2
$\begingroup$

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 $N\ge 2$.

$\endgroup$

1 Answer 1

4
$\begingroup$

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 $${\rm var}\,s=\frac{1}{\pi^2}\int_{a_-}^{a_+}d\lambda\int_{a_-}^{a_+}d\mu\ln|\lambda-\mu|\frac{d}{d\mu}\left(\frac{\sqrt{(\mu-a_-)(a_+-\mu)}}{\sqrt{(\lambda-a_-)(a_+-\lambda)}}\frac{1}{1+\alpha\lambda}\frac{d}{d\mu}\frac{1}{1+\alpha\mu}\right).$$

$\endgroup$
3
  • $\begingroup$ You wrote that the support of the eigenvalue density $\rho(x)$ is a specific (finite) interval, but the Marchenko-Pastur theorem gives us only limiting results, that is, the distribution converges to one with finite support. In practice, it could be that for any $M,N$, the support is infinite, right? Specifically, I am interested in proving that $\mathbb{E}\lambda_max$ is finite (for large enough $N,M$), and I can't seem to find such a result. $\endgroup$
    – Student88
    Feb 4 at 11:08
  • $\begingroup$ the probability density function of the largest eigenvalue is given by the Tracy-Widom law; it is sharply peaked around the large-$N$ limits $a_\pm$, with a standard deviation that scales as $N^{-1/6}$. $\endgroup$ Feb 4 at 11:41
  • $\begingroup$ Sorry, I did not clarify enough, I meant $\lambda_{max}$ of the inverse-Wishart, not the Wishart. That is, I am interested at $\mathbb{E} \lambda_{max}(W_n^{-1}(m,I))$ $\endgroup$
    – Student88
    Feb 4 at 17:26

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.