MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let $X$ be a Gaussian vector in dimension $n$, with $0$ mean and covariance identity. Let $A$ be a square matrix of size $n$, and $Y = A X$. Let $N$ be the square of $Y$ euclidean norm: $N = \sum Y_i^2$. One computes easily the mean of $N$: $E[N] = \text{Tr}(A A')$. But what about its variance?

share|cite|improve this question
This is simplified a lot by using the singular value decomposition to reduce to the case in which $A$ is diagonal. If my arithmetic is right the variance is $2 \sum \sigma_i^4$, where $\sigma_i$ are the singular values of $A$. – Mark Meckes Sep 16 '13 at 12:36
@Ricardo Andrade: Is there a reason you felt it necessary to capitalize Gaussian but not euclidean? – Mark Meckes Sep 16 '13 at 12:37
Dear @Mark: That is a very good point. I am afraid the capitalization was a very silly lapse on my part. I was editing the tags and noticed the occurrence of 'gaussian' in the title. Unfortunately, I completely missed the occurrence of 'euclidean'. Now the title looks unbalanced. I apologize for my poor edit. /// Dear msfr: I am sorry for my poor edit. Please undo or correct my partial capitalization. – Ricardo Andrade Sep 16 '13 at 17:22
Dear Mark and Ricardo. Actually, I wanted to avoid svd ! – msfr Sep 16 '13 at 19:22
@msfr: If you just want, say, an upper or lower bound then one can find more easily computable expressions which may or may not be sharp enough for your purposes, but the variance is what it is. – Mark Meckes Sep 16 '13 at 20:46

Apparently, one possible answer is $$ E[N^2] = 3\ \text{Tr} (A A' A A') + 2 \sum_{\text{all } A_i 2\times 2 \text{ submatrices}} \text{Det} (A_i)^2. $$ Does this sound familiar to anyone ?

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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