I have a conditional Laplace prior:

$$ \pi(\boldsymbol{\beta}|\sigma^2) = \prod\limits_{j=1}^{p}\frac{\lambda}{2\sqrt{\sigma^2}}e^{-\lambda|\beta_j|/\sqrt{\sigma^2}} $$

and a marginal prior on $\sigma^2$, $\pi(\sigma^2)$.

I want to decompose this Laplace prior for a hierarchical representation of a model.

I know that that the following relations will be necessary: \begin{equation} f(x|z)=\int_{Y}f(x|y)f(y|z)dy \end{equation}

and $$ \frac{a}{2}e^{-a|z|} = \int_0^\infty \frac{1}{\sqrt{2\pi s}}e^{-z^2/(2s)}\frac{a^2}{2}e^{-a^2s/2} ds, \ \ \ \ \ \ a>0. $$

However, I have problems with some details. According to the paper Casella, Park (2008) I should get that, this \begin{align*} \boldsymbol{\beta} | \sigma^2, \tau_1^2, \dots, \tau_p^2 &\sim N_p(\mathbf{0}_p, \sigma^2\mathbf{D}_{\boldsymbol{\tau}}), \\ \mathbf{D}_{\boldsymbol{\tau}} &= \text{diag}(\tau_1^2, \dots, \tau_p^2), \\ \sigma^2, \tau_1^2, \dots, \tau_p^2 &\sim \pi(\sigma^2) d\sigma^2 \prod\limits_{j=1}^{p}\frac{\lambda^2}{2}e^{-\lambda^2\tau_j^2/2}d\tau_j^2 \end{align*}

represents the conditional Laplace prior, however I still end up with something slightly different. Can someone tell me how to get there (e.g. by telling me, plug $z = \beta_j$ etc.)? Thank you for your help!


1 Answer 1


There are a few errors here. Basically you put the $\sigma^2$ in the wrong place.

Note that technically speaking you should really have $f(x|yz)$ and not $f(x|y)$ in the integral formula. Although we will see that for your specific representation the result as you have written it is still true using $f(x|y)$.

The distribution $p(\beta|\sigma^2,\tau_1^2,\dots,\tau_p^2)$ and $p(\sigma^2,\tau_1^2,\dots,\tau_p^2)$ are both inconsistent with the integral formula you provide. The first mistake is that the latent variable distribution should also be conditional on $\sigma^2$, so you should have $p(\tau_1^2,\dots,\tau_p^2|\sigma^2)$. In relating this to the general integral forumla you have $z=\sigma^2,y=\tau_j^2,x=\beta_j$ for $j=1,...,p$.

In your definition of $p(\beta|\sigma^2)$, the value of $a$ should be $a=\frac{\lambda}{\sqrt{\sigma^2}}=\frac{\lambda}{\sigma}$ in order for you to use the integral result. Now note that $a$ does not appear in the term $\frac{z^2}{2s}$ inside your integral result. This is the normal distribution part so you can remove $\sigma^2$ from the distribution $p(\beta|\sigma^2,\tau_1^2,\dots,\tau_p^2)$. So you have:

$$ \begin{align*} \boldsymbol{\beta} | \sigma^2, \tau_1^2, \dots, \tau_p^2 &\sim N_p(\mathbf{0}_p, \mathbf{D}_{\boldsymbol{\tau}}), \\ \mathbf{D}_{\boldsymbol{\tau}} &= \text{diag}(\tau_1^2, \dots, \tau_p^2) \end{align*} $$

This confirms the previous note that $f(x|z)=\int_{Y}f(x|y)f(y|z)$ holds for your specific representation. For a different representation where this isn't true, make the change of variables $s=ra$ in the integral. The other term is straigh-forward, and an exponential density with rate parameter $\frac{a^2}{2}=\frac{\lambda^2}{2\sigma^2}$. So the product should be:



On having a bit more think about it, the end result you have is actually correct, because if $\tau_j^2\sim Expo\left(\frac{\lambda^2}{2\sigma^2}\right)$ then we also have that $\sigma^2\tau_j^2\sim Expo\left(\frac{\lambda^2}{2}\right)$. This shows that the result is correct. You can also make a change of variables in your integral definition of $s=r\sigma^2$ and you have:

$$\frac{a}{2}e^{-a|z|} = \int_0^\infty \frac{1}{\sqrt{2\pi r\sigma^2}}e^{-z^2/(2r\sigma^2)}\frac{a^2\sigma^2}{2}e^{-a^2r\sigma^2/2} dr, \ \ \ \ \ \ a>0$$

This is consistent with your answer, by setting $r=\tau_j^2$

  • $\begingroup$ Does it mean that the result in the paper is wrong? $\endgroup$
    – far away
    Dec 19, 2012 at 16:43
  • $\begingroup$ Yes, my bad. To be precise, I should have written $\int_Y f(x|y,z)f(y|z)$. Thanks for noting that. This is exactly what I don't get... how they managed to get $\sigma^2$ in the normal part of the "decomposition". $\endgroup$
    – far away
    Dec 19, 2012 at 17:14
  • $\begingroup$ Thank you very much for the precise answer and for your time! $\endgroup$
    – far away
    Dec 20, 2012 at 18:03

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