diff --git a/inputs/lecture_03.tex b/inputs/lecture_03.tex index 960d9f3..4f0a0c4 100644 --- a/inputs/lecture_03.tex +++ b/inputs/lecture_03.tex @@ -1,38 +1,46 @@ -\lecture{3}{}{} -\todo{My battery died during this lecture, so this still needs to be finished} +\lecture{3}{2023-04-13}{} \begin{notation} Let $\cB_n$ denote $\cB(\R^n)$. \end{notation} \begin{goal} Suppose we have a probability measure $\mu_n$ on $(\R^n, \cB(\R^n))$ - for each $n$. - We want to show that there exists a unique probability measure $\bP^{\otimes}$ - on $(\R^\infty, \cB_\infty)$ (where $\cB_{\infty}$ still needs to be defined), - such that $\bP^{\otimes}\left( \prod_{n \in \N} B_n \right) = \prod_{n \in \N} \mu_n(B_n)$ + for each $n \in \N$. + We want to show that there exists a unique probability measure + $\bP^{\otimes}$ on $(\R^\infty, \cB_\infty)$ + (where the $\sigma$-algebra $\cB_{\infty}$ still needs to be defined), + such that + \[ + \bP^{\otimes}\left( \prod_{n \in \N} B_n \right) + = \prod_{n \in \N} \mu_n(B_n) + \] for all $\{B_n\}_{n \in \N}$, $B_n \in \cB_1$. \end{goal} -% $\bP_n = \mu_1 \otimes \ldots \otimes \mu_n$. \begin{remark} $\prod_{n \in \N} \mu_n(B_n)$ converges, since $0 \le \mu_n(B_n) \le 1$ for all $n$. \end{remark} First we need to define $\cB_{\infty}$. -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 -$\sigma$-algebra -Let \[\cB_\infty \coloneqq \sigma \left( \left\{\prod_{n \in \N} B_n : B_n \in \cB(\R)\right\} \right).\] +This $\sigma$-algebra must contain all ``boxes'' $\prod_{n \in \N} B_n$ +for $B_i \in \cB_1$. +We simply take the smallest $\sigma$-algebra with this property: +\begin{definition} + \[ + \cB_\infty \coloneqq \sigma \left( \left\{\prod_{n \in \N} B_n : + \forall n .~ B_n \in \cB(\R)\right\} \right). + \] +\end{definition} \begin{question} What is there in $\cB_\infty$? - Can we identify sets in $\cB_\infty$ for which we can define the product measure - easily? + Can we identify sets in $\cB_\infty$ + for which we can define the desired product measure easily? \end{question} Let $\cF_n \coloneqq \{ C \times \R^{\infty} | C \in \cB_n\}$. It is easy to see that $\cF_n \subseteq \cF_{n+1}$ and using that $\cB_n$ is a $\sigma$-algebra, we can show that $\cF_n$ is also a $\sigma$-algebra. Now, for any $C \subseteq \R^n$ let $C^\ast \coloneqq C \times \R^{\infty}$. - +Note that $C \in \cB_n \implies C^\ast \in \cF_n$. Thus $\cF_n = \{C^\ast : C \in \cB_n\}$. Define $\lambda_n : \cF_n : \to [0,1]$ by $\lambda_n(C^\ast) \coloneqq (\mu_1 \otimes \ldots \otimes \mu_n)(C)$. It is easy to see that $\lambda_{n+1} \defon{\cF_n} = \lambda_n$ (\vocab{consistency}). @@ -49,6 +57,9 @@ Recall the following theorem from measure theory: Then $\bP$ extends uniquely to a probability measure on $(\Omega, \cF)$, where $\cF = \sigma(\cA)$. \end{theorem} +\begin{proof} + See theorem 2.3.3 in Stochastik. +\end{proof} 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 @@ -63,7 +74,7 @@ We want to prove: \end{claim} \begin{claim} \label{claim:lambdacountadd} - $\lambda$ as defined above is countably additive on $\cF$. + $\lambda$ is countably additive on $\cF$. \end{claim} \begin{refproof}{claim:sF=Binfty} @@ -86,14 +97,17 @@ We want