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".

  • $\begingroup$ (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 $\endgroup$
    – whuber
    Sep 18, 2015 at 17:00
  • 1
    $\begingroup$ @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. $\endgroup$ Sep 18, 2015 at 17:20

1 Answer 1


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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.