tail probabilities for the sum of independent Laplace random variables How might I find tail probabilities (pr X>x), or a reasonable approximation, for a variable that is the sum of independent Laplace random variables? 
 A: I'd suggest using the characteristic funciton of the Laplace Distribution. The characteristic function of the sum of N independent Laplace r.v.'s is the product of their characteristic functions. Then, you can invert the characteristic function to get the density function of the sum. Integrate this result and subtract from 1 to get what you want. 
If you don't feel like getting the exact value, you can use Cantelli's Inequality to estimate the tail probability. If $X_i\sim$ Laplace($\mu_i \, b_i)$, and $Y=\sum X_i$, then you can apply Cantelli's inequality by setting $\mu_Y=\sum \mu_i$ and $\sigma_Y^2=\sum 2b_i^2$ and $\lambda = x$ to get:  
$P(Y-\mu_Y\geq x)\leq \frac{\sigma_Y^2}{\sigma_Y^2+x^2}$ if $x\geq0$ OR
$P(Y-\mu_Y\geq x)\geq 1-\frac{\sigma_Y^2}{\sigma_Y^2+x^2}$ if $x<0$
A: A simple bound that is often useful is to use the Chernoff bound:
From Markov's inequality, we know for a non-negative r.v. $X$, $P(X \geq c) \leq \frac{E[X]}c$. Now, introduce a non-negative parameter $t$, and note $P(t X \geq c t) = P(e^{t X} \geq e^{c t}) \leq \frac{E[e^{t X}]}{c t} = \frac{M_X(t)}{c t}$. You can now optimize the right hand side over $t>0$ to tighten the bound.
This ineqality is highly useful and forms the basis of proof for a lot of other inequalities. You can apply this to a sum as well (let $X = \sum_i Y_i$ where $\{Y_i\}$ is a collection of R.V.'s - note that when $\{Y_i\}$ is an independent collection, $M_X(t) = \prod_i M_{Y_i}(t)$).
In the case of an i.i.d. sum of $\{X_i\}_{i=1}^n$ with common moment generating function $M_X$, the Chernoff bound for the sum takes the form $P(\sum_i X_i \geq c) \leq \inf_{t >0} \frac{(M_X(t))^n}{c t}$.
