# Calculate mean and autocovariance function to check stationarity

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.

• Hi there. Please add the [self-study] tag. – Jim May 21 '18 at 14:19

## 1 Answer

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.

• 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? – makome May 25 '18 at 7:37
• EDIT: Now I have the same solution, I did some more steps between yours. And used the trigonometric addition theorems. Thanks! – makome May 25 '18 at 8:04