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I have a (I hope) simple question! If I had a linear regression,

$Y_t = \alpha + \beta X_t + \epsilon_t$

with $\epsilon_t \sim N(0,\sigma^2)$

and I assume a Cauchy prior for $\sigma$, is it possible to get a conditional conjugate posterior, that I could embed in a Gibbs sampler without having to rely on a MH step? I am aware of the papers from Gelman and Polson, but I do not think they help here... Actually what I am trying to do is a bit more complicated (put Cauchy priors on the variances of latent states in a state space model), but if I know the posterior for the linear regression model, I could adapt it easily.

Thanks!

EDIT:

Ok, so just in case, yes, I meant the half Cauchy prior. I understand what Gelman does in the paper, but I am not sure I can apply the same parameter expansion to a model like this one, which is a simple state space model:

$Y_t = \alpha + \beta X_t + \epsilon_t$

$X_t = X_{t-1} + \eta_t$

I would like to put the half Cauchy prior on the variance of $\eta_t$. Ideally, I would pass a Kalman filter/smoother to get the distribution $p(X^T|Y^T,\beta ,\alpha ,\sigma^\epsilon , \sigma^\eta ) $ . Then, to build the Gibbs sampler, I would like to obtain the distribution $p(\sigma^\eta|X^T,Y^T,\beta ,\alpha ,\sigma^\epsilon ) $ . If I put an inverted gamma prior, given a draw from $p(X^T|...)$, the distribution is also inverted gamma as usual. But with a half Cauchy, I have no clue, to be honest. So, any help is more than welcome!

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    $\begingroup$ Did you really mean to write "conjugate"? $\endgroup$
    – whuber ♦
    Commented Jul 5, 2016 at 17:04
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    $\begingroup$ I think you mean half-Cauchy, since a Cauchy distribution has support over the reals and a valid variance must be nonnegative. $\endgroup$
    – Sycorax ♦
    Commented Jul 5, 2016 at 17:16
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    $\begingroup$ I think those papers do tell you how to construct the conditional conjugate by parameter expansion (introducing a third parameter - I think e.g. "Prior distributions for variance parameters in hierarchical models(Comment on Article by Browne and Draper)" by Gelman has the details, pages 519-520). $\endgroup$
    – Björn
    Commented Jul 5, 2016 at 18:30
  • $\begingroup$ Thanks guys! I edited the question, just in case it is more clear $\endgroup$
    – Chucky
    Commented Jul 5, 2016 at 19:25
  • $\begingroup$ Given the hierarchical nature of your model, the first equation is irrelevant for the simulation of $\sigma^\eta$. And once the $X_t$'s have been simulated, you end up with an iid sample from a Normal(0,\sigma^\eta)$, that is, the simplest possible case. $\endgroup$
    – Xi'an
    Commented Jul 6, 2016 at 16:23

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