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包含所有课后习题答案，非常详尽！ 《时间序列分析及应用（R语言）（原书第2版）》以易于理解的方式讲述了时间序列模型及其应用，内容包括趋势、平稳时间序列模型、非平稳时间序列模型、模型识别、参数估计、模型诊断、预测、季节模型、时间序列回归模型、异方差模型、谱分析入门、谱估计和门限模型。对所有的思想和方法，都用真实数据集和模拟数据集进行了说明。 《时间序列分析及应用（R语言）（原书第2版）》的一大特点是采用R语言来作图和分析数据，书中的所有图表和实证结果都是用R命令得到的。作者还为《时间序列分析及应用（R语言）（原书第2版）》制作了大量新增或增强的-函数。《时间序列分析及应用（R语言）（原书第2版）》的另一特点是包含很多有用的附录．例如，回顾了有关期望、方差、协方差、相关系数等概念．筒述了条件期望的性质以及最小均方误差预测等内容，这些附录有利于关心技术细节的读者深入了解相关内容。
1
Solutions Manual to Accompan
y
T
ime Series Analysis
with Applications in R, Second Edition
b
y Jonathan D
. Cr
ye
r and Kung-Sik Chan
Solutions b
y Jonathan Cr
y
er and Xuemiao Hao
, updated 7/28/08
C
HAPTER
1
Exercise 1.1
Use software to produce the time series plot sho
wn in
Exh
ibit (1.2), page 2. The follo
wing R code will
produce the graph.
> library(TSA); data(larain); win.gr
aph(width=3,height=3,pointsize=8)
> plot(y=larain,x=zlag(larain),ylab=
'Inches',xlab='Previous Year Inches')
Exercise 1.2
Produce the time series plot displayed in Exhibit (1.3), page 3. Use the R code
> data(color); plot(color,ylab='Colo
r Property',xlab='Batch',type='o')
Exercise 1.3
Simulate a completely random process of length 48 wi
th indepen
dent, normal v
a
lues. Repeat this exer-
cise sev
eral times with a new simula
tion, that is, a ne
w seed, each time.
> plot(ts(rnorm(n=48)),type='o')
# If you repeat this command R will use
a new “random
numbers” each time. If you want to r
eproduce the same simulation first use the
command set.seed(#########) where ##
####### is an integer of your choice.
Exercise 1.4
Simulate a completely random process of length 48
with independent, chi-square distributed v
alues
each with 2 degrees of freedom. Use the same R code as in the solution of
Exercise 1.3
but replace
rnorm(n=48)
with
rchisq(n=48,df=2).
Exercise 1.5
Simulate a completely random process of length
4
8 with independent,
t
-distributed v
alues each with 5
degrees of freedom. Construct the time series plot
. Use the sam
e R code as in the solution of
Exer
cise 1.3
bu
t
replace
rnorm(n=48)
with
rt(n=48,df=5).
Exercise 1.6
Construct a time series plot wi
th monthly plotting symbol
s for the Dub
u
que temperature series as in
Exhibit (1.7), page 6. (Make the plot full
screen so that you can see all of
detail.)
> data(tempdub); plot(tempdub,ylab='
Temperature')
> points(y=tempdub,x=time(tempdub),
pch=as.vector(season(tempdub)))
C
HAPTER
2
Exercise 2.1
Suppose
E
(
X
) = 2,
Va
r
(
X
) = 9,
E
(
Y
) = 0,
Va
r
(
Y
) = 4, and
Corr
(
X
,
Y
) = 0.25. Find:
(a)
Va
r
(
X
+
Y
) =
Va
r
(
X
) +
Va
r
(
Y
) +2
Cov
(
X
,
Y
) = 9 + 4 + 2(3*2*0.25) = 16
(b)
Cov
(
X
,
X
+
Y
) =
Co
v
(
X
,
X
) +
Co
v
(
X
,
Y
)
= 9 + ((3*2*0.25) = 9 + 3/2 =
10.5
(c)
Corr
(
X
+
Y
,
X
−
Y
). As in part (a),
Va
r
(
X
−
Y
) = 10. Then
Cov
(
X
+
Y
,
X
−
Y
) =
Cov
(
X
,
X
)
−
Co
v
(
Y
,
Y
) +
Cov
(
X
,
Y
)
−
Cov
(
X
,
Y
) =
Va
r
(
X
)
−
Va
r
(
Y
) = 9
−
4 = 5. So
Exercise 2.2
If
X
and
Y
are dependent b
ut
Va
r
(
X
) =
Va
r
(
Y
), f
ind
Cov
(
X
+
Y
,
X
−
Y
).
