josia-notes/s23-probability-theory
1
0
Fork 0
Code Issues Pull Requests Actions Packages Projects Releases Wiki Activity

some cleanup

Browse Source
This commit is contained in:
Josia Pietsch 2023-06-29 22:18:23 +02:00
parent c39ae924fe
commit 47219b0424
Signed by untrusted user who does not match committer: josia
GPG Key ID: E70B571D66986A2D
12 changed files with 155 additions and 106 deletions
Show Stats Download Patch File Download Diff File Expand all files Collapse all files

6
inputs/lecture_13.tex
View File

@ -46,7 +46,7 @@ We will not prove the \autoref{lindebergclt} or \autoref{lyapunovclt}
in this lecture. However, they are quite important.
We will now sketch the proof of \autoref{levycontinuity},
details can be found in the notes.\todo{Complete this}
details can be found in the notes.\notes
A generalized version of \autoref{levycontinuity} is the following:
\begin{theorem}[A generalized version of Levy's continuity \autoref{levycontinuity}]
\label{genlevycontinuity}
@ -120,7 +120,7 @@ A generalized version of \autoref{levycontinuity} is the following:
We will prove \autoref{levycontinuity} assuming
\autoref{lec10_thm1}.
\autoref{lec10_thm1} will be shown in the notes.\todo{TODO}
\autoref{lec10_thm1} will be shown in the notes.\notes
We will need the following:
\begin{lemma}
\label{lec13_lem1}
@ -239,7 +239,7 @@ We still need to show that $\mu_n \implies \mu$.
Then $a_n \to a$.
\end{fact}
\begin{subproof}
\todo{in the notes}
\notes
\end{subproof}
Assume that $\mu_n$ does not converge to $\mu$.
By \autoref{lec10_thm1}, pick a continuity point $x_0$ of $F$,

7
inputs/lecture_14.tex
View File

@ -87,7 +87,8 @@ We now want to generalize this to arbitrary random variables.
\]
\end{definition}
\paragraph{Plan}
\subsection{Existence of conditional probability}
We will give two different proves of \autoref{conditionalexpectation}.
The first one will use orthogonal projections.
The second will use the Radon-Nikodym theorem.
@ -108,7 +109,7 @@ and then do the harder proof.
\end{enumerate}
\end{lemma}
\begin{proof}
\todo{Notes}
\notes
\end{proof}
\begin{refproof}{conditionalexpectation}
@ -162,7 +163,7 @@ and then do the harder proof.
$0 \overset{\text{a.s.}}{\le} Z_n \uparrow$.
\end{claim}
\begin{subproof}
\todo{Notes}
\notes
\end{subproof}
Define $Z(\omega) \coloneqq \limsup_{n \to \infty} Z_n(\omega)$.

