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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 $α = \mu_1 − \mu_2$?

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

I know how to find the MLEs of $\mu_1$ and $\mu_2$, but I don't know how to use these to find $α$ for the first part of this question. I don't even know where to start on minimising the variance.

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  • $\begingroup$ en.wikipedia.org/wiki/Welch%27s_t-test $\endgroup$ – lion Nov 22 '17 at 18:14
  • $\begingroup$ @lion I'm not sure this helps me with finding the MLE, I've taken a read through but it hasn't helped me with this question. (It did really help with some issues I had with hypothesis testing though, so thank you for that!) $\endgroup$ – Clong123 Nov 22 '17 at 21:35
  • $\begingroup$ In this case, the difference of the mean would be an estimator of $\alpha$, and its distribution is given in that link, so you can calculate the variance etc. But I agree this estimator may not be the mle. $\endgroup$ – lion Nov 22 '17 at 21:42
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To be strict. the parameters can be $\alpha,\sigma_1, \mu_2,\sigma_2$, so $\mu_1=\alpha+\mu_2$. We can write the negative of log likelihood as:

$\sum_1^{n_1}{\frac{(X_i-\mu_2-\alpha)^2}{2\sigma_1^2}}+\sum_1^{n_2}{\frac{(Y_i-\mu_2)^2}{2\sigma_2^2}}$

take FOC we have:

$-\sum_1^{n_1}{\frac{X_i-\mu_2-\alpha}{\sigma_1^2}}=0 \rightarrow \sum_1^{n_1}({X_i-\mu_2-\alpha})=0$

$-\sum_1^{n_1}{\frac{X_i-\mu_2-\alpha}{\sigma_1^2}}-\sum_1^{n_2}{\frac{Y_i-\mu_2}{\sigma_2^2}}=0 \rightarrow \sum_1^{n_2}({Y_i-\mu_2})=0$

So $\hat\mu_2=\bar{Y},\hat\alpha=\bar{X}-\bar{Y} $

$var(\hat\alpha)=var(\bar{X}-\bar{Y})=var(\bar{X})+var(\bar{Y})$ if $\sigma_1, \sigma_2$ are known, it would be easy to minimize.Otherwise use the t-distribution in my comments.

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