125 lines
5.6 KiB
TeX
125 lines
5.6 KiB
TeX
\todo{Lecture 3 needs to be finished}
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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 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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\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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\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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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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Recall the following theorem from measure theory:
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\begin{theorem}[Caratheodory's extension theorem] % 2.3.3 in the notes
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\label{caratheodory}
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Suppose $\cA$ is an algebra (i.e.~closed under finite union)
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und $\Omega \neq \emptyset$.
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Suppose $\bP$ is countably additive on $\cA$ (i.e.~if $(A_n)_{n}$
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are pairwise disjoint and $\bigcup_{n \in \N} A_n \subseteq \cA $
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then $\bP\left( \bigcup_{n \in \N} A_n \right) = \sum_{n \in \N} \bP(A_n)$).
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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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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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$\lambda(A) = \lambda_n(A)$ for any $n$ where this is well defined,
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then $\lambda$ is countably additive on $\cF$.
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Using \autoref{caratheodory}, $\lambda$ will extend uniquely to a probability measure on $\sigma(\cF)$.
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We want to prove:
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\begin{claim}
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\label{claim:sF=Binfty}
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$\sigma(\cF) = \cB_\infty$.
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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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\end{claim}
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\begin{refproof}{claim:sF=Binfty}
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Consider an infinite dimensional box $\prod_{n \in \N} B_n$.
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We have
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\[
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\left( \prod_{n=1}^N B_n \right)^\ast \in \cF_n \subseteq \cF
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\]
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thus
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\[
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\prod_{n \in \N} B_n = \bigcap_{N \in \N} \left( \prod_{n=1}^N B_n \right)^\ast \in \sigma(\cF).
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\]
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Since $\sigma(\cF)$ is a $\sigma$-algebra, $\cB_\infty \subseteq \sigma(\cF)$. This proves ``$\supseteq$''.
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For the other direction we'll show $\cF_n \subseteq \cB_\infty$ for all $n$.
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Let $\cC \coloneqq \{ Q \in \cB_n | Q^\ast \in \cB_\infty\}$.
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For $B_1,\ldots,B_n \in \cB$, $B_1 \times \ldots \times B_n \in \cB_n$
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and $(B_1 \times \ldots \times B_n)^\ast \in \cB_\infty$.
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We have $B_1 \times \ldots \times B_n \in \cC$.
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And $\cC$ is a $\sigma$-algebra, because:
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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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\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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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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\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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and suppose $\bP: \cA \to [0,1]$ is a finitely additive
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probability measure.
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Suppose whenever $\{B_n\}_n$ is a sequence of sets from $\cA $
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decreasing to $\emptyset$ it is the case that
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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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\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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We have $\lambda(\R^\infty) = \lambda_1(\R^\infty) = 1$ and
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$\lambda(\emptyset) = \lambda_1(\emptyset) = 0$.
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Suppose that $A_1, A_2 \in \cF$ are disjoint.
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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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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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