\documentclass[11pt]{scrartcl} \usepackage{evan} \usepackage{mdframed} \makeatletter \let\theorem\undefined \let\c@theorem\undefined \let\lemma\undefined \let\c@lemma\undefined \makeatother \newmdtheoremenv[outerlinewidth=2,leftmargin=20, rightmargin=20,backgroundcolor=white, outerlinecolor=blue,innertopmargin=0.5\topskip, splittopskip=\topskip,skipbelow=\baselineskip, skipabove=\baselineskip] {theorem} {Theorem} \newmdtheoremenv[outerlinewidth=2,leftmargin=20, rightmargin=20,backgroundcolor=white, outerlinecolor=blue,innertopmargin=0.5\topskip, splittopskip=\topskip,skipbelow=\baselineskip, skipabove=\baselineskip] {lemma} {Lemma} \lhead{} \rhead{Basics in Probabitlity Theory} \begin{document} \title{Basics in Probability Theory} \author{Joel Antonio-V\'asquez\plusemail{hello@joelantonio.me}} \date{\today} \maketitle %\setcounter{section}{-1} %\section{Abstract Nonsense} \begin{definition*} Let $\Omega$ be a set. A \emph{boolean algebra} is a collection $\SF$ of subsets from $\Omega$ such that \begin{itemize} \item{$\emptyset \in \Omega$}, \item{If $E, F \in \SF$, then $E \cup F$ and $E \cap F$ lie in $\SF$ (closed under union and intersection)}, \item{If $F \in \SF$, so does $\Omega \backslash F$ (closed under complement).} \end{itemize} \end{definition*} \begin{definition*} A $\sigma$-algebra is a boolean algebra $\SF \in \Omega$ and closed under countable unions and intersections (i.e. Let $E_{1}, \dots, E_{n} \in \SF$, then $\bigcup\limits_{i=1}^{n} E_{i}$ and $\bigcap\limits_{i=1}^{n} E_{i}$ also lie $\SF$). \end{definition*} We define the Borel $\sigma$-algebra on $\Omega$ to be the $\sigma$-algebra generated by the open sets of $\Omega$ (can be also generated by the compact subsets of $\Omega$). Example, let $\SF$ be the collection of all the open subsets from the interval $[0, 1]$, then the Borel $\sigma$-algebra is the smallest $\sigma$-algebra that contains $\SF$. \textbf{Proof:} Let $\SB$ be the set generated by $\SF$, \textbf{any} union and intersection from a $\sigma$-algebra is a $\sigma$-algebra. Finally let $\SA$ be a $\sigma$-algebra that contains $\SF$, then $\SB \subset \SA$ and claim follows. \begin{definition*} A measurable space is a pair $(\Omega, \SF)$, where $\SF$ is a $\sigma$-algebra from the set $\Omega$. \end{definition*} \begin{definition*}[Probability theory] Let $(\Omega, \SF)$ be a measureable space and let $\mu$ a measure on that space such that $\mu:\SF \longrightarrow [0, +\infty]$. A probability space is a triple $(\Omega, \SF, \mu)$ obeying the \textbf{Kolmogorov axioms for probability} \begin{itemize} \item{${\bf P}(\emptyset)=0$}, \item{${\bf P}(\overline{\emptyset})=1$}, \item{If $E_ {1}, E_{2}, \dots$ are disjoint events, then ${\bf P}\left( \bigvee_{n=1}^{\infty}E_{n} \right) = \sum_{n=1}^{\infty} {\bf P}(E_{n})$.} \end{itemize} \end{definition*} \section{Expectation} \begin{lemma}[Fatou's Lemma] Let $(\Omega, \SF, \mu)$ be a measure space, and let $f_{1}, f_{2}, \dots: \Omega \longrightarrow [0, +\infty]$ be a sequence of unsigned measurable function. Then $$\int_{\Omega} \liminf_{n \to \infty} f_{n} d\mu \leq \liminf_{n \to \infty} \int_{\Omega} f_{n} d\mu.