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 $N\ge 2$.


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, 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 $$\rho(x)=\frac{1}{2\pi}\frac{N}{yx}(x-a_-)^{1/2}(a_+-x)^{1/2},$$ 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 $y$,

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

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


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