# Autoregressive vs integrated time series model

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.

• How is my second equation incorrect? – Frank Vel Jul 19 '18 at 20:32
• It suggests that what remains after differencing $X_t$ is white noise, which is not necessarily true if $X_t$ is ARIMA. – Kevin Li Jul 19 '18 at 20:35
• It's supposed to be the integrated series, which I thought would be ARIMA(0,d,0)? – Frank Vel Jul 19 '18 at 20:56
• 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. – Kevin Li Jul 20 '18 at 14:05
• Isn't I(d) = ARIMA(0,d,0)? How would you write an integrated series? – Frank Vel Jul 20 '18 at 21:26