lectures 1 - 10
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inputs/intro.tex
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inputs/intro.tex
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These are my notes on the lecture probability theory, taught by Prof.~\textsc{Chiranjib Mukherjee}
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in the summer term 2023 at the University Münster.
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\begin{warning}
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This is not an official script.
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The official lecture notes can be found on the Learnweb.
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\end{warning}
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These notes contain errors almost surely.
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If you find some error or want to improve something, send me a message: \texttt{probabilitytheory@jrpie.de}.
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Topics of this lecture:
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\begin{enumerate}[(1)]
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\item Limit theorems: Laws of large numbers and the central limit theorem for i.i.d.~sequences,
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\item Conditional expectation and conditional probabilities,
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\item Martingales,
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\item Markov chains.
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\end{enumerate}
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inputs/lecture_1.tex
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inputs/lecture_1.tex
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% Lecture 1 - 2023-04-04
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First, let us recall some basic definitions:
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\begin{definition}
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A \vocab{probability space} is a triplet $(\Omega, \cF, \bP)$,
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such that
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\begin{itemize}
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\item $\Omega \neq \emptyset$,
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\item $\cF$ is a $\sigma$-algebra over $\Omega$, i.e.~$\cF \subseteq \cP(\Omega)$ and
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\begin{itemize}
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\item $\emptyset, \Omega \in \cF$,
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\item $A \in \cF \implies A^c \in \cF$,
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\item $A_1, A_2,\ldots \in \cF \implies \bigcup_{i \in \N} A_i \in \cF$.
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\end{itemize}
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The elements of $\cF$ are called \vocab[Event]{events}.
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\item $\bP$ is a \vocab{probability measure}, i.e.~$\bP$ is a function $\bP: \cF \to [0,1]$
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such that
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\begin{itemize}
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\item $\bP(\emptyset) = 1$, $\bP(\Omega) = 1$,
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\item $\bP\left( \bigsqcup_{n \in \N} A_n \right) = \sum_{n \in \N} \bP(A_n)$
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for mutually disjoint $A_n \in \cF$.
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\end{itemize}
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\end{itemize}
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\end{definition}
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\begin{definition}
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A \vocab{random variable} $X : (\Omega, \cF) \to (\R, \cB(\R))$
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is a measurable function, i.e.~for all $B \in \cB(\R)$ we have $X^{-1}(B) \in \cF$.
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(Equivalently $X^{-1}\left( (a,b] \right) \in \cF$ for all $a < b \in \R$ ).
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\end{definition}
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\begin{definition}
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$F: \R \to \R_+$ is a \vocab{distribution function} iff
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\begin{itemize}
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\item $F$ is monotone non-decreasing,
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\item $F$ is right-continuous,
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\item $\lim_{x \to -\infty} F(x) = 0$ and $\lim_{x \to \infty} F(x) = 1$.
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\end{itemize}
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\end{definition}
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\begin{fact}
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Let $\bP$ be a probability measure on $(\R, \cB(\R))$.
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Then $F(x) \coloneqq\bP\left( (-\infty, x] \right)$
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is a probability distribution function.
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(See lemma 2.4.2 in the lecture notes of Stochastik)
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\end{fact}
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The converse to this fact is also true:
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\begin{theorem}[Kolmogorov's existence theorem / basic existence theorem of probability theory]
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\label{kolmogorovxistence}
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Let $\cF(\R)$ be the set of all distribution functions on $\R$
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and let $\cM(\R)$ be the set of all probability measures on $\R$.
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Then there is a one-to-one correspondence between $\cF(\R)$ and $\cM(\R)$
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given by
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\begin{IEEEeqnarray*}{rCl}
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\cM(\R) &\longrightarrow & \cF(\R)\\
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\bP &\longmapsto & \begin{pmatrix*}[l]
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\R &\longrightarrow & \R_+ \\
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x &\longmapsto & \bP((-\infty, x]).
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\end{pmatrix*}
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\end{IEEEeqnarray*}
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\end{theorem}
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\begin{proof}
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See theorem 2.4.3 in Stochastik.
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\end{proof}
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\begin{example}[Some important probability distribution functions]\hfill
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\begin{enumerate}[(1)]
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\item \vocab{Uniform distribution} on $[0,1]$:
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\[
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F(x) = \begin{cases}
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0 & x \in (-\infty, 0],\\
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x & x \in (0,1],\\
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1 & x \in (1,\infty).\\
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\end{cases}
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\]
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\item \vocab{Exponential distribution}:
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\[
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F(x) = \begin{cases}
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1 - e^{-\lambda x} & x \ge 0,\\
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0 & x < 0.
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\end{cases}
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\]
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\item \vocab{Gaussian distribution}:
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\[
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\Phi(x) \coloneqq \frac{1}{\sqrt{2\pi}} \int_{-\infty}^x e^{-\frac{y^2}{2}} dy.
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\]
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\item $\bP[X = 1] = \bP[X = -1] = \frac{1}{2}$ :
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\[
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F(x) = \begin{cases}
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0 & x \in (-\infty, -1),\\
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\frac{1}{2} & x \in [-1,1),\\
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1 & x \in [1, \infty).
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\end{cases}
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\]
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\end{enumerate}
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\end{example}
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inputs/lecture_10.tex
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% lecture 10 - 2023-05-09
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% RECAP
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First, we will prove some of the most important facts about Fourier transforms.
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We consider $(\R, \cB(\R))$.
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\begin{notation}
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By $M_1 (\R)$ we denote the set of all probability measures on $\left( \R, \cB(\R) \right)$.
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\end{notation}
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For all $\bP \in M_1(\R)$ we define $\phi_{\bP}(t) = \int_{\R} e^{\i t x}d\bP(x)$.
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If $X: (\Omega, \cF) \to (\R, \cB(\R))$ is a random variable, we write
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$\phi_X(t) \coloneqq \bE[e^{\i t X}] = \phi_{\mu}(t)$,
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where $\mu = \bP X^{-1}$.
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\begin{refproof}{inversionformula}
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We will prove that the limit in the RHS of \autoref{invf}
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exists and is equal to the LHS.
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Note that the term on the RHS is integrable, as
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\[
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\lim_{t \to 0} \frac{e^{-\i t b} - e^{-\i t a}}{- \i t} \pi(t) = a - b
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\]
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and note that $\phi(0) = 1$ and $|\phi(t)| \le 1$.
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% TODO think about this
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We have
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\begin{IEEEeqnarray*}{rCl}
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&&\lim_{T \to \infty} \frac{1}{2 \pi} \int_{-T}^T \int_{\R} \frac{e^{-\i t b}- e^{-\i t a}}{-\i t} e^{\i t x} d \bP(x)\\
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&\overset{\text{Fubini for $L^1$}}{=}& \lim_{T \to \infty} \frac{1}{2 \pi} \int_{\R} \int_{-T}^T \frac{e^{-\i t b}- e^{-\i t a}}{-\i t} e^{\i t x} d \bP(x)\\
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&=& \lim_{T \to \infty} \frac{1}{2 \pi} \int_{\R} \int_{-T}^T \frac{e^{\i t (b-x)}- e^{\i t (x-a)}}{-\i t} d \bP(x)\\
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&=& \lim_{T \to \infty} \frac{1}{2 \pi} \int_{\R} \underbrace{\int_{-T}^T \left[ \frac{\cos(t (x-b)) - \cos(t(x-a))}{-\i t}\right] d \bP(x)}_{=0 \text{, as the function is odd}}
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\\&&
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+ \lim_{T \to \infty} \frac{1}{2\pi} \int_{\R}\int_{-T}^T \frac{\sin(t ( x - b)) - \sin(t(x-a))}{-t} dt d\bP(x)\\
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&=& \lim_{T \to \infty} \frac{1}{\pi} \int_\R \int_{0}^T \frac{\sin(t(x-a)) - \sin(t(x-b))}{t} dt d\bP(x)\\
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&\overset{\text{\autoref{fact:intsinxx}, dominated convergence}}{=}& \frac{1}{\pi} \int -\frac{\pi}{2} \One_{x < a} + \frac{\pi}{2} \One_{x > a }
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- (- \frac{\pi}{2} \One_{x < b} + \frac{\pi}{2} \One_{x > b}) d\bP(x)\\
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&=& \frac{1}{2} \bP(\{a\} ) + \frac{1}{2} \bP(\{b\}) + \bP((a,b))\\
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&=& \frac{F(b) + F(b-)}{2} - \frac{F(a) - F(a-)}{2}
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\end{IEEEeqnarray*}
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\end{refproof}
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\begin{fact}
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\label{fact:intsinxx}
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\[
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\int_0^\infty \frac{\sin x}{x} dx = \frac{\pi}{2}
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\]
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where the LHS is an improper Riemann-integral.
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Note that the LHS is not Lebesgue-integrable.
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It follows that
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\begin{IEEEeqnarray*}{rCl}
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\lim_{T \to \infty} \int_0^T \frac{\sin(t(x-a))}{x} dt &=&
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\begin{cases}
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- \frac{\pi}{2}, &x < a,\\
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0, &x = a,\\
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\frac{\pi}{2}, & \frac{\pi}{2}
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\end{cases}
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\end{IEEEeqnarray*}
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\end{fact}
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\begin{theorem} % Theorem 3
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\label{thm:lec10_3}
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Let $\bP \in M_1(\R)$ such that $\phi_\R \in L^1(\lambda)$.
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Then $\bP$ has a continuous probability density given by
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\[
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f(x) = \frac{1}{2 \pi} \int_{\R} e^{-\i t x} \phi_{\R(t) dt}.
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\]
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\end{theorem}
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\begin{example}
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\begin{itemize}
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\item Let $\bP = \delta_{\{0\}}$.
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Then
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\[
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\phi_{\R}(t) = \int e^{\i t x} d \delta_0(x) = e^{\i t 0 } = 1
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\]
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\item Let $\bP = \frac{1}{2} \delta_1 + \frac{1}{2} \delta_{-1}$.
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Then
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\[
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\phi_{\R}(t) = \frac{1}{2} e^{\i t} + \frac{1}{2} e^{- \i t} = \cos(t)
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\]
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\end{itemize}
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\end{example}
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\begin{refproof}{thm:lec10_3}
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Let $f(x) \coloneqq \frac{1}{2 \pi} \int_{\R} e^{ - \i t x} \phi(t) dt$.
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\begin{claim}
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If $x_n \to x$, then $f(x_n) \to f(x)$.
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\end{claim}
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\begin{subproof}
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If $e^{-\i t x_n} \phi(t) \xrightarrow{n \to \infty} e^{-\i t x } \phi(t) $ for all $t$.
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Then
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\[
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|e^{-\i t x} \phi(t)| \le |\phi(t)|
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\]
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and $\phi \in L^1$, hence $f(x_n) \to f(x)$
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by the dominated convergence theorem.
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\end{subproof}
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We'll show that for all $a < b$ we have
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\[
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\bP\left( (a,b] \right) = \int_a^b (x) dx.\label{thm10_3eq1}
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\]
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Let $F$ be the distribution function of $\bP$.
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It is enough to prove \autoref{thm10_3eq1}
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for all continuity points $a $ and $ b$ of $F$.
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We have
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\begin{IEEEeqnarray*}{rCl}
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RHS &\overset{\text{Fubini}}{=}& \frac{1}{2 \pi} \int_{\R} \int_{a}^b e^{-\i t x} \phi(t) dx dt\\
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&=& \frac{1}{2 \pi} \int_\R \phi(t) \int_a^b e^{-\i t x} dx dt\\
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&=& \frac{1}{2\pi} \int_{\R} \phi(t) \left( \frac{e^{-\i t b} - e^{-\i t a}}{- \i t} \right) dt\\
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&\overset{\text{dominated convergence}}{=}& \lim_{T \to \infty} \frac{1}{2\pi} \int_{-T}^{T} \phi(t) \left( \frac{e^{-\i t b} - e^{- \i t a}}{- \i t} \right) dt
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\end{IEEEeqnarray*}
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By \autoref{inversionformula}, the RHS is equal to $F(b) - F(a) = \bP\left( (a,b] \right)$.
