6
$\begingroup$

I have a sequence of independent random variables $X_1, X_2,...$ such that $P(X_n = 1) = \frac{1}{n}$ and $P(X_n = 0) = 1 - \frac{1}{n}$. Using the second Borel-Cantelli Lemma, we have $\sum P(X_n \neq 0) = \sum P(X_n = 1) = \sum \frac{1}{n} = \infty$ implies that $X_n$ does not converge to 0 a.s.

But I feel like we can construct an example where almost convergence occurs. Let $\Omega = (0, 1)$, and let the probability measure be the Lebesgue measure restricted to $(0, 1)$. Finally, let $X_n = \mathbb{I}\{(0, \frac{1}{n})\}$. Then for any $\omega \in \Omega$, $\underset{n \to \infty}{\lim} X_n(\omega) = 0$, so $X_n$ converges to 0 a.s.

I don't understand why I'm getting contradicting conclusions. Any help would be much appreciated. Thank you!

$\endgroup$
5
$\begingroup$

The random variables $X_n := 1_{(0,1/n)}$ are not independent. This follows from the fact that

$$\lambda(X_n=1,X_m=1) = \lambda(X_m=1) = \frac{1}{m} \neq \frac{1}{n} \frac{1}{m} = \lambda(X_n=1) \lambda(X_m=1)$$

for any $m \leq n$. (Here $\lambda$ denotes the Lebesgue measure restricted to $(0,1)$.)

Consequence: The assumptions of the Borel-Cantelli Lema are not satisfied. Therefore, the almost surely convergence of the sequence is no contradiction to the Borel-Cantelli Lemma.

$\endgroup$
  • $\begingroup$ Thank you so much for the reply! I feel so stupid now. :P I'm sorry I couldn't up-vote because of my low reputation. $\endgroup$ – user207886 Jan 14 '15 at 18:16
  • $\begingroup$ @user207886 You are welcome. $\endgroup$ – saz Jan 14 '15 at 18:36

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.