Cov
(
X
+
Y
,
X
−
Y
) =
Cov
(
X
,
X
)
−
Cov
(
Y
,
Y
) +
Cov
(
X
,
Y
)
−
Cov
(
Y
,
X
) =
Va
r
(
X
)
−
Va
r
(
(
Y
) = 0
Exercise 2.3
Let
X
ha
ve a distribution with mean
μ
and v
ariance
σ
2
and let
Y
t
=
X
for all
t
.
(a)
Show that {
Y
t
} is strictly and weakly stationary
. Let
t
1
,
t
2
,
…
,
t
n
be any set of time points an
d
k
an
y time lag.
Then
Corr
X
Y
X
Y
–
,
+
()
Cov
X
Y
X
Y
–
,
+
()
Var
X
Y
+
()
Var
X
Y
–
()
-----------------
------------------------------
------------
-
5
16
10
×
-----------------
----
-
5
41
0
------------
-
0
.
3
9
5
2
8
4
7
1
==
=
=
2
as required for strict stationarity
. Since the autoco
variance cl
early exis
ts, (see part (b)), the process is also weakly
stationary
.
(b)
Find the autocov
ariance function for {
Y
t
}.
Cov
(
Y
t
,
Y
t
−
k
) =
Cov
(
X
,
X
) =
σ
2
for all
t
and
k
, free of
t
(and
k
).
(c)
Sketch a “typical” time plot of
Y
t
. The plot will be a horizontal “line” (really a discrete-time horizontal line)
at the height of the observed
X
.
Exercise 2.4
Let {
e
t
} be a zero mean white noise processes.
Suppose tha
t
the obse
rv
ed process is
Y
t
=
e
t
+
θ
e
t
−
1
where
θ
is either 3 or 1/3.
(a)
Find the autocorrel
ation function for {
Y
t
} both when
θ
= 3 and when
θ
=1/3.
E
(
Y
t
) =
E
(
e
t
+
θ
e
t
−
1
) = 0.
Also
Va
r
(
Y
t
) =
Va
r
(
e
t
+
θ
e
t
−
1
) =
σ
2
+
θ
2
σ
2
=
σ
2
(1 +
θ
2
).
Also
Cov
(
Y
t
,
Y
t
−
1
) =
Cov
(
e
t
+
θ
e
t
−
1
,
e
t
−
1
+
θ
e
t
−
2
) =
θσ
2
free of
t
. Now for
k
> 1,
Cov
(
Y
t
,
Y
t
−
k
) =
Cov
(
e
t
+
θ
e
t
−
1
,
e
t
−
k
+
θ
e
t
−
k
−
1
) = 0 since all of these error terms are
uncorrelated. So
But 3/(1+3
2
) = 3/10 and (1/3)/[1+(1/3)
2
] = 3/10. So the autocorrela
tion functions are identical.
(b)
Y
ou should hav
e discovered that the time series
is stati
onary re
gardless of the value of
θ
and that the autocor-
relation functions are the same for
θ
= 3 and
θ
=
1/3. F
or simpl
icity
, suppose
that the process mean is kno
wn
to be zero and the variance
of
Y
t
is known to be 1. Y
ou ob
serv
e the seri
es {
Y
t
} for
t
= 1, 2,
...
,
n
an
d suppose
that you can produce good est
ima
tes of the autocorrelations
ρ
k
. Do you think that yo
u could determine
which value of
θ
is
correct (3 or 1/3) based on the estimate of
ρ
k
? Why or why not?
Exercise 2.5
Suppose
Y
t
= 5 + 2
t
+
X
t
where {
X
t
} is a zero mean stationary series with autoco
variance function
γ
k
.
(a)
Find the mean function for {
Y
t
}.
E
(
Y
t
) =
E
(5 + 2
t
+
X
t
) = 5 + 2
t
+
E
(
X
t
) = 5 + 2
t
.
