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Convergence

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4
inputs/intro.tex
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@ -22,3 +22,7 @@ please send me a message:\\
\item Martingales,
\item Markov chains.
\end{enumerate}
This notes follow the way the material was presented in the lecture rather
closely. Additions (e.g.~from exercise sheets)
and slight modifications have been marked with $\dagger$.

212
inputs/prerequisites.tex
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@ -2,14 +2,14 @@ This section provides a short recap of things that should be known
from the lecture on stochastic.
\subsection{Notions of Convergence}
\begin{definition}
\begin{definition}+
\label{def:convergence}
Fix a probability space $(\Omega,\cF,\bP)$.
Let $X, X_1, X_2,\ldots$ be random variables.
\begin{itemize}
\item We say that $X_n$ converges to $X$
\vocab[Convergence!almost surely]{almost surely}
($X_n \xrightarrow{a.s.} X$)
($X_n \xrightarrow{\text{a.s.}} X$)
iff
\[
\bP(\{\omega | X_n(\omega) \to X(\omega)\}) = 1.
@ -29,34 +29,58 @@ from the lecture on stochastic.
\[
\bE[|X_n - X|^p] \xrightarrow{n \to \infty} 0.
\]
\item We say that $X_n$ converges to $X$
\vocab[Convergence!in distribution]{in distribution}%
\footnote{
This notion of convergence was actually
defined during the course of the lecture,
but has been added here for completeness;
see \autoref{def:weakconvergence}.
}
($X_n \xrightarrow{\text{dist}} X$)
iff for every continuous, bounded $f: \R \to \R$
\[
\bE[f(X_n)] \xrightarrow{n \to \infty} \bE[f(X)].
\]
\end{itemize}
\end{definition}
% TODO Connect to AnaIII
\begin{theorem}
\begin{theorem}+
\label{thm:convergenceimplications}
\vspace{10pt}
Let $X$ be a random variable and $X_n, n \in \N$ a sequence of random variables.
Let $1 \le p < q < \infty$.
Then
\begin{figure}[H]
\centering
\begin{tikzpicture}
\node at (0,1.5) (as) { $X_n \xrightarrow{a.s.} X$};
\node at (0,1.5) (as) { $X_n \xrightarrow{\text{a.s.}} X$};
\node at (1.5,0) (p) { $X_n \xrightarrow{\bP} X$};
\node at (3,1.5) (L1) { $X_n \xrightarrow{L^1} X$};
\node at (1.5,-1.5) (w) { $X_n \xrightarrow{\text{dist}} X$};
%\node at (3,1.5) (L1) { $X_n \xrightarrow{L^1} X$};
\node at (3,1.5) (Lp) { $X_n \xrightarrow{L^p} X$};
\node at (3,3) (Lq) { $X_n \xrightarrow{L^q} X$};
\draw[double equal sign distance, -implies] (as) -- (p);
\draw[double equal sign distance, -implies] (L1) -- (p);
\draw[double equal sign distance, -implies] (p) -- (w);
% \draw[double equal sign distance, -implies] (L1) -- (p);
% \draw[double equal sign distance, -implies] (Lp) -- (L1);
\draw[double equal sign distance, -implies] (Lp) -- (p);
\draw[double equal sign distance, -implies] (Lq) -- (Lp);
\end{tikzpicture}
\end{figure}
and none of the other implications hold.
and none of the other implications hold (apart from the transitive closure).
\end{theorem}
\begin{proof}
\begin{refproof}{thm:convergenceimplications}
\begin{claim}
$X_n \xrightarrow{a.s.} X \implies X_n \xrightarrow{\bP} X$.
$X_n \xrightarrow{\text{a.s.}} X \implies X_n \xrightarrow{\bP} X$.
\end{claim}
\begin{subproof}
$\Omega_0 \coloneqq \{\omega \in \Omega : \lim_{n\to \infty} X_n(\omega) = X(\Omega)\} $.
Let $\epsilon > 0$ and consider $A_n \coloneqq \bigcup_{m \ge n} \{\omega \in \Omega: |X_m(\omega) - X(\Omega)| > \epsilon\}$.
$\Omega_0 \coloneqq
\{\omega \in \Omega : \lim_{n\to \infty} X_n(\omega) = X(\omega)\}$.
Let $\epsilon > 0$ and consider
$A_n \coloneqq \bigcup_{m \ge n}
\{\omega \in \Omega: |X_m(\omega) - X(\omega)| > \epsilon\}$.
Then $A_n \supseteq A_{n+1} \supseteq \ldots$
Define $A \coloneqq \bigcap_{n \in \N} A_n$.
Then $\bP[A_n] \xrightarrow{n\to \infty} \bP[A]$.
@ -67,28 +91,68 @@ from the lecture on stochastic.
\bP[\{\omega \in \Omega | ~|X_n(\omega) - X(\omega)| > \epsilon\}] < \bP[A_n] \to 0.
