*Bounty: 50*

*Bounty: 50*

The mean of the Gamma distribution is $alpha/beta$, while the mean of the Inverse Gamma is $beta/(alpha-1)$. Similarly, the mode of the Gamma is $(alpha-1)/beta$, but the mode of the Inverse Gamma is $alpha/(beta+1)$.

How does this relate to the title of the question?

Well, if we are given data $X$ assumed to be normally distributed, the population variance is given by:

$$sigma_{pop}^2=frac{sum(x_i-bar x)^2}{n}equivfrac{s_n^2}{n}$$

However, if we take a Bayesian approach and choose a non-informative Normal-Inverse-Gamma conjugate prior for the variance (i.e. $alpha_0rightarrow0, beta_0rightarrow0$), we have that the marginal distribution of $sigma^2$ is also Inverse-Gamma distributed, with $alpha=n/2, beta=s_n^2/2$ and mean:

$$E[sigma^2] = frac{s_n^2/2}{n/2-1}neqsigma_{pop}^2 quad(!)$$

On the other hand, if one uses an uninformative Normal-Gamma prior:

$$E[tau] = frac{n/2}{s_n^2/2} = frac{1}{sigma_{pop}^2}$$

Assuming the above is correct, I have a couple of questions:

- I realize that $E[1/X]neq1/E[X]$, yet I’m not sure why $E[tau]$ specifically should yield the "correct" result. What’s wrong with using $E[sigma^2]$?
- The frequentist approach would lead to using the sample variance, which seems to match neither approaches with an uninformative prior. What if any, is the significance of the particular prior that would result in the sample variance for both $tau$ and $sigma$?
- What is the significance of the mode, i.e. the MAP estimator of $sigma^2$ or $tau$? again, they are both different, and I don’t believe I’ve seen either being used in practice.