(b)
Find the autocov
ariance function for {
Y
t
}.
Cov
(
Y
t
,
Y
t
−
k
) =
Cov
(5 + 2
t
+
X
t
, 5 + 2(
t
−
k)
+
X
t
−
k
) =
Cov
(
X
t
,
X
t
−
k
) =
γ
k
free of
t
.
(c)
Is {
Y
t
} stationary? (Why or why not?) In spite of part (b), Th
e
process {
Y
t
} is
not
stationary since its mean
varies with time.
Exercise 2.6
Let {
X
t
} be a stationary time series and define
(a)
Show that
is free of
t
for all lags
k
.
Cov
(
Y
t
,
Y
t
−
k
) =
Cov
(
X
t
+ 3,
X
t
−
k
+ 3) =
Cov
(
X
t
,
X
t
−
k
) is free of
t
since {
X
t
} is stationary
.
(b)
Is {
Y
t
} stationary? {
Y
t
} is
not
stationary since
E
(
Y
t
) =
E
(
X
t
) =
μ
X
for
t
odd but
E
(
Y
t
) =
E
(
X
t
+ 3) =
μ
X
+ 3 for
t
even
.
Exercise 2.7
Suppose that {
Y
t
} is stationary with autoco
variance function
γ
k
.
(a)
Show that
W
t
=
∇
Y
t
=
Y
t
−
Y
t
−
1
is stationary by f
inding th
e mean and autoco
variance function for {
W
t
}.
E
(
W
t
) =
E
(
Y
t
−
Y
t
−
1
) =
E
(
Y
t
)
−
E
(
Y
t
−
1
) = 0 since {
Y
t
} is stationary
. Also
Cov
(
Y
t
,
Y
t
−
k
) =
Cov
(
Y
t
−
Y
t
−
1
,
Y
t
−
k
−
Y
t
−
k
−
1
) =
Cov
(
Y
t
,
Y
t
−
k
)
−
Cov
(
Y
t
,
Y
t
−
k
−
1
)
−
Cov
(
Y
t
−
1
,
Y
t
−
k
) +
Cov
(
Y
t
−
1
,
Y
t
−
k
−
1
)
=
γ
k
−
γ
k+
1
−
γ
k
−
1
+
γ
k
= 2
γ
k
−
γ
k+
1
−
γ
k
−
1
, free of
t
.
(b)
Show that
U
t
=
∇
2
Y
t
=
∇
[
Y
t
−
Y
t
−
1
] =
Y
t
−
2
Y
t
−
1
+
Y
t
−
2
is stationary
. (Y
ou need not
find the mean and auto-
cov
ariance function for {
U
t
}.)
U
t
is the first dif
ference of the process{
∇
Y
t
}. By part (a), {
∇
Y
t
} is stationary
.
So
U
t
is the dif
ference of a stationary process a
nd, again by part (a), is itself
st
ationary
.
Pr
Y
t
1
y
t
1
<
Y
t
2
y
t
2
≤…
Y
t
n
y
t
n
≤
,,
,
()
Pr
X
y
t
1
<
Xy
t
2
≤…
Xy
t
n
≤
,,
,
()
=
Pr
Y
t
1
k
–
y
t
1
<
Y
t
2
k
–
y
t
2
≤…
Y
t
n
k
–
y
t
n
≤
,,
,
()
=
Corr
Y
t
Y
tk
–
,
()
Cov
Y
t
Y
tk
–
,
()
Var
Y
t
()
Var
Y
tk
–
()
-----------------
------------------------------
---
-
1
for
k
0
=
θσ
2
σ
2
1
θ
2
+
()
------------------
-------
-
θ
1
θ
2
+
--------------
-
=
for
k
1
=
0f
o
r
k
1
>
⎩
⎪
⎪
⎨
⎪
⎪
⎧
==
Y
t
X
t
X
t
3
+
⎩
⎨
⎧
=
for
t
odd
for
t
ev
en
Cov
Y
t
Y
tk
–
,
()
3
Exercise 2.8
Suppose that {
Y
t
} is stationary with auto
co
variance function
γ
k
. Show that for an
y fixed positi
ve inte
ger
n
and any constants
c
1
,
c
2
,
...
