# Convergence of exponential random variables

How to find necessary and sufficient condition when $X_i \text{ ~ } \text{Exp} (\lambda_i)$ sequence of independent random variables converge to zero in distribution and almost surely.

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benny: Surely you tried something, you scribbled some estimates on a piece of paper. Please show these. – Did Apr 14 '12 at 19:24
I am afraid I have no valuable results. I would be grateful if you could help me. – benny Apr 14 '12 at 22:01
@benny Are these random variables independent? If so, this is crucial information that you left out of the question. – Byron Schmuland Apr 15 '12 at 15:34
benny: Do not be shy! Surely this question does not come out from nowhere... Surely you solved similar, simpler ones and/or were given some tools, or related results, whatever. As it is, you make it look like you are only interested in a full written solution, and not at all in thinking about this problem. Come on, just prove me wrong on this! – Did Apr 15 '12 at 15:44

Let's see the first part, convergence in distribution means that $F_{X_n}(x) \to_{n \to \infty} F_X(x) \;\;\; \forall x$.

Thus: $F_{X_i}(x) \to F_0(x) \iff 1-e^{-\lambda_ix} \to 1\text{ if } x> 0 \iff \lambda_i \to \infty$.

So we found the necessary and sufficient condition.

Let's see the other part: almost surely convergence means that $\mathbb{P}(\lim_{n \to \infty} X_n=X)=1$.

Consider the following: $\mathbb{P}(\neg(X_i\to 0))=\mathbb{P}(\bigcup_{m=1}^{\infty}\{\text{for infinitely many } i \;\;|X_i|>\frac1m\}) \leq \sum_{m=1}^{\infty}\mathbb{P}(\text{for infinitely many } i \;\;|X_i|>\frac1m)$

But: $\mathbb{P}(X_i > \frac1m) \leq \frac{\mathbb{E}(X_i)}{\frac1m}=\frac{m}{\lambda_i}$ (We know it from Markov inequality.)

Thus: $\sum_{m=1}^{\infty}\mathbb{P}(X_i > \frac1m)\leq \sum_{m=1}^{\infty}\frac{m}{\lambda_i}<\infty \iff \sum_{m=1}^{\infty}\mathbb{P}(\text{for infinitely many } i \;\;|X_i|>\frac1m)=0$ (It follows from Borel-Cantelli lemma.)

Thus: $\mathbb{P}(\neg(X_i\to 0))=0 \Rightarrow \mathbb{P}(\text{for infinitely many } i \;\;|X_i|>\frac1m))=0 \iff \sum_{i=1}^{\infty}\frac{1}{\lambda_i}<\infty$ (It follows from Borel-Cantelli lemma.)

So we found the necessary and sufficient condition.

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This only shows that, if $\sum\limits_i1/\lambda_i$ converges, then $X_i\to0$ almost surely, not the other implication. – Did Apr 16 '12 at 13:14
@Dawson there something wrong here. Firstly in the second line you have used the tail cdf in stead of the proper cdf. It is clear that $\lambda_i$ should goes to infinity. In the second part I think the condition should follow from $\sum_n e^{-\lambda_n \varepsilon}<+ \infty$. For example $\lambda_n =n$ satisfies the convergence. – Kolmo Apr 16 '12 at 13:23
@Kolmo Whoops! I made those changes in error. I will change them back. – Byron Schmuland Apr 16 '12 at 15:18

Part 1 is answered by Dawson. For part 2, we need to use both Borel Cantelli lemmas:

• Use the first BC lemma, for any $\epsilon>0$ we have $\sum P(X_i>\epsilon) < \infty\implies\sum e^{-\lambda_i\epsilon} < \infty$ is the sufficient condition as $P(\lim_{n\to \infty} X_n>\epsilon)=0$.
• Use the second BC lemma, as the sequence is independent. $\sum P(X_i>\epsilon) = \infty\implies \sum e^{-\lambda_i\epsilon} = \infty$ is the condition for $P(\lim_{n\to \infty} X_n>\epsilon)=1$.

This means $\sum e^{-\lambda_i\epsilon}< \infty$ is both necessary and sufficient.

PS: As noted in a comment, it is better not to use Markov inequality in such proofs as it is a weak inequality.

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