Finbarr Timbers

ARIMA, ARMA, what's the difference?

I'm working through TSA, and I noticed that some of my classmates are struggling to understand the difference between an ARIMA process, an AR process, and a MA process, not to mention seasonal version of the above.

Using \(B\) as the lag operator, i.e. \(BX_t = X_{t-1}\), an ARIMA(p, d, q) process is a discrete time stochastic process of the form

\[\phi(B) (1 - B)^d X_t = \theta(B)w_t,\]

where \(\phi\) is a polynomial of degree p, and \(\theta\) is a polynomial of degree q. An AR(p) process is an ARIMA(p, 0, 0) process, and a MA(q) process is an ARIMA(0, 0, q) process. To make life even more complicated, we introduce the notion of seasonality:

An ARIMA\((p, d, q) \times (P, D, Q)_s\) model is a s.p. of the form

\[\Phi(B^s) \phi(B) (1 - B^s)^D (1 - B)^d X_t = \Theta(B^s)\theta(B)w_t,\]

where \(\Phi(B)\) is a polynomial of degree \(P\), and \(\Theta(B)\) is a polynomial of degree \(Q\).

Example

Suppose we have the stochastic process

\[X_t = \frac 1 2 X_{t-1} + X_{t-4} - \frac 1 2 X_{t-5} + w_t - \frac 1 4 w_{t-4}.\]

How can we write this as an ARIMA\((p, d, q) \times (P, D, Q)_s\) model? Note that

\[(1 - B^4) X_t = \frac 1 2 X_{t-1} - \frac 1 2 X_{t-5} + w_t - \frac 1 4 w_{t-4}.\]

We can rewrite this as

\[(1 - B^4) X_t - \frac{1}{2} B (1 - B^4)X_t = (1 - \frac 1 4 B^4) w_t,\]

or, more concisely,

\[(1 - B^4) (1 - \frac 1 2 B) X_t = (1 - \frac 1 4 B^4) w_t.\]

Consequently, we can see that \(X_t\) is an ARIMA\((1, 0, 0) \times (0, 1, 1)_4\) process.