lecture 21

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Josia Pietsch 2023-06-29 20:31:46 +02:00
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5 changed files with 162 additions and 21 deletions

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@ -16,7 +16,7 @@ First, let us recall some basic definitions:
\item $\bP$ is a \vocab{probability measure}, i.e.~$\bP$ is a function $\bP: \cF \to [0,1]$
such that
\begin{itemize}
\item $\bP(\emptyset) = 1$, $\bP(\Omega) = 1$,
\item $\bP(\emptyset) = 0$, $\bP(\Omega) = 1$,
\item $\bP\left( \bigsqcup_{n \in \N} A_n \right) = \sum_{n \in \N} \bP(A_n)$
for mutually disjoint $A_n \in \cF$.
\end{itemize}

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@ -2,7 +2,7 @@
By the tower property (\autoref{cetower})
it is clear that $(\bE[X | \cF_n])_n$
is a martingale.
First step:
Assume that $X$ is bounded.
Then, by \autoref{cejensen}, $|X_n| \le \bE[|X| | \cF_n]$,
@ -84,7 +84,7 @@ we need the following theorem, which we won't prove here:
L^p &\longrightarrow & (L^q)^\ast \\
f &\longmapsto & (g \mapsto \int g f \dif d\bP)
\end{IEEEeqnarray*}
We also have $(L^1)^\ast \cong L^\infty$,
however $ (L^\infty)^\ast \not\cong L^1$.
\end{fact}
@ -95,7 +95,7 @@ we need the following theorem, which we won't prove here:
$(X_{n_k})_k$ such that for all $Y \in L^q$ ($\frac{1}{p} + \frac{1}{q} = 1$ )
\[
\int X_{n_k} Y \dif \bP \to \int XY \dif \bP
\]
\]
(Note that this argument does not work for $p = 1$,
because $(L^\infty)^\ast \not\cong L^1$).
@ -116,14 +116,14 @@ we need the following theorem, which we won't prove here:
\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},
A random variable $T: \Omega \to \N_0 \cup \{\infty\}$ 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}
@ -131,21 +131,21 @@ we need the following theorem, which we won't prove here:
\end{example}
\begin{example}[Hitting times]
For an adapted process $(X_n)_n$
For an adapted process $(X_n)_n$
with values in $\R$ and $A \in \cB(\R)$, the \vocab{hitting time}
\[
T \coloneqq \inf \{n \in \N : X_n \in A\}
T \coloneqq \inf \{n \in \N : X_n \in A\}
\]
is a stopping time,
as
\[
\{T \le n \} = \bigcup_{k=1}^n \{X_k \in A\} \in \cF_n.
\]
\]
However, the last exit time
\[
T \coloneqq \sup \{n \in \N : X_n \in A\}
\]
T \coloneqq \sup \{n \in \N : X_n \in A\}
\]
is not a stopping time.
\end{example}
@ -158,7 +158,7 @@ we need the following theorem, which we won't prove here:
Then
\[
T \coloneqq \inf \{n \in \N : S_n \ge A \lor S_n \le B\}
\]
\]
is a stopping time.
\end{example}
@ -173,11 +173,11 @@ we need the following theorem, which we won't prove here:
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 $T$
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 $T$,
@ -193,22 +193,22 @@ we need the following theorem, which we won't prove here:
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 $X^T$
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}
@ -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
\[
\bE[|X^T_n|] \le \sum_{k=1}^{n} \bE[|X_k|] < \infty.
\]
\]
We have
\begin{IEEEeqnarray*}{rCl}
@ -259,7 +259,7 @@ we need the following theorem, which we won't prove here:
Then $\bP[T < \infty] = 1$, but
\[
1 = \bE[S_T] \neq \bE[S_0] = 0.
\]
\]
\end{example}
\begin{theorem}[Optional Stopping]

