3
$\begingroup$

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.

$\endgroup$
  • 1
    $\begingroup$ I think it's just that $P(X_n > a_n) = P(X_1 > a_n)$. $\endgroup$ – Trevor Gunn Nov 27 '17 at 1:18
  • $\begingroup$ 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? $\endgroup$ – pkenneth81 Nov 27 '17 at 1:37
3
$\begingroup$

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

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

$\endgroup$
  • $\begingroup$ Thanks for the detailed answer. Could you please help me understanding why is it that $P(X_n > a_n) = P(X_1 > a_n)$ ? $\endgroup$ – pkenneth81 Nov 27 '17 at 1:54
  • $\begingroup$ @htennek2k They have the same distribution. If their cdf is $F$ then these probabilities are both $1 - F(a_n)$. $\endgroup$ – Trevor Gunn Nov 27 '17 at 1:58
  • $\begingroup$ Thanks. I see that. I guess my question is whether $a_n = k$ (some $k$) for all $n$. ? $\endgroup$ – pkenneth81 Nov 27 '17 at 2:01
  • $\begingroup$ @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. $\endgroup$ – Trevor Gunn Nov 27 '17 at 2:05
  • $\begingroup$ 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? $\endgroup$ – pkenneth81 Nov 27 '17 at 2:10

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.