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Josia Pietsch 2023-07-12 15:27:13 +02:00
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inputs/lecture_06.tex
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@ -63,9 +63,9 @@ In order to prove \autoref{thm2}, we need the following:
|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\}.
|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 $\bigcup_{1 \le i \le n} A_i$.
@ -179,7 +179,7 @@ In order to prove \autoref{thm2}, we need the following:
since $\{N_t \ge n\} = \{X_1 + \ldots+ X_n \le t\}$.
\begin{claim}
$\bP[\frac{S_n}{n} \xrightarrow{n \to \infty} m
$\bP[\frac{S_n}{n} \xrightarrow{n \to \infty} m
\land N_t \xrightarrow{t \to \infty} \infty] = 1$.
\end{claim}
\begin{subproof}