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 Oct 12 '12 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 Oct 12 '12 at 19:01
  • $\begingroup$ But wait this "density" would only make sense for connected arcs, wouldn't it? $\endgroup$ – AndreasS Oct 12 '12 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 Oct 12 '12 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 Oct 12 '12 at 20:37

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 Oct 17 '15 at 12:56

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.