,
c
n
, the process {
W
t
} defined b
y
is station-
ary
. First
free of
t
. Also
Exercise 2.9
Suppose
Y
t
=
β
0
+
β
1
t
+
X
t
where {
X
t
} is a zero mean stationary
seri
es with autocovariance function
γ
k
and
β
0
and
β
1
are constants.
(a)
Show that {
Y
t
} is not stationary but that
W
t
=
∇
Y
t
=
Y
t
−
Y
t
−
1
is stationary
. {
Y
t
} is
not
stationary since its
mean,
β
0
+
β
1
t
, varies with
t
. Ho
wev
er,
E
(
W
t
) =
E
(
Y
t
−
Y
t
−
1
) = (
β
0
+
β
1
t
)
−
(
β
0
+
β
1
(
t
−
1)) =
β
1
, free of
t.
The argument
in the solution of Exercise
2.7 shows that the cov
a
riance function for {
W
t
} is free of
t
.
(b)
In general, show that if
Y
t
=
μ
t
+
X
t
where {
X
t
} is a zero mean stationary series and
μ
t
is a polynomial
in
t
of
degree
d
, then
∇
m
Y
t
=
∇
(
∇
m
−
1
Y
t
) is stationary for
m
≥
d
and nonstationary for 0
≤
m
<
d
.
Use part (a) and proceed b
y induction.
Exercise 2.10
Let {
X
t
} be a zero-mean, unit-v
ariance stationary
process with autocorrelation function
ρ
k
. Suppose
that
μ
t
is a nonconstant
function and that
σ
t
is a positi
ve-valued nonconstant function. The observed series is formed
as
Y
t
=
μ
t
+
σ
t
X
t
(a)
Find the mean and cov
ari
ance function for the {
Y
t
} process.
Notice that
Cov
(
X
t
,
X
t
−
k
) =
Corr
(
X
t
,
X
t
−
k
) since {
X
t
} has unit variance.
E
(
Y
t
) =
E
(
μ
t
+
σ
t
X
t
) =
μ
t
+
σ
t
E
(
X
t
) =
μ
t
.
Now
Cov
(
Y
t
,
Y
t
−
k
) =
Cov
(
μ
t
+
σ
t
X
t
,
μ
t
−
k
+
σ
t
−
k
X
t
−
k
) =
σ
t
σ
t
−
k
Cov
(
X
t
,
X
t
−
k
) =
σ
t
σ
t
−
k
ρ
k
. Notice that
Va
r
(
Y
t
) =
(
σ
t
)
2
.
(b)
Sho
w that autocorrelat
i
on function for the {
Y
t
} process depends only on time lag. Is the {
Y
t
} process station-
ary?
Corr
(
Y
t
,
Y
t
−
k
) =
σ
t
σ
t
−
k
ρ
k
/[
σ
t
σ
t
−
k
] =
ρ
k
but
{
Y
t
} is not necessarily stationary since
E
(
Y
t
) =
μ
t
.
(c)
Is it possible t
o
h
av
e a time seri
es with a constant mean and with
Corr
(
Y
t
,
Y
t
−
k
) free of
t
but
wit
h {
Y
t
}
not
stationary? If
μ
t
is constant but
σ
t
va
ri
es
wi
th
t
,
this will be the case.
Exercise 2.11
Suppose
Cov
(
X
t
,
X
t
−
k
) =
γ
k
is free of
t
b
ut that
E
(
X
t
) = 3
t
.
(a)
Is {
X
t
} stationary? No since
E
(
X
t
) varies with
t
.
(b)
Let
Y
t
= 7
−
3
t
+
X
t
. Is {
Y
t
} stationary? Y
es, since the co
v
ariances are unchanged but now
E
(
X
t
) = 7
−
3
t
+
3
t
= 7, free of
t
.
Exercise 2.12
Suppose that
Y
t
=
e
t
−
e
t
−
12
. Show that {
Y
t
} is stationary and that, for
k
> 0, its autocorrelation func-
tion is nonzero onl
y for lag
k
= 12.
E
(
Y
t
) =
E
(
e
t
−
e
t
−
12
) = 0. Also
Cov
(
Y
t
,
Y
t
−
k
) =
Co
v
(
e
t
−
e
t
−
12
,e
t
−
k
−
e
t
−
12
−
k
) =
−
Cov
(
e
t
−
12
,e
t
−
k
)
= −
(
σ
e
)
2
when
k
= 12. It is nonzero only for
k
= 12 since, otherwise, all of the
error terms in
volved are
uncorrelated.
