Let $X_1,X_2,\dots$ be a sequence of i.i.d. random variables with $EX_1=0$ and $EX_1^2=1$. Show that
$$\max\left(\frac{|X_1|}{\sqrt{n}},\frac{|X_2|}{\sqrt{n}}, \dots\right) \longrightarrow0 $$
in distribution.
This exercise is on the section about Central Limit Theorem on Shiryaev's book of Probability.
So, under these conditions, I know that
$$\frac{S_n}{\sqrt{n}}\longrightarrow\cal{N}(0,1)$$
in distribution, but I am not sure about how to proceed…
Best Answer
$\sum_n P(|X_n| >\epsilon \sqrt n)= \sum_n P(|X_1| >\epsilon \sqrt n)=\sum_n P(X_n^{2} >\epsilon^{2} n) <\infty$ because $E\frac {X_1^{2}} {\epsilon ^{2}} <\infty$. By Borel -Cantelli Lemma this implies that $\frac {|X_n|} {\sqrt n} \leq \epsilon$ for all $n$ sufficiently large, with probability $1$. Can you finish the proof?