Mutual or pairwise independence needed? Variance of a sum.

Question

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.

Answer 1

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:

$$\text{Var}\left(\sum_{i=1}^{n}X_{i}\right)=\sum_{i=1}^{n}\text{Var}X_{i}$$