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\end{refproof}
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However, Fourier analysis is not only useful for continuous probability density functions:
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\begin{theorem}[Bochner's formula for the mass at a point]\label{bochnersformula} % Theorem 4
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Let $\bP \in M_1(\lambda)$.
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Then
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\[
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\forall x \in \R ~ \bP\left( \{x\} \right) = \lim_{T \to \infty} \frac{1}{2 T} \int_{-T}^T e^{-\i t x } \phi(t) dt.
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\]
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\end{theorem}
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\begin{refproof}{bochnersformula}
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We have
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\begin{IEEEeqnarray*}{rCl}
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RHS &=& \lim_{T \to \infty} \frac{1}{2 T} \int_{-T}^T e^{-\i t x} \int_{\R} e^{\i t y} d \bP(y) \\
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&\overset{\text{Fubini}}{=}& \lim_{T \to \infty} \frac{1}{2 T} \int_\R \bP(dy) \int_{-T}^T \underbrace{e^{-\i t (y - x)}}_{\cos(t ( y - x)) + \i \sin(t (y-x))} dt\\
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&=& \lim_{T \to \infty} \frac{1}{2T} \int_{\R} d\bP(y) \int_{-T}^T \cos(t(y - x)) dt\\
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&=& \lim_{T \to \infty} \frac{1}{2 T }\int_{\R} \frac{2 \sin(T (y-x)}{T (y-x)} d \bP(y)\\
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\end{IEEEeqnarray*}
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Furthermore
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\[
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\lim_{T \to \infty} \frac{\sin(T(x-y)}{T (y- x)} = \begin{cases}
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1, &y = x,\\
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0, &y \neq x.
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\end{cases}
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\]
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Hence
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\begin{IEEEeqnarray*}{rCl}
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\lim_{T \to \infty} \frac{1}{2 T }\int_{\R} \frac{2 \sin(T (y-x)}{T (y-x)} d \bP(y) &=& \bP\left( \{x\}\right)
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\end{IEEEeqnarray*}
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% TODO by dominated convergence?
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\end{refproof}
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\begin{theorem} % Theorem 5
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\label{thm:lec_10thm5}
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Let $\phi$ be the characteristic function of $\bP \in M_1(\lambda)$.
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Then
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\begin{enumerate}[(a)]
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\item $\phi(0) = 1$, $|\phi(t)| \le t$ and $\phi(\cdot )$ is continuous.
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\item $\phi$ is a \vocab{positive definite function},
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i.e.~
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\[\forall t_1,\ldots, t_n \in \R, (c_1,\ldots,c_n) \in \C^n ~ \sum_{j,k = 1}^n c_j \overline{c_k} \phi(t_j - t_k) \ge 0
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\]
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(equivalently, the matix $(\phi(t_j- t_k))_{j,k}$ is positive definite.
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\end{enumerate}
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\end{theorem}
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\begin{refproof}{thm:lec_10thm5}
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Part (a) is obvious.
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% TODO
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For part (b) we have:
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\begin{IEEEeqnarray*}{rCl}
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\sum_{j,k} c_j \overline{c_k} \phi(t_j - t_k) &=& \sum_{j,k} c_j \overline{c_k} \int_\R e^{\i (t_j - t_k) x} d \bP(x)\\
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&=& \int_{\R} \sum_{j,k} c_j \overline{c_k} e^{\i t_j x} \overline{e^{\i t_k x}} d\bP(x)\\
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&=& \int_{\R}\sum_{j,k} c_j e^{\i t_j x} \overline{c_k e^{\i t_k x}} d\bP(x)\\
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&=& \int_{\R} \left| \sum_{l} c_l e^{\i t_l x}\right|^2 \ge 0
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\end{IEEEeqnarray*}
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\end{refproof}
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\begin{theorem}[Bochner's theorem]\label{bochnersthm}
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The converse to \autoref{thm:lec_10thm5} holds, i.e.~
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any $\phi: \R \to \C$ satisfying (a) and (b) of \autoref{thm:lec_10thm5}
|
||||||
|
must be the Fourier transform of a probability measure $\bP$
|
||||||
|
on $(\R, \cB(\R))$.
|
||||||
|
\end{theorem}
|
||||||
|
Unfortunately, we won't prove \autoref{bochnersthm} in this lecture.
|
||||||
|
|
||||||
|
|
||||||
|
\begin{definition}[Convergence in distribution / weak convergence]
|
||||||
|
We say that $\bP_n \subseteq M_1(\R)$ \vocab[Convergence!weak]{converges weakly} towards $\bP \in M_1(\R)$ (notation: $\bP_n \implies \bP$), iff
|
||||||
|
\[
|
||||||
|
\forall f \in C_b(\R)~ \int f d\bP_n \to \int f d\bP.
|
||||||
|
\]
|
||||||
|
Where
|
||||||
|
\[
|
||||||
|
C_b(\R) \coloneqq \{ f: \R \to \R \text{ continuous and bounded}\}
|
||||||
|
\]
|
||||||
|
|
||||||
|
In analysis, this is also known as $\text{weak}^\ast$ convergence.
|
||||||
|
\end{definition}
|
||||||
|
\begin{remark}
|
||||||
|
This notion of convergence makes $M_1(\R)$ a separable metric space. We can construc a metric on $M_1(\R)$ that turns $M_1(\R)$ into a complete
|
||||||
|
and separable metric space:
|
||||||
|
|
||||||
|
Consider the sets
|
||||||
|
\[
|
||||||
|
\{\bP \in M_1(\R): \forall i=1,\ldots,n ~ \int f d \bP - \int f_i d\bP < \epsilon \}
|
||||||
|
\]
|
||||||
|
for any $f,f_1,\ldots, f_n \in C_b(\R)$.
|
||||||
|
These sets form a basis for the topology on $M_1(\R)$.
|
||||||
|
More of this will follow later.
|
||||||
|
\end{remark}
|
||||||
|
\begin{example}
|
||||||
|
\begin{itemize}
|
||||||
|
\item Let $\bP_n = \delta_{\frac{1}{n}}$.
|
||||||
|
Then $\int f d \bP_n = f(\frac{1}{n}) \to f(0) = \int f d \delta_0$
|
||||||
|
for any continuous, bounded function $f$.
|
||||||
|
Hence $\bP_n \to \delta_0$.
|
||||||
|
\item $\bP_n \coloneqq \delta_n$ does not converge weakly,
|
||||||
|
as for example
|
||||||
|
\[
|
||||||
|
\int \cos(\pi x) d\bP_n(x)
|
||||||
|
\]
|
||||||
|
does not converge.
|
||||||
|
|
||||||
|
\item $\bP_n \coloneqq \frac{1}{n} \delta_n + (1- \frac{1}{n}) \delta_0$.
|
||||||
|
Let $f \in C_b(\R)$ arbitrary.
|
||||||
|
Then
|
||||||
|
\[
|
||||||
|
\int f d\bP_n = \frac{1}{n}(n) + (1 - \frac{1}{n}) f(0) \to f(0)
|
||||||
|
\]
|
||||||
|
since $f$ is bounded.
|
||||||
|
Hence $\bP_n \implies \delta_0$.
|
||||||
|
\item $\bP_n \coloneqq \frac{1}{\sqrt{2 \pi n}} e^{-\frac{x^2}{2n}}$.
|
||||||
|
This ``converges'' towards the $0$-measure, which is not a probability measure. Hence $\bP_n$ does not converge weakly.
|
||||||
|
(Exercise) % TODO
|
||||||
|
\end{itemize}
|
||||||
|
\end{example}
|
||||||
|
\begin{definition}
|
||||||
|
We say that a series of random variables $X_n$
|
||||||
|
\vocab[Convergence!in distribution]{converges in distribution}
|
||||||
|
to $X$ (notation: $X_n \xrightarrow{\text{dist}} X$), iff
|
||||||
|
$\bP_n \implies \bP$, where $\bP_n$ is the distribution of $X_n$
|
||||||
|
and $\bP$ is the distribution of $X$.
|
||||||
|
\end{definition}
|
||||||
|
\begin{example}
|
||||||
|
Let $X_n \coloneqq \frac{1}{n}$
|
||||||
|
and $F_n$ the distribution function, i.e.~$F_n = \One_{[\frac{1}{n},\infty)}$.
|
||||||
|
Then $\bP_n = \delta_{\frac{1}{n}} \implies \delta_0$
|
||||||
|
which is the distribution of $X \equiv 0$.
|
||||||
|
But $F_n(0) \centernot\to F(0)$.
|
||||||
|
\end{example}
|
||||||
|
\begin{theorem}
|
||||||
|
$X_n \xrightarrow{\text{dist}} X$ iff
|
||||||
|
$F_n(t) \to F(t)$ for all continuity points $t$ of $F$.
|
||||||
|
\end{theorem}
|
||||||
|
\begin{theorem}[Levy's continuity theorem]\label{levycontinuity}
|
||||||
|
$X_n \xrightarrow{\text{dist}} X$ iff
|
||||||
|
$\phi_{X_n}(t) \to \phi(t)$ for all $t \in \R$.
|
||||||
|
\end{theorem}
|
||||||
|
We will assume these two theorems for now and derive the central limit theorem.
|
||||||
|
The theorems will be proved later.
|
|
@ -1,13 +1,7 @@
|
||||||
|
|
||||||
$(\Omega, \cF, \bP)$ Probability Space, $X : ( \Omega, \cF) \to (\R, \cB(\R))$ random variable.
|
|
||||||
Then $\Q(\cdot) = \bP [ x\in \cdot ]$ is the distribution of $X$ under $\bP$.
|
|
||||||
|
|
||||||
|
|
||||||
\section{Independence and product measures}
|
\section{Independence and product measures}
|
||||||
|
|
||||||
In order to define the notion of independence, we first need to construct
|
In order to define the notion of independence, we first need to construct
|
||||||
product measures in order to be able to consider several random variables
|
product measures.
|
||||||
at the same time.
|
|
||||||
|
|
||||||
The finite case of a product is straightforward:
|
The finite case of a product is straightforward:
|
||||||
\begin{theorem}{Product measure (finite)}
|
\begin{theorem}{Product measure (finite)}
|
||||||
|
@ -44,37 +38,47 @@ We now want to construct a product measure for infinite products.
|
||||||
|
|
||||||
\begin{example}
|
\begin{example}
|
||||||
Suppose we throw a dice twice. Let $A \coloneqq \{\text{first throw even}\}$,
|
Suppose we throw a dice twice. Let $A \coloneqq \{\text{first throw even}\}$,
|
||||||
$B \coloneqq \{second throw even\}$
|
$B \coloneqq \{\text{second throw even}\}$
|
||||||
and $C \coloneqq \{\text{sum even}\} $.
|
and $C \coloneqq \{\text{sum even}\} $.
|
||||||
|
|
||||||
Are $\One_A, \One_B, \One_C$ mutually independent random variables?
|
It is easy the see, that the random variables are pairwise independent,
|
||||||
|
but not mutually independent.
|
||||||
\end{example}
|
\end{example}
|
||||||
It is easy the see, that the random variables are pairwise independent,
|
|
||||||
but not mutually independent.
|
|
||||||
|
|
||||||
The definition of mutual independence can be rephrased as follos:
|
\begin{definition}
|
||||||
Let $X_1, X_2, \ldots, X_n$ r.v.s. Let $\bP[(X_1,\ldots, X_n) \in \cdot ] \text{\reflectbox{$\coloneqq$}} \Q^{\otimes}(\cdot )$.
|
Let $(\Omega, \cF, \bP)$ be a probability space
|
||||||
|
and $X : ( \Omega, \cF) \to (\R, \cB(\R))$ a random variable.
|
||||||
|
Then $\Q(\cdot) \coloneqq \bP [ X \in \cdot ]$ is called the \vocab{distribution}
|
||||||
|
of $X$ under $\bP$.
