josia-notes/s23-probability-theory
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fixed bugs

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Josia Pietsch 2023-06-29 01:24:35 +02:00
parent 62fc7f892d
commit 888b670668
Signed by untrusted user who does not match committer: josia
GPG key ID: E70B571D66986A2D
1 changed files with 5 additions and 5 deletions
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inputs/lecture_20.tex
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@ -246,7 +246,7 @@ we need the following theorem, which we won't prove here:
we get from the above that
\[
\bE[X_T] \overset{n \ge M}{=} \bE[X^T_n] \begin{cases}
\le \bE[X_0] & \text{ supermartingale},
\le \bE[X_0] & \text{ supermartingale},\\
= \bE[X_0] & \text{ martingale}.
\end{cases}
\]
@ -269,7 +269,7 @@ we need the following theorem, which we won't prove here:
taking values in $\N$.
If one of the following holds
\begin{itemize}[(i)]
\begin{enumerate}[(i)]
\item $T \le M$ is bounded,
\item $(X_n)_n$ is uniformly bounded
and $T < \infty$ a.s.,
@ -277,7 +277,7 @@ we need the following theorem, which we won't prove here:
and $|X_n(\omega) - X_{n-1}(\omega)| \le K$
for all $n \in \N, \omega \in \Omega$ and
some $K > 0$,
\end{itemize}
\end{enumerate}
then $\bE[X_T] \le \bE[X_0]$.
If $(X_n)_n$ even is a martingale, then