10.1 Definition

When a process is non-stationary, the non-stationary sources of variation must be removed. A common approach, which goes back to Box and Jenkins, is to difference the series. If in the ARMA model \[ X_t = \alpha _1X_{t-1} + \dots + \alpha _pX_{t-p} + Z_t + \beta _1Z_{t-1} + \dots + \beta _qZ_{t-q} \] \(X_t\) is replaced by its \(d\)th difference \(\nabla ^dX_t\), the model is capable of describing a certain type of non-stationary series. Writing \[ W_t = \nabla ^dX_t = (1 - B)^dX_t , \] the model is the ARMA process \[ W_t = \alpha _1W_{t-1} + \dots + \alpha _pW_{t-p} + Z_t + \beta _1Z_{t-1} + \dots + \beta _qZ_{t-q} , \] or \(\phi (B)W_t = \theta (B)Z_t\), or, in terms of the original series, \[ \phi (B)(1 - B)^dX_t = \theta (B)Z_t . \] Thus \(W_t\) is a stationary ARMA(\(p, q\)) process, while \(X_t\), described through its \(d\)th difference, is an autoregressive integrated moving average process of order \((p, d, q)\), ARIMA(\(p, d, q\)). The name “integrated” is explained by the case \(d = 1\): then \(X_t = X_0 + W_1 + W_2 + \dots + W_t\), a sum (the discrete analogue of an integral) of the stationary \(W\)’s.

The model for \(X_t\) is non-stationary: its generalised AR operator \(\phi (B)(1 - B)^d\) has \(d\) roots on the unit circle, at \(B = 1\), rather than outside it. In practice \(d\) is usually \(1\), occasionally \(2\). The second difference is \begin {align*} \nabla ^2X_t &= \nabla (X_t - X_{t-1}) \\ &= (X_t - X_{t-1}) - (X_{t-1} - X_{t-2}) \\ &= X_t - 2X_{t-1} + X_{t-2} , \end {align*}

which is \((1 - B)^2X_t = (1 - 2B + B^2)X_t\), as the binomial expansion says. Special cases are familiar: ARIMA(0,1,0) is the random walk of Section 5, and ARIMA(0,0,\(q\)) and ARIMA(\(p\),0,0) are the MA and AR processes.

Example 10.1.1. Write the ARIMA(1,1,0) model \((1 - 0.6B)(1 - B)X_t = Z_t\) as a difference equation in \(X_t\), and find the roots of its generalised AR operator. Given \(x_{n-1} = 50\) and \(x_n = 54\), find the value of \(X_{n+1}\) if the next shock is \(0\).

Solution. \((1 - 0.6B)(1 - B) = 1 - 1.6B + 0.6B^2\), so \[ X_t = 1.6X_{t-1} - 0.6X_{t-2} + Z_t . \] The roots of \(1 - 1.6B + 0.6B^2 = 0\) are \(B = 1\) and \(B = 1/0.6 = 1.67\): one unit root, from the differencing, and one stationary root. As an AR(2) this would fail the triangle condition, since \(\alpha _1 + \alpha _2 = 1.6 - 0.6 = 1\), exactly on the boundary. With \(Z_{n+1} = 0\), \(X_{n+1} = 1.6(54) - 0.6(50) = 56.4\). In terms of differences: the last change was \(W_n = 4\), the next is \(0.6 \times 4 = 2.4\), and \(54 + 2.4 = 56.4\). The changes die away geometrically, but the level never returns to any fixed mean.

Questions on this section

Stuck on something here? Ask below and it stays attached to this topic.