2
$\begingroup$

I am interested in drawing samples from a total variation prior: $$\pi_{\mathrm{pr}}(\boldsymbol{x})=\left(\frac{\alpha}{2}\right)^n\exp\left(-\alpha\sum_{j=0}^{n-1}\vert x_{j+1}-x_{j}\vert\right)=\left(\frac{\alpha}{2}\right)^n\exp\left(-\alpha\Vert\mathbf{L}\boldsymbol{x}\Vert_1\right)=\mathrm{Laplace}(\Vert\mathbf{L}\boldsymbol{x}\Vert_1;0,\alpha^{-1}),$$ where $\mathbf{L}$ is a difference matrix and $\alpha$ is the inverse of the scale parameter of a Laplace distribution.

My initial guess was to generate sample paths of a process such that: (i) it starts at 0, (ii) the increments are independent, and (iii) the increments are distributed according to $\mathrm{Laplace}(\cdot;0,\alpha^{-1})$. Is this procedure correct? Or is there any other way to generate samples from such distribution (other than MCMC)?

Thanks in advance!

$\endgroup$
1
  • $\begingroup$ But after doing so, I guess one still end up using MCMC (component-wise MH or Gibbs) $\endgroup$
    – Felipe
    Apr 1 '20 at 9:58
2
$\begingroup$

The joint density $$\pi(\boldsymbol{x})=\overbrace{\left(\frac{\alpha}{2}\right)^n}^{\substack{ \text{correct}\\ \text{constant}}}\exp\left(-\alpha\sum_{j=0}^{n}\vert x_{j+1}-x_{j}\vert\right)$$ assuming $\boldsymbol{x}=(x_1,\ldots,x_n)$ writes as $$\pi(\boldsymbol{x})=\frac{\alpha}{2}\exp\left(-\alpha\vert x_{1}-x_{0}\vert\right)\frac{\alpha}{2}\exp\left(-\alpha\vert x_{2}-x_{1}\vert\right)\cdots\frac{\alpha}{2}\exp\left(-\alpha\vert x_{n}-x_{n-1}\vert\right)$$ and can be expressed as a product of conditional Laplace densities $$\pi(\boldsymbol{x})=f_\alpha(x_1|x_0)f_\alpha(x_2|x_1)\cdots f_\alpha(x_n|x_{n-1})$$meaning it can be simulated as

  • simulate $X_1\sim \mathfrak{L}(x_0,\alpha)$
  • simulate $X_2|X_1=x_1\sim \mathfrak{L}(x_1,\alpha)$
  • $\qquad\vdots$
  • simulate $X_n|X_{n-1}=x_{n-1}\sim \mathfrak{L}(x_{n-1},\alpha)$
$\endgroup$
0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.