Exercise 2.13
Let
. For this ex
ercise, assume that the white noise series is nor
mally distrib
u
ted.
(a)
Find the autocorrelation function for {
Y
t
}. First recall that for a zer
o-mean norm
al distrib
ution
and
. Then
which is constant in
t
and
Also
All other co
variances are also zero.
(b)
Is {
Y
t
} stationary? Y
es, in fact, it is a non-normal whit
e noise in disguise!
W
t
c
1
Y
t
c
2
Y
t
1
–
…
c
n
Y
tn
–1
+
++
+
=
EW
t
()
c
1
EY
t
c
2
EY
t
1
–
…
c
n
EY
tn
–1
+
++
+
c
1
c
2
…
c
n
++
+
()
μ
Y
==
Cov
W
t
W
tk
–
,
()
Cov
c
1
Y
t
c
2
Y
t
1
–
…
c
n
Y
tn
–1
+
++
+
c
1
Y
tk
–
c
2
Y
t
1
–
k
–
…
c
n
Y
tn
–1
k
–
+
++
+
,
()
=
c
j
c
i
Cov
Y
tj
–
Y
tk
–
i
–
,
()
i
0
=
n
∑
j
0
=
n
∑
=
c
j
c
i
γ
jk
–
i
–
i
0
=
n
∑
j
0
=
n
∑
=
free of
t
.
Y
t
e
t
θ
e
t
1
–
2
–
=
Ee
t
1
–
3
()
0
=
Ee
t
1
–
4
()
3
σ
e
4
=
EY
t
()
θ
Var
e
t
1
–
()
–
θσ
e
2
–
==
Var
Y
t
()
Var
e
t
()
θ
2
Var
e
t
1
–
2
()
+
σ
e
2
θ
2
Ee
t
1
–
4
()
Ee
t
1
–
2
()
[]
2
–
{}
+
==
σ
e
2
θ
2
3
σ
e
4
σ
e
2
[]
2
–
{}
+
=
σ
e
2
2
θ
2
σ
e
4
+
=
Cov
Y
t
Y
t
1
–
,
()
Cov
e
t
θ
e
t
1
–
2
–
e
t
1
–
θ
e
t
2
–
2
–
,
()
Cov
θ
e
t
1
–
2
–
e
t
1
–
,
()
θ
Ee
t
1
–
3
()
–0
==
=
=
4
Exercise 2.14
Ev
aluate the mean and covariance
function for each of the follo
wing
processes. In each case determine
whether or not the process i
s stationary
.
(a)
Y
t
=
θ
0
+
te
t
. The mean is
θ
0
but it is not stationary since
Va
r
(
Y
t
) =
t
2
Va
r
(
e
t
) =
t
2
σ
2
is not free of
t
.
(b)
W
t
=
∇
Y
t
where
Y
t
is as giv
e
n in part (a).
W
t
=
∇
Y
t
= (
θ
0
+
te
t
)
−
(
θ
0
+ (
t
−
1
)e
t
−
1
) =
te
t
−
(
t
−
1
)e
t
−
1
So the mean of
W
t
is zero. Howe
ver
,
Va
r
(
W
t
) = [
t
2
+ (
t
−
1)
2
](
σ
e
)
2
which depends on
t
and
W
t
is not stationary
.
(c)
Y
t
=
e
t
e
t
−
1
. (Y
ou m
ay assume that
{
e
t
} is normal white noise.)
The mean of
Y
t
is clearly zero. Lag one is the
only lag at which there might be correlation. Howe
ver
,
Cov
(
Y
t
,
Y
t
−
1
) =
E
(
e
t
e
t
−
1
e
t
−
1
e
t
−
2
)
= E
(
e
t
)
E
[
e
t
−
1
]
2
E
(
e
t
−
2
) = 0. So the process
Y
t
=
e
t
e
t
−
1
is stationary and is a non-normal white noise!
Exercise 2.15
Suppose that
X
is a random v
ariable wit
h zero mean. Def
ine a time series by
Y
t
=(
−
1)
t
X.
