Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let $\{X_n\}$ be a sequence of independent Exponential random variables with mean $$ E(X_n)=\frac{1}{\lambda_n}, $$ where $$ 0 < \lambda_n < \infty. $$ If $$ \sum \frac{1}{\lambda_n} = \infty, $$ show $$ P\left(\sum X_n = \infty\right) = 1. $$

It looks like a job for the Borel-Cantelli lemma, but I don't see which sequence of events will give the result.

Let $$ X=\sum X_n, $$ then $$ \sum \frac{1}{\lambda_n}=E(X) \ge \sum P(X>n), $$ but this is in the wrong direction.

share|cite|improve this question
up vote 2 down vote accepted
  1. The event $\sum_j X_j=\infty$ is a tail event, hence its probability is $0$ or $1$.
  2. Assume it is $0$; then the series $\sum_j X_j$ is almost surely convergent, hence it is supposed to converge in law.

  3. Computing the characteristic function of partial sums, and in particular their modulus, one can see this can't happen.

share|cite|improve this answer
Whoa! Of very great level. Thanks. – Nicolas Essis-Breton Apr 29 '13 at 20:25
You are welcome. – Davide Giraudo Apr 29 '13 at 20:30

I can't reach the contradiction. I get the characteristic function of the partial sums to be $\phi_{n}(\theta) = \prod_{i=1}^{n} \left(\frac{1}{1-i\frac{\theta}{\lambda_{i}}} \right)$. So $|\phi_{n}(\theta)| = \prod_{i=1}^{n} \frac{1}{\sqrt{1 + (\frac{\theta}{\lambda_{i}})^2}} \leq 1$. What am I doing wrong?

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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