A2-S-431F08 - SOLUTIONS TO ASSIGNMENT 2 ACTSC 431/831, FALL...

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SOLUTIONS TO ASSIGNMENT 2 – ACTSC 431/831, FALL 2008 1. (a) i. E ( Y A ) = 1 2 ( E ( X 1 ) + E ( X 2 )) = 30 and E ( Y B ) = 1 2 ( E ( X 1 ) + E ( X 2 )) = 30. ii. E ( Y 2 A ) = 1 2 ( E ( X 2 1 ) + E ( X 2 2 )) = 1306 . 5 and thus V ar ( Y A ) = 1306 . 5 - 30 2 = 406 . 5. V ar ( Y B ) = 1 4 ( V ar ( X 1 ) + V ar ( X 2 )) = 13 / 4 = 3 . 25. iii. Let the pdfs of F 1 and F 2 be f 1 and f 2 , respectively. Then, the mgf of Y A is given by E ( e tY A ) = Z -∞ e ty 1 2 ( f 1 ( y ) + f 2 ( y )) dy = 1 2 ( M X 1 ( t ) + M X 2 ( t )) = 1 2 ( e 10 t + 5 2 t 2 + e 50 t + 8 2 t 2 ) . If Y A has a normal distribution, the mgf of Y A must be identical to e 30 t + 406 . 5 2 t 2 . Obviously, e 30 t + 406 . 5 2 t 2 is not identical to 1 2 ( e 10 t + 5 2 t 2 + e 50 t + 8 2 t 2 ). Hence, the distribution of Y A is not normal distribution. iv. The mgf of Y B is given by M Y B ( t ) = M X 1 2 ( t ) M X 2 2 ( t ) = e 10 t 2 + 5 2 ( t 2 ) 2 e 50 t 2 + 8 2 ( t 2 ) 2 = e 30 t + 13 4 t 2 2 , which is the mgf of the normal distribution
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A2-S-431F08 - SOLUTIONS TO ASSIGNMENT 2 ACTSC 431/831, FALL...

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