# Orthogonal transformations of random vectors and statistical independence

In this old CV post, there is the statement

"(...) I have also shown the transformations to preserve the independence, as the transformation matrix is orthogonal."

It refers to the $$k$$-dimensional linear transformation $$\mathbf y = \mathbf A \mathbf x$$ with the (normally distributed) random variables in $$\mathbf x$$ being assumed independent (the "orthogonal matrix" is $$\mathbf A$$).

• Does the statement mean that the elements of $$\mathbf y$$ are jointly independent? If not, what?
• Does the result hinges on the normality of the $$\mathbf x$$'s?
• Can somebody provide a proof and/or a literature reference for this result (even if it is restricted to linear transformations of normals?)

Some thoughts: Assume zero means. The variance-covariance matrix of $$\mathbf y$$ is

$${\rm Var}(\mathbf y) = \mathbf A \mathbf \Sigma \mathbf A'$$

where $$\Sigma$$ is the diagonal variance-covariance matrix of the $$\mathbf x$$'s. Now, if the variables in $$\mathbf x$$ have the same variance, $$\sigma^2$$, and so $$\Sigma = \sigma^2 I$$, then

$${\rm Var}(\mathbf y) = \sigma^2 \mathbf A \mathbf A' = \sigma^2 I$$

due to orthogonality of $$\mathbf A$$.

If moreover the variables in $$\mathbf x$$ are normally distributed, then the diagonal variance-covariance matrix of $$\mathbf y$$ is enough for joint independence.

Does then the result holds only in this special case (same variance - normally distributed), or it can be generalized, I wonder... my hunch is that the "same variance" condition cannot be dropped but the "normally distributed" condition can be generalized to "any joint distribution where zero covariance implies independence".

• (1) Are you making a distinction between "jointly independent" and "independent"? (2) Your second and third bullets are answered in many places on this site--a search might help. That's fine, because it narrows your question to the one in the last paragraph. (+1) That sounds remarkably close to assertions made by the Herschel-Maxwell theorem
– whuber
Sep 18, 2015 at 17:00
• @whuber. I think that's the relevant theorem here indeed, thanks. I think I will prepare an answer that details the theorem. it appears to be perhaps the most "natural" characterization of the normal distribution. Sep 18, 2015 at 17:20

## 1 Answer

It generalizes to the case where variances are not the same (heteroskadestic). In that case, the matrix $\Sigma=DD$ where D is a diagonal matrix. You can then eventually reach the conclusion that $Var(y)=\Sigma$.

This result also holds if uncorrelatedness implies independence.