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