TITLE: Mutual or pairwise independence needed? Variance of a sum.
QUESTION [7 upvotes]: This is a simple question:
Do we need mutual independence or only pairwise independence in order to state that
$$\mathrm{Var}\left[\sum_{i=1}^n X_i\right] = \sum_{i=1}^n \mathrm{Var}\left[X_i\right]?$$
As I do not know what uncorrelated means (I know that this is the actual condition), I am not sure whether it is enough for each pair to be independent.
Thanks for your help.

REPLY [9 votes]: Edit (with mainly the same content and some adaptions)
Pairwise independence is enough here. In general:
$$\text{Var}\left(\sum_{i=1}^{n}X_{i}\right)=\sum_{i=1}^{n}\sum_{j=1}^{n}\text{Cov}\left(X_{i},X_{j}\right)$$
If $X_{i}$ and $X_{j}$ are independent then $\mathbb{E}X_{i}X_{j}=\mathbb{E}X_{i}\mathbb{E}X_{j}$
and consequently $$\text{Cov}\left(X_{i},X_{j}\right):=\mathbb{E}X_{i}X_{j}-\mathbb{E}X_{i}\mathbb{E}X_{j}=0$$
Or in words: $X_{i}$ and $X_{j}$ are uncorrelated. This leads to:
$$\text{Var}\left(\sum_{i=1}^{n}X_{i}\right)=\sum_{i=1}^{n}\text{Var}X_{i}$$