38
inputs/lecture_15.tex
View File

@ -1,4 +1,5 @@
\lecture{15}{2023-06-06}{}
\subsection{Properties of conditional expectation}
We want to derive some properties of conditional expectation.
@ -12,16 +13,20 @@ We want to derive some properties of conditional expectation.
\begin{proof}
Apply (b) from the definition for $G = \Omega \in \cG$.
\end{proof}
\begin{theorem} % Thm 2
\begin{theorem}
\label{ceprop2}
If $X$ is $\cG$-measurable, then $X = \bE[X | \cG]$ a.s..
If $X$ is $\cG$-measurable, then $X \overset{\text{a.s.}}{=}\bE[X | \cG]$.
\end{theorem}
\begin{proof}
Suppose $\bP[X \neq Y] > 0$.
Without loss of generality $\bP[X > Y] > 0$.
Hence $\bP[ X > Y + \frac{1}{n}]> 0$ for some $n \in \N$.
Let $A \coloneqq \{X > Y + \frac{1}{n}\}$.
% TODO
Then
\[
\int_A X \dif \bP \ge \frac{1}{n}\bP(A) + \int_A Y \dif \bP,
\]
contradicting property (b) from \autoref{conditionalexpectation}.
\end{proof}
\begin{example}
@ -40,7 +45,7 @@ We want to derive some properties of conditional expectation.
\]
\end{theorem}
\begin{proof}
Trivial % TODO
trivial\todo{add details}
\end{proof}
\begin{theorem}[Positivity]
@ -52,7 +57,7 @@ We want to derive some properties of conditional expectation.
\begin{proof}
Let $W $ be a version of $\bE[X | \cG]$.
Suppose $\bP[ W < 0] > 0$.
Then $G \coloneqq \{W < -\frac{1}{n}\} \in \cG$
Then \[G \coloneqq \{W < -\frac{1}{n}\} \in \cG.\]
For some $n \in \N$, we have $\bP[G] > 0$.
However it follows that
\[
@ -106,7 +111,7 @@ We want to derive some properties of conditional expectation.
\]
\end{theorem}
\begin{proof}
\todo{in the notes}
\notes
\end{proof}
\begin{theorem}[Conditional dominated convergence theorem]
\label{ceprop7}
@ -118,14 +123,14 @@ We want to derive some properties of conditional expectation.
\end{theorem}
\begin{proof}
\todo{in the notes}
\notes
\end{proof}
Recall
\begin{theorem}[Jensen's inequality]
\begin{fact}[Jensen's inequality]
If $c : \R \to \R$ is convex and $\bE[|c \circ X|] < \infty$,
then $\bE[c \circ X] \ge c(\bE[X])$.
\end{theorem}
then $\bE[c \circ X] \overset{\text{a.s.}}{\ge} c(\bE[X])$.
\end{fact}
For conditional expectation, we have
\begin{theorem}[Conditional Jensen's inequality]
@ -150,7 +155,7 @@ For conditional expectation, we have
Hence
\[
\bE[c(X) | \cG] \ge a_n \bE[X | \cG] + \bE[b_n | \cG]
= a_n \bE[X | \cG] + b_n \text{a.s.}
= a_n \bE[X | \cG] + b_n \text{ a.s.}
\]
for all $n$.
Using that a countable union of sets o f measure zero has measure zero,
@ -162,14 +167,14 @@ For conditional expectation, we have
\end{refproof}
Recall
\begin{theorem}[Hölder's inequality]
\begin{fact}[Hölder's inequality]
Let $p,q \ge 1$ such that $\frac{1}{p} + \frac{1}{q} = 1$.
Suppose $X \in L^p(\bP)$ and $Y \in L^q(\bP)$.
Then
\[
\bE(X Y) \le \underbrace{\bE(|X|^p)^{\frac{1}{p}}}_{\text{\reflectbox{$\coloneqq$}} \|X\|_{L^p}} \bE(|Y|^q)^{\frac{1}{q}}.
\]
\end{theorem}
\end{fact}
\begin{theorem}[Conditional Hölder's inequality]
\label{ceprop9}
@ -193,7 +198,7 @@ Recall
Suppose $\cF \supset \cG \supset \cH$ are sub-$\sigma$-algebras.
Then
\[
\bE\left[\bE[X | \cG] \mid \cH\right] = \bE[X | \cH].
\bE\left[\bE[X | \cG] \mid \cH\right] \overset{\text{a.s.}}{=} \bE[X | \cH].
\]
\end{theorem}
\begin{proof}
@ -219,7 +224,8 @@ Assume $Y = \One_B$, then $Y$ simple, then take the limit (using that $Y$ is bou
\begin{definition}
Let $\cG$ and $\cH$ be $\sigma$-algebras.
We call $\cG$ and $\cH$ \vocab[$\sigma$-algebra!independent]{independent},
\todo{TODO}
if $\bP(G \cap H) = \bP(G) \bP(H)$
for all events $G \in \cG$, $H \in \cH$.
\end{definition}
\begin{theorem}[Role of independence]
@ -251,7 +257,7 @@ Assume $Y = \One_B$, then $Y$ simple, then take the limit (using that $Y$ is bou
For $\bE[S_{n+1} | \cF_n]$ we obtain
\begin{IEEEeqnarray*}{rCl}
\bE[S_{n+1} | \cF_n] &\overset{\autoref{celinearity}}{=}&
\bE[S_{n+1} | \cF_n] &\overset{\text{\autoref{celinearity}}}{=}&
\bE[S_n | \cF_n] + \bE[X_{n+1} | \cF_n]\\
&\overset{\text{a.s.}}{=}& S_n + \bE[X_{n+1} | \cF_n]\\
&\overset{\text{\autoref{ceprop12}}}{=}& S_{n} + \bE[X_n]\\

