lecture 6

inputs/lecture_06.tex

@@ -1,61 +1,157 @@
-\lecture{6}{}{}
+\lecture{6}{}{Proof of SLLN}
-\todo{Large parts of lecture 6 are missing}
 \begin{refproof}{lln}
     We want to deduce the SLLN (\autoref{lln}) from \autoref{thm2}.
-    W.l.o.g.~let us assume that $\bE[X_i] = 0$ (otherwise define $X'_i \coloneqq  X_i - \bE[X_i]$).
+    W.l.o.g.~let us assume that $\bE[X_i] = 0$
+    (otherwise define $X'_i \coloneqq  X_i - \bE[X_i]$).
     We will show that $\frac{S_n}{n} \xrightarrow{a.s.} 0$.
     Define $Y_i \coloneqq \frac{X_i}{i}$.
     Then the $Y_i$ are independent and we have $\bE[Y_i] = 0$
     and $\Var(Y_i) = \frac{\sigma^2}{i^2}$.
     Thus $\sum_{i=1}^\infty \Var(Y_i) < \infty$.
-    From \autoref{thm2} we obtain that $\sum_{i=1}^\infty Y_i < \infty$ a.s.
+    From \autoref{thm2} we obtain that $\sum_{i=1}^\infty Y_i$ converges a.s.
     \begin{claim}
-        Let $(a_n)$ be a sequence in $\R$ such that $\sum_{n=1}^{\infty} \frac{a_n}{n}$, then $\frac{a_1 + \ldots + a_n}{n} \to  0$.
+        Let $(a_n)$ be a sequence in $\R$
+        such that $\sum_{n=1}^{\infty} \frac{a_n}{n}$ converges,
+        then $\frac{a_1 + \ldots + a_n}{n} \to  0$.
     \end{claim}
     \begin{subproof}
         Let $S_m \coloneqq \sum_{n=1}^\infty \frac{a_n}{n}$.
         By assumption, there exists $S \in \R$
-        such that $S_m \to  S$ as $m \to  \infty$.
+        such that $S_m \xrightarrow{m \to \infty} S$.
         Note that $j \cdot (S_{j} - S_{j-1}) = a_j$.
         Define $S_0 \coloneqq  0$.
-        Then $a_1 + \ldots + a_n = (S_1 - S_0) + 2(S_2 - S_1) + 3(S_3 - S_2) +
+        Then
-        \ldots + n (S_n - S_{n-1})$.
+        \begin{IEEEeqnarray*}{rCl}
-        Thus $a_1 + \ldots + a_n = n S_n - (S1 $ % TODO
+            a_1 + \ldots + a_n &=&
-   
+            (S_1 - S_0) + 2(S_2 - S_1) + \ldots + n(S_n - S_{n-1})\\
+                               &=& n S_n - (S_1 + S_2 + \ldots + S_{n-1}).
+        \end{IEEEeqnarray*}
+        Thus
+        \begin{IEEEeqnarray*}{rCl}
+            \frac{a_1 + \ldots + a_n}{n}
+            &=& S_n - \frac{S_1 + \ldots + S_{n-1}}{n}\\
+            &=& \underbrace{S_n}_{\to S}
+               - \underbrace{\left( \frac{n-1}{n} \right)}_{\mathclap{\to 1}}
+                \cdot \underbrace{\frac{S_1 + \ldots +  S_{n-1}}{n-1}}_{\to S}\\
+            &\to & 0,
+        \end{IEEEeqnarray*}
+        where we have used
+        \begin{fact}
+            \[
+            \lim_{n \to \infty} S_n = \lim_{n \to \infty} \frac{1}{n}\sum_{i=1}^{n} S_i
+        \].
+        \end{fact}
     \end{subproof}
     The SLLN follows from the claim.
 \end{refproof}
 
