This edition is intended for use outside of the U.S. only, with content that may be different from the U.S. Edition. This may not be resold, copied,
or distributed without the prior consent of the publisher.
78
CHAPTER 14
SOLUTIONS TO PROBLEMS
14.1
First, for each
t
> 1, Var(
Δ
u
it
) = Var(
u
it
–
u
i,t-
1
) = Var(
u
it
) + Var(
u
i,t-
1
) =
2
2
u
σ
, where we use
the assumptions of no serial correlation in {
u
t
} and constant variance.
Next, we find the
covariance between
Δ
u
it
and
Δ
u
i,t+
1
.
Because these each have a zero mean, the covariance is
E(
Δ
u
it
⋅Δ
u
i,t+
1
) = E[(
u
it
–
u
i,t
-1
)(
u
i,t+
1
–
u
it
)] = E(
u
it
u
i,t+
1
) – E(
2
it
u
) – E(
u
i,t-
1
u
i,t+
1
) + E(
u
i,t-
1
u
it
) =
−
E(
2
it
u
) =
2
u
−
because of the no serial correlation assumption.
Because the variance is constant
across
t
, by Problem 11.1, Corr(
Δ
u
it
,
Δ
u
i,t+
1
) = Cov(
Δ
u
it
,
Δ
u
i,t+
1
)/Var(
∆
u
it
) =
22
/(2
)
uu
−
=
−
.5.
14.3
(i) E(
e
it
) = E(
v
it
−
i
v
λ
) = E(
v
it
)
−
E(
i
v
) = 0 because E(
v
it
) = 0 for all
t
.
(ii) Var(
v
it
−
i
v
) = Var(
v
it
) +
2
Var(
i
v
)
−
2
⋅
Cov(
v
it
,
i
v
) =
2
v
+
2
E(
2
i
v
)
−
2
⋅
E(
v
it
i
v
).
Now,
2
2
E( )
vi
t
a
u
v
σσ
==
+
and E(
v
it
i
v
) =
1
1
()
T
it is
s
TE
v
v
−
=
∑
=
1
T
−
[
2
a
+
2
a
+
…
+ (
2
a
+
2
u
) +
…
+
2
a
] =
2
a
+
2
u
/
T
.
Therefore, E(
2
i
v
) =
1
1
T
it i
t
v
v
−
=
∑
=
2
a
+
2
u
/
T
.
Now, we can collect
terms:
Var(
v
it
−
i
v
)
=
2
(
/
)
2
(
/
)
au
TT
++
+
−
+
.
Now, it is convenient to write
= 1
−
/
η
γ
, where
≡
2
u
/
T
and
≡
2
a
+
2
u
/
T
.
Then
Var(
v
it
−
i
v
)
=
(
2
a
+
2
u
)
−
2
(
2
a
+
2
u
/
T
) +
2
(
2
a
+
2
u
/
T
)
= (
2
a
+
2
u
)
−
2(1
−
/
)
+ (1
−
/
)
2
= (
2
a
+
2
u
)
−
2
+ 2
⋅
+ (1
−
2
/
+
/
)
= (
2
a
+
2
u
)
−
2
+ 2
⋅
+ (1
−
2
/
+
/
)
= (
2
a
+
2
u
)
−
2
+ 2
⋅
+
−
2
⋅
+
= (
2
a
+
2
u
)
+
−
=
2
u
.
This is what we wanted to show.