Posterior computation for Laplace distribution I am dealing with being Bayesian and looking for a closed form for a posterior for the scale parameter $\tau$ of a Laplace distribution, such that I can derive a full conditional in my Gibbs sampler.
I do not think I could exploit conjugacy for the Laplace, but any closed-form computation for the posterior would be useful. 
Any ideas on how to choose the prior and obtain a closed form posterior from which it is known how to sample from?
 A: Conditioning on $\lambda$, the posterior on $\mu$ can be expressed in a reasonably closed form:
\begin{align}
\pi(\mu|x_1,\ldots,x_n) &\propto \pi(\mu) \exp\left(-\frac{1}{\lambda}\sum_{i=1}^n|x_i-\mu|\right)\\
&= \pi(\mu) \exp\left(-\frac{1}{\lambda}\sum_{i=1}^n|x_{(i)}-\mu|\right)\\
&= \pi(\mu) \sum_{j=0}^n \mathbb{I}_{(x_{(j)},x_{(j+1)})}(\mu) \exp\left(-\frac{1}{\lambda}\sum_{i=1}^n|x_{(i)}-\mu|\right)\\
&= \pi(\mu) \sum_{j=0}^n \mathbb{I}_{(x_{(j)},x_{(j+1)})}(\mu) \exp\left(\frac{1}{\lambda}\sum_{i=1}^j[x_{(i)}-\mu|]\right)\exp\left(\frac{1}{\lambda}\sum_{i=j+1}^n[\mu-x_{(i)}|]\right)\\
&=\sum_{j=0}^n \exp\left(\frac{1}{\lambda}\sum_{i=1}^jx_{(i)}-\frac{1}{\lambda}\sum_{i=j+1}^nx_{(i)}\right)\pi(\mu)\exp\left(\frac{2j-n}{\lambda}\mu\right)\mathbb{I}_{(x_{(j)},x_{(j+1)})}(\mu)
\end{align}
since using a Normal prior $\pi(\mu)$ returns a mixture of truncated Normals. For instance, if the prior is a ${\cal N}(0,\sigma^2)$, then
\begin{align}
\pi(\mu|x_1,\ldots,x_n) &\propto
\sum_{j=0}^n \overbrace{\exp\left(\frac{1}{\lambda}\sum_{i=1}^jx_{(i)}-\frac{1}{\lambda}\sum_{i=j+1}^nx_{(i)}\right)}^{\omega_j}\exp\left(\frac{2j-n}{\lambda}\mu-\frac{\mu^2}{2\sigma^2}\right)\mathbb{I}_{(x_{(j)},x_{(j+1)})}(\mu)\\
&\propto \sum_{j=0}^n \omega_j\exp\left(2\sigma^2\frac{2j-n}{\lambda}\frac{\mu}{2\sigma^2}-\frac{\mu^2}{2\sigma^2}\right)\mathbb{I}_{(x_{(j)},x_{(j+1)})}(\mu)\\
&\propto \sum_{j=0}^n \omega_j\exp\left(\sigma^2\frac{(2j-n)^2}{\lambda^2}\right)\exp\left(\frac{-1}{2\sigma^2}\left[\mu-\frac{(2n-j)\sigma^2}{\lambda}\right]^2\right)
\mathbb{I}_{(x_{(j)},x_{(j+1)})}(\mu)\\
\end{align}
a mixture of $n+1$ truncated Normal distributions in $\mu$, truncated respectively to the intervals $(x_{(j)},x_{(j+1)})$ with original mean $(2n-j)\sigma^2/\lambda$ and original variance $\sigma^2$. (While obvious, the weights of the mixture are cumbersome in that they imply the coverage probability of $(x_{(j)},x_{(j+1)})$ by the Normal distribution with mean $(2n-j)\sigma^2/\lambda$ and variance $\sigma^2$.)
And the conditional posterior on $\lambda$ associated with an Inverse Gamma ${\cal IG}(a,b)$ is indeed an Inverse Gamma$${\cal IG}\left(a+n,b+\sum_i|x_i-\mu|\right)$$which implies that a two-step Gibbs sampler can be implemented.
A: The likelihood for $n$ iid observations looks like:
$ f(x_1,...x_n|\lambda,\mu) \propto \frac{1}{\lambda^n} exp(-\frac{1}{\lambda}\sum_{i=1}^n|x_i-\mu|)$
Hence a conjugate prior for $\lambda$ with $\mu, x$ known must (thinking only about the algebra) look like:
$ f(\lambda) \propto \frac{1}{\lambda^a} exp(-\frac{b}{\lambda})$
As suggested by marmle, this is an Inverse Gamma, although to be nice we'd need to change $a\rightarrow a-1$ and let $a>0, b>0$.
EDIT: To get the updated parameters for $\lambda$:
$ f(\lambda|x_1,...x_n, \mu) \propto f(\lambda)f(x_1,...x_n|\lambda,\mu) $
$ \propto \frac{1}{\lambda^{a-1}} exp(-\frac{b}{\lambda}) \frac{1}{\lambda^n} 
exp(-\frac{1}{\lambda}\sum_{i=1}^n|x_i-\mu|)$
$ \propto \frac{1}{\lambda^{n+a-1}} exp(-\frac{1}{\lambda}(b+\sum_{i=1}^n|x_i-\mu|)) $
