Conditional distribution of $\exp(-|x|-|y|-a \cdot |x-y|)$ I am trying to use Gibbs sampling or Metropolis-Hastings to draw samples from the joint distribution$$f(x,y)\propto\exp(-|x|-|y|-a \cdot |x-y|)$$ For this I need the conditional distributions of $x$ and $y$ ($a$ is known and constant). How do I obtain these? Is there a trick / shortcut?
 A: 
Disclaimer: although there is nothing to complain about Ben's answer (!), except maybe that the normalising constant of the conditional is not of direct use, here is what I
  wrote while being off-line, so I may as well post it!

The full conditional of $X$ given $Y$ has a density that is proportional to
\begin{align}
f(x|y) &\propto \exp\{ -|x|-a|y-x|\}\\
&=\begin{cases}
\exp\{ -x-a(y-x)\} &\text{ if } 0\le x\le y\\
\exp\{ x-a(y-x)\} &\text{ if } x\le \min(y,0)\\
\exp\{ -x+a(y-x)\} &\text{ if } x\ge \max(0,y)\\
\exp\{ x+a(y-x)\} &\text{ if } y\le x\le 0\\
\end{cases} \\
&=\begin{cases}
\exp\{ -(1-a)x-ay\} &\text{ if } 0\le x\le y\\
\exp\{ (1-a)x-ay\} &\text{ if } x\le \min(y,0)\\
\exp\{ -(a+1)x+ay\} &\text{ if } x\ge \max(0,y)\\
\exp\{ (1-a)x+ay\} &\text{ if } y\le x\le 0\\
\end{cases} \\
\end{align}
with one of the two conditions $0\le x\le y$ or $y\le x\le 0$ being obviously empty. Each of the (lhs) terms can be normalised into a (rhs) density:
\begin{align}\exp\{ -(1-a)x\} \mathbb{I}_{0\le x\le y} &\propto \dfrac{(a-1)\exp\{ -(1-a)x\}}{\exp\{ (a-1)y\}-1}\mathbb{I}_{0\le x\le y}\\
\exp\{ (1+a)x\} \mathbb{I}_{x\le \min(y,0)} &\propto \dfrac{(1+a)\exp\{ (1+a)x\}}{\exp\{ (1+a)\min(y,0)\}} \mathbb{I}_{x\le \min(y,0)}\\
\exp\{ -(1+a)x\} \mathbb{I}_{x\ge \max(0,y)}&\propto \dfrac{(1+a)\exp\{ (1+a)x\}}{\exp\{ (1+a)\max(0,y)\}}\mathbb{I}_{x\ge \max(0,y)}\\
\exp\{ (1-a)x\} \mathbb{I}_{y\le x\le 0} &\propto \dfrac{(1-a)\exp\{ (1-a)x\}}{1-\exp\{ (1-a)y\}}\mathbb{I}_{y\le x\le 0}\\
\end{align}
which means that it is directly possible to simulate from each of these densities, by inverse cdf, picking one of the four (three if accounting fo) with probabilities
$$\dfrac{\exp\{-ay\}[\exp\{ -(1-a)y\}-1]}{a-1}\mathbb{I}_{y\ge 0}, \dfrac{\exp\{-ay\}\exp\{ (1+a)\min(y,0)\}}{1+a}, \dfrac{\exp\{ay\}\exp\{ (1+a)\max(0,y)\}}{1+a}, \dfrac{\exp\{ay\}[1-\exp\{ (1-a)y\}]}{1-a}\mathbb{I}_{y\le 0}$$
(Obviously, a basic accept-reject algorithm based on a double exponential $\pm\mathcal{E}(1)$ proposal avoids this painful derivation since, when simulating $X\sim\pm\mathcal{E}(1)$ the accept-reject ratio
$$\dfrac{\exp\{ -|x|-a|y-x|\}}{\exp\{-|x|\}}$$ is $\exp\{-a|y-x|\}\le 1$ and hence can be compared with a Uniformn draw.) Here is a Gibbs outcome for $a=10$ and $10^3$ Gibbs steps (using accept-reject sampling):

