The Normal-Inverse-Gamma distribution is often written as

$N(\phi | \mu, \sigma^2 \Sigma) IG(\sigma^2 | \alpha, \beta),$

and used as a conjugate prior for a linear model given observations

$y_t \sim N(y_t | x_t \phi, \sigma^2).$

However, the model would still be conjugate even if the prior would be written as

$N(\phi | \mu, \Sigma) IG(\sigma^2 | \alpha, \beta),$

which begs the question of why is the variance of the Normal part of the prior scaled by $\sigma^2$ in the first place, as this makes some derivations longer and more problematic. Most importantly, is there some clear benefit regarding the quality of the inferences?


The prior below is not conjugate

$$ N(\phi|\mu,\Sigma)IG(\sigma^2|\alpha,\beta) $$

since the posterior does not have the form

$$ N(\phi|\mu',\Sigma')IG(\sigma^2|\alpha',\beta'). $$

So the answer to the question "why is the variance of the Normal part of the prior scaled by $\sigma^2$?" is the prior

$$ N(\phi|\mu,\sigma^2\Sigma)IG(\sigma^2|\alpha,\beta) $$

is used because it is conjugate, i.e. the posterior can be put in the form

$$ N(\phi|\mu',\sigma^2\Sigma')IG(\sigma^2|\alpha',\beta'). $$

The benefit of using this prior is that it is conjugate.

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    $\begingroup$ At present, this is more of a comment than an answer. Can you expand it? Why wouldn't it be conjugate? $\endgroup$ – gung - Reinstate Monica Sep 30 '15 at 15:11
  • $\begingroup$ I guess that if you want to keep the variance of the posterior Normal part clearly as a product of observation and model variances, it makes sense to use the first formulation. However, if you mostly care about the posterior inferences do you think that there is any essential difference? $\endgroup$ – akangasr Oct 1 '15 at 11:00

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