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better proof of convergence in L^1 => convergence in measure

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Josia Pietsch 2023-07-13 16:32:47 +02:00
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inputs/prerequisites.tex
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@ -101,7 +101,7 @@ from the lecture on stochastic.
We have We have
\begin{IEEEeqnarray*}{rCl} \begin{IEEEeqnarray*}{rCl}
\|X_n - X\|_{L^p} &=& \|1 \cdot (X_n - X)\|_{L^p}\\ \|X_n - X\|_{L^p} &=& \|1 \cdot (X_n - X)\|_{L^p}\\
&\overset{\text{Hölder}}{\le }& \|1\|_{L^r} \|X_n - X\|_{L^q}\\ &\overset{\text{Hölder}}{\le}& \|1\|_{L^r} \|X_n - X\|_{L^q}\\
&=& \|X_n - X\|_{L^q} &=& \|X_n - X\|_{L^q}
\end{IEEEeqnarray*} \end{IEEEeqnarray*}
Hence $\bE[|X_n - X|^q] \xrightarrow{n\to \infty} 0 \implies \bE[|X_n - X|^p] \xrightarrow{n\to \infty} 0$. Hence $\bE[|X_n - X|^q] \xrightarrow{n\to \infty} 0 \implies \bE[|X_n - X|^p] \xrightarrow{n\to \infty} 0$.
@ -114,7 +114,7 @@ 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}}{\ge}& \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 $X_n \xrightarrow{\bP} X$.