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fixed chebyshev

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Josia Pietsch 2023-07-17 12:36:38 +02:00
parent 3e816f515e
commit 28bc88cbc5
Signed by untrusted user who does not match committer: jrpie
GPG key ID: E70B571D66986A2D
2 changed files with 3 additions and 2 deletions
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1
inputs/lecture_16.tex
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@ -197,6 +197,7 @@ Typically $\cF_n = \sigma(X_1, \ldots, X_n)$ for a sequence of random variables.
\footnote{In this form it means, that there is some filtration, \footnote{In this form it means, that there is some filtration,
that we don't explicitly specify}. that we don't explicitly specify}.
Then $(f(X_n))_n$ is a sub-martingale. Then $(f(X_n))_n$ is a sub-martingale.
Likewise, if $f$ is concave, then $((f(X_n))_n$ is a super-martingale.
\end{corollary} \end{corollary}
\begin{proof} \begin{proof}
Apply \autoref{cjensen}. Apply \autoref{cjensen}.

4
inputs/prerequisites.tex
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@ -280,7 +280,7 @@ This is taken from section 6.1 of the notes on Stochastik.
Let $X$ be a random variable and $a > 0$. Let $X$ be a random variable and $a > 0$.
Then Then
\[ \[
\bP[|X - \bE(X)| \ge a] \le \frac{\Var(X)}{a}. \bP[|X - \bE(X)| \ge a] \le \frac{\Var(X)}{a^2}.
\] \]
\end{theorem} \end{theorem}
\begin{proof} \begin{proof}
@ -288,7 +288,7 @@ This is taken from section 6.1 of the notes on Stochastik.
\begin{IEEEeqnarray*}{rCl} \begin{IEEEeqnarray*}{rCl}
\bP[|X-\bE(X)| \ge a] \bP[|X-\bE(X)| \ge a]
&=& \bP[|X - \bE(X)|^2 \ge a^2]\\ &=& \bP[|X - \bE(X)|^2 \ge a^2]\\
&\overset{\text{Markov}}{\ge}& \frac{\bE[|X - \bE(X)|^2]}{a^2}. &\overset{\text{Markov}}{\le}& \frac{\bE[|X - \bE(X)|^2]}{a^2}.
\end{IEEEeqnarray*} \end{IEEEeqnarray*}
\end{proof} \end{proof}