Code for Laplace prior

[Patrik Waldmann]

Hello,

 

I understand that it is possible to define own priors and wonder if someone has R code for a Laplace (double exponential) prior?

 

Best,

 

Patrik Waldmann

[Daniel Simpson]

Hi Patrik,
The example here is fairly straightforward to modify
http://www.math.ntnu.no/inla/r-inla.org/doc/prior/expression.pdf

I'm not 100% sure that it will work - priors are assumed to be twice
differentiable in order to compute the laplace approximation, and the
Laplace prior is not differentiable at zero (which, i guess, is the
point).

Best,
Dan

[Patrik Waldmann]

What about the scale mixtures of normals with exponential mixing density approach used by Park and Casella (2008) in their Bayesian Lasso paper? Would that be better?

 

Patrik

[Patrik Waldmann]

The following works, but I wonder if the log(dens) is correct? And how do I assign a hyperprior to b?

laplace = "expression:
mean = 0;
b = 1;
dens = 1/(2*b) *  exp(-abs(x-mean)/b);
logdens = log(dens);
return(logdens)"

 

Patrik

 

Den tisdagen den 17:e juli 2012 kl. 14:48:17 UTC+2 skrev Daniel Simpson:

[Daniel Simpson]

The prior specification is correct, but it will not do what you want.

I am assuming that you wish to fit the Bayesian Lasso model

y|beta,sigma ~ X*beta + eps
beta_j|sigma ~ Laplace( beta_j; mean=0, scale=sigma^2)

This cannot be fitted in INLA because the beta's are not Gaussian.

To do this in INLA, you need to use the Hierarchical model specification


Assume that there are p beta's. Then

y|beta,sigma ~ N(X*beta, sigma^2 *I)
beta|sigma, tau_1,..., tau_p ~ N(0_p, sigma^2 * D_p)
where D_p = diag(tau_1^2, ..., tau_p^2)
p(tau_j) = lambda^2/2*exp(-lambda^2*tau_j^2/2

So the prior that you need to set is just normally distributed with
variance lambda. You cannot put a prior on lambda within the INLA
framework.

As said in a previous email, you cannot have p > 10 (15 at the
absolute maximum).

I hope this helps,

Dan