$$ \end{lemma} \textbf{Proof:} Let $g$ be an unsigned simple function (i.e. $g = \sum_{i=1}^{k} a_{i}1_{E_{i}}$ where $a_{i} > 0$ and $1_{E_{i}}$ be the indicator function for disjoint events $E_{i}$) such that $g \leq \liminf\limits_{n \to \infty} f_{n}$. It's enough to show that $$(1 - \eps) \int_{\Omega} g \leq \liminf_{n \to \infty} \int_{\Omega} f_{n} d\mu$$ by definition of the unsigned integral for any $0 < \eps < 1$. Writting as a sumation $$\int_{\Omega} f_{N} d\mu \geq \sum_{i=1}^{k} (1 - \eps)a_{i}\mu(E_{i, N})$$ Where $N$ goes to $+\infty$ and $1_{E_{i, N}} = \mu(E_{i, N})$, taking the limit inferior $$\liminf_{N \to \infty} \int_{\Omega} f_{N} d\mu \geq \sum_{i=1}^{k} (1 - \eps)a_{i}\mu(E_{i})$$ becuase $\mu(E_{i, N}) \to \mu(E_{i})$ as $N \to \infty$. Since the right-hand side is $(1 - \eps) \int_{\Omega} g d\mu$, and the claim follows. \\ \\ Rewriting the Fatou's lemma for random variables \begin{lemma}[Fatou's lemma for random variables] Let $X_{1}, X_{2}, \dots$ be a sequence of unsigned random variables. Then $${\bf E} \liminf_{n \to \infty} X_{n} \leq \liminf_{n \to \infty} {\bf E}X_{n}.$$ \end{lemma} Now, let's establish Monotone convergence theorem \begin{theorem}[Monotone convergence theorem] Let $(\Omega, \SF, \mu)$ be a measure space, and let $f_{1}, f_{2}, \dots: \Omega \longrightarrow [0, +\infty]$ be a sequence of unsigned measurables functions which is monotone increasing such that $f_{n}(\omega) \leq f_{n+1}(\omega)$for all $n$ and $\omega \in \Omega$. Then $$\int_{\Omega} \lim_{n \to \infty} f_{n} d\mu = \lim_{n \to \infty} \int_{\Omega} f_{n} d\mu.$$ \end{theorem} \\ \\ \textbf{Proof:} Recalling the Fatou's lemma (only for the limit case), we have $$\int_{\Omega} \lim_{n \to \infty} f_{n} d\mu \leq \lim_{n \to \infty} \int_{\Omega} f_{n} d\mu.$$ By continuity below (i.e. if $E_{1} \supset E_{2} \supset \dots$ are measurable, then $\mu(\bigcup_{n=1}^{\infty} E_{n}) = \lim_{n \to \infty} \mu(E_{n})$), then there is a $m$ that goes to $+\infty$ such that $$\int_{\Omega} \lim_{n \to \infty} f_{n} d\mu \geq \int_{\Omega} f_{m} d\mu,$$ since $f_{n}$ are indicator functions, then claim follows. \\ \\ One more time, rewriting for random variables \begin{theorem}[Monotone convergence theorem for random variables] Let $0 \leq X_{1} \leq X_{2} \leq \dots$ be a monotone non-decreasing sequence of unsigned random variables. Then $${\bf E} \lim_{n \to \infty} X_{n} = \lim_{n \to \infty} {\bf E} X_{n}.$$ \end{theorem} \\ \\ Suppose that $f_{i}: \Omega \longrightarrow \CC$ for $i = 1, 2, \dots$ and let $g: \Omega \longrightarrow [0, +\infty]$ which \emph{dominates} $f_{n}$ in the sense of $|f_{n}(\omega)| \leq |g(\omega)|$ for all $n$ and all $\omega$, What can one say about it? (see Problem 3). \section{Products measures} Let $(\Omega_{i}, \SF_{i}, \mu_{i})$ for $i= 1,2$ be probability spaces. Let $\SE_{1} \in \SF_{1}$ and $\SE_{2} \in \SF_{2}$, then $\mu_{1} \times \mu_{2}(\SE_{1} \times \SE_{2}) = \mu_{1}(\SE_{1})\mu_{2}(\SE_{2})$. \begin{theorem}[Tonelli's theorem] Let $f: \Omega_{1} \times \Omega_{2} \longrightarrow [0, +\infty]$ be a measurable function, then \begin{enumerate}[1.] \item{$f_{\omega_{1}}$ is measurable for almost all $\omega_{1}$ and $f_{\omega_{2}}$ is measurable for almost all $\omega_{2}$.