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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)

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  • $\begingroup$ 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 $P (\limsup_n \Lambda_n ) = 0$. $\endgroup$ May 16, 2022 at 14:19
  • $\begingroup$ @JoseAvilez so you mean from the beginning I should use the second Borel-Cantelli lemma? I.e. I need to show that $\sum_n P(\limsup_n \Lambda_n)=\infty$? $\endgroup$
    – user123234
    May 16, 2022 at 14:27
  • $\begingroup$ That's correct. See below. $\endgroup$ May 16, 2022 at 14:35

1 Answer 1

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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 $P\left( \limsup_{n \to \infty} \Lambda_n \right) = 1$.

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  • $\begingroup$ sorry I don't see why the first equation with $E(Y)$ holds. Because we have defined $E(Y)=\int_\Omega Y(\omega) \Bbb{P}(d\omega)$ $\endgroup$
    – user123234
    May 16, 2022 at 15:19
  • $\begingroup$ @Wave It's a consequence of Fubini's theorem. See math.stackexchange.com/questions/64186/… $\endgroup$ May 16, 2022 at 15:26

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