Time Series Analysis
Practice test with solutions
Q1. Consider the time series
xt = −2t + wt + 0.1wt−1 ,
where xt are the observations and wt ∼ iid N (0, 1).
(a) Is this time series stationary? Justify your answer. (2 pts)
Answer:
E(xt ) = E[−2t] + E[wt ] + 0.1E[wt−1 ] = −2t
The mean is not constant and depends on t. No, xt is not stationary.
(b) Prove that the first difference series 5xt = xt − xt−1 is stationary. (12 pts)
Answer:
5xt = −2t + wt + 0.1wt−1 + 2(t − 1) − wt−1 − 0.1wt−2
= −2 + wt − 0.9wt−1 − 0.1wt−2
First we compute the mean:
E[5xt ] = E[−2] + E[wt ] − 0.9E[wt−1 ] − 0.1E[wt−2 ] = −2
Second we compute the autocovariance:
γ5x (h) = cov(5xt+h , 5xt )
= cov(−2 + wt+h − 0.9wt+h−1 − 0.1wt+h−2 , −2 + wt − 0.9wt−1 − 0.1wt−2 )
= cov(wt+h , wt ) − 0.9cov(wt+h , wt−1 ) − 0.1cov(wt+h , wt−2 )
−0.9cov(wt+h−1 , wt ) + 0.81cov(wt+h−1 , wt−1 ) + 0.09cov(wt+h−1 , wt−2 )
−0.1cov(wt+h−2 , wt ) + 0.09cov(wt+h−2 , wt−1 ) + 0.01cov(wt+h−2 , wt−2 )
By independence of wt ,we have:
γ5x (h) =



1.82






−0.81



−0.1





0
1
h=0
h = ±1
h = ±2
elsewhere.
The mean is constant and the auto-covariance only depends on s, t through
their difference h, so we conclude that 5xt is stationary.
Q2. Let {xt , t = 0, 1, 2, . . .} be a random walk with constant drift µ, defined by x0 = 0
and
xt = µ + xt−1 + wt ,
t = 1, 2, 3, . . .
where xt are the observations and wt ∼ iid N (0, σw2 ).
(a) What are the mean function and autocovariance function of this process {xt }?
Is this time series stationary? Justify your answer. (6 pts)
Answer:
First we compute the mean:
xt = µ + wt + xt−1 = µ + wt + µ + wt−1 + xt−2
xt = wt + wt−1 + 2µ + xt−2
...
xt =
t−1
X
wt−j + tµ + x0 =
j=0
t−1
X
wt−j + tµ
j=0
" t−1
#
X
E[xt ] = E
wt−j + E[tµ] = tµ
j=0
Second we compute the autocovariance:
γ(t, t+h) = cov
t−1
X
j=0
γ(t, t+h) =
t−1
X
wt−j + tµ,
t+h−1
X
!
wt+h−i + (t + h)µ
i=0
cov (wt−j , wt−j ) = tσw2
j=0
xt is not stationary because the mean and the autocovariance function depends
on t.
(b) Prove that the first difference series 5xt = xt − xt−1 is stationary. (6 pts)
Answer:
5xt = xt − xt−1 = µ + wt
2
First we compute the mean:
E[5xt ] = E[µ] + E[wt ] = µ
Second we compute the autocovariance:
γ5x (h) = cov(5xt+h , 5xt ) = cov(µ + wt+h , µ + wt )
By independence of wt ,we have:
γ5x (h) =


σw2
h=0

0
|h| > 0
The mean is constant and the auto-covariance only depends on s, t through
their difference h, so we conclude that 5xt is stationary.
Q3. Consider the two series
xt = w t
yt = wt − 0.5wt−1 + t ,
where wt and t are independent white noise series with variances 2 and 3, respectively.
(a) Determine the cross-covariance function (CCF), ρxy (h) relating xt and yt . (10
pts)
Answer:
First we compute the cross-covariance function of xt and yt .
γxy (h) = cov(xt+h , yt ) = cov(wt+h , wt − 0.5wt−1 + t )
Therefore our cross-covariance function is:




2
h=0



γxy (h) = −1
h = −1





0
h 6= 0, −1
Second we compute the autocovariances of x and y:
γx (h) = cov(xt+h , xt ) = cov(wt+h , wt )
3
By independence of wt ,we have:
γx (h) =


2
h=0

0
|h| ≥ 1
γy (h) = cov(yt+h , yt ) = cov(wt+h − 0.5wt+h−1 + t+h , wt − 0.5wt−1 + t )
Again, by independence of wt and t ,we have:




5.5
h=0



γy (h) = −1
|h| = 1





0
|h| ≥ 2
Now, we compute the cross-correlation function of xt and yt .
γxy (h)
γxy (h)
ρxy (h) = p
= √
11
γx (0)γy (0)

√



2/
11



√
ρxy (h) = −1/ 11





0
h=0
h = −1
h 6= 0, −1
(b) Show that xt and yt are jointly stationary. (14 pts)
Answer:
Two time series, say, xt and yt , are said to be jointly stationary if they are each
stationary, and the cross-covariance function
γxy (h) = cov(xt+h , yt )
is a function only of lag h.
First we compute the mean of xt and yt :
E[xt ] = E[wt ] = 0
E[yt ] = E[wt ] − 0.5E[wt−1 ] + E[t ] = 0
Second the autocovariance functions of x and y by part(a):


2
h=0
γx (h) =

0
|h| ≥ 1
4




5.5



γy (h) = −1





0
h=0
|h| = 1
|h| ≥ 2
The mean is constant and the auto-covariance only depends on s, t through
their difference h, so we conclude that xt and yt are stationary. Also, the
cross-covariance function of xt and yt




2



γxy (h) = −1





0
h=0
h = −1
h 6= 0, −1
is a function only of lag h. Therefore, xt , yt are each stationary and the crosscovariance function only depends on the lag h, we can conclude that xt and yt
are jointly stationary.
5