35
inputs/lecture_16.tex
View File

@ -68,7 +68,8 @@ The Radon Nikodym theorem is the converse of that:
Such a $Z$ is called the \vocab{Radon-Nikodym derivative}.
\end{theorem}
\begin{definition}
Whenever the property $\forall A \in \cF, \mu(A)= 0 \implies \nu(A) = 0$,
Whenever the property $\forall A \in \cF, \mu(A) = 0 \implies \nu(A) = 0$
holds for two measures $\mu$ and $\nu$,
we say that $\nu$ is \vocab{absolutely continuous}
w.r.t.~$\mu$.
This is written as $\nu \ll \mu$.
@ -86,16 +87,19 @@ of conditional expectation:
\begin{refproof}{radonnikodym}
We will only sketch the proof. A full proof can be found in the notes.
We will only sketch the proof.
A full proof can be found in the official notes.
\paragraph{Step 1: Uniqueness}
See notes.
\paragraph{Step 1: Uniqueness} \notes
\paragraph{Step 2: Reduction to the finite measure case}
See notes.
\notes
\paragraph{Step 3: Getting hold of $Z$}
Assume now that $\mu$ and $\nu$ are two finite measures.
Let
\[\cC \coloneqq \{f: \Omega \to [0,\infty] | \forall A \in \cF.~\int_A f \dif \mu \le \nu(A)\}.\]
\[
\cC \coloneqq \left\{f: \Omega \to [0,\infty] \middle| %
\forall A \in \cF.~\int_A f \dif \mu \le \nu(A)\right\}.
\]
We have $\cC \neq \emptyset$ since $0 \in \cC$.
The goal is to find a maximal function $Z$ in $\cC$.
@ -145,7 +149,7 @@ of conditional expectation:
but it need not satisfy (ii).
The tricky part is to make $A$ smaller such that it also
satisfies (ii).\todo{Copy from notes}
satisfies (ii).\notes
\end{subproof}
\end{refproof}
@ -153,6 +157,8 @@ of conditional expectation:
\section{Martingales}
\subsection{Definition}
We have already worked with martingales, but we will define them
rigorously now.
@ -168,15 +174,16 @@ Typically $\cF_n = \sigma(X_1, \ldots, X_n)$ for a sequence of random variables.
Let $(\cF_n)$ be a filtration and
$X_1,\ldots,X_n$ be random variables such that $X_i \in L^1(\bP)$.
Then we say that $(X_n)_{n \ge 1}$ is an $(\cF_n)_n$-\vocab{martingale}
if
if the following hold:
\begin{itemize}
\item $X_n$ is $\cF_n$-measurable for all $n$
($X_n$ is \vocab[Sequence!adapted to a filtration]{adapted to the filtration} $\cF_n$ ).
\item $X_n$ is $\cF_n$-measurable for all $n$.
($X_n$ is \vocab[Sequence!adapted to a filtration]{adapted to the filtration} $\cF_n$ ).
\item $\bE[X_{n+1} | \cF_n] \overset{\text{a.s.}}{=} X_n$
for all $n$.
\end{itemize}
$(X_n)$ is called a \vocab{sub-martingale},
$(X_n)_n$ is called a \vocab{sub-martingale},
if it is adapted to $\cF_n$ but
\[
\bE[X_{n+1} | \cF_n] \overset{\text{a.s.}}{\ge} X_n.
@ -185,12 +192,12 @@ Typically $\cF_n = \sigma(X_1, \ldots, X_n)$ for a sequence of random variables.
if it is adapted but $\bE[X_{n+1} | \cF_n] \overset{\text{a.s.}}{\le} X_n$.
\end{definition}
\begin{corollary}
Suppose that $f: \R \to \R$ is a convex function such that $f(xn) \in L^1(\bP)$.
Suppose that $(X_n)_n$ is a martingal\footnote{In this form it means, that there is some filtration, that we don't explicitly specify}.
Suppose that $f: \R \to \R$ is a convex function such that $f(X_n) \in L^1(\bP)$.
Suppose that $(X_n)_n$ is a martingale\footnote{In this form it means, that there is some filtration, that we don't explicitly specify}.
Then $(f(X_n))_n$ is a sub-martingale.
\end{corollary}
\begin{proof}
Apply \autoref{cejensensinequality}.
Apply \autoref{cjensen}.
\end{proof}
\begin{corollary}

