lecture 20

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Josia Pietsch 2023-06-27 18:08:38 +02:00
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@ -219,7 +219,8 @@ Let $(\Omega, \cF, \bP)$ as always and let $(\cF_n)_n$ always be a filtration.
\begin{theorem}
\label{ceismartingale}
Let $X \in L^p$ for some $p \ge 1$.
Let $X \in L^p$ for some $p \ge 1$
and $\bigcup_n \cF_n \to \cF$.
Then $X_n \coloneqq \bE[X | \cF_n]$ defines a martingale which converges
to $X$ in $L^p$.
\end{theorem}

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

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@ -43,6 +43,7 @@
\input{inputs/lecture_17.tex}
\input{inputs/lecture_18.tex}
\input{inputs/lecture_19.tex}
\input{inputs/lecture_20.tex}
\cleardoublepage