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I'm stucked with this exercise.

Let $X_1,X_2,\ldots$ be i.i.d. random variables with $E(X_1)=0$ and $Var(X_1)=\infty$ Prove that$$P(\limsup\limits_{n\to\infty}\{|X_n|\geq \sqrt{n}\})=1$$

I need to show $$\sum\limits_{n=1}^\infty P(|X_n|\geq \sqrt{n})=1$$ so I can use Borel-Cantelli.

I know from the law of large numbers that $$\sum\limits_{n=1}^\infty \frac{X_n}{n}=E(X_1)=0$$ So this means that $$\lim\limits_{n\to \infty}\frac{X_n}{c_n}=0\quad \text{or} \quad P(\lim\limits_{n\to \infty}\frac{X_n}{c_n}=0)=1$$

I don't know how to go further.

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First of all, note that you have to show

$$\sum_{n \geq 1} \mathbb{P}(|X_n| \geq \sqrt{n}) = \color{red}{\infty} \tag{1}$$

in order to apply the Borel-Cantelli-lemma.


Suppose that $(1)$ does not hold true, i.e. $\sum_{n \geq 1} \mathbb{P}(|X_n| \geq \sqrt{n}) < \infty$. Since the random variables are identically distributed, this implies

$$\sum_{n \geq 1} \mathbb{P}(|X_1|^2 \geq n)< \infty. \tag{2}$$

Now recall that for any non-negative random variable $Y$ we have

$$\mathbb{E}(Y) \leq \sum_{n \geq 0} \mathbb{P}(Y \geq n)$$

(this is a direct consequence of the pointwise inequality $Y \leq \sum_{n \geq 0} 1_{\{Y \geq n\}}$). Applying this for $Y= |X_1|^2$, we conclude from $(2)$ that

$$\mathbb{E}(X_1^2) \leq \sum_{n \geq 0} \mathbb{P}(|X_1|^2 \geq n)<\infty$$

in contradiction to our assumption that $X_1$ has infinite variance.

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  • $\begingroup$ @MarcZ You are welcome. $\endgroup$ – saz Apr 21 '16 at 16:25
  • $\begingroup$ Your equation $Y = \Sigma_{n \geq 1} 1_{Y \geq n}$ is only true for non-negative integer valued random variables. I think the question doesn't assume this? $\endgroup$ – Lorenzo Oct 31 '16 at 17:51
  • $\begingroup$ @AreaMan Yeah, but I did mention in my answer that the equation holds only for non-negative $Y$.... and obviously $Y = |X_1|^2$ is non-negative, so there is no trouble using this identity the way I did. $\endgroup$ – saz Oct 31 '16 at 23:34
  • $\begingroup$ I was complaining about integer values or not... $\endgroup$ – Lorenzo Oct 31 '16 at 23:34
  • $\begingroup$ @AreaMan Ah, sorry, I was too rush. You are right, the equation holds only for non-negative integer valued random variables, but there is a similar inequality for non-negative (real-valued) random variables; see my edited answer. $\endgroup$ – saz Nov 1 '16 at 6:34

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