Department of Statistical Sciences
Time Series Analysis —– STA457H1/STA2202H
Fall 2020
Quiz 6
Due: Oct. 28th at 10.00 am EST
Q1. For the AR(2) series shown below, use the results of Example 3.10 to determine a
set of difference equations that can be used to find the ACF ρ(h), h = 0, 1, . . .; solve
for the constants in the ACF using the initial conditions. Then calculate the ACF
values to lag 2.
(a) xt − 1.1xt−1 + .3xt−2 = wt.
Answer (8 points):
> z1 < −c(1,−1.1, 0.3)
> (roots1 < −polyroot(z1))
#[1]1.666667 + 0i 2.000000− 0i
Both roots are outside the unit circle, so we know the model is causal. When
z1 and z2 are real and distinct, then
ρ(h) = c1z
−h
1 + c2z
−h
2
= c1(1.6667)
−h + c2(2)−h
ρ(0) = c1 + c2 = 1 and ρ(1) =
φ1
(1− φ2) =
1.1
1 + 0.3
= 0.8462.
⇒ 0.8462 = c1(1.6667)−1 + c2(2)−1
⇒ c1 = −32
13
and c2 =
45
13
.
Therefore the ACF is ρ(h) =
45
13
· (1.6667)−h − 32
13
· (2)−h.
We have ρ(1) = 0.8461123 and ρ(2) = 0.6307194.
1
(b) xt − xt−1 + 0.25xt−2 = wt
Answer (2 points):
> z2 < −c(1,−1, 0.25)
> (roots2 < −polyroot(z2))
#[1]2 + 0i 2− 0i
Both roots are outside the unit circle, so we know the model is causal. Whenz1 =
z2 = z0 are real and equal, then
γ(h) = z−h0 (c1 + c2h)
= (2)−h(c1 + c2h)
ρ(0) = c1 = 2
ρ(1) =
φ1
(1− φ2) =
1
1 + 0.25
= 0.80
⇒ 0.80 = 2−1(1 + c2)⇒ c2 = 0.60
Therefore the ACF is ρ(h) = (2)−h(1 + 0.60 h).
We have ρ(1) = 0.80 and ρ(2) = 0.55.
Q2. The actual forecasts are calculated as
xnn+m = 0.2x
n
n+m−1 − 0.2xnn+m−2 + 0.6xnn+m−3,
for n = 10 and m = 1, 2, 3, 4. We wish to forecast using the following values:
x1 = 3.1, x2 = 0.93, x3 = −2.3, x4 = −3.2, x5 = −1.3
x6 = −2.5, x7 = 1.6, x8 = 3.0, x9 = −3.5, x10 = −0.74.
Use the best linear predictor to compute the forecast m = 4. If σ2w = 1 and ψ0 = 1,
give a 95% prediction interval for the value xnn+4. Assume z0.025 = −1.96 and
z0.975 = 1.96.
Answer (6 points):
xn11 = 0.2x
n
10 − 0.2xn9 + 0.6xn8 = 0.2(−0.74)− 0.2(−3.5) + 0.6(3) = 2.352
xn12 = 0.2x
n
11 − 0.2xn10 + 0.6xn9 = 0.2(2.352)− 0.2(−0.74) + 0.6(−3.5) = −1.4816
xn13 = 0.2x
n
12−0.2xn11 +0.6xn10 = 0.2(−1.4816)−0.2(2.352)+0.6(−0.74) = −1.21072
xn14 = 0.2x
n
13 − 0.2xn12 + 0.6xn11 = 0.2(−1.21072) − 0.2(−1.4816) + 0.6(2.352) =
1.465376.
2
Now we need to compute P nn+4 mean square prediction error to find the predic-
tion interval xn14 ± 1.96
√
P nn+4.
We know that P nn+m = σ
2
w
∑m−1
j=0 ψ
2
j with initial conditions ψk =
∑k
j=1 φjψk−j for
k < p = 3. When k ≥ p, we have ψk =
∑3
j=1 φjψk−j, where φ1 = 0.2, φ2 = −0.2,
φ3 = 0.6 are the coefficients to the given autoregressive polynomial.
ψ0 = 1
ψ1 = φ1ψ0 = 0.20
ψ2 = φ1ψ1 + φ2ψ0 = 0.04− 0.20 = −0.16
ψ3 = φ1ψ2 + φ2ψ1 + φ3ψ0 == .528
Now, we can compute P nn+m = σ
2
w
∑m−1
j=0 ψ
2
j
= ψ20 + ψ
2
1 + ψ
2
2 + ψ
2
3 = 1.344384
Now the prediction interval is: xn14 ± 1.96
√
P nn+4
= 1.465376 ± 1.96√1.344384
= (−0.8072, 3.7379)
3

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