20
inputs/lecture_17.tex
View File

@ -1,5 +1,8 @@
\lecture{17}{2023-06-15}{}
\subsection{Doob's martingale convergence theorem}
\begin{definition}[Stochastic process]
A \vocab{stochastic process} is a collection of random
variables $(X_t)_{t \in T}$ for some index set $T$.
@ -64,6 +67,7 @@ such that $X_{s_j}(\omega) < a$ and $X_{t_j}(\omega) > b$ for all $1 \le j \le
Clearly $U_N^X([a,b]) \uparrow$ as $N$ increases.
It follows that the monotonic limit $U_\infty([a,b]) \coloneqq \lim_{N \to \infty} U_N([a,b])$ exists pointwise.
\begin{lemma} % Lemma 1
\label{lec17l1}
\[
\{\omega | \liminf_{N \to \infty} Z_N(\omega) < a < b <
\limsup_{N \to \infty} Z_N(\omega)\} \subseteq
@ -73,8 +77,9 @@ It follows that the monotonic limit $U_\infty([a,b]) \coloneqq \lim_{N \to \inf
\end{lemma}
\begin{lemma} % 2
\label{lec17l2}
Let $Y_n(\omega) \coloneqq \sum_{j=1}^n C_j(X_j - X_{j-1})$.
Then $Y_N \ge (b -a) U_N([a,b]) - (X_N - a)^{-}$.
Then \[Y_N \ge (b -a) U_N([a,b]) - (X_N - a)^{-}.\]
\end{lemma}
\begin{proof}
Every upcrossing of $[a,b]$ increases the value of $Y$ by $(b-a)$,
@ -82,21 +87,24 @@ It follows that the monotonic limit $U_\infty([a,b]) \coloneqq \lim_{N \to \inf
\end{proof}
\begin{lemma} %3
\label{lec17l3}
Suppose $(X_n)_n$ is a supermartingale.
Then in the above setup, $(b-a) \bE[U_N([a,b])] \le \bE[(X_n - a)^-]$.
Then in the above setup
\[(b-a) \bE[U_N([a,b])] \le \bE[(X_n - a)^-].\]
\end{lemma}
\begin{proof}
Obvious from lemma 2 % TODO REF
This is obvious from \autoref{lec17l2}
and the supermartingale property.
\end{proof}
\begin{corollary}
Let $(X_n)_n$ be a \vocab[Supermartingale!bounded]{supermartingale bounded in $L^1(\bP)$ },
i.e.~$\sup_n \bE[|X_n| ] < \infty$.
Let $(X_n)_n$ be a
\vocab[Supermartingale!bounded]{supermartingale bounded in $L^1(\bP)$},
i.e.~$\sup_n \bE[|X_n|] < \infty$.
Then $(b-a) \bE(U_\infty) \le |a| + \sup_n \bE(|X_n|)$.
In particular, $\bP[U_\infty = \infty] = 0$.
\end{corollary}
\begin{proof}
By lemma 3 % TODO REF
By \autoref{lec17l3}
we have that
\[(b-a) \bE[U_N([a,b])] \le \bE[ | X_N| ] + |a| \le \sup_n \bE[|X_n|] + |a|.\]
Since $U_N(\cdot) \ge 0$ and $U_N(\cdot ) \uparrow U_\infty(\cdot )$,

