SARIMAX

257. SARIMAX

Seasonal ARIMA with exogenous regressors

$SARIMAX (𝑝, 𝑑, 𝑞) {(𝑃, 𝐷, 𝑄)}_{𝑚}$

𝜑 (𝐵) Φ (𝐵^{𝑚}) {(1 - 𝐵)}^{𝑑} {(1 - 𝐵^{𝑚})}^{𝐷} 𝑦_{𝑡} = 𝑐 + \sum_{𝑖 = 1}^{𝑘} 𝛽_{𝑖} 𝑥_{𝑖, 𝑡} + 𝜃 (𝐵) Θ (𝐵^{𝑚}) 𝜀_{𝑡}

$𝑦_{𝑡}$ = endogenous (target) series
$𝑥_{𝑖, 𝑡}$ = exogenous regressors, $𝑖 = 1, \dots, 𝑘$
$𝛽_{𝑖}$ = regression coefficients on exogenous variables
The $𝑋$ adds linear regression on external predictors to SARIMA.

Parameters: $𝜑_{1}, \dots, 𝜑_{𝑝}$ , $Φ_{1}, \dots, Φ_{𝑃}$ , $𝜃_{1}, \dots, 𝜃_{𝑞}$ , $Θ_{1}, \dots, Θ_{𝑄}$ , $𝛽_{1}, \dots, 𝛽_{𝑘}$ , $𝑐$ , $𝜎^{2}$
Orders: $𝑝$ , $𝑃$ , $𝑑$ , $𝐷$ , $𝑞$ , $𝑄$ , $𝑚$ , $𝑘$ (regressors)

Example:

SARIMAX (1, 0, 0) {(1, 0, 0)}_{4} + 1 exog

Given

Orders: $𝑝 = 1$ , $𝑃 = 1$ , $𝑑 = 0$ , $𝐷 = 0$ , $𝑞 = 0$ , $𝑄 = 0$ , $𝑚 = 4$ , $𝑘 = 1$
Parameters: $𝜑_{1} = 0.5$ , $Φ_{1} = 0.5$ , $𝛽_{1} = 0.5$ , $𝑐 = 0$
Endogenous data $𝑦_{𝑡} = 𝑥_{𝑡}$ :

$𝑡$	1	2	3	4	5	6	7	8	9	10	11	12	13	14	15	16
$y_{𝑡}$	12	10	8	11	14	12	9	13	16	14	11	15	18	16	13	17

Exogenous regressor $𝑧_{𝑡}$ :

$𝑡$	1	2	3	4	5	6	7	8	9	10	11	12	13	14	15	16
$z_{𝑡}$	1	2	3	4	5	6	7	8	9	10	11	12	13	14	15	16

Step 1 — formula

Substitute orders into the SARIMAX recursion. With $𝑑 = 𝐷 = 0$ and $𝑞 = 𝑄 = 0$ :

(1 - 𝜑_{1} 𝐵) (1 - Φ_{1} 𝐵^{𝑚}) 𝑦_{𝑡} = 𝑐 + 𝛽_{1} 𝑧_{𝑡} + 𝜀_{𝑡}

Expand the operator product (same as SARIMA, plus the exogenous term):

𝑦_{𝑡} - 𝜑_{1} 𝑦_{𝑡 - 1} - Φ_{1} 𝑦_{𝑡 - 𝑚} + 𝜑_{1} Φ_{1} 𝑦_{𝑡 - 𝑚 - 1} = 𝑐 + 𝛽_{1} 𝑧_{𝑡} + 𝜀_{𝑡}

Forecast (set $𝜀_{𝑡} = 0$ ):

{\hat{𝑦}}_{𝑡} = 𝑐 + 𝛽_{1} 𝑧_{𝑡} + 𝜑_{1} 𝑦_{𝑡 - 1} + Φ_{1} 𝑦_{𝑡 - 𝑚} - 𝜑_{1} Φ_{1} 𝑦_{𝑡 - 𝑚 - 1}

Innovation:

𝜀_{𝑡} = 𝑦_{𝑡} - {\hat{𝑦}}_{𝑡}

Pre-compute the cross term: $𝜑_{1} Φ_{1} = 0.5 \cdot 0.5 = 0.25$ .

