Probability density of a ring delta plus Gaussian

Question

I'm trying to find the pdf for a ring delta plus a complex Gaussian rv. I'm really not sure of the best way to approach this. The density of the ring delta is $$ \frac{1}{2\pi r_0} \delta(r - r_0) $$ in polar coordinates. The density of the Gaussian random variable is $$ f_{X,Y}(x,y) = \frac{1}{\pi\sigma^2} e^{-(x^2 + y^2)/\sigma^2} $$

My initial thought was to convert from polar to cartesian coordinates for the delta, and then use the sifting property. However, I am having trouble knowing how to properly do that conversion. Maybe that's not a good way to approach it.

Any help would be appreciated.

Answer 1

While looking through Papoulis' Systems and Transforms with Applications in Optics, I came across the equation: $$ f(r)\ast \ast \delta(r-r_0) = r_o \int_{0}^{2\pi} f\left ( \sqrt{r^2 + r_0^2 - 2rr_0\cos\theta}\right ) d\theta $$ where $\ast\ast$ is used to denote the double convolution. The pdf of the Gaussian random variable can be rewritten as \begin{align} f_{X,Y} (x,y) &= \frac{1}{\pi \sigma^2} e^{-(x^2 + y^2)/\sigma^2} \\ &= \frac{1}{\pi \sigma^2} e^{-r^2/\sigma^2} \\ &= f_{X,Y}(r) \end{align} using $r = \sqrt{x^2 + y^2}$. One important note here is that this is not the density of the amplitude of the random variable (which would follow a Rayleigh distribution). So, putting these together:

\begin{align} f_{X,Y}(r)\ast \ast \frac{1}{2\pi r_0}\delta(r-r_0) &= r_o \int_{0}^{2\pi} \frac{1}{2\pi r_0}f\left ( \sqrt{r^2 + r_0^2 - 2rr_0\cos\theta}\right ) d\theta \\ &= r_0 \frac{1}{2\pi r_0}\int_{0}^{2\pi}\frac{1}{\pi \sigma^2} e^{-\frac{1}{\sigma^2}(r^2 + r_0^2 - 2rr_0\cos\theta)} d\theta \\ &= \frac{1}{2\pi^2\sigma^2}e^{-\frac{1}{\sigma^2}(r^2 + r_0^2)}\int_{0}^{2\pi}e^{\frac{2rr_0}{\sigma^2}\cos\theta} d\theta \\ &= \frac{1}{2\pi^2\sigma^2}e^{-\frac{1}{\sigma^2}(r^2 + r_0^2)} 2\pi \text{I}_0\left (\frac{2rr_0}{\sigma^2} \right ) \\ & = \frac{1}{\pi\sigma^2}e^{-\frac{1}{\sigma^2}(r^2 + r_0^2)} \text{I}_0\left (\frac{2rr_0}{\sigma^2} \right ) \end{align}

Where $\text{I}_0(z)$ is the zeroth order, modified Bessel function of the first kind. The integral form that I used is found here.

It can be rewritten in terms of $x$ and $y$ as $$ g_{X,Y}(x,y) = \frac{1}{\pi\sigma^2}e^{-\frac{1}{\sigma^2}(x^2 + y^2 + r_0^2)} \text{I}_0\left (\frac{2\sqrt{x^2 + y^2}r_0}{\sigma^2} \right ) $$

I tested this out for $\sigma^2=1$ and $r_0=2$. Here's the code to generate the samples:

r0 = 2;
x=-5:0.01:5;
[xx,yy] = ndgrid(x);
numSamples = 1e6;
varN = 1;
% Make ring and add a unit variance complex normal
ringSamps = r0*exp(1j*2*pi*rand(1,numSamples));
awgnSamps = sqrt(varN/2)*complex(randn(1,numSamples), randn(1,numSamples));
ringPlusAWGN = ringSamps + awgnSamps;

Here's an image of the samples

And here's an emperical pdf using matlab's histogram2 function, along with the analytic pdf.

The code to generate the above figures is

% Use a histogram to estimate the pdf
figure
histogram2(real(ringPlusAWGN), imag(ringPlusAWGN), ...
    'Normalization','pdf', 'DisplayStyle','tile')
colorbar
title('Empirical PDF','FontSize',14)
axis equal

% Plot the analytic pdf
figure
xi = -5:0.1:5;
[xxi, yyi] = ndgrid(xi);

g_xy = 1/(pi*varN)*besseli(0,2*sqrt(xxi.^2 + yyi.^2)*r0/varN).*...
    exp(-1/varN*(xxi.^2 + yyi.^2 + r0^2));
imagesc(xi,xi, g_xy)
title('Analytic PDF','FontSize',14)
axis equal
colorbar