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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.

Robert L.
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3 Answers3

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

AlexTP
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    i think this is correct. – robert bristow-johnson Nov 15 '18 at 00:02
  • 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 :) – Maxtron Nov 15 '18 at 00:14
  • 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. – Matt L. Nov 15 '18 at 09:57
  • @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. – AlexTP Nov 15 '18 at 10:29
  • I cannot see how you approached into your last formula, form the definition of $V$? And what is it to do with standard normal RVs $X$, $Y$? isn't it enough to say that circularly symmetric uniform distribution along a circular curve in x-y plane indicates a circular impulse on x-y plane. Hence your last formula. – Fat32 Nov 15 '18 at 11:03
  • @Fat32 $V$ and $Z$ have the same distribution ($V \stackrel{d}{=} Z$), and $Z_\rho = \rho Z \stackrel{d}{=} \rho V$. Yes, you can use that "circulary symmetric" argument and it is enough. Normal (not necessarily standard Normal) is to be circularly symmetric and the denominator is to make sure the norm equals constant. This proof is the long version of your argument. However, some people prefer talking long ;) – AlexTP Nov 15 '18 at 12:16
  • hmm ok then ;-) – Fat32 Nov 15 '18 at 14:13
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    @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. – Matt L. Nov 15 '18 at 16:31
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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.

Jason R
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    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 $1/2\pi$ ? – Fat32 Nov 14 '18 at 22:57
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    I corrected the typos you referred to, thanks. – Jason R Nov 14 '18 at 23:08
  • @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. – Jason R Nov 15 '18 at 01:50
  • My experience is that you have to be careful with scale factors on Dirac deltas when you have a function in the argument-- this answer seems to use the least amount of "hand waving" to get there. (I don't know for sure, but I'll go ahead and mark this one as "answered"). All the answers are well thought-out though. – Robert L. Nov 15 '18 at 18:06
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    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? – AlexTP Nov 16 '18 at 00:14
  • @AlexTP I think the singularity is actually what the behavior should be, no? – Robert L. Nov 16 '18 at 16:39
  • @AlexTP: I think that $f_{r,i}(1, 0) = \infty$. Note that for these values of $r$ and $i$, the argument to the first delta function in the numerator is zero, so the result is infinite. The idea is that at any point in the $ri$ plane, the pdf is either $0$ or $\infty$. – Jason R Nov 16 '18 at 20:55
  • @JasonR sorry for bogging down, but the weight of the ring-impulse should be same at any point of the circle on the im-re plane for the pdf to be symmetric. Being infinite does not suffice, as we are dealing with impulses, whose weights are the actual concern. So in your derivation the denominator $\sqrt{ 1- i^2}$ creates an i dependent weighting on the impulse which is changing for every different i on the circle, and this is not symmetric? Am I wrong on this, please correct me? i=0, gives a weight of 1, i=0.5 gives a weight of 1.1547, (apart from $2\pi$) etc... – Fat32 Nov 16 '18 at 22:44
  • @Fat32: I see what you're saying, but I think the condition is that the pdf, integrated over any differential patch $dr\ di$ that touches the unit circle, should have the same value. This is very hand-wavy and mathematically imprecise, but it seems the weight is higher for areas on the unit circle where $\frac{dr}{di}$ is greater; that is, for a particular value of $i$, the circle is "curving away" from the corresponding value of $r$ faster. As if, the "more circle" that is in the neighborhood of that value of $i$, the less of a weight you need to make it integrate to the same value. – Jason R Nov 17 '18 at 00:04
  • @Fat32: I'm not saying I'm confident that this analysis is bulletproof, but I'm not sure where any error is. – Jason R Nov 17 '18 at 00:04
  • I have only one argument to object: the weight of the impulse should be independent of individual im or re, but now it depends... But I cannot also find the flaw in your derivation. Probably it's right... Derivation seems right but the result not... confusing... probably I cannot see it but it seems MattL found yours as correct too. Anyway. – Fat32 Nov 17 '18 at 00:24
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    everyone, maybe a quick simulation would be useful showing your points :) – AlexTP Nov 17 '18 at 11:07
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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 $Y$.

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 $\theta$.

Matt L.
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  • I think the determinant of Jacobian transform should be $r$ then $$\int_{\theta=0}^{2\pi} \int_{r=0}^{+\infty} f_Z(r) r dr d\theta \neq 1$$ – AlexTP Nov 15 '18 at 16:45
  • @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. – Matt L. Nov 15 '18 at 16:58
  • 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. – AlexTP Nov 15 '18 at 17:43
  • Yes, but if you integrate your pdf (for $\rho=0$), what do you get (and how would you do it)? – Matt L. Nov 15 '18 at 17:48
  • 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. – AlexTP Nov 15 '18 at 20:50
  • @AlexTP: Would you agree that the marginal density $f_R(r)$ is obtained from $f_{R,\Theta}(r,\theta)$ via integrating $\int_{0}^{2\pi}f_{R,\Theta}(r,\theta)d\theta$? For your formula, the result would be $f_R(r)=\delta(r-\rho)/\rho$, which integrates to $1/\rho$, but $f_R(r)$ should integrate to $1$, without multiplication of $r$ inside the integral. – Matt L. Nov 16 '18 at 07:46
  • Got it. My formula is $f_{X,Y}(\rho\cos(\theta),\rho\sin(\theta))$ which is not PDF in polar coordinate. The PDF in polar coordinate should be $f_{R,\Theta}(r, \theta) = f_{X,Y}(\rho\cos(\theta),\rho\sin(\theta)) \times \det(\textrm{Jacobian})$ which is yours. What do you think? – AlexTP Nov 16 '18 at 08:34
  • @AlexTP: I think that's it, we finally got 3 equivalent formulas! I will edit my answer to show that Jason's answer is identical with mine, just looking a bit more complicated. – Matt L. Nov 16 '18 at 09:08
  • cool. Could you please also explain my doubt in the last comment of Jason's answer in your edit? Its pleasant to discuss with you people. – AlexTP Nov 16 '18 at 09:11
  • As in my first comment to JasonR's answer, the associated pdf should be circulary symmetric and of impulse type which is the one in your answer. (and Alex's as well). Now more interesting thing is why Jason's method $f_{ri}(r,i) = f_{r|i} f_i $ does not provide the correct (circularly symmetric) answer ? The denominator of $\sqrt{1-i^2}$ makes it a-symmteric... – Fat32 Nov 16 '18 at 10:09
  • @Fat32: Jason's answer is correct, it is circularly symmetrical, he just hid it very well :) I'll edit my answer to show that. – Matt L. Nov 16 '18 at 10:18
  • Ok. I'll wait for that but I really wonder how $f_{ri}(r,i) = \frac{ \delta(r^2+i^2-1) }{\sqrt{ 1-i^2 }}$ will turn about to be symmetric on the r-i plane? (r is real, i imaginary parts of z) – Fat32 Nov 16 '18 at 10:29