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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?

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  • $\begingroup$ Hi: that's an MA(1) with $\theta = 4$. So, I would look up info on the MA(1) because I could try to explain it ( obtaining a solution ) here but it won't be as good as what you should find. The short of it is that, $\theta=4$ means that you can't write $\epsilon_t$ as a function of the past ( which is what I think they mean by covariance stationary solution ) but you can write it as a function of the future ( but this solution I don't think is considered covariance stationary ) If you can't find anything that explains this, let me know and I'll try to find something. $\endgroup$
    – mark leeds
    Jan 24 at 5:17
  • $\begingroup$ Not sure if this is relevant, but all MA processes are covariance stationary. I wonder what the definition of "covariance stationary solution" is. Could you recommend a textbook? $\endgroup$ Jan 24 at 8:21

1 Answer 1

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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.

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  • $\begingroup$ Thanks Alecos. That's the solution that I was thinking of in my mind but the gory details are appreciated. $\endgroup$
    – mark leeds
    Jan 26 at 4:08
  • $\begingroup$ @markleeds You're welcome Mark. $\endgroup$ Jan 26 at 8:39

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