# Borel Zero-One Law for a sequence of constants

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.

• I think it's just that $P(X_n > a_n) = P(X_1 > a_n)$. – Trevor Gunn Nov 27 '17 at 1:18
• Thanks for your reply. Any intuition that may support your claim? I think since all the X_n are iid, it's not necessarily true that i.e. $P[X_1>a_1 ] = P[X_2 > a_2]$ considering that the sequence $\{ a_n\}$ can be any sequence of different constants. I don't think $a_n = k, \forall n$. However, If that is the case, then since all $X_n$ are iid, $P[X_n > a_n] = P[X_1 > a_n]$. But then why they used $a_n$? They could had used simply k. What do you think? – pkenneth81 Nov 27 '17 at 1:37

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

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

• Thanks for the detailed answer. Could you please help me understanding why is it that $P(X_n > a_n) = P(X_1 > a_n)$ ? – pkenneth81 Nov 27 '17 at 1:54
• @htennek2k They have the same distribution. If their cdf is $F$ then these probabilities are both $1 - F(a_n)$. – Trevor Gunn Nov 27 '17 at 1:58
• Thanks. I see that. I guess my question is whether $a_n = k$ (some $k$) for all $n$. ? – pkenneth81 Nov 27 '17 at 2:01
• @htennek2k that's a special case. Another one is where $a_n = n$ and $X_n$ is non-negative. Then $\sum_n P(X_1 > n) < \infty$ iff $E(X_n) < \infty$. This is a good exercise and basically amounts to the integral test for convergence of series. – Trevor Gunn Nov 27 '17 at 2:05
• Thanks again for your reply. Just please, need a little more help. Lets take the case you mentioned $a_n = n$, then, It's hard for me to see that $P(X_n > n) = P(X_{n+k} > n+k)$. Unless... we think of very large n. Is that the trick? – pkenneth81 Nov 27 '17 at 2:10