|
||||||
|
\end{definition}
|
||||||
|
|
||||||
|
Let $X_1, \ldots, X_n$ be random variables and $\Q^{\otimes}(\cdot ) \coloneqq \bP[(X_1,\ldots, X_n) \in \cdot ]$
|
||||||
|
their \vocab{joint distribution}.
|
||||||
Then $\Q^{\otimes}$ is a probability measure on $\R^n$.
|
Then $\Q^{\otimes}$ is a probability measure on $\R^n$.
|
||||||
|
|
||||||
|
The definition of mutual independence can be rephrased as follows:
|
||||||
\begin{fact}
|
\begin{fact}
|
||||||
$X_1,\ldots, X_n$ are mutually independent iff $\Q^{\otimes} = \Q_1 \otimes \ldots \otimes \Q_n$.
|
$X_1,\ldots, X_n$ are mutually independent iff $\Q^{\otimes} = \Q_1 \otimes \ldots \otimes \Q_n$,
|
||||||
|
where $\Q_i$ is the distribution of $X_i$.
|
||||||
|
In this setting, $\Q_i$ is called the \vocab{marginal distribution} of $X_i$.
|
||||||
\end{fact}
|
\end{fact}
|
||||||
By constructing an infinite product, we can thus extend the notion of independence
|
By constructing an infinite product, we can thus extend the notion of independence
|
||||||
to an infinite number of r.v.s.
|
to an infinite number of random variables.
|
||||||
|
|
||||||
\begin{goal}
|
\begin{goal}
|
||||||
Can we construct infinitely many independent random variables?
|
Can we construct infinitely many independent random variables?
|
||||||
\end{goal}
|
\end{goal}
|
||||||
|
|
||||||
\begin{definition}[Consistent family of random variables]
|
\begin{definition}[Consistent family of random variables]
|
||||||
|
\label{def:consistentfamily}
|
||||||
Let $\bP_n, n \in \N$ be a family of probability measures on $(\R^n, \cB(\R^n))$.
|
Let $\bP_n, n \in \N$ be a family of probability measures on $(\R^n, \cB(\R^n))$.
|
||||||
The family is called \vocab{consistent} if if
|
The family is called \vocab{consistent} if
|
||||||
$\bP_{n+1}[B_1 \times B_2 \times \ldots \times B_n \times \R] = \bP_n[B_1 \times \ldots \times B_n]$
|
$\bP_{n+1}[B_1 \times B_2 \times \ldots \times B_n \times \R] = \bP_n[B_1 \times \ldots \times B_n]$
|
||||||
for all $n \in \N, B_i \in B(\R)$.
|
for all $n \in \N, B_i \in B(\R)$.
|
||||||
|
|
||||||
\end{definition}
|
\end{definition}
|
||||||
|
|
||||||
\begin{theorem}[Kolmogorov extension / consistency theorem]
|
\begin{theorem}[Kolmogorov extension / consistency theorem]
|
||||||
|
\label{thm:kolmogorovconsistency}
|
||||||
Informally:
|
Informally:
|
||||||
``Probability measures are determined by finite-dimensional marginals
|
``Probability measures are determined by finite-dimensional marginals
|
||||||
(as long as these marginals are nice)''
|
(as long as these marginals are nice)''
|
||||||
|
@ -96,8 +100,7 @@ to an infinite number of r.v.s.
|
||||||
index sets. However this requires a different notion of consistency.
|
index sets. However this requires a different notion of consistency.
|
||||||
\end{remark}
|
\end{remark}
|
||||||
|
|
||||||
\begin{example}of a consistent family:
|
\begin{example}[A consistent family]
|
||||||
|
|
||||||
Let $F_1, \ldots, F_n$ be probability distribution functions
|
Let $F_1, \ldots, F_n$ be probability distribution functions
|
||||||
and let $\bP_n$ be the probability measure on $\R^n$ defined
|
and let $\bP_n$ be the probability measure on $\R^n$ defined
|
||||||
by
|
by
|
|
@ -1,3 +1,4 @@
|
||||||
|
\todo{Lecture 3 needs to be finished}
|
||||||
\begin{notation}
|
\begin{notation}
|
||||||
Let $\cB_n$ denote $\cB(\R^n)$.
|
Let $\cB_n$ denote $\cB(\R^n)$.
|
||||||
\end{notation}
|
\end{notation}
|
||||||
|
@ -19,7 +20,7 @@ First we need to define $\cB_{\infty}$.
|
||||||
This $\sigma$-algebra must contain all sets $\prod_{n \in \N} B_n$
|
This $\sigma$-algebra must contain all sets $\prod_{n \in \N} B_n$
|
||||||
for all $B_n \in \cB_1$. We simply define $\cB_{\infty}$ to be the
|
for all $B_n \in \cB_1$. We simply define $\cB_{\infty}$ to be the
|
||||||
$\sigma$-algebra
|
$\sigma$-algebra
|
||||||
Let $\cB_\infty \coloneqq \sigma \left( \{\prod_{n \in \N} B_n | B_n \in \cB(\R)\} \right)$.
|
Let \[\cB_\infty \coloneqq \sigma \left( \left\{\prod_{n \in \N} B_n : B_n \in \cB(\R)\right\} \right).\]
|
||||||
\begin{question}
|
\begin{question}
|
||||||
What is there in $\cB_\infty$?
|
What is there in $\cB_\infty$?
|
||||||
Can we identify sets in $\cB_\infty$ for which we can define the product measure
|
Can we identify sets in $\cB_\infty$ for which we can define the product measure
|
||||||
|
@ -41,25 +42,30 @@ Recall the following theorem from measure theory:
|
||||||
\label{caratheodory}
|
\label{caratheodory}
|
||||||
Suppose $\cA$ is an algebra (i.e.~closed under finite union)
|
Suppose $\cA$ is an algebra (i.e.~closed under finite union)
|
||||||
und $\Omega \neq \emptyset$.
|
und $\Omega \neq \emptyset$.
|
||||||
Suppose $\bP$ is countably additive on $\cA$ (i.e.~if $(A_n)_{n}$
|
Suppose $\bP$ is countably additive on $\cA$ (i.e.~if $(A_n)_{n}$
|
||||||
are pairwise disjoint and $\bigcup_{n \in \N} A_n \subseteq \cA $
|
are pairwise disjoint and $\bigcup_{n \in \N} A_n \subseteq \cA $
|
||||||
then $\bP\left( \bigcup_{n \in \N} A_n \right) = \sum_{n \in \N} \bP(A_n)$).
|
then $\bP\left( \bigcup_{n \in \N} A_n \right) = \sum_{n \in \N} \bP(A_n)$).
|
||||||
Then $\bP$ extends uniquely to a probability measure on $(\Omega, \cF)$,
|
Then $\bP$ extends uniquely to a probability measure on $(\Omega, \cF)$,
|
||||||
where $\cF = \sigma(\cA)$.
|
where $\cF = \sigma(\cA)$.
|
||||||
\end{theorem}
|
\end{theorem}
|
||||||
|
|
||||||
Define $\cF = \bigcup_{n \in \N} \cF_n$. Check that $\cF$ is an algebra.
|
Define $\cF = \bigcup_{n \in \N} \cF_n$. Then $\cF$ is an algebra.
|
||||||
We'll show that if we define $\lambda: \cF \to [0,1]$ with
|
We'll show that if we define $\lambda: \cF \to [0,1]$ with
|
||||||
$\lambda(A) = \lambda_n(A)$ for any $n$ where this is well defined,
|
$\lambda(A) = \lambda_n(A)$ for any $n$ where this is well defined,
|
||||||
then $\lambda$ is countably additive on $\cF$.
|
then $\lambda$ is countably additive on $\cF$.
|
||||||
Using \autoref{caratheodory} $\lambda$ will extend uniquely to a probability measure on $\sigma(\cF)$.
|
Using \autoref{caratheodory}, $\lambda$ will extend uniquely to a probability measure on $\sigma(\cF)$.
|
||||||
|
|
||||||
We want to prove:
|
We want to prove:
|
||||||
\begin{enumerate}[(1)]
|
\begin{claim}
|
||||||
\item $\sigma(\cF) = \cB_\infty$,
|
\label{claim:sF=Binfty}
|
||||||
\item $\lambda$ as defined above is countably additive on $\F$.
|
$\sigma(\cF) = \cB_\infty$.
|
||||||
\end{enumerate}
|
\end{claim}
|
||||||
\begin{proof}[Proof of (1)]
|
\begin{claim}
|
||||||
|
\label{claim:lambdacountadd}
|
||||||
|
$\lambda$ as defined above is countably additive on $\cF$.
|
||||||
|
\end{claim}
|
||||||
|
|
||||||
|
\begin{refproof}{claim:sF=Binfty}
|
||||||
Consider an infinite dimensional box $\prod_{n \in \N} B_n$.
|
Consider an infinite dimensional box $\prod_{n \in \N} B_n$.
|
||||||
We have
|
We have
|
||||||
\[
|
\[
|
||||||
|
@ -85,7 +91,7 @@ We want to prove:
|
||||||
Hence $\cF_n \subseteq \cB_\infty$ for all $n$,
|
Hence $\cF_n \subseteq \cB_\infty$ for all $n$,
|
||||||
thus $\cF \subseteq \cB_\infty$. Since $\cB_\infty$ is a $\sigma$-algebra,
|
thus $\cF \subseteq \cB_\infty$. Since $\cB_\infty$ is a $\sigma$-algebra,
|
||||||
$\sigma(\cF) \subseteq \cB_\infty$.
|
$\sigma(\cF) \subseteq \cB_\infty$.
|
||||||
\end{proof}
|
\end{refproof}
|
||||||
We are going to use the following
|
We are going to use the following
|
||||||
\begin{fact}
|
\begin{fact}
|
||||||
\label{fact:finaddtocountadd}
|
\label{fact:finaddtocountadd}
|
||||||
|
@ -97,14 +103,13 @@ We are going to use the following
|
||||||
$\bP(B_n) \to 0$. Then $\bP$ must be countably additive.
|
$\bP(B_n) \to 0$. Then $\bP$ must be countably additive.
|
||||||
\end{fact}
|
\end{fact}
|
||||||
\begin{proof}
|
\begin{proof}
|
||||||
Exercise
|
Exercise. % TODO
|
||||||
|
|
||||||
\end{proof}
|
\end{proof}
|
||||||
\begin{proof}[Proof of (2)]
|
\begin{refproof}{claim:lambdacountadd}
|
||||||
Let's prove that $\lambda$ is finitely additive.
|
Let us prove that $\lambda$ is finitely additive.
|
||||||
$\lambda(\R^\infty) = \lambda_1(\R^\infty) = 1$.
|
We have $\lambda(\R^\infty) = \lambda_1(\R^\infty) = 1$ and
|
||||||
$\lambda(\emptyset) = \lambda_1(\emptyset) = 0$.
|
$\lambda(\emptyset) = \lambda_1(\emptyset) = 0$.
|
||||||
Suppose $A_1, A_2 \in \cF$ are disjoint.
|
Suppose that $A_1, A_2 \in \cF$ are disjoint.
|
||||||
Then pick some $n$ such that $A_1, A_2 \in \cF_n$.
|
Then pick some $n$ such that $A_1, A_2 \in \cF_n$.
|
||||||
Take $C_1, C_2 \in \cB_n$ such that $C_1^\ast = A_1$
|
Take $C_1, C_2 \in \cB_n$ such that $C_1^\ast = A_1$
|
||||||
and $C_2^\ast = A_2$.
|
and $C_2^\ast = A_2$.
|
||||||
|
@ -113,7 +118,8 @@ We are going to use the following
|
||||||
by the definition of the finite product measure.
|
by the definition of the finite product measure.
|
||||||
|
|
||||||
In order to use \autoref{fact:finaddtocountadd},
|
In order to use \autoref{fact:finaddtocountadd},
|
||||||
we need to show that if $B_n \in \cF$ with $B_n \to \emptyset \implies \lambda(B_n) \to 0$.