(a)
Find the mean function for {
Y
t
}.
E
(
Y
t
) =
(
−
1)
t
E
(
X
) = 0.
(b)
Find the cov
arian
ce
functi
on for {
Y
t
}.
Co
v
(
Y
t
,
Y
t
−
k
) =
Cov
[(
−
1)
t
X,
(
−
1)
t
−
k
X
]
=
(
−
1)
2t
−
k
Cov
(
X,X
) =
(
−
1)
k
(
σ
X
)
2
(c)
Is {
Y
t
} stationary? Y
es, the mean is constant and the cov
ariance only depends on lag.
Exercise 2.16
Suppose
Y
t
=
A
+
X
t
where {
X
t
} is stationary and
A
is random
b
ut independent o
f {
X
t
}. Find the mean
and cov
ariance function for {
Y
t
} in terms of the mean and autocov
ariance function for {
X
t
} and the mean and v
ari-
ance of
A
. First
E
(
Y
t
) =
E
(
A
) +
E
(
X
t
) =
μ
A
+
μ
X
, free of
t
. Also, since {
X
t
} and
A
are independent,
Exercise 2.17
Let {
Y
t
} be stationary with
autocov
ariance function
γ
k
. Let
. Show that
No
w make the change of variable
t
−
s
=
k
and
t
=
j
in the double sum. The
range of the summation
{1
≤
t
≤
n
,
1
≤
s
≤
n
} is transfo
rmed into {1
≤
j
≤
n
, 1 ≤
j
−
k
≤
n
} =
{
k
+ 1
≤
j
≤
n
+
k
, 1
≤
j
≤
n
} which may be
written .
Thus
Use
γ
k
=
γ
−
k
to get the f
irst expression in the ex
ercise.
Exercise 2.18
Let {
Y
t
} be stationary with
autocov
ariance function
γ
k
. Define the sample v
ariance as
.
(a)
First show that
.
(b)
Use part (a) to show that
.
(Use the results of Exercise (2
.17) for the last expression.)
Cov
Y
t
Y
tk
–
,
()
Cov
A
X
t
+
AX
tk
–
+
,
()
Cov
A
A
,
()
Cov
X
t
X
tk
–
,
()
+
Var
A
()
γ
k
X
+
=
=
=
free of
t
·
Y
_
1
n
--
-
Y
t
t
1
=
n
∑
=
Var
Y
_
()
γ
0
n
----
-
2
n
--
-
1
k
n
--
-
–
⎝⎠
⎛⎞
γ
k
k
1
=
n
1
–
∑
+
=
1
n
--
-
1
k
n
----
-
–
⎝⎠
⎛⎞
γ
k
kn
–1
+
=
n
1
–
∑
=
Var
Y
_
()
1
n
2
----
-
Var
Y
t
t
1
=
n
∑
[]
1
n
2
----
-
Cov
Y
t
t
1
=
n
∑
Y
s
s
1
=
n
∑
,
[]
1
n
2
----
-
γ
ts
–
s
1
=
n
∑
t
1
=
n
∑
==
=
k
0
k
1
jn
≤≤
+
,
>
{}
k
01
jnk
+
≤≤
,
≤
{}
∪
Var
Y
_
()
1
n
2
----
-
γ
k
jk
1
+
=
n
∑
k
1
=
n
1
–
∑
γ
k
j
1
=
nk
+
∑
kn
–1
+
=
0
∑
+
[]
=
1
n
2
----
-
nk
–
()
γ
k
k
1
=
n
1
–
∑
nk
+
()
γ
k
kn
–1
+
=
0
∑
+
[]
=
1
n
--
-
1
k
n
----
-
–
⎝⎠
⎛⎞
γ
k
kn
–1
+
=
n
1
–
∑
=
S
2
1
n
1
–
-----------
-
Y
t
Y
_
–
()
2
t
1
=
n
∑
=
Y
t
μ
–
()
2
t
1
=
n
∑
Y
t
Y
_
–
()
2
t
1
=
n
∑
nY
_
μ
–
()
2
+
=
Y
t
μ
–
()
2
t
1
=
n
∑
Y
t
Y
_
–
Y
_
μ
–
+
()
2
t
1
=
n
∑
Y
t
Y
_
–
()
2
t
1
=
n
∑
Y
_
μ
–
()
2
t
1
=
n
∑
2
Y
t
Y
_
–
()
Y
_
μ
–
()
t
1
=
n
∑
++
==
Y
t
Y
_
–
()
2
t
1
=
n
∑
nY
_
μ
–
()
2
2
Y
_
μ
–
()
Y
t
Y
_
–
()
t
1
=
n
∑
++
=
Y
t
Y
_
–
()
2
t
1
=
n
∑
nY
_
μ
–
()
2
+
=
ES
2
()
n
n
1
–
-----------
-
γ
0
n
n
1
–
-----------
-
Var
Y
_
()
–
γ
0
2
n
1
–
-----------
-
1
k
n
--
-
–
⎝⎠
⎛⎞
γ
k
k
1
=
n
1
–
∑
–
==
5
(c)
If {
Y
t
} is a white noise process with variance
γ
0
, show that
E
(
S
2
) =
γ
0
. This follows since for white noise
γ
k
= 0 for
k
> 0.
