Sum of Independent Exponential Random Variables Can we prove a sharp concentration result on the sum of independent exponential random variables, i.e. Let $X_1, \ldots X_r$ be independent random variables such that $Pr(X_i &lt; x) = 1 - e^{-x/\lambda_i}$. Let $Z = \sum X_i$. Can we prove bounds of the form $Pr(|Z-\mu_Z|&gt;t) &lt; e^{-t^2/\sum (\lambda_i)^2}$. This follows directly if we use the variance form of chernoff bounds and hence I believe is true , but the bounds that I read require bounded-ness or have some dependence on bounded-ness of the variables. Could someone point to me to a proof of the above ?
For concreteness, say that the pdf of the r.v. $X_i$ is 

$$p(X_i = x) = \frac{1}{2} \lambda_i e^{-\lambda_i|x|}.$$ 

This is the Laplace distribution, or the double exponential distribution. Its variance is $\frac{2}{\lambda_i^2}$. The cdf is

$$
\Pr[X_i \leq x] = 1 - \frac{1}{2}e^{-\lambda_i x}
$$
for $x \geq 0$. 

The moment generating function of $X_i$ is 

$$
\mathbf{E}\ e^{uX_i} = \frac{1}{1 - u^2/\lambda_i^2},
$$
for $|u| &lt; \lambda_i$. Using this fact and the exponential moment method which is standard in the proof of Chernoff bounds, you get that for $X = \sum_i X_i$ and $\sigma^2 = 2\sum_i \lambda_i^{-2}$, the following inequality holds

$$
\Pr[X &gt; t\sigma] &lt; e^{-t^2/4},
$$
as long as $t \leq 2\sigma \min_{i}{\lambda_i}$. You can find a detailed derivation in the proof of Lemma 2.8 of this paper.