Suppose $\mathbf{u}\in\mathbb{C}^n$ is a complex random vector with circular symmetry, uniformly distributed on the unit complex $n$-sphere, so we have $\|\mathbf{u}\|=1$. In other words, $\mathbf{u}$ is a uniform random unit vector in $\mathbb{C}^n$. Can we say $\mathbf{x}={\left[\begin{matrix}\operatorname{Re}\{\mathbf{u}\}\\\operatorname{Im}\{\mathbf{u}\}\end{matrix}\right]}\in\mathbb{R}^{2n}$ is uniformly distributed on the real $2n$-sphere?

Clearly, $\|\mathbf{x}\|=1$, but is this enough?

Thanks for any help or hint.

  • $\begingroup$ Some details are not quite right. What exactly do you mean by a "complex $n$-sphere", given that you are working inside an $n$-dimensional Complex space to begin with? Is "$H$" the conjugate or the Hermitian conjugate? Why don't you write each coordinate of $u$ in terms of its real and imaginary parts and show us specifically what "$u^Hu$" is in terms of them. I believe that in so doing you will discover what your question should be asking as well as the answer to it. $\endgroup$ – whuber Jan 25 '16 at 22:31
  • $\begingroup$ I made some changes. I hope its clear now. $\endgroup$ – user298935 Jan 26 '16 at 16:00
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    $\begingroup$ It's still not clear, because it's mathematically impossible for an element of $\mathbb{C}^n$ to be part of a "complex $n$-sphere." It sounds like you're talking about a vector that has a uniform distribution on the $2n-1$-sphere $S^{2n-1}\subset \mathbb{R}^{2n}\approx \mathbb{C}^n$, in which case there's nothing to be shown: you're just stating the same thing in two clearly equivalent ways. $\endgroup$ – whuber Jan 26 '16 at 16:05
  • $\begingroup$ Sorry for the informal presentation. I study engineering. $\endgroup$ – user298935 Jan 26 '16 at 23:25

I came up with an answer!

We know that $\mathbf{u}=\frac{\mathbf{n}}{\|\mathbf{n}\|}$ is uniformly distributed on the complex $n$-sphere, where $\mathbf{n}$ is a white complex Gaussian vector. Since we can write $\mathbf{x}={{\left[\begin{matrix}\operatorname{Re}\{\mathbf{n}\}\\\operatorname{Im}\{\mathbf{n}\}\end{matrix}\right]}}/{\|\mathbf{n}\|}$, it readily follows that $\mathbf{x}$ is uniformly distributed on the real $2n$-sphere.

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    $\begingroup$ granted I had to google 'complex normal,' but this looks a lot like begging the question $\endgroup$ – Taylor Jan 25 '16 at 22:50

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