Solved – Why use ${1/\sigma^2}$ as a prior for $\sigma^2$

bayesianprior

In a lot of cases, the prior for $\sigma^2$ is chosen so that it is proportional to ${1/\sigma^2}$. I have a few queries re this:

  1. What is the intuition for this choosing this prior?
  2. What is the information conveyed by this prior? Does it mean that a higher value for $\sigma^2$ is less likely?
  3. I know that this is an improper prior, but is it non-informative? Sorry, I am not entirely sure how non-informative priors are different from improper priors.

Best Answer

First of all, $p(\sigma^2) \propto 1/\sigma^2$ is the Jeffreys prior (http://en.wikipedia.org/wiki/Jeffreys_prior) for a scale parameter. It also coincides with the reference prior under certain conditions.

1) and 2): Intuitively, it can be understood as the only prior expressing correclty that $\sigma$ is a scale parameter: let $X$ distributed according to $1/\sigma .f(x/\sigma)$ then $Y=c.X$ has the same distribution than $X$ but with a different scale, $p(\sigma^2) \propto 1/\sigma^2$ also fits this property.

3): Jeffreys priors are generally improper. However, to my knowledge improperness is not related to informativeness (but simply to integrability considerations).