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For MA(1) process, it is easy to show how one can convert it into AR($\infty$). However, how can we really show that MA(2), giving its characteristics roots lie outside unit circle, can have invertibility?

Consider $Y_t=\theta_1\varepsilon_{t-1}+\theta_2\varepsilon_{t-2}+\varepsilon_{t}$, we can write it using lag operator just like MA(1) case, $$Y_t=(1+\theta_1L+\theta_2L^2)\varepsilon_t$$ $$\frac{Y_t}{(1+\theta_1L+\theta_2L^2)}=\varepsilon_t$$

I do not know how to continue from here to show invertibility. Is it true to claim $$Y_t + (\theta_1L+\theta_2L^2)Y_t+(\theta_1L+\theta_2L^2)^2Y_t+\dots=\varepsilon_t$$

Then, how can we show from these equations the MA(2) invertibility conditions for $\theta_1$ and $\theta_2$ ?

Jacky
  • 115

1 Answers1

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Factor the polynomial into $$(1 + \theta_1L + \theta_2L^2) = (1 - \phi_1L)(1 - \phi_2L)$$

Notice that:

$$(1-\phi_iL)^{-1} = \sum_{k=0}^\infty \phi_i^kL^k$$

which follows by the rules of a geometric series.

So, the rules of convergence of the above follows by the rules of convergence of geometric series. Namely $\vert\phi_1\vert<1$ and $\vert\phi_2\vert<1$.

If so, then you can write $$\epsilon_t = (1-\phi_2L)^{-1}(1-\phi_1L)^{-1}y_t$$

This is a pretty good intro to lag/difference operators, if interested.

measure_theory
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