|
we need to show that for any sequence $B_n \in \cF$ with $B_n \xrightarrow{n\to \infty} \emptyset$
|
||||||
|
we have $\lambda(B_n) \xrightarrow{n \to \infty} 0$.
|
||||||
|
\todo{Finish this}
|
||||||
%TODO
|
%TODO
|
||||||
\end{proof}
|
\end{refproof}
|
||||||
|
|
1
inputs/lecture_4.tex
Normal file
1
inputs/lecture_4.tex
Normal file
|
@ -0,0 +1 @@
|
||||||
|
\todo{Lecture 4 missing}
|
|
@ -1,3 +1,7 @@
|
||||||
|
% Lecture 5 2023-04-21
|
||||||
|
\subsection{The laws of large numbers}
|
||||||
|
|
||||||
|
|
||||||
We want to show laws of large numbers:
|
We want to show laws of large numbers:
|
||||||
The LHS is random and represents ``sane'' averaging.
|
The LHS is random and represents ``sane'' averaging.
|
||||||
The RHS is constant, which we can explicitly compute from the distribution of the RHS.
|
The RHS is constant, which we can explicitly compute from the distribution of the RHS.
|
|
@ -1,3 +1,4 @@
|
||||||
|
\todo{Large parts of lecture 6 are missing}
|
||||||
\begin{refproof}{lln}
|
\begin{refproof}{lln}
|
||||||
We want to deduce the SLLN (\autoref{lln}) from \autoref{thm2}.
|
We want to deduce the SLLN (\autoref{lln}) from \autoref{thm2}.
|
||||||
W.l.o.g.~let us assume that $\bE[X_i] = 0$ (otherwise define $X'_i \coloneqq X_i - \bE[X_i]$).
|
W.l.o.g.~let us assume that $\bE[X_i] = 0$ (otherwise define $X'_i \coloneqq X_i - \bE[X_i]$).
|
||||||
|
@ -21,8 +22,7 @@
|
||||||
Thus $a_1 + \ldots + a_n = n S_n - (S1 $ % TODO
|
Thus $a_1 + \ldots + a_n = n S_n - (S1 $ % TODO
|
||||||
|
|
||||||
\end{subproof}
|
\end{subproof}
|
||||||
The claim implies SLLN.
|
The SLLN follows from the claim.
|
||||||
|
|
||||||
\end{refproof}
|
\end{refproof}
|
||||||
|
|
||||||
We need the following inequality:
|
We need the following inequality:
|
||||||
|
@ -58,7 +58,7 @@ We need the following inequality:
|
||||||
|
|
||||||
|
|
||||||
|
|
||||||
\paragraph{Application of SLLN}
|
\subsubsection{Application: Renewal Theorem}
|
||||||
|
|
||||||
\begin{theorem}[Renewal theorem]
|
\begin{theorem}[Renewal theorem]
|
||||||
Let $X_1,X_2,\ldots$ i.i.d.~random variables with $X_i \ge 0$, $\bE[X_i] = m > 0$. The $X_i$ model waiting times.
|
Let $X_1,X_2,\ldots$ i.i.d.~random variables with $X_i \ge 0$, $\bE[X_i] = m > 0$. The $X_i$ model waiting times.
|
179
inputs/lecture_7.tex
Normal file
179
inputs/lecture_7.tex
Normal file
|
@ -0,0 +1,179 @@
|
||||||
|
% TODO \begin{goal}
|
||||||
|
% TODO We want to drop our assumptions on finite mean or variance
|
||||||
|
% TODO and say something about the behaviour of $ \sum_{n \ge 1} X_n$
|
||||||
|
% TODO when the $X_n$ are independent.
|
||||||
|
% TODO \end{goal}
|
||||||
|
\begin{theorem}[Theorem 3, Kolmogorov's three-series theorem] % Theorem 3
|
||||||
|
\label{thm3}
|
||||||
|
Let $X_n$ be a family of independent random variables.
|
||||||
|
\begin{enumerate}[(a)]
|
||||||
|
\item Suppose for some $C \ge 0$, the following three series
|
||||||
|
of numbers converge:
|
||||||
|
\begin{itemize}
|
||||||
|
\item $\sum_{n \ge 1} \bP(|X_n| > C)$,
|
||||||
|
\item $\sum_{n \ge 1} \underbrace{\int_{|X_n| \le C} X_n d\bP}_{\text{\vocab{truncated mean}}}$,
|
||||||
|
\item $\sum_{n \ge 1} \underbrace{\int_{|X_n| \le C} X_n^2 d\bP - \left( \int_{|X_n| \le C} X_n d\bP \right)^2}_{\text{\vocab{truncated variance} }}$.
|
||||||
|
\end{itemize}
|
||||||
|
Then $\sum_{n \ge 1} X_n$ converges almost surely.
|
||||||
|
\item Suppose $\sum_{n \ge 1} X_n$ converges almost surely.
|
||||||
|
Then all three series above converge for every $C > 0$.
|
||||||
|
\end{enumerate}
|
||||||
|
\end{theorem}
|
||||||
|
For the proof we'll need a slight generalization of \autoref{thm2}:
|
||||||
|
\begin{theorem}[Theorem 4] % Theorem 4
|
||||||
|
\label{thm4}
|
||||||
|
Let $\{X_n\}_n$ be independent and \vocab{uniformly bounded}
|
||||||
|
(i.e. $\exists M < \infty : \sup_n \sup_\omega |X_n(\omega)| \le M$).
|
||||||
|
Then $\sum_{n \ge 1} X_n$ converges almost surely
|
||||||
|
$\iff$ $\sum_{n \ge 1} \bE(X_n)$ and $\sum_{n \ge 1} \Var(X_n)$
|
||||||
|
converge.
|
||||||
|
\end{theorem}
|
||||||
|
\begin{refproof}{thm3}
|
||||||
|
Assume, that we have already proved \autoref{thm4}.
|
||||||
|
We prove part (a) first.
|
||||||
|
Put $Y_n = X_n \cdot \One_{\{|X_n| \le C\}}$.
|
||||||
|
Since the $X_n$ are independent, the $Y_n$ are independent as well.
|
||||||
|
Furthermore, the $Y_n$ are uniformly bounded.
|
||||||
|
By our assumption, the series
|
||||||
|
$\sum_{n \ge 1} \int_{|X_n| \le C} X_n d\bP = \sum_{n \ge 1} \bE[Y_n]$
|
||||||
|
and $\sum_{n \ge 1} \int_{|X_n| \le C} X_n^2 d\bP - \left( \int_{|X_n| \le C} X_n d\bP \right)^2 = \sum_{n \ge 1} \Var(Y_n)$
|
||||||
|
converges.
|
||||||
|
By \autoref{thm4} it follows that $\sum_{n \ge 1} Y_n < \infty$
|
||||||
|
almost surely.
|
||||||
|
Let $A_n \coloneqq \{\omega : |X_n(\omega)| > C\}$.
|
||||||
|
Since the first series $\sum_{n \ge 1} \bP(A_n) < \infty$,
|
||||||
|
by Borel-Cantelli, $\bP[\text{infinitely many $A_n$ occcur}] = 0$.
|
||||||
|
|
||||||
|
|
||||||
|
For the proof of (b), suppose $\sum_{n\ge 1} X_n(\omega) < \infty$
|
||||||
|
for almost every $\omega$.
|
||||||
|
Fix an arbitrary $C > 0$.
|
||||||
|
Define
|
||||||
|
\[
|
||||||
|
Y_n(\omega) \coloneqq \begin{cases}
|
||||||
|
X_n(\omega) & \text{if} |X_n(\omega)| \le C,\\
|
||||||
|
C &\text{if } |X_n(\omega)| > C.
|
||||||
|
\end{cases}
|
||||||
|
\]
|
||||||
|
Then the $Y_n$ are independent and $\sum_{n \ge 1} Y_n(\omega) < \infty$
|
||||||
|
almost surely and the $Y_n$ are uniformly bounded.
|
||||||
|
By \autoref{thm4} $\sum_{n \ge 1} \bE[Y_n]$ and $\sum_{n \ge 1} \Var(Y_n)$
|
||||||
|
converge.
|
||||||
|
Define
|
||||||
|
\[
|
||||||
|
Z_n(\omega) \coloneqq \begin{cases}
|
||||||
|
X_n(\omega) &\text{if } |X_n| \le C,\\
|
||||||
|
-C &\text{if } |X_n| > C.
|
||||||
|
\end{cases}
|
||||||
|
\]
|
||||||
|
Then the $Z_n$ are independent, uniformly bounded and $\sum_{n \ge 1} Z_n(\omega) < \infty$
|
||||||
|
almost surely.
|
||||||
|
By \autoref{thm4} we have
|
||||||
|
$\sum_{n \ge 1} \bE(Z_n) < \infty$
|
||||||
|
and $\sum_{n \ge 1} \Var(Z_n) < \infty$.
|
||||||
|
|
||||||
|
We have
|
||||||
|
\begin{IEEEeqnarray*}{rCl}
|
||||||
|
\bE(Y_n) &=& \int_{|X_n| \le C} X_n d \bP + C \bP(|X_n| \ge C),\\
|
||||||
|
\bE(Z_n) &=& \int_{|X_n| \le C} X_n d \bP - C \bP(|X_n| \ge C).
|
||||||
|
\end{IEEEeqnarray*}
|
||||||
|
Since $\bE(Y_n) + \bE(Z_n) = 2 \int_{|X_n| \le C} X_n d\bP$
|
||||||
|
the second series converges,
|
||||||
|
and since
|
||||||
|
$\bE(Y_n) - \bE(Z_n)$ converges, the first series converges.
|
||||||
|
For the third series, we look at
|
||||||
|
$\sum_{n \ge 1} \Var(Y_n)$ and
|
||||||
|
$\sum_{n \ge 1} \Var(Z_n)$ to conclude that this series converges
|
||||||
|
as well.
|
||||||
|
\end{refproof}
|
||||||
|
Recall \autoref{thm2}.
|
||||||
|
We will see, that the converse of \autoref{thm2} is true if the $X_n$ are uniformly bounded.
|
||||||
|
More formally:
|
||||||
|
\begin{theorem}[Theorem 5]
|
||||||
|
\label{thm5}
|
||||||
|
Let $X_n$ be a series of independent variables with mean $0$,
|
||||||
|
that are uniformly bounded.
|
||||||
|
If $\sum_{n \ge 1} X_n < \infty$ almost surely,
|
||||||
|
then $\sum_{n \ge 1} \Var(X_n) < \infty$.
|
||||||
|
\end{theorem}
|
||||||
|
\begin{refproof}{thm4}
|
||||||
|
Assume we have proven \autoref{thm5}.
|
||||||
|
|
||||||
|
``$\impliedby$'' Assume $\{X_n\} $ are independent, uniformly bounded
|
||||||
|
and $\sum_{n \ge 1} \bE(X_n) < \infty$ as well as $\sum_{n \ge 1} \Var(X_n) < \infty$.
|
||||||
|
We need to show that $\sum_{n \ge 1} X_n < \infty$ a.s.
|
||||||
|
Let $Y_n \coloneqq X_n - \bE(X_n)$.
|
||||||
|
Then the $Y_n$ are independent, $\bE(Y_n) = 0$ and $\Var(Y_n) = \Var(X_n)$.
|
||||||
|
By \autoref{thm2} $\sum_{n \ge 1} Y_n < \infty$ a.s.
|
||||||
|
Thus $\sum_{n \ge 1} X_n < \infty$ a.s.
|
||||||
|
|
||||||
|
``$\implies$'' We assume that $\{X_n\}$ are independent, uniformly bounded
|
||||||
|
and $\sum_{n \ge 1} X_n(\omega) < \infty$ a.s.
|
||||||
|
We have to show that $\sum_{n \ge 1} \bE(X_n) < \infty$
|
||||||
|
and $\sum_{n \ge 1} \Var(X_n) < \infty$.
