You observe n data pairs (Xi, Yi), i=1,..,n from simple linear regression model Yi=Bo+B1Xi+ei where ei's satisfy Gauss Markov Theorem. Let bo and b1 be least square estimates. Now let Yi' = (Yi-a)/c, where a,c are constants. Show that:
a) slope estimate b1' = b1/c for data (Xi,Yi), i=1,..,n
b) intercept estimate bo'=(bo-a)/c for data (Xi,Yi), i=1,..,n
Linear Regression
2010-10-20 4:07 pm
回答 (2)
2010-10-21 6:46 am
✔ 最佳答案
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2010-10-20 22:47:02 補充：
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2010-10-20 7:28 pm
Given two upper triangular matrices,
A = [ a1 * . . * ], B = [ b1 * . . * ]
0 a2 . . * 0 b2 . . *
: : : : : :
0 0 . . an 0 0 . . bn
their multiplication is still upper triangular, with the diagonal entries simply multiplied together (the other entries are more complicated).
AB = [ a1b1 * . . * ]
0 a2b2 . . *
: : :
0 0 . . anbn
The similar statement is true for the multiplication of two lower triangular matrices.
For two diagonal matrices,
A = [ a1 0 . . 0 ], B = [ b1 0 . . 0 ]
0 a2 . . 0 0 b2 . . 0
: : : : : :
0 0 . . an 0 0 . . bn
their multiplication is still diagonal, with the diagonal entries simply multiplied together.
AB = [ a1b1 0 . . 0 ]
0 a2b2 . . 0
: : :
0 0 . . anbn
A = [ a1 * . . * ], B = [ b1 * . . * ]
0 a2 . . * 0 b2 . . *
: : : : : :
0 0 . . an 0 0 . . bn
their multiplication is still upper triangular, with the diagonal entries simply multiplied together (the other entries are more complicated).
AB = [ a1b1 * . . * ]
0 a2b2 . . *
: : :
0 0 . . anbn
The similar statement is true for the multiplication of two lower triangular matrices.
For two diagonal matrices,
A = [ a1 0 . . 0 ], B = [ b1 0 . . 0 ]
0 a2 . . 0 0 b2 . . 0
: : : : : :
0 0 . . an 0 0 . . bn
their multiplication is still diagonal, with the diagonal entries simply multiplied together.
AB = [ a1b1 0 . . 0 ]
0 a2b2 . . 0
: : :
0 0 . . anbn
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