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.

  • $\begingroup$ 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... $\endgroup$
    – fgp
    Commented Oct 12, 2012 at 18:42
  • $\begingroup$ 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? $\endgroup$
    – AndreasS
    Commented Oct 12, 2012 at 19:01
  • $\begingroup$ But wait this "density" would only make sense for connected arcs, wouldn't it? $\endgroup$
    – AndreasS
    Commented Oct 12, 2012 at 19:04
  • $\begingroup$ 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... $\endgroup$
    – fgp
    Commented Oct 12, 2012 at 19:19
  • $\begingroup$ 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 $e^{i\theta}$? $\endgroup$
    – AndreasS
    Commented Oct 12, 2012 at 20:37

1 Answer 1


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 $S^1$.

  • $\begingroup$ Another interesting thing of the equation $$E[f(Z)]=\frac{1}{2\pi i}\int_{S^1}{\frac{f(z)}{z}}dz$$ is, that it is quite near the Cauchy's integral formula, which says that for a function $f$ holomorphic in a neighborhood of $S^1$ $$f(0)=\frac{1}{2\pi i}\int_{S^1}{\frac{f(z)}{z}}dz$$ However, this might be just because both formulas rely on the parametrization of the unit circle... $\endgroup$
    – AndreasS
    Commented Oct 17, 2015 at 12:56

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