0
$\begingroup$

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.

$\endgroup$
2
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.