# Second lemma of Borel-Cantelli: Normal distribution

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.

• @saz You are right. I have corrected it. – Quasar Oct 1 '17 at 8:33
• 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! – saz Oct 1 '17 at 8:43

## 1 Answer

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 $$N$$.