Covariance stationary solution to the equation

Question

Is there a covariance stationary solution to the equation?

$X_t = 4 \varepsilon_{t-1} + \varepsilon_t, \quad \text{where } \{ \varepsilon_t \} \in WN(0, \sigma^2), t \in \mathbb{Z}$.

Unfortunately I completely do not know how to handle this exercise. My only thought was to check if $X_t$ is covariance stationary. So is it true that

1. $E(X_t) = const$,
2. $\gamma(t,t+l) = Cov(X_t, X_{t+1}) = \gamma(s, s+l)$.

But is a good idea?

Answer 1

The process in question is a Moving-Average process of order one,

$$\varepsilon_t = \frac {1} {1+\theta L} X_t = \frac {1} {1-(-\theta) L} X_t, \qquad \theta =4.$$

Here $L$ is the lag operater, $L^k(z_t) = z_{t-k}$. The process is covariance-stationary irrespective of the value of $\theta$, but it is not invertible since $\theta \geq 1$, so it does not have an autoregressive $AR(\infty)$ representation.

Since $\theta \geq 1$ we can solve it "forward", as follows: First, define the forward operator $F^k \equiv L^{-k}, F^k(z_t) = z_{t+k}$.

Second, note that

$$\frac {1} {1-(-\theta) L} = \frac{1}{\theta L[-(-\theta^{-1})L^{-1} +1]} = \frac{\theta^{-1}L^{-1}}{1-(-\theta)^{-1}L^{-1}]} = \frac{\theta^{-1}F}{1-(-\theta^{-1})F}$$

Then, since $\theta^{-1} <1$,

$$\frac {1} {1-(-\theta) L} = \theta^{-1}F\cdot [1-\theta^{-1}F + \theta^{-2}F^{2}-\theta^{-3}F^{3}+...]$$

So $$\varepsilon_t = X_t \cdot \theta^{-1}F\cdot [1-\theta^{-1}F + \theta^{-2}F^{2}-\theta^{-3}F^{3}+...]$$

$$\implies \varepsilon_t = \sum_{j=1}^{\infty}\frac {(-1)^{j+1}}{\theta^j}X_{t+j}$$

This is the "forward" solution, when the MA process is not invertible.

I guess the OP can now determine whether this solution is covariance-stationary.