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Econometric take home APPS_Part_40

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

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= 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 .

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