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Assume two independent random complex vectors (with real and imaginary parts): $$ \vec{\dot{Z}}=(\dot{z}_1,\dot{z}_2,\ldots,\dot{z}_n)~~~~~~~~~~ \begin{cases} \Re(\dot{z}_i)&=x_i \\ \Im(\dot{z}_i)&=y_i \end{cases}\\ \text{and} \\ \vec{\dot{\zeta}}=(\dot{\zeta}_1,\dot{\zeta}_2,\ldots,\dot{\zeta}_n)~~~~~~~~~~ \begin{cases} \Re(\dot{\zeta}_i)&=\xi_i \\ \Im(\dot{\zeta}_i)&=\eta_i \end{cases} $$ can be viewed as quadrature components of some real signal.

Let them be governed by the multivariate complex normal distributions $\vec{\dot{Z}}\sim\ \mathcal{CN}(\vec{\dot{\mu}}_{z},\, \Gamma_{z},\, C_{z})$, $\vec{\dot{\zeta}}\sim\ \mathcal{CN}(\vec{\dot{\mu}}_{\zeta},\, \Gamma_{\zeta},\, C_{\zeta})$ with mean vector $\vec{\dot{\mu}}$, covariance matrix $\Gamma$, and the relation matrix $C$.
Let me form the variable $\varphi$ as the argument of the scalar product between $\vec{\dot{Z}}$ and complex conjugate of $\vec{\dot{\zeta}}$, i.e. $$\varphi=\arg(\vec{\dot{Z}}\!.\!\vec{\dot{\zeta^*}})$$ where $\cdot^*$ is a conjugate transpose operator.

I would like to know, is there any way to find the distribution of $\varphi$? Or even its absolute value $|\varphi|$?

Up to now I haven't seen the solution of that problem in the textbooks or scientific papers available to me. I tried to look at it as the part of some bilinear form and tried A.M. Mathai and S.B. Provost "Quadratic Forms in Random Variables: Theory and Applications", but failed because

  1. it mainly deals with quadratic forms,
  2. frankly speaking I am not that good at that kind of transformation.

However there are some similar scientific papers, but they all again deal with quadratic forms and usually simplify the problem (zero mean or zero relation matrix etc.).

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  • $\begingroup$ Welcome to the site, @Caran-d'Ache. I have tweaked your formatting somewhat for greater readability. Please ensure it still says what you want it to say. $\endgroup$ – gung Jun 16 '13 at 17:19
  • $\begingroup$ @gung well, thank you, I guess I was a bit wordy. It's better now. $\endgroup$ – Caran-d'Ache Jun 17 '13 at 3:44

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