Borel Zero-One Law for a sequence of constants

Question

I'm working on a problem and need some intuition to get unstuck, thanks in advance.

So, we are given a sequence of iid random variables $ \{X_n, n \geq 1\}$. Then we have $\{a_n\}$ is a sequence of constants.

We need to show:

$$P \{ [X_n > a_n] \text{ i.o.} \} = \begin{cases} 0 & \text{iff } \sum_n P[X_1 >a_n] < \infty \\ 1 & \text{iff } \sum_n P[X_1 >a_n] = \infty \end{cases} $$ My intuition:

It is clear that this problem is related to Borel Zero-One Law.

So, I think of defining the event $A_n = X_1 > a_n$ and then, claim that by the one-zero law:

$$P \{ [A_n] \text{ i.o.} \} = \begin{cases} 0 & \text{iff } \sum_n P(A_n) < \infty \\ 1 & \text{iff } \sum_n P(A_n) = \infty \end{cases} $$ However, I think I'm missing something or ignoring something by not considering that what I'm asked to show involves the sum of terms having $X_1$ only: i.e. $\sum_n P[X_1 > a_n]$.

Any thought you may have about it?

Thanks.

Answer 1

Let $A_n = \{X_n > a_n\}$. Then you'll agree that what we are trying to calculate is $P(A_n \text{ i.o})$. These events are independent (since the random variables are independent) so by the Borell-Cantelli Lemmas, $P(A_n \text{ i.o})$ is $0$ or $1$ depending on whether or not the sum

$$ \sum_{n = 1}^\infty P(A_n) $$

converges. But since $P(A_n) = P(X_n > a_n) = P(X_1 > a_n)$ what we have is

$$ \sum_{n = 1}^\infty P(A_n) = \sum_{n = 1}^\infty P(X_1 > a_n). $$