Step 2 — apply at $𝑡 = 6$ (first usable step: needs $𝑦_{𝑡 - 𝑚 - 1} = 𝑦_{1}$ )

Plug in $𝛽_{1} = 0.5$ , $𝜑_{1} = 0.5$ , $Φ_{1} = 0.5$ , $𝑧_{6} = 6$ , $𝑦_{5} = 14$ , $𝑦_{2} = 10$ , $𝑦_{1} = 12$ :

{\hat{𝑦}}_{6} = 0 + 0.5 \cdot 6 + 0.5 (14) + 0.5 (10) - 0.25 (12)

{\hat{𝑦}}_{6} = 3 + 7 + 5 - 3 = 12

𝜀_{6} = 𝑦_{6} - {\hat{𝑦}}_{6} = 12 - 12 = 0

Step 3 — iterate

Each row adds the exogenous contribution $𝛽_{1} 𝑧_{𝑡}$ on top of the SARIMA forecast.

$𝑡$	$𝑦_{𝑡}$	$𝛽_{1} 𝑧_{𝑡}$	SARIMA part: $0.5 𝑦_{𝑡 - 1} + 0.5 𝑦_{𝑡 - 4} - 0.25 𝑦_{𝑡 - 5}$	${\hat{𝑦}}_{𝑡}$	$𝜀_{𝑡}$
6	12	$0.5 (6) = 3$	$0.5 (14) + 0.5 (10) - 0.25 (12) = 9$	$3 + 9 = 12$	$0$
7	9	$0.5 (7) = 3.5$	$0.5 (12) + 0.5 (8) - 0.25 (10) = 7.5$	$3.5 + 7.5 = 11$	$- 2$
8	13	$0.5 (8) = 4$	$0.5 (9) + 0.5 (11) - 0.25 (8) = 8$	$4 + 8 = 12$	$1$
9	16	$0.5 (9) = 4.5$	$0.5 (13) + 0.5 (14) - 0.25 (11) = 10.75$	$4.5 + 10.75 = 15.25$	$0.75$
10	14	$0.5 (10) = 5$	$0.5 (16) + 0.5 (12) - 0.25 (14) = 10.5$	$5 + 10.5 = 15.5$	$- 1.5$
11	11	$0.5 (11) = 5.5$	$0.5 (14) + 0.5 (9) - 0.25 (12) = 8.5$	$5.5 + 8.5 = 14$	$- 3$
12	15	$0.5 (12) = 6$	$0.5 (11) + 0.5 (13) - 0.25 (9) = 9.75$	$6 + 9.75 = 15.75$	$- 0.75$
13	18	$0.5 (13) = 6.5$	$0.5 (15) + 0.5 (16) - 0.25 (13) = 12.25$	$6.5 + 12.25 = 18.75$	$- 0.75$
14	16	$0.5 (14) = 7$	$0.5 (18) + 0.5 (14) - 0.25 (16) = 12$	$7 + 12 = 19$	$- 3$
15	13	$0.5 (15) = 7.5$	$0.5 (16) + 0.5 (11) - 0.25 (14) = 10$	$7.5 + 10 = 17.5$	$- 4.5$
16	17	$0.5 (16) = 8$	$0.5 (13) + 0.5 (15) - 0.25 (11) = 11.25$	$8 + 11.25 = 19.25$	$- 2.25$
17	—	$0.5 (17) = 8.5$	$0.5 (17) + 0.5 (18) - 0.25 (15) = 13.75$	$8.5 + 13.75 = 22.25$	—

Here $𝑧_{𝑡} = 𝑡$ is collinear with $𝑦$ ’s linear trend, so $𝛽_{1} 𝑧_{𝑡}$ adds an extra drift on top of the AR/seasonal structure — forecasts now overshoot. With a less collinear regressor, $𝛽_{1}$ would correct part of $𝑦$ that AR/seasonal lags miss.