62
inputs/lecture_18.tex
View File

@ -1,10 +1,9 @@
\lecture{18}{2023-06-20}{}
Recall our key lemma for supermartingales from last time:
Recall our key lemma \ref{lec17l3} for supermartingales from last time:
\[
(b-a) \bE[U_N([a,b])] \le \bE[(X_n - a)^-].
\]
% TODO Ref
What happens for submartingales?
If $(X_n)_{n \in \N}$ is a submartingale, then $(-X_n)_{n \in \N}$ is a supermartingale.
@ -14,11 +13,14 @@ Hence the same holds for submartingales, i.e.
a.s.~to a finite limit, which is a.s.~finite.
\end{lemma}
\subsection{Doob's $L^p$ inequality}
\begin{question}
What about $L^p$ convergence of martingales?
\end{question}
\begin{example}[Branching process]
\begin{example}[\vocab{Branching process}]
Let $ (Z_n)_{n \ge 1}$ be i.i.d.~$\pm 1$ with
$\bP[Z_n = 1] = p \in (0,1)$.
@ -26,11 +28,11 @@ Hence the same holds for submartingales, i.e.
Define $X_{n+1} \coloneqq u^{Z_{n+1}} X_{n}$.
\paragraph{Exercise}
Given $u \ge 0$, find $p = p(u)$
such that $(X_n)_n$ is a martingale w.r.t.~the canonical filtration.
% TODO
\begin{exercise}
Given $u \ge 0$, find $p = p(u)$
such that $(X_n)_n$ is a martingale w.r.t.~the canonical filtration.
\end{exercise}
\todo{TODO}
By \autoref{doobmartingaleconvergence}, there is an
@ -51,10 +53,11 @@ Hence the same holds for submartingales, i.e.
$N_0(\epsilon)$ (possibly random)
such that for all $n > N_0(\epsilon)$
\[
\left( \frac{X_n}{x} \right)^{\frac{1}{n}} \le u^{2p - 1}(1 + \epsilon) \implies x [\underbrace{u^{2p - 1} (1+\epsilon)}_{<1}]^n \xrightarrow{n \to \infty} 0.
\left( \frac{X_n}{x} \right)^{\frac{1}{n}} \le u^{2p - 1}(1 + \epsilon) %
\implies x [\underbrace{u^{2p - 1} (1+\epsilon)}_{<1}]^n \xrightarrow{n \to \infty} 0.
\]
Hence it can not converge in $L^1$.
% TODO Confusion
Thus it can not converge in $L^1$.
% TODO Make this less confusing
\end{example}
@ -75,16 +78,21 @@ consider $L^2$.
Since $\bE[X_n | \cF_{n-1}] = X_{n-1}$ a.s.,
by induction $\bE[X_n | \cF_{k}] = X_k$ a.s.~for all $k \le n$.
Play with conditional expectation.
% TODO Exercise
\todo{Exercise}
\end{proof}
\begin{fact}[Parallelogram identity]
% TODO
\begin{fact}[\vocab{Parallelogram identity}]
Let $X, Y \in L^2$.
Then
\[
2 \bE[X^2] + 2 \bE[Y^2] = \bE[(X+Y)^2] + \bE[(X-Y)^2].
\]
\end{fact}
\begin{theorem}
\begin{theorem}\label{martingaleconvergencel2}
Suppose that $(X_n)_n$ is a martingale bounded in
$L^2$, i.e.~$\sup_n \bE[X_n^2] < \infty$.
$L^2$,\\
i.e.~$\sup_n \bE[X_n^2] < \infty$.
Then there is a random variable $X_\infty$ such that
\[
X_n \xrightarrow{L^2} X_\infty.
@ -100,7 +108,7 @@ consider $L^2$.
\bE[X_n^2] = \bE[X_0^2] + \sum_{j=1}^{n} \bE[Y_j^2]
\]
by \autoref{martingaleincrementsorthogonal}
(this is known as the \vocab{parallelogram identity}). % TODO Move
% (this is known as the \vocab{parallelogram identity}). % TODO how exactly is this used here?
In particular,
\[
\sup_n \bE[X_n^2] < \infty \iff \sum_{j=1}^{\infty} \bE[Y_j^2] < \infty.
@ -141,7 +149,7 @@ First, we need a very important inequality:
\end{enumerate}
\end{theorem}
We first need
In order to prove \autoref{dooblp}, we first need
\begin{lemma}
\label{dooplplemma}
Let $p > 1$ and $X,Y$ non-negative random variable
@ -176,7 +184,7 @@ We first need
Suppose now $Y \not\in L^p$.
Then look at $Y_M = Y \wedge M$.
Apply the case of $Y \in L^p$ and use the monotone convergence theorem.
Apply the above to $Y_M \in L^p$ and use the monotone convergence theorem.
\end{proof}
\begin{refproof}{dooblp}
@ -185,9 +193,13 @@ We first need
\[
E_j = \{|X_1| \le \ell, |X_2| \le \ell, \ldots, |X_{j-1}| \le \ell, |X_j| \ge \ell\}.
\]
Then $\bP[E_j] \overset{\text{Markov}}{\le } \frac{1}{\ell} \int_{E_j} |X_j| \dif \bP (\ast\ast)$.
Then
\begin{equation}
\bP[E_j] \overset{\text{Markov}}{\le } \frac{1}{\ell} \int_{E_j} |X_j| \dif \bP
\label{lec18eq2star}
\end{equation}
Since $(X_n)_n$ is a sub-martingale, $(|X_n|)_n$ is also a sub-martingale
(by \autoref{jensen}).
(by \autoref{cjensen}).
Hence
\begin{IEEEeqnarray*}{rCl}
\bE[\One_{E_j}(|X_n| - |X_{j}|) | \cF_j]
@ -196,14 +208,14 @@ We first need
\end{IEEEeqnarray*}
By the law of total expectation, \autoref{totalexpectation},
it follows that
\[
\bE[\One_{E_j} (|X_n| - |X_j|)] \ge 0 (\ast\ast\ast).
\]
\begin{equation}
\bE[\One_{E_j} (|X_n| - |X_j|)] \ge 0. \label{lec18eq3star}
\end{equation}
Now
\begin{IEEEeqnarray*}{rCl}
\bP(E) &=& \sum_{j=1}^n \bP(E_j)\\
&\overset{(\ast\ast) (\ast\ast\ast)}{\le }& \frac{1}{\ell} \left( \int_{E_1} |X_n| \dif \bP + \ldots + \int_{E_n} |X_n| \dif \bP \right)\\
&\overset{\eqref{lec18eq2star}, \eqref{lec18eq3star}}{\le }& \frac{1}{\ell} \left( \int_{E_1} |X_n| \dif \bP + \ldots + \int_{E_n} |X_n| \dif \bP \right)\\
&=& \frac{1}{\ell} \int_E |X_n| \dif \bP
\end{IEEEeqnarray*}

