Let $X(t) = Acos(2\pi f_c t)$ be a random process where $A$ is a uniform random variable within $(-1,1)$. I'm trying to prove this is a weakly(i.e. wide sense) stationary process. I need to show two conditions

  1. Constant mean.
  2. Autocorrelation $R_X(t_1,t_2)$ only depends on the time difference $t_2-t_1$.

First one is easy. However, I can't prove the second condition. I've tried the following,

$R_X(t_1,t_2) = E[X(t_1)X(t_2)] = E[ Acos(2\pi f_c t_1) * Acos(2\pi f_c t_2)] = cos(2\pi f_c t_1)* cos(2\pi f_c t_2)*E[A^2]$

Using trig identities,

$=1/2(cos(2\pi f_c (t_1+t_2)-cos(2\pi f_c (t_1-t_2))*E[A^2] $

Where $E[.]$ is the expectation operator.

I'm stuck here, how can I do this?

  • $\begingroup$ You are trying to prove a statement that is false. The process $X(t) = A\cos(2\pi f_c t)$ is not a WSS process regardless of the distribution of $A$. See Why is $A\cos(2\pi f_ct)$ a non-stationary process? and its answers for details. I vote to close. $\endgroup$ Apr 24 '20 at 3:19
  • $\begingroup$ @DilipSarwate In the linked answer, the conditions you list seem to be for strictly stationary processes, whereas I'm interested to see if it is a weakly stationary process. Am I wrong? $\endgroup$
    – zeke
    Apr 24 '20 at 6:21