Estimating the average of two gaussians' mean with minimal squared error This is a follow-up to my previous question.
Assume that $X\sim \mathcal N(\mu_1,\sigma_1^2)$ and $Y\sim \mathcal N(\mu_2,\sigma_2^2)$.
I want to estimate $\frac{\mu_1+\mu_2}{2}$ after observing $X,Y$.
In my setting, $\sigma_1,\sigma_2$ are known and we want to estimate the average of the means with the goal of minimizing the expected squared error.

Intuitively, if $\sigma_1<\sigma_2$, we should have an estimate that's closer to $X$ than to $Y$. How can we formalize this?
 A: $\newcommand\si\sigma$Clearly, the best estimator of $\mu_1$ is $X$, no matter what $\si_1$ and $\si_2$ are. Similarly, the best estimator of $\mu_2$ is $Y$, no matter what $\si_1$ and $\si_2$ are. So, one may argue, a good estimator of $(\mu_1+\mu_2)/2$ is the substitution estimator $(X+Y)/2$.
As was shown in the previous answer, $(X+Y)/2$ is indeed the maximum likelihood estimator of $(\mu_1+\mu_2)/2$. It is also easy to see that $(X+Y)/2$ is the only unbiased linear estimator (of the form $aX+bY$) of $(\mu_1+\mu_2)/2$, as well as the minimax linear estimator of $(\mu_1+\mu_2)/2$ (with respect to the quadratic loss function).
On the other hand, one may argue that, if $\si_1<\si_2$, then the uncertainly about $\mu_1$ is less than that about $\mu_2$, and so $X$ has to be given a greater weight than $Y$. However, if there are no restrictions on $\mu_1,\mu_2$, then $|\mu_1|$ and $|\mu_2|$ can be much greater than both $\si_1$ and $\si_2$, so that $\si_1$ and $\si_2$ will matter little, if at all.
Thus, as it was said in a comment, without restrictions on $\mu_1,\mu_2$, you will hardly get the desired result.
One way to impose (soft/fuzzy) restrictions on $\mu_1,\mu_2$ is to suppose that $(\mu_1,\mu_2)$ has the prior bivariate normal distribution with means $0,0$, variances $b^2,b^2$, and correlation $0$. Then the Bayes estimator of $(\mu_1,\mu_2)$ (assuming the quadratic loss function) is
$$(\hat\mu_1,\hat\mu_2):=\Big(\frac X{1+\sigma_1^2/b^2},\frac Y{1+\sigma_2^2/b^2}\Big).$$
Then the corresponding estimator of $(\mu_1+\mu_2)/2$ is
$$\frac{\hat\mu_1+\hat\mu_2}2=\frac12\,\Big(\frac X{1+\sigma_1^2/b^2}+\frac Y{1+\sigma_2^2/b^2}\Big).$$
Here, as desired, the weight/coefficient of $X$ is greater than that of $Y$ if $\sigma_1<\sigma_2$.
If you now let $b\to\infty$, thus making the prior knowledge (i.e., the restrictions on $\mu_1,\mu_2$) insignificant, then you get back the nice old estimator $(X+Y)/2$ from the previous answer.