18
inputs/lecture_19.tex
View File

@ -54,7 +54,7 @@ However, some subsets can be easily described, e.g.
where we have applied Markov's inequality. % TODO REF
Since $\sup_n \bE[|X_n|^{1+\delta}] < \infty$,
we have that $\sup_n \bE[|X_n|] < \infty$ by Jensen. % TODO REF
we have that $\sup_n \bE[|X_n|] < \infty$ by Jensen (\autoref{cjensen}).
Hence, choose $k$ large enough to make the relevant
term less than $\epsilon$.
\end{proof}
@ -94,7 +94,7 @@ However, some subsets can be easily described, e.g.
&\ge & \epsilon
\end{IEEEeqnarray*}
where the assumption that $X$ is in $L^1$ was used to apply
the reverse of Fatou's lemma.
the reverse of Fatou's lemma. % TODO reverse fatou
This yields a contradiction since $\bP(F) = 0$.
\item We want to apply part (a) to $F = \{ |X| > k\}$.
@ -120,7 +120,10 @@ However, some subsets can be easily described, e.g.
\begin{proof}
Fix $\epsilon > 0$.
Choose $\delta > 0$ such that
\[\forall F \in \cF.~ \bP(F) < \delta \implies \bE[|X| \One_F] <\epsilon. (\ast)\]
\begin{equation}
\forall F \in \cF.~ \bP(F) < \delta \implies \bE[|X| \One_F] <\epsilon.
\label{lec19eqstar}
\end{equation}
Let $Y = \bE[X | \cG]$ for some sub-$\sigma$-algebra $\cG$.
Then, by \autoref{condjensen}, $|Y| \le \bE[ |X| | \cG]$.
Hence $\bE[|Y|] \le \bE[|X|]$.
@ -132,7 +135,7 @@ However, some subsets can be easily described, e.g.
\begin{IEEEeqnarray*}{rCl}
\bE[|Y| \One_{\{|Y| > k\} }] &<& \epsilon
\end{IEEEeqnarray*}
by $(\ast)$, since $\bP[|Y| > k] < \delta$.
by \eqref{lec19eqstar}, since $\bP[|Y| > k] < \delta$.
\end{proof}
\begin{theorem}
@ -155,7 +158,7 @@ However, some subsets can be easily described, e.g.
\end{cases}
\end{IEEEeqnarray*}
$\phi$ is $1$-Lipshitz. % TODO
$\phi$ is $1$-Lipschitz. % TODO
We have
\begin{IEEEeqnarray*}{rCl}
@ -168,7 +171,7 @@ However, some subsets can be easily described, e.g.
\autoref{lec19f4} part (b).
Similarly $\int_{|X| > k} |X - \phi(X)| \dif \bP < \epsilon$.
Since $\phi$ is Lipshitz,
Since $\phi$ is Lipschitz,
$ X_n \xrightarrow{\bP} X \implies \phi(X_n) \xrightarrow{\bP} \phi(X)$.
By the bounded convergence theorem % TODO
$|\phi(X_n)| \le k \implies \int | \phi(X_n) - \phi(X)| \dif \bP \to 0$.
@ -232,6 +235,3 @@ Let $(\Omega, \cF, \bP)$ as always and let $(\cF_n)_n$ always be a filtration.
Then there exists a random variable $X \in L^p$, such that
$X_n = \bE[X | \cF_n]$ for all $n$.
\end{theorem}

