For every $\alpha >1$, show that $\frac {\max\{X_1, \ldots ,X_n\}}{n^a} \rightarrow 0$ with probability 1 If $\{X_n\}_{n\in\mathbb{N}}$ is a sequence of independent and identically distributed random variables with $E(X^2_1)<+\infty$, then for every $\alpha >1$ show that $\frac {\max\{X_1, \ldots ,X_n\}}{n^a} \rightarrow 0$ ,with probability 1.
I am trying to prove this but with no luck yet.
I would appreciate any kind of hints.
Thanks for your time.
 A: We have that $\max \{ X_ 1, \cdots , X_n\} \leq \sum_{k = 1}^{n}|X_k|$. By the strong form of the Law of Large Numbers:
$$
\frac{1}{n}\sum_{k = 1}^n |X_k| \to \mathbb{E}[|X|]<\infty \quad \text{ almost surely}
$$
Thus:$$\frac{\max \{ X_ 1, \cdots , X_n\}}{n^\alpha} = 
\frac{\max \{ X_ 1, \cdots , X_n\}}{n \cdot n^{\alpha-1}} \leq \frac{1}{n^{\alpha - 1}} \frac{1}{n}\sum_{k = 1}^n |X_k| \to0 \quad \text{Almost surely}
$$
A: Can someone see if my solution is on the right track?
Let $f$ be the common pdf shared among the random variables stated in the problem, and suppose $X \sim f$. We will first show $E(|X|)$ exists. Notice how $$0\leq\int_{|x|>1}|x|f(x)dx \leq \int _{|x|>1}x^2f(x)dx\leq\int_{-\infty}^{\infty}x^2f(x)dx=E(X^2)<\infty$$
This proves $\int_{|x|>1}|x|f(x)dx$ converges, and since $$0\leq\int_{|x| \leq 1}|x|f(x)dx \leq \int_{|x|\leq 1}f(x)dx=P(|X| \leq 1)<\infty$$ we see that$$0\leq\int_{|x| \leq 1}|x|f(x)dx+\int_{|x|>1}|x|f(x)dx=\int_{-\infty}^{\infty}|x|f(x)dx=E(|X|)$$ exists. This also implies $V(|X|)$ exists since $$V(|X|)=E(|X|^2)-(E(|X|))^2=E(X^2)-(E(|X|))^2<\infty$$
Set $\mu=E(|X|)$ and $\sigma^2=V(|X|)$, and choose $\epsilon>0$ arbitrarily. Since $$|\max\{X_1,..,X_n\}|\leq \max\{|X_1|,...,|X_n|\}\leq \sum_{j=1}^n|X_j|$$ we have by the central limit theorem for large $n$ that $$P\Bigg(\frac{|\max\{X_1,...,X_n\}|}{n^{\alpha}}\leq \epsilon\Bigg)\geq P\Bigg(\frac{1}{n}\sum_{j=1}^n|X_j|\leq\epsilon n^{\alpha-1}\Bigg) \approx P\Bigg(Z\leq\frac{\sqrt{n}(\epsilon n^{\alpha-1}-\mu)}{\sigma}\Bigg)$$ Here $Z \sim N(0,1)$. Take $n \rightarrow \infty$ to get $$P\Bigg(\frac{|\max\{X_1,...,X_n\}|}{n^{\alpha}}\leq \epsilon\Bigg) \rightarrow 1$$ Am I on the right track?
