question related to something in karlin and taylor stochastic processes one text

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.

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.

• 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. Jan 3, 2020 at 15:07
• @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. Jan 3, 2020 at 15:54
• 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. Jan 4, 2020 at 3:52
• @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 $\omega$. Jan 4, 2020 at 9:07
• Gotcha. I'm not sure what I was thinking of there !!!!!!! and thanks for patience. Jan 4, 2020 at 13:46