# Expectation value of inverse covariance matrix when sampling from unit sphere

Let $$X \sim \operatorname{Unif}S_{d-1}$$, so $$X\in\mathbb{R}^d$$ and is distributed uniformly on the unit sphere.

Then let $$X_1, \dots, X_n \sim X$$ iid and define the matrix $$\mathbf{X}\in\mathbb{R}^{n\times d}$$ by $$\mathbf{X}_{ij} = (X_i)_j$$

Is there a name for the distribution of $$(\mathbf{X}^\top \mathbf{X})^+$$?

The + denotes Moore-Penrose Pseudo-Inverse. This would be to $$\operatorname{Unif}S_{d-1}$$ as the (generalised) Wishart is to the normal distribution.

• the elements of $X_i$ have for large $d$ approximately a Gaussian distribution with zero mean and variance $1/d$; so I would think that the distribution of $X^\top X$ will be close to the usual Wishart distribution (and its inverse close to the inverse Wishart distribution) Jan 14 at 18:10
• True, but I ultimately want to look at the first moment of this object in terms of $n$ and $d$. By isotropy, we know that $\mathbb{E}[(\mathbf{X}^\top\mathbf{X})^+] = r(n, d) I$ for some scalar $r(n, d)$, I just don't know which scalar... (For wishart this is known) Jan 14 at 18:26
• so my best guess would be $\mathbb{E}[(\mathbf{X}^\top\mathbf{X})^+] = d(n-d-1)^{-1} I$ for $n>d-1\gg 1$ -- you might want to check this numerically. Jan 14 at 21:05
• Thanks. Quite similar to the Gaussian case. How did you formulate your guess? I find it difficult to believe that no one has studied this, but maybe it’s the case… Jan 15 at 15:19
• I have worked out this estimate below. Jan 15 at 18:25

First note that the vector $$x$$ distributed uniformly on the $$d$$-dimensional hypersphere can be constructed from a vector $$y$$ with i.i.d. normal elements $$y_1,y_2,\ldots y_d$$, via $$x=\left(\sum_{i=1}^d y_i^2\right)^{-1/2} y.$$ Consider a set of $$n$$ vectors $$y^{(1)},y^{(2)}\ldots y^{(n)}$$ and construct the $$n\times d$$ dimensional matrix $$Y$$ with elements $$Y_{ij}=y^{(i)}_j$$. The matrix product $$Y^TY$$ has a Wishart distribution, and $$(Y^TY)^+$$ has the inverse Wishart distribution, with expectation value $$\mathbb{E}(Y^TY)^+=\frac{I}{n-d-1},\;\;n>d+1.$$

Also construct the $$n\times n$$ diagonal matrix $$D$$ with elements $$D_{ij}=\delta_{ij}\left(\sum_{k=1}^d (y_k^{(i)})^2\right)^{-1/2}$$. The matrix $$X$$ in the OP is related to the matrix $$Y$$ by $$X=DY$$.

We seek the expectation value of the scalar $$R\equiv \frac{1}{d}\,{\rm tr}\,(X^TX)^+=\frac{1}{d}{\rm tr}\,(Y^T D^2Y)^+.$$ For $$d\gg 1$$ the matrix $$D^2$$ selfaverages to $$1/d$$ times the identity, hence we estimate $$r_{n,d}\equiv\mathbb{E}(R)\approx\frac{d}{n-d-1},\;\;n>d+1\gg 1.$$

I have performed some numerical checks, averaging over 500 realizations.

d,n d/(n-d-1) numerics
100,105 25 25.1
100,110 11.11 11.0
100,120 5.26 5.15
50,100 1.02 1.00
50,200 0.336 0.332

The difference between the numerical value and the estimate $$d/(n-d-1)$$ is within the statistical uncertainty.