# Is $A\cos(\omega t+\theta)$ a Gaussian random process?

$$Z(t) = A\cos(\omega t+\theta)$$ 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.

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. 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). 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.

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
hist(sin(omega*t))

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

# Third diagram
qqnorm(A)
qqline(A, col = 2)

# Fourth diagram
qqnorm(Z)
qqline(Z, col = 2)

• Hey, this is great, thank you for your comment Any thoughts on the SSS thing? Jan 19 '16 at 7:38
• @user3921 : I don't believe it is stationary at all, because it is cyclostationary.
– Peter K.
Jan 19 '16 at 17:41