CHAPTER 18
SOLUTIONS TO PROBLEMS
18.1
With
z
t
1
and
z
t
2
now in the model, we should use one lag each as instrumental variables,
z
t
1,1
and
z
t
1,2
.
This gives one overidentifying restriction that can be tested.
18.3
For
δ
≠
β
,
y
t
–
z
t
=
y
t
–
z
t
+ (
–
)
z
t
, which is an I(0) sequence (
y
t
–
z
t
) plus an I(1)
sequence.
Since an I(1) sequence has a growing variance, it dominates the I(0) part, and the
resulting sum is an I(1) sequence.
18.5
Following the hint, we have
y
t
–
y
t
1
=
x
t
–
x
t
1
+
x
t
1
–
y
t
1
+
u
t
or
Δ
y
t
=
Δ
x
t
– (
y
t
1
–
x
t
1
) +
u
t
.
Next, we plug in
Δ
x
t
=
γ
Δ
x
t
1
+
v
t
to get
Δ
y
t
=
(
Δ
x
t
1
+
v
t
) – (
y
t
1
–
x
t
1
) +
u
t
=
βγ
Δ
x
t
1
– (
y
t
1
–
x
t
1
) +
u
t
+
v
t
≡
1
Δ
x
t
1
+
(
y
t
1
–
x
t
1
) +
e
t
,
where
1
=
,
= –1, and
e
t
=
u
t
+
v
t
.
18.7
If
unem
t
follows a stable AR(1) process, then this is the null model used to test for Granger
causality:
under the null that
gM
t
does not Granger cause
unem
t
, we can write
unem
t
=
β
0
+
β
1
unem
t
1
+
u
t
E(
u
t

unem
t
1
,
gM
t
1
,
unem
t
2
,
gM
t
2
,
…
)
=
0
and 
1
 < 1.
Now, it is up to us to choose how many lags of
gM
to add to this equation.
The
simplest approach is to add
gM
t
1
and to do a
t
test.
But we could add a second or third lag (and
probably not beyond this with annual data), and compute an
F
test for joint significance of all
lags of
gM
t
.
18.9
Let
be the forecast error for forecasting
y
n
+1
, and let
1
ˆ
n
e
+
1
ˆ
n
a
+
be the forecast error for
forecasting
Δ
y
n
+1
.
By definition,
=
y
n
+1
−
1
ˆ
n
e
+
ˆ
n
f
=
y
n
+1
– (
+
y
n
) = (
y
n
+1
–
y
n
)
−
=
Δ
y
n
+1
−
=
, where the last equality follows by definition of the forecasting error for
Δ
y
n
+1
.
ˆ
n
g
ˆ
n
g
ˆ
n
g
1
+
ˆ
n
a
110