Let $Y(t) = Ae^{j\phi} + W(t)$, where $\phi \sim unif[-\pi,\pi]$ and $W \sim \mathcal{N}(0,\sigma^2)$. What is the probability distribution of $|Y(t)|$ ?

If $\phi$ was deterministic, i.e. a constant value, the distribution of $|Y(t)|$ would be $|Y(t)| \sim Rice(A,\sigma^2) $. However, in my case $\phi$ is not a deterministic value, so the PDF might change.

I have simulated in MATLAB this, comparing the case where $\phi$ is a uniformly distributed random variable and three more cases where $\phi = \{ \phi_1, \phi_2, \phi_3 \}$ are constant values.

Matlab code:

phi = 0:pi*1e-3:1e3*pi; % phase vector
phi1 = phi(randi(length(phi)));
phi2 = phi(randi(length(phi)));
phi3 = phi(randi(length(phi)));

A = 2.8e-6; % signal amplitude
x = A*exp(1i*phi); % signal vector
x1 = A*exp(1i*phi1); % signal vector 1
x2 = A*exp(1i*phi2); % signal vector 2
x3 = A*exp(1i*phi3); % signal vector 3

sigma = 2.5e-5; % noise variance
w = sigma/sqrt(2)*(randn(1,length(phi)) + 1i*randn(1,length(phi))); % noise signal

y_th = x+w; % received signal
y_th1 = x1+w; % received signal 1
y_th2 = x2+w; % received signal 2
y_th3 = x3+w; % received signal 3

subplot(2,2,1); hist(abs(y_th),200); title('\phi uniformly distirbuted');
subplot(2,2,2); hist(abs(y_th1),200); title('\phi = \phi_1');
subplot(2,2,3); hist(abs(y_th2),200); title('\phi = \phi_2');
subplot(2,2,4); hist(abs(y_th3),200); title('\phi = \phi_3');

Matlab results:

enter image description here

As you can see in the final results the PDFs of the signal with $\phi$ as a random variable and of the signals with $\phi$ as deterministic values are really close to. Are they equivalent? Is it fine to assume that?

Really appreciate your help :)

  • $\begingroup$ Assume that the distribution of |Y| for each phi has PDF f_phi then the distribution of |Y| for some random phi has PDF f = E(f_phi). In your case f_phi does not depend on phi hence indeed, for every distribution of phi the distribution of |Y| is the same. $\endgroup$ – Did May 6 '15 at 16:58
  • $\begingroup$ So: $\phi = \{\phi_1, \phi_2, ...\}$, then $f_{|Y|}(|Y|) = \mathbb{E}_\phi\{f_{|Y|}(|Y|,\phi)\} = f_{|Y|}(|Y|,\phi) $, right? Thanks again @Did $\endgroup$ – lucasrodesg May 6 '15 at 17:27

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