|
||||||
|
|
||||||
|
Consider the product space $(\Omega, \cF, \bP) \otimes (\Omega, \cF, \bP)$.
|
||||||
|
On this product space, we define
|
||||||
|
$Y_n \left( (\omega, \omega') \right) \coloneqq X_n(\omega)$
|
||||||
|
and $Z_n \left( (\omega, \omega') \right) \coloneqq X_n(\omega')$.
|
||||||
|
\begin{claim}
|
||||||
|
For every fixed $n$, $Y_n$ and $Z_n$ are independent.
|
||||||
|
\end{claim}
|
||||||
|
\begin{subproof}
|
||||||
|
This is obvious, but well prove it carefully here.
|
||||||
|
\begin{IEEEeqnarray*}{rCl}
|
||||||
|
&&(\bP \otimes \bP) [Y_n \in (a,b) , Z_n \in (a',b') ]\\
|
||||||
|
&=& (\bP\otimes\bP) \left( (\omega, \omega') : X_n(\omega) \in (a,b) \land X_n(\omega') \in (a',b') \right)\\
|
||||||
|
&=& (\bP\otimes\bP)(A \times A') \text{where }
|
||||||
|
A \coloneqq X_n^{-1}\left( (a,b)\right) \text{ and } A' \coloneqq X_n^{-1}\left( (a',b') \right)\\
|
||||||
|
&=& \bP(A)\bP(A')
|
||||||
|
\end{IEEEeqnarray*}
|
||||||
|
\end{subproof}
|
||||||
|
Now $\bE[Y_n - Z_n] = 0$ (by definition) and $\Var(Y_n - Z_n) = 2\Var(X_n)$.
|
||||||
|
Obviously, $(Y_n - Z_n)_{n \ge 1}$ is also uniformly bounded.
|
||||||
|
\begin{claim}
|
||||||
|
$\sum_{n \ge 1} (Y_n - Z_n) < \infty$ almost surely
|
||||||
|
on $(\Omega \otimes \Omega, \cF \otimes\cF, \bP \otimes\bP)$.
|
||||||
|
\end{claim}
|
||||||
|
\begin{subproof}
|
||||||
|
Suppose $\Omega_0 = \{\omega: \sum_{n \ge 1} X_n(\omega) < \infty\}$.
|
||||||
|
Then $\bP(\Omega_0) = 1$.
|
||||||
|
Thus $(\bP\otimes\bP)(\Omega_0 \otimes \Omega_0) = 1$.
|
||||||
|
Furthermore
|
||||||
|
$\sum_{n \ge 1} \left(Y_n(\omega, \omega') - Z_n(\omega, \omega') \right)= \sum_{n \ge 1} \left(X_n(\omega) - X_n(\omega')\right)$.
|
||||||
|
Thus $\sum_{n \ge 1} \left( Y_n(\omega, \omega') - Z_n(\omega, \omega') \right) < \infty$ a.s.~on $\Omega_0\otimes\Omega_0$.
|
||||||
|
\end{subproof}
|
||||||
|
By \autoref{thm5}, $\sum_{n} \Var(X_n) = \frac{1}{2}\sum_{n \ge 1} \Var(Y_n - Z_n) < \infty$ a.s.
|
||||||
|
Define $U_n \coloneqq X_n - \bE(X_n)$.
|
||||||
|
Then $\bE(U_n) = 0$ and the $U_n$ are independent
|
||||||
|
and uniformly bounded.
|
||||||
|
We have $\sum_{n} \Var(U_n) = \sum_{n} \Var(X_n) < \infty$.
|
||||||
|
Thus $\sum_{n} U_n$ converges a.s.~by \autoref{thm2}.
|
||||||
|
Since by assumption $\sum_{n} X_n < \infty$ a.s.,
|
||||||
|
it follows that $\sum_{n} \bE(X_n) < \infty$.
|
||||||
|
\end{refproof}
|
||||||
|
\begin{remark}
|
||||||
|
In the proof of \autoref{thm4}
|
||||||
|
``$\impliedby$'' is just a trivial application of \autoref{thm2}
|
||||||
|
and uniform boundedness was not used.
|
||||||
|
The idea of `` $\implies$ '' will lead to coupling. % TODO ?
|
||||||
|
\end{remark}
|
||||||
|
|
||||||
|
% TODO Proof of thm5 in the notes
|
||||||
|
\begin{example}[Application of \autoref{thm5}]
|
||||||
|
The series $\sum_{n} \frac{1}{n^{\frac{1}{2} + \epsilon}}$
|
||||||
|
does not converge for $\epsilon < \frac{1}{2}$.
|
||||||
|
However
|
||||||
|
\[
|
||||||
|
\sum_{n} X_n \frac{1}{n^{\frac{1}{2} + \epsilon}}
|
||||||
|
\]
|
||||||
|
where $\bP[X_n = 1] = \bP[X_n = -1] = \frac{1}{2}$
|
||||||
|
converges almost surely for all $\epsilon > 0$.
|
||||||
|
And
|
||||||
|
\[
|
||||||
|
\sum_{n} X_n \frac{1}{n^{\frac{1}{2} - \epsilon}}
|
||||||
|
\]
|
||||||
|
does not converge.
|
||||||
|
|
||||||
|
|
||||||
|
\end{example}
|
150
inputs/lecture_8.tex
Normal file
150
inputs/lecture_8.tex
Normal file
|
@ -0,0 +1,150 @@
|
||||||
|
% Lecture 8 2023-05-02
|
||||||
|
\subsection{Kolmogorov's 0-1-law}
|
||||||
|
Some classes of events always have probability $0$ or $1$.
|
||||||
|
One example of such a 0-1-law is the Borel-Cantelli Lemma
|
||||||
|
and its inverse statement.
|
||||||
|
|
||||||
|
We now want to look at events that capture certain aspects of long term behaviour
|
||||||
|
of sequences of random variables.
|
||||||
|
|
||||||
|
\begin{definition}
|
||||||
|
Let $X_n, n \in \N$ be a sequence of random variables
|
||||||
|
on a probability space $(\Omega, \cF, \bP)$.
|
||||||
|
Let $\cT_i \coloneqq \sigma(X_i, X_{i+1}, \ldots )$
|
||||||
|
be the $\sigma$-algebra generated by $X_i, X_{i+1}, \ldots$.
|
||||||
|
Then the \vocab{tail-$\sigma$-algebra} is defined as
|
||||||
|
\[
|
||||||
|
\cT \coloneqq \bigcap_{i \in \N} \cT_i.
|
||||||
|
\]
|
||||||
|
The events $A \in \cT \subseteq \cF$ are called \vocab[Tail event]{tail events}.
|
||||||
|
\end{definition}
|
||||||
|
\begin{remark}
|
||||||
|
\begin{enumerate}[(i)]
|
||||||
|
\item Since intersections of arbitrarily many $\sigma$-algebras
|
||||||
|
is again a $\sigma$-algebra, $\cT$ is indeed a $\sigma$-algebra.
|
||||||
|
\item We have
|
||||||
|
\[
|
||||||
|
\cT = \{A \in \cF ~|~ \forall i \exists B \in \cF^{\otimes \N} : A = \{\omega | (X_i(\omega), X_{i+1}(\omega), \ldots) \in B\} \}. % TODO?
|
||||||
|
\]
|
||||||
|
\end{enumerate}
|
||||||
|
\end{remark}
|
||||||
|
\begin{example}[What are tail events?]
|
||||||
|
Let $X_n, n \in \N$ be a sequence of independent random variables on a probability
|
||||||
|
space $(\Omega, \cF, \bP)$. Then
|
||||||
|
\begin{enumerate}[(i)]
|
||||||
|
\item $\left\{\omega | \sum_{n \in \N} X_n(\omega) \text{ converges} \right\}$ is a tail event,
|
||||||
|
since for all $\omega \in \Omega$ we have
|
||||||
|
\begin{IEEEeqnarray*}{rCl}
|
||||||
|
&& \sum_{i=1}^\infty X_i(\omega) \text{ converges}\\
|
||||||
|
&\iff& \sum_{i=2}^\infty X_i(\omega) \text{ converges}\\
|
||||||
|
&\iff& \ldots \\
|
||||||
|
&\iff& \sum_{i=k}^\infty X_i(\omega) \text{ converges}.\\
|
||||||
|
\end{IEEEeqnarray*}
|
||||||
|
(Since the $X_i$ are independent, the convergence
|
||||||
|
of $\sum_{n \in \N} X_n$ is not influenced by $X_1,\ldots, X_k$
|
||||||
|
for any $k$.)
|
||||||
|
\item $\left\{\omega | \sum_{n \in \N} X_n(\omega) = c\right\} $
|
||||||
|
for some $c \in \R$
|
||||||
|
is not a tail event,
|
||||||
|
because $\sum_{n \in \N} X_n$ depends on $X_1$.
|
||||||
|
\item $\{\omega | \lim_{n \to \infty} \frac{1}{n} \sum_{i=1}^{n} X_i(\omega) = c\}$
|
||||||
|
is a tail event, since
|
||||||
|
\[
|
||||||
|
c = \lim_{n \to \infty} \sum_{i=1}^{n} X_i = \underbrace{\lim_{n \to \infty} \frac{1}{n} X_1}_{= 0} + \lim_{n \to \infty} \frac{1}{n} \sum_{i=2}^n X_i = \ldots = \lim_{n \to \infty} \frac{1}{n} \sum_{i=k}^n X_i.
|
||||||
|
\]
|
||||||
|
\end{enumerate}
|
||||||
|
\end{example}
|
||||||
|
So $\cT$ includes all long term behaviour of $X_n, n \in \N$,
|
||||||
|
which does not depend on the realisation of the first $k$ random variables
|
||||||
|
for any $k \in \N$.
|
||||||
|
|
||||||
|
\begin{theorem}[Kolmogorov's 0-1 law]
|
||||||
|
\label{kolmogorov01}
|
||||||
|
Let $X_n, n \in \N$ be a sequence of independent random variables
|
||||||
|
and let $\cT$ denote their tail-$\sigma$-algebra.
|
||||||
|
Then $\cT$ is \vocab{$\bP$-trivial}, i.e.~$\bP[A] \in \{0,1\}$
|
||||||
|
for all $A \in \cT$.
|
||||||
|
\end{theorem}
|
||||||
|
\begin{idea}
|
||||||
|
The idea behind proving, that a $\cT$ is $\bP$-trivial is to show that
|
||||||
|
for any $A, B \in \cF$ we have
|
||||||
|
\[
|
||||||
|
\bP[A \cap B] = \bP[A] \cdot \bP[B].
|
||||||
|
\]
|
||||||
|
Taking $A = B$, it follows that $\bP[A] = \bP[A]^2$, hence $\bP[A] \in \{0,1\}$.
|
||||||
|
\end{idea}
|
||||||
|
\begin{refproof}{kolmogorov01}
|
||||||
|
Let $\cF_n \coloneqq \sigma(X_1,\ldots,X_n)$
|
||||||
|
and remember that $\cT_{n} = \sigma(X_{n}, X_{n+1},\ldots)$.
|
||||||
|
The proof rests on two claims:
|
||||||
|
\begin{claim}
|
||||||
|
For all $n \ge 1$, $A \in \cF_n$ and $B \in \cT_{n+1}$
|
||||||
|
we have $\bP[A \cap B] = \bP[A]\bP[B]$.
|
||||||
|
\end{claim}
|
||||||
|
\begin{subproof}
|
||||||
|
This follows from the independence of the $X_i$.
|
||||||
|
It is
|
||||||
|
\[
|
||||||
|
\sigma\left( X_1,\ldots,X_n \right) = \sigma\left(\underbrace{\{X_{1}^{-1}(B_1) \cap \ldots \cap X_n^{-1}(B_n)\} | B_1,\ldots,B_n \in \cB(\R)\}}_{\text{\reflectbox{$\coloneqq$}}\cA} \right).
