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.
Thank you very much for your help in advance!
 A: 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: $\sum_{m=1}^{\infty}\mathbb{P}(X_i > \frac1m)\leq \sum_{m=1}^{\infty}\frac{m}{\lambda_i}&lt\infty \iff \sum_{m=1}^{\infty}\mathbb{P}(\text{for infinitely many } i \;\;|X_i|>\frac1m)=0$ (It follows from Borel-Cantelli lemma.)
Thus: $\mathbb{P}(\neg(X_i\to 0))=0 \Rightarrow \mathbb{P}(\text{for infinitely many } i \;\;|X_i|>\frac1m))=0 \iff \sum_{i=1}^{\infty}\frac{1}{\lambda_i}&lt\infty$ (It follows from Borel-Cantelli lemma.)
So we found the necessary and sufficient condition.
A: 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 $\sum P(X_i>\epsilon) &lt \infty\implies\sum e^{-\lambda_i\epsilon} &lt \infty$ 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 $\sum e^{-\lambda_i\epsilon}&lt \infty$ 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.
