lecture 4

inputs/lecture_03.tex

@@ -1,38 +1,46 @@
-\lecture{3}{}{}
+\lecture{3}{2023-04-13}{}
-\todo{My battery died during this lecture, so this still needs to be finished}
 \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 $n$.
+    for each $n \in \N$.
-    We want to show that there exists a unique probability measure $\bP^{\otimes}$
+    We want to show that there exists a unique probability measure
-    on $(\R^\infty, \cB_\infty)$ (where $\cB_{\infty}$ still needs to be defined),
+    $\bP^{\otimes}$ on $(\R^\infty, \cB_\infty)$
-    such that $\bP^{\otimes}\left( \prod_{n \in \N}  B_n \right) = \prod_{n \in \N} \mu_n(B_n)$
+    (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 $\prod_{n \in \N} B_n$
+This $\sigma$-algebra must contain all ``boxes'' $\prod_{n \in \N} B_n$
-for all $B_n \in  \cB_1$. We simply define $\cB_{\infty}$ to be the
+for $B_i \in  \cB_1$.
-$\sigma$-algebra
+We simply take the smallest $\sigma$-algebra with this property:
-Let \[\cB_\infty \coloneqq  \sigma \left( \left\{\prod_{n \in \N} B_n : B_n \in  \cB(\R)\right\}  \right).\]
+\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
+    Can we identify sets in $\cB_\infty$
-    easily?
+    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 $A_1 \cup  A_2 = (C_1 \cup  C_2)^\ast$.
+    Then $C_1$ and $C_2$ are disjoint
-    $\lambda(A_1 \cup  A_2) = \lambda_n(A_1 \cup  A_2) = (\mu_1 \otimes \ldots \otimes \mu_n)(C_1 \cup  C_2) = \lambda_n(C_1) + \lambda_n(C_2)$
+    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}

inputs/lecture_04.tex

@@ -1,2 +1,196 @@
-\lecture{4}{}{}
+\lecture{4}{}{End of proof of Kolmogorov's consistency theorem}
-\todo{Lecture 4 missing}
+
+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}.

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