The problem:

Let $(X_n)_{n\geq 1}$ be a real-valued sequence of i.i.d. random variables and let $c > 0$. Use Borel-Cantelli's lemma to show that

$$\sum_{n=1}^\infty P(X_n^2 > n) < \infty \Rightarrow P(|X_n|\geq c\sqrt{n} \hspace{7pt} \text{i.o.} \hspace{7pt} )=0.$$

My attempt: So from Borel-Cantelli we have $$P(X_n^2 > n \hspace{7pt}\text{i.o.}\hspace{7pt})=0$$

and using the definition of "infinitely often":

$$\bigcap_{m=1}^\infty \bigcup_{n=1}^\infty (X_n^2 > n)=\bigcap_{m=1}^\infty \bigcup_{n=1}^\infty (|X_n| > \sqrt{n})$$

But I don't see how I get the inclusion into the event containing $c$.

  • $\begingroup$ This sounds like a homework question so I'll only give a hint. You want to show that $\sum_{n=1}^\infty P( X_n^2 > cn) = \infty$ but you only know that $\sum_{n=1}^\infty P(X_n^2 > n) = \infty$. Try choosing $k > c$ and dividing the latter sum into $k$ pieces, corresponding to the $k$ residue classes mod $k$. The sums are going to be 'approximately' equal therefore they are all infinite. $\endgroup$ Oct 21 '18 at 23:16
  • 1
    $\begingroup$ I don't have a complete solution, but I suspect one way forward might be to actually not use the given condition directly. In particular I think $\sum^{\infty} P(X_n^2 > n) < \infty$ implies that $E(X_n^2)<\infty$, which you might be able to bound with Chebyshev's inequality and then in turn feed to the BC lemma to. That might resolve the difficulty with $c$. Just an idea. $\endgroup$
    – gogurt
    Oct 21 '18 at 23:47

The reasoning in the opening post gives indeed the result when $0\lt c\leqslant 1$. In order to extend it to all the values of $c$, we use the following facts:

  1. Since the sequence $(X_n)_{n\geqslant 1}$ is identically distributed, $\Pr\{X_n^2\gt n\}=\Pr\{X_1^2\gt n\}$;
  2. If $Y$ is a non.negative integrable random variable, them $\sum_{n\geqslant 1}\Pr\{Y\gt n\}\leqslant \mathbb E[Y]$.

Consequently, for all positive $c$, the series $\sum_{n\geqslant 1}\Pr\{X_n^2\gt cn\}$ converges.

Notice that we only use the fact that the sequence $(X_n)_{n\geqslant 1}$ is identically distributed. Its independence is not needed.


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