1
$\begingroup$

In the univariate case ($\chi^2$ distribution), I know we can expand the pdf into power series of the variance $\sigma^2$ with Laguerre polynomials. Indeed, since the Laguerre polynomials are related to the derivatives w.r.t. the variance $\sigma^2$, this expansion is exactly the Taylor expansion.

Does anyone know if we can do the same for the general multivariate case, the Wishart distribution? Maybe now we need a matrix variate orthogonal polynomials.

If this is possible, does the expansion still works for the singular case? (i.e., the sample size n is smaller than the dimension p)

$\endgroup$
  • 2
    $\begingroup$ you might consult Graph presentations for moments of noncentral Wishart distributions and their applications $\endgroup$ – Carlo Beenakker Aug 8 '17 at 8:24
  • 1
    $\begingroup$ @CarloBeenakker Carlo, I know that Wishart's expansion can be represented in terms of Laguerre polynomials, but what is the insight the graph representation you provided here? $\endgroup$ – Henry.L Aug 8 '17 at 22:03
  • $\begingroup$ In Appendix A of this paper you will find multivariate Laguere polynomials, although I am not sure this will help you: arxiv.org/pdf/1510.02390.pdf $\endgroup$ – Marcel Aug 8 '17 at 23:47
  • $\begingroup$ @Henry.L, can you give a reference for how this can be done? Thanks! $\endgroup$ – Felix Y. Aug 9 '17 at 23:01
  • $\begingroup$ @FelixY. See if my answer helps, thanks $\endgroup$ – Henry.L Aug 12 '17 at 14:20
2
$\begingroup$

Note that the Laguerre orthogonal polynomials are in form of [1](bearing combinatoric interpretation) and [3]

\begin{align} & L_n^\nu(x)=(-1)^n\sum_{m=0}^n \binom n m \prod_{i=1}^m (\nu+2(n-i))(-x)^{n-m} \\[8pt] = {} & \sum_{m=0}^n \frac{\Gamma(\nu+n+1)\frac{\Gamma(-n+m)}{\Gamma(-n)}} {n!m!(\nu+m+1)} x^k=\sum_{m=0}^n c_m(n,\nu) x^k \end{align}

The connection between Wishart matrix and the Laguerre polynomial is that the eigenvalues of the Wishart matrix $M_s = \frac{1}{s} Z_s Z_s^T$ where $Z_s\sim \left[\operatorname{Normal}(0,1)\right]_{n\times s}$ are the zeros of appropriately scaled generalized Laguerre polynomials as explained in [2]. Therefore the spectral density of the Wishart distribution can be written in terms of Laguerre polynomials, and we know that the moments of distribution can be directly derived from its spectral density. Now we obtain relation inbetween moments of Wishart distribution $\beta=1,2,4$ and corresponding Laguerre polynomials in [3] when the Wishart distribution is central. For noncentral case, the derivation in [3] may also apply but with slight modification of the intermediate quantity

$$ \mathcal{Q}(r;m,\ell;\alpha):={ \int_0^\infty dx \, x^r e^{-x} L_m^\alpha (x) L_\ell^\alpha (x)}, $$

which is no longer of the simple form

$$\sum_{k=0,\cdots m,k'=0\cdots\ell} c_k(m,\alpha) c_{k^{'}} (\ell,\alpha)\Gamma(1+r+k+k').$$

Reference

[1]Kuriki, Satoshi, and Yasuhide Numata. "Graph presentations for moments of noncentral Wishart distributions and their applications." Annals of the Institute of Statistical Mathematics 62.4 (2010): 645-672.

This reference is pointed out to us by Carlo.

[2]Dette, Holger. "Strong approximation of eigenvalues of large dimensional Wishart matrices by roots of generalized Laguerre polynomials." Journal of Approximation Theory 118.2 (2002): 290-304.

[3]Livan, Giacomo, and Pierpaolo Vivo. "Moments of Wishart-Laguerre and Jacobi ensembles of random matrices: application to the quantum transport problem in chaotic cavities." arXiv preprint arXiv:1103.2638 (2011).

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ Thanks for your answer! The reason that I think there could be an expression for such expansion is that the Wishart distribution belongs to the natural exponential family and from the general theory, we know such orthogonal polynomials exist can can be determined by taking derivatives. What do you think? $\endgroup$ – Felix Y. Aug 19 '17 at 18:39
  • $\begingroup$ @FelixY I think you raised some reasonable points but digging further requires more efforts and carefulness. :) $\endgroup$ – Henry.L Aug 19 '17 at 19:09

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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