How to obtain a confidence interval for the location of a Laplace distribution? I have the following p.d.f: 
$$f\left( x|\theta  \right) =\frac { 1 }{ 2 } \exp\left\{ |x-\theta | \right\}$$
with $x,\theta \in \mathbb{R}$
I need a confidence interval for $θ$, but I have no idea how to start. What I can do with that absolute value?
(I'm working with random sample with n=1)
 A: A common strategy for finding confidence intervals is first to identify a pivotal quantity.  This is a random variable that is a function of $X$ and $\theta$ but whose distribution does not depend on $\theta$.  From there it amounts to finding some region to which that pivotal quantity is likely to belong, and then restating this as an interval that is likely to contain $\theta$.
You should be able to show without too much difficulty that the distribution of $|X - \theta|$ does not depend on $\theta$ (you may even be able to guess the distribution).  From there find a suitably large quantile $q$ of this distribution (the $95^\text{th}$ percentile, say) and then "unfold" the event inside $P(|X - \theta| \leq q)$ to where you end up with an interval depending on $X$ and $q$ that contains $\theta$.
A: To start with, since the exponent is an absolute value, what is the least it can be? When theta equals x (assuming x positive), your pdf is equal to 1/2, and thus you know something about the joint distribution. This would be the lower bound of your pdf. If x is negative, then you want theta equal to the opposite of x, and this allows you to form upper bounds on theta. 
