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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.

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    $\begingroup$ Not every distribution has a conjugate prior. First you need to know it exists before you talk about a form for it. $\endgroup$ – Michael R. Chernick Sep 10 '19 at 13:41
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    $\begingroup$ 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!). $\endgroup$ – Glen_b Sep 11 '19 at 3:22
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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 $(0,\tau_a^{-2})$.

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