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Let $X_1$ and $X_2$ be independent, normal distributed random variables with equal mean $\mu$ but non-equal standard deviations $\sigma_1$ and $\sigma_2$.

Suppose I know $\sigma_1$ and $\sigma_2$ and I have $n$ samples $x_{11},\ldots,x_{1n}$ from $X_1$ and m samples $x_{21}, \ldots, x_{2m}$ from $X_2$, what is the best estimator for $\mu$? What is its distribution?

(edit: I'm mainly interested in the n=1, m=1 case)

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    $\begingroup$ Is this homework? $\endgroup$
    – John
    Sep 6, 2011 at 21:12
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    $\begingroup$ This question for the case $n=1$, $m=1$ is answered at stats.stackexchange.com/questions/12251/… . The reply shows how to obtain the answer in general. $\endgroup$
    – whuber
    Sep 6, 2011 at 21:40
  • $\begingroup$ @John: I need it at my job - why? $\endgroup$ Sep 7, 2011 at 5:12

1 Answer 1

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This is analogous to fixed-effect meta-analysis. The best (minimum-variance unbiased) estimator would be the inverse-variance weighted mean $$\hat{\mu} = \frac{ \frac{1}{\sigma_1^2} \sum x_{1i} + \frac{1}{\sigma_2^2} \sum x_{2i} }{n/\sigma_1^2 + m/\sigma_2^2}$$

When $n=m=1$ that reduces to $$\hat{\mu} = \frac{ x_{11} / \sigma_1^2 + x_{21} / \sigma_2^2 }{1/\sigma_1^2 + 1/\sigma_2^2}$$ The distribution of $\hat\mu$ is normal with mean $\mu$ and variance $\frac{ 1 }{n/\sigma_1^2 + m/\sigma_2^2}$.

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