Let $X\sim\mathcal{N}(0,I_{d})$. I would like to compute the the following quantity: \begin{equation} \mathbb{E}\bigg[\frac{XX^{\top}}{\|X\|_{2}^{2}}\bigg]. \end{equation}

Letting $B=\frac{XX^{\top}}{\|X\|_{2}^{2}}$, one can see that \begin{equation} B_{ii}=\frac{X{i}^{2}}{X_{1}^{2}+\cdots+X_{d}^{2}} \end{equation} which is a ratio of chi-squared random variables. This is a beta random variable with parameters $1/2$ and $(d-1)/2$, and the expectation is $1/d$. This takes care of the diagonal entries.

It's the off-diagonal entries where I'm stuck. We have \begin{equation} B_{ij}=\frac{X_{i}X_{j}}{X_{1}^{2}+\cdots+X_{d}^{2}}. \end{equation} For this, I thought of conditioning on $X_{i}$ and $X_{j}$ first. That basically gives me $\mathbb{E}[1/(c+Z)]$, where $c=x_{i}^{2}+x_{j}^{2}$ and $Z\sim\chi_{d-2}^{2}$. Any ideas on how to compute this expectation? Any other approaches are also welcome.


Your off diagonal entries are all $0$. The distribution of $X$ is invariant with respect to rotations, and your matrix $B$ should reflect that fact: $O^TBO=B$ for all orthogonal $O$. Or you could note that the distribution of $X$ is invariant with respect to replacing $X_i$ with $-X_i$, for any particular $i$.

Arguing from $\pm$ symmetry of $X_1$, conditional on everything else, you have $EX_1X_2/\|X\|_2^2$ = $-EX_1X_2/\|X\|_2^2$. Since the marginal density function of $X_1$ is symmetric with respect to sign change, you are integrating an odd function against an even density. If you believe $\int_{\mathbb R} z \varphi(z)dz = 0$ because of symmetry, hold that thought and apply it here as well.

  • $\begingroup$ Yes, I expect them to be zero. It should be possible through this entrywise calculation; the conditioning brings out $X_{i}$ and $X_{j}$ and they are zero mean. But I don't know what exactly $\mathbb{E}(1/c+Z)$ looks like $\endgroup$ – nemo Apr 16 at 14:28
  • $\begingroup$ That works, thanks! $\endgroup$ – nemo Apr 17 at 5:18

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