Let $Q$ be a (say 4x4) unitary matrix, distributed according to the Haar distribution. Denote the upper left 2x2 submatrix of $Q$ as $Q\_{1:2,1:2}$. I am interested in the following expectation:


$E(I - Q\_{1:2,1:2}^HQ\_{1:2,1:2})^{-1}$, where $H$ is Hermitian transpose and the expectation is across $Q\_{1:2,1:2}$. Anyone that knows how to do it (or can point me to a reference)?
Thanks!



Expected value of the inverse of a random, truncated Haar matrix
I will first consider the case of a $4\times 4$ matrix $Q$, and generalize to a higher dimensional $Q$ at the end.


**A. The four-by-four case.**


It is helpful to start from the polar decomposition


$$Q=
\left(\begin{array}{cc}U'&0\\ 0&V'\end{array}\right)
\left(\begin{array}{cc}\sqrt{1-T}&\sqrt{T}\\
\sqrt{T}&-\sqrt{1-T}
\end{array}\right)
\left(\begin{array}{cc}U&0\\ 0&V\end{array}\right)
$$
where $U,V,U',V'$ are four unitary matrices and
$$
T=\left(\begin{array}{cc}T\_1&0\\ 0&T\_2\end{array}\right)$$
is a diagonal matrix with diagonal elements $0\leq T\_n\leq 1$.


You seek the expectation value of the matrix 
$$M=[I-U^{\rm H}(1-T)U]^{-1}=U^{\rm H}T^{-1}U$$
(I have used that $U^{\rm H}U=I$.)


The Haar distribution for $Q$ implies a Haar distribution for $U$, and moreover implies for $T\_1,T\_2$ the following distribution [\*]


$$P(T\_1,T\_2)=6(T\_1-T\_2)^2,\;\;0\leq T\_n\leq 1$$


The marginal distribution for $T\_1$ is


$$P(T\_1)=2-6T\_1+6T\_1^2$$


So you see that the expectation value of $M$ diverges: $E(M)=E(1/T\_1)\,I=\infty$.


**B. The higher-dimensional case.** 


Now let's consider an $N\times N$ dimensional unitary matrix $Q$, with $N\geq 4$. The $2\times 2$ upper-left block is still of the form $U'TU$, with unitary $U'$, $U$, so we still have $M=U^{\rm H}T^{-1}U$. The Haar distribution of $Q$ still implies a Haar distribution for $U$, what changes is the distribution of $T\_1$ and $T\_2$ [\*]


$$P(T\_1,T\_2)=C(T\_1-T\_2)^2 (T\_1 T\_2)^{N-4}$$


with normalization constant $C=\frac{1}{2}(N-2)^2(N^2-4N+3)$. 
The marginal distribution of $T\_1$ is


$$P(T\_1)=CT\_1^{N-4}\left(\frac{1}{N-1}-\frac{2T\_1}{N-2}+\frac{T\_1^2}{N-3}\right)$$


Now the expectation of $M$ converges,


$$E(M)=\frac{N-2}{N-4}I$$




---


[\*] See Equation (2.10) of this [review.](http://arxiv.org/abs/cond-mat/9612179)


