This is a follow-up to my [previous question](https://mathoverflow.net/questions/406653/estimating-the-average-of-two-gaussians-mean).


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.




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Intuitively, if $\sigma\_1<\sigma\_2$, we should have an estimate that's closer to $X$ than to $Y$. How can we formalize this?



Estimating the average of two gaussians' mean with minimal squared error
$\newcommand\si\sigma$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](https://mathoverflow.net/a/406655/36721), $(X+Y)/2$ 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](https://mathoverflow.net/a/406655/36721).


