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I'm trying to obtain the mean and variance of a function given by $f=\sum_{i=1}^{N} a \lambda_i$, where $a$ is a constant and $\lambda_i$ are the ordered eigenvalues of a Wishart matrix given by $\mathbf{W}=\mathbf{H}\mathbf{H}^H$, where the elements of $H$ are complex Gaussian distributed with zero-mean and unitary variance.

I know that when $N$ is large, the distribution of the eigenvalues of $\mathbf{W}$ is a Marchenko-Pastur distribution $p(\lambda)$ (I have validated this already). Therefore, following distribution on the inverse Wishart matrix eigenvalues summation , I'm considering $f$ as a Gaussian RV.

Although I can obtain the mean of $f$ with success, I'm not able to obtain its variance. I'm considering eq. (17) of https://arxiv.org/abs/cond-mat/9310010, but the numerical integration gives me a singularity.

Can someone help me with this? Thank you very much.

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    $\begingroup$ Enough to consider $a=1$. Then $f$ has a $\chi^2$ distribution, since is is the trace of $W$, that is, the sum of squares of the entries of $H$. From this you can compute whatever moments you need. $\endgroup$ Nov 6 '20 at 10:49
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As pointed out by Ofer Zeitouni in a comment, in this case there is an exact result $2a^2(NM-1)$ for the variance of $\sum_n a\lambda_n$ in the ensemble of $N\times M$ Wishart matrices. The calculation below serves as a check that the large-$N$ approach mentioned in the OP leads to the same answer.


According to the formula in the cited paper, the variance of $X=\sum_{i=1}^Nf(\lambda_i/M)$ in the large-$N,M$ limit (at fixed $y=N/M\leq 1$) is given by the principal-value integral $${\rm var}\,X=\frac{1}{\pi^2}\int_{a_-}^{a_+}d\lambda\int_{a_-}^{a_+}d\mu\frac{\sqrt{(\mu-a_-)(a_+-\mu)}}{\sqrt{(\lambda-a_-)(a_+-\lambda)}}\frac{1}{\lambda-\mu}f(\lambda)\frac{d}{d\mu}f(\mu),$$ where the Marchenko-Pastur eigenvalue distribution has support $(a_-,a_+)$ with $a_\pm=(1\pm\sqrt y)^2$. For the case in the OP one has $f(\lambda)=aM\lambda$, and then the integral can be evaluated in closed form, $${\rm var}\,X=\tfrac{1}{8}a^2M^2(a_+-a_-)^2=2a^2NM.$$

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