I have a Bayesian version of a linear regression with 3 covariates. The model is given by \begin{align*} Y\sim N(\mu,\tau)\end{align*} \begin{align*} \mu=\alpha + \sum\beta_{i}x_{i}\end{align*}

where $Y=log(Y)$ and $\beta_{i}$ has the following flat priors

\begin{align*} \alpha \sim N(0,0.00001)\end{align*} \begin{align*} \beta_{1} \sim N(0,0.00001)\end{align*} \begin{align*} \beta_{2} \sim N(0,0.00001)\end{align*} \begin{align*} \beta_{3} \sim N(0,0.00001)\end{align*} \begin{align*}\tau \sim \Gamma(\alpha=0.001,\beta=0.001)\end{align*}

Hence if $\tau$ has a gamma distribution as prior and keeping in mind that $\frac{1}{\tau}$ = $\sigma^{2}$, $\sigma^{2}$ must have an inverse gamma distribution as prior.

1) How do I show that $\sigma^2$ has a inverse gamma distribution as a prior?

2) From the above, how do I derive the prior distribution of $log(\sigma^{2})$? (Essentially, this question is about using change of variable, it is not related to estimation. We have a random variable $\sigma^{2}$ which has an inverse gamma distribution and we want to find the density function of $log(\sigma^{2})$ which is a function of $\sigma^{2}$).

  • $\begingroup$ Are you sure it is OLS and not just a linear regression..? If yes, how do you use least squares in Bayesian approach? $\endgroup$ – Tim Jan 4 '15 at 22:30
  • $\begingroup$ I have edited the post to specify linear regression. Thanks $\endgroup$ – user13441344 Jan 4 '15 at 22:47
  • $\begingroup$ Ok, however after re-reading your question I don't understand the questions. What do you mean by "showing" the model? In you description you also don't describe the $log(\sigma^2)$ so it is not clear. How do you want to estimate the model? Also consider that the priors you use may not be the best ones for this case (e.g. Gelman, 2006). $\endgroup$ – Tim Jan 5 '15 at 7:29
  • $\begingroup$ Ok, I edited the question to specify that i want to show that $\sigma^{2}$ has inverse Gamma distribution. This is a derivation question. Since the response is logged, the $log(\sigma^{2})$ comes into play I think. $\endgroup$ – user13441344 Jan 5 '15 at 10:56
  • $\begingroup$ You are assuming that $\sigma^2$ has an inverse-Gamma distribution, so what do you mean by "showing" ? $\endgroup$ – Stéphane Laurent Jan 5 '15 at 11:08

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