Given a $N \times M$ matrix $X$ comprised of standard normal entries ($M > N$), I'm interested in approximating $E[trace((XX^T\frac{\gamma}{M} + I)^{-1}]$ in terms of $N, M$ and $\gamma$. Unfortunately, I can't necessarily assume $\gamma$ is small. I've had no luck in coming up with any kind of approximation. Thanks!


For context, this problem relates to the effective degrees of freedom for ridge regression.



Expectation of the trace of an inverse of a random matrix
let $\lambda\_{1},\lambda\_2,\ldots\lambda\_N$ be the eigenvalues of $M^{-1}XX^{T}$; including for convenience a factor $1/N$, the quantity you seek is


$$N^{-1}E[{\rm Tr}(XX^{T}\gamma/M)+I)^{-1}]=\int d\lambda \rho(\lambda)(\lambda\gamma+1)^{-1}$$


where $\rho(\lambda)=E[N^{-1}\sum\_n\delta(\lambda-\lambda\_n)]$ is the eigenvalue density of Wishart matrices; this quantity is known in closed form for $N,M\rightarrow\infty$ at finite ratio $N/M=r\in(0,1]$ ([Marchenko-Pastur distribution](http://en.wikipedia.org/wiki/Marchenko-Pastur_distribution)). I find in this way the answer


$$\lim\_{N,M\rightarrow\infty}N^{-1}E[{\rm Tr}(XX^{T}\gamma/M)+I)^{-1}]=(2r\gamma)^{-1}\left(-1-\gamma\sqrt{ab}+\sqrt{(1+a\gamma)(1+b\gamma)}\right)$$


with $a=(1+\sqrt r)^2$ and $b=(1-\sqrt r)^2$. As a check, you can take the limit $\gamma\rightarrow 0$ of this expression and obtain $1$, as it should.


