AR

252. AR

Autoregressive

AR (𝑝)

𝜑 (𝐵) 𝑥_{𝑡} = 𝑐 + 𝜀_{𝑡}

𝜑 (𝐵) = 1 - 𝜑_{1} 𝐵 - 𝜑_{2} 𝐵^{2} - \dots - 𝜑_{𝑝} 𝐵^{𝑝}

i.e.,

𝑥_{𝑡} = 𝑐 + 𝜑_{1} 𝑥_{𝑡 - 1} + \dots + 𝜑_{𝑝} 𝑥_{𝑡 - 𝑝} + 𝜀_{𝑡}

Stationary if all roots of $𝜑 (𝐵) = 0$ lie outside the unit circle

Parameters: $𝜑_{1}, \dots, 𝜑_{𝑝}$ , $𝑐$ , $𝜎^{2}$
Orders: $𝑝$

Example:

AR (1)

Given

Order: $𝑝 = 1$
Parameters: $𝜑_{1} = 0.5$ , $𝑐 = 0$
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 $𝑝 = 1$ into the AR $(𝑝)$ recursion:

𝑥_{𝑡} = 𝑐 + 𝜑_{1} 𝑥_{𝑡 - 1} + 𝜀_{𝑡}

Forecast (set $𝜀_{𝑡} = 0$ ):

{\hat{𝑥}}_{𝑡} = 𝑐 + 𝜑_{1} 𝑥_{𝑡 - 1}

Innovation:

𝜀_{𝑡} = 𝑥_{𝑡} - {\hat{𝑥}}_{𝑡}

Step 2 — apply at $𝑡 = 2$

Plug in $𝑐 = 0$ , $𝜑_{1} = 0.5$ , and $𝑥_{1} = 12$ :

{\hat{𝑥}}_{2} = 0 + 0.5 \cdot 12 = 6

𝜀_{2} = 𝑥_{2} - {\hat{𝑥}}_{2} = 10 - 6 = 4

Step 3 — iterate

$𝑡$	$𝑥_{𝑡}$	${\hat{𝑥}}_{𝑡} = 0.5 \cdot 𝑥_{𝑡 - 1}$	$𝜀_{𝑡} = 𝑥_{𝑡} - {\hat{𝑥}}_{𝑡}$
2	10	$0.5 \cdot 12 = 6$	$10 - 6 = 4$
3	8	$0.5 \cdot 10 = 5$	$8 - 5 = 3$
4	11	$0.5 \cdot 8 = 4$	$11 - 4 = 7$
5	14	$0.5 \cdot 11 = 5.5$	$14 - 5.5 = 8.5$
6	12	$0.5 \cdot 14 = 7$	$12 - 7 = 5$
7	9	$0.5 \cdot 12 = 6$	$9 - 6 = 3$
8	13	$0.5 \cdot 9 = 4.5$	$13 - 4.5 = 8.5$
9	16	$0.5 \cdot 13 = 6.5$	$16 - 6.5 = 9.5$
10	14	$0.5 \cdot 16 = 8$	$14 - 8 = 6$
11	11	$0.5 \cdot 14 = 7$	$11 - 7 = 4$
12	15	$0.5 \cdot 11 = 5.5$	$15 - 5.5 = 9.5$
13	18	$0.5 \cdot 15 = 7.5$	$18 - 7.5 = 10.5$
14	16	$0.5 \cdot 18 = 9$	$16 - 9 = 7$
15	13	$0.5 \cdot 16 = 8$	$13 - 8 = 5$
16	17	$0.5 \cdot 13 = 6.5$	$17 - 6.5 = 10.5$
17	—	$0.5 \cdot 17 = 8.5$	—

Residuals are large because AR(1) cannot capture the trend or seasonality in $𝑥$ — that is the role of ARIMA / SARIMA / ETS, applied to the same data in their own examples.