Variance of MLE for the common mean of two normals

I have the following problem.

$$X_i \overset{IID}{\sim} Normal(\mu, \sigma_1^2)$$, $$Y_j \overset{IID}{\sim} Normal(\mu, \sigma_2^2), i = 1, \cdots, m, j=1, \cdots n$$

Find the MLE for the $$\hat{\mu}^{MLE}$$, where both $$\sigma_1^2, \sigma_2^2$$ are unknown.

I bumped into the following old paper : Estimating the Common Mean of Several Normal Populations.

I understand they suggest

$$\hat{\mu}^{MLE} = \frac {m \bar{x}/s_1^2 + n \bar{y}/s_2^2} {1/s_1^2 + 1/s_2^2}$$

where $$s_1^2 = \frac{\sum_{i=1}^m (x_i - \bar{x})^2}{m-1}, s_2^2 = \frac{\sum_{j=1}^n (y_j - \bar{y})^2}{n-1}$$, the sample variance of each.

I understand $$\mathbb{E}(\hat{\mu}^{MLE}) = \mu$$, because sample mean and sample variance is are independent in IID normals, we have

$$\mathbb{E}(\hat{\mu}^{MLE}) = \mathbb{E}\left[ \mathbb{E}(\hat{\mu}^{MLE}| s_1^2, s_2^2) \right] = \mathbb{E}\left[ \mu \right] = \mu$$

But the variance of $$\hat{\mu}^{MLE}$$ was not suggested.

My attempt $$\mathrm{Var}(\hat{\mu}^{MLE}) = \mathbb{E}(\mathrm{Var}(\hat{\mu}^{MLE}|s_1^2, s_2^2)) + \mathrm{Var} (\mathbb{E}(\hat{\mu}^{MLE}|s_1^2, s_2^2)))$$

The second term is zero, because conditional mean yields constant. But the mean of variance part, I have variance

$$\mathrm{Var}(\hat{\mu}^{MLE}|s_1^2, s_2^2) = A^2 m \sigma_1^2 + (1-A)^2 n \sigma_2^2$$

where $$A = \frac {1/s_1^2} {1/s_1^2 + 1/s_2^2}$$.

But I can't proceed. Is there anyone to help me with algebra/Any papers to refer to?

• Is this not covered by the answer to your earlier question stats.stackexchange.com/questions/370946/… ?? – Glen_b Oct 11 '18 at 5:35
• @Glen_b Well, I think I found MLE, but even after some algebraic justfications, there still remains the problem of interval estimation. I'm asking if there is any closed form for the variance of the above MLE. So I don't think this post to be duplicate. – moreblue Oct 11 '18 at 5:45