Exercise 2.19
Let
Y
1
=
θ
0
+
e
1
and then for
t
> 1 define
Y
t
recursi
vely by
Y
t
=
θ
0
+
Y
t
−
1
+
e
t
. Here
θ
0
is a constant.
The process {
Y
t
} is called a
random walk with drift
.
(a)
Show that
Y
t
may be rewritten as
. Substitute
Y
t
−
1
=
θ
0
+
Y
t
−
2
+
e
t
−
1
into
Y
t
=
θ
0
+
Y
t
−
1
+
e
t
and repeat until
you get back to
e
1
.
(b)
Find the mean function for
Y
t
.
(c)
Find the autocov
ariance function for
Y
t
.
Exercise 2.20
Consider the standard
random walk model where
Y
t
=
Y
t
−
1
+
e
t
with
Y
1
=
e
1
.
(a)
Use the abov
e representation of
Y
t
t
o s
how
th
at
μ
t
=
μ
t
−
1
for
t
> 1 with i
nitial condition
μ
1
=
E
(
e
1
) = 0. Hence
sho
w that
μ
t
= 0 for all
t
. Clearly
,
μ
1
=
E
(
Y
1
) =
E
(
e
1
) = 0. Then
E
(
Y
t
) =
E
(
Y
t
−
1
+
e
t
) =
E
(
Y
t
−
1
) +
E
(
e
t
) =
E
(
Y
t
−
1
) or
μ
t
=
μ
t
−
1
for
t
> 1 and the result follows b
y inductio
n.
(b)
Similarly
, show that
Va
r
(
Y
t
) =
Va
r
(
Y
t
−
1
) +
, for
t
> 1 with
Va
r
(
Y
1
) =
, and, hence
Va
r
(
Y
t
) =
t.
Va
r
(
Y
1
) =
is immediate. Then
.
Recursion or induct
ion on
t
yields
Va
r
(
Y
t
) =
t
.
(c)
For
0
≤
t
≤
s
, us
e
Y
s
=
Y
t
+
e
t
+1
+
e
t
+2
++
e
s
to show that
Co
v
(
Y
t
,
Y
s
) =
Va
r
(
Y
t
) and, hence, that
Cov
(
Y
t
,
Y
s
) = min(
t
,
s
)
.
For 0
≤
t
≤
s
,
and hence the result.
Exercise 2.21
A random walk with random starting value. Let
for
t
> 0 where
Y
0
has
a distribution with mean
μ
0
and v
ariance
. Suppose further that
Y
0
,
e
1
,
...
,
e
t
are independent.
(a)
Show that
E
(
Y
t
) =
μ
0
for all
t
.
.
(b)
Show that
Va
r
(
Y
t
) =
t
+.
(c)
Show that
Co
v
(
Y
t
,
Y
s
) = min(
t
,
s
)
+
. Let
t
be less than
s
. Then, as in the pre
vious exercise,
(d)
Sho
w that
. Just use the results of parts (b) and (c).
Exercise 2.22
Let {
e
t
} be a zero-mean white noise process and let
c
be a constant with |
c
| < 1. Def
ine
Y
t
recursi
vely
by
Y
t
=
cY
t
−
1
+
e
t
with
Y
1
=
e
1
.