-We need the fol]
+In order to prove \autoref{thm2}, we need the following:
 \begin{theorem}[Kolmogorov's inequality]
+    \label{thm:kolmogorovineq}
     If $X_1,\ldots, X_n$ are independent with $\bE[X_i] = 0$
     and $\Var(X_i) = \sigma_i^2$, then
     \[
-    \bP\left[\max_{1 \le  i \le  n} \left| \sum_{j=1}^{i}  X_j \right| > \epsilon \right] \le  \frac{1}{\epsilon ^2} \sum_{i=1}^m \sigma_i^2 % TODO
+    \bP\left[\max_{1 \le  i \le  n} \left| \sum_{j=1}^{i}  X_j \right| > \epsilon \right]
+    \le  \frac{1}{\epsilon^2} \sum_{i=1}^m \sigma_i^2.
     \]
 \end{theorem}
 \begin{proof}
-    Let $A_1 \coloneqq  \{\omega : |X_1(\omega)| > \epsilon\}, \ldots,
+    Let 
-    A_i := \{\omega: |X_1(\omega)| \le \epsilon, |X_1(\omega) + X_2(\omega)| \le  \epsilon, \ldots, |X_1(\omega) + \ldots + X_{i-1}(\omega)| \le \epsilon,
+    \begin{IEEEeqnarray*}{rCl}
-    |X_1(\omega) + \ldots + X_i(\omega)| > \epsilon\}$.
+        A_1 &\coloneqq&  \{\omega : |X_1(\omega)| > \epsilon\},\\
+        A_2 &\coloneqq & \{\omega: |X_1(\omega)| \le \epsilon,
+            |X_1(\omega) + X_2(\omega)| > \epsilon \},\\
+        \ldots\\
+            A_i &\coloneqq& \{\omega: |X_1(\omega)| \le \epsilon,
+                |X_1(\omega) + X_2(\omega)| \le  \epsilon, \ldots,
+                |X_1(\omega) + \ldots + X_{i-1}(\omega)| \le \epsilon,
+                |X_1(\omega) + \ldots + X_i(\omega)| > \epsilon\}.
+    \end{IEEEeqnarray*}
+    It is clear, that the $A_i$ are disjoint.
     We are interested in $\bigcup_{1 \le  i \le n} A_i$.
 
     We have
     \begin{IEEEeqnarray*}{rCl}
-        &&\int_{A_i} (\underbrace{X_1 + \ldots + X_i}_C + \underbrace{X_{i+1} + \ldots + X_n}_D)^2 d \bP\\
+        &&\int_{A_i} (\underbrace{X_1 + \ldots + X_i}_C
-        &=& \int_{A_i} C^2 d\bP + \underbrace{\int_{A_i} D^2 d \bP}_{\ge  0} + 2 \int_{A_i} CD d\bP\\
+            + \underbrace{X_{i+1} + \ldots + X_n}_D)^2 d \bP\\
-                                                                                                       &\ge & \int_{A_i}  \underbrace{C^2}_{\ge \epsilon^2} d \bP + 2 \int \underbrace{\One_{A_i} (X_1 + \ldots + X_i)}_E \underbrace{(X_{i+1} + \ldots + X_n)}_D d \bP\\
+        &=& \int_{A_i} C^2 d\bP
-                                                                                                       &\ge& \int_{A_i} \epsilon^2 d\bP
+            + \underbrace{\int_{A_i} D^2 d \bP}_{\ge  0}
+            + 2 \int_{A_i} CD d\bP\\
+        &\ge& \int_{A_i}  \underbrace{C^2}_{\ge \epsilon^2} d \bP
+        + 2 \int \underbrace{\One_{A_i} (X_1 + \ldots + X_i)}_E \underbrace{(X_{i+1} + \ldots + X_n)}_D d \bP\\
+        &\ge& \int_{A_i} \epsilon^2 d\bP,
     \end{IEEEeqnarray*}
-    (By the independence of $X_1,\ldots, X_n$ and therefore that of $E$ and $D$ and $\bE(X_{i+1}) = \ldots = \bE(X_n) = 0$ we have $\int D E d\bP = 0$.)
+    since by the independence of $E$ and $D$,
-
+    and $\bE(X_{i+1}) = \ldots = \bE(X_n) = 0$ we have $\int D E d\bP = 0$.
-    % TODO
 
