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Suppose $X_{t} = B\sin(\omega t) + w_{t}$ and $\omega$ is between $(0,\pi)$, $B$ has mean 0 and variance 1 $w_{t}$ is $N(0,1)$ and $w_{t}$ is independent of $B$. Show that $X_{t}$ is weakly stationary.

Attempt: Normally, this series would not be stationary but with the addition of the constant A, the mean is 0.

For $\gamma_{x}(h)$ we need to find $E[X_t X_{t+h}] = E[(B\sin(\omega t) + w_{t})(B\sin(\omega(t+h) + w_{t+h})]$. I am having difficulty simplifying this function. Would this be equal to $B^2 \sin^{2}(wt\cdot w(t+h) + w_t w_{t+h}$?

Michael Hardy
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phil12
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1 Answers1

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$$E[X_tX_{t+h}]=\sin(\omega t)\sin(\omega(t+h))+E[w_tw_{t+h}]$$

Did
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  • How did the B's cancel? Also would the $\gamma_{x}(h) = 1 +sin(\omegat)sin(\omega(t+h))$ for h = 1 and $sin(\omegat)sin(\omega(t+h))$ for h =0? – phil12 May 01 '13 at 00:23
  • @phil12 The mixed term vanishs since $\mathbb{E}B=0$ (note that $w_t$ and $B$ are independent!). The $B^2$ cancels because of $\mathbb{E}(B^2)=1$. – saz May 01 '13 at 05:55
  • But the expectation is still a function of time implying the series is not stationary? Is this correct? – phil12 May 01 '13 at 06:38
  • Would this function be stationary? – phil12 May 01 '13 at 16:17
  • Already the variance of $X_t$ is not constant so one wonders how $(X_t)$ could be stationary. – Did May 01 '13 at 17:56