An autoregressive series is of the form

$$ X_t = c + \varphi_1 X_{t-1} + \varphi_2 X_{t-2} + \cdots + \varepsilon_t $$

An integrated series is of the form

$$ (1-B)^d X_t = \varepsilon_t $$

However, it seems to me that this is just a special case of the autoregressive series with $\phi_k = (-1)^k\binom{n}{k}$ and $c=0$. Is this really the case? If so, why is worth naming?

For context: I can only find any mention of integrated models in ARIMA models, so I was wondering why they appear there but nowhere else.


The 'I' in ARIMA stands for integrated!

And your second equation isn't exactly correct. Suppose we have a process $ X_t $, and we define the following:

$$ Y_t := (1 - B)^d X_t $$

where $ B $ is the backshift operator.

If $ Y_t $ is an ARMA(p, q) process, then $ X_t $ is said to be an ARIMA(p, d, q) process.

It's not exactly a "special case" of AR because $ Y_t $ is required to be ARMA.

  • $\begingroup$ How is my second equation incorrect? $\endgroup$ – Frank Vel Jul 19 '18 at 20:32
  • $\begingroup$ It suggests that what remains after differencing $ X_t $ is white noise, which is not necessarily true if $ X_t $ is ARIMA. $\endgroup$ – Kevin Li Jul 19 '18 at 20:35
  • $\begingroup$ It's supposed to be the integrated series, which I thought would be ARIMA(0,d,0)? $\endgroup$ – Frank Vel Jul 19 '18 at 20:56
  • $\begingroup$ No, in your equation the integrated series is $ X_t $. After differencing, it should become a stationary ARMA(p, q) process, of general p and q. $ \epsilon_t $ is generally used to represent a white noise series; here, it should be a general ARMA(p, q) series. $\endgroup$ – Kevin Li Jul 20 '18 at 14:05
  • $\begingroup$ Isn't I(d) = ARIMA(0,d,0)? How would you write an integrated series? $\endgroup$ – Frank Vel Jul 20 '18 at 21:26

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