Expected value of orthogonal projection $X^{+}X$ Let $X\in\mathbb{R}^{m\times n}$, where $m<n$, be a random matrix where the rows $x_i$ ($i=1,...,m$) are sampled i.i.d. from Gaussian distribution with mean $0$ and covariance $\Sigma$, i.e. $x_i\sim N(0,\Sigma)$.
How to calculate the expected value $\mathbb{E}[X^{+}X]$ where $X^{+}$ is the Moore–Penrose inverse of $X$ ?
Thank you.
 A: Assume that $\det\Sigma\ne0$. Then the random matrix $X$ is of rank $m$ almost surely (a.s.). So, a.s. the Moore--Penrose inverse of $X$ is $X^+=X^\top(XX^\top)^{-1}$ and hence
$$X^+X=X^\top(XX^\top)^{-1}X.$$
It appears that $EX^+X=EX^\top(XX^\top)^{-1}X$ cannot be expressed in closed form, even in the fully specified case when $m=2$, $n=3$, and $\Sigma=\left(
\begin{array}{ccc}
 1 & 1 & 0 \\
 1 & 2 & 0 \\
 0 & 0 & 1 \\
\end{array}
\right)$.
Indeed, in this case $\Sigma=A^\top A$ for $A:=\left(
\begin{array}{ccc}
 1 & 1 & 0 \\
 0 & 1 & 0 \\
 0 & 0 & 1 \\
\end{array}
\right)$. So,
for the rows $x_i$ of $X$ we can write $x_i=z_i A$, where the $z_i$'s are iid rows of iid standard normal random variables $z_{i,j}$.
In the image of a Mathematica notebook below, the expression of even the $(1,1)$-entry (P11) of the matrix $P:=X^+X$ in terms of the $z_{i,j}$'s looks very formidable, and Mathematica cannot do anything for the expectation of P11, leaving it unevaluated after working on it for more than an hour (click on the image to enlarge it):

