6 Invertibility

This imposes restrictions on the \(\{\beta _i\}\) to ensure that the process satisfies the invertibility conditions.

Consider the first order MA(1) \[\text {A}: X_t = Z_t + \theta Z_{t-1}\] \[\text {B}: X_t = Z_t + \frac {1}{\theta } Z_{t-1}\] these functions have the same auto correlation function. Thus we can’t classify many MA uniquely from a given auto correlation function.

We express A and B in terms of \(X_t\) and \(X_{t-1}\) \[\text {A}: Z_t = X_t - \theta X_{t-1} + \theta ^2 X_{t-2}\] \[\text {B}: Z_t = X_t - \frac {1}{\theta } X_{t-1} + \frac {1}{\theta ^2} X_{t-2}\] If \(|\theta | < 1\) the process A converges while the series B diverges.
Thus, if \(|\theta | < 1\) the model A is convertible while B is not.
The impression of the invertibility condition ensures that there is a unique process for a given auto correlation.
To compute the invertibility condition for MA. We express it by the use of the backward shift given by

\(\beta ^j X_t = X_{t-j}, \hspace {1cm}\forall j\)

\(\beta ^2 X_t = X_{t-2}\), \(\hspace {1cm}\beta X_t = X_{t-1}\)

Thus \(\hspace {0.2cm} X_t = \beta _0Z_t + \beta _1Z_{t-1} + \cdots \cdots + \beta _qZ_{t-q}\hspace {0.2cm}\) can be written as \begin {align*} X_t &= (\beta _0 + \beta _1B + \beta _2B^2 + \cdots \cdots + \beta _qB^q)Z_t\\ &= \theta (B)Z_t\\ \end {align*}

Where \(\theta (B)\) is a polynomial of order q in B.
An MA(q) is invertible if the roots of the equation \(\theta (B) = 0\)
i.e \(\hspace {0.2cm}\beta _0 + \beta _1B + \beta _2B^2 + .......... + \beta _qB_q = 0\) , lie outside a unit circle.

EXAMPLE:
MA(1) , \(\hspace {0.2cm} X_t = Z_t + \theta Z_{t-1}= (1 + \theta B)Z_t\) or \(\theta (B) = 1 + \theta B\)
The roots of \(\theta (B) = 0\) are \(B = \dfrac {-1}{\theta }\) , the root is outside the unit circle provided \(|\theta | < 1\).

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