Problem5

2014 Qual Exam Appl Problem 5 Tsung-Hung Yao May 25, 2018 Problem 5. (a) l = - 1 2 σ 2 1 n 1 X i =1 ( y 1 i - β 10 - β 1 x 1 i ) 2 - 1 2 σ 2 2 n 2 X i =1 ( y 2 i - β 20 - β 1 x 2 i ) 2 ∂l ∂β j, 0 = 1 σ 2 j n j X i =1 ( y ji - β j 0 - β 1 x ji ) = 0 ⇒ β j 0 = ¯ y j - β 1 ¯ x j ∂l ∂β 1 = 1 σ 2 1 ( n 1 X i =1 ( y 1 i - β 10 - β 1 x 1 i ) x 1 i ) + 1 σ 2 2 ( n 2 X i =1 ( y 2 i - β 20 - β 1 x 2 i ) x 2 i ) = 1 σ 2 1 ( n 1 X i =1 y 1 i x 1 i - n 1 β 10 ¯ x 1 - β 1 n 1 X i =1 x 2 1 i ) + 1 σ 2 2 ( n 2 X i =1 y 2 i x 2 i - n 2 β 20 ¯ x 2 - β 1 n 2 X i =1 x 2 2 i ) Since n j X i =1 y ji x ji - n j β j 0 ¯ x j - β 1 n j X i =1 x 2 ji = ( n j X i =1 y ji x ji - n j ¯ x j ¯ y j ) - β 1 ( n j X i =1 x 2 ji - n j ¯ x 2 j ) = S j,xy - β 1 S j,xx ∂l ∂β 1 = S 1 ,xy - β 1 S 1 ,xx σ 2 1 + S 2 ,xy - β 1 S 2 ,xx σ 2 2 = 0 ˆ β 1 = S 1 ,xy /σ 2 1 + S 2 ,xy /σ 2 2 S 1 ,xx /σ 2 1 + S 2 ,xx /σ 2 2 Also, since ˆ β k, 1 = S k,xy S k,xx , V k, 1 = σ 2 k S k,xx ˆ β 1 = ˆ β 1 , 1 /V 1 , 1 + ˆ β 2 , 1 /V 2 , 1 1 /V 1 , 1 + 1 /V 2 , 1 = ˆ β 1 , 1 V 2 , 1 + ˆ β 2 , 1 V 1 , 1 V 1 , 1 + V 2 , 1 (b) ˆ β k, 1 ∼ N ( β 1 , V k, 1 ) ˆ β 1 = V 2 , 1 V 1 , 1 + V 2 , 1 ˆ β 1 , 1 + V 1 , 1 V 1 , 1 + V 2 , 1 ˆ β 2 , 1 ∼ N ( β 1 , V 1 , 1 V 2 , 1 V 1 , 1 + V 2 , 1 ) 1