lecture 17
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@ -221,8 +221,7 @@ Typically $\cF_n = \sigma(X_1, \ldots, X_n)$ for a sequence of random variables.
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\begin{example}
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\begin{example}
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\begin{itemize}
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The simple random walk:
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\item The simple random walk:
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Let $\xi_1, \xi_2, ..$ iid,
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Let $\xi_1, \xi_2, ..$ iid,
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$\bP[\xi_i = 1] = \bP[\xi_i = -1] = \frac{1}{2}$,
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$\bP[\xi_i = 1] = \bP[\xi_i = -1] = \frac{1}{2}$,
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@ -231,7 +230,8 @@ Typically $\cF_n = \sigma(X_1, \ldots, X_n)$ for a sequence of random variables.
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Then $X_n$ is $\cF_n$-measurable.
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Then $X_n$ is $\cF_n$-measurable.
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Showing that $(X_n)_n$ is a martingale
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Showing that $(X_n)_n$ is a martingale
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is left as an exercise.
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is left as an exercise.
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\item See exercise sheet 9.
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\end{example}
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\item The branching process (next lecture).
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\begin{example}
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\end{itemize}
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See exercise sheet 9.
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\todo{Copy}
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\end{example}
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\end{example}
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inputs/lecture_17.tex
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inputs/lecture_17.tex
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\lecture{17}{2023-06-15}{}
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\begin{definition}[Stochastic process]
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% TODO
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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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Consider a stochastic process $(X_n)_{n \in \N}$.
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Note that the increments $X_{n+1} - X_n$ can be thought of as the win
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or loss per round of a game.
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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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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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\[
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Y_n \coloneqq \sum_{j=1}^n C_j(X_j - X_{j-1})
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\]
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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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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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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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\end{lemma}
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\begin{proof}
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Exercise. \todo{Copy}
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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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\end{remark}
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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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We play according to the following principle:
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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{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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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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\begin{lemma} % Lemma 1
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\[
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\{\omega | \liminf_{N \to \infty} Z_N(\omega) < a < b <
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\limsup_{N \to \infty} Z_N(\omega)\} \subseteq
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\{\omega: U^{Z}_\infty([a,b])(\omega) = \infty\}
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\]
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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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Let $Y_n(\omega) \coloneqq \sum_{j=1}^n C_j(X_j - X_{j-1})$.
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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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\end{proof}
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\begin{lemma} %3
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Suppose $(X_n)_n$ is a supermartingale.
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Then in the above setup, $(b-a) \bE[U_N([a,b])] \le \bE[(X_n - a)^-]$.
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\end{lemma}
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\begin{proof}
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Obvious from lemma 2 % TODO REF
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and the supermartingale property.
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\end{proof}
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\begin{corollary}
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Let $(X_n)_n$ be a \vocab[Supermartingale!bounded]{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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\end{corollary}
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\begin{proof}
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By lemma 3 % TODO REF
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we have that
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\[(b-a) \bE[U_N([a,b])] \le \bE[ | X_N| ] + |a| \le \sup_n \bE[|X_n|] + |a|.\]
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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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\[
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\bE(U_n([a,b])] \uparrow \bE[U_\infty([a,b])].
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\]
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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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bounded in $L^1(\bP)$.
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Let
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\[
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\Lambda \coloneqq \{\omega | X_n(\omega) \text{ does not converge to anything in $[-\infty,\infty]$}\}.
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\]
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We have
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\begin{IEEEeqnarray*}{rCl}
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\Lambda &=& \{\omega | \liminf_N X_N(\omega) < \limsup_N X_N(\omega)\}\\
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&=& \{\omega | \liminf_N X_N(\omega) < a < b \limsup_N X_N(\omega)\} \\
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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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\begin{claim}
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$\bP[X_\infty \in \{\pm \infty\}] = 0$.
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\end{claim}
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\begin{subproof}
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It suffices to show that $\bE[|X_\infty|] < \infty$.
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We have.
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\begin{IEEEeqnarray*}{rCl}
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\bE[|X_\infty|] &=& \bE[\liminf_{n \to \infty} |X_n|]\\
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&\overset{\text{Fatou}}{\le }& \liminf_n \bE[|X_n|]\\
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&\le & \sup_n \bE[|X_n|]\\
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&<& \infty.
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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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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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\end{claim}
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\begin{subproof}
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We need to show $\sup_n \bE(|X_n|) < \infty$.
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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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\begin{example}[Branching process]
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% TODO
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\end{example}
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@ -40,6 +40,7 @@
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\input{inputs/lecture_14.tex}
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\input{inputs/lecture_14.tex}
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\input{inputs/lecture_15.tex}
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\input{inputs/lecture_15.tex}
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\input{inputs/lecture_16.tex}
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\input{inputs/lecture_16.tex}
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\input{inputs/lecture_17.tex}
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\cleardoublepage
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\cleardoublepage
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