346 lines
13 KiB
TeX
346 lines
13 KiB
TeX
This section provides a short recap of things that should be known
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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 $(\Omega,\cF,\bP)$.
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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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($X_n \xrightarrow{\text{a.s.}} X$)
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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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\item We say that $X_n$ converges to $X$
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\vocab[Convergence!in distribution]{in distribution}%
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\footnote{
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This notion of convergence was actually
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defined during the course of the lecture,
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but has been added here for completeness;
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see \yaref{def:weakconvergence}.
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}
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($X_n \xrightarrow{\text{d}} X$)
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iff for every continuous, bounded $f: \R \to \R$
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\[
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\bE[f(X_n)] \xrightarrow{n \to \infty} \bE[f(X)].
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\]
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\end{itemize}
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\end{definition}
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% TODO Connect to AnaIII
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\pagebreak
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\begin{theorem}+
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% \footnote{see exercise 3.4}
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\label{thm:convergenceimplications}
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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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Let $1 \le p < q < \infty$.
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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) { $X_n \xrightarrow{\text{a.s.}} X$};
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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) { $X_n \xrightarrow{L^1} X$};
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\node at (3,1.5) (Lp) { $X_n \xrightarrow{L^p} X$};
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\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);
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% \draw[double equal sign distance, -implies] (L1) -- (p);
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% \draw[double equal sign distance, -implies] (Lp) -- (L1);
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\draw[double equal sign distance, -implies] (Lp) -- (p);
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\draw[double equal sign distance, -implies] (Lq) -- (Lp);
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\end{tikzpicture}
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\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}
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\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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$A_n \coloneqq \bigcup_{m \ge n}
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\{\omega \in \Omega: |X_m(\omega) - X(\omega)| > \epsilon\}$.
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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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\[\forall \omega \in \Omega_0 .~ \exists n \in \N .~
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\forall m \ge n.~ |X_m(\omega) - X(\omega)| < \epsilon.\]
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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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Let $1 \le p < q < \infty$.
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Then $X_n \xrightarrow{L^q} X \implies X_n \xrightarrow{L^p} X$.
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\end{claim}
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\begin{subproof}
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Take $r$ such that $\frac{1}{p} = \frac{1}{q} + \frac{1}{r}$.
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We have
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\begin{IEEEeqnarray*}{rCl}
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\|X_n - X\|_{L^p} &=& \|1 \cdot (X_n - X)\|_{L^p}\\
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&\overset{\text{Hölder}}{\le}& \|1\|_{L^r} \|X_n - X\|_{L^q}\\
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&=& \|X_n - X\|_{L^q}
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\end{IEEEeqnarray*}
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Hence $\bE[|X_n - X|^q] \xrightarrow{n\to \infty} 0 \implies \bE[|X_n - X|^p] \xrightarrow{n\to \infty} 0$.
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\end{subproof}
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\begin{claim}
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\label{claim:convimpll1p}
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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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Suppose $\bE[|X_n - X|] \to 0$.
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Then for every $\epsilon > 0$
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\begin{IEEEeqnarray*}{rCl}
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\bP[|X_n - X| \ge \epsilon]
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&\overset{\text{Markov}}{\le}& \frac{\bE[|X_n - X|]}{\epsilon}\\
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&\xrightarrow{n \to \infty} & 0,
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\end{IEEEeqnarray*}
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hence $X_n \xrightarrow{\bP} X$.
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\end{subproof}
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\begin{claim} %+
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$X_n \xrightarrow{\bP} X \implies X_n \xrightarrow{\text{d}} X$.
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\end{claim}
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\begin{subproof}
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Let $F$ be the distribution function of $X$
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and $(F_n)_n$ the distribution functions of $(X_n)_n$.
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By \yaref{lec10_thm1}
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it suffices to show that $F_n(t) \to F(t)$ for all continuity
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points $t$ of $F$.
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Let $t$ be a continuity point of $F$.
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Take some $\epsilon > 0$.
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Then there exists $\delta > 0$ such that
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$|F(t) - F(t')| < \frac{\epsilon}{2}$
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for all $t'$ with $|t - t'| \le \delta$.
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For all $n$ large enough,
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we have
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$\bP[|X_n - X| > \delta] < \frac{\epsilon}{2}$.
