ACF & PACF of AR (2) Model Lecture 16

 

ACF & PACF of AR (2) Model

Lecture 16

Consider an AR(2) model.

yt = β0 + β1 yt-1 + β2 yt-2 + ϵt

ignoring intercept

yt = β1 yt-1 + β2 yt-2 + ϵt

We use Yule-Walker equations

 Multiply the AR (2) by its immediate lag yt-1, and take expectations and divide by the variance of yt.

yt x yt-1 = β1 yt-1 x yt-1 + β2 yt-2 x yt-1 + ϵt x yt-1

E (yt x yt-1) = β1E (yt-1 x yt-1) + β2 E (yt-2 x yt-1) + E (ϵt x yt-1)

Cov (yt , yt-1) = β1Var (yt-1) + β2 Cov (yt-2, yt-1) 

Now divide by the variance of yt

Cov (yt , yt-1)/ Var(yt) = β1Var (yt-1) / Var(yt) + β2 Cov (yt-2 x yt-1) / Var(yt)

 ${\rho }_{\mathrm{1\; = }}$β1 + ρ1β2

ρ1-ρ1β2 ${\mathrm{= }}_{}$β1 

ρ1(1- β2) ${\mathrm{= }}_{}$β1 

ρ1 ${\mathrm{= }}_{}$β1/(1- β2) 

Now AR(2) model is multiplied by yt-2 and their expectation is taken.

yt x yt-2 = β1 yt-1 x yt-2 + β2 yt-2 x yt-2 + ϵt x yt-2

E (yt x yt-2) = β1E (yt-1 x yt-2) + β2 E (yt-2 x yt-2) + E (ϵt x yt-2)

Cov (yt, yt-2) = β1Cov (yt-1, yt-2) + β2 Var (yt-2) 

Now divide by the variance of yt

Cov (yt, yt-2) / Var(yt) = β1 Cov (yt-1, yt-2) / Var(yt) + β2 Var(yt-2) / Var(yt)

We know that Var(yt) = Var(yt-2)

Cov (yt, yt-2) / Var(yt) = β1 Cov (yt-1, yt-2) / Var(yt) + β2 

ρ₂ = ρ₁β₁ + β₂

As ρ1 ${\mathrm{= }}_{}$β1/(1- β2) 

ρ₂ = β₁β₁/(1-β₂) + β₂

Example: Consider the AR (2) model

yt = 0.75 yt-1 - 0.25 yt-2 + ut

Find the first- and second-order ACFs.

Solution: From the model

β₁ = 0.75,  β₂ =  -0.25

ρ1 ${\mathrm{= }}_{}$β1/(1- β2) 

ρ1 ${\mathrm{=\; 0.75}}_{}$/(1+ 0.25) 

ρ1 ${\mathrm{=\; 0.75}}_{}$/1.25

 ρ1 ${\mathrm{=\; 0.60}}_{}$

${}_{}$

ACF & PACF of AR (p) model

Consider the AR(p) model

yt = β1 yt-1 + β2 yt-2 + β3 yt-3 + ... + βp yt-p + ϵt

Using the Yule–Walker equation

${}_{}$

The AR(p) model is multiplied by yt-j, take the expectation and divided by Var(yt).

yt = β1 yt-1 + β2 yt-2 + β3 yt-3 + ... + βp yt-p + ϵt

yt x yt-j = β1 yt-1 x yt-j + β2 yt-2 x yt-j + β3 yt-3 x yt-j + ... + βp yt-p x yt-j + ϵt x yt-j

E (yt x yt-j) = β1 E (yt-1 x yt-j) + β2 E (yt-2 x yt-j) + β3 E (yt-3 x yt-j) + ... + βp E (yt-p x yt-j) + E (ϵt x yt-j)

Cov(yt, yt-j) = β1 Cov(yt-1, yt-j) + β2 Cov (yt-2, yt-j)   β3 Cov(yt-3, yt-j) + ... + βp Cov(yt-p, yt-j) 

j = 1, 2, 3, ...

Using Cramer's rule to obtain the estimates of β's.

The first, second order of ACF

For third-order PACF

Example

Find the ACF and PACF using SPSS. 

Month – year	Covid-19Cases  (per million)
Mar-20 Apr-20 May-20 Jun-20 Jul-20 Aug-20 Sep-20 Oct-20 Nov-20 Dec-20 Jan-21 Feb-21 Mar-21 Apr-21 May-21 Jun-21 Jul-21 Aug-21 Sep-21 Oct-21 Nov-21 Dec-21	0.001938 0.015525 0.069496 0.209337 0.278305 0.295636 0.312263 0.332993 0.398026 0.479775 0.544813 0.579973 0.667957 0.820823 0.918936 0.957371 1.029811 1.160119 1.245127 1.272345 1.28484 1.292728

Solution:

The ACF and PACF of Covid 19 spread from March, 2020 to December 2021.

Autocorrelations
Series: VAR00001
Lag	Autocorrelation	Std. Errora	Box-Ljung Statistic
			Value	df	Sig.b
1	.871	.203	18.329	1	.000
2	.721	.198	31.541	2	.000
3	.566	.193	40.139	3	.000
4	.432	.188	45.442	4	.000
5	.313	.182	48.402	5	.000
6	.191	.176	49.580	6	.000
7	.064	.170	49.720	7	.000
8	-.055	.164	49.833	8	.000
9	-.147	.158	50.704	9	.000
10	-.222	.151	52.870	10	.000
11	-.292	.144	57.000	11	.000
12	-.354	.137	63.733	12	.000
13	-.392	.129	73.033	13	.000
14	-.400	.120	84.075	14	.000
15	-.395	.111	96.659	15	.000
16	-.387	.102	111.145	16	.000

 

Partial Autocorrelations
Series: VAR00001
Lag	Partial Autocorrelation	Std. Error
1	.871	.218
2	-.158	.218
3	-.100	.218
4	-.007	.218
5	-.042	.218
6	-.117	.218
7	-.122	.218
8	-.070	.218
9	-.012	.218
10	-.065	.218
11	-.107	.218
12	-.072	.218
13	-.010	.218
14	.008	.218
15	-.065	.218
16	-.071	.218

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