SARIMA

270. SARIMA

Seasonal ARIMA

$SARIMA (𝑝, 𝑑, 𝑞) {(𝑃, 𝐷, 𝑄)}_{𝑚}$

𝜑 (𝐵) Φ (𝐵^{𝑚}) {(1 - 𝐵)}^{𝑑} {(1 - 𝐵^{𝑚})}^{𝐷} 𝑥_{𝑡} = 𝑐 + 𝜃 (𝐵) Θ (𝐵^{𝑚}) 𝜀_{𝑡}

Φ (𝐵^{𝑚}) = 1 - Φ_{1} 𝐵^{𝑚} - \dots - Φ_{𝑃} 𝐵^{𝑃 𝑚}

Θ (𝐵^{𝑚}) = 1 + Θ_{1} 𝐵^{𝑚} + \dots + Θ_{𝑄} 𝐵^{𝑄 𝑚}

Seasonal differencing $(1 - 𝐵^{𝑚}) 𝑦_{𝑡} = 𝑦_{𝑡} - 𝑦_{𝑡 - 𝑚}$ removes annual/periodic patterns. Period $𝑚$ (e.g., 12 for monthly).

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

Example:

SARIMA (1, 0, 0) {(1, 0, 0)}_{4}

Given

Orders: $𝑝 = 1$ , $𝑃 = 1$ , $𝑑 = 0$ , $𝐷 = 0$ , $𝑞 = 0$ , $𝑄 = 0$ , $𝑚 = 4$
Parameters: $𝜑_{1} = 0.5$ , $Φ_{1} = 0.5$ , $𝑐 = 0$
Data:

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

Step 1 — formula

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

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

Expand the operator product:

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

𝑥_{𝑡} - 𝜑_{1} 𝑥_{𝑡 - 1} - Φ_{1} 𝑥_{𝑡 - 𝑚} + 𝜑_{1} Φ_{1} 𝑥_{𝑡 - 𝑚 - 1} = 𝑐 + 𝜀_{𝑡}

Forecast (set $𝜀_{𝑡} = 0$ ):

{\hat{𝑥}}_{𝑡} = 𝑐 + 𝜑_{1} 𝑥_{𝑡 - 1} + Φ_{1} 𝑥_{𝑡 - 𝑚} - 𝜑_{1} Φ_{1} 𝑥_{𝑡 - 𝑚 - 1}

Innovation:

𝜀_{𝑡} = 𝑥_{𝑡} - {\hat{𝑥}}_{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 $𝑥_{𝑡 - 𝑚 - 1} = 𝑥_{1}$ )

Plug in $𝜑_{1} = 0.5$ , $Φ_{1} = 0.5$ , $𝑥_{5} = 14$ , $𝑥_{2} = 10$ , $𝑥_{1} = 12$ :

{\hat{𝑥}}_{6} = 0.5 \cdot 14 + 0.5 \cdot 10 - 0.25 \cdot 12 = 7 + 5 - 3 = 9

𝜀_{6} = 𝑥_{6} - {\hat{𝑥}}_{6} = 12 - 9 = 3

Step 3 — iterate

Each row uses lag-1 ( $𝑥_{𝑡 - 1}$ ), lag- $𝑚$ ( $𝑥_{𝑡 - 4}$ ), and lag- $(𝑚 + 1)$ ( $𝑥_{𝑡 - 5}$ ).

$𝑡$	$𝑥_{𝑡}$	$𝑥_{𝑡 - 1}$ , $𝑥_{𝑡 - 4}$ , $𝑥_{𝑡 - 5}$	${\hat{𝑥}}_{𝑡} = 0.5 𝑥_{𝑡 - 1} + 0.5 𝑥_{𝑡 - 4} - 0.25 𝑥_{𝑡 - 5}$	${\hat{𝑥}}_{𝑡}$	$𝜀_{𝑡}$
6	12	$14, 10, 12$	$0.5 (14) + 0.5 (10) - 0.25 (12) = 7 + 5 - 3$	$9$	$3$
7	9	$12, 8, 10$	$0.5 (12) + 0.5 (8) - 0.25 (10) = 6 + 4 - 2.5$	$7.5$	$1.5$
8	13	$9, 11, 8$	$0.5 (9) + 0.5 (11) - 0.25 (8) = 4.5 + 5.5 - 2$	$8$	$5$
9	16	$13, 14, 11$	$0.5 (13) + 0.5 (14) - 0.25 (11) = 6.5 + 7 - 2.75$	$10.75$	$5.25$
10	14	$16, 12, 14$	$0.5 (16) + 0.5 (12) - 0.25 (14) = 8 + 6 - 3.5$	$10.5$	$3.5$
11	11	$14, 9, 12$	$0.5 (14) + 0.5 (9) - 0.25 (12) = 7 + 4.5 - 3$	$8.5$	$2.5$
12	15	$11, 13, 9$	$0.5 (11) + 0.5 (13) - 0.25 (9) = 5.5 + 6.5 - 2.25$	$9.75$	$5.25$
13	18	$15, 16, 13$	$0.5 (15) + 0.5 (16) - 0.25 (13) = 7.5 + 8 - 3.25$	$12.25$	$5.75$
14	16	$18, 14, 16$	$0.5 (18) + 0.5 (14) - 0.25 (16) = 9 + 7 - 4$	$12$	$4$
15	13	$16, 11, 14$	$0.5 (16) + 0.5 (11) - 0.25 (14) = 8 + 5.5 - 3.5$	$10$	$3$
16	17	$13, 15, 11$	$0.5 (13) + 0.5 (15) - 0.25 (11) = 6.5 + 7.5 - 2.75$	$11.25$	$5.75$
17	—	$17, 18, 15$	$0.5 (17) + 0.5 (18) - 0.25 (15) = 8.5 + 9 - 3.75$	$13.75$	—

Forecasts now sit much closer to the data than AR(1) — the seasonal lag $𝑥_{𝑡 - 4}$ tracks the within-cycle pattern, and lag-1 picks up the trend.