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In moving-average time series, I was told that the condition for a MA series $Y_t=\Theta(B)Z_t$ to be invertible is for all the roots of $\Theta(B)=0$ lying outside the unit circle. However I only found the proof for MA(1). I wonder what is the general proof for higher order $\Theta$.

I understand that being invertible the actual shock $Z_t$ can be recovered from the observed $\{Y_t\}$. This would hypothetically require the existence and convergence of $\Theta^{-1}(B)$ since $$Z_t=\Theta^{-1}(B)Y_t.$$ However I cannot explicitly get the expression of $\Theta^{-1}$ not to mention deciding the condition of its convergence.

Could anyone help explain this, thanks!

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1 Answers1

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Well sorry I think I put the wrong tag under this question. After some research I decide the following might be a brief proof:

Proof:

Let's note the roots of $\Theta(x)$ as $\zeta_1,\zeta_2,\ldots,\zeta_q$. Suppose that the roots are now ordered by their magnitude: $1<|\zeta_1|\le|\zeta_2|\le\ldots\le|\zeta_q|$. Write $|\zeta_1|=1+\epsilon$ where $\epsilon>0$.

Obviously for any $|z|<1+\epsilon$, $\Theta(z)\neq0$. Hence, $$ \frac1{\Theta(z)}=\sum\limits_{i=0}^{\infty}\psi_iz^i $$ Should be a convergent series.

Now choose $0<\delta<\epsilon$ and plug $z'=1+\delta$ in the above equation: $$ \frac1{\Theta(1+\delta)}=\sum\limits_{i=0}^{\infty}\psi_i(1+\delta)^i<\infty; $$ Therefore $\exists M>0, |\psi_i(1+\delta)^i|<M$, that is, for all $i$, $$ |\psi_i|<M(1+\delta)^{-i}. $$ Hence $Z_t=\frac{Y_t}{\Theta(B)}=\sum\limits_{i=0}^{\infty}\psi_iY_{t-i}$ converges therefore invertible.

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