Distribution of $e^{j\theta}$

Question

Is there a canonical or analytic expression for the probability distribution for the circularly-symmetric complex random variable $Z$: $$ Z = e^{j\theta}, $$ where $\theta \sim \mathcal U(0, 2\pi)$?

Side notes:

It is known that the real and imaginary parts, i.e.: $$ \Re(Z) = \cos \theta \\ \Im(Z) = \sin \theta $$ have marginal densities given by: $$ f_{\Re(Z)}(z) = f_{\Im(Z)}(z) = \frac{1}{\pi\sqrt{1-z^2}}, \quad -1 < z < 1, $$ but because they are not independent, computing their joint PDF is nontrivial.

EDIT: $Z$ is different from a complex normal in that here, the amplitude $|Z|$ is deterministic and identically 1, whereas if $Z$ were complex normal, $|Z|$ would be Rayleigh distributed.

Answer 1

Since the real and imaginary parts are very much dependent on one another (if you have the value of one, you know the value of the other exactly), it seems like you could apply the marginal pdf of the real part $r$, given a value of the imaginary part $i$:

$$ f_{ri}(r, i) = f_{r | i}(r\ |\ i) f_i(i) $$

You noted the pdf of the real and imaginary parts individually:

$$ f_r(z) = f_i(z) = \frac{1}{\pi\sqrt{1-z^2}} $$

That leaves the marginal pdf $f_{r | i}(r\ |\ i)$. Remember that for a given realization of the random variable $Z$, the two components are deterministically related:

$$ r^2 + i^2 = \cos^2(\theta) + \sin^2(\theta) = 1 $$

Given this relationship, we can solve for $r$ in terms of $i$:

$$ r^2 = 1 - i^2 $$ $$ r = \pm \sqrt{1-i^2} $$

Therefore, the marginal pdf of $r$ given a value of $i$ is a pair of impulses:

$$ f_{r | i}(r\ |\ i) = \frac{1}{2}\delta\left(r - \sqrt{1 - i^2}\right) + \frac{1}{2}\delta\left(r + \sqrt{1 - i^2}\right) $$

Putting these together would yield:

$$ f_{ri}(r, i) = \frac{\delta\left(r - \sqrt{1 - i^2}\right) + \delta\left(r + \sqrt{1 - i^2}\right)}{2\pi\sqrt{1-i^2}} $$

Thinking about this geometrically, for every horizontal line $i = i_0$ (for $i_0 \in [-1, 1]$) in the $ri$ plane, there are only two points $r_0 = \pm \sqrt{1 - i_0^2}$ that are nonzero, and the pdf has has infinite height at those points. As we might expect, those points of intersection (i.e. points where the pdf is nonzero) are where the horizontal line intersects with the unit circle!

This means that the joint pdf is zero-valued, except along the unit circle, where it takes on infinite height. That aligns with intuition, as the definition of the random variable $Z$ ensures that can only take on values that are on the unit circle.

There's nothing special about the specific way I laid this out; you could also transpose the problem and look at vertical lines in the $ri$ plane of the form $r = r_0$ and you would find the same relationship due to the close coupling of the two random variables.

I believe this formulation is equivalent to that in AlexTP's answer, but his derivation is probably more intuitive.

Answer 2

Avoid complicated calculations, let $X$ and $Y$ be i.i.d. standard normal random variables, your random variable $Z$ has the same distribution of $V$ $$V \triangleq \left(\frac{X}{\sqrt{X^2+Y^2}},\frac{Y}{\sqrt{X^2+Y^2}} \right)$$ (easy to see $\lVert V\rVert=1$ and the angle of $V$ is equivalent to the angle of a circularly symmetric Normal hence uniform).

This kind of $V$ is one of the constructions of a point uniformly distributed on circle (which can be generalized to $(n-1)$-sphere, see Sphere Point Picking and for example this answer).

Thus the PDF of $Z$ is simply the reciprocal of the unit circle's circumference. For $Z_\rho = \rho e^{j\Theta}$ with fixed $\rho$ and uniform $\Theta$,

in polar coordinates (where infinitesimal area is $r dr d\theta$), $$f_{R,\Theta}(r,\theta)=\frac{1}{2\pi} \delta(r - \rho)$$

Answer 3

Based on the existing answers, which opened my eyes for what's going on here, I would like to present yet another very simple expression for the solution, which is just slightly different from the one in AlexTP's answer (and which turned out to be equivalent to the one given in Jason R's answer, as shown below in the EDIT-part).

[EDIT: now that AlexTP has edited his answer, our expressions for the PDF are identical; so all three answers finally agree with each other].

Let the complex random variable $Z=X+jY$ be defined as

$$Z=\rho e^{j\theta}\tag{1}$$

where the radius $\rho$ is deterministic and given, whereas the angle $\theta$ is random and uniformly distributed on $[0,2\pi)$. I state without further proof that $Z$ is circularly symmetrical, from which it follows that its probability density function (PDF) must satisfy

$$f_Z(z)=f_Z(x+jy)=f_Z(r),\qquad\textrm{with}\quad r=\sqrt{x^2+y^2}\tag{2}$$

i.e., it can be written as a function of the radius (magnitude) $r$.

Since the PDF must be zero everywhere except for $r=\rho$, and since it must integrate to unity (when integrated over the 2-dimensional plane), the only possible PDF is

$$f_Z(r)=\frac{1}{2\pi}\delta(r-\rho)\tag{3}$$

It can be shown that $(3)$ leads to the correct marginal densities for the random variables $X$ and $Y$.

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EDIT:

After some very useful discussion in the comments it appears that we've managed to agree on one solution to the problem. I will show in the following that the unassuming formula $(3)$ is actually equivalent to the more involved looking formula in Jason R's answer. Note that I use $r$ for the magnitude (radius) of the complex RV $Z$, whereas in Jason's answer $r$ denotes the real part of $Z$. I will use $x$ and $y$ for the real and imaginary parts, respectively. Here we go:

$$f_Z(r)=\frac{1}{2\pi}\delta(r-\rho)=\frac{1}{2\pi}\delta\left(\sqrt{x^2+y^2}-\rho\right)\tag{4}$$

We know that $\delta\big(g(x)\big)$ is given by

$$\delta\big(g(x)\big)=\sum_i\frac{\delta(x-x_i)}{|g'(x_i)|}\tag{5}$$

where $x_i$ are the (simple) roots of $g(x)$. We have

$$g(x)=\sqrt{x^2+y^2}-\rho\quad\textrm{and}\quad g'(x)=\frac{x}{\sqrt{x^2+y^2}}=\frac{x}{r}\tag{6}$$

The two roots $x_i$ are

$$x_{1,2}=\pm\sqrt{\rho^2-y^2}\tag{7}$$

Consequently,

$$|g'(x_1)|=|g'(x_2)|=\frac{\sqrt{\rho^2-y^2}}{\rho}=\sqrt{1-\left(\frac{y}{\rho}\right)^2}\tag{8}$$

With $(5)$-$(8)$, Eq. $(4)$ can be written as

$$f_{X,Y}(x,y)=\frac{1}{2\pi\sqrt{1-\left(\frac{y}{\rho}\right)^2}}\left[\delta\left(x-\sqrt{\rho^2-y^2}\right)+\delta\left(x+\sqrt{\rho^2-y^2}\right)\right]\tag{9}$$

For $\rho=1$, Eq. $(9)$ is identical to the expression given in Jason R's answer.

I think we can now agree that Eq. $(3)$ is a correct (and very simple) expression for the PDF of the complex RV $Z=\rho e^{j\theta}$ with deterministic $\rho$ and uniformly distributed $\theta$.