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I was trying to prove that for a standard complex Gaussian variable $Z$ it holds that $|Z|^2$ is exponentially distributed with parameter 1, $\frac{Z}{|Z|}$ is uniformly distributed on the unit circle $S^1:=\{z\in\mathbb{C} | |z|=1\}$ and that the two are independent.

At some point I began asking myself:

How does one describe the uniform distribution on the unit circle $S^1$?

I resolved to say that it is the complex r.v. $e^{i\theta}$ where $\theta$ is uniformly distributed on $[0,2\pi]$. This seemed to work out fine (c.f. Byron's answer to this question).

However, if this is correct then this small argument will go through:

Let $f:S^1 \rightarrow \mathbb{R}$ be bounded. Then $$E[f(Z)]=\int_{0}^{2\pi}{f(e^{i\theta})\frac{1}{2\pi}}d\theta=\frac{1}{2\pi i}\int_{S^1}{\frac{f(z)}{z}}dz,$$

where for the last equation $z=e^{i\theta}$ and thus $\frac{dz}{d\theta}=ie^{i\theta}$ i.e. $\frac{dz}{iz}=\frac{dz}{ie^{i\theta}}={d\theta}$. So:

Is $\frac{1}{2\pi i z}$ some kind of density for a uniformly distributed random variable on $S^1$?

(I write "some kind" as it cannot be one because the unit circle has Lebesque-measure 0 and hence the induced probability measure cannot be absolutely continuous to it.)

Thanks for clearing my lack of clarity.

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The unit circle only has measure $0$ as a subset of $\mathbb{C}$. But you're looking only at functions defined on the unit circle, so it becomes the base set of your measure space, and as such can have measure $\geq 0$. Regarding $\frac{1}{2\pi iz}$ - how can the density of a probability distribution be complex? You'd have to define what that means first... –  fgp Oct 12 '12 at 18:42
    
By "density" I mean that $\mathbb{P}(Z \in B)=\int_{B}{\frac{1}{2\pi i z}}$ for any arc $B$ on the unit circle. I believe (but am not sure) that this always gives a real number. The unit circle has measure zero so: $\mathbb{P}(Z\in S^1)=1$ but $\lambda_{\mathbb{C}}(S^1)=0$. So "$\mathbb{P}(Z\in \bullet) << \lambda_{\mathbb{C}}$" doesn't hold, does it? –  AndreasS Oct 12 '12 at 19:01
    
But wait this "density" would only make sense for connected arcs, wouldn't it? –  AndreasS Oct 12 '12 at 19:04
    
It always gives a real number because you shows that it's actually just a funny way to write an integral over the unit circle for a function with domain $\mathbb{R}$. I still don't understand what the lesbegue measure on $\mathbb{C}$ has to do with it - you're only looking at the unit circle, and your "density" is defined only on the unit circle... –  fgp Oct 12 '12 at 19:19
    
Ok, I realize now that the "$dz$" indicates that the integral is something different to the usual Lebesque-measure idea I had in mind. It is a line integral. That it was a fancy way of writing the integral came also to my mind but I wondered if there is something more in this presentation... However, can I now go on and say that - looking only on the unit circle - this gives me some sort of "density"? Or do I just mix up the relatively simple idea that a r.v. uniformly distributed on $S^1$ is just of the form $e^{i\theta}$? –  AndreasS Oct 12 '12 at 20:37

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