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probability-theory/inputs/prerequisites.tex

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This section provides a short recap of things that should be known
from the lecture on stochastic.
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\subsection{Notions of Convergence}
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\begin{definition}+
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\label{def:convergence}
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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}
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($X_n \xrightarrow{\text{a.s.}} X$)
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iff
\[
\bP(\{\omega | X_n(\omega) \to X(\omega)\}) = 1.
\]
\item We say that $X_n$ converges to $X$
\vocab[Convergence!in probability]{in probability}
($X_n \xrightarrow{\bP} X$)
iff
\[
\lim_{n \to \infty}\bP[|X_n - X| > \epsilon] = 0
\]
for all $\epsilon > 0$.
\item We say that $X_n$ converges to $X$
\vocab[Convergence!in mean]{in the $p$-th mean}
($X_n \xrightarrow{L^p} X$ )
iff
\[
\bE[|X_n - X|^p] \xrightarrow{n \to \infty} 0.
\]
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\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}.
}
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($X_n \xrightarrow{\text{d}} X$)
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iff for every continuous, bounded $f: \R \to \R$
\[
\bE[f(X_n)] \xrightarrow{n \to \infty} \bE[f(X)].
\]
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\end{itemize}
\end{definition}
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% TODO Connect to AnaIII
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\pagebreak
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\begin{theorem}+
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\label{thm:convergenceimplications}
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\vspace{10pt}
Let $X$ be a random variable and $X_n, n \in \N$ a sequence of random variables.
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Let $1 \le p < q < \infty$.
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Then
\begin{figure}[H]
\centering
\begin{tikzpicture}
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\node at (0,1.5) (as) { $X_n \xrightarrow{\text{a.s.}} X$};
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\node at (1.5,0) (p) { $X_n \xrightarrow{\bP} X$};
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\node at (1.5,-1.5) (w) { $X_n \xrightarrow{\text{d}} X$};
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%\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) { $X_n \xrightarrow{L^q} X$};
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\draw[double equal sign distance, -implies] (as) -- (p);
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\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);
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\end{tikzpicture}
\end{figure}
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and none of the other implications hold (apart from the transitive closure).
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\end{theorem}
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\begin{refproof}{thm:convergenceimplications}
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\begin{claim}
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$X_n \xrightarrow{\text{a.s.}} X \implies X_n \xrightarrow{\bP} X$.
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\end{claim}
\begin{subproof}
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Let $\Omega_0 \coloneqq
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\{\omega \in \Omega : \lim_{n\to \infty} X_n(\omega) = X(\omega)\}$.
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Fix some $\epsilon > 0$ and consider
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$A_n \coloneqq \bigcup_{m \ge n}
\{\omega \in \Omega: |X_m(\omega) - X(\omega)| > \epsilon\}$.
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Then $A_n \supseteq A_{n+1} \supseteq \ldots$
Define $A \coloneqq \bigcap_{n \in \N} A_n$.
Then $\bP[A_n] \xrightarrow{n\to \infty} \bP[A]$.
Since $X_n \xrightarrow{a.s.} X$ we have that
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\[\forall \omega \in \Omega_0 .~ \exists n \in \N .~
\forall m \ge n.~ |X_m(\omega) - X(\omega)| < \epsilon.\]
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We have $A \subseteq \Omega_0^{c}$, hence $\bP[A_n] \to 0$.
Thus \[
\bP[\{\omega \in \Omega | ~|X_n(\omega) - X(\omega)| > \epsilon\}] < \bP[A_n] \to 0.
\]
\end{subproof}
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\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}\\
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&=& \|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}
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\begin{claim}
\label{claim:convimpll1p}
$X_n \xrightarrow{L^1} X \implies X_n\xrightarrow{\bP} X$.
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\end{claim}
\begin{subproof}
Suppose $\bE[|X_n - X|] \to 0$.
Then for every $\epsilon > 0$
\begin{IEEEeqnarray*}{rCl}
\bP[|X_n - X| \ge \epsilon]
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&\overset{\text{Markov}}{\le}& \frac{\bE[|X_n - X|]}{\epsilon}\\
&\xrightarrow{n \to \infty} & 0,
\end{IEEEeqnarray*}
hence $X_n \xrightarrow{\bP} X$.