|
||||||
|
\]
|
||||||
|
$\cA$ is a semi-algebra, since
|
||||||
|
\begin{enumerate}[(i)]
|
||||||
|
\item $\emptyset, \Omega \in \cA$,
|
||||||
|
\item $A, B \in \cA \implies A \cap B \in \cA$,
|
||||||
|
\item for $A \in \cA$, $A^c = \bigsqcup_{i=1}^n A_i$
|
||||||
|
for disjoint sets $A_1,\ldots,A_n \in \cA$.
|
||||||
|
\end{enumerate}
|
||||||
|
Hence it suffices to show the claim for sets $A \in \cA$.
|
||||||
|
Similarly
|
||||||
|
\[
|
||||||
|
\sigma(\cT_{n+1}) = \sigma \left( \underbrace{ \{X_{n+1}^{-1}(M_1) \cap \ldots \cap X_{n+k}^{-1}(M_k) | k \in \N, M_1,\ldots, M_k \in \cB(\R)\}}_{\text{\reflectbox{$\coloneqq$}} \cB} \right).
|
||||||
|
\]
|
||||||
|
Again, $\cB$ is closed under intersection.
|
||||||
|
|
||||||
|
So let $A \in \cA$ and $B \in \cB$.
|
||||||
|
Then
|
||||||
|
\[
|
||||||
|
\bP[A \cap B] = \bP[A] \cdot \bP[B)
|
||||||
|
\]
|
||||||
|
by the independence of $\{X_1,\ldots,X_{n+k}\}$,
|
||||||
|
and since $A$ only depends on $\{X_1,\ldots,X_n\}$
|
||||||
|
and $B$ only on $\{X_{n+1},\ldots, X_{n+k}\}$.
|
||||||
|
\end{subproof}
|
||||||
|
\begin{claim}
|
||||||
|
$\bigcup_{n \in \N} \cF_n$ is an algebra
|
||||||
|
and
|
||||||
|
\[
|
||||||
|
\sigma\left( \bigcup_{n \in \N} \cF_n \right) = \sigma(X_1,X_2,\ldots) = \cT_1.
|
||||||
|
\]
|
||||||
|
\end{claim}
|
||||||
|
\begin{subproof}
|
||||||
|
``$\supseteq$ '' If $A_n \in \sigma(X_n)$, then $A_n \in \cF_n$.
|
||||||
|
Hence $A_n \in \bigcup_{n \in \N} \cF_n$.
|
||||||
|
|
||||||
|
Since $\sigma(X_1,X_2,\ldots)$ is generated by $\{A_n \in \sigma(X_n) : n \in \N\}$,
|
||||||
|
this also means $\sigma(X_1,X_2,\ldots) \subseteq\sigma\left( \bigcup_{n \in \N} \cF_n \right)$.
|
||||||
|
|
||||||
|
``$\subseteq$ '' Since $\cF_n = \sigma(X_1,\ldots,X_n)$,
|
||||||
|
obviously $\cF_n \subseteq \sigma(X_1,\ldots,X_n)$
|
||||||
|
for all $n$.
|
||||||
|
It follows that $\bigcup_{n \in \N} \cF_n \subseteq \sigma(X_1,X_2,\ldots)$.
|
||||||
|
Hence $\sigma\left( \bigcup_{n \in \N} \cF_n \right) \subseteq\sigma(X_1,X_2,\ldots)$.
|
||||||
|
\end{subproof}
|
||||||
|
|
||||||
|
Now let $T \in \cT$.
|
||||||
|
Then $T \in \cT_{n+1}$ for any $n$.
|
||||||
|
Hence $\bP[A \cap T] = \bP[A] \bP[T]$
|
||||||
|
for all $A \in \cF_n$ by the first claim.
|
||||||
|
|
||||||
|
It follows that the same folds for all $A \in \bigcup_{n \in \N} \cF_n$,
|
||||||
|
hence for all $A \in \sigma\left( \bigcup_{n \in \N} \cF_n \right)$,
|
||||||
|
and by the second claim for all $A \in \sigma(X_1,X_2,\ldots) = \cT_1$.
|
||||||
|
But since $T \in \cT$, in particular $T \in \cT_1$,
|
||||||
|
so by choosing $A = T$, we get
|
||||||
|
\[
|
||||||
|
\bP[T] = \bP[T \cap T] = \bP[T]^2
|
||||||
|
\]
|
||||||
|
hence $\bP[T] \in \{0,1\}$.
|
||||||
|
\end{refproof}
|
||||||
|
|
||||||
|
|
146
inputs/lecture_9.tex
Normal file
146
inputs/lecture_9.tex
Normal file
|
@ -0,0 +1,146 @@
|
||||||
|
\subsubsection{Application: Percolation}
|
||||||
|
|
||||||
|
|
||||||
|
We will now discuss another application of Kolmogorov's $0-1$-law, percolation.
|
||||||
|
|
||||||
|
\begin{definition}[\vocab{Percolation}]
|
||||||
|
Consider the graph with nodes $\Z^d$, $d \ge 2$, where edges from the lattice are added with probability $p$. The added edges are called \vocab[Percolation!Edge!open]{open};
|
||||||
|
all other edges are called
|
||||||
|
\vocab[Percolation!Edge!closed]{closed}.
|
||||||
|
|
||||||
|
More formally, we consider
|
||||||
|
\begin{itemize}
|
||||||
|
\item $\Omega = \{0,1\}^{\bE_d}$, where $\bE_d$ are all edges in $\Z^d$,
|
||||||
|
\item $\cF \coloneqq \text{product $\sigma$-algebra}$,
|
||||||
|
\item $\bP \coloneqq \left(p \underbrace{\delta_{\{1\} }}_{\text{edge is open}} + (1-p) \underbrace{\delta_{\{0\} }}_{\text{edge is absent closed}}\right)^{\otimes \bE_d}$.
|
||||||
|
\end{itemize}
|
||||||
|
\end{definition}
|
||||||
|
\begin{question}
|
||||||
|
Starting at the origin, what is the probability, that there exists
|
||||||
|
an infinite path (without moving backwards)?
|
||||||
|
\end{question}
|
||||||
|
\begin{definition}
|
||||||
|
An \vocab{infinite path} consists of an infinite sequence of distinct points
|
||||||
|
$x_0, x_1, \ldots$
|
||||||
|
such that $x_n$ is connected to $x_{n+1}$, i.e.~the edge $\{x_n, x_{n+1}\}$ is open.
|
||||||
|
\end{definition}
|
||||||
|
|
||||||
|
Let $C_\infty \coloneqq \{\omega | \text{an infinite path exists}\}$.
|
||||||
|
|
||||||
|
\begin{exercise}
|
||||||
|
Show that changing the presence / absence of finitely many edges
|
||||||
|
does not change the existence of an infinite path.
|
||||||
|
Therefore $C_\infty$ is an element of the tail $\sigma$-algebra.
|
||||||
|
Hence $\bP(C_\infty) \in \{0,1\}$.
|
||||||
|
\end{exercise}
|
||||||
|
Obviously, $\bP(C_\infty)$ is monotonic with respect to $p$.
|
||||||
|
For $d = 2$ it is known that $p = \frac{1}{2}$ is the critical value.
|
||||||
|
For $d > 2$ this is unknown.
|
||||||
|
|
||||||
|
% TODO: more in the notes
|
||||||
|
We'll get back to percolation later.
|
||||||
|
|
||||||
|
|
||||||
|
\section{Characteristic functions, weak convergence and the central limit theorem}
|
||||||
|
|
||||||
|
Characteristic functions are also known as the \vocab{Fourier transform}.
|
||||||
|
Weak convergence is also known as \vocab{convergence in distribution} / \vocab{convergence in law}.
|
||||||
|
We will abbreviate the central limit theorem by \vocab{CLT}.
|
||||||
|
|
||||||
|
So far we have dealt with the average behaviour,
|
||||||
|
\[
|
||||||
|
\frac{\overbrace{X_1 + \ldots + X_n}^{\text{i.i.d.}}}{n} \to \bE(X_1).
|
||||||
|
\]
|
||||||
|
We now want to understand \vocab{fluctuations} from the average behaviour,
|
||||||
|
i.e.\[
|
||||||
|
X_1 + \ldots + X_n - n \cdot \bE(X_1).
|
||||||
|
\]
|
||||||
|
% TODO improve
|
||||||
|
The question is, what happens on other timescales than $n$?
|
||||||
|
An example is
|
||||||
|
\[
|
||||||
|
\frac{X_1 + \ldots + X_n - n \bE(X_1)}{\sqrt{n} } \xrightarrow{n \to \infty} hv \cN(0, \Var(X_i)) (\ast)
|
||||||
|
\]
|
||||||
|
Why is $\sqrt{n}$ the right order? (Handwavey argument)
|
||||||
|
|
||||||
|
Suppose $X_1, X_2,\ldots$ are i.i.d. $\cN(0,1)$.
|
||||||
|
The mean of the l.h.s.~is $0$ and for the variance we get
|
||||||
|
\[
|
||||||
|
\Var(\frac{X_1 + \ldots + X_n - n \bE(X_1)}{\sqrt{n} }) = \Var\left( \frac{X_1+ \ldots + X_n}{\sqrt{n} } \right) = \frac{1}{n} \left( \Var(X_1) + \ldots + \Var(X_n) \right) = 1
|
||||||
|
\]
|
||||||
|
For the r.h.s.~we get a mean of $0$ and a variance of $1$.
|
||||||
|
So, to determine what $(\ast)$ could mean, it is necessary that $\sqrt{n}$
|
||||||
|
is the right scaling.
|
||||||
|
To define $(\ast)$ we need another notion of convergence.
|
||||||
|
This will be the weakest notion of convergence, hence it is called
|
||||||
|
\vocab{weak convergence}.
|
||||||
|
This notion of convergence will be defined in terms of characteristic functions of Fourier transforms.
|
||||||
|
|
||||||
|
\subsection{Characteristic functions and Fourier transform}
|
||||||
|
|
||||||
|
Consider $(\R, \cB(\R), \bP)$.
|
||||||
|
For every $t \in \R$ define a function $\phi(t) \coloneqq \phi_\bP(t) \coloneqq \int_{\R} e^{\i t x} \bP(dx)$.
|
||||||
|
We have
|
||||||
|
\[
|
||||||
|
\phi(t) = \int_{\R} \cos(tx) \bP(dx) + \i \int_{\R} \sin(tx) \bP(dx).
|
||||||
|
\]
|
||||||
|
\begin{itemize}
|
||||||
|
\item Since $|e^{\i t x}| \le 1$ the function $\phi(\cdot )$ is always defined.
|
||||||
|
\item We have $\phi(0) = 1$.
|
||||||
|
\item $|\phi(t)| \le \int_{\R} |e^{\i t x} | \bP(dx) = 1$.
|
||||||
|
\end{itemize}
|
||||||
|
We call $\phi_{\bP}$ the \vocab{characteristic function} of $\bP$.
|
||||||
|
|
||||||
|
\begin{remark}
|
||||||
|
Suppose $(\Omega, \cF, \bP)$ is an arbitrary probability space and
|
||||||
|
$X: (\Omega, \cF) \to (\R, \cB(\R))$ is a random variable.
|
||||||
|
Then we can define
|
||||||
|
\[
|
||||||
|
\phi_X(t) \coloneqq \bE[e^{\i t x}] = \int e^{\i t X(\omega)} \bP(d \omega) = \int_{\R} e^{\i t x} \mu(dx) = \phi_\mu(t)
|
||||||
|
\]
|
||||||
|
where $\mu = \bP x^{-1}$.
|
||||||
|
\end{remark}
|
||||||
|
|
||||||
|
\begin{theorem}[Inversion formula] % thm1
|
||||||
|
\label{inversionformula}
|
||||||
|
Let $(\Omega, \cB(\R), \bP)$ be a probability space.
|
||||||
|
Let $F$ be the distribution function of $\bP$
|
||||||
|
(i.e.~$F(x) = \bP((-\infty, x])$ for all $x \in \R$ ).
