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 Lebesgue-measure 0 and hence the induced probability measure cannot be absolutely continuous to it.)
Thanks for clearing my lack of clarity.
Best Answer
I try to reformulate the commentaries of fgp as an answer.
First consider a probability space $(\Omega, \mathcal{A}, \mathbb{P}')$ and a random variable uniformly distributed on $[0,2\pi]$:
$$X: (\Omega, \mathcal{A}, \mathbb{P}') \rightarrow ([0,2\pi],\mathcal{B}([0,2\pi])).$$
Furthermore consider the parametrization of the unit circle
$$p: [0,2\pi] \rightarrow S^1; \quad x \mapsto e^{i\cdot x}$$
which is continuous.
Now consider the space $(S^1,\mathcal{B}(S^1))$ where $\mathcal{B}(S^1)$ is the $\sigma$-algebra generated by the open sets of $S^1$.
We define $\mathbb{P}$ to be the probability measure induced by the map
$$p\circ X: (\Omega, \mathcal{A}, \mathbb{P}') \rightarrow (S^1,\mathcal{B}(S^1)); \quad \omega \mapsto e^{i X(\omega)}.$$
Then we have
\begin{equation} P(Z\in A) = \frac{1}{2\pi i}\int_{A}{\frac{1}{z}}dz \qquad \forall A \in \mathcal{B}(S^1) \end{equation}
or more generally
$$E[f(Z)]=\frac{1}{2\pi i}\int_{S^1}{\frac{f(z)}{z}}dz$$
for any measurable, bounded function $f: S^1 \rightarrow \mathbb{R}$
So we must be careful what the measurable space really is. In the case of the "density" in the question we are considering the probability space $(S^1,\mathcal{B}(S^1), \mathbb{P})$ and on this we get can give the value of $\mathbb{P}$ via the above formula.
Hence the above is not a density with respect to the Lebesgue-measure on $\mathbb{C}$, but a way of formulating the value with the help of the parametrization of (or - to be more precise - a contour integral along) the unit circle $S^1$.