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This question is essentially a question about something in Karlin and Taylor's Stochastic Processes One text in the spectral chapter. Since this is a DSP list, Karlin and Taylor may not be so popular so I will describe the setting which I think I've seen in other texts also.

They define the random variable $X_n = A \cos(\omega n) + B \sin(\omega n)$.

So, $A$ and $B$ are zero mean normal random variables. $\omega$ is fixed at some frequency between $0$ and $\pi$. The process is on the integers so the process start at some value of $-k$ and continues through to positive $k$. So $k$ can be any integer and $X_n$ represents a discrete stochastic process.

They then say that the $X_n$ process is stationary in the mean. Evidently this is so obvious that there's no need to explain it. I don't get it. Well, I sort of half get it.

A and B are mean zero so, okay the mean will always be zero no matter what the values of the cos or sin term. But since $n$ is increasing by 1 as the process proceeds, isn't the trajectory of $X_n$ changing at each $n$ because it multiplies omega ? So, because A and B are zero, stationary in the mean works. But, say A and B were both 1.0 say. Would the $X_n$ process still be stationary in the mean ? Since $X_n$ and $X_{n+1}$ will be at a different frequencies, $\omega n$ and $\omega (n+1)$ respectively, doesn't that make the mean different at times $n$ and $n + 1$ ? so it can't be a stationary process.

So, I guess my question is: For this process, does stationary in mean only hold because $A$ and $B$ are zero in mean or would it work no matter what their means were ? If the latter is true, then I'm totally not understanding the process. Thanks.

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The mean, i.e., the expected value of $X_n$ is

$$E\{X_n\}=E\{A\cos(\omega n)+B\sin(\omega n)\}\tag{1}$$

Since the expectation operator is linear, $(1)$ is equivalent to

$$E\{X_n\}=E\{A\cos(\omega n)\}+E\{B\sin(\omega n)\}\tag{2}$$

and since $\cos(\omega n)$ and $\sin(\omega n)$ are deterministic you finally get

$$E\{X_n\}=E\{A\}\cos(\omega n)+E\{B\}\sin(\omega n)\tag{3}$$

which equals zero if both $E\{A\}$ and $E\{B\}$ are zero. If at least one of the two random variables $A$ and $B$ has a non-zero mean, then $(3)$ will generally depend on $n$ (except for very specific choices of $\omega$), and, consequently, the mean of $X_n$ will also depend on $n$, i.e., $X_n$ cannot be stationary.

As a side note, a constant mean does of course generally not imply stationarity of $X_n$. For the given process, however, if $A$ and $B$ are identically distributed and $E\{A\}=E\{B\}=0$, and under the additional assumption that $E\{AB\}=0$ holds, $X_n$ is a wide-sense stationary (WSS) process.

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  • $\begingroup$ Hi Matt. yes, I was being sloppy on the non-mean part but they proved that explicitly so I didn't include it. here's the part I'm confused about: yes, sin and cos deterministic but, since they are functions of $n$, it seems like the both A and B being zero case is the only case where the stationarity could possibly hold ? And even then, i still can't get my hands around what the process is at each step ? The arguments of the sin and cos are getting multiples like in a DFT ? t Thanks. $\endgroup$ – mark leeds Jan 3 at 15:07
  • $\begingroup$ @markleeds: Yes, that's what I tried to explain in the text following Eq. (3). If $A$ or $B$ have a non-zero mean, then the mean of $X_n$ depends on $n$, i.e., $X_n$ can't be stationary. I'm not sure what you mean with your last question when you refer to the DFT. $\endgroup$ – Matt L. Jan 3 at 15:54
  • $\begingroup$ Hi Matt. That helps. My last question is what's going on in terms of $\omega$ becoming a multiple of itself as n increases ? Does that mean that, as time passes discretely, the process is jumping from one frequency to a higher frequency ? Thanks. $\endgroup$ – mark leeds Jan 4 at 3:52
  • $\begingroup$ @markleeds: No, the sequence $\cos(\omega n)$ has a constant frequency $\omega$, $n$ is just the (time) index. Just like a continuous-time sinusoid $\cos(\omega t)$ having a constant frequency $\omega$. $\endgroup$ – Matt L. Jan 4 at 9:07
  • $\begingroup$ Gotcha. I'm not sure what I was thinking of there !!!!!!! and thanks for patience. $\endgroup$ – mark leeds Jan 4 at 13:46

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