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In my textbook is an example of an application of Borel-Cantelli's lemma which I don't understand.

Let $X_n$, $n\geq 1$, be a sequence of independent $N(0, \sigma^2)$- distributed random variables, with $\sigma > 0$. From the second lemma of Borel-Cantelli it follows that: P- almost surely (P-a.s.) $\limsup_n X_n = \infty$.

For sake of completeness: $N(0, \sigma^2$) denotes the normal distribution with mean $=0$ and variance $\sigma$. Our 2nd lemma of Borel Cantelli says:

Let $A_n$, $n \geq 1$, be a sequence of independent events on a probability space. Then: $$\sum_n P(A_n) = \infty \Rightarrow P\left[\limsup A_n\right]=1.$$

What I don't see is why the normal distribution of $X_n$ implies that $\sum_nP(\{X_n \leq x\}) = \infty$ (or does it not?) and how from this we then can apply Borel Cantelli's lemma.

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  • $\begingroup$ @saz You are right. I have corrected it. $\endgroup$ – Quasar Oct 1 '17 at 8:33
  • $\begingroup$ You are interested in $X_n$ having large values, right? So it would make more sense to study $\sum_n P(\{X_n \color{red}{\geq} x\})$ ... Use that the random variables are identically distributed! $\endgroup$ – saz Oct 1 '17 at 8:43
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Let us fix a positive integer $N$ and define the (independent) events $A_n:=\left\{X_n\geqslant N\right\}$. Since the random variable $X_n$ has the same distribution as $X_1$, $\mathbb P\left(A_n\right)=\mathbb P\left(A_1\right)\gt 0$, which implies that the series $\sum_{n\geqslant 1}\mathbb P\left(A_n\right)$ is divergent. By the second Borel-Cantelli lemma, $\mathbb P\left(\limsup_{n\to +\infty}A_n\right)=1$, which means that there exists a set $\Omega_N$ of probability one such that for all $\omega\in\Omega_N$, the set $\{n\in\mathbb N, X_n(\omega)\geqslant N\}$ is finite. This implies that $$ \forall \omega\in\Omega_N, \limsup_{n\to +\infty}X_n(\omega)\geqslant N. $$ Let $\Omega':=\bigcap_{N\geqslant 1}\Omega_N$. Then $\Omega'$ has probability one and $\limsup_{n\to +\infty}X_n(\omega)=+\infty$ for all $\omega\in\Omega'$.

Remark that we do not really need the $X_n$ to have a normal distribution: it suffices that the $X_n$ have the same distribution and that $\mathbb P\left(\left\{X_1\geqslant N\right\}\right)\gt 0$ for all $N$.

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