This question is motivated by the fact that a linear transformation of an IID standard normal vector gives the multivariate Gaussian distribution, and that the statistical dependence of such transformed variables is sufficiently described by the covariance matrix.

Suppose I have a random vector $\vec X \in \mathbb{R}^n$ of identical and independent non-normal densities $f_j$ for $j \in \{1, \cdots, n \}$. Now I apply an $n\times n$ linear transformation $T$ to obtain $\vec Y$:

$$\vec Y := T \vec X$$

Will the statistical dependence of the variables in $\vec Y$ be sufficiently described by their covariance matrix? That is to say more precisely: $Y_i$ and $Y_j$ are independent iff $\operatorname{Cov}[Y_i, Y_j] = 0$?

  • $\begingroup$ I am not sure I understand your question: If your density is not normal, you cannot in general deduce independence for vanishing covariance. This is independent of whether you apply a transformation $T$ or not. $\endgroup$
    – frank
    Sep 23 at 6:04
  • $\begingroup$ @frank I don't understand how you don't understand the question when your comment seems to answer the question; what a strange coincidence if that is the case. I'd be happy to accept it as an answer if you post it with an explanation of why such a linear transform does not make the covariance sufficient. $\endgroup$
    – Galen
    Sep 23 at 6:10
  • $\begingroup$ Does this answer your question: stats.stackexchange.com/questions/179511/… $\endgroup$ Sep 23 at 13:45

1 Answer 1


For each nonconstant random vector $\mathbf X$, no matter whether the components $X_i$ are independent or not, there are linear transformations which can create random vectors $\mathbf Y$ that have zero covariances as well as those that have non-zero covariances.

For zero covariances, just consider the trivial linear transformation that sends everything to zero.

For non-zero covariances, consider the linear map $\mathbf Y = (\mathbf X, \mathbf X)$.

Again, this is independent of the dependencies within $\mathbf X$.


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