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.

  • $\begingroup$ How is this different from circular-symmetric complex normal? $\endgroup$
    – Maxtron
    Nov 14, 2018 at 21:09
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    $\begingroup$ @Maxtron the amplitude here is 1, whereas complex normal has an amplitude that itself is a random variable with a Rayleigh distribution. $\endgroup$
    – Robert L.
    Nov 14, 2018 at 21:10
  • $\begingroup$ @OlliNiemitalo had an answer for something similar as I remember... $\endgroup$
    – Fat32
    Nov 14, 2018 at 21:14
  • $\begingroup$ @Fat32 Yes this rings a bell. The other question is: What is the distribution of it? $\endgroup$ Nov 16, 2018 at 9:22
  • $\begingroup$ @OlliNiemitalo Yes that was the question! $\endgroup$
    – Fat32
    Nov 16, 2018 at 10:02

3 Answers 3


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.

  • 1
    $\begingroup$ is there a typo in $r = 1- i^2$ ; possibly you mean $r^2=1-i^2$ ? Also can we show that the joint pdf integrates to 1 in x-y (or eqv r-i) plane. Again also the joint pdf $f_{ri}(r,i)$ is not circularly symmetric? (the denominator?). Cannot we simply use the logic that given $z = e^{j \theta} = \cos(\theta) + j \sin(\theta) = x + jy$ with $\theta$ uniform in [$0,2\pi$] and the constraint $x^2+y^2=1$ simply offers a circularly symmetric joint pdf of the type $f_{xy}(x,y) = K \delta_2(x^2+y^2-1)$ or $f_{R,\theta} = K \delta_2(R-1)$ ; a ring impulse on x-y plane? where K is possibly $1/2\pi$ ? $\endgroup$
    – Fat32
    Nov 14, 2018 at 22:57
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    $\begingroup$ I corrected the typos you referred to, thanks. $\endgroup$
    – Jason R
    Nov 14, 2018 at 23:08
  • $\begingroup$ @Fat32: I made some fixes that led me to what sounds like what you were getting at in your comment above. AlexTP's answer still is probably more intuitively pleasing though. $\endgroup$
    – Jason R
    Nov 15, 2018 at 1:50
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    $\begingroup$ There is a typo in copying $f_{r|i}(r|i)$ to $f_{r,i}(r,i)$. And I observe that $f_{r,i}(0,1) = \infty$ is different from $f_{r,i}(1,0) = \frac{1}{2\pi}$. I think they should have the same value, shouldn't they? $\endgroup$
    – AlexTP
    Nov 16, 2018 at 0:14
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    $\begingroup$ everyone, maybe a quick simulation would be useful showing your points :) $\endgroup$
    – AlexTP
    Nov 17, 2018 at 11:07

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 $r dr d\theta$), $$f_{R,\Theta}(r,\theta)=\frac{1}{2\pi} \delta(r - \rho)$$

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    $\begingroup$ i think this is correct. $\endgroup$ Nov 15, 2018 at 0:02
  • $\begingroup$ The intuition seems to be so simple behind the proof. All you need to do is pick points from the surface of a unit sphere :) $\endgroup$
    – Maxtron
    Nov 15, 2018 at 0:14
  • $\begingroup$ Can't we just remove the $||z||$ in the denominator? Otherwise, how would you evaluate the expression for $||z||\rightarrow 0$? Of course, we would like $f_Z(0)=0$ to hold. $\endgroup$
    – Matt L.
    Nov 15, 2018 at 9:57
  • $\begingroup$ @MattL. you are right, it is wrong having $\lVert z \rVert$ in the denominator, not due to the evaluation at limits but because it is simply wrong. I have fixed the final result. $\endgroup$
    – AlexTP
    Nov 15, 2018 at 10:29
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    $\begingroup$ @AlexTP: I still think that there shouldn't be any variable in the denominator. Have a look at my answer and please tell me if/where I go wrong. $\endgroup$
    – Matt L.
    Nov 15, 2018 at 16:31

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


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


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:


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


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




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


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 $\theta$.

  • $\begingroup$ I think the determinant of Jacobian transform should be $r$ then $$\int_{\theta=0}^{2\pi} \int_{r=0}^{+\infty} f_Z(r) r dr d\theta \neq 1$$ $\endgroup$
    – AlexTP
    Nov 15, 2018 at 16:45
  • $\begingroup$ @AlexTP: Take as an extreme example $\rho=0$, which makes the RV $Z$ deterministic, it's always zero. It's PDF should then be a Dirac impulse at $r=z=y=0$, which is what I get with the formula I suggested. With the formula in your answer we get an undefined term in that case. $\endgroup$
    – Matt L.
    Nov 15, 2018 at 16:58
  • $\begingroup$ it's true, but I think the value of PDF can be infinity and only the CDF should be bounded. In other words, the Dirac impulse matters only inside integrals. $\endgroup$
    – AlexTP
    Nov 15, 2018 at 17:43
  • $\begingroup$ Yes, but if you integrate your pdf (for $\rho=0$), what do you get (and how would you do it)? $\endgroup$
    – Matt L.
    Nov 15, 2018 at 17:48
  • $\begingroup$ I would do like this $$\lim_{\rho \to 0^+}\int_\theta \int_{r=0}^{+\infty}\frac{1}{2\pi\rho}\delta(r-\rho) r dr d\theta = \lim_{\rho \to 0^+} \int_\theta \frac{1}{2\pi\rho} \rho d\theta = \int_\theta \frac{1}{2\pi} d\theta = 1$$ The integral is finite while the PDF is not thanks to the Jacobian determinant. My point is that you need that Jacobian determinant $r$ whenever you integrate the PDF in polar coordinate. Tell me if you find it unconvinced. $\endgroup$
    – AlexTP
    Nov 15, 2018 at 20:50

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