What would be the probability density function (pdf) of the complex random variable given below?

$$Z = \sum_{i=1}^{M}{x_{i}^{*}y_{i}}$$

where $x_i, y_i$ are independent r.v.'s with $\mathcal{CN}(0,c)$.

  • $\begingroup$ The title might be clearer if you just said dot product. Also, you need to specify the joint PDF of $x$ and $y$, e.g. are they independent? $\endgroup$ – GeoMatt22 Jan 1 '17 at 21:46
  • $\begingroup$ Does someone know what would be the pdf of $x_i$ and $y_i$, that would be a good starting point. $\endgroup$ – Felipe Augusto de Figueiredo Jan 4 '17 at 8:34
  • $\begingroup$ They are independent complex gaussian r.v.'s, as far as I'm concerned, $Z$ won't have a chi-squared distribution. $\endgroup$ – Felipe Augusto de Figueiredo Jan 4 '17 at 15:09
  • $\begingroup$ The problem is that they are both complex r. v.'s. $\endgroup$ – Felipe Augusto de Figueiredo Jan 4 '17 at 18:11
  • $\begingroup$ I think that is not possible as the r.v.'s are complex but I'm not sure. Could someone else advise on that matter, please? $\endgroup$ – Felipe Augusto de Figueiredo Jan 4 '17 at 19:40

Note that a complex multivariate normal random variable in $\mathbb{C}^n$ is equivalent to a real multivariate normal in $\mathbb{R}^{2n}$.

You have not specified the joint distributions of 1) the real/imaginary components of each variable, 2) the different variables. If the variables are jointly normal, then for $z,w\in\mathbb{C}^n$, with $z=x+iy$ and $w=u+iv$, we have $[x,y,u,v]\sim\mathrm{N}_{0,C}$ for some covariance matrix $C$.

If we assume that $C=I_{4n\times{4n}}$, then all of the components are independent of each other, and some progress can be made.

For the $n=1$ case, the dot product is $$\bar{z}w=(xu+yv)+i(xv-yu)$$ As the components $x,y,u,v$ are independent and standard normal, this means that the dot-product components $\Re[\bar{z}w]$ and $\Im[\bar{z}w]$ will each have a standard Laplace distribution.

For $n>1$ the dot-product components will each be a sum of $n$ independent Laplace-distributed variables. Eventually this will tend to normal, by the central limit theorem. But for finite $n$ the distribution will have no clean expression (e.g. see here), although computing its characteristic function should be straightforward.

  • $\begingroup$ I think you should have a look at this paper: odonoughue.org/files/2012-TSP.pdf. I guess it`s not so simple like you put it. $\endgroup$ – Felipe Augusto de Figueiredo Jan 5 '17 at 8:36
  • $\begingroup$ I am not sure the paper you cite disagrees with my answer above. 1) The author states they are not satisfied with the "characteristic function" solution, but does not say it is incorrect. 2) The author wishes to derive a PDF parameterized in terms of complex polar coordinates $|z|,\mathrm{arg}[z]$ rather than $\Re[z],\Im[z]$. 3) The paper concerns the case where the variables are not zero-mean. Your question did not specify any of these caveats. If you feel the paper answers what you intended to ask, then I suggest you post an answer of your own and accept it. $\endgroup$ – GeoMatt22 Jan 5 '17 at 11:28
  • $\begingroup$ Could you simulate your answer? If it fits $Z$ then it must be correct, that's is the only way I see if it's correct. $\endgroup$ – Felipe Augusto de Figueiredo Jan 6 '17 at 7:45
  • $\begingroup$ That's what I did before posting originally. I noticed the Laplace pattern that way, then confirmed on Wikipedia that it was expected. You should try it yourself, as I am still not sure we are using the same definitions. $\endgroup$ – GeoMatt22 Jan 6 '17 at 14:24
  • $\begingroup$ Have a look at this answer: mathoverflow.net/questions/258999/… $\endgroup$ – Felipe Augusto de Figueiredo Jan 8 '17 at 20:57

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