# Pairwise vs. Mutual independence for normal random vector

Lets say that I have a vector $X$ for which the marginal distribution of each element $x_1,x_2,...$ is normal with variance = 1. Additionally, assume that $X$ is pairwise independent.

Does this prove that the covariance matrix of $X$ is the identity matrix? What is the difference in the distribution of $X$ if instead I assume that $X$ is mutually independent? Is the distribution any different?

Thanks,

Clark

• @Clark Under the assumption of mutual independence, you can get $f(x_1, \ldots, x_p) = f_{X_1}(x_1) \cdots f_{X_p}(x_p)$, so that multivariate normality holds. – Zhanxiong Nov 21 '15 at 6:40