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.

  • 1
    $\begingroup$ 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) $\endgroup$ Jan 14 at 18:10
  • $\begingroup$ 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) $\endgroup$
    – user27182
    Jan 14 at 18:26
  • $\begingroup$ 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. $\endgroup$ Jan 14 at 21:05
  • $\begingroup$ 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… $\endgroup$
    – user27182
    Jan 15 at 15:19
  • $\begingroup$ I have worked out this estimate below. $\endgroup$ Jan 15 at 18:25

1 Answer 1


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.


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