136
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@ -0,0 +1,136 @@
\lecture{21}{2023-06-29}{}
% TODO: replace bf
This is the last lecture relevant for the exam.
(Apart from lecture 22 which will be a repetion).
\begin{goal}
We want to see an application of the
optional stopping theorem \ref{optionalstopping}.
\end{goal}
\begin{notation}
Let $E$ be a complete, separable metric space (e.g.~$E = \R$).
Suppose that for all $x \in E$ we have a probability measure
$\bfP(x, \dif y)$ on $E$.
% i.e. $\mu(A) \coloneqq \int_A \bP(x, \dif y)$ is a probability measure.
Such a probability measure is a called
a \vocab{transition probability measure}.
\end{notation}
\begin{examle}
$E =\R$,
\[\bfP(x, \dif y) = \frac{1}{\sqrt{2 \pi} } e^{- \frac{(x-y)^2}{2}} \dif y\]
is a transition probability measure.
\end{examle}
\begin{example}[Simple random walk as a transition probability measure]
$E = \Z$, $\bfP(x, \dif y)$
assigns mass $\frac{1}{2}$ to $y = x+1$ and $y = x -1$.
\end{example}
\begin{definition}
For every bounded, measurable function $f : E \to \R$,
$x \in E$
define
\[
(\bfP f)(x) \coloneqq \int_E f(y) \bfP(x, \dif y).
\]
This $\bfP$ is called a \vocab{transition operator}.
\end{definition}
\begin{fact}
If $f \ge 0$, then $(\bfP f)(\cdot ) \ge 0$.
If $f \equiv 1$, we have $(\bfP f) \equiv 1$.
\end{fact}
\begin{notation}
Let $\bfI$ denote the \vocab{identity operator},
i.e.
\[
(\bfI f)(x) = f(x)
\]
for all $f$.
Then for a transition operator $\bfP$ we write
\[
\bfL \coloneqq \bfI - \bfP.
\]
\end{notation}
\begin{goal}
Take $E = \R$.
Suppose that $A^c \subseteq \R$ is a bounded domain.
Given a bounded function $f$ on $\R$,
we want a function $u$ which is bounded,
such that
$Lu = 0$ on $A^c$ and $u = f$ on $A$.
\end{goal}
We will show that $u(x) = \bE_x[f(X_{T_A})]$
is the unique solution to this problem.
\begin{definition}
Let $(\Omega, \cF, \{\cF_n\}_n, \bP_x)$
be a filtered probability space, where for every $x \in \R$,
$\bP_x$ is a probability measure.
Let $\bE_x$ denote expectation with respect to $\bfP(x, \cdot )$.
Then $(X_n)_{n \ge 0}$ is a \vocab{Markov chain} starting at $x \in \R$
with \vocab[Markov chain!Transition probability]{transition probability}
$\bfP(x, \cdot )$ if
\begin{enumerate}[(i)]
\item $\bP_x[X_0 = x] = 1$,
\item for all bounded, measurable $f: \R \to \R$,
\[\bE_x[f(X_{n+1}) | \cF_n] \overset{\text{a.s.}}{=}%
\bE_{x}[f(X_{n+1}) | X_n] = %
\int f(y) \bfP(X_n, \dif y).\]
\end{enumerate}
(Recall $\cF_n = \sigma(X_1,\ldots, X_n)$.)
\end{definition}
\begin{example}
Suppose $B \in \cB(\R)$ and $f = \One_B$.
Then the first equality of (ii) simplifies to
\[
\bP_x[X_{n+1} \in B | \cF_n] = \bP_x[X_{n+1} \in B | \sigma(X_n)].
\]
\end{example}
\begin{definition}[Conditional probability]
\[
\bP[A | \cG] \coloneqq \bE[\One_A | \cG].
\]
\end{definition}
\begin{example}
Let $\xi_i$ be i.i.d.~with$\bP[\xi_i = 1] = \bP[\xi_i = -1] = \frac{1}{2}$
and define $X_n \coloneqq \sum_{i=1}^{n} \xi_i$.
Intuitively, conditioned on $X_n$, $X_{n+1}$ should
be independent of $\sigma(X_1,\ldots, X_{n-1})$.
For a set $B$, we have
\[
\bP_0[X_{n+1} \in B| \sigma(X_1,\ldots, X_n)]
= \bE[\One_{X_n + \xi_{n+1} \in B} | \sigma(X_1,\ldots, X_n)]
= \bE[\One_{X_n + \xi_{n+1} \in B} | \sigma(X_n)].
\]
\begin{claim}
$\bE[\One_{X_{n+1} \in B} | \sigma(X_1,\ldots, X_n)] = \bE[\One_{X_{n+1} \in B} | \sigma(X_n)]$.
\end{claim}
\begin{subproof}
The rest of the lecture was very chaotic...
\end{subproof}
\end{example}
%TODO
{ \huge\color{red}
New information after this point is not relevant for the exam.
}
Stopping times and optional stopping are very relevant for the exam,
the Markov property is not.

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@ -40,6 +40,10 @@
\RequirePackage{mkessler-faktor}
\RequirePackage{mkessler-mathsymb}
\RequirePackage[extended]{mkessler-mathalias}
% \makeatletter
% \expandafter\MakeAliasesForwith\expandafter\mathbf\expandafter{\expandafter bf\expandafter}\expandafter{\mkessler@mathalias@all}
% \makeatother
\RequirePackage{mkessler-refproof}
% mkessler-mathfont has already been imported

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@ -44,6 +44,7 @@
\input{inputs/lecture_18.tex}
\input{inputs/lecture_19.tex}
\input{inputs/lecture_20.tex}
\input{inputs/lecture_21.tex}
\cleardoublepage