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fixed typo

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Josia Pietsch 2023-07-13 18:35:33 +02:00
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inputs/prerequisites.tex
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@ -77,16 +77,17 @@ from the lecture on stochastic.
$X_n \xrightarrow{\text{a.s.}} X \implies X_n \xrightarrow{\bP} X$. $X_n \xrightarrow{\text{a.s.}} X \implies X_n \xrightarrow{\bP} X$.
\end{claim} \end{claim}
\begin{subproof} \begin{subproof}
$\Omega_0 \coloneqq Let $\Omega_0 \coloneqq
\{\omega \in \Omega : \lim_{n\to \infty} X_n(\omega) = X(\omega)\}$. \{\omega \in \Omega : \lim_{n\to \infty} X_n(\omega) = X(\omega)\}$.
Let $\epsilon > 0$ and consider Fix some $\epsilon > 0$ and consider
$A_n \coloneqq \bigcup_{m \ge n} $A_n \coloneqq \bigcup_{m \ge n}
\{\omega \in \Omega: |X_m(\omega) - X(\omega)| > \epsilon\}$. \{\omega \in \Omega: |X_m(\omega) - X(\omega)| > \epsilon\}$.
Then $A_n \supseteq A_{n+1} \supseteq \ldots$ Then $A_n \supseteq A_{n+1} \supseteq \ldots$
Define $A \coloneqq \bigcap_{n \in \N} A_n$. Define $A \coloneqq \bigcap_{n \in \N} A_n$.
Then $\bP[A_n] \xrightarrow{n\to \infty} \bP[A]$. Then $\bP[A_n] \xrightarrow{n\to \infty} \bP[A]$.
Since $X_n \xrightarrow{a.s.} X$ we have that Since $X_n \xrightarrow{a.s.} X$ we have that
$\forall \omega \in \Omega_0 \exists n \in \N \forall m \ge n |X_m(\omega) - X(\omega)| < \epsilon$. \[\forall \omega \in \Omega_0 .~ \exists n \in \N .~
\forall m \ge n.~ |X_m(\omega) - X(\omega)| < \epsilon.\]
We have $A \subseteq \Omega_0^{c}$, hence $\bP[A_n] \to 0$. We have $A \subseteq \Omega_0^{c}$, hence $\bP[A_n] \to 0$.
Thus \[ Thus \[
\bP[\{\omega \in \Omega | ~|X_n(\omega) - X(\omega)| > \epsilon\}] < \bP[A_n] \to 0. \bP[\{\omega \in \Omega | ~|X_n(\omega) - X(\omega)| > \epsilon\}] < \bP[A_n] \to 0.
@ -114,7 +115,7 @@ from the lecture on stochastic.
Then for every $\epsilon > 0$ Then for every $\epsilon > 0$
\begin{IEEEeqnarray*}{rCl} \begin{IEEEeqnarray*}{rCl}
\bP[|X_n - X| \ge \epsilon] \bP[|X_n - X| \ge \epsilon]
&\overset{\text{Markov}}{\ge}& \frac{\bE[|X_n - X|]}{\epsilon}\\ &\overset{\text{Markov}}{\le}& \frac{\bE[|X_n - X|]}{\epsilon}\\
&\xrightarrow{n \to \infty} & 0, &\xrightarrow{n \to \infty} & 0,
\end{IEEEeqnarray*} \end{IEEEeqnarray*}
hence $X_n \xrightarrow{\bP} X$. hence $X_n \xrightarrow{\bP} X$.