Estimation of joint errors Suppose we have multiple normal random variables, each which is a normal distribution with it's own mean but joint variance:
$$X_i \sim N(\mu_i, \sigma^2)$$
Now suppose we collect data from these variables, $D=\{x_{i,j}\}$. 
What methods can one use to estimate the common parameter $\sigma^2$?
One can assume we have some priors $\mu_i\sim N(0, \sigma_m^2), \sigma \sim Ga(\alpha, \beta)$, and one could then write down the posterior by integrating over all $\mu_i$, but it seems there should be an easier "trick" that I'm missing. 
 A: For independent samples from two such populations with the same variance, consider using the pooled variance $S_p^2 = \frac{(n_1 - 1)S_1^2 + (n_2-1)S_2^2}{n_1 - n_2 - 2}$ of the pooled 2-sample t test. 
For $g$ independent samples use the obvious generalization, which is $S_w^2 =$ MS(Error) in an ordinary (unbalanced) one-way ANOVA. 
A confidence interval for $\sigma^2$ can be found using $\frac{S_w^2}{\sigma^2} \sim \mathsf{Chisq}(\nu),$ where $\nu = \sum_{i=1}^g n_i - g =$ DF(Error)}.
Because
$$0.95 = P\left(L \le \frac{\nu S_w^2}{\sigma^2} \le U\right) =
P\left(\frac{\nu S_w^2}{U}\le\sigma^2\le\frac{\nu S_w^2}{L}\right)$$
a 95% CI for $\sigma^2$ is of the form 
$$\left(\frac{\nu S_w^2}{U},\,\frac{\nu S_w^2}{L}\right),$$
where $L$ and $U$ cut probability $0.025$ from the lower and upper tails,
respectively, of $\mathsf{Chisq}(\nu).$
Here is a numerical example in R with three groups of different sizes $(n_1 = 10, n_2 = 8, n_3 = 15)$ all from
normal populations with $\sigma^2 = 15^2 = 256,$ yielding the 95%
confidence interval $(154.95,\, 433.54)$ for $\sigma^2$ or, upon
taking square roots of endpoints, a 95% CI $(12.45,20.82)$ for $\sigma.$
set.seed(811)
x1 = rnorm(10, 100, 15);  x2 = rnorm(8, 130, 15);  x3 = rnorm(15, 80, 15)
sse = 9*var(x1) + 7*var(x2) + 14*var(x3)
dfe = 9 + 7 + 14
mse = sse/dfe
sse/qchisq(c(.975,.025), dfe)
[1] 154.9513 433.5414
sqrt(sse/qchisq(c(.975,.025), dfe))
[1] 12.44794 20.82166

