Solution
STAT 758: Homework #6
Due on Wednesday, 11 April, 2012
Zaliapin, 1:00pm
Tracy Backes
1
Tracy Backes
STAT 758 (Zaliapin): HW #6
Problem #1
We assume below that Zt ∼ W N (0, σ 2 ), B is a backshift operator. 6.1 For the model (1 − B)(1 − 0.2B)Xt = (1 − 0.5B)Zt : a) Classify the model as an ARIMA(p, d, q) process (i.e. find p, d, q). ARIMA(1,1,1) b) Determine whether the process is stationary, causal, invertible. • The process is stationary if all roots of ϕ(z) are off of the unit circle. ϕ(z) = (1 − z)(1 − 0.2z) = 0 =⇒ z = 1, 5
Because ϕ(z) has root z = 1, which lies on the unit circle, the process is not stationary . • The process is causal if all roots of ϕ(z) are outside of the unit circle. Because ϕ(z) has one root z = 1, which lies on the unit circle, the process is not causal . • The process is invertible if all roots of θ(z) are outside of the unit circle. θ(B) = (1 − 0.5z) = 0 =⇒ B=z
Because the only root of θ(z), z = 2, lies outside the unit circle, the process is invertible . c) Evaluate the first three ψ-weights of the model when expressed as an AR(∞) model. ∞ Xt can be expressed as an AR(∞) process of the form Xt = k=0 ψk Zt−k Xt = θ(B) Zt = (1 − 0.5B) ϕ(B)
∞ ∞
(1.25 − 0.25(0.2)k )B k k=0 Zt
= Zt + k=1 0.625 + 0.075(0.2)k−1 Zt−k
= Zt + 0.7Zt−1 + 0.64Zt−2 + 0.628Zt−3 + · · · d) Evaluate the first four π-weights of the model when expressed as an MA(∞) model. ∞ Zt can be expressed as a MA(∞) process of the form Zt = k=0 πk Xt−k Zt = ϕ(B) Xt = (1 − 1.2B + 0.2B 2 ) θ(B)
∞ ∞
0.5k B k k=0 Xt
= Xt − 0.7Xt−1 + k=2 −0.15(0.5)k−2 Xt−k
= Xt − 0.7Xt−1 − 0.15Xt−2 − 0.075Xt−3 − 0.0375Xt−4 + · · · Discuss how the behavior of the weights is related to the properties of the model found in (b). ∞ An ARMA model is only causal if k=0 |ψk | < ∞, where ψk are the coefficients from the AR(∞) process found in (c). Note that in out case, this infinite sum is given by
∞
1+ k=1 |0.625 + 0.075(0.2)k−1 |
which does
Due on Wednesday, 11 April, 2012
Zaliapin, 1:00pm
Tracy Backes
1
Tracy Backes
STAT 758 (Zaliapin): HW #6
Problem #1
We assume below that Zt ∼ W N (0, σ 2 ), B is a backshift operator. 6.1 For the model (1 − B)(1 − 0.2B)Xt = (1 − 0.5B)Zt : a) Classify the model as an ARIMA(p, d, q) process (i.e. find p, d, q). ARIMA(1,1,1) b) Determine whether the process is stationary, causal, invertible. • The process is stationary if all roots of ϕ(z) are off of the unit circle. ϕ(z) = (1 − z)(1 − 0.2z) = 0 =⇒ z = 1, 5
Because ϕ(z) has root z = 1, which lies on the unit circle, the process is not stationary . • The process is causal if all roots of ϕ(z) are outside of the unit circle. Because ϕ(z) has one root z = 1, which lies on the unit circle, the process is not causal . • The process is invertible if all roots of θ(z) are outside of the unit circle. θ(B) = (1 − 0.5z) = 0 =⇒ B=z
Because the only root of θ(z), z = 2, lies outside the unit circle, the process is invertible . c) Evaluate the first three ψ-weights of the model when expressed as an AR(∞) model. ∞ Xt can be expressed as an AR(∞) process of the form Xt = k=0 ψk Zt−k Xt = θ(B) Zt = (1 − 0.5B) ϕ(B)
∞ ∞
(1.25 − 0.25(0.2)k )B k k=0 Zt
= Zt + k=1 0.625 + 0.075(0.2)k−1 Zt−k
= Zt + 0.7Zt−1 + 0.64Zt−2 + 0.628Zt−3 + · · · d) Evaluate the first four π-weights of the model when expressed as an MA(∞) model. ∞ Zt can be expressed as a MA(∞) process of the form Zt = k=0 πk Xt−k Zt = ϕ(B) Xt = (1 − 1.2B + 0.2B 2 ) θ(B)
∞ ∞
0.5k B k k=0 Xt
= Xt − 0.7Xt−1 + k=2 −0.15(0.5)k−2 Xt−k
= Xt − 0.7Xt−1 − 0.15Xt−2 − 0.075Xt−3 − 0.0375Xt−4 + · · · Discuss how the behavior of the weights is related to the properties of the model found in (b). ∞ An ARMA model is only causal if k=0 |ψk | < ∞, where ψk are the coefficients from the AR(∞) process found in (c). Note that in out case, this infinite sum is given by
∞
1+ k=1 |0.625 + 0.075(0.2)k−1 |
which does