If I have a normal distributed variable $N(\mu,\sigma^2)$ so with fixed $\mu$ the conjugate prior for $\lambda:=\frac{1}{\sigma^2}$ is given by the gamma distribution $\propto \lambda^{\alpha-1}exp{-\lambda\beta}$ (therefore $\sigma^2\sim\sigma^{2(1-\alpha)}exp{\frac{-\beta}{\sigma^2}}$). And the posterior is given by $Ga(\lambda|\alpha_n,\beta_n)$ with $\alpha_n=\alpha+\frac{n}{2}$ and $\beta_n=\beta+\frac{1}{2}\sum^n_{i=1}\limits{(x_i-\mu)^2}$.

That means with increasing sample size the $\sigma^2$ decreases, right? Or is that observation wrong?

In my special case I do not expect a decrease in $\sigma^2$ for some reasons. I expect the opposite for $\sigma^2$. That means $\sigma^2$ should converge against a value greater than zero ($\approx 50$).

Is there a trick or a similar prior which reflects the described behavior? That the variance can be far away from zero?

(2) I forgot... I have a lot of prior knowledge so I can simplify the issue as follows:

$z~(\mu,\tau\times\sigma_{fix}^2)$ In this notation I "just" have to estimate $\tau$. But it is enough to estimate it on a logarithmic scale $(0.001, 0.1, 1 ,10, 100 ,1000)$ What could be the prior distribution for $\tau$ in this case?

  • $\begingroup$ Point estimates (e.g. posterior mean or median) for $\sigma^2$ will not go to zero. The Gamma prior is perfectly capable of "dealing with" the situation, where $\sigma^2$ is some positive value and the estimates should concentrate around the true value. Or am I misunderstanding the question? $\endgroup$ – Björn Nov 4 '15 at 13:04
  • $\begingroup$ Yes, that was the question. Can you briefly explain why they concentrate around the real variance? Is the best way to draw the precision from $Ga(\alpha,\beta)$ while setting $Ga(x|\alpha_n,\beta_n)$ as the posterior in the acceptance ratio? $\endgroup$ – JohnScott Nov 4 '15 at 13:59

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