Assume that $X\sim \mathcal N(\sigma\_1,\mu\_1)$ and $Y\sim \mathcal N(\sigma\_2,\mu\_2)$.


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 (e.g., what is the MLE of it?).




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In the special case where we know that $\mu=\mu\_1=\mu\_2$ (i.e., we estimate the same quantity from observations with different variances), [it is known](https://arxiv.org/pdf/2010.11537.pdf) that the MLE is:


$$
X\cdot \frac{\sigma\_2^2}{\sigma\_1^2+\sigma\_2^2} + Y\cdot \frac{\sigma\_1^2}{\sigma\_1^2+\sigma\_2^2}.
$$


How does this generalize to arbitrary $\mu\_1,\mu\_2$?



Estimating the average of two gaussians' mean
The maximum likelihood estimator (MLE) for $(\mu\_1,\mu\_2)$ is $(X,Y)$. So, by the [functional invariance of the MLE](https://en.wikipedia.org/wiki/Maximum_likelihood_estimation#Functional_invariance) (that is, simply by definition), the MLE of $g(\mu\_1,\mu\_2):=(\mu\_1+\mu\_2)/2$ is $g(X,Y):=(X+Y)/2$, which also, obviously, maximizes the [profile likelihood](https://en.wikipedia.org/wiki/Likelihood_function#Profile_likelihood)
$$L\_{X,Y}(\mu):=\sup\{L\_{X,Y}(\mu\_1,\mu\_2)\colon(\mu\_1+\mu\_2)/2=\mu\}$$
in $\mu$, where $L\_{X,Y}(\mu\_1,\mu\_2)$ is the likelihood.




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One may note that the MLE $(X+Y)/2$ of $(\mu\_1+\mu\_2)/2$ does not depend on $(\sigma\_1,\sigma\_2)$.


