Interpret ARIMA models in plain english

Question

What is the proper plain-English interpretation of these models (unit of measure is month)?

1. $\text{ARIMA}(4,1,1)$
2. $\text{ARIMA}(3,1,3)(1,0,2)_{12}$
3. $\text{ARIMA}(0,1,1)$
4. $\text{ARIMA}(1,1,0)$

My attempt:

1. The value of the current month depends on the values of the last four months along with the prediction error of the last month.
2. (not sure)
3. The value of the current month depends only on the prediction error of the last month
4. The value of the current month depends only on the value from the last month

Answer 1

1, 3 and 4 can be interpreted in plain English. 2... I'm unsure about.

Note first that all your ARIMA processes are integrated of order 1 (that's the middle 1). We therefore are not discussing each month's value itself, but its increment over the previous month's value.

We will translate your ARIMA processes into formulas, then into plain English. For the first step, we use the backshift operator $B$:

$$ By_t = y_{t-1} \quad\text{(and }Be_t=e_{t-1}).$$

And we will use an abbreviation:

$$ \Delta y_t := (1-B)y_t = y_t-y_{t-1}.$$

Note that

$$ B\Delta y_t = B(1-B)y_t = B(y_t-y_{t-1}) = y_{t-1}-y_{t-2}=\Delta y_{t-1}.$$

I'll go through your processes not in the order you gave them, but in the order in which they are easiest to understand. For 1, 3 and 4, I recommend section 8.5 of Hyndman & Athanasopoulos, Forecasting: Principles and Practice.

4. $\text{ARIMA}(1,1,0)$: we have

   $$ (1-\phi_1B)(1-B)y_t = c+e_t $$

   or

   $$ \Delta y_t = c + \phi_1 B\Delta y_t + e_t. $$

   Thus, the increment month-over-month $\Delta y_t$ depends on the previous increment $\phi_1 B\Delta y_t=\phi_1\Delta y_{t-1}$, plus a constant $c$ and an innovation $e_t$.

   3. $\text{ARIMA}(0,1,1)$: we have

   $$ (1-B)y_t = c+(1-\theta_1B)e_t $$

   or

   $$ \Delta y_t = c + e_t + \theta_1 Be_t. $$

   Thus, the increment depends only on the current innovation $e_t$ and the previous innovation $\theta_1Be_t = \theta_1e_{t-1}$, plus a constant $c$.

   1. $\text{ARIMA}(4,1,1)$: we have

   $$ (1-\phi_1B^1-\phi_2B^2-\phi_3B^3-\phi_4B^4)(1-B)y_t = c + (1-\theta_1B)e_t $$

   or

   $$ \Delta y_t = c + \phi_1\Delta y_{t-1} + \phi_2\Delta y_{t-2} + \phi_3\Delta y_{t-3} + \phi_4\Delta y_{t-4} + e_t+\theta_1 e_{t-1}. $$

   Thus, the increment depends on the previous four increments, plus the innovation and the previous innovation. And a constant.

Your last ARIMA process is seasonal. Refer to section 8.9 of the book, on seasonal ARIMA models.

2. $\text{ARIMA}(3,1,3)(1,0,2)_{12}$: we have

   $$ (1-\phi_1B-\phi_2B^2-\phi_3B^3)(1-\Phi_1B^{12})(1-B)y_t = (1+\theta_1B+\theta_2B^2+\theta_3B^3)(1+\Theta_1B^{12}+\Theta_2B^{24})e_t. $$

   The seasonal parts here are the ones with lags of order 12 or 24. If you want to "simplify" this, go ahead. I don't think you'll find a plain English interpretation lurking anywhere here.

If you have read this far, you may wonder whether I didn't forget a constant in the last process. No, I didn't. There are different conventions on whether ARIMA processes include a constant or not. For instance, section 8.5 of the book assumes a constant, while section 8.9 doesn't. So it's always a good idea to indicate whether your process includes a constant or not. (And note that a constant in an integrated process of course means that the increment has a non-zero constant $c$ - i.e., that the original series has a trend of slope $c$.)