Let us assume we have several samples of unknown size but known mean value $\mu_{i}(x)$ and known variance $\sigma_{i}^{2}$. Now we want to calculate the mean value and variance for the total population.
The mean values should IMHO just be weighted with the reciprocal variances:
$w_{i}=\frac{1}{\sigma_{i}^{2} \cdot \sum \frac{1}{\sigma_{i}^{2}}} \Rightarrow \mu(x) = \sum \frac{1}{\sigma_{i}^{2} \cdot \frac{1}{\sigma_{i}^{2}}} \mu_{i}(x)$
This looks and feels right and logical and I have seen similar answers to similar questions.
However I am not sure whether it is correct to calculate the new variance in the same simple way and I have not yet been able to find an answer with a decent explanation.
My Idea:
$\sigma^{2} = \mu(x^2) - \mu(x)^2 = \sum \frac{1}{\sigma_{i}^{2} \cdot \sum\frac{1}{\sigma_{i}^{2}}} \mu_{i}(x)^2 - [\sum \frac{1}{\sigma_{i}^{2} \cdot \sum\frac{1}{\sigma_{i}^{2}}} \mu_{i}(x)]^2$
Apart from the expressions probably not being completely simplified is this statistically sound or did I make some stupid mistake?