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Josia Pietsch 2023-06-29 01:24:35 +02:00
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inputs/lecture_20.tex
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@ -222,7 +222,7 @@ we need the following theorem, which we won't prove here:
It is also clear that $X^T_n$ is integrable since It is also clear that $X^T_n$ is integrable since
\[ \[
\bE[|X^T_n|] \le \sum_{k=1}^{n} \bE[|X_k|] < \infty. \bE[|X^T_n|] \le \sum_{k=1}^{n} \bE[|X_k|] < \infty.
\] \]
We have We have
\begin{IEEEeqnarray*}{rCl} \begin{IEEEeqnarray*}{rCl}
@ -246,7 +246,7 @@ we need the following theorem, which we won't prove here:
we get from the above that we get from the above that
\[ \[
\bE[X_T] \overset{n \ge M}{=} \bE[X^T_n] \begin{cases} \bE[X_T] \overset{n \ge M}{=} \bE[X^T_n] \begin{cases}
\le \bE[X_0] & \text{ supermartingale}, \le \bE[X_0] & \text{ supermartingale},\\
= \bE[X_0] & \text{ martingale}. = \bE[X_0] & \text{ martingale}.
\end{cases} \end{cases}
\] \]
@ -259,7 +259,7 @@ we need the following theorem, which we won't prove here:
Then $\bP[T < \infty] = 1$, but Then $\bP[T < \infty] = 1$, but
\[ \[
1 = \bE[S_T] \neq \bE[S_0] = 0. 1 = \bE[S_T] \neq \bE[S_0] = 0.
\] \]
\end{example} \end{example}
\begin{theorem}[Optional Stopping] \begin{theorem}[Optional Stopping]
@ -269,7 +269,7 @@ we need the following theorem, which we won't prove here:
taking values in $\N$. taking values in $\N$.
If one of the following holds If one of the following holds
\begin{itemize}[(i)] \begin{enumerate}[(i)]
\item $T \le M$ is bounded, \item $T \le M$ is bounded,
\item $(X_n)_n$ is uniformly bounded \item $(X_n)_n$ is uniformly bounded
and $T < \infty$ a.s., and $T < \infty$ a.s.,
@ -277,7 +277,7 @@ we need the following theorem, which we won't prove here:
and $|X_n(\omega) - X_{n-1}(\omega)| \le K$ and $|X_n(\omega) - X_{n-1}(\omega)| \le K$
for all $n \in \N, \omega \in \Omega$ and for all $n \in \N, \omega \in \Omega$ and
some $K > 0$, some $K > 0$,
\end{itemize} \end{enumerate}
then $\bE[X_T] \le \bE[X_0]$. then $\bE[X_T] \le \bE[X_0]$.
If $(X_n)_n$ even is a martingale, then If $(X_n)_n$ even is a martingale, then