Minimizing the variance of my MLE

Question

A random sample of size $n_1$ is to be drawn from a normal population with mean $\mu_1$ and variance $\sigma^2_1$.

A second random sample of size $n_2$ is to be drawn from a normal population with mean $\mu_2$ and variance $\sigma^2_2$. The two samples are independent.

What is the maximum likelihood estimator of $\alpha =\mu_1-\mu_2$?

Assuming that the total sample size $n=n_1-n_2$ is fixed, how should the observations be divided between the two populations in order to minimise the variance of α̂ ?

I already came up with my MLE of $\alpha =\mu_1-\mu_2$. Now I need to divide my n observations, provided that $n=n_1-n_2$, in such a way that I can minimize the variance of α. Any leads on how I could go about this?

Answer 1

Hints:

• Find the maximum likelihood estimator of $\mu_1-\mu_2$ given $n_1$ and $n_2$; here this is is also the natural estimator, and is the same as the MLE of $\mu_1$ minus the MLE of $\mu_2$
• Find the variance of the MLE of $\mu_1-\mu_2$ given $n_1$ and $n_2$; it will be a function of $n_1, n_2, \sigma^2_1, \sigma^2_2$ and you can use the independence of the to parts of the MLE
• Minimise that variance of the MLE subject to $n_1+n_2=n$; probably easiest using calculus to find $n_1$ as a function of $\sigma^2_1, \sigma^2_2$ though note that $n_1$ should be an integer