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I know that in order for a stochastic process to be a second-order weakly stationary process.
Then for every $t$, the following conditions should hold: $$E(Z(t)) = \mu$$ $$D(Z(t)) = \sigma$$

and $$\operatorname{cov}(Z(t), Z(t+p)) = \gamma(p)$$

but I just cannot show it mathematically like $$E(z(t)) = E (A \cos(bt) + B \sin(bt) ) = A (E(\cos(bt))) + B (E(\sin(bt)))$$

and I have no clue after that. So, any help will be appreciated on solving this

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  • $\begingroup$ You need to know the mean/variance of $A$ and $B$, and also their covariance. Also, is $b$ random or fixed? $\endgroup$
    – gunes
    Oct 25, 2019 at 20:35
  • $\begingroup$ yeah b is a constant. but E(cos(bt)) = 0 ? is that true ? and nope I have no clue about the covariance and how to find it. Can someone show me the steps ? $\endgroup$ Oct 26, 2019 at 7:12
  • $\begingroup$ As i said you need to know mean/variance about A and B. neither b nor t is random. $\endgroup$
    – gunes
    Oct 26, 2019 at 7:16
  • $\begingroup$ i think A and B are constant as well so the covariance would be zero I suppose $\endgroup$ Oct 26, 2019 at 7:28

2 Answers 2

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When all $A$, $B$ and $b$ are constant, the whole expression will be a function of $t$. No uncertainty remains in it, and although it is still valid, there is no reason to call it as stochastic. In this case, the mean is also a function of $t$, so the process is not mean stationary, which means it is also not WSS: $$E[Z(t)]=Z(t)=A\cos(bt)+B\sin(bt)=\mu(t)$$ The covariance will be $0$ because both $Z(t),Z(t+p)$ are constants. Typically, what I encounter is $A,B$ being random. In that case, for the mean being stationary, you should have $E[A]=E[B]=0$ (for this specific case), since: $$E[A\cos(bt)+B\sin(bt)]=E[A]\cos(bt)+E[B]\sin(bt)$$ and there is no combination of constants to make this expression constant, other than zero-means.

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  • $\begingroup$ okay so when A and B are random variable. can you tell me what happens to the covariance Z(t) and Z(t+p). I mean how do we calculate it ? $\endgroup$ Oct 26, 2019 at 8:26
  • $\begingroup$ That is where you need to know something about $E[AB]$ $\endgroup$
    – gunes
    Oct 26, 2019 at 8:27
  • $\begingroup$ just found out a case where A and B are random variable like you mentioned with mean 0 and variance σ² So I suppose now Cov(A, A) = σ² and cov(B, B) = σ² $\endgroup$ Oct 26, 2019 at 8:38
  • $\begingroup$ Certain cases exist where $Z(t)$ is WSS, depending on variance and covariance of $A,B$. You'll be using some trigonometric identities to show it. $\endgroup$
    – gunes
    Oct 26, 2019 at 8:46
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Assuming that $A$ and $B$ are independent random variables (and $b$ is a constant), and using standard notation for the moments of these random variables, you have the mean function: $$\begin{align} \mu(t) &\equiv \mathbb{E}(Z(t)) \\[6pt] &= \mathbb{E}(A \cos(bt) + B \sin(bt)) \\[6pt] &= \mathbb{E}(A) \cos(bt) + \mathbb{E}(B) \sin(bt) \\[6pt] &= \mu_A \cos(bt) + \mu_B \sin(bt), \\[6pt] \end{align} \quad \quad \quad \quad \quad \quad \quad$$

and autocovariance function:

$$\begin{align} \quad \quad \quad \quad \gamma(t,k) &\equiv \mathbb{C}(Z(t),Z(t+k)) \\[6pt] &= \mathbb{C}(A \cos(bt) + B \sin(bt), A \cos(bt+bk) + B \sin(bt+bk)) \\[6pt] &= \mathbb{V}(A) \cos(bt)\cos(bt+bk) + \mathbb{V}(B) \sin(bt)\sin(bt+bk) \\[6pt] &= \sigma_A^2 \cos(bt)\cos(bt+bk) + \sigma_B^2 \sin(bt)\sin(bt+bk) \\[6pt] &= \sigma_A^2 \cos(bk) + [\sigma_B^2 - \sigma_A^2] \sin(bt)\sin(bt+bk). \\[6pt] \end{align}$$

Second-order weak stationarity occurs when neither of these functions depend on $t$. Assuming that $b \neq 0$ (giving a non-trivial problem), a sufficient condition for this is to have $\mu_A = \mu_B = 0$ and $\sigma_A^2 = \sigma_B^2 \equiv \sigma^2$, which gives $\mu(t) = 0$ and $\gamma(t,k) = \sigma^2 \cos(bk)$. (Note also that weak-sense stationarity is only achieved when these amplitude values are random; once you condition on the amplitude values the series is no longer stationary.)

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