|
||||||
|
Then for every $a < b$ we have
|
||||||
|
\begin{eqnarray}
|
||||||
|
\frac{F(b) + F(b-)}{2} - \frac{F(a) + F(a-)}{2} = \lim_{T \to \infty} \frac{1}{2 \pi} \int_{-T}^T \frac{e^{-\i t b} - e^{- \i t a}}{- \i t} \phi(t) dt
|
||||||
|
\label{invf}
|
||||||
|
\end{eqnarray}
|
||||||
|
where $F(b-)$ is the left limit.
|
||||||
|
\end{theorem}
|
||||||
|
% TODO!
|
||||||
|
We will prove this later.
|
||||||
|
|
||||||
|
\begin{theorem}[Uniqueness theorem] % thm2
|
||||||
|
\label{charfuncuniqueness}
|
||||||
|
Let $\bP$ and $\Q$ be two probability measures on $(\R, \cB(\R))$.
|
||||||
|
Then $\phi_\bP = \phi_\Q \implies \bP = \Q$.
|
||||||
|
|
||||||
|
Therefore, probability measures are uniquely determined by their characteristic functions.
|
||||||
|
Moreover, \eqref{invf} gives a representation of $\bP$ (via $F$)
|
||||||
|
from $\phi$.
|
||||||
|
\end{theorem}
|
||||||
|
\begin{refproof}{charfuncuniqueness}
|
||||||
|
Assume that we have already shown \autoref{inversionformula}.
|
||||||
|
Suppose that $F$ and $G$ are the distribution functions of $\bP$ and $\Q$.
|
||||||
|
Let $a,b \in \R$ with $a < b$.
|
||||||
|
Assume that $a $ and $b$ are continuity points of both $F$ and $G$.
|
||||||
|
By \autoref{inversionformula} we have
|
||||||
|
\begin{IEEEeqnarray*}{rCl}
|
||||||
|
F(b) - F(a) = G(b) - G(a) \label{eq:charfuncuniquefg}
|
||||||
|
\end{IEEEeqnarray*}
|
||||||
|
|
||||||
|
Since $F$ and $G$ are monotonic, \autoref{eq:charfuncuniquefg}
|
||||||
|
holds for all $a < b$ outside a countable set.
|
||||||
|
|
||||||
|
Take $a_n$ outside this countable set, such that $a_n \ssearrow -\infty$.
|
||||||
|
Then, \autoref{eq:charfuncuniquefg} implies that
|
||||||
|
$F(b) - F(a_n) = G(b) - G(a_n)$ hence $F(b) = G(b)$.
|
||||||
|
Since $F$ and $G$ are right-continuous, it follows that $F = G$.
|
||||||
|
\end{refproof}
|
||||||
|
|
|
@ -1,3 +1,142 @@
|
||||||
|
% This section provides a short recap of things that should be known
|
||||||
|
% from the lecture on stochastics.
|
||||||
|
|
||||||
|
\subsection{Notions of convergence}
|
||||||
|
\begin{definition}
|
||||||
|
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}
|
||||||
|
($X_n \xrightarrow{a.s.} X$)
|
||||||
|
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
|
||||||
|
\[
|
||||||
|
\bE[|X_n - X|^p] \xrightarrow{n \to \infty} 0.
|
||||||
|
\]
|
||||||
|
\end{itemize}
|
||||||
|
\end{definition}
|
||||||
|
% TODO Connect to ANaIII
|
||||||
|
|
||||||
|
\begin{theorem}
|
||||||
|
\vspace{10pt}
|
||||||
|
Let $X$ be a random variable and $X_n, n \in \N$ a sequence of random variables.
|
||||||
|
Then
|
||||||
|
\begin{figure}[H]
|
||||||
|
\centering
|
||||||
|
\begin{tikzpicture}
|
||||||
|
\node at (0,1.5) (as) { $X_n \xrightarrow{a.s.} X$};
|
||||||
|
\node at (1.5,0) (p) { $X_n \xrightarrow{\bP} X$};
|
||||||
|
\node at (3,1.5) (L1) { $X_n \xrightarrow{L^1} X$};
|
||||||
|
\draw[double equal sign distance, -implies] (as) -- (p);
|
||||||
|
\draw[double equal sign distance, -implies] (L1) -- (p);
|
||||||
|
\end{tikzpicture}
|
||||||
|
\end{figure}
|
||||||
|
and none of the other implications hold.
|
||||||
|
\end{theorem}
|
||||||
|
\begin{proof}
|
||||||
|
\begin{claim}
|
||||||
|
$X_n \xrightarrow{a.s.} X \implies X_n \xrightarrow{\bP} X$.
|
||||||
|
\end{claim}
|
||||||
|
\begin{subproof}
|
||||||
|
$\Omega_0 \coloneqq \{\omega \in \Omega : \lim_{n\to \infty} X_n(\omega) = X(\Omega)\} $.
|
||||||
|
Let $\epsilon > 0$ and consider $A_n \coloneqq \bigcup_{m \ge n} \{\omega \in \Omega: |X_m(\omega) - X(\Omega)| > \epsilon\}$.
|
||||||
|
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
|
||||||
|
$\forall \omega \in \Omega_0 \exists n \in \N \forall m \ge n |X_m(\omega) - X(\omega)| < \epsilon$.
|
||||||
|
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}
|
||||||
|
\begin{claim}
|
||||||
|
$X_n \xrightarrow{L^1} X \implies X_n\xrightarrow{\bP} X$
|
||||||
|
\end{claim}
|
||||||
|
\begin{subproof}
|
||||||
|
We have $\bE[|X_n - X|] \to 0$.
|
||||||
|
Suppose there exists an $\epsilon > 0$ such that
|
||||||
|
$\lim_{n \to \infty} \bP[|X_n - X| > \epsilon] = c > 0$.
|
||||||
|
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\\
|
||||||
|
&=& \epsilon \cdot c > 0 \lightning
|
||||||
|
\end{IEEEeqnarray*}
|
||||||
|
\todo{Improve this with Markov}
|
||||||
|
\end{subproof}
|
||||||
|
\begin{claim}
|
||||||
|
$X_n \xrightarrow{\bP} X \notimplies X_n\xrightarrow{L^1} X$
|
||||||
|
\end{claim}
|
||||||
|
\begin{subproof}
|
||||||
|
Take $([0,1], \cB([0,1 ]), \lambda)([0,1], \cB([0,1 ]), \lambda)$
|
||||||
|
and define $X_n \coloneqq n \One_{[0, \frac{1}{n}]}$.
|
||||||
|
We have $\bP[|X_n| > \epsilon] = \frac{1}{n}$
|
||||||
|
for $n$ large enough.
|
||||||
|
|
||||||
|
However $\bE[|X_n|] = 1$.
|
||||||
|
\end{subproof}
|
||||||
|
|
||||||
|
\begin{claim}
|
||||||
|
$X_n \xrightarrow{a.s.} X \notimplies X_n\xrightarrow{L^1} X$.
|
||||||
|
\end{claim}
|
||||||
|
\begin{subproof}
|
||||||
|
We can use the same counterexample as in c).
|
||||||
|
|
||||||
|
$\bP[\lim_{n \to \infty} X_n = 0] \ge \bP[X_n = 0] = 1 - \frac{1}{n} \to 0$.
|
||||||
|
We have already seen, that $X_n$ does not converge in $L_1$.
|
||||||
|
\end{subproof}
|
||||||
|
|
||||||
|
\begin{claim}
|
||||||
|
$X_n \xrightarrow{L^1} X \notimplies X_n\xrightarrow{a.s.} X$.
|
||||||
|
\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
|
||||||
|
\[
|
||||||
|
\bE[|X_n|] = \int_{\Omega}|X_n| d\bP = \frac{1}{2^k} \to 0.
|
||||||
|
\]
|
||||||
|
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]
|
||||||
|
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}
|
||||||
|
|
||||||
|
|
||||||
|
|
||||||
|
\iffalse
|
||||||
|
\todo{Add more stuff here}
|
||||||
|
\subsection{Some inequalities}
|
||||||
|
% TODO: Markov
|
||||||
|
|
||||||
|
|
||||||
\begin{theorem}[Chebyshev's inequality] % TODO Proof
|
\begin{theorem}[Chebyshev's inequality] % TODO Proof
|
||||||
Let $X$ be a r.v.~with $\Var(x) < \infty$.
|
Let $X$ be a r.v.~with $\Var(x) < \infty$.
|
||||||
Then $\forall \epsilon > 0 : \bP \left[ \left| X - \bE[X] \right| > \epsilon\right] \le \frac{\Var(x)}{\epsilon^2}$.
|
Then $\forall \epsilon > 0 : \bP \left[ \left| X - \bE[X] \right| > \epsilon\right] \le \frac{\Var(x)}{\epsilon^2}$.
|
||||||
|
@ -8,16 +147,6 @@
|
||||||
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 $X_i$.
|
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 $X_i$.
|
||||||
|
|
||||||
|
|
||||||
How do we prove that something happens almost surely?
|
|
||||||
\begin{lemma}[Borel-Cantelli]
|
|
||||||
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: $\bP[\limsup_{n \to \infty} A_n] = 0$).
|
|
||||||
|
|
||||||
The converse also holds for independent events $A_n$.
|
|
||||||
|
|
||||||
\end{lemma}
|
|
||||||
|
|
||||||
|
|
||||||
Modes of covergence: $L^p$, in probability, a.s.
|
Modes of covergence: $L^p$, in probability, a.s.
|
||||||
|
\fi
|
||||||
|
|
|
@ -1,91 +0,0 @@
|
||||||
% TODO \begin{goal}
|
|
||||||
% TODO We want to drop our assumptions on finite mean or variance
|
|
||||||
% TODO and say something about the behaviour of $ \sum_{n \ge 1} X_n$
|
|
||||||
% TODO when the $X_n$ are independent.
|
|
||||||
% TODO \end{goal}
|
|
||||||
\begin{theorem}[Theorem 3, Kolmogorov's three-series theorem] % Theorem 3
|
|
||||||
\label{thm3}
|
|
||||||
Let $X_n$ be a family of independent random variables.
|
|
||||||
\begin{enumerate}[(a)]
|
|
||||||
\item Suppose for some $C \ge 0$, the following three series
|
|
||||||
of numbers converge:
|
|
||||||
\begin{itemize}
|
|
||||||
\item $\sum_{n \ge 1} \bP(|X_n| > C)$,
|
|
||||||
\item $\sum_{n \ge 1} \underbrace{\int_{|X_n| \le C} X_n d\bP}_{\text{\vocab{truncated mean}}}$,
|
|
||||||
\item $\sum_{n \ge 1} \underbrace{\int_{|X_n| \le C} X_n^2 d\bP - \left( \int_{|X_n| \le C} X_n d\bP \right)^2}_{\text{\vocab{truncated variance} }}$.
|
|
||||||
\end{itemize}
|
|
||||||
Then $\sum_{n \ge 1} X_n$ converges almost surely.
|
|
||||||
\item Suppose $\sum_{n \ge 1} X_n$ converges almost surely.
|
|
||||||
Then all three series above converge for every $C > 0$.
|
|
||||||
\end{enumerate}
|
|
||||||
\end{theorem}
|
|
||||||
For the proof we'll need a slight generalization of \autoref{thm2}:
|
|
||||||
\begin{theorem}[Theorem 4] % Theorem 4
|
|
||||||
\label{thm4}
|
|
||||||
Let $\{X_n\}_n$ be independent and \vocab{uniformly bounded}
|
|
||||||
(i.e. $\exists M < \infty : \sup_n \sup_\omega |X_n(\omega)| \le M$).
|
|
||||||
Then $\sum_{n \ge 1} X_n$ converges almost surely
|
|
||||||
$\iff$ $\sum_{n \ge 1} \bE(X_n)$ and $\sum_{n \ge 1} \Var(X_n)$
|
|
||||||
converge.
