From aaae7145952eb83f930483e4f615891b93bf9eeb Mon Sep 17 00:00:00 2001 From: Vincent Tam Date: Sat, 15 Dec 2018 12:36:23 +0100 Subject: [PATCH] Post WLLN Ex: fixed math in direct quote --- content/post/2018-12-02-weak-lln-practice.md | 4 ++-- 1 file changed, 2 insertions(+), 2 deletions(-) diff --git a/content/post/2018-12-02-weak-lln-practice.md b/content/post/2018-12-02-weak-lln-practice.md index cf70dab..504adbc 100644 --- a/content/post/2018-12-02-weak-lln-practice.md +++ b/content/post/2018-12-02-weak-lln-practice.md @@ -23,9 +23,9 @@ unsure how to do so. https://math.stackexchange.com/q/3021650/290189, I'll assume the independence of $(X_n)$. By Chebylshev's inequality, -> $$P(|\bar{X}_n|>\epsilon) = P(|\sum_{k=1}^n X_k|> n\epsilon) \le +>$$P(|\bar{X}_n|>\epsilon) = P(|\sum_{k=1}^n X_k|> n\epsilon) \le \frac{\sum_{k=1}^n var(X_k)}{n^2 \epsilon^2} = \frac{\sum_{k=1}^n -(\ln(k))^2}{n^2 \epsilon^2} \le \frac{(\ln(n))^2}{n \epsilon} \to 0$$+(\ln(k))^2}{n^2 \epsilon^2} \le \frac{(\ln(n))^2}{n \epsilon} \to 0$$
Recall: $(\ln(n))^p/n \to 0$ whenever $p \ge 1$. To see this, make a change of variables $n = e^x$, so that it becomes $x^p / e^x$. -- 2.21.0