Commit aaae7145 authored by Vincent Tam's avatar Vincent Tam

Post WLLN Ex: fixed math in direct quote

parent 3ebc403d
Pipeline #40253575 passed with stages
in 35 seconds
......@@ -23,9 +23,9 @@ unsure how to do so., I'll assume the independence of
$(X_n)$. By Chebylshev's inequality,
> $$P(|\bar{X}_n|>\epsilon) = P(|\sum_{k=1}^n X_k|> n\epsilon) \le
> <div>$$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$$
(\ln(k))^2}{n^2 \epsilon^2} \le \frac{(\ln(n))^2}{n \epsilon} \to 0$$</div>
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 $x^p / e^x$.
Markdown is supported
0% or
You are about to add 0 people to the discussion. Proceed with caution.
Finish editing this message first!
Please register or to comment