This ex
ercise can be solved us
ing the recursi
ve definition of
Y
t
or by expressing
Y
t
explicitly using repeated substi-
tution as
. Parts (c), (d), and (e) essen-
tially assume you are working with the recursiv
e version of
Y
t
but the
y can
al
so be solv
ed using th
is
explicit
representation.
ES
2
()
E
1
n
1
–
-----------
-
Y
t
Y
_
–
()
2
t
1
=
n
∑
⎝⎠
⎜⎟
⎜⎟
⎛⎞
1
n
1
–
-----------
-
EY
t
μ
–
()
2
t
1
=
n
∑
nY
_
μ
–
()
2
–
⎝⎠
⎜⎟
⎜⎟
⎛⎞
1
n
1
–
-----------
-
EY
t
μ
–
()
2
[]
t
1
=
n
∑
nE
Y
_
μ
–
()
2
–
==
=
1
n
1
–
-----------
-
n
γ
0
nVar
Y
_
()
–
[]
=
1
n
1
–
-----------
-
n
γ
0
n
γ
0
n
----
-
2
n
--
-
1
k
n
--
-
–
⎝⎠
⎛⎞
γ
k
k
1
=
n
1
–
∑
+
⎩⎭
⎪⎪
⎨⎬
⎪⎪
⎧⎫
–
γ
0
2
n
1
–
-----------
-
1
k
n
--
-
–
⎝⎠
⎛⎞
γ
k
k
1
=
n
1
–
∑
–
==
Y
t
t
θ
0
e
t
e
t
1
–
…
e
1
++
+
+
=
EY
t
()
Et
θ
0
e
t
e
t
1
–
…
e
1
++
+
+
()
t
θ
0
==
Cov
Y
t
Y
tk
–
,
()
Cov
t
θ
0
e
t
e
t
1
–
…
e
1
++
+
+
tk
–
()
θ
0
e
tk
–
e
t
1
–
k
–
…
e
1
++
+
+
,
[]
=
Cov
e
tk
–
e
t
1
–
k
–
…
e
1
++
+
e
tk
–
e
t
1
–
k
–
…
e
1
++
+
,
[]
=
Var
e
tk
–
e
t
1
–
k
–
…
e
1
++
+
()
=
tk
–
()
σ
e
2
=
for
tk
≥
σ
e
2
σ
e
2
σ
e
2
σ
e
2
Var
Y
t
()
Var
Y
t
1
–
e
t
+
()
Var
Y
t
1
–
()
Var
e
t
()
+
Var
Y
t
1
–
()
σ
e
2
+
==
=
σ
e
2
…
σ
e
2
Cov
Y
t
Y
s
,
()
Cov
Y
t
Y
t
e
t
1
+
e
t
2
+
…
e
s
+++
+
,
()
Cov
Y
t
Y
t
,
()
Var
Y
t
()
t
σ
e
2
==
=
=
Y
t
Y
0
e
t
e
t
1
–
…
e
1
++
+
+
=
σ
0
2
EY
t
()
EY
0
e
t
e
t
1
–
…
e
1
++
+
+
()
EY
0
()
Ee
t
()
Ee
t
1
–
()
…
Ee
1
()
++
+
+
EY
0
()
μ
0
==
=
=
σ
e
2
σ
0
2
Var
Y
t
()
Var
Y
0
e
t
e
t
1
–
…
e
1
++
+
+
()
Var
Y
1
()
Var
e
t
()
Var
e
t
1
–
()
…
Var
e
1
()
++
+
+
σ
0
2
t
σ
e
2
+
==
=
σ
e
2
σ
0
2
Cov
Y
t
Y
s
,
()
Cov
Y
t
Y
t
e
t
1
+
e
t
2
+
…
e
s
+++
+
,
()
Var
Y
t
()
σ
0
2
t
σ
e
2
+
==
=
Corr
Y
t
Y
s
,
()
t
σ
a
2
σ
0
2
+
s
σ
a
2
σ
0
2
+
------------------
---
-
=
for 0
ts
≤≤
Y
t
cc
Y
t
2
–
e
t
1
–
+
()
e
t
+
…
e
t
ce
t
1
–
c
2
e
t
2
–
…
c
t
1
–
e
1
++
+
+
==
=
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