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It is
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\begin{IEEEeqnarray*}{rCl}
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|F_n(t) - F(t)|
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&=& |\bP[X_n \le t] - F(t)|\\
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&\le& \max(|\frac{\epsilon}{2} + \bP[X \le t + \delta] - F(t)|,
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|\bP[X \le t -\delta] - F(t)|)\\
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&\le& \max(|\frac{\epsilon}{2} + F(t + \delta) - F(t)|,
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|F(t-\delta) -F(t)|)\\
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&\le& \epsilon,
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\end{IEEEeqnarray*}
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hence $F_n(t) \to F(t)$.
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\end{subproof}
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\begin{claim}
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\label{claim:convimplpl1}
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$X_n \xrightarrow{\bP} X \notimplies X_n\xrightarrow{L^1} X$.%
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\footnote{Note that the implication holds under certain assumptions,
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see \yaref{thm:l1iffuip}.}
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\end{claim}
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\begin{subproof}
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Take $([0,1], \cB([0,1 ]), \lambda)$
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and define $X_n \coloneqq n \One_{[0, \frac{1}{n}]}$.
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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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$X_n \xrightarrow{\text{a.s.}} X \notimplies X_n\xrightarrow{L^1} X$.
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\end{claim}
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\begin{subproof}
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We can use the same counterexample as in \yaref{claim:convimplpl1}
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$\bP[\lim_{n \to \infty} X_n = 0] \ge \bP[X_n = 0] = 1 - \frac{1}{n} \to 0$.
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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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$X_n \xrightarrow{L^1} X \notimplies X_n\xrightarrow{\text{a.s.}} X$.
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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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\[
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\bE[|X_n|] = \int_{\Omega}|X_n| \dif\bP = \frac{1}{2^k} \to 0.
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\]
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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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\begin{claim}
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$X_n \xrightarrow{\text{d}} X \notimplies X_n \xrightarrow{\bP} X$.
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\end{claim}
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\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 $X_1,X_2,\ldots$
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For example, take some $p \in (0,1)$ and let
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$X$, $X_1, X_2,\ldots$ 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
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\[
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\bP[|X_n - X| = 1]
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= \bP[X_n = 0]\bP[X = 1] + \bP[X_n = 1]\bP[X = 0]
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= 2p(1-p).
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\]
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\end{subproof}
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\begin{claim}
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Let $1 \le p < q < \infty$. Then
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$X_n \xrightarrow{L^p} X \notimplies X_n \xrightarrow{L^q} X$.
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\end{claim}
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\begin{subproof}
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Consider $\Omega = [0,1]$, $\cF = \cB([0,1])$, $\bP = \lambda\upharpoonright [0,1]$
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and $X_n(\omega) = \frac{1}{n \sqrt[q]{\omega}}$.
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Then $\|X_0(\omega)\|_{L^p} < \infty$, since $p < q$.
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Thus $X_n \xrightarrow{L_p} 0$.
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However $\|X_n(\omega)\|_{L^q} = \infty$ for all $n$.
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\end{subproof}
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\end{refproof}
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\begin{theorem}[Bounded convergence theorem]
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\yalabel{Bounded Convergence Theorem}{Bounded convergence}{thm:boundedconvergence}
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Suppose that $X_n \xrightarrow{\bP} X$ and there exists
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some $K$ such that $|X_n| \le K$ for all $n$.
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Then $X_n \xrightarrow{L^1} X$.
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\end{theorem}
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\begin{proof}
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Note that $|X| \le K$ a.s.~since
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\[\bP[|X| \ge K + \epsilon] \le \bP[|X_n - X| > \epsilon]
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\xrightarrow{n \to \infty} 0.\]
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Hence
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\begin{IEEEeqnarray*}{rCl}
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\int |X_n - X| \dif \bP &\le& \int_{|X_n - X| \ge \epsilon} |X_n - X| \dif \bP + \epsilon\\
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&\le &2K\bP[|X_n - X| \ge \epsilon] +\epsilon.
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\end{IEEEeqnarray*}
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\end{proof}
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\subsection{Some Facts from Measure Theory}
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\begin{fact}+[Finite measures are {\vocab[Measure]{regular}}, Exercise 3.1]
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Let $\mu$ be a finite measure on $(\R, \cB(\R))$.
