diff --git a/inputs/intro.tex b/inputs/intro.tex index e2dc4ed..a33f50b 100644 --- a/inputs/intro.tex +++ b/inputs/intro.tex @@ -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 $\dagger$. diff --git a/inputs/prerequisites.tex b/inputs/prerequisites.tex index c092688..ac622d2 100644 --- a/inputs/prerequisites.tex +++ b/inputs/prerequisites.tex @@ -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} +