Simultaneous Equations Models

1. Consider the following two equation model:

y = Yy + Pn*i + P21X2 + P31X3 + si y2 = Y2yi + P12X1 + P22X2 + P32X3 + 62.

(a) Verify that as stated, neither equation is identified.

(b) Establish whether or not the following restrictions are sufficient to identify (or partially identify) the model: (i) p21 = p32 = 0,

Since nothing is excluded from either equation and there are no other restrictions, neither equation passes the order condition for identification.

(1) We use (15-12) and the equations which follow it. For the first equation, [A3' ,A5' ] = P22, a scalar which has rank M-1 = 1 unless P22 = 0. For the second, [A3' ,A5' ] = P31. Thus, both equations are identified.

(2) This restriction does not restrict the first equation, so it remains unidentified. The second equation is now identified, as [A3' ,A5' ] = [P11,P21] has rank 1 if either of the two ceofficients are nonzero.

(3) If y1 equals 0, the model becomes partially recursive. The first equation becomes a regression which can be estimated by ordinary least squares. However, the second equation continues to fail the order condition. To see the problem, consider that even with the restriction, any linear combination of the two equations has the same variables as the original second eqation.

(4) We know from above that if P32 = 0, the second equation is identifiable. If it is, then y2 is identified. We may treat it as known. As such, y1 is known. By regressing y1 - Y1y2 on the xs, we would obtain estimates of the remaining parameters, so these restrictions identify the model. It is instructive to analyze this from the standpoint of false structures as done in the text. A false structure which incorporates

the known restrictions would be

1

- Y

-X

1

'11

P12

'21

P 22

'31

then f11

f11 f21

f12 f22.

If the false structure is to obey the restrictions,

■ Yfa P31 f12 = 0. It follows then thatf12 = 0 so f11 = 1. Then, f21 --Y or f21 = (f11 - 1)y so thatf11 - Y2(/11 - 1) = 1. This can only hold for all values of y iff11 = 1 and, = 0. Therefore, F = I which establishes identification.

(5) If p31 = 0, the first equation is identified by the usual rank and order conditions. Consider, then, the off-diagonal element of E = F QT. Q is identified since it is the reduced form covariance matrix. The off-diagonal element is ct12 = ra11 + ra22 - (y1 + Y2)ro12 = 0. Since Y1 is zero, y2 = ro12/(ro11 + ro22). With y2 known, the remaining parameters are estimable by least squares regression of (y2 - Y2y1) on the xs. Therefore, the restrictions identify the model.

(6) Since this is only a single restriction, it will not likely identify the entire model. Consider again the false structure. The restrictions implied by the theory are f11 - Y2f21 = 1, f22 - Yf12 = 1, Pf + P22f21 =

P21/12 + P22/22- The three restrictions on four unknown elements of F do not serve to pin down any of them. This restriction does not even partially identify the model.

(7) The last four restrictions remove x2 and x3 from the model. The remaining model is not identified by the usual rank and order conditions. From part (5), we see that the first restriction implies cti2 = ran + ra22 - (yj + y2)ra12 = 0. But, with neither yj nor y2 specified, this does not identify either parameter.

(8) The first equation is identified by the conventional rank and order conditions. The second equation fails the order condition. But, the restriction ct12 = 0 provides the necessary additional information needed to identify the model. For simplicity, write the model with the restrictions imposed as yj = yy + sj and y2 = yy + Px + 62. The reduced form is y1 = n1x + v1 and y2 = n2x + v2

where n1 = y1p/A and n2 = p/A with A = (1 - y1y2), and v1 = (sj + y162)/A and v2 = (s2 + y261)/A. The reduced form variances and covariances are ran = (y12a22 + a11)/A2, ra22 = (y22a11 + a22)/A2, ra12 = (y1a22 + y2CTn)/A2 All reduced form parameters are estimable directly by using least squares, so the reduced form is identified in all cases. Now, y1 = n1/n2. ctjj is the residual variance in the euqation (y1 - y1y2) = 61, so ctjj must be estimable (identified) if y1 is. Now, with a bit of manipulation, we find that y1ro12 - ra11 = -ctjj/A. Therefore, with ctjj and y1 "known" (identified), the only remaining unknown is y2, which is therefore identified. With y1 and y2 in hand, p may be deduced from n2. With y2 and p in hand, ct22 is the residual variance in the equation (y2 - Px -y2y1) = s2, which is directly estimable, therefore, identified.

2. Verify the rank and order conditions for identification of the second and third behavioral equation in Klein's Model I. [Hint: See Example 15.6.]

Following the method in Example 15.6, for identification of the investment equation, we require that

the matrix

(1)

(2)

(3)

(4)

(5)

(6)

(7)

(8) (9)

-1

a 3

0

0

a 3

0

0

0 0

0

-1

y 1

0

0

0

0

y 3 y 2

0

0

-1

0

0

1

0

0 0

0

-1

1

0

0

0

-1

0 0

0

0

0

1

0

0

0

0 0

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. 3. Check the identifiability of the parameters of the following model:

(i)

(2)

(3)

(4)

(5)

(6)

(7)

(8)

(9)

(10)

-1

0

ai

0

a 3

0

0

0

a 2

0

0

-1

Pi

0

0

0

0

0

P2

P3

1

1

0

0

0

01

0

0

0

0

0

0

-1

0

0

0

-1

0

0

0

0

1

0

-1

0

0

0

0

0

1

[y1 y2 y3 y4]

1

y 12

0

0

y 21

1

y 23

y 24

0

y 32

1

y 34

y 4i

y 42

0

1

+ [x1 x2 x3 x4 x5]

0

P12

P13

Pl4

P 21

1

0

P24

P 31

P 32

P 33

0

0

0

P 43

P 44

0

P52

0

Y 32

1

Y 34 "

" 1

Y12

0 "

ß 12

ß 13

ß14

, [0 ß 43 ß44 ],

Y 41

Y 42

1

0

ß 43

ß 4

ß 21

1

0

ß32

0

0

0

ß52

Y 12

Identification requires that the rank of each matrix be M-1 = 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 y1 = Yy + Pn*i + P21x2 + P31x3 + s1 y2 = Y2V1 + P12X1 + P22X2 + P32X3 + S2.

Therefore, r =

Y2 1

and £ is unrestricted. The reduced form is q = (r"1)'e(r"1) =

Y 2CT11 + Y22 Y 2^11 +CT 22 _+(Y1 + Y 2)CT12 + 2y (6) The model is y1 = P11x1 + P21x2 + P31x3 + s1

y2 = Y2y1 + P12X1 + P22X2 + P32X3 + S2 The first equation is already a reduced form. Substituting it into the second provides the second reduced form.

The coefficient matrix is P=

CT11 Y 2CT11 Y 2 CT11 Y 2ct11 +CT 2

0 0

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