s23-probability-theory/inputs/lecture_03.tex

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\lecture{3}{2023-04-13}{}
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\begin{notation}
Let $\cB_n$ denote $\cB(\R^n)$.
\end{notation}
\begin{goal}
Suppose we have a probability measure $\mu_n$ on $(\R^n, \cB(\R^n))$
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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
\[
\bP^{\otimes}\left( \prod_{n \in \N} B_n \right)
= \prod_{n \in \N} \mu_n(B_n)
\]
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for all $\{B_n\}_{n \in \N}$, $B_n \in \cB_1$.
\end{goal}
\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 $\cB_{\infty}$.
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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}
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\begin{question}
What is there in $\cB_\infty$?
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Can we identify sets in $\cB_\infty$
for which we can define the desired product measure easily?
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\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 $C^\ast \coloneqq C \times \R^{\infty}$.
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Note that $C \in \cB_n \implies C^\ast \in \cF_n$.
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Thus $\cF_n = \{C^\ast : C \in \cB_n\}$.
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}).
Recall the following theorem from measure theory:
\begin{theorem}[Caratheodory's extension theorem] % 2.3.3 in the notes
\label{caratheodory}
Suppose $\cA$ is an algebra (i.e.~closed under finite union)
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 $
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 $(\Omega, \cF)$,
where $\cF = \sigma(\cA)$.
\end{theorem}
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\begin{proof}
See theorem 2.3.3 in Stochastik.
\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
$\lambda(A) = \lambda_n(A)$ for any $n$ where this is well defined,
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}
\label{claim:sF=Binfty}
$\sigma(\cF) = \cB_\infty$.
\end{claim}
\begin{claim}
\label{claim:lambdacountadd}
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$\lambda$ is countably additive on $\cF$.
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\end{claim}
\begin{refproof}{claim:sF=Binfty}
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Consider an infinite dimensional box $\prod_{n \in \N} B_n$.
We have
\[
\left( \prod_{n=1}^N B_n \right)^\ast \in \cF_n \subseteq \cF
\]
thus
\[
\prod_{n \in \N} B_n = \bigcap_{N \in \N} \left( \prod_{n=1}^N B_n \right)^\ast \in \sigma(\cF).
\]
Since $\sigma(\cF)$ is a $\sigma$-algebra, $\cB_\infty \subseteq \sigma(\cF)$. This proves ``$\supseteq$''.
For the other direction we'll show $\cF_n \subseteq \cB_\infty$ for all $n$.
Let $\cC \coloneqq \{ Q \in \cB_n | Q^\ast \in \cB_\infty\}$.
For $B_1,\ldots,B_n \in \cB$, $B_1 \times \ldots \times B_n \in \cB_n$
and $(B_1 \times \ldots \times B_n)^\ast \in \cB_\infty$.
We have $B_1 \times \ldots \times B_n \in \cC$.
And $\cC$ is a $\sigma$-algebra, because:
\begin{itemize}
\item $\cB_n$ is a $\sigma$-algebra
\item $\cB_\infty$ is a $\sigma$-algebra,
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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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Since $\cC \subseteq \cB_n$ is a $\sigma$-algebra
and contains all rectangles,
it holds that $\cC = \cB_n$.
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Hence $\cF_n \subseteq \cB_\infty$ for all $n$,
thus $\cF \subseteq \cB_\infty$. Since $\cB_\infty$ is a $\sigma$-algebra,
$\sigma(\cF) \subseteq \cB_\infty$.
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\end{refproof}
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For the proof of \autoref{claim:lambdacountadd},
we are going to use the following:
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\begin{fact}
\label{fact:finaddtocountadd}
Suppose $\cA$ is an algebra on $\Omega \neq \emptyset$,
and suppose $\bP: \cA \to [0,1]$ is a finitely additive
probability measure.
Suppose whenever $\{B_n\}_n$ is a sequence of sets from $\cA $
decreasing to $\emptyset$ it is the case that
$\bP(B_n) \to 0$. Then $\bP$ must be countably additive.
\end{fact}
\begin{proof}
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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
\[
\bP\left[\bigcup_{i \in \N} A_i\right]
= \bP[A]
= \sum_{i \in \N} \bP[A_i].
\]
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\end{proof}
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\begin{refproof}{claim:lambdacountadd}
Let us prove that $\lambda$ is finitely additive.
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 $A_1, A_2 \in \cF_n$.
Take $C_1, C_2 \in \cB_n$ such that $C_1^\ast = A_1$
and $C_2^\ast = A_2$.
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Then $C_1$ and $C_2$ are disjoint
and $A_1 \cup A_2 = (C_1 \cup C_2)^\ast$.
Hence
\[
\lambda(A_1 \cup A_2) = \lambda_n(A_1 \cup A_2)
= (\mu_1 \otimes \ldots \otimes \mu_n)(C_1 \cup C_2)
= \lambda_n(C_1) + \lambda_n(C_2)
\]
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by the definition of the finite product measure.
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\phantom\qedhere
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\end{refproof}