to prove: \begin{itemize} \item $\cB_n$ is a $\sigma$-algebra \item $\cB_\infty$ is a $\sigma$-algebra, - \item $\phi^\ast \phi$, $(\R^n \setminus Q)^\ast = \R^{\infty} \setminus Q^\ast$, $\bigcup_{i \in I} Q_i^\ast = (\bigcup_{i \in I} Q_i)^\ast$. + \item $\emptyset^\ast = \emptyset$, $(\R^n \setminus Q)^\ast = \R^{\infty} \setminus Q^\ast$, $\bigcup_{i \in I} Q_i^\ast = (\bigcup_{i \in I} Q_i)^\ast$. \end{itemize} - Thus $\cC \subseteq \cB_n$ is a $\sigma$-algebra and contains all rectangles, hence $\cC = \cB_n$. + Since $\cC \subseteq \cB_n$ is a $\sigma$-algebra + and contains all rectangles, + it holds that $\cC = \cB_n$. Hence $\cF_n \subseteq \cB_\infty$ for all $n$, thus $\cF \subseteq \cB_\infty$. Since $\cB_\infty$ is a $\sigma$-algebra, $\sigma(\cF) \subseteq \cB_\infty$. \end{refproof} -We are going to use the following +For the proof of \autoref{claim:lambdacountadd}, +we are going to use the following: \begin{fact} \label{fact:finaddtocountadd} Suppose $\cA$ is an algebra on $\Omega \neq \emptyset$, @@ -104,7 +118,24 @@ We are going to use the following $\bP(B_n) \to 0$. Then $\bP$ must be countably additive. \end{fact} \begin{proof} - Exercise. % TODO + Let $(A_n)_{n\in \N}$ be a sequence of disjoint, + measurable sets with + $A \coloneqq \bigcup_{n} A_n \in \cA$. + Let $A'_n \coloneqq A \setminus \bigcup_{i=1}^n A_i$. + Then we have $ \bP[A] = \bP[A'_n] + \sum_{i=1}^{n} \bP[A_i]$ for all $n$. + Thus + \[ + \bP[A] - \lim_{n \to \infty} \bP[A'_n] + = \lim_{n \to \infty} \sum_{i=1}^{n} \bP[A_i]. + \] + Since $\bigcap_{n \in \N} A'_n = \emptyset$, we have + $\lim_{n \to \infty} \bP[A'_n] = 0$, + hence + \[ + \bP\left[\bigcup_{i \in \N} A_i\right] + = \bP[A] + = \sum_{i \in \N} \bP[A_i]. + \] \end{proof} \begin{refproof}{claim:lambdacountadd} Let us prove that $\lambda$ is finitely additive. @@ -114,13 +145,13 @@ We are going to use the following Then pick some $n$ such that $A_1, A_2 \in \cF_n$. Take $C_1, C_2 \in \cB_n$ such that $C_1^\ast = A_1$ and $C_2^\ast = A_2$. - Then $C_1$ and $C_2$ are disjoint and $A_1 \cup A_2 = (C_1 \cup C_2)^\ast$. - $\lambda(A_1 \cup A_2) = \lambda_n(A_1 \cup A_2) = (\mu_1 \otimes \ldots \otimes \mu_n)(C_1 \cup C_2) = \lambda_n(C_1) + \lambda_n(C_2)$ + Then $C_1$ and $C_2$ are disjoint + and $A_1 \cup A_2 = (C_1 \cup C_2)^\ast$. + Hence + \[ + \lambda(A_1 \cup A_2) = \lambda_n(A_1 \cup A_2) + = (\mu_1 \otimes \ldots \otimes \mu_n)(C_1 \cup C_2) + = \lambda_n(C_1) + \lambda_n(C_2) + \] by the definition of the finite product measure. - - In order to use \autoref{fact:finaddtocountadd}, - 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 \end{refproof} diff --git a/inputs/lecture_04.tex b/inputs/lecture_04.tex index 7cdc7eb..fb244d7 100644 --- a/inputs/lecture_04.tex +++ b/inputs/lecture_04.tex @@ -1,2 +1,196 @@ -\lecture{4}{}{} -\todo{Lecture 4 missing} +\lecture{4}{}{End of proof of Kolmogorov's consistency theorem} + +To finish the proof of \autoref{claim:lambdacountadd}, +we need the following: +\begin{fact} + \label{lec4fact1} + Suppose $\{x_k^{(n)}\}_{n \in \N}$ + is a bounded sequence of real numbers for each $k \in \N$. + Then there exists a strictly increasing sequence + of natural