1
$\begingroup$

I want to compute the expected norm of a vector-matrix multiplication. I have a vector $x \in \mathbb{R}^n$ with norm one and a matrix $M \in \mathbb{R}^{n \times n}$, whose entries are iid taken from the Gaussian distribution with mean zero and variance $2/n$: $M_{i,j} \sim \mathcal{N}(0,2/n)$.

I need to compute the expected value of the norm of the product, i.e. $ \mathbb{E}[||xM||].$

I have computed $\mathbb{E}[||xM||^2] = 2$, but i have no idea on how to get rid of that square sign.

I could use the chi or Gamma functions, but I'm somehow stuck:

I'd have $$\sqrt{\sum_i\sum_j^n x_i^2 M_{i,j}^2 + \sum_i \sum_j \sum_k x_i x_j M_{i,j} M_{i,k}}$$

I know I can use the gamma function for the first sum, and that the expected value of the second part goes to zero. The problem is that I can't compute the expected value of just the second part since it's under the square root. Any suggestions?

Thank you!

$\endgroup$
2
$\begingroup$

The key is the simple observation that, for any orthogonal matrix $Q$, the matrix $QM$ equals $M$ in distribution. Let now $Q$ be any orthogonal matrix such that $xQ=e_1:=[1,0,\dots,0]$. Then $xM$ equals \begin{equation} xQM=e_1M=[M_{1,1},\dots,M_{1,n}] \end{equation} in distribution. So, $\|xM\|$ equals $\sqrt{\frac2n\,X}$ in distribution, where $X\sim\chi^2_n=\text{Gamma}(\frac n2,2)$. It follows that \begin{equation} E\|xM\|=\frac{2\Gamma(\frac{n+1}2)}{\sqrt n\,\Gamma(\frac n2)}, \end{equation} which is close to $\sqrt 2=\sqrt{E\|xM\|^2}$ for large $n$, as should be expected, in view of the measure concentration phenomenon.

$\endgroup$
  • $\begingroup$ Thank you very much! $\endgroup$ – Alfred Jan 31 at 14:44

Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

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