54
inputs/lecture_20.tex
View File

@ -1,3 +1,4 @@
\lecture{20}{2023-06-27}{}
\begin{refproof}{ceismartingale}
By the tower property (\autoref{cetower})
it is clear that $(\bE[X | \cF_n])_n$
@ -5,10 +6,11 @@
First step:
Assume that $X$ is bounded.
Then, by \autoref{cejensen}, $|X_n| \le \bE[|X| | \cF_n]$,
Then, by \autoref{cjensen}, $|X_n| \le \bE[|X| | \cF_n]$,
hence $\sup_{\substack{n \in \N \\ \omega \in \Omega}} | X_n(\omega)| < \infty$.
Thus $(X_n)_n$ is a martingale in $L^{\infty} \subseteq L^2$.
By the convergence theorem for martingales in $L^2$ % TODO REF
By the convergence theorem for martingales in $L^2$
(\autoref{martingaleconvergencel2})
there exists a random variable $Y$,
such that $X_n \xrightarrow{L^2} Y$.
@ -42,28 +44,31 @@
\begin{IEEEeqnarray*}{rCl}
\int | X - X'|^p \dif \bP &=& \int_{\{|X| > M\} } |X|^p \dif \bP \xrightarrow{M \to \infty} 0
\end{IEEEeqnarray*}
as $\bP$ is \vocab{regular}, \todo{Definition?}
i.e.~$\forall \epsilon > 0 \exists k . \bP[|X|^p \in [-k,k] \ge 1-\epsilon$.
as $\bP$ is \vocab[Measure!regular]{regular}, \todo{Make this a definition?}
i.e.~$\forall \epsilon > 0 . ~\exists k . ~\bP[|X|^p \in [-k,k] \ge 1-\epsilon$.
% Take some $\epsilon > 0$ and $M$ large enough such that
% \[
% \int |X - X'| \dif \bP < \epsilon.
% \]
Take some $\epsilon > 0$ and $M$ large enough such that
\[
\int |X - X'| \dif \bP < \epsilon.
\]
% Let $(X_n')_n$ be the martingale given by $(\bE[X' | \cF_n])_n$.
% Then $X_n' \xrightarrow{L^p} X'$ by the first step.
Let $(X_n')_n$ be the martingale given by $(\bE[X' | \cF_n])_n$.
Then $X_n' \xrightarrow{L^p} X'$ by the first step.
% It is
% \begin{IEEEeqnarray*}{rCl}
% \|X_n - X_n'\|_{L^p}^p &=& \bE[\bE[X - X' | \cF_n]^{p}]\\
% &\overset{\text{Jensen}}{\le}& \bE[\bE[(X- X')^p | \cF_n]\\
% &=& \|X - X'\|_{L^p}^p\\
% &<& \epsilon.
% \end{IEEEeqnarray*}
It is
\begin{IEEEeqnarray*}{rCl}
\|X_n - X_n'\|_{L^p}^p
&=& \bE[\bE[X - X' | \cF_n]^{p}]\\
&\overset{\text{Jensen}}{\le}& \bE[\bE[(X - X')^p | \cF_n]]\\
&=& \|X - X'\|_{L^p}^p\\
&<& \epsilon.
\end{IEEEeqnarray*}
Hence
\[
\|X_n - X\|_{L^p} \le |X_n - X_n'|_{L^p} + |X_n' - X'|_{L^p} + | X - X'|_{L^p} \le 3 \epsilon.
\|X_n - X\|_{L^p} %
\le \|X_n - X_n'\|_{L^p} + \|X_n' - X'\|_{L^p} + \|X - X'\|_{L^p} %
\le 3 \epsilon.
\]
Thus $X_n \xrightarrow{L^p} X$.
\end{refproof}
@ -226,15 +231,16 @@ we need the following theorem, which we won't prove here:
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^T_n - X^T_{n-1} | \cF_{n-1}]\\
&=& \bE\left[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\}})\right.\\
&&\left.+ \sum_{k=1}^{n-2} X_k \One_{\{T = k\} } \middle| \cF_{n-1}\right]\\
&=& \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}
&=& \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{cases}
\end{IEEEeqnarray*}
\end{proof}

