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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!

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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.

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  • $\begingroup$ Sorry, I'm trying to plug numbers inside the equation you provided, but even for large $n$, the final output is not close to $\sqrt2$. How is this possible? Thanks! $\endgroup$ – Alfred May 28 '19 at 10:55
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    $\begingroup$ @Alfred : I guess you made a mistake somewhere. I have rechecked the limit manually using Striling's formula, and it is $\sqrt2$ indeed. Look also at the section "Series expansion at $n=\infty$" at wolframalpha.com/input/… $\endgroup$ – Iosif Pinelis May 28 '19 at 17:17
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    $\begingroup$ @Alfred : You can also look at the numerics at wolframalpha.com/input/… $\endgroup$ – Iosif Pinelis May 28 '19 at 17:23
  • $\begingroup$ I did make a mistake, thank you again! $\endgroup$ – Alfred May 29 '19 at 9:46

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