I am interested in the mean and variance over n random variables $x$ where each $x_{i}$ is independently distributed with an unique mean and standard deviation, $x_{i} ~ \sim \mathcal{N}(\mu_{i}, \sigma_{i})$.
For example, suppose I measure the length of n persons multiple times to get for each person an average length $l_{i}$ with uncertainty $\sigma_{l_{i}}$. I then would like to compute the average length of these subjects and the corresponding variance.
Intuitively I would say the answer would be something like:
$$\bar{\mu} = \frac1n \sum_i \mu_i$$
$$\sigma^2 = \frac1n \sum_i \sigma_i^2 + \frac1n \sum_i (\mu_i-\bar{\mu})^2$$
i.e., the total variance is the sum of the individual variances plus the variance over the group. I unfortunately can't seem to derive it, perhaps it is wrong. I found an answer similar to this on the math stackexchange, https://math.stackexchange.com/questions/1210491/standard-deviation-of-mean-of-a-set-of-numbers-which-are-imprecise but the first answer also shows no complete proof.
Another idea I have is that I could use a weighted arithmic mean, like stated on wikipedia: https://en.wikipedia.org/wiki/Weighted_arithmetic_mean :
$$ \bar{\mu} = \frac{ \sum_{i=1}^n \left( x_i \sigma_i^{-2} \right)}{\sum_{i=1}^n \sigma_i^{-2}} $$
and the uncertainty on the weighted mean:
$$\sigma^2 = \frac{ 1 }{\sum_{i=1}^n \sigma_i^{-2}}$$
where $w_i = \frac{1}{\sigma_i^2}$.
Which of these methods is the one I need?
I found an answer to a similar question here: Average of Average Quantities and Their Associated Error but this only uses the variances of the individual measurements. If I would not have those then the variance of the average would be 0, which seems not very useful as I would still have the variance over the group, no?