number $\{n_i\}_{i\in \N}$ + such that for all $k \in \N$ + the series $\{x_k^{(n_i)}\}_{i \in \N}$ + converges. +\end{fact} +\begin{proof} + We'll use a diagonalization argument. + For $S \subseteq \N$ infinite, + we say that a sequence of real number, $(x_n)_{n \in \N}$, + \vocab[Convergence along a subset]{converges along $S$}, + if + \[ + \lim_{\substack{n \to \infty\\n \in S}} x_n + \] + exists. + + Let $S_1$ be such that $\{x_1^{(n)}\}_{n \in \N}$ + converges along $S_1$. Such an $S_1$ exists by + Bolzano-Weierstraß. + We proceed recursively. + Suppose we have already chosen $S_1,\ldots, S_{k-1}$. + Consider $\{x_k^{(n)}\}_{n \in S_{k-1}}$. + By Bolzano-Weierstraß, there exists $S_k \subseteq S_{k-1}$ + such that $\{x_{k}^{(n)}\}_{n \in S_{k-1}}$ + converges along $S_k$. + For an infinite subset $T \subseteq \N$ and $\nu \in \N$ + let $\# \nu(T)$ denote the $\nu$-th smallest + element of $T$. + Let + \[ + S \coloneqq \{\#\nu(S_k) : k \in \N\}. + \] + Since $S_{k+1} \subseteq S_k$, + we have $\#(k+1)(S_{k+1}) > \#k(S_{k+1}) \ge \# k (S_k)$. + Hence $S$ is infinite. + Each $\{x_k^{(n)}\}_{n \in \N}$ + converges along $S$, + since all but finitely many elements of $S$ + belong to $S_k$. +\end{proof} + +\begin{lemma} + \label{lem:intersectioncompactsets} + Let $\{K_n\}_{n \in \N}$ be a sequence of compact sets + $K_n \subseteq \R^{l_n}$ for some $l_n$. + Suppose for all $n$ + \[ + \bigcap_{i=1}^n K_i^\ast \neq \emptyset. + \] + Then + \[ + \bigcap_{i\in \N} K_i^\ast \neq \emptyset. + \] + +\end{lemma} +\begin{refproof}{lem:intersectioncompactsets} + We know from analysis + that if $\{K_n\}_{n \in \N}$ is a sequence of compact sets + such that the intersection of finitely many of them is non-empty, + then + \[ + \bigcap_{n \in \N} K_n \neq \emptyset. + \] + Here, different $K_n$ may have different dimensions $l_n$, + but we can view them as subsets of $\R^\infty$ + by applying ${}^\ast$. + For each $n$, choose $x^{(n)} \in \bigcap_{i =1}^n K_i^\ast$. + We can assume $x ^{(n)}_k = 0$ for $k > \max \{l_1,\ldots, l_n\}$. + For all $k \in \N$ we will show that $\{x_k^{(n)}\}$ + is bounded. + \begin{itemize} + \item Case 1: Suppose every $l_n \le k$. + Then $\{x_k^{(n)}\}_n$ only contains zeros. + \item Case 2: + Suppose some $l_{n_0} \ge k$. + Let $Z$ be the projection of $K_{n_0} \subseteq \R^{l_{n_0}}$ + onto its $k$-th component. + $Z$ is a compact subset of $\R$. + Hence it is bounded. + For all $n \ge n_0$, we have $x^{(n)} \in K_{n_0}^\ast$ + and $x_k^{(n)} \in Z$, + so $\{x_k^{(n)}\}_n$ is bounded. + \end{itemize} + + By \autoref{lec4fact1}, + there is an infinite set $S \subseteq \N$, + such that $\{x_k^{(n)}\}_{n \in S}$ + converges for every $k$. + Let $x_k \coloneqq \lim_{\substack{n \to \infty\\n \in S}} x_k^{(n)}$. + Now let $x = (x_1,x_2,\ldots)\in \R^{\infty}$. + \begin{claim} + $x \in \bigcap_{i \in \N} K_i^\ast$. + \end{claim} + \begin{subproof} + Consider $x^{(n)}$ for $n > i$ and $n \in S$. + Then $(x^{(n)}_1, \ldots, x^{(n)}_{l_i}) \in K_i$ + and + \[ + \lim_{\substack{n \to \infty\\n \in S}} + (x^{(n)}_1, \ldots, x^{(n)}_{l_i}) + = (x_1,\ldots, x_{l_i}). + \] + Since $K_i$ is compact, + it follows that $x \in