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\end{subproof}
\begin{claim} %+
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$X_n \xrightarrow{\bP} X \implies X_n \xrightarrow{\text{d}} X$.
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\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
\begin{IEEEeqnarray*}{rCl}
|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,
\end{IEEEeqnarray*}
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hence $F_n(t) \to F(t)$.
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\end{subproof}
\begin{claim}
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\label{claim:convimplpl1}
$X_n \xrightarrow{\bP} X \notimplies X_n\xrightarrow{L^1} X$.%
\footnote{Note that the implication holds under certain assumptions,
see \autoref{thm:l1iffuip}.}
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\end{claim}
\begin{subproof}
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Take $([0,1], \cB([0,1 ]), \lambda)$
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and define $X_n \coloneqq n \One_{[0, \frac{1}{n}]}$.
We have $\bP[|X_n| > \epsilon] = \frac{1}{n}$
for $n$ large enough.
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However $\bE[|X_n|] = 1$.
\end{subproof}
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\begin{claim}
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$X_n \xrightarrow{\text{a.s.}} X \notimplies X_n\xrightarrow{L^1} X$.
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\end{claim}
\begin{subproof}
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We can use the same counterexample as in \autoref{claim:convimplpl1}
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$\bP[\lim_{n \to \infty} X_n = 0] \ge \bP[X_n = 0] = 1 - \frac{1}{n} \to 0$.
We have already seen, that $X_n$ does not converge in $L_1$.
\end{subproof}
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\begin{claim}
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$X_n \xrightarrow{L^1} X \notimplies X_n\xrightarrow{\text{a.s.}} X$.
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\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
\[
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\bE[|X_n|] = \int_{\Omega}|X_n| \dif\bP = \frac{1}{2^k} \to 0.
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\]
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}
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\begin{claim}
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$X_n \xrightarrow{\text{d}} X \notimplies X_n \xrightarrow{\bP} X$.
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\end{claim}
\begin{subproof}
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Note that $X_n \xrightarrow{\text{d}} X$ only makes a statement
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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 $X \sim \Bin(1,p)$.
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Trivially $X_n \xrightarrow{\text{d}} X$.
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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}
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\begin{theorem}[Bounded convergence theorem]
\label{thm:boundedconvergence}
Suppose that $X_n \xrightarrow{\bP} X$ and there exists
some $K$ such that $|X_n| \le K$ for all $n$.
Then $X_n \xrightarrow{L^1} X$.
\end{theorem}
\begin{proof}
Note that $|X| \le K$ a.s.~since
\[\bP[|X| \ge K + \epsilon] \le \bP[|X_n - X| > \epsilon]
\xrightarrow{n \to \infty} 0.\]
Hence
\begin{IEEEeqnarray*}{rCl}
\int |X_n - X| \dif \bP &\le& \int_{|X_n - X| \ge \epsilon} |X_n - X| \dif \bP + \epsilon\\
&\le &2K\bP[|X_n - X| \ge \epsilon] +\epsilon.
\end{IEEEeqnarray*}
\end{proof}
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\subsection{Some Facts from Measure Theory}
\begin{fact}+[Finite measures are {\vocab[Measure]{regular}}, Exercise 3.1]
Let $\mu$ be a finite measure on $(\R, \cB(\R))$.
Then for all $\epsilon > 0$,
there exists a compact set $K \in \cB(\R)$ such that
$\mu(K) > \mu(\R) - \epsilon$.
\end{fact}
\begin{proof}
We have $[-k,k] \uparrow \R$, hence $\mu([-k,k]) \uparrow \mu(\R)$.
\end{proof}
\begin{theorem}[Riemann-Lebesgue]
\label{riemann-lebesgue}
Let $f: \R \to \R$ be integrable.
Then
\[
\lim_{n \to \infty} \int_{\R} f(x) \cos(n x) \lambda(\dif x) = 0.
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\]
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\end{theorem}
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\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
\[
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\bP[|X - \bE(X)| \ge a] \le \frac{\Var(X)}{a^2}.
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\]
\end{theorem}
\begin{proof}
We have
\begin{IEEEeqnarray*}{rCl}
\bP[|X-\bE(X)| \ge a]
&=& \bP[|X - \bE(X)|^2 \ge a^2]\\
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&\overset{\text{Markov}}{\le}& \frac{\bE[|X - \bE(X)|^2]}{a^2}.
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\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}
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