lecture 20
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@ -219,7 +219,8 @@ Let $(\Omega, \cF, \bP)$ as always and let $(\cF_n)_n$ always be a filtration.
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\begin{theorem}
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\label{ceismartingale}
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Let $X \in L^p$ for some $p \ge 1$.
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Let $X \in L^p$ for some $p \ge 1$
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and $\bigcup_n \cF_n \to \cF$.
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Then $X_n \coloneqq \bE[X | \cF_n]$ defines a martingale which converges
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to $X$ in $L^p$.
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\end{theorem}
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@ -22,8 +22,8 @@
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&\overset{A \in \cF_n}{=}& \lim_{\substack{n \to \infty\\n \ge m}} \bE[X \One_A]\\
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\end{IEEEeqnarray*}
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Hence $\int_A Y \dif \bP = \int_A X \dif \bP$ for all $m \in \N, A \in \cF_m$.
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Since $\cF = \sigma\left( \bigcup \cF_n \right)$
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this holds for all $A \in \cF$.
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Since $\sigma(X) = \bigcup \cF_n$
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this holds for all $A \in \sigma(X)$.
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Hence $X = Y$ a.s., so $X_n \xrightarrow{L^2} X$.
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Since $(X_n)_n$ is uniformly bounded, this also means
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$X_n \xrightarrow{L^p} X$.
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@ -111,8 +111,203 @@ we need the following theorem, which we won't prove here:
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Hence $X_n = \bE[X | \cF_m]$ by the uniqueness of conditional expectation
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and by \autoref{ceismartingale},
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we get the convergence.
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\end{refproof}
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\subsection{Stopping times}
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\begin{definition}[Stopping time]
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A random variable $T: \Omega \to \N \cup \{\infty\}$ on a filtered probability space $(\Omega, \cF, \{\cF_n\}_n, \bP)$ is called a \vocab{stopping time},
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if
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\[
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\{T \le n\} \in \cF_n
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\]
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for all $n \in \N$.
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Equivalently, $\{T = n\} \in \cF_n$ for all $n \in \N$.
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\end{definition}
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\begin{example}
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A constant random variable $T = c$ is a stopping time.
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\end{example}
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\begin{example}[Hitting times]
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For an adapted process $(X_n)_n$
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with values in $\R$ and $A \in \cB(\R)$, the \vocab{hitting time}
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\[
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T \coloneqq \inf \{n \in \N : X_n \in A\}
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\]
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is a stopping time,
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as
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\[
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\{T \le n \} = \bigcup_{k=1}^n \{X_k \in A\} \in \cF_n.
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\]
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However, the last exit time
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\[
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T \coloneqq \sup \{n \in \N : X_n \in A\}
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\]
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is not a stopping time.
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\end{example}
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\begin{example}
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Consider the simple random walk, i.e.
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$X_n$ i.i.d.~with $\bP[X_n = 1] = \bP[X_n = -1] = \frac{1}{2}$.
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Set $S_n \coloneqq \sum_{i=1}^{n} X_n$.
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Then
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\[
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T \coloneqq \inf \{n \in \N : S_n \ge A \lor S_n \le B\}
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\]
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is a stopping time.
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\end{example}
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\begin{example}
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If $T_1, T_2$ are stopping times with respect to the same filtration,
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then
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\begin{itemize}
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\item $T_1 + T_2$,
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\item $\min \{T_1, T_2\}$ and
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\item $\max \{T_1, T_2\}$
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\end{itemize}
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are stopping times.
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Note that $T_1 - T_2$ is not a stopping time.
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\end{example}
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\begin{remark}
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There are two ways to interpret the interaction between a stopping time $T$
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and a stochastic process $(X_n)_n$.
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\begin{itemize}
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\item The behaviour of $ X_n$ until $T$,
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i.e.~looking at the \vocab{stopped process}
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\[
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X^T \coloneqq \left(X_{T \wedge n}\right)_{n \in \N}
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\].
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\item The value of $(X_n)_n)$ at time $T$,
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i.e.~looking at $X_T$.
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\end{itemize}
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\end{remark}
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\begin{example}
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If we look at a process
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\[
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S_n = \sum_{i=1}^{n} X_i
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\]
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for some $(X_n)_n$, then
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\[
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S^T = (\sum_{i=1}^{T \wedge n} X_i)_n
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\]
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and
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\[
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S_T = \sum_{i=1}^{T} X_i.
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\]
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\end{example}
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\begin{theorem}
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If $(X_n)_n$ is a supermartingale and $T$ is a stopping time,
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then $X^T$ is also a supermartingale,
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and we have $\bE[X_{T \wedge n}] \le \bE[X_0]$ for all $n$.
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If $(X_n)_n$ is a martingale, then so is $X^T$
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and $\bE[X_{T \wedge n}] \le \bE[X_0]$.