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Then for all $\epsilon > 0$,
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there exists a compact set $K \in \cB(\R)$ such that
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$\mu(K) > \mu(\R) - \epsilon$.
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\end{fact}
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\begin{proof}
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We have $[-k,k] \uparrow \R$, hence $\mu([-k,k]) \uparrow \mu(\R)$.
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\end{proof}
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\begin{theorem}+[Change of variables formula]
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Let $X$ be a random variable with a continuous density $f$,
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and let $g: \R \to \R$ be continuous such that
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$g(X)$ is integrable.
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Then
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\[
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\bE[g(X)] = \int g \circ X \dif \bP
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= \int_{-\infty}^\infty g(y) f(y) \lambda(\dif y)
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= \int_{-\infty}^\infty g(y) f(y) \dif y.
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\]
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\end{theorem}
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\begin{theorem}+[Riemann-Lebesgue]
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%\footnote{see exercise 3.3}
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\yalabel{Riemann-Lebesgue}{Riemann-Lebesgue}{riemann-lebesgue}
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Let $f: \R \to \R$ be integrable.
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Then
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\[
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\lim_{n \to \infty} \int_{\R} f(x) \cos(n x) \lambda(\dif x) = 0.
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\]
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\end{theorem}
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\begin{theorem}+[Fubini-Tonelli]
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\yalabel{Fubuni-Tonelli}{Fubini}{thm:fubini}
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%\footnote{exercise sheet 1}
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Let $(\Omega_{i}, \cF_i, \bP_i), i \in \{0,1\}$
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be probability spaces and $\Omega \coloneqq \Omega_0 \otimes \Omega_1$,
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$\cF \coloneqq \cF_1 \otimes\cF_2$,
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$\bP \coloneqq \bP_0 \otimes \bP_1$.
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Let $f \ge 0$ be $(\Omega, \cF)$-measurable,
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then
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\[
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\Omega_0 \ni x \mapsto \int_{\Omega_{2}} f(x,y) \bP_2(\dif y)
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\]
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and
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\[
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\Omega_1 \ni y \mapsto \int_{\Omega_1} f(x,y) \bP_1(\dif x)
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\]
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are measurable, and
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\[
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\int f \dif \bP = \int_{\Omega_1} \int_{\Omega_2} f(x,y) \bP_2(\dif y) \bP_1(\dif x)(\dif x)
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= \int_{\Omega_2} \int_{\Omega_1} f(x,y) \bP_1(\dif x) \bP_2(\dif y).
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\]
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\end{theorem}
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\subsection{Inequalities}
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This is taken from section 6.1 of the notes on Stochastik.
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\begin{theorem}[Markov's inequality]
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\yalabel{Markov's Inequality}{Markov}{thm:markov}
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Let $X$ be a random variable and $a > 0$.
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Then
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\[
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\bP[|X| \ge a] \le \frac{\bE[|X|]}{a}.
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\]
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\end{theorem}
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\begin{proof}
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We have
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\begin{IEEEeqnarray*}{rCl}
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\bE[|X|] &\ge & \int_{|X| \ge a} |X| \dif \bP\\
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&=& a \int_{|X| \ge a} \dif \bP = a\bP[|X| \ge a].
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\end{IEEEeqnarray*}
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\end{proof}
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\begin{theorem}[Chebyshev's inequality]
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\yalabel{Chebyshev's Inequality}{Chebyshev}{thm:chebyshev}
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Let $X$ be a random variable and $a > 0$.
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Then
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\[
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\bP[|X - \bE(X)| \ge a] \le \frac{\Var(X)}{a^2}.
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\]
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\end{theorem}
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\begin{proof}
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We have
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\begin{IEEEeqnarray*}{rCl}
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\bP[|X-\bE(X)| \ge a]
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&=& \bP[|X - \bE(X)|^2 \ge a^2]\\
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&\overset{\yaref{thm:markov}}{\le}& \frac{\bE[|X - \bE(X)|^2]}{a^2}.
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\end{IEEEeqnarray*}
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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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\yalabel{Borel-Cantelli}{Borel-Cantelli}{thm: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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