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Silly question: From what I understand, an MA process is invertible when it can be represented as an AR($\infty$) process.

When using lag operator, it is somewhat clear that $|\theta|<1$ is required. But I am unable to see this in following derivation.

Consider the following MA(1) process:

\begin{align} x_t &= \epsilon_t + \theta \epsilon_{t-1} \\ &= \epsilon_t + \theta (x_{t -1}- \theta \epsilon_{t-2}) \\ &= \epsilon_t + \theta x_{t -1}- \theta^2 (x_t- \theta\epsilon_{t-3})\\ &=\epsilon_t - \sum\limits_{j=1}^{\infty}(-\theta)^jx_{t-j} \end{align} and so on.

So we are able to express MA(1) process as an AR($\infty$).

At which step am I using $|\theta|<1$?

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It's not a silly question --- this is a common misconception in the recursive method, and I've seen heaps of people make the same mistake. The problem here is that your "and so on" glosses over the fact that the recursive application of that substitution yields a limiting term at the end:$^\dagger$

$$\begin{align} x_t &= \epsilon_t + \theta \epsilon_{t-1} \\[6pt] &= \epsilon_t + \theta (x_{t -1}- \theta \epsilon_{t-2}) \\[6pt] &= \epsilon_t + \theta x_{t -1}- \theta^2 (x_t- \theta\epsilon_{t-3})\\[6pt] &=\epsilon_t - \sum\limits_{j=1}^{\infty}(-\theta)^jx_{t-j} - \underbrace{\lim_{j \rightarrow \infty} (-\theta)^j \epsilon_{t-j}}_\text{You left this out}. \end{align}$$

In order to get the equation you want, you need that last term to disappear (i.e., converge stochastically to zero), and one way to do that is to have $|\theta|<1$ and a bounded variance on the series of error terms (a fixed finite error variance is sufficient here). If $|\theta| = 1$ then that final limiting term fails to disappear and if $|\theta| > 1$ the final limiting term explodes.


$^\dagger$ Just to show this more formally, observe that the application of the recursive substitution $N-1$ times gives you the equation:

$$\begin{align} x_t &= \epsilon_t - \sum\limits_{j=1}^{N-1} (-\theta)^jx_{t-j} - (-\theta)^{N} \epsilon_{t-N}. \end{align}$$

Taking the limit as $N \rightarrow \infty$ then gives the equation shown above.

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    $\begingroup$ thanks a lot. I was thinking exactly this that there is a problem in 'so on' but couldn't figure out the limit thing. $\endgroup$
    – Dayne
    Jul 22 at 12:27
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$x_{t} = \epsilon_t + \theta \epsilon_{t-1} $

$\frac{x_{t}}{1 + \theta L} = \epsilon_t $

But the division on the left hand side so that one obtains an infinite geometric series is invalid if $abs(\theta)$ is not less than 1.0.

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  • $\begingroup$ my doubt is in the above derivation. Where is this condition being used if I invert the way I have done above? $\endgroup$
    – Dayne
    Jul 22 at 11:42
  • $\begingroup$ i have edited the title of question to avoid confusion. $\endgroup$
    – Dayne
    Jul 22 at 12:11
  • $\begingroup$ I'm sorry Dayne. I clearly mis-understood your question. Ben cleared it up beautifully. $\endgroup$
    – mlofton
    Jul 23 at 13:25
  • $\begingroup$ there's absolutely no need to be sorry. About your answer, let me add few interesting points. The reason I preferred the recursive approach was because I find the operator explanation more difficult to explain. You see $L$ here is not a usual number that can be added like this. The whole idea of a series of L and its convergence needs to be qualified for which appropriate space needs to be defined. To avoid all that explanation (say in an exam), I simply prefer recursive approach. $\endgroup$
    – Dayne
    Jul 23 at 17:01
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    $\begingroup$ Dayne: I think the use of L and the proof that you can view it as a number gets into operator theory and functional analysis ( which is not my thing ). So, I agree that the recursive approach is more intuitive. It's just that I use lagged dependent variables quite often so it's quicker to use the L notation once you accept that it's allowed. Ben's answer was educational for me so thanks for question. $\endgroup$
    – mlofton
    Jul 24 at 2:19
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This is actually somewhat subtle. When $|\theta|>1$, you can still express $\epsilon_t$ as AR($\infty$) on $x_t, x_{t+1}, x_{t+2}, ...$. But $\epsilon_t$ is no longer the fundamental innovation on $x_t$, defined as $x_t - E(x_t|x_{t-1}, x_{t-2}, ...)$.

So if being "invertible" means a MA process on the "fundamental" innovations, then root conditions like $|\theta|<1$ is needed.

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  • $\begingroup$ Hi can you explain this a little more? How is $|\theta|<1$ related to 'fundamental' property? $\endgroup$
    – Dayne
    Jul 25 at 2:01
  • $\begingroup$ When $|\theta| < 1$, $\epsilon_t$ can be expressed as MA on $x_t$ and its past values. Re-arranging the expression, $x_t$ is expressed as AR on its past values, with the same $\epsilon_t$ as the disturbance. It is in this sense that $\epsilon_t$ is "fundamental". $\endgroup$ Jul 25 at 2:12
  • $\begingroup$ But where is $|\theta|<1$ getting used. Please expand your answer for more details as this looks like a different approach. $\endgroup$
    – Dayne
    Jul 25 at 2:15
  • $\begingroup$ When $|\theta|<1$, $\epsilon_t$ can be expressed as MA on $x_t$ and its past values. When $|\theta|>1$, $\epsilon_t$ can be expressed as MA on $x_t$ and its future values. If you still need more details, I will be happy to recommend the texts I used. $\endgroup$ Jul 25 at 2:42
  • $\begingroup$ yes i want to know exactly how we can express error as MA of future values and how is theta>1 used in such a derivation. More relevant to my question, how to write error as sum of past values if x and where to use theta<1 in such derivation. $\endgroup$
    – Dayne
    Jul 25 at 2:46

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