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\end{theorem}
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\begin{proof}
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First, we need to show that $X^T$ is adapted.
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This is clear since
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\begin{IEEEeqnarray*}{rCl}
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X^T_n &=& X_T \One_{T < n} + X_n \One_{T \ge n}\\
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&=& \sum_{k=1}^{n-1} X_k \One_{T = k} + X_n \One_{T \ge n}.
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\end{IEEEeqnarray*}
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It is also clear that $X^T_n$ is integrable since
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\[
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\bE[|X^T_n|] \le \sum_{k=1}^{n} \bE[|X_k|] < \infty.
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\]
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We have
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\begin{IEEEeqnarray*}{rCl}
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\bE[X^T_n - X^T_{n-1} | \cF_{n-1}]
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&=& \bE[X_n \One_{\{T \ge n\}} + \sum_{k=1}^{n-1} X_k \One_{\{ T = k\} } - X_{n-1}(\One_{T \ge n} + \One_{\{T = n-1\}})
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+ \sum_{k=1}^{n-2} X_k \One_{\{T = k\} } | \cF_{n-1}]\\
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&=& \bE[(X_n - X_{n-1}) \One_{\{ T \ge n\} } | \cF_{n-1}]\\
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&=& \One_{\{ T \ge n\}} (\bE[X_n | \cF_{n-1}] - X_{n-1})\\
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&& \begin{cases}
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\le 0\\
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= 0 \text{ if $(X_n)_n$ is a martingale}.
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\end{cases}.
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\end{IEEEeqnarray*}
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\end{proof}
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\begin{remark}
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\label{roptionalstoppingi}
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We now want a similar statement for $X_T$.
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In the case that $T \le M$ is bounded,
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we get from the above that
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\[
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\bE[X_T] \overset{n \ge M}{=} \bE[X^T_n] \begin{cases}
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\le \bE[X_0] & \text{ supermartingale},
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= \bE[X_0] & \text{ martingale}.
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\end{cases}
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\]
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However if $T$ is not bounded, this does not hold in general.
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\end{remark}
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\begin{example}
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Let $(S_n)_n$ be the simple random walk
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and take $T \coloneqq \inf \{n : S_n = 1\}$.
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Then $\bP[T < \infty] = 1$, but
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\[
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1 = \bE[S_T] \neq \bE[S_0] = 0.
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\]
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\end{example}
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\begin{theorem}[Optional Stopping]
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\label{optionalstopping}
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Let $(X_n)_n$ be a supermartingale
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and let $T$ be a stopping time
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taking values in $\N$.
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If one of the following holds
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\begin{itemize}[(i)]
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\item $T \le M$ is bounded,
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\item $(X_n)_n$ is uniformly bounded
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and $T < \infty$ a.s.,
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\item $\bE[T] < \infty$
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and $|X_n(\omega) - X_{n-1}(\omega)| \le K$
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for all $n \in \N, \omega \in \Omega$ and
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some $K > 0$,
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\end{itemize}
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then $\bE[X_T] \le \bE[X_0]$.
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If $(X_n)_n$ even is a martingale, then
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under the same conditions
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$\bE[X_T] = \bE[X_0]$.
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\end{theorem}
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\begin{proof}
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(i) was dealt with in \autoref{roptionalstoppingi}.
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(ii): Since $(X_n)_n$ is bounded, we get that
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\begin{IEEEeqnarray*}{rCl}
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\bE[|X_T - X_0|] &\overset{\text{dominated convergence}}{=}& \lim_{n \to \infty} \bE[|X_{T \wedge n} - X_0|]\\
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&\overset{\text{part (i)}}{\le}& 0.
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\end{IEEEeqnarray*}
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(iii): It is
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\begin{IEEEeqnarray*}{rCl}
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|X_{T \wedge n}- X_0| &\le& | \sum_{k=1}^{T \wedge n} X_k - X_{k-1}|\\
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&\le & (T \wedge n) \cdot K\\
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&\le & T \cdot K < \infty.
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\end{IEEEeqnarray*}
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Hence, we can apply dominated convergence and obtain
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\begin{IEEEeqnarray*}{rCl}
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\bE[X_T - X_0] &=& \lim_{n \to \infty} \bE[X_{T \wedge n} - X_0].
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\end{IEEEeqnarray*}
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Thus, we can apply (ii).
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The statement about martingales follows from
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applying this to $(X_n)_n$ and $(-X_n)_n$,
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which are both supermartingales.
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\end{proof}
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@ -43,6 +43,7 @@
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\input{inputs/lecture_17.tex}
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\input{inputs/lecture_18.tex}
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\input{inputs/lecture_19.tex}
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\input{inputs/lecture_20.tex}
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\cleardoublepage
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