lecture 4
This commit is contained in:
parent
0a5131f313
commit
f543fd1de6
GPG Key ID:
E70B571D66986A2D
@ -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}
|
||||
|
||||
@ -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}.
|
||||
|
||||
@ -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
|
||||
|
||||
Loading…
Reference in New Issue
Block a user