7.2 The backward shift operator
The invertibility condition for a general MA(\(q\)) is stated with the backward shift operator \(B\), defined by \[ BX_t = X_{t-1} , \qquad B^jX_t = X_{t-j} \quad \text {for all } j . \] So \(B^2X_t = X_{t-2}\), and the MA(\(q\)) process can be written \[ X_t = (\beta _0 + \beta _1B + \beta _2B^2 + \dots + \beta _qB^q)Z_t = \theta (B)Z_t , \] where \(\theta (B) = \beta _0 + \beta _1B + \dots + \beta _qB^q\) is a polynomial of order \(q\) in \(B\). Operators like this can be added, multiplied and factorised like ordinary polynomials.
Theorem 7.2.1. The MA(\(q\)) process \(X_t = \theta (B)Z_t\) is invertible if all the roots of the equation \[ \theta (B) = \beta _0 + \beta _1B + \beta _2B^2 + \dots + \beta _qB^q = 0 \] lie outside the unit circle, that is, have modulus greater than \(1\).
Proof. Let \(r_1, \dots , r_q\) be the roots, possibly complex. With \(\beta _0 = 1\), \(\theta (B) = (1 - B/r_1)(1 - B/r_2)\cdots (1 - B/r_q)\), since both sides are polynomials of degree \(q\) with the same roots and the same constant term \(1\). For each factor, \(|1/r_j| < 1\), so as in Proposition 7.1.2 (with \(\beta = -1/r_j\)), \((1 - B/r_j)^{-1} = \sum _{m=0}^\infty (B/r_j)^m\) is an operator with absolutely summable weights. Applying the \(q\) inverses in turn to \(X_t = \theta (B)Z_t\) gives \(Z_t = \pi (B)X_t\), where \(\pi (B)\), a product of \(q\) series with absolutely summable weights, again has absolutely summable weights. □
For the MA(1), \(\theta (B) = 1 + \beta B\), with the root \(B = -1/\beta \), which lies outside the unit circle exactly when \(|\beta | < 1\), in agreement with Proposition 7.1.2. Invertibility matters for two reasons: it makes the model unique for a given ac.f., and it allows the past shocks to be recovered from the data, which is how forecasts are computed (Section 14).
Example 7.2.2. Determine whether each MA(2) process is invertible:
- (a)
- \(X_t = Z_t - 1.3Z_{t-1} + 0.4Z_{t-2}\);
- (b)
- \(X_t = Z_t + 2.5Z_{t-1} + Z_{t-2}\).
For any that is not, find an invertible MA(2) process with the same autocovariance function.
Solution. (a) \(\theta (B) = 1 - 1.3B + 0.4B^2 = (1 - 0.5B)(1 - 0.8B)\), with roots \(B = 2\) and \(B = 1.25\). Both exceed \(1\) in modulus, so the process is invertible.
(b) \(\theta (B) = 1 + 2.5B + B^2 = (1 + 0.5B)(1 + 2B)\), with roots \(B = -2\) and \(B = -0.5\). The second lies inside the unit circle, so the process is not invertible.
To find an equivalent invertible process, replace the offending factor \((1 + 2B)\) by \((1 + 0.5B)\), its root reflected from \(-0.5\) to \(-2\), and multiply the noise variance by \(2^2 = 4\), as in the MA(1) case. The candidate is \[ X_t = (1 + 0.5B)^2Z_t^* = Z_t^* + Z_{t-1}^* + 0.25Z_{t-2}^* , \qquad \Var (Z_t^*) = 4\sigma _Z^2 . \] Check with Theorem 6.1.2. The original has \(\gamma (0) = (1 + 6.25 + 1)\sigma _Z^2 = 8.25\sigma _Z^2\), \(\gamma (1) = (2.5 + 2.5)\sigma _Z^2 = 5\sigma _Z^2\), \(\gamma (2) = \sigma _Z^2\). The new one has \(\gamma (0) = 4\sigma _Z^2(1 + 1 + 0.0625) = 8.25\sigma _Z^2\), \(\gamma (1) = 4\sigma _Z^2(1 + 0.25) = 5\sigma _Z^2\), \(\gamma (2) = 4\sigma _Z^2(0.25) = \sigma _Z^2\). The autocovariances agree, and both roots of \((1 + 0.5B)^2\) equal \(-2\), outside the unit circle.
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