|
|
||||||
\end{theorem}
|
|
||||||
\begin{refproof}{thm3}
|
|
||||||
Assume, that we have already proved \autoref{thm4}.
|
|
||||||
We prove part (a) first.
|
|
||||||
Put $Y_n = X_n \cdot \One_{\{|X_n| \le C\}}$.
|
|
||||||
Since the $X_n$ are independent, the $Y_n$ are independent as well.
|
|
||||||
Furthermore, the $Y_n$ are uniformly bounded.
|
|
||||||
By our assumption, the series
|
|
||||||
$\sum_{n \ge 1} \int_{|X_n| \le C} X_n d\bP = \sum_{n \ge 1} \bE[Y_n]$
|
|
||||||
and $\sum_{n \ge 1} \int_{|X_n| \le C} X_n^2 d\bP - \left( \int_{|X_n| \le C} X_n d\bP \right)^2 = \sum_{n \ge 1} \Var(Y_n)$
|
|
||||||
converges.
|
|
||||||
By \autoref{thm4} it follows that $\sum_{n \ge 1} Y_n < \infty$
|
|
||||||
almost surely.
|
|
||||||
Let $A_n \coloneqq \{\omega : |X_n(\omega)| > C\}$.
|
|
||||||
Since the first series $\sum_{n \ge 1} \bP(A_n) < \infty$,
|
|
||||||
by Borel-Cantelli, $\bP[\text{infinitely many $A_n$ occcur}] = 0$.
|
|
||||||
|
|
||||||
|
|
||||||
For the proof of (b), suppose $\sum_{n\ge 1} X_n(\omega) < \infty$
|
|
||||||
for almost every $\omega$.
|
|
||||||
Fix an arbitrary $C > 0$.
|
|
||||||
Define
|
|
||||||
\[
|
|
||||||
Y_n(\omega) \coloneqq \begin{cases}
|
|
||||||
X_n(\omega) & \text{if} |X_n(\omega)| \le C,\\
|
|
||||||
C &\text{if } |X_n(\omega)| > C.
|
|
||||||
\end{cases}
|
|
||||||
\]
|
|
||||||
Then the $Y_n$ are independent and $\sum_{n \ge 1} Y_n(\omega) < \infty$
|
|
||||||
almost surely and the $Y_n$ are uniformly bounded.
|
|
||||||
By \autoref{thm4} $\sum_{n \ge 1} \bE[Y_n]$ and $\sum_{n \ge 1} \Var(Y_n)$
|
|
||||||
converge.
|
|
||||||
Define
|
|
||||||
\[
|
|
||||||
Z_n(\omega) \coloneqq \begin{cases}
|
|
||||||
X_n(\omega) &\text{if } |X_n| \le C,\\
|
|
||||||
-C &\text{if } |X_n| > C.
|
|
||||||
\end{cases}
|
|
||||||
\]
|
|
||||||
Then the $Z_n$ are independent, uniformly bounded and $\sum_{n \ge 1} Z_n(\omega) < \infty$
|
|
||||||
almost surely.
|
|
||||||
By \autoref{thm4} we have
|
|
||||||
$\sums_{n \ge 1} \bE(Z_n) < \infty$
|
|
||||||
and $\sums_{n \ge 1} \Var(Z_n) < \infty$.
|
|
||||||
|
|
||||||
We have
|
|
||||||
\[
|
|
||||||
\bE(Y_n) &=& \int_{|X_n| \le C} X_n d \bP + C \bP(|X_n| \ge C)\\
|
|
||||||
\bE(Z_n) &=& \int_{|X_n| \le C} X_n d \bP - C \bP(|X_n| \ge C)\\
|
|
||||||
\]
|
|
||||||
Since $\bE(Y_n) + \bE(Z_n) = 2 \int_{|X_n| \le C} X_n d\bP$
|
|
||||||
the second series converges,
|
|
||||||
and since
|
|
||||||
$\bE(Y_n) - \bE(Z_n)$ converges, the first series converges.
|
|
||||||
For the third series, we look at
|
|
||||||
$\sum_{n \ge 1} \Var(Y_n)$ and
|
|
||||||
$\sum_{n \ge 1} \Var(Z_n)$ to conclude that this series converges
|
|
||||||
as well.
|
|
||||||
\end{refproof}
|
|
||||||
|
|
||||||
|
|
||||||
|
|
|
@ -1,18 +1,47 @@
|
||||||
\documentclass[10pt,ngerman,a4paper, fancyfoot, git]{mkessler-script}
|
\documentclass[10pt,a4paper, fancyfoot, git, english]{mkessler-script}
|
||||||
|
|
||||||
\course{Probability Theory}
|
\course{Probability Theory}
|
||||||
\lecturer{}
|
\lecturer{Prof.~Chiranjib Mukherjee}
|
||||||
\author{}
|
\author{Josia Pietsch}
|
||||||
|
|
||||||
\usepackage{wtheo}
|
\usepackage{wtheo}
|
||||||
|
|
||||||
\begin{document}
|
\begin{document}
|
||||||
|
|
||||||
\maketitle
|
\maketitle
|
||||||
|
%\frontmatter
|
||||||
|
|
||||||
|
|
||||||
\cleardoublepage
|
\cleardoublepage
|
||||||
|
|
||||||
\tableofcontents
|
\tableofcontents
|
||||||
\cleardoublepage
|
\cleardoublepage
|
||||||
|
|
||||||
|
%\mainmatter
|
||||||
|
|
||||||
|
\input{inputs/intro.tex}
|
||||||
|
|
||||||
|
\section*{Prerequisites}
|
||||||
|
|
||||||
|
\input{inputs/lecture_1.tex}
|
||||||
|
\input{inputs/prerequisites.tex}
|
||||||
|
|
||||||
|
\input{inputs/lecture_2.tex}
|
||||||
|
\input{inputs/lecture_3.tex}
|
||||||
|
\input{inputs/lecture_5.tex}
|
||||||
|
\input{inputs/lecture_6.tex}
|
||||||
|
\input{inputs/lecture_7.tex}
|
||||||
|
\input{inputs/lecture_8.tex}
|
||||||
|
\input{inputs/lecture_9.tex}
|
||||||
|
\input{inputs/lecture_10.tex}
|
||||||
|
|
||||||
|
\cleardoublepage
|
||||||
|
|
||||||
|
%\backmatter
|
||||||
|
%\chapter{Appendix}
|
||||||
|
|
||||||
|
\cleardoublepage
|
||||||
|
\printvocabindex
|
||||||
|
|
||||||
|
|
||||||
\end{document}
|
\end{document}
|
||||||
|
|
94
wtheo.sty
94
wtheo.sty
|
@ -1 +1,95 @@
|
||||||
\ProvidesPackage{wtheo}[2022/02/10 - Style file for notes of Probability Theory]
|
\ProvidesPackage{wtheo}[2022/02/10 - Style file for notes of Probability Theory]
|
||||||
|
\usepackage[english]{babel}
|
||||||
|
\usepackage[cache, number in = section]{fancythm}
|
||||||
|
\usepackage{mkessler-mathfont}
|
||||||
|
\usepackage{centernot}
|
||||||
|
\usepackage{enumerate}
|
||||||
|
\usepackage{mkessler-todo}
|
||||||
|
\usepackage[index]{mkessler-vocab}
|
||||||
|
\usepackage{mkessler-code}
|
||||||
|
\usepackage{jrpie-math}
|
||||||
|
\usepackage[normalem]{ulem}
|
||||||
|
\usepackage{pdflscape}
|
||||||
|
\usepackage{longtable}
|
||||||
|
\usepackage{xcolor}
|
||||||
|
\usepackage{dsfont}
|
||||||
|
\usepackage{csquotes}
|
||||||
|
\usepackage{tikz}
|
||||||
|
\usepackage{tikz-cd}
|
||||||
|
\usetikzlibrary{arrows}
|
||||||
|
%\usepackage{wrapfig}
|
||||||
|
\usepackage{listings}
|
||||||
|
\usepackage{multirow}
|
||||||
|
\usepackage{float}
|
||||||
|
%\usepackage{algorithmicx}
|
||||||
|
|
||||||
|
\newcounter{subsubsubsection}[subsubsection]
|
||||||
|
\renewcommand\thesubsubsubsection{\thesubsubsection.\arabic{subsubsubsection}}
|
||||||
|
\newcommand\subsubsubsection[1]
|
||||||
|
{
|
||||||
|
\stepcounter{subsubsubsection}
|
||||||
|
\medskip
|
||||||
|
\textbf{\thesubsubsubsection~#1}
|
||||||
|
\medskip
|
||||||
|
}
|
||||||
|
|
||||||
|
\newcommand\todoimg[1]
|
||||||
|
{
|
||||||
|
\todo{FEHLENDES BILD: #1}
|
||||||
|
}
|
||||||
|
|
||||||
|
\usepackage{siunitx}
|
||||||
|
|
||||||
|
% Wenn auf die Klausurrelevanz EXPLIZIT hingewiesen wurde
|
||||||
|
\newcommand\klausurrelevant{
|
||||||
|
\footnote{\color{red}klausurrelevant!}
|
||||||
|
}
|
||||||
|
|
||||||
|
\usepackage{acro}
|
||||||
|
|
||||||
|
\def\alert#1{{\color{red} #1}}
|
||||||
|
|
||||||
|
|
||||||
|
\usepackage{imakeidx}
|
||||||
|
\makeindex[name = ccode, title = \texttt{C} functions and macros]
|
||||||
|
|
||||||
|
\usepackage{hyperref}
|
||||||
|
|
||||||
|
\usepackage[quotation]{knowledge}[22/02/12]
|
||||||
|
|
||||||
|
\newcommand\main[1]{\underline{#1}}
|
||||||
|
\newcommand\usage[1]{\textit{#1}}
|
||||||
|
\renewcommand\i{\mathrm{\mathbf{i}}}
|
||||||
|
\newcommand\notimplies{\centernot\implies}
|
||||||
|
|
||||||
|
\knowledgestyle{ccode}{color=purple!30!black, index style = usage, wrap = \code}
|
||||||
|
\knowledgestyle{ccode unknown}{ wrap = \code, color = brown}
|
||||||
|
\knowledgestyle{ccode unknown cont}{ wrap = \code}
|
||||||
|
\knowledgestyle{ccode intro}{color=blue, boldface, index style = main, wrap = \code}
|
||||||
|
|
||||||
|
|
||||||
|
\knowledgestyle{autoref link}{autoref link}
|
||||||
|
\knowledgestyle{autoref target}{autoref target}
|
||||||
|
|
||||||
|
\knowledgenewvariant\cc{
|
||||||
|
default style = {autoref link, ccode},
|
||||||
|
unknown style = {ccode unknown},
|
||||||
|
unknown style cont = {ccode unknown cont},
|
||||||
|
% unknown warning = false,
|
||||||
|
% unknown diagnose = false,
|
||||||
|
}
|
||||||
|
|
||||||
|
\knowledgenewvariant\ccintro {
|
||||||
|
auto knowledge = {autoref, scope=document, also now, index, index name = ccode, wrap = \code},
|
||||||
|
default style = {autoref target, ccode intro},
|
||||||
|
unknown style = ccode unknown,
|
||||||
|
unknown style cont = ccode unknown
|
||||||
|
}
|
||||||
|
|
||||||
|
\knowledgevariantmodifier{\intro*\cc}\ccintro
|
||||||
|
\knowledgevariantmodifier{\cintro*\cc}\ccintro
|
||||||
|
|
||||||
|
\hypersetup{colorlinks, citecolor=violet, urlcolor=blue!80!black, linkcolor=red!50!black, pdfauthor=\@author, pdftitle=\ifdef{\@course}{\@course}{\@title}}
|
||||||
|
|
||||||
|
\NewFancyTheorem[thmtools = { style = thmredmargin} , group = { big } ]{warning}
|
||||||
|
\DeclareSimpleMathOperator{Var}
|
||||||
|
|
Loading…
Reference in a new issue