} \item{$\omega_{1} \mapsto \int_{\Omega_{2}} f(\omega_{1}, \omega_{2})d\mu_{2}(\omega_{2})$ is measurable as a function of $\omega_{1}$}. \item{$\omega_{2} \mapsto \int_{\Omega_{1}} f(\omega_{1}, \omega_{2})d\mu_{1}(\omega_{1})$ is measurable as a function of $\omega_{2}$}. \item{} $$\int_{\Omega_{1} \times \Omega_{2}} f d(\mu_{1} \times \mu_{2}) = \int_{\Omega_{1}} \left( \int_{\Omega_{2}} f(\omega_{1}, \omega_{2}) d\mu_{2}(\omega_{2}) \right) d\mu_{1}(\omega_{1})$$ \end{enumerate} For each $\omega_{i} \in \Omega_{i}$ for $i= 1, 2$. The equality holds if we invert the integrals at (4). \end{theorem} \\ \\ \textbf{Proof:} Since $\mu_{1}, \mu_{2}$ are $\sigma$-algebra so is $(\mu_{1} \times \mu_{2})$. Let $S_{n}$ be a simple function, replacing with $S_{n}1_{\Omega_{1} \times \Omega_{2}}$ such that increases to $f$ for all $n$. For monotone convergence theorem claim follows. \\ \\ \section{Problems} \begin{enumerate} \ii Show the linearity for the complex case when $f, g: \Omega \longrightarrow \CC$ $$\int_{\Omega} f + g \ d\mu = \int_{\Omega} f d\mu + \int_{\Omega} g d\mu,$$ and for a $c \in \CC$, then $$\int_{\Omega} cf \ d\mu = c\int_{\Omega} fd\mu.$$ \ii Show that $$\int_{\Omega} \min(f, n) d\mu \to \int_{\Omega} f d\mu.$$ as $n \to \infty$ and $f: \Omega \to [0, +\infty]$ is unsigned measurable. \ii (Dominated convergence theorem) Let $(\Omega, \SF, \mu)$ be a measure space and let $f_{1}, f_{2}, \dots: \Omega \longrightarrow \CC$ be measurables functions which converges pointwise to some limit. Let $g: \Omega \longrightarrow [0, +\infty]$ which dominates $f_{n}$ in the sense that $|f_{n}(\omega)| \leq |g(\omega)|$ for all $n$ and all $\omega$. Show that $$\int_{\Omega} \lim_{n \to \infty} f_{n} d\mu = \lim_{n \to \infty} \int_{\Omega} f_{n} d\mu.$$ \ii (Fubini's theorem) If $f: \Omega_{1} \times \Omega_{2} \longrightarrow \CC$ is absoutely integrable, and \begin{enumerate}[I.] \ii{$f_{\omega_{1}}$ is measurable for almost all $\omega_{1}$ and $f_{\omega_{2}}$ is measurable for almost all $\omega_{2}$,} \ii{$\omega_{1} \mapsto \int_{\Omega_{2}} f(\omega_{1}, \omega_{2})d\mu_{2}(\omega_{2})$ is measurable as a function of $\omega_{1}$}, \ii{$\omega_{2} \mapsto \int_{\Omega_{1}} f(\omega_{1}, \omega_{2})d\mu_{1}(\omega_{1})$ is measurable as a function of $\omega_{2}$}. \end{enumerate} Then, prove that $$\int_{\Omega_{1} \times \Omega_{2}} f d(\mu_{1} \times \mu_{2}) = \int_{\Omega_{1}} \left( \int_{\Omega_{2}} f(\omega_{1}, \omega_{2}) d\mu_{2}(\omega_{2}) \right) d\mu_{1}(\omega_{1})$$ \ii Let $(\Omega_{i}, \SF_{i}, \mu_{i})_{i \in A}$ be a collection of probability spaces. Show that $$\prod_{i \in A} \mu_{i} = \left( \prod_{i \in A_{1}} u_{i} \right) \times \left( \prod_{i \in A_{2}} u_{i} \right)$$ for any partition $A = A_{1} \uplus A_{2}$. \end{enumerate} \section{Further Links} \begin{itemize} \ii Terence Tao's notes on Probability Theory: \url{https://terrytao.wordpress.com/category/teaching/275a-probability-theory/} \ii Christopher King's notes on Probability Theory: \url{http://www.hamilton.ie/ollie/Downloads/ProbMain.pdf} \end{itemize} \end{document}