A: An alternative to the painful simulation from the exact full conditional distributions is to ressort to slice sampling, that is, to express the density in $(x,y)$ as the marginal of a density in $(x,y,u_1,u_2,u_3)$ as follows:
\begin{align*}f(x,y)&\propto\exp(-|x|-|y|-a \cdot |x-y|)\\&=\int_0^\infty\mathbb{I}_{u_1\le\exp\{-|x|\}}\text{d}u_1\int_0^\infty\mathbb{I}_{u_2\le\exp\{-|y|\}}\text{d}u_2\int_0^\infty\mathbb{I}_{u_3\le\exp\{-a|x-y|\}}\text{d}u_3\end{align*}
The full conditionals are then all uniforms
\begin{align*}
x|y,u_1,u_2,u_3&\sim f(x|y,u_1,u_2,u_3)\propto\mathbb{I}_{u_1\le\exp\{-|x|\}}\mathbb{I}_{u_3\le\exp\{-a|x-y|\}}\\
y|x,u_1,u_2,u_3&\sim f(y|x,u_1,u_2,u_3)\propto\mathbb{I}_{u_2\le\exp\{-|y|\}}\mathbb{I}_{u_3\le\exp\{-a|x-y|\}}\\
u_1|x&\sim\mathcal U(0,\exp\{-|x|\})\\
u_3|x,y&\sim\mathcal U(0,\exp\{-a|x-y|\})\\
u_2|y&\sim\mathcal U(0,\exp\{-|x|\})\\
\end{align*}
hence straightforward to simulate:
xz=yz=rep(pi,1e3)
a=10
for (t in 2:1e3){
 u1=runif(1)*exp(-abs(xz[t-1]))
 u2=runif(1)*exp(-abs(yz[t-1]))
 u3=runif(1)*exp(-a*abs(xz[t-1]-yz[t-1]))
 xz[t]=runif(1,log(u1),-log(u1))
 while (a*abs(xz[t]-yz[t-1])>-log(u3)) xz[t]=runif(1,log(u1),-log(u1))
 yz[t]=runif(1,log(u2),-log(u2))
 while (a*abs(xz[t]-yz[t])>-log(u3)) yz[t]=runif(1,log(u2),-log(u2))
}

$\qquad\qquad$
A: The conditional density kernels are:
$$\begin{equation} \begin{aligned}
f(x|y) &\propto \exp(-|x|-a \cdot |x-y|), \\[6pt]
f(y|x) &\propto \exp(-|y|-a \cdot |x-y|). \\[6pt]
\end{aligned} \end{equation}$$
The difficulty here is to derive the actual densities that go with these kernels, which takes a bit of algebra.  Integration over the full range of values gives the appropriate constant-of-integration:
$$C(a,y) \equiv \int \limits_\mathbb{R} \exp(-|x|-a \cdot |x-y|) \ dx.$$
To evaluate this constant you need to break your integral up into pieces so that you can remove the absolute value signs.  I will not do all of this, but I will show you part of it, which will show how it can be done.  (For simplicity I am going to assume that $0<a<1$.)  In the case where $y>0$ we have:
$$\begin{equation} \begin{aligned}
C(a,y) 
&= \int \limits_{-\infty}^{0} \exp(x-a (y-x)) \ dx \\ 
&\quad + \int \limits_{0}^{y} \exp(-x-a (y-x)) \ dx \\
&\quad + \int \limits_{y}^{\infty} \exp(-x-a (x-y)) \ dx \\[6pt]
&= \frac{1}{1+a} \Bigg[ \exp((1+a)x-ay)) \Bigg]_{x \rightarrow -\infty}^{x=0} \\ 
&\quad - \frac{1}{1-a} \Bigg[ \exp((a-1)x-ay)) \Bigg]_{x=0}^{x=y}  \\
&\quad - \frac{1}{1+a} \Bigg[ \exp(-(1+a)x + ay)) \Bigg]_{x=y}^{x \rightarrow \infty} \\[6pt]
&= \frac{1}{1+a} \cdot \exp(-ay) \\ 
&\quad - \frac{1}{1-a} \cdot \exp(-y) + \frac{1}{1-a} \cdot \exp(-ay) \\
&\quad + \frac{1}{1+a} \cdot \exp(-y) \\[6pt]
&= \Big( \frac{1}{1+a} + \frac{1}{1-a} \Big) \exp(-ay)
 + \Big( \frac{1}{1+a} - \frac{1}{1-a} \Big) \exp(-y) \\[6pt]
&= \frac{2}{1-a^2} \cdot \exp(-ay) - \frac{2a}{1-a^2} \cdot \exp(-y). \\[6pt]
\end{aligned} \end{equation}$$
So for $y>0$ you have:
$$f(x|y) = \frac{(1-a^2) \exp(-|x|-a \cdot |x-y|)}{2 \exp(-ay) - 2a \exp(-y)}.$$
You will need to derive the corresponding case where $y \leqslant 0$ and then you will have the full density.  Both the conditional densities in this problem are the same, so you will then be able to implement the Gibbs sampler by generating from this density.
