Spring 2008 STA 4853/5856 HW 3 Partial Solutions Page: 1 of 2Time Series STA 4953/5856

◦ π(z) from Slide 10 on Feb 6 (2 points) Write π(z) = 1−.9z1+.5z

as an infinite series.

• Using the geometric series for the denominator, we have

π(z) = (1− .9z)∞∑

j=0

(−.5)jzj

= (1− .9z)(1− .5z + .52z2 − .53z3 + · · · )= 1− (.5 + .9)z + (.5(.9) + .52)z2 − (.52(.9) + .53) + · · ·= 1− (.5 + .9)z + .5(.9 + .5)− .52(.9 + .5) + · · ·

= 1− 1.4∞∑

j=1

(−.5)j−1zj

◦ #3.14(a) (4 points) Verify if the following models are stationary, invertible, or both.

(i) (1−B)Zt = (1− 1.5B)at

(ii) (1− .8B)Zt = (1− .5B)at

(iv) (1− .6B)Zt = (1− 1.2B + .2B2)at

• For stationary (resp. invertibility) the roots of φ(B) (resp. θ(B)) should lie outside the unit disk.

(i) φ(B) = 1 − B has a unit root (since φ(1) = 0) and therefore the process is not stationary.The root of θ(B) = 1 − 1.5B is 2/3 which is smaller than one in absolute value, and thereforethis process is not invertible either.

(ii) This part is not graded.

(iv) The polynomial φ(B) = 1− .6B has root 1/.6 = 5/3 which is bigger than one in absolute value,and therefore this process is stationary. The θ(B) polynomial factors as

θ(B) =1

5(B − 1)(B − 5)

and because one of the roots is 1, the process is not invertible.

◦ #3.14(b,c) for (i) Consider the ARMA(1,1) process

(1−B)Zt = (1− 1.5B)at

(a) Express the model in an MA representation if it exists.

(b) Express the model in an AR representation if it exists.

• This process is neither stationary nor invertible, so neither representation exists.

◦ Exponential Smoothing (2 points) Show that the process

Zt = µ+ α∑j<t

aj + at

is ARIMA(0,1,1).

Spring 2008 STA 4853/5856 HW 3 Partial Solutions Page: 2 of 2Time Series STA 4953/5856

• Computing the difference of Zt gives

∇Zt = Zt − Zt−1

= µ+ α∑j<t

+at −

(µ+ α

∑j<t−1

aj + at

)= αat−1 + at − at−1

= at + (α− 1)at−1

which is a MA(1) process. Therefore Zt is ARIMA(0,1,1).

◦ #4.5 (2 points) Consider the simple white noise process, Zt = at. Discuss the consequences by exam-ining the ACF, PACF, and AR representation of the differenced series, Wt = Zt − Zt−1.

• Only the ACF of Wt is graded. The differenced series Zt = at − at−1 is MA(1) and therefore has anACF given by

ρk =

{−12, k = 1

0, k > 0

(with the usual ρ0 = 1 and ρ−k = ρk).