K_i^\ast$. + \end{subproof} +\end{refproof} + +\begin{refproof}{claim:lambdacountadd} + In order to apply \autoref{fact:finaddtocountadd}, + we need the following: + \begin{claim} + For any sequence $B_n \in \cF$ + with $B_n \xrightarrow{n\to \infty} \emptyset$ + we have $\lambda(B_n) \xrightarrow{n \to \infty} 0$. + \end{claim} + \begin{subproof} + Suppose that $B_1^\ast \supseteq B_2^\ast \supseteq \ldots$ + is a decreasing sequence such that + $\lim_{n \to \infty} \lambda(B_n^\ast) = \epsilon > 0$. + For each $n$, + let $l_n$ be such that $B_n \in \cB_{l_n}$. + By regularity of Borel probability measures, % + % TODO see the proof of Caratheodory extension theorem + given $\epsilon > 0$, + there exists a compact set $L_n \subseteq B_n$, + such that + \[ + (\mu_1 \otimes \ldots \otimes\mu_n)(B_n \setminus L_n) + < \frac{\epsilon}{2^{n+1}} + \] + We have + \[ + B_n^\ast \setminus \bigcap_{k=1}^n L_k^\ast % + \subseteq \bigcup_{k=1}^n \left( B_k^\ast \setminus L_k^\ast \right). + \] + Hence + \begin{IEEEeqnarray*}{rCl} + \lambda\left( B_n^\ast \setminus \bigcap_{k=1}^n L_k^\ast \right) + &\le & \lambda\left(\bigcup_{k=1}^n % + B_k^\ast \setminus L_k^\ast \right) \\ + &\le & \sum_{k=1}^{n} \lambda(B_k^\ast \setminus L_k^\ast)\\ + &\le& \sum_{k=1}^{n} \frac{\epsilon}{2^{k+1}}\\ + &\le & \frac{\epsilon}{2}. + \end{IEEEeqnarray*} + By our assumption, + $\lambda(B_n^\ast) \downarrow \epsilon > 0$. + Hence $\lambda(B_n^\ast) \ge \epsilon$ for all $n$. + Thus + \[ + \lambda\left( \bigcap_{k=1}^n L_k^\ast \right) % + \ge \epsilon - \frac{\epsilon}{2} = \frac{\epsilon}{2}. + \] + In particular, for all $n$ + \[ + \bigcap_{k=1}^n L_k^\ast \neq \emptyset. + \] + By \autoref{lem:intersectioncompactsets}, + it follows that + \[ + \bigcap_{k \in \N} L_k^\ast \neq \emptyset. + \] + Since + \[ + \bigcap_{k \in \N} B_k^\ast \supseteq \bigcap_{k \in \N} L_k^\ast, + \] + we have $\bigcap_{k \in \N} B_k^\ast \neq \emptyset$. + \end{subproof} +\end{refproof} + +The measure $\lambda$ is as desired: +For all $n \in \N$ take some $B_n \in \cB_1$ +and let $C_n \coloneqq \prod_{i=1}^n B_i$. +Then $C_n^\ast \downarrow \prod_{i=1}^{\infty} B_i$, +hence +\begin{IEEEeqnarray*}{rCl} + \lambda\left( \prod_{i=1}^{\infty} B_i \right) + &\overset{\text{continuity}}{=}& \lim_{N \to \infty} \lambda(C_N^\ast)\\ + &=& \lim_{N \to \infty} \lambda_N(C_N^\ast)\\ + &=& \lim_{N \to \infty} \prod_{n=1}^{N} \mu_n(B_n)\\ + &=& \prod_{n \in \N} \mu_n(B_n). +\end{IEEEeqnarray*} + +For the definition of $\lambda$ +as well as the proof of \autoref{claim:lambdacountadd} +we have only used that $(\lambda_n)_{n \in \N}$ +is a consistent family. +Hence we have in fact shown \autoref{thm:kolmogorovconsistency}. diff --git a/probability_theory.tex b/probability_theory.tex index 8bbbf5b..c398eb6 100644 --- a/probability_theory.tex +++ b/probability_theory.tex @@ -28,6 +28,7 @@ \input{inputs/lecture_02.tex} \input{inputs/lecture_03.tex} +\input{inputs/lecture_04.tex} \input{inputs/lecture_05.tex} \input{inputs/lecture_06.tex} \input{inputs/lecture_07.tex} @@ -54,6 +55,7 @@ \section{Appendix} \input{inputs/a_0_distributions.tex} \end{landscape} +\input{inputs/lecture_23.tex} \cleardoublepage \printvocabindex