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I'm reading a book on time series and I started scratching my head in the following part:

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Could someone explain the intuition for me? I couldn't get it from this text. Why do we need the process to be invertible? What is the big picture here? Thank you for any help. I'm new on this stuff so if you could be kind to use student-level terms when explaining this :)

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  • $\begingroup$ Besides the invertibility is important if the time series observation or disturbance term is convergent.This is important for forecasting. If the process does not satisfy the condition of invertiblity it is impossible to forecast. $\endgroup$ – Narain Sinha Mar 26 '17 at 14:20
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In the AR($\infty$) representation, the most recent error can be written as a linear function of current and past observations: $$w_t = \sum_{j=0}^\infty (-\theta)^j x_{t-j}$$ For an invertible process, $|\theta|<1$ and so the most recent observations have higher weight than observations from the more distant past. But when $|\theta| > 1$, the weights increase as lags increase, so the more distant the observations the greater their influence on the current error. When $|\theta|=1$, the weights are constant in size, and the distant observations have the same influence as the recent observations. As neither of these situations make much sense, we prefer the invertible processes.

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Time series is invertible if erros can be inverted into a representation of past observations.

For the time series data, the error ($\epsilon$) at time $t$ can be represented as:

$$\epsilon_t = \sum\limits_{i=0}^{\infty}(-\theta)^i \, y_{t-i}$$

With every lagged value ($y_{t-i})$, its coefficient is $i^{th}$ power of $\theta$ term. So, the infinite series converges to a finite value only if $|\theta|<1$, which also means that recent past observations are given more weight than distant past observations.

Therefore, time series is invertible if $|\theta|<1$ (possibility of representing errors as linear combination of past observations)

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