γ
1
"known" (identified), the only remaining unknown is
γ
2
, which is therefore identified.
With
γ
1
and
γ
2
in
hand,
β
may be deduced from
π
2
.
With
γ
2
and
β
in hand,
σ
22
is the residual variance in the equation (
y
2

β
x

γ
2
y
1
) =
ε
2
, which is directly estimable, therefore, identified.
±
2.
Following the method in Example 13.6, for identification of the investment equation, we require that the
matrix
have rank 5.
Columns (1), (4), (6), (7), and (8) each
have one element in a different row, so they are linearly independent.
Therefore, the matrix has rank five.
For
the third equation, the required matrix is
.
Columns
(4), (6), (7), (9), and (10) are linearly independent.
±
() () () () () () () () ()
123456789
1
00
0000
01
0
0
0
0
1
001000
1
0
0
0
0
000100000
33
1
−
−
−
−−
⎡
⎣
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎤
⎦
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
αα
γγ
3
2
γ
3
β
12
41
42
21
52
γ
β
β
⎡
⎣
⎢
⎢
⎢
⎢
⎤
⎦
⎥
⎥
⎥
⎥
10
12
31
32
33
52
γ
βββ
β
⎡
⎣
⎢
⎢
⎢
⎤
⎦
⎥
⎥
⎥
() () () () () () () () () ( )
1234567891
0
0
0 0 0
0
0
0
0
0
0
11000
0
1
000
0
0 0 10 0 0 10 0 0
010
1
00000
1
13
2
12
−
−
−
⎡
⎣
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎤
⎦
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
α
ββ
3.
We find [
A
3
′
,
A
5
′
]
′
for each equation.
(1)
(2)
(3)
(4)
β
32
34
12
13
14
43
4
32
1
0
⎡
⎣
⎢
⎢
⎢
⎢
⎤
⎦
⎥
⎥
⎥
⎥
,
[]
,
,
0
43
44
1
0
Identification requires that the rank of each matrix be M1 = 3.
The second is obviously not identified.
In (1),
none of the three columns can be written as a linear combination of the other two, so it has rank 3.
(Although
the second and last columns have nonzero elements in the same positions, for the matrix to have short rank,
we would require that the third column be a multiple of the second, since the first cannot appear in the linear
combination which is to replicate the second column.)
By the same logic, (3) and (4) are identified.
±
4.
Obtain the reduced form for the model in Exercise 1 under each of the assumptions made in parts (a) and
(b1), (b6), and (b9).
(1).
The model is
y
1
=
γ
1
y
2
+
β
11
x
1
+
β
21
x
2
+
β
31
x
3
+
ε
1
y
2
=
γ
2
y
1
+
β
12
x
1
+
β
22
x
2
+
β
32
x
3
+
ε
2
.
Therefore,
Γ
=
and
B
=
and
Σ
is unrestricted.
The reduced form is
1
1
2
1
−
−
⎡
⎣
⎢
⎤
⎦
⎥
γ
γ
−
−
⎡
⎣
⎢
⎢
⎢
⎤
⎦
⎥
⎥
⎥
β
β
11
12
22
31
0
0
Π
=
1
1
11
1 21
2 11
12
2
2
2
31
2 31
−
++
⎡
⎣
⎢
⎢
⎢
⎤
⎦
⎥
⎥
⎥
βγ
γβ
β
β
and
91