From 0e6e7ad0e3b66e7bf713ecaec5658c1a2c762009 Mon Sep 17 00:00:00 2001 From: Josia Pietsch Date: Thu, 13 Jul 2023 16:32:47 +0200 Subject: [PATCH] better proof of convergence in L^1 => convergence in measure --- inputs/prerequisites.tex | 4 ++-- 1 file changed, 2 insertions(+), 2 deletions(-) diff --git a/inputs/prerequisites.tex b/inputs/prerequisites.tex index 713cd6f..6c825bf 100644 --- a/inputs/prerequisites.tex +++ b/inputs/prerequisites.tex @@ -101,7 +101,7 @@ from the lecture on stochastic. We have \begin{IEEEeqnarray*}{rCl} \|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} \end{IEEEeqnarray*} 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 $\epsilon > 0$ \begin{IEEEeqnarray*}{rCl} \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, \end{IEEEeqnarray*} hence $X_n \xrightarrow{\bP} X$.