153 lines
5.5 KiB
TeX
153 lines
5.5 KiB
TeX
% This section provides a short recap of things that should be known
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% from the lecture on stochastics.
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\subsection{Notions of convergence}
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\begin{definition}
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Fix a probability space $(\Omega,\cF,\bP)$.
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Let $X, X_1, X_2,\ldots$ be random variables.
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\begin{itemize}
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\item We say that $X_n$ converges to $X$
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\vocab[Convergence!almost surely]{almost surely}
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($X_n \xrightarrow{a.s.} X$)
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iff
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\[
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\bP(\{\omega | X_n(\omega) \to X(\omega)\}) = 1.
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\]
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\item We say that $X_n$ converges to $X$
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\vocab[Convergence!in probability]{in probability}
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($X_n \xrightarrow{\bP} X$)
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iff
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\[
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\lim_{n \to \infty}\bP[|X_n - X| > \epsilon] = 0
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\]
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for all $\epsilon > 0$.
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\item We say that $X_n$ converges to $X$
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\vocab[Convergence!in mean]{in the $p$-th mean}
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($X_n \xrightarrow{L^p} X$ )
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iff
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\[
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\bE[|X_n - X|^p] \xrightarrow{n \to \infty} 0.
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\]
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\end{itemize}
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\end{definition}
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% TODO Connect to ANaIII
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\begin{theorem}
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\vspace{10pt}
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Let $X$ be a random variable and $X_n, n \in \N$ a sequence of random variables.
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Then
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\begin{figure}[H]
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\centering
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\begin{tikzpicture}
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\node at (0,1.5) (as) { $X_n \xrightarrow{a.s.} X$};
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\node at (1.5,0) (p) { $X_n \xrightarrow{\bP} X$};
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\node at (3,1.5) (L1) { $X_n \xrightarrow{L^1} X$};
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\draw[double equal sign distance, -implies] (as) -- (p);
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\draw[double equal sign distance, -implies] (L1) -- (p);
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\end{tikzpicture}
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\end{figure}
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and none of the other implications hold.
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\end{theorem}
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\begin{proof}
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\begin{claim}
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$X_n \xrightarrow{a.s.} X \implies X_n \xrightarrow{\bP} X$.
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\end{claim}
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\begin{subproof}
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$\Omega_0 \coloneqq \{\omega \in \Omega : \lim_{n\to \infty} X_n(\omega) = X(\Omega)\} $.
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Let $\epsilon > 0$ and consider $A_n \coloneqq \bigcup_{m \ge n} \{\omega \in \Omega: |X_m(\omega) - X(\Omega)| > \epsilon\}$.
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Then $A_n \supseteq A_{n+1} \supseteq \ldots$
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Define $A \coloneqq \bigcap_{n \in \N} A_n$.
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Then $\bP[A_n] \xrightarrow{n\to \infty} \bP[A]$.
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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 $\bP[A_n] \to 0$.
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Thus \[
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\bP[\{\omega \in \Omega | ~|X_n(\omega) - X(\omega)| > \epsilon\}] < \bP[A_n] \to 0.
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\]
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\end{subproof}
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\begin{claim}
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$X_n \xrightarrow{L^1} X \implies X_n\xrightarrow{\bP} X$
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\end{claim}
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\begin{subproof}
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We have $\bE[|X_n - X|] \to 0$.
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Suppose there exists an $\epsilon > 0$ such that
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$\lim_{n \to \infty} \bP[|X_n - X| > \epsilon] = c > 0$.
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We have
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\begin{IEEEeqnarray*}{rCl}
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\bE[|X_n - X|] &=& \int_\Omega |X_n - X | d\bP\\
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&=& \int_{|X_n - X| > \epsilon} |X_n - X| d\bP + \underbrace{\int_{|X_n - X| \le \epsilon} |X_n - X | d\bP}_{\ge 0}\\
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&\ge& \epsilon \int_{|X_n -X | > \epsilon} d\bP\\
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&=& \epsilon \cdot c > 0 \lightning
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\end{IEEEeqnarray*}
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\todo{Improve this with Markov}
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\end{subproof}
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\begin{claim}
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$X_n \xrightarrow{\bP} X \notimplies X_n\xrightarrow{L^1} X$
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\end{claim}
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\begin{subproof}
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Take $([0,1], \cB([0,1 ]), \lambda)([0,1], \cB([0,1 ]), \lambda)$
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and define $X_n \coloneqq n \One_{[0, \frac{1}{n}]}$.
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We have $\bP[|X_n| > \epsilon] = \frac{1}{n}$
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for $n$ large enough.
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However $\bE[|X_n|] = 1$.
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\end{subproof}
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\begin{claim}
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$X_n \xrightarrow{a.s.} X \notimplies X_n\xrightarrow{L^1} X$.
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\end{claim}
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\begin{subproof}
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We can use the same counterexample as in c).
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$\bP[\lim_{n \to \infty} X_n = 0] \ge \bP[X_n = 0] = 1 - \frac{1}{n} \to 0$.
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We have already seen, that $X_n$ does not converge in $L_1$.
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\end{subproof}
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\begin{claim}
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$X_n \xrightarrow{L^1} X \notimplies X_n\xrightarrow{a.s.} X$.
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\end{claim}
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\begin{subproof}
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Take $\Omega = [0,1], \cF = \cB([0,1]), \bP = \lambda$.
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Define $A_n \coloneqq [j 2^{-k}, (j+1) 2^{-k}]$ where $n = 2^k + j$.
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We have
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\[
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\bE[|X_n|] = \int_{\Omega}|X_n| d\bP = \frac{1}{2^k} \to 0.
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\]
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However $X_n$ does not converge a.s.~as for all $\omega \in [0,1]$
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the sequence $X_n(\omega)$ takes the values $0$ and $1$ infinitely often.
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\end{subproof}
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\end{proof}
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How do we prove that something happens almost surely?
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The first thing that should come to mind is:
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\begin{lemma}[Borel-Cantelli]
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\label{borelcantelli}
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If we have a sequence of events $(A_n)_{n \ge 1}$
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such that $\sum_{n \ge 1} \bP(A_n) < \infty$,
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then $\bP[ A_n \text{for infinitely many $n$}] = 0$
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(more precisely: $\bP[\limsup_{n \to \infty} A_n] = 0$).
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For independent events $A_n$ the converse holds as well.
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\end{lemma}
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\iffalse
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\todo{Add more stuff here}
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\subsection{Some inequalities}
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% TODO: Markov
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\begin{theorem}[Chebyshev's inequality] % TODO Proof
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Let $X$ be a r.v.~with $\Var(x) < \infty$.
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Then $\forall \epsilon > 0 : \bP \left[ \left| X - \bE[X] \right| > \epsilon\right] \le \frac{\Var(x)}{\epsilon^2}$.
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\end{theorem}
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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 $X_i$.
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Modes of covergence: $L^p$, in probability, a.s.
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\fi
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