17
inputs/lecture_21.tex
View File

@ -1,4 +1,5 @@
\lecture{21}{2023-06-29}{}
\subsection{An Application of the Optional Stopping Theorem}
This is the last lecture relevant for the exam.
(Apart from lecture 22 which will be a repetion).
@ -60,7 +61,7 @@ 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 $A$.
$\mathbf{L}u = 0$ on $A^c$ and $u = f$ on $A$.
\end{goal}
We will show that $u(x) = \bE_x[f(X_{T_A})]$
@ -98,12 +99,18 @@ is the unique solution to this problem.
Intuitively, conditioned on $X_n$, $X_{n+1}$ should
be independent of $\sigma(X_1,\ldots, X_{n-1})$.
\begin{claim}
\begin{claim*}
For a set $B$, we have
$\bE[\One_{X_{n+1} \in B} | \sigma(X_1,\ldots, X_n)] = \bE[\One_{X_{n+1} \in B} | \sigma(X_n)]$.
\end{claim}
\[
\bE[\One_{X_{n+1} \in B} | \sigma(X_1,\ldots, X_n)] %
= \bE[\One_{X_{n+1} \in B} | \sigma(X_n)].\]
\end{claim*}
\begin{subproof}
The rest of the lecture was very chaotic...
\todo{TODO}
% We have $\sigma(\One_{X_{n+1} \in B}) \subseteq \sigma(X_{n}, \xi_{n+1})$.
% $\sigma(X_1,\ldots,X_{n-1})$
% is independent of $\sigma( \sigma(\One_{X_{n+1} \in B}), X_n)$.
% Hence the claim follows from \autoref{ceroleofindependence}.
\end{subproof}
\end{example}

2
inputs/lecture_3.tex
View File

@ -1,5 +1,5 @@
\lecture{3}{}{}
\todo{Lecture 3 needs to be finished}
\todo{My battery died during this lecture, so this still needs to be finished}
\begin{notation}
Let $\cB_n$ denote $\cB(\R^n)$.
\end{notation}

1
inputs/prerequisites.tex
View File

@ -121,6 +121,7 @@
How do we prove that something happens almost surely?
The first thing that should come to mind is:
\begin{lemma}[Borel-Cantelli]
\label{borelcantelli}
If we have a sequence of events $(A_n)_{n \ge 1}$
such that $\sum_{n \ge 1} \bP(A_n) < \infty$,
then $\bP[ A_n \text{for infinitely many $n$}] = 0$

1
wtheo.sty
View File

@ -104,3 +104,4 @@
\newcommand*\dif{\mathop{}\!\mathrm{d}}
\newcommand\lecture[3]{\hrule{\color{darkgray}\hfill{\tiny[Lecture #1, #2]}}}
\newcommand\notes{\todo{TODO: copy from official notes}}