Expectation of the determinant of the inverse of non-central Wishart matrix Let $A$ be $(n,n)$ central Wishart matrix with $k$ degrees of freedom.
my question is there is a way to estimate the expectation of:
\begin{align}
E[det(I+(I+A)^{-1})]
\end{align}
 A: The joint distribution of the eigenvalues $\lambda_i$, $i=1,2,\ldots n$ of $A$ is known,
$$P(\lambda_1,\lambda_2,\ldots\lambda_n)=c_{k,n}\prod_{i<j}|\lambda_i-\lambda_j|\prod_m e^{-\lambda_m/2}\lambda_m^{(k-n-1)/2},$$
with $c_{k,n}$ a normalization constant.
The desired expectation value is given by
$$U_{n,k}=\mathbb E[\det(I+(I+A)^{-1})]=\int_0^\infty d\lambda_1\int_0^\infty d\lambda_2\cdots \int_0^\infty d\lambda_n\,\prod_{i<j}|\lambda_i-\lambda_j|\prod_m e^{-\lambda_m/2}\lambda_m^{(k-n-1)/2}\left(1+(1+\lambda_m)^{-1}\right).$$
For small $n$ the integrals can be done by quadrature, but the integrals quickly become unwieldy.
For example, for $n=1$, $k>0$ I find
$$U_{1,k}=2^{-\frac{k}{2}} \left[2^{k/2}+\sqrt{e} \,\Gamma \left(1-\tfrac{1}{2}k,\tfrac{1}{2}\right)\right],$$
with $\Gamma$ the incomplete Gamma function.
If you are satisfied with the expectation value of the logarithm of the determinant, then you can use the Marchenko-Pastur distribution to obtain an accurate result for large $n$.
A: You did not specify whether you have a relation between $n$ and $k$ and whether you care about asymptotics. I will assume that $k/n\to 1$ and that $n\to\infty$, and that your scaling is such that the eigenvalues are of order $1$. (The case $k/n\to\alpha$ can be handled similarly; also, if you meant that the typical eigenvalues are of order $k$, then the situation is much simpler).
Let $\lambda_i$ denote the eigenvalues of $A$. Let $L_k=k^{-1} \sum \delta_{\lambda_i}$ denote the empirical measure. Then
$\log det(I+(I+A)^{-1})=\sum \log (1+(1+\lambda_i))^{-1}= k  \langle L_k,g\rangle$ where
$g(x)=\log(1+(1+x)^{-1})$ is continuous and bounded on $R_+$. Now, $L_k$ satisfies a LDP
at speed $k^2$, which means that the probability that $L_k$ is not in a small neighborhood of the Pastur-Marchenko law $\mu$
is exponentially small at scale $k^2$. You then have that
$$k^{-1}\log E \det(I+(I+A)^{-1}) =k^{-1} \log E(e^{k \langle L_k,g\rangle})\to \langle \mu,g\rangle $$
