# “Normalized” covariance matrix of a Gaussian random vector

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

Letting $$B=\frac{XX^{\top}}{\|X\|_{2}^{2}}$$, one can see that $$$$B_{ii}=\frac{X{i}^{2}}{X_{1}^{2}+\cdots+X_{d}^{2}}$$$$ which is a ratio of chi-squared random variables. This is a beta random variable with parameters $$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 $$$$B_{ij}=\frac{X_{i}X_{j}}{X_{1}^{2}+\cdots+X_{d}^{2}}.$$$$ For this, I thought of conditioning on $$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.
• 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 – nemo Apr 16 at 14:28