Hierarchical Bayesian Regression, Can an Inverse-Gamma distributed Variance look Normal or t?

Question

Using Peter Hoff's book, A First Course in Bayesian Statistical Methods, I used some of my own data to fit a Hierarchical Bayesian Regression following his example. In his book, he utilized a Gibbs sampler to solve for the posterior distribution $p(\beta_1,...,\beta_m,\theta,\Sigma,\sigma^2|\mathbf{X}_1,...,\mathbf{X}_m,\boldsymbol{y}_1,...,\boldsymbol{y}_m)$ and used as the full conditional distribution for $\sigma^2$:

$\sigma^2 \sim \text{inverse-gamma}([\nu_0 + \sum n_j]/2, [\nu_0 \sigma_0 ^2 + >\text{SSR}]/2)$ where $\text{SSR} = \sum_{j=1}^{m} \sum_{i=1}^{n_j} (y_{i,j} - >\boldsymbol{\beta_j^{T}}\boldsymbol{x_{i,j}} $

I believe I ran the Gibbs sampler exactly as he outlined, using 20000 samples and discarding the first 2000 and got the following trace plot for the variance $\sigma^2$

The problem is that this does not look like an Inverse-Gamma distribution to me. Is it possible to get something that looks like this, despite drawing from an Inverse-Gamma distribution? Thanks!

Answer 1

Yes, it's possible. Here's a histogram of a sample from an (explicitly) inverse gamma random variable that looks somewhat similar to yours:

With a sufficiently large shape parameter it will look close to symmetric. In my case I used a shape parameter, $\alpha$ of $10^5$.

Indeed, both the moment skewness and kurtosis will approach that for a normal in the limit, and the shape looks increasingly normal; the QQ plot for the above data looks like this: