Variance of MLE for the common mean of two normals

by moreblue   Last Updated October 11, 2018 05:19 AM

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?

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