Doob L1 typo
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@ -8,10 +8,21 @@ Exercise 4.3
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10.2
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10.2
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Martingales converging a.s.~but not in $L^1$.
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\begin{example}[Martingale not converging in $L^1$]
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Let $\Omega = [0,1]$, $\bP = \lambda\upharpoonright [0,1]$.
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Define $X_n \coloneqq 2^n \cdot \One_{[0,2^n]}$,
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and let $(\cF_n)_n$ be the canonical filtration.
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Then $(X_n)_{n}$ is a Martingale
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with $\bE[X_0] = 1$,
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but $X_n \xrightarrow{a.s.} 0$.
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\end{example}
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Stopping times
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Stopping times
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\begin{example}[{Martingale such that $\bE[X_T] \neq \bE[X_0]$}]
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Consider the simple random walk and $T = \inf \{n : X_n \ge 1\}$.
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Obviously $X_T = 1$.
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\end{example}
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@ -193,7 +193,9 @@ Typically $\cF_n = \sigma(X_1, \ldots, X_n)$ for a sequence of random variables.
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\end{definition}
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\end{definition}
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\begin{corollary}
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\begin{corollary}
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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 $f: \R \to \R$ 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\footnote{In this form it means, that there is some filtration, that we don't explicitly specify}.
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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 sub-martingale.
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Then $(f(X_n))_n$ is a sub-martingale.
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\end{corollary}
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\end{corollary}
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\begin{proof}
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\begin{proof}
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@ -140,7 +140,7 @@ First, we need a very important inequality:
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Let $X_n^\ast \coloneqq \max \{|X_1|, |X_2|, \ldots, |X_n|\}$
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Let $X_n^\ast \coloneqq \max \{|X_1|, |X_2|, \ldots, |X_n|\}$
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denote the \vocab{running maximum}.
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denote the \vocab{running maximum}.
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\begin{enumerate}[(1)]
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\begin{enumerate}[(1)]
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\item Then \[ \forall \ell > 0 .~\bP[X_n^\ast \ge \ell] \le \frac{1}{\ell} \int_{\{X_n^\ast \ell\}} |X_n| \dif \bP \le \frac{1}{\ell} \bE[|X_n|]. \]
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\item Then \[ \forall \ell > 0 .~\bP[X_n^\ast \ge \ell] \le \frac{1}{\ell} \int_{\{X_n^\ast \ge \ell\}} |X_n| \dif \bP \le \frac{1}{\ell} \bE[|X_n|]. \]
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(Doob's $L^1$ inequality).
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(Doob's $L^1$ inequality).
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\item Fix $p > 1$. Then \[
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\item Fix $p > 1$. Then \[
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\bE[(X_n^\ast)^p] \le \left( \frac{p}{p-1} \right)^p \bE[|X_n|^p].
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\bE[(X_n^\ast)^p] \le \left( \frac{p}{p-1} \right)^p \bE[|X_n|^p].
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