Econometric take home APPS_Part_40 - x1 y1 x1 y2 x y x y 1...

=
∂ ∂ ∂ ∂ ∂ ∂
∂ ∂ ∂ ∂ ∂ ∂
∂ ∂ ∂ ∂ ∂ ∂
x y x y x y
x y x y x y
x y x y x y
1 1 1 2 1 3
2 1 2 2 2 3
3 1 3 2 3 3
/ / /
/ / /
/ / /
⎡
⎣
⎢
⎢
⎢
⎤
⎦
⎥
⎥
⎥
x
x
x
1
2
3
0 0
0 0
0 0
⎡
⎣
⎢
⎢
⎢
⎤
⎦
⎥
⎥
⎥
1 1 2 1 2
0 1 2 1 2
1 1 0
−
−
⎡
⎣
⎢
⎢
⎢
⎤
⎦
⎥
⎥
⎥
/ /
/ /
1 0 0
0 1 0
0 0 1
1
2
3
/
/
/
.
y
y
y
⎡
⎣
⎢
⎢
⎢
⎤
⎦
⎥
⎥
⎥
The determinant of the product matrix is the product of the three determinants. The determinant of the center
matrix is -1/2. The determinants of the diagonal matrices are the products of the diagonal elements.
Therefore, the Jacobian is J = abs(| ∂ x / ∂ y ′ |)= ½( x
1
x
2
x
3
)/( y
1
y
2
y
3
) = 2( y
1
/ y
2
) (after making the substitutions for
x
i
).
20. Prove that exchanging two columns of a square matrix reverses the sign of its determinant. ( Hint: use a
permutation matrix. See Exercise 6.)
Exchanging the first two columns of a matrix is equivalent to postmultiplying it by a permutation
matrix B = [ e
2
, e
1
, e
3
, e
4
,...] where e
i
is the i th column of an identity matrix. Thus, the determinant of the matrix
is | AB | = | A | | B |. The question turns on the determinant of B . Assume that A and B have n columns. To obtain
the determinant of B , merely expand it along the first row. The only nonzero term in the determinant is (-1)| I
n -
1
| = -1, where I
n -1
is the ( n -1) ( n -1) identity matrix. This completes the proof.
×
21. Suppose x = x ( z ) where z is a scalar. What is ∂ [( x ′ Ax )/( x ′ Bx )]/ z ?
The required derivatives are given in Exercise 16. Let g = ∂ x / ∂ z and let the numerator and
denominator be a and b , respectively. Then,
∂ ( a / b )/ ∂ z = [ b ( ∂ a / ∂ z ) - a ( ∂ b / ∂ z )]/ b
2
= [ x ′ Bx (2 x ′ Ag ) - x ′ Ax (2 x ′ Bg )] / ( x ′ Bx )
2
= 2[ x ′ Ax / x ′ Bx ][ x ′ Ag / x ′ Ax - x ′ Bg / x ′ Bx ].
22. Suppose y is an n × 1 vector and X is an n × K matrix. The projection of y into the column space of X is
defined in the text after equation (2-55), = Xb . Now, consider the projection of y
*
= c y into the column
space of X
*
= XP where c is a scalar and P is a nonsingular K
ˆ y
×
K matrix. Find the projection of y
*
into the
column space of X
*
. Prove that the cosine of the angle between y
*
and its projection into the column space of
X
*
is the same as that between y and its projection into the column space of X . How do you interpret this
result?
The projection of y
*
into the column space of X
*
is X
*
b
*
where b
*
is the solution to the set of
equations X
*
′ y
*
= X
*
′ X
*
b
*
or P ′ X ′ ( c y ) = P ′ X ′ XPb
*
. Since P is nonsingular, P ′ has an inverse.
Premultiplying the equation by ( P ′ )
-1
, we have c X ′ y = X ′ X ( Pb
*
) or X ′ y = X ′ X [(1/ c ) Pb
*
]. Therefore, in
terms of the original y and X , we see that b = (1/ c ) Pb
*
which implies b
*
= c P
-1
b . The projection is X
*
b
*
=
( XP )( c P
-1
b ) = c Xb . We conclude, therefore, that the projection of y
*
into the column space of X
*
is a
multiple c of the projection of y into the space of X . This makes some sense, since, if P is a nonsingular
matrix, the column space of X
*
is exactly the same as the same as that of X . The cosine of the angle between
y
*
and its projection is that between c y and c Xb .