MATH461HW2
Vinh Pham
1.13
a)
Since
y
t
is a combination of independent normal random variables, the variance of
y
t
is
σ
2
w
+
θ
2
σ
2
w
+
σ
2
u
for
any
t
. For any
y
t
, y
s
, the ACF or
ρ
(
t, s
)
can be calculated as
Cov
(
Y
t
, Y
s
)
qqqqqqq
V ar
(
Y
t
)
×
V ar
(
Y
s
)
The numerator becomes
Cov
(
w
t
−
θw
t
−
1
+
µ
t
, w
s
−
θw
s
−
1
+
µ
s
)
.
Clearly, if

s
−
t
 ≥
2
, then
Cov
(
y
t
, y
s
) = 0
, which means that
ρ
(
t, s
) = 0
.
Denote
h
=
t
−
s
, then
ρ
(
h
) = 0
if

h
 ≥
2
.
If
h
= 0
, the ACF becomes
1
.
If
h
= 1
, then
ρ
(
h
) =
−
θσ
2
w
σ
2
w
+
θ
2
σ
2
w
+
σ
2
u
.
b)
p
xy
(
s, t
) =
γ
xy
(
s, t
)
qqqqqqq
γ
x
(
s, s
)
×
γ
y
(
t, t
)
=
E
[(
x
s
−
E
[
x
s
])(
y
t
−
E
[
y
t
])]
σ
w
qqqqqqq
(1 +
θ
)
2
σ
2
w
+
σ
2
u
=
E
[
w
s
(
w
t
−
θw
t
−
1
+
u
t
)]
σ
w
qqqqqqq
(1 +
θ
)
2
σ
2
w
+
σ
2
u
Therefore,
p
xy
(
s, t
) =
999
?????
?????
=====
?????
?????
;;;;
σ
w
√
(1+
θ
)
2
σ
2
w
+
σ
2
u
if
s
=
t
−
θσ
w
√
(1+
θ
)
2
σ
2
w
+
σ
2
u
if
s
=
t
±
1
0
otherwise
Thus,
p
xy
(
h
) =
999
?????
?????
=====
?????
?????
;;;;
σ
w
√
(1+
θ
)
2
σ
2
w
+
σ
2
u
if
h
= 0
−
θσ
w
√
(1+
θ
)
2
σ
2
w
+
σ
2
u
if
h
±
1
0
otherwise
1