Discussion
Let \(\{X_{t}\}\) be a time series defined as \(X_{t} = A\sin(\omega t + B)\), where \(E(A) = 0\), \(\mathrm{Var}(A) = 1\), \(B \sim\) Uniform \((-\pi, \pi)\) and A and B are independent. Then, \(h\)-lag covariance function \(r(h)\) is :
Answer: A
\(r(h) = \) \(\tfrac{1}{2}\cos(\omega h)\).
Since \(E(A) = 0\), the mean is 0, and by independence
\[ \begin{aligned} r(h) &= E(A^{2}) \\ &\quad \times E[\sin(\omega t + B)\sin(\omega t + \omega h + B)] \end{aligned} \]
with \(E(A^{2}) = 1\). Using \(\sin u \sin v = \tfrac{1}{2}[\cos(u - v) - \cos(u + v)]\), the first term is \(\tfrac{1}{2}\cos(\omega h)\) and the second, \(\tfrac{1}{2}E\cos(2\omega t + \omega h + 2B)\), averages to 0 over B uniform on \((-\pi, \pi)\).
So \(r(h) = \tfrac{1}{2}\cos\omega h\), which depends only on the lag, confirming stationarity. An autocovariance must be even in \(h\) and equal to the variance at \(h = 0\), which rules out the sine forms and the shifted cosine.
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