# Calculation of the variance in the Wishart random matrix ensemble

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.

• 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. Nov 6 '20 at 10:49

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, $${\rm var}\,X=\tfrac{1}{8}a^2M^2(a_+-a_-)^2=2a^2NM.$$