2
$\begingroup$

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?).

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 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 $\mu_1,\mu_2$?

Improve this question
$\endgroup$
1
  • $\begingroup$ In standard usage, $\operatorname N(a,b)$ or $\mathcal N(a,b)$ means normal with EXPECTED VALUE $a$ and VARIANCE $b. \qquad$ $\endgroup$
    – Michael Hardy
    Commented Oct 21, 2021 at 19:37

1 Answer 1

Reset to default
4
$\begingroup$

The maximum likelihood estimator (MLE) for $(\mu_1,\mu_2)$ is $(X,Y)$. So, by the functional invariance of the MLE (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 $$L_{X,Y}(\mu):=\sup\{L_{X,Y}(\mu_1,\mu_2)\colon(\mu_1+\mu_2)/2=\mu\}$$ in $\mu$, where $L_{X,Y}(\mu_1,\mu_2)$ is the likelihood.

One may note that the MLE $(X+Y)/2$ of $(\mu_1+\mu_2)/2$ does not depend on $(\sigma_1,\sigma_2)$.

Improve this answer
$\endgroup$
7
  • $\begingroup$ You're right, and the MLE probably doesn't capture what I intended (but that's the closest thing I have found). I have some difficulty phrasing the question; what I ultimately want to estimate $\frac{\mu_1+\mu_2}{2}$ in a way that would minimize the expected squared error with respect to the drawing of $X,Y$. $\endgroup$
    – R B
    Commented Oct 20, 2021 at 18:28
  • $\begingroup$ To put it differently: I want a method for estimating the average so that if we run it many times (drawing different $X,Y$ each time) it would minimize the expected squared error. $\endgroup$
    – R B
    Commented Oct 20, 2021 at 18:30
  • $\begingroup$ @RB : The estimator $(X+Y)/2$ is also the only unbiased linear estimator (of the form $aX+bY$) of $(\mu_1+\mu_2)/2$. Moreover, this estimator is the minimax linear estimator of $(\mu_1+\mu_2)/2$ with respect to the quadratic loss function. However, all that is for another question. To keep things in good order, you may want to post such "expected squared error" questions separately. $\endgroup$
    – Iosif Pinelis
    Commented Oct 20, 2021 at 18:50
  • $\begingroup$ Thanks again. I am not necessarily interested in unbiased estimates (i.e., I expect the estimate to be closer to $X$, if $\sigma_1<\sigma_2$, and vice versa). I will ask a separate question. $\endgroup$
    – R B
    Commented Oct 21, 2021 at 8:12
  • $\begingroup$ "Invariance" is a frequently used misnomer for this. It's actually equivariance. $\endgroup$
    – Michael Hardy
    Commented Oct 21, 2021 at 19:39

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .