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Let us say $X_t = \sum_{i=0}^{\infty}\phi_j Z_{t-i}$, where $Z$ is white noise.

Is $MA(\infty)$ process invertible?

I don't know how to show this, because $\theta(z) = 1 + \phi_1 z + \phi_2 z^2 +... $ is infinite.

If this process is invertible it can be shown that $\theta(z)\neq 0$ for $|z|\leq 1$.

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  • $\begingroup$ After writing more than half of the answer I realized that the question may come from class exercise. However I did not want to waste my effort, so the result is much more detailed answer, than it should have been otherwise. $\endgroup$
    – mpiktas
    Apr 13, 2015 at 12:24

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The following statement is very useful.

Suppose $\{Z_j, j\in \mathbb{Z}\}$ is a time series. Then if $\sup_jE|Z_j|<\infty$ and $\sum_{j}|\psi_j|<\infty$, then $X_t=\sum_j\psi_jZ_{t-j}$ converges absolutely, almost surely and in mean. If $Z_j$ is stationary, then $X_t$ is stationary too.

$MA(\infty)$ process is invertible if there exists such a set of coefficients $\alpha_j$, such that

$$\sum_{j=0}^\infty \alpha_j X_{t-j}=Z_t,$$

and

$$\sum_{j=0}|\alpha_j|<\infty.$$

If we employ the backwards operator $B$ ($BX_t=X_{t-1}$), the requirement translates to

$$\alpha(B)\theta(B)=1,$$

where $\alpha(B)=\sum_{j=0}^\infty\alpha_jB^j$ and $\theta(B)=\sum_{j=0}^\infty \phi_jB^j$. (Here I used the notation from the question).

Now we can employ the trick where we exchange $B$ with $z\in \mathbb{C}$. We can do that because of the definition of multiplication of power series. This change however gives us another way of looking at the problem. The power series $\theta(z)=\sum_{j=0}\theta_jz^j$ is Laurent series with no negative powers of $z$ and $\sum_{j=0}|\theta_j|<\infty$, hence it is analytic in a neighborhood of a disc $|z|\le 1$.

Having established that $\theta(z)$ is analytic we can turn our attention to $\alpha(z)$. We have that

$$\alpha(z)=\frac{1}{\theta(z)}.$$

So $\alpha(z)$ is analytic whenever $\theta(z)\neq 0$, so it is analytic in a neighborhood of a disc $|z|\le 1$. However we need that $\sum_{j=0}^\infty|\alpha_j|<\infty$ which means that we need $\alpha(z)$ to be analytic in $|z|<R$, where $R>1$. This can only happen if $\theta(z)\neq 0$ for $|z|\le 1$.

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