# Do linear transforms of IID random vectors make the resulting covariances sufficient to describe statistical dependence?

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 $$\operatorname{Cov}[Y_i, Y_j] = 0$$?

• 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. Sep 23 at 6:04
• @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. Sep 23 at 6:10
• Does this answer your question: stats.stackexchange.com/questions/179511/… Sep 23 at 13:45

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 non-zero covariances, consider the linear map $$\mathbf Y = (\mathbf X, \mathbf X)$$.
Again, this is independent of the dependencies within $$\mathbf X$$.