Is $(-1)^t A$ a stationary process?

Question

If $A$ is a random variable with $E(A)=0$ and $Var(A)=1$, is $(-1)^t A$ a stationary process?

Well if $t$ is even $Z_t=A$ and if $t$ is odd $Z_t=-A$, so the process change with time. Is it enough to show that a process is not stationary or I need to show that one of the conditions of a stationary process not hold?

Answer 1

If the distribution of $A$ is symmetric, that is, the same as the distribution of $-A$, then your process is stationary. Otherwise not.

Answer 2

Let $X(t)=(-1)^t A$. If $t$ is an integer (or equivalently, $X(t)$ takes on real values), then $X(t)$ is a second-order stationary process.

First, $$\mathbb{E}[X(t)]=(-1)^t \mathbb{E}[A]=0.$$ Second, $$\mathbb{E}[X(t)X(t+\tau)]=\mathbb{E}[(-1)^{2t+\tau} A^2]=(-1)^{\tau}.$$

Therefore, the process is a second-order stationary process.