Is the series $X_j = Z_0 \cos(c j)$ stationary?

Question

Let

• $Z$ be a gaussian white noise with mean $0$ and variance $1$
• $c \in \mathbb{R}$ constant

Is the time series stationary? I compute the mean and variance and they look constant, am I right? if so what is the value of the auto-covariance function, $\mathbb{C}(X_j, X_{j+l})$ ?

$$\mathbb{E}(X_j) = \mathbb{E}(Z_0 \cos(cj)) = \cos(cj) \mathbb{E}(Z_0) = 0 $$

$$ \mathbb{V}(X_j) = \cos^2(cj) \mathbb{V}(Z_0) = \cos^2(cj)$$

Answer 1

The equality $$ \cos(cj) \mathbb{E}(Z_0) = \cos(cj) Z_0 $$ is not correct since the right hand side is random, contrarily to the left hand side.

But the computation $\mathbb E\left[X_j\right]=\cos(cj)\mathbb E\left[X_j\right]$ is correct. Therefore, the only change for $\left(X_j\right)_{j\geqslant 1}$ to be stationary in the weak or strong sense is that $\cos(cj)=1$ for all integer $j\geqslant 0$ hence $c$ should be a multiple of $2\pi$. Conversely, if $c=2k_0\pi$ for some integer $k_0$, then $X_j=Z_0$ which is stationary.