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lecture 6

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\lecture{6}{}{}
\todo{Large parts of lecture 6 are missing}
\lecture{6}{}{Proof of SLLN}
\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 $a_1 + \ldots + a_n = (S_1 - S_0) + 2(S_2 - S_1) + 3(S_3 - S_2) +
\ldots + n (S_n - S_{n-1})$.
Thus $a_1 + \ldots + a_n = n S_n - (S1 $ % TODO
Then
\begin{IEEEeqnarray*}{rCl}
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 $A_1 \coloneqq \{\omega : |X_1(\omega)| > \epsilon\}, \ldots,
A_i := \{\omega: |X_1(\omega)| \le \epsilon, |X_1(\omega) + X_2(\omega)| \le \epsilon, \ldots, |X_1(\omega) + \ldots + X_{i-1}(\omega)| \le \epsilon,
|X_1(\omega) + \ldots + X_i(\omega)| > \epsilon\}$.
Let
\begin{IEEEeqnarray*}{rCl}
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} C^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
&&\int_{A_i} (\underbrace{X_1 + \ldots + X_i}_C
+ \underbrace{X_{i+1} + \ldots + X_n}_D)^2 d \bP\\
&=& \int_{A_i} C^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 $\int D E d\bP = 0$.)
% TODO
since by the independence of $E$ and $D$,
and $\bE(X_{i+1}) = \ldots = \bE(X_n) = 0$ we have $\int D E d\bP = 0$.
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}