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Assume that $z$ is a random vector. Also, assume that $$z = A(\|y\|)$$ where A is whitening function and $\|\cdot\|$ is complex modulus (norm) such that if

$$y_j = a_j + ib_j,$$ then $$\|y_j\|=\sqrt{a_j^2+b_j^2}$$ and $y$ is a complex vector which is derived from a real vector $x$ using some invertible projection matrix $W$ $$x = Wy$$ Considering $$z\sim\mathcal{N}(0, I),$$ what can we say about distribution of $x|z$?

My problem is with inverting the modulus operator. If that was not present, I could have said

$$x|y\sim\mathcal{N}(Wy, \sigma^2I)$$ Any suggestions would be appreciated.

Blade
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    Could you help me with the modulus of a complex vector giving a vector. Is that taking the modulus component by component? – Kitter Catter Jun 17 '19 at 02:00
  • What is a "modulus operator"? I observe that are trying to define $y$ in terms of $x$ but your equation $x=Wy$ defines $x$ in terms of $y$, which leaves $y$ ambiguous if $W$ has a nontrivial null space. – Michael Jun 17 '19 at 03:36
  • I also observe that you use norm notation to define $z$ (you write $z=||y||$). Now one thing about norms is that they are nonnegative, and hence it is impossible for $z$ to be Gaussian as claimed. [You are not calling $||y||$ a "norm" you are calling it a "modulus," you don't bother to define that, but I suspect that your "modulus" is something that is also nonnegative, perhaps it is $\sqrt{\sum_{i=1}^n |y_i|^2}$.] – Michael Jun 17 '19 at 03:41
  • @KitterCatter I revised the question – Blade Jun 17 '19 at 15:45
  • @Michael I revised the question after your comment. I clarified that the matrix projection is invertible, what modulus operator is, and that I'm whitening out the output and that is why it has standard normal distribution. – Blade Jun 17 '19 at 15:51

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