Let $\lambda\_1>\lambda\_2>....>\lambda\_N$ be the ordered eigenvalues of Wishart matrix my objective is to find if it is possible the distribution of:


\begin{align}
 s = \sum\limits\_{i = 1}^N {\frac{1}{{1 + a{\lambda \_i}}}} 
\end{align}
where a is positive and $N\ge 2$.



distribution on the inverse Wishart matrix eigenvalues summation
For small $N$ explicit expressions are cumbersome. For $N\gg 1$ the distribution $P(s)$ is a Gaussian. The mean is given by integration with the Marchenko-Pastur distribution, the variance is given by integration with a formula given in [arXiv:9310010](https://arxiv.org/abs/cond-mat/9310010), Equation 17. Let me work this out:


The Wishart matrix is $X=WW^T$, with $W$ an $N\times M$ real matrix, $N\leq M$ and $y=N/M$. I rescale the eigenvalues $\lambda\_i$ of $X$ by $x\_i=\lambda\_i/M$, and define $\alpha=aM$. We seek the distribution of $s=\sum\_{i}(1+\alpha x\_i)^{-1}$. The support of the eigenvalue density $\rho(x)$ is the interval $(a\_-,a\_+)$, with $a\_\pm=(1\pm\sqrt y)^2$. For $N\gg 1$ one has the [Marchenko-Pastur distribution](https://en.wikipedia.org/wiki/Marchenko%E2%80%93Pastur_distribution)
$$\rho(x)=\frac{1}{2\pi}\frac{N}{yx}(x-a\_-)^{1/2}(a\_+-x)^{1/2},$$
normalized to $\int\rho(x)dx=N$. The mean of the Gaussian is then equal to
$$\mathbb{E}[s]=\int\_{a\_-}^{a\_+}\frac{\rho(x)}{1+\alpha x}\,dx=N\frac{\sqrt{\alpha^2 (y-1)^2+2 \alpha (y+1)+1}+\alpha (y-1)-1}{2 \alpha y}.$$


For the variance we apply Eq. 17 of the cited paper,
$${\rm var}\,s=\frac{1}{\pi^2}\int\_{\alpha\_-}^{a\_+}d\lambda\int\_{a\_-}^{a\_+}d\mu\frac{\sqrt{(\mu-a\_-)(a\_+-\mu)}}{\sqrt{(\lambda-a\_-)(a\_+-\lambda)}}\frac{1}{\lambda-\mu}\frac{1}{1+\alpha\lambda}\frac{d}{d\mu}\frac{1}{1+\alpha\mu},$$
the integrals being Cauchy principal values.  

I have not succeeded in evaluating the integrals in closed form. Here is a numerical calculation of the variance for $\alpha=1$ as a function of $y$,


![](https://ilorentz.org/beenakker/MO/variance.png)


and a plot of the variance for $y=0.5$ as a function of $\alpha$,


![](https://ilorentz.org/beenakker/MO/variance2.png)


For the numerical evaluation it is convenient to rewrite the integral in the form
$${\rm var}\,s=\frac{1}{\pi^2}\int\_{a\_-}^{a\_+}d\lambda\int\_{a\_-}^{a\_+}d\mu\ln|\lambda-\mu|\frac{d}{d\mu}\left(\frac{\sqrt{(\mu-a\_-)(a\_+-\mu)}}{\sqrt{(\lambda-a\_-)(a\_+-\lambda)}}\frac{1}{1+\alpha\lambda}\frac{d}{d\mu}\frac{1}{1+\alpha\mu}\right).$$


