# If $E(X_n^2) = \infty$ then $\limsup \frac{|X_n|}{\sqrt{n}} \geq a$ almost surely.

We have given $$X_1,X_2,\ldots$$ an i.i.d. sequence of random variables such that $$\Bbb{E}(X_1^2)=\infty$$ I claim that for all $$a>0$$ $$\Bbb{P}\left(\limsup_{n\rightarrow \infty} \frac{|X_n|}{\sqrt{n}}\geq a\right)=1$$

My idea was to use Borel-Cantelli, but somehow I'm a bit confused since I never used that $$\Bbb{E}(X_1^2)=\infty$$.

I wanted to do this as follows:

Let $$\Lambda_n=\left\{\frac{|X_n|}{\sqrt{n}}\geq a\right\}$$ then $$\sum_{n=1}^\infty \Bbb{P}(\Lambda_n)=\sum_{n=1}^\infty \Bbb{P}\left(\frac{|X_n|}{\sqrt{n}}\geq a\right)=\sum_{n=1}^\infty 1-\Bbb{P}\left(\frac{|X_n|}{\sqrt{n}}\leq a\right)$$Now if $$\sum_{n=1}^\infty 1-\Bbb{P}\left(\frac{|X_n|}{\sqrt{n}}\leq a\right)<\infty$$ then it would mean that for infinitely many $$n\in \Bbb{N}$$ $$\Bbb{P}\left(\frac{|X_n|}{\sqrt{n}}\leq a\right)=1$$ Here I think I need some argument to show that this is not possible right?

If this works I then could apply Borel-Cantelli and would be done.

I'm not sure if this is correct so.

(I also thought about the central limit theorem but I don't think this is useful here)

• You may wish to use the second Borel-Cantelli lemma, as the one you are citing (i.e. with summable $P(\Lambda_n)$) would give $P (\limsup_n \Lambda_n ) = 0$. May 16, 2022 at 14:19
• @JoseAvilez so you mean from the beginning I should use the second Borel-Cantelli lemma? I.e. I need to show that $\sum_n P(\limsup_n \Lambda_n)=\infty$? May 16, 2022 at 14:27
• That's correct. See below. May 16, 2022 at 14:35

Note that for a non-negative random variable $$Y$$, we have $$E(Y) = \int_0^\infty P(Y > y) dy$$ Since $$S(y) = P(Y > y) = 1 - F_Y(y)$$ is a decreasing function in $$y$$, we have the following Riemann sum approximation: $$E(Y) = \int_0^\infty P(Y > y) dy \leq \sum_{n=1}^\infty P(Y \geq n)$$ Define the events $$\Lambda_n$$ as you did: i.e. $$\Lambda_n = \{ |X_n| \geq a \sqrt{n} \} = \left\lbrace \left(\frac{X_n}{a} \right)^2 \geq n\right\rbrace$$. Then, \begin{align*} \sum_{n=1}^\infty P(\Lambda_n) &= \sum_{n=1}^\infty P\left(\left\lbrace \left(\frac{X_n}{a} \right)^2 \geq n\right\rbrace \right)\\ &= \sum_{n=1}^\infty P\left(\left\lbrace \left(\frac{X_1}{a} \right)^2 \geq n\right\rbrace \right) \\ &\geq \int P(Y >y)dy & \text{where }Y = \frac{X_1^2}{a^2} \\ &= E(Y) \\ &= \infty \end{align*} Since the events $$\Lambda_n$$ are independent, the second Borel-Cantelli lemma allows you to conclude that $$P\left( \limsup_{n \to \infty} \Lambda_n \right) = 1$$.
• sorry I don't see why the first equation with $E(Y)$ holds. Because we have defined $E(Y)=\int_\Omega Y(\omega) \Bbb{P}(d\omega)$ May 16, 2022 at 15:19