# The distribution of the condition number for complex Wishart matrices.

I am trying to derive the distribution of the condition number for centered uncorrelated complex Wishart matrices $n\times n$ with $m$ degrees of freedom.
The problem is with the solution I got (it's numeric verification).
First of all one begins with the joined distribution of the ordered eigenvalues $\vec{\Lambda}\!:\!\Lambda_1\!>\!\ldots\!>\!\Lambda_n$ $$\begin{eqnarray} f_{\vec{\Lambda}}(\vec{\lambda})&=&K_{m,n}|V(\vec{\lambda})|^2\prod_{l=1}^n e^{-\lambda_l}\lambda_{l}^{m-n}=K_{m,n}\left(\sum_{\vec{\mu}}\text{sgn}(\vec{\mu})\prod_{i=1}^n[V(\vec{\lambda}]_{\mu_i,i}\right)^2\prod_{l=1}^n e^{-\lambda_l}\lambda_{l}^{m-n}=\\ &=&K_{m,n}\sum_{\vec{\mu}}\sum_{\vec{\nu}}\text{sgn}(\vec{\mu})\text{sgn}(\vec{\nu})\prod_{i=1}^n\prod_{j=1}^n[V(\vec{\lambda}]_{\mu_i,i}[V(\vec{\lambda}]_{\nu_j,j}\prod_{l=1}^n e^{-\lambda_l}\lambda_{l}^{m-n}\\ \end{eqnarray}$$ Where $|V(\vec{\lambda})|$ - is the Vandermonde determinant, $K_{m,n}=\frac{1}{\prod_{1}^n (n-i)!(m-i)!}$ - normalization constant and $\vec{\mu}, \vec{\nu}$ - are vectors of all permutations of numbers $1$ through $n$.
Using the fact that $[V(\vec{\lambda}]_{i,j}=\lambda_j^{i-1}$: $$\begin{eqnarray} f_{\vec{\Lambda}}(\vec{\lambda})&=&K_{m,n}\sum_{\vec{\mu},\vec{\nu}}\text{sgn}(\vec{\mu})\text{sgn}(\vec{\nu})\prod_{l=1}^n e^{-\lambda_l}\lambda_{l}^{m-n+\mu_l+\nu_l-2}\\ \end{eqnarray}$$ In order to get the pdf of the condition number first I integrate out $\lambda_2\ldots\lambda_{n-1}$: $$\begin{eqnarray} f_{\vec{\Lambda}}(\vec{\lambda})&=&K_{m,n}\sum_{\vec{\mu},\vec{\nu}}\text{sgn}(\vec{\mu})\text{sgn}(\vec{\nu})e^{-(\lambda_l+\lambda_n)}\lambda_{1}^{m-n+\mu_1+\nu_1-2}\lambda_{n}^{m-n+\mu_n+\nu_n-2}\prod_{l=2}^{n-1} e^{-\lambda_l}\lambda_{l}^{m-n+\mu_l+\nu_l-2}\\ \end{eqnarray}$$ $$\begin{eqnarray} f_{\Lambda_1,\Lambda_n}(\lambda_1,\lambda_n)&=&K_{m,n}\sum_{\vec{\mu},\vec{\nu}}\text{sgn}(\vec{\mu})\text{sgn}(\vec{\nu})e^{-(\lambda_l+\lambda_n)}\lambda_{1}^{m-n+\mu_1+\nu_1-2}\lambda_{n}^{m-n+\mu_n+\nu_n-2}\times\\ &\mbox{}\mbox{}\mbox{} &\times\prod_{l=2}^{n-1}\int_0^{\infty}\lambda_{l}^{m-n+\mu_l+\nu_l-2} e^{-\lambda_l} \ \text{d}\lambda_{l}=\\ &=&K_{m,n}\sum_{\vec{\mu},\vec{\nu}}\text{sgn}(\vec{\mu})\text{sgn}(\vec{\nu})e^{-(\lambda_l+\lambda_n)}\lambda_{1}^{u_1}\lambda_{n}^{u_n}\prod_{l=2}^{n-1}\Gamma(u_l+1) \end{eqnarray}$$ Here I denoted $$\cases{u_1=m-n-2+\mu_1+\nu_1, \\ u_n=m-n-2+\mu_n+\nu_n,\\ u_l=m-n-2+\mu_l+\nu_l}.$$ At last using the standard manipulation to get the pdf of $\eta=\frac{\lambda_1}{\lambda_n}$ $$f_{H}(\eta)=\int_0^\infty f_{\Lambda_1,\Lambda_n}(\eta\lambda_n,\lambda_n) \ \lambda_n \ \rm d\lambda_n$$ $$\begin{eqnarray} f_{H}(\eta)&=&K_{m,n}\sum_{\vec{\mu},\vec{\nu}}\text{sgn}(\vec{\mu})\text{sgn}(\vec{\nu})\prod_{l=2}^{n-1}\Gamma(u_l+1)\eta^{u_1}\int_0^\infty e^{-((1+\eta)\lambda_n)}\lambda_{n}^{u_1+u_n+1}\ \rm d\lambda_n=\\ &=& K_{m,n}\sum_{\vec{\mu},\vec{\nu}}\text{sgn}(\vec{\mu})\text{sgn}(\vec{\nu})\prod_{l=2}^{n-1}\Gamma(u_l+1)\frac{\eta^{u_1}}{(1+\eta)^{u_1+u_n+2}}\Gamma(u_1+u_n+2) \end{eqnarray}$$ Seems like this is the final result but then I go on with plotting the obtained solution for various $m$ and $n$ and get counterintuitive results:

1. $f_{H}(\eta)\neq 0$ for $\eta<1$ (even for negative $\eta$). (I think I can manage with it by using Heaviside theta function).
2. $\int_0^\infty f_{H}(\eta)\ \rm d\eta\neq 1$
3. The simulation shows that the solution works great for some $m$ and $n$ for example if $m=4,\ n=2$:

but if $m=15,\ n=3$ I get

So where's the catch?
• This paper might be relevant mediatum.ub.tum.de/doc/957123/957123.pdf – Pierpaolo Vivo Apr 30 '16 at 11:01
• @PierpaoloVivo Thank you, but they derive exact pdf for the case $n=2, m\geq 2$. This is the case when my solution works just perfectly, but for $n\neq 2$ it fails. – Caran-d'Ache Apr 30 '16 at 11:24
• It seems to me they also have a general formula (9) [quite cumbersome], but you should check that your formula agrees with their (9)... – Pierpaolo Vivo Apr 30 '16 at 11:29