\]
\end{subproof}
\begin{claim}
Let $1 \le p < q < \infty$.
Then $X_n \xrightarrow{L^q} X \implies X_n \xrightarrow{L^p} X$.
\end{claim}
\begin{subproof}
Take $r$ such that $\frac{1}{p} = \frac{1}{q} + \frac{1}{r}$.
We have
\begin{IEEEeqnarray*}{rCl}
\|X_n - X\|_{L^p} &=& \|1 \cdot (X_n - X)\|_{L^p}\\
&\overset{\text{Hölder}}{\le }& \|1\|_{L^r} \|X_n - X\|_{L^q}\\
&=& \|X_n - X\|_{L^q}
\end{IEEEeqnarray*}
Hence $\bE[|X_n - X|^q] \xrightarrow{n\to \infty} 0 \implies \bE[|X_n - X|^p] \xrightarrow{n\to \infty} 0$.
\end{subproof}
\begin{claim}
$X_n \xrightarrow{L^1} X \implies X_n\xrightarrow{\bP} X$
\end{claim}
\begin{subproof}
We have $\bE[|X_n - X|] \to 0$.
Let $\bE[|X_n - X|] \to 0$.
Suppose there exists an $\epsilon > 0$ such that
$\limsup\limits_{n \to \infty} \bP[|X_n - X| > \epsilon] = c > 0$.
W.l.o.g.~$\lim_{n \to \infty} \bP[|X_n - X| > \epsilon] = c$,
otherwise choose an appropriate subsequence.
We have
\begin{IEEEeqnarray*}{rCl}
\bE[|X_n - X|] &=& \int_\Omega |X_n - X | d\bP\\
&=& \int_{|X_n - X| > \epsilon} |X_n - X| d\bP + \underbrace{\int_{|X_n - X| \le \epsilon} |X_n - X | d\bP}_{\ge 0}\\
&\ge& \epsilon \int_{|X_n -X | > \epsilon} d\bP\\
\bE[|X_n - X|] &=& \int_\Omega |X_n - X | \dif\bP\\
&=& \int_{|X_n - X| > \epsilon} |X_n - X| \dif\bP
+ \underbrace{\int_{|X_n - X| \le \epsilon}|X_n-X|\dif\bP}_{\ge 0}\\
&\ge& \epsilon \int_{|X_n -X | > \epsilon} \dif\bP\\
&=& \epsilon \cdot c > 0 \lightning
\end{IEEEeqnarray*}
\todo{Improve this with Markov}
\todo{Counter examples}
\todo{weak convergence}
\todo{$L^p$ convergence}
\end{subproof}
\begin{claim} %+
$X_n \xrightarrow{\bP} X \implies X_n \xrightarrow{\text{dist}} X$.
\end{claim}
\begin{subproof}
Let $F$ be the distribution function of $X$
and $(F_n)_n$ the distribution functions of $(X_n)_n$.
By \autoref{lec10_thm1}
it suffices to show that $F_n(t) \to F(t)$ for all continuity
points $t$ of $F$.
Let $t$ be a continuity point of $F$.
Take some $\epsilon > 0$.
Then there exists $\delta > 0$ such that
$|F(t) - F(t')| < \frac{\epsilon}{2}$
for all $t'$ with $|t - t'| \le \delta$.
For all $n$ large enough,
we have
$\bP[|X_n - X| > \delta] < \frac{\epsilon}{2}$.
It is
\[|F_n(t) - F(t)|
= |\bP[X_n \le t] - F(t)|
\le \max(|\frac{\epsilon}{2} + \bP[X \le t + \delta] - F(t)|,
|\bP[X \le t -\delta] - F(t)|)\\
\le \max(|\frac{\epsilon}{2} + F(t + \delta) - F(t)|, |F(t-\delta) -F(t)|
\le \epsilon,
\]
hence $F_n(t) \to F(t)$.
\end{subproof}
\begin{claim}
\label{claim:convimplpl1}
$X_n \xrightarrow{\bP} X \notimplies X_n\xrightarrow{L^1} X$
\end{claim}
\begin{subproof}
@ -101,60 +165,58 @@ from the lecture on stochastic.
\end{subproof}
\begin{claim}
$X_n \xrightarrow{a.s.} X \notimplies X_n\xrightarrow{L^1} X$.
$X_n \xrightarrow{\text{a.s.}} X \notimplies X_n\xrightarrow{L^1} X$.
\end{claim}
\begin{subproof}
We can use the same counterexample as in c).
We can use the same counterexample as in \autoref{claim:convimplpl1}
$\bP[\lim_{n \to \infty} X_n = 0] \ge \bP[X_n = 0] = 1 - \frac{1}{n} \to 0$.
We have already seen, that $X_n$ does not converge in $L_1$.
\end{subproof}
\begin{claim}
$X_n \xrightarrow{L^1} X \notimplies X_n\xrightarrow{a.s.} X$.
