Sum of Independent Exponential Random Variables
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.