Given i.i.d. random variables $\{X_n\}$ with finite second moment. Prove $n\cdot P\left(\left|X_{1}\right|\geq\epsilon\sqrt{n}\right)\rightarrow0$ Given i.i.d. random variables $\{X_n\}$ with finite second moment. How to prove $n\cdot P\left(\left|X_{1}\right|\geq\epsilon\sqrt{n}\right)\rightarrow0$?
I've tried Chebyshev inequality:
$$n\cdot P\left(\left|X_{1}\right|\geq\epsilon\sqrt{n}\right)\leq n\frac{Var(X_1)}{\epsilon^2n}=\frac{Var(X_1)}{\epsilon^2}$$
but it didn't work because we only have finite second order moment. Is there any inequalities that are more delicate than Chebyshev inequality?
 A: $nP(|X_1| \geq \epsilon \sqrt n) \leq n\frac { \int_{E_n} |X_1|^{2}dP} {\epsilon n}$ where $E_n=(|X_1| \geq \epsilon \sqrt n)$. Use the fact that $\int_{E_n} |X_1|^{2}dP \to 0$ since the events $\int_{E_n} |X_1|^{2}$ decrease to empty set and $E|X_1|^{2} <\infty$.
A: I will prove the following lemma from which your answer will follow.
Let $X$ be a non-negative real-valued random variable such that $\mathbb E(X)<\infty$. Then $$n \mathbb P[X>n ]\rightarrow 0 \text{ as }n\uparrow \infty$$
Proof: $\mathbb E(X)=\mathbb E(X\mathbb 1_{X\leq n})+\mathbb E(X\mathbb 1_{X>n })$.
Since $X\mathbb 1_{X<n}\uparrow X$ as $n\uparrow \infty$ and all the random variables are non-negative, by Monotone Convergence Theorem we have $$\lim_{n\uparrow \infty }\mathbb E(X\mathbb 1_{X<n})=\mathbb E(X)$$It follows therefore that $$\lim_{n\rightarrow \infty }\mathbb E(X\mathbb 1_{X>n}) =0$$
Since $0\leq n\mathbb 1_{X>n}\leq X\mathbb1 _{X>n}$, we get $$0\leq \mathbb E(n\mathbb 1_{X>n})\leq \mathbb E(X\mathbb1 _{X>n})$$ $$\implies 0\leq n \mathbb P[ X>n ] \leq \mathbb E(X\mathbb1 _{X>n})\rightarrow 0 \text{ as }n\rightarrow \infty$$
Use Sandwich Theorem to conclude.
Finally in your problem look at $Z:=\frac{|X_1 |}{\epsilon}$
