Hey I need some help with this exercise:

Assume $z_{ t }$ be a sequence of independent normal random variables, each with mean 0 and time independent variance $\sigma^{ 2 }$, and let c be a constant. Is the following process for a time series stationary? If it is stationary specify the mean and the autocovariance function:

$$ x_{ t }=z_{ 1 }*cos\left( ct \right)+z_{ 2 }*sin\left( ct \right) $$

So for a weak stationarity time series the mean and covariance function should be independent of time. I'm not sure how to calculate them by hand without R.

Thanks for help.

  • $\begingroup$ Hi there. Please add the [self-study] tag. $\endgroup$
    – Jim
    Commented May 21, 2018 at 14:19

1 Answer 1


The model is $$x_t = z_1 \cdot \cos(ct) + z_2 \cdot \sin(ct).$$ Thus the mean function of the series will be $$m(t) = E(z_1)\cdot \cos(ct) + E(z_2) \cdot \sin(ct) = 0$$ and the covariance function,

\begin{align} cov(x_j, x_t) &= cov\left(z_1 \cdot \cos(cj) + z_2 \cdot \sin(cj), ~z_1 \cdot \cos(ct) + z_2 \cdot \sin(ct)\right)\\ &= \cos(cj)\cdot \cos(ct)\cdot var(z_1) + \sin(cj)\cdot\sin(ct) \cdot var(z_2)\\ &=\sigma^2 \cdot cos(c(j-t)). \end{align}

We can see the mean function is independent of time and the covariance function depends on the time difference implying the series weak stationary.

  • $\begingroup$ Thanks, for the mean I have the same solution. But for the cov not. My formula is $cov\left( x_{ t+h }, x_{ t } \right)$. For my calculation I used the linearity property of the covariance. I don't understand how you come from step one to step two and from step two to the solution? $\endgroup$
    – makome
    Commented May 25, 2018 at 7:37
  • $\begingroup$ EDIT: Now I have the same solution, I did some more steps between yours. And used the trigonometric addition theorems. Thanks! $\endgroup$
    – makome
    Commented May 25, 2018 at 8:04

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.