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I'm trying to find the following:

$$\mathbb{E}\left[\left(\mathbf{W}+\lambda\mathbf{I}\right)^{-1}\right]$$

where $\mathbf{W}\sim\mathcal{W}_{p}(n,\mathbf{\Sigma})$ is Wishart-distributed, $\lambda\ge 0$ is a constant, and $\mathbf{I}$ is the identity matrix. It is known (e.g. Das Gupta, 1968) that $\mathbb{E}\left[\mathbf{W}^{-1}\right]=\frac{\mathbf{\Sigma}^{-1}}{n-p-1}$, and the expected value of the inverse of a non-central Wishart matrix is also known (Hillier, 2019). However, it's not clear how $\mathbf{W}+\lambda\mathbf{I}$ could be expressed as a non-central Wishart-distributed matrix.

Approximations would also be useful, and I have tried

$$\mathbb{E}\left[\left(\mathbf{W}+\lambda\mathbf{I}\right)^{-1}\right]\approx\mathbb{E}\left[\mathbf{W}^{-1}\right]-\lambda\mathbb{E}\left[\mathbf{W}^{-2}\right]+\lambda^{2}\mathbb{E}\left[\mathbf{W}^{-3}\right]+\mathcal{O}(\lambda^{3})$$

The higher order moments of the inverted Wishart are known (Von Rosen, 1988), so this can be calculated, but the estimate is only valid for very small $\lambda$, especially when $p$ is close to $n$.

References:

Das Gupta, S., "Some aspects of discrimination function coefficients," The Indian Journal of Statistics, Vol. 30, No. 4, 1968.

Hillier, G. et al., "Properties of the inverse of a noncentral Wishart matrix," Available at SSRN 3370864, 2019.

Von Rosen, D., "Moments for the inverted Wishart distribution," Scandinavian Journal of Statistics, Vol. 15, No. 2, pp. 97-109, 1988.

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    $\begingroup$ I am inclined to doubt that $\mathbf{W}+\lambda\mathbf{I}$ is a non-central Wishart-distributed random matrix, and I won't be surprised it it can easily and quickly be proved that it's not. $\endgroup$
    – Michael Hardy
    Commented Dec 2, 2020 at 21:10
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    $\begingroup$ Michael: I believe you are correct (although, I haven't proven it). I guess that means that a different approach will need to be used to find the expected value of $\left(\mathbf{W}+\lambda\mathbf{I}\right)^{-1}$ $\endgroup$
    – Cory Merkel
    Commented Dec 2, 2020 at 21:15
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    $\begingroup$ $$ \begin{align} & \lambda^{-1} \left( \frac 1 \lambda \mathbf{W}+\mathbf{I}\right)^{-1} \\ {} \\ = {} & \lambda^{-1} \left( \mathbf I - \frac 1 \lambda\mathbf W + \frac 1 {\lambda^2} \mathbf W^2 - \frac 1 {\lambda^3} \mathbf W^3 + \cdots \right) \text{ ?} \end{align} $$ I don't know whether this will work, nor what to do with $\operatorname E(\mathbb W^n). \qquad$ $\endgroup$
    – Michael Hardy
    Commented Dec 2, 2020 at 21:18
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    $\begingroup$ Michael: Interesting idea. For large(ish) $\lambda$ this could be truncated to only deal with the mean and variance of $\mathbf{W}$, which are fairly simple. However, for $\lambda<1$, the higher-order terms would need to be accounted for as well... $\endgroup$
    – Cory Merkel
    Commented Dec 2, 2020 at 21:24

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