+    Hence
+    \[
+    \bP(A_i)
+       \le \frac{1}{\epsilon^2} \int_{A_i} (X_1 + \ldots + X_n)^2 \dif \bP.
+    \]
+    Since the $A_i$ are disjoint, we obtain
+    \begin{IEEEeqnarray*}{rCl}
+        \bP\left( \bigcup_{i \in \N} A_i  \right)
+        &\le & \frac{1}{\epsilon^2}
+        \int_{\bigcup_{i \in \N} A_i} (X_1 + \ldots + X_n)^2 \dif \bP\\
+        &\le & \frac{1}{\epsilon^2}
+        \int_{\Omega} (X_1 + \ldots + X_n)^2 \dif \bP\\
+        &\overset{\text{independence}}{=}&
+        \frac{1}{\epsilon^2}(\bE[X_1^2] + \ldots + \bE[X_n^2])\\
+        &\overset{\bE[X_i] = 0}{=}& \frac{1}{\epsilon^2}
+        \left( \Var(X_1) + \ldots + \Var(X_n)\right).
+    \end{IEEEeqnarray*}
 \end{proof}
 
 \begin{refproof}{thm2}
-    % TODO
+    Let $S_n \coloneqq x_1 + \ldots + x_n$.
+    We'll show that $\{S_n(\omega)\}_{n \in \N}$
+    is a Cauchy sequence
+    for almost every $\omega$.
 
+    Let
+    \[
+    a_m(\omega) \coloneqq  \sup_{k \in \N}
+    \{ | S_{ m+k}(\omega) - S_m(\omega)|\}
+    \]
+    and
+    \[
+    a(\omega) \coloneqq  \inf_{m \in \N} a_m(\omega).
+    \]
+    Then $\{S_n(\omega)\}_{n \in \R}$
+    is a Cauchy sequence iff $a(\omega) = 0$.
+
+    We want to show that $\bP[a(\omega) > 0] = 0$.
+    For this, it suffices to show that $\bP(a(\omega) > \epsilon] = 0$
+    for all $\epsilon > 0$.
+    For a fixed $\epsilon > 0$, we obtain:
+    \begin{IEEEeqnarray*}{rCl}
+        \bP[a_m > \epsilon]
+        &=& \bP[ \sup_{k \in \N} | S_{m+k} - S_m| > \epsilon]\\
+        &=& \lim_{l \to \infty} \bP[%
+        \underbrace{\sup_{k \le l} |S_{m+k} - S_m| > \epsilon}_{%
+        \text{\reflectbox{$\coloneqq$}} B_l \uparrow%
+        B \coloneqq \{\sup_{k \in \N} |S_{m+k} - S_m| > \epsilon\}}%
+        ]
+    \end{IEEEeqnarray*}
+
+    Now,
+    \begin{IEEEeqnarray*}{rCl}
+        &&\max \{|S_{m+1} - S_m|, |S_{m+2} - S_m|, \ldots, |S_{m+l} - S_m|\}\\
+        &=& \max \{|X_{m+1}|, |X_{m+1} + X_{m+2}|, \ldots, |X_{m+1} + X_{m+2} + \ldots + X_{m+l}|\}\\
+        &\overset{\text{\autoref{thm:kolmogorovineq}}}{\le}&
+        \frac{1}{\epsilon^2} \sum_{i=m}^{l} \Var(X_i)\\
+        &\le & \frac{1}{\epsilon^2} \sum_{i=m}^\infty \Var(X_i)
+        \xrightarrow{m \to \infty} 0,
+    \end{IEEEeqnarray*}
+    since by our assumption, $\sum_{n \in \N} \Var(X_i) < \infty$.
+
+    Hence
+    \[
+    \bP[a_m > \epsilon] \xrightarrow{m \to \infty} 0.
+    \]
+    It follows that $\bP[a > \epsilon] = 0$,
+    as claimed.
 \end{refproof}