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