$X_n \xrightarrow{L^1} X \notimplies X_n\xrightarrow{\text{a.s.}} X$.
\end{claim}
\begin{subproof}
Take $\Omega = [0,1], \cF = \cB([0,1]), \bP = \lambda$.
Define $A_n \coloneqq [j 2^{-k}, (j+1) 2^{-k}]$ where $n = 2^k + j$.
We have
\[
\bE[|X_n|] = \int_{\Omega}|X_n| d\bP = \frac{1}{2^k} \to 0.
\bE[|X_n|] = \int_{\Omega}|X_n| \dif\bP = \frac{1}{2^k} \to 0.
\]
However $X_n$ does not converge a.s.~as for all $\omega \in [0,1]$
the sequence $X_n(\omega)$ takes the values $0$ and $1$ infinitely often.
\end{subproof}
\end{proof}
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$
(more precisely: $\bP[\limsup_{n \to \infty} A_n] = 0$).
\begin{claim}
$X_n \xrightarrow{\text{dist}} X \notimplies X_n \xrightarrow{\bP} X$.
\end{claim}
\begin{subproof}
Note that $X_n \xrightarrow{\text{dist}} X$ only makes a statement
about the distributions of $X$ and $X_1,X_2,\ldots$
For example, take some $p \in (0,1)$ and let
$X$, $X_1, X_2,\ldots$ be i.i.d.~with $X \sim \Bin(1,p)$.
Trivially $X_n \xrightarrow{\text{dist}} X$.
However
\[
\bP[|X_n - X| = 1]
= \bP[X_n = 0]\bP[X = 1] + \bP[X_n = 1]\bP[X = 0]
= 2p(1-p).
\]
\end{subproof}
\begin{claim}
Let $1 \le p < q < \infty$. Then
$X_n \xrightarrow{L^p} X \notimplies X_n \xrightarrow{L^q} X$.
\end{claim}
\begin{subproof}
Consider $\Omega = [0,1]$, $\cF = \cB([0,1])$, $\bP = \lambda\upharpoonright [0,1]$
and $X_n(\omega) = \frac{1}{n \sqrt[q]{\omega}}$.
Then $\|X_0(\omega)\|_{L^p} < \infty$, since $p < q$.
Thus $X_n \xrightarrow{L_p} 0$.
However $\|X_n(\omega)\|_{L^q} = \infty$ for all $n$.
\end{subproof}
\end{refproof}
For independent events $A_n$ the converse holds as well.
\end{lemma}
\iffalse
\todo{Add more stuff here}
\subsection{Some inequalities}
% TODO: Markov
\begin{theorem}[Chebyshev's inequality] % TODO Proof
Let $X$ be a r.v.~with $\Var(x) < \infty$.
Then $\forall \epsilon > 0 : \bP \left[ \left| X - \bE[X] \right| > \epsilon\right] \le \frac{\Var(x)}{\epsilon^2}$.
\end{theorem}
We used Chebyshev's inequality. Linearity of $\bE$, $\Var(cX) = c^2\Var(X)$ and $\Var(X_1 +\ldots + X_n) = \Var(X_1) + \ldots + \Var(X_n)$ for independent $X_i$.
\fi
\subsection{Some Facts from Measure Theory}
\begin{fact}+[Finite measures are {\vocab[Measure]{regular}}, Exercise 3.1]
@ -178,5 +240,51 @@ We used Chebyshev's inequality. Linearity of $\bE$, $\Var(cX) = c^2\Var(X)$ and
\subsection{Inequalities}
This is taken from section 6.1 of the notes on Stochastik.
\begin{theorem}[Markov's inequality]
Let $X$ be a random variable and $a > 0$.
Then
\[
\bP[|X| \ge a] \le \frac{\bE[|X|]}{a}.
\]
\end{theorem}
\begin{proof}
We have
\begin{IEEEeqnarray*}{rCl}
\bE[|X|] &\ge & \int_{|X| \ge a} |X| \dif \bP\\
&=& a \int_{|X| \ge a} \dif \bP = a\bP[|X| \ge a].
\end{IEEEeqnarray*}
\end{proof}
\begin{theorem}[Chebyshev's inequality]
Let $X$ be a random variable and $a > 0$.
Then
\[
\bP[|X - \bE(X)| \ge a] \le \frac{\Var(X)}{a}.
\]
\end{theorem}
\begin{proof}
We have
\begin{IEEEeqnarray*}{rCl}
\bP[|X-\bE(X)| \ge a]
&=& \bP[|X - \bE(X)|^2 \ge a^2]\\
&\overset{\text{Markov}}{\ge}& \frac{\bE[|X - \bE(X)|^2]}{a^2}.
\end{IEEEeqnarray*}
\end{proof}
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$
(more precisely: $\bP[\limsup_{n \to \infty} A_n] = 0$).
For independent events $A_n$ the converse holds as well.
\end{lemma}