Dr. Hackney STA Solutions pg 191

Dr. Hackney STA Solutions pg 191 - Chapter 12 Regression...

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Chapter 12 Regression Models 12.1 The point (ˆ x 0 , ˆ y 0 ) is the closest if it lies on the vertex of the right triangle with vertices ( x 0 ,y 0 ) and ( x 0 ,a + bx 0 ). By the Pythagorean theorem, we must have h x 0 - x 0 ) 2 + ( ˆ y 0 - ( a + bx 0 ) ) 2 i + h x 0 - x 0 ) 2 +(ˆ y 0 - y 0 ) 2 i = ( x 0 - x 0 ) 2 + ( y 0 - ( a + bx 0 )) 2 . Substituting the values of ˆ x 0 and ˆ y 0 from (12.2.7) we obtain for the LHS above " ± b ( y 0 - bx 0 - a ) 1+ b 2 ² 2 + ± b 2 ( y 0 - bx 0 - a ) 1+ b 2 ² 2 # + " ± b ( y 0 - bx 0 - a ) 1+ b 2 ² 2 + ± y 0 - bx - a ) 1+ b 2 ² 2 # = ( y 0 - ( a + bx 0 )) 2 " b 2 + b 4 + b 2 +1 (1+ b 2 ) 2 # = ( y 0 - ( a + bx 0 )) 2 . 12.3 a. Diﬀerentiation yields ∂f/∂ξ i = - 2( x i - ξ i ) - 2 λβ [ y i - ( α + βξ i )] set = 0 ξ i (1 + λβ 2 ) = x i - λβ ( y i - α ), which is the required solution. Also, 2 f/∂ξ 2 = 2(1 + λβ 2 ) > 0, so this is a minimum. b. Parts i), ii), and iii) are immediate. For iv) just note that D is Euclidean distance between ( x 1 , λy 1 ) and ( x 2 , λy 2 ), hence satisﬁes the triangle inequality. 12.5 Diﬀerentiate log L , for L in (12 . 2 . 17), to get ∂σ 2 δ log L = - n σ 2 δ + 1 2( σ 2 δ ) 2 λ 1+ ˆ β 2 n X i =1 h y i - α + ˆ βx i ) i 2 . Set this equal to zero and solve for σ 2 δ . The answer is (12 . 2 . 18). 12.7 a. Suppressing the subscript i and the minus sign, the exponent is
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This note was uploaded on 02/03/2012 for the course STA 1014 taught by Professor Dr.hackney during the Spring '12 term at UNF.

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