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Suppose one is given $n$ random normal variables, all with the same variance but with different means: $$X_i\sim N(\mu_i, \sigma^2)$$

Now suppose we observe $m_i$ observations $\{x_{ij}\}_{j=1}^{m_i}$ for each variable $X_i$. How would one calculate the posterior distribution of $\sigma^2$?


I thought of using the Normal Inverse Gamma as a conjugate prior, but I don't see this working for multiple different means. I also thought of modeling this as a multidimensional distribution, but it's not clear how to enforce $\bf \Sigma$ to be equal to $\sigma I$.

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  • $\begingroup$ Surely a Bayesian ANOVA would do the trick? $\endgroup$ – jcken Jul 28 '20 at 20:52
  • $\begingroup$ @jcken - I was looking for an explicit formula, please feel free to vet my answer below. $\endgroup$ – nbubis Jul 28 '20 at 22:26
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Working out the math explicitly, and assuming an uninformative prior $1/\sigma^4$, I believe one should get that:

$$P(\sigma^2)\sim \left(\frac{1}{\sigma^2}\right)^{1+\sum m_i/2 -n +1 }\exp\left(\frac{1}{2\sigma^2}\sum_{i=1}^n\left[ \frac{1}{m_i}\left(\sum_{j=1}^{m_i}x_{ij}\right)^2-\sum_{j=1}^{m_i} x_{ij}^2 \right]\right)$$

This is unsurprisingly an inverse gamma distribution, $\text{IG}(\alpha,\beta)$, with: $$\alpha = \sum m_i/2 -n +1, \ \beta=\frac{1}{2}\sum_{i=1}^n\left[ \sum_{j=1}^{m_i} x_{ij}^2-\frac{1}{m_i}\left(\sum_{j=1}^{m_i}x_{ij}\right)^2 \right]$$ $$\longleftarrow-----\longrightarrow$$

As a sanity check, when $n=1$, we have that: $$\alpha=m/2,\ \beta=\frac{1}{2}\left(\sum x_j^2-m{\bar x}^2\right)=\frac{1}{2}\sum\left(x_j-\bar x\right)^2\ ,$$ exactly as we would get using a Normal-Inverse Gamma non-informative prior.

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