Homework12 - y 1 ) f ( y 2 ), where Y 1 Gamma(2 , 2) and Y...

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HOMEWORK PROBLEMS #12 SOLUTIONS TO EVEN PROBLEMS 8. (Extra credit) (a) Z 1 0 Z 1 0 ky 1 y 2 dy 1 dy 2 = k Z 1 0 y 2 y 2 1 2 ± ± ± ± 1 0 dy 2 = k 2 Z 1 0 y 2 dy 2 = k 2 y 2 2 2 ± ± ± ± 1 0 = k 4 . Since this inte- gral must equal 1, it must be that k = 4. (b) (i) y 1 < 0 or y 2 < 0: F ( y 1 ,y 2 ) = 0. (ii) y 1 > 1 and y 2 > 1: F ( y 1 ,y 2 ) = 1. (iii) y 1 > 1 and 0 y 2 1: F ( y 1 ,y 2 ) = Z 1 0 Z y 2 0 4 t 1 t 2 dt 2 dt 1 = Z 1 0 4 t 1 y 2 2 2 dt 1 = y 2 2 . (iv) y 2 > 1 and 0 y 1 1: F ( y 1 ,y 2 ) = y 2 1 . (v) 0 y 1 1 and 0 y 2 1: F ( y 1 ,y 2 ) = Z y 1 0 Z y 2 0 4 t 1 t 2 dt 2 dt 1 = Z y 1 0 4 t 1 y 2 2 2 dt 1 = y 2 1 y 2 2 . (c) P ( Y 1 1 / 2 ,Y 2 3 / 4) = F (1 / 2 , 3 / 4) = ( 1 2 ) 2 ( 3 4 ) 2 = 9 64 . 102. C = 3 Y 1 + 5 Y 2 , E ( C ) = 3 E ( Y 1 ) + 5 E ( Y 2 ) = 3(40) + 5(65) = 445, and since Y 1 and Y 2 are independent, we have that Var( C ) = 9 Var( Y 1 ) + 25 Var( Y 2 ) = 9(4) + 25(8) = 236. 108. Using integration by parts we get that: E ( Y 1 - Y 2 ) = Z 0 Z y 2 ( y 1 - y 2 ) e - y 1 dy 1 dy 2 = Z 0 e - y 2 dy 2 = 1. E [( Y 1 - Y 2 ) 2 ] = Z 0 Z y 2 ( y 1 - y 2 ) 2 e - y 1 dy 1 dy 2 = 2 Z 0 Z y 2 ( y 1 - y 2 ) e - y 1 dy 1 dy 2 = 2(1) = 2. Var( Y 1 - Y 2 ) = 2 - 1 2 = 1. P ( Y 1 - Y 2 > 4) = Z 0 Z y 2 +4 e - y 1 dy 1 dy 2 = Z 0 e - ( y 2 +4) dy 2 = e - 4 0 . 02. Therefore, it is not very likely that a randomly selected customer would spend more than 4 minutes at the service window, we only expect that to happen about 2% of the time. 112. For y 1 > 0 and y 2 > 0 note that f ( y 1 ,y 2 ) = 1 4 y 1 e - y 1 / 2 · 1 2 e - y 2 / 2 = f (
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Unformatted text preview: y 1 ) f ( y 2 ), where Y 1 Gamma(2 , 2) and Y 2 Exp(2) and Y 1 and Y 2 are independent. Therefore, E ( Y 1 ) = 4, Var( Y 1 ) = 8, E ( Y 2 ) = 2 and Var( Y 2 ) = 4. So we have that: E ( C ) = 50+2 E ( Y 1 )+4 E ( Y 2 ) = 50 + 2(4) + 4(2) = 66 and Var( C ) = 4Var( Y 1 ) + 16Var( Y 2 ) = 4(8) + 16(4) = 96. 138. Y | Poisson( ) and Gamma( , ), so E ( Y | ) = , Var( Y | ) = , E ( ) = and Var( ) = 2 . Therefore, (a) E ( Y ) = E [ E ( Y | )] = E ( ) = . (b) Var( Y ) = E [Var( Y | )]+ Var[ E ( Y | )] = E ( )+ Var( ) = + 2 , so Y = p + 2 . 1...
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This note was uploaded on 02/23/2011 for the course MATH 444 taught by Professor Any during the Fall '10 term at Roosevelt.

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