# 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}

The joint distribution of the eigenvalues $$\lambda_i$$, $$i=1,2,\ldots n$$ of $$A$$ is known, $$P(\lambda_1,\lambda_2,\ldots\lambda_n)=c_{k,n}\prod_{i with $$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 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 $$n$$.
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 $$k^{-1}\log E \det(I+(I+A)^{-1}) =k^{-1} \log E(e^{k \langle L_k,g\rangle})\to \langle \mu,g\rangle$$