# Gibbs sampling: Conjugate prior for a two component known-unknown variance?

If I have a normally distributed variable $$\mathcal{N}(\mu,\frac{1}{\tau})$$ with fixed $$\mu$$ then the conjugate prior for an unknown $$\tau$$ is then $$\mathcal{Ga}(\frac{n}{2}+\alpha, \beta + \sum_i \frac{(x_i - \mu)^2}{2})$$.

If my precision is instead $$\tau = \frac{1}{\sigma_a^2 + \sigma_b^2}$$, where $$\sigma_a^2$$ is known and $$\sigma_b^2$$ is unknown, is there a form of the conjugate prior for $$\sigma_a^2$$?

It seems as though this should be simple but I have struggled with it. The application of this is to sample $$\sigma_b^2$$ in a Gibbs sampler.

• Not every distribution has a conjugate prior. First you need to know it exists before you talk about a form for it. – Michael R. Chernick Sep 10 '19 at 13:41
• In your first sentence you appear to be showing the formula for the posterior but you're calling it a prior (it has data in it!). – Glen_b Sep 11 '19 at 3:22

Since $$\tau = \frac{1}{\sigma_a^2 + \sigma_b^2}$$ is in one-to-one correspondence with $$\sigma_a^2$$, setting a prior on $$\tau$$ is equivalent to setting a prior on $$\sigma^2_a$$. The Gamma prior of $$\tau$$ continues to be conjugate in this setting, except that it is now restricted to $$(0,\tau_a^{-2})$$.