I'll give the lecture note file right away so that you can see the questions. Document Preview: Sta457 Time Series AnalysisHomework 1Jan. 18, 2013Due Jan. 29, 2012 in class You should work out...

{ Xt } is said to be stationary time series if its mean as well as covariance function is independent of the
time point t.
Since {Zt} is a sequence of independent normal random variables so,
( ) cov(Zt,Zt+h) = 0 for all h>0
= for h=0
This implies, E(ZtZs) = 0 for all t ≠ s.
a) E(Xt) = a, since E(Zt) = 0
Cov(Xt,Xt+h) = E(Xt-E(Xt))(Xt+h- E(Xt+h))
= E( bZt+cZt-2)(bZt+h+cZt+h-2)
=E(b2 ZtZt+h +bc Zt Zt+h-2 +bc Zt+h Zt-2 +c
2 Zt-2 Zt+h-2)
= b2 ( ) +bc ( )+bc ( )+c
2 ( )
= (b2+c2 ) if h=0
bc if h= 2 or -2
0 otherwise
So we can see the mean and the covariance function is independent of t thus it is a stationary
process.
b) Here E(Xt) = 0
Cov(Xt,Xt+h) = E(XtXt+h) = E(Z1cos(ct)+Z2sin(ct)) (Z1cos(c(t+h))+Z2sin(c(t+h)))
= E(Z1
2cos(ct) cos c(t+h) + Z1Z2 (sin(ct)cos c(t+h)+cos(ct) sin c(t+h)) + Z2
2sin(ct) sin c(t+h))
= for h=0
(cos(ct) cos c(t+h)+ sin(ct) sin c(t+h)) = cos(ch) for h>0
Thus this one is also a stationary process.
c) Here also E(Xt) = 0
Cov(Xt,Xt+h)
= E(XtXt+h) = E(Ztcos(ct)+Zt-1sin(ct)) (Zt+hcos(c(t+h))+Zt+h-1sin(c(t+h)))
= E(ZtZt+hcos(ct) cos c(t+h) + Zt-1Zt+hsin(ct)cos c(t+h)+ ZtZt+h-1cos(ct) sin c(t+h) + Zt-1Zt+h-1sin(ct) sin
c(t+h))
= for h=0
sin(ct) cos c(t-1) for h=-1
So we can see the covariance function is not independent of t thus it’s not a stationary series.
d) E(Xt) = a
Cov(Xt,Xt+h) = E(b
2Z0
2)
= b2 for any h.
Thus its independent of t and hence stationary process.
e) Here also E(Xt) = 0
Cov(Xt,Xt+h) = E(Z0
cos(ct) Z0
cos(c(t+h)))
= cos(ct) cos(c(t+h))
So if h>0 it depends on t thus not stationary series.
f) Here E(Xt) = E(Zt Zt-1) = 0
Cov(Xt,Xt+h) = E(Zt
Zt-1
Zt+h
Zt+h-1
)
= ...