Variance of the mean of independent (not IID) normal random variables Say you've got $i=1:N$ observations, $X_i$.  Each observation is the mean of a Gaussian with an associated variance $Var(X_i)$. The variances are NOT identical.  The expectation of $X$ should simply be their mean $N^{-1}\displaystyle\sum X$.  The variance of the mean should be 
$$
Var\left(N^{-1}\displaystyle\sum X_i\right) = N^{-2}\displaystyle\sum Var(X_i)
$$
Is this correct?  I'm following the basic formula on Wikipedia, but I'm getting implausibly small variances.  Am I missing something?  The above formula is usually used in the context of equal variances, but it doesn't say anything about requiring equal variances.
 A: Yes, this is so.   The random variables do not need to be identically distributed.   However the result requires the random variables to be independent.
Let ${\{(X_i, \mu_i, \sigma_i^2)\}}_{i=1}^N$ be a series of mutually independent random variables, their means, and variances.
Let $\bar X:=\tfrac 1 N \sum\limits_{i=1}^N X_i$ be the mean of the series (ie: the sample mean).
Then indeed the expectation of the sample mean is: $\mathsf E(\bar X) = \tfrac 1 N\sum\limits_{i=1}^N \mu_i$
And the variance of the sample mean is:$$\begin{align}
\mathsf{Var}(\bar X) ~=~& \mathsf {Cov}(\tfrac 1 N \sum\limits_{i=1}^N X_i,\tfrac 1 N \sum\limits_{j=1}^N X_j) 
\\[1ex] =~&\tfrac 1{N^2} \sum_{i=1}^N\sum_{j=1}^N \mathsf{Cov}(X_i,X_j)
\\[1ex] =~& \tfrac 1{N^2} \sum_{i=1}^N \mathsf{Var}(X_i) & ^\dagger
\\[1ex] =~& \tfrac {1}{N^2} \sum_{i=1}^N \sigma_i^2
\end{align}$$
(† because $\forall i\forall j: \big[i\neq j ~\to~ \mathsf{Cov}(X_i,X_j)=0\big]$ due to independence.)
And indeed, as the count $N$ increases infinitely the variance of the sample mean will decrease vanishingly.   $~\lim\limits_{N\to\infty} \mathsf{Var}(\bar X)=0$ 

PS: This is not the mean of the sample variance, if that is what is confusing you.
