$Z(t) = A\cos(\omega t+\theta)$ where $A$~$\mathcal N(0,\sigma ^2) $ and $\theta $~$\mathcal U(0,2\pi)$ are independent.

I'm trying to figure out if $Z(t)$ is a Gaussian random process and whether it is strict sense stationary. It is easy to see that it is WSS, but I can't figure out about the SSS and whether it is a Gaussian process. It does look like to me that for every chosen time $t$ I get some kind of a normal random variable multiplied by a constant per $\theta$ and therefore it is a Gaussian process, but I can't seem to prove nor disprove it. Can someone clarify this for me? Thanks.


1 Answer 1


I'm pretty sure that even algebraically, $Z$ will not be Gaussian. You're multiplying a Gaussian by a random variable that has a distribution that is effectively the histogram of a (co)sine function.

enter image description here

Let's do it empirically, though, following the example here.

Looking first at the densities, you can see that the density estimate for $Z$ (red) is much more central than that of $A$ (green).

enter image description here

Plotting the qqnorm and qqline plots for both, are below. As you can see, the one for $Z$ does not coincide much with the red line.

First, for $A$: Plot for $A$

Then, for $Z$: Plot for $Z$

Regarding stationarity: the signal $Z$ is not stationary at all because it is cyclostationary.

R Code Below

# 28338
T <- 10000
sigma  <- 1

A <- rnorm(T,0,sigma)
theta <- runif(1,0, 2 * pi)

fs <- 100
t <- (1:T)/fs
omega <- 2* pi * 0.0892987
Z <- A * cos(omega*t + theta)

# First diagram

# Second diagram
dA <- density(A)
dZ <- density(Z)
plot(dZ, col="red", lwd=5)
lines(dA, col="green", lwd=5)

# Third diagram
qqline(A, col = 2)

# Fourth diagram
qqline(Z, col = 2)
  • $\begingroup$ Hey, this is great, thank you for your comment Any thoughts on the SSS thing? $\endgroup$
    – user3921
    Jan 19, 2016 at 7:38
  • $\begingroup$ @user3921 : I don't believe it is stationary at all, because it is cyclostationary. $\endgroup$
    – Peter K.
    Jan 19, 2016 at 17:41

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