_N(\cdot ) \uparrow U_\infty(\cdot )$, + by the monotone convergence theorem + \[ + \bE(U_n([a,b])] \uparrow \bE[U_\infty([a,b])]. + \] +\end{proof} + +Assume now, that our process $(X_n)_{n \ge 1}$ is a supermartingale +bounded in $L^1(\bP)$. +Let +\[ +\Lambda \coloneqq \{\omega | X_n(\omega) \text{ does not converge to anything in $[-\infty,\infty]$}\}. +\] +We have +\begin{IEEEeqnarray*}{rCl} + \Lambda &=& \{\omega | \liminf_N X_N(\omega) < \limsup_N X_N(\omega)\}\\ + &=& \{\omega | \liminf_N X_N(\omega) < a < b \limsup_N X_N(\omega)\} \\ + &=& \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 $X_n \xrightarrow{a.s.} X_\infty$. + +\begin{claim} + $\bP[X_\infty \in \{\pm \infty\}] = 0$. +\end{claim} +\begin{subproof} + It suffices to show that $\bE[|X_\infty|] < \infty$. + We have. + \begin{IEEEeqnarray*}{rCl} + \bE[|X_\infty|] &=& \bE[\liminf_{n \to \infty} |X_n|]\\ + &\overset{\text{Fatou}}{\le }& \liminf_n \bE[|X_n|]\\ + &\le & \sup_n \bE[|X_n|]\\ + &<& \infty. + \end{IEEEeqnarray*} +\end{subproof} + +We have thus shown +\begin{theorem}[Doob's martingale convergence theorem] + \label{doobmartingaleconvergence} + 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 $L^1$. +\end{claim} +\begin{subproof} + We need to show $\sup_n \bE(|X_n|) < \infty$. + 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} + + + +\begin{example}[Branching process] + % TODO + +\end{example} + + diff --git a/probability_theory.tex b/probability_theory.tex index 87de2d8..fc9e70f 100644 --- a/probability_theory.tex +++ b/probability_theory.tex @@ -40,6 +40,7 @@ \input{inputs/lecture_14.tex} \input{inputs/lecture_15.tex} \input{inputs/lecture_16.tex} +\input{inputs/lecture_17.tex} \cleardoublepage