Assignment3

Assignment3 - r-1)! 1 and ( y-r )! 1, so y-1 r-1 = ( y-1)!...

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ASSIGNMENT #3 SOLUTIONS (1) We know that 1 = Z 1 0 cy (1 - y ) dy = c Z 1 0 ( y - y 2 ) dy = c y 2 2 - y 3 3 ± ± ± ± 1 0 ! = c 6 , so c = 6. (2) (a) F ( y ) = Z y -∞ f ( t ) dt = 0 , y ≤ - 1 1 - y 2 2 , - 1 < y 0 1+ y 2 2 , 0 < y 1 1 , y > 1 (b) P ( - 0 . 5 < Y 0 . 5) = F (0 . 5) - F ( - 0 . 5) = 1+0 . 5 2 2 - 1 - ( - 0 . 5) 2 2 = 0 . 25. (c) P ( Y 0 . 5) = 0 . 5 2 + 0 . 5 2 2 = 0 . 375. (d) E ( Y ) = Z 0 - 1 - y 2 dy + Z 1 0 y 2 dy = - y 3 3 ± ± ± ± 0 - 1 + y 3 3 ± ± ± ± 1 0 = ( - 1) 3 3 + 1 3 3 = 0. (e) E ( Y 2 ) = Z 0 - 1 - y 3 dy + Z 1 0 y 3 dy = - y 4 4 ± ± ± ± 0 - 1 + y 4 4 ± ± ± ± 1 0 = ( - 1) 4 4 + 1 4 4 = 1 2 . Therefore, Var( Y ) = E ( Y 2 ) - [ E ( Y )] 2 = 1 2 . (3) Y Poisson(0 . 4 × 16 = 6 . 4). Therefore P ( Y < 3) = P ( Y = 0) + P ( Y = 1) + P ( Y = 2) = 6 . 4 0 0! e - 6 . 4 + 6 . 4 1 1! e - 6 . 4 + 6 . 4 2 2! e - 6 . 4 0 . 046. (4) (a) E ( Y ) = m 0 (0) = 45 120 e t + 40 120 e 2 t + 30 120 e 3 t + 5 120 e 5 t ± ± 0 = 120 120 = 1. E ( Y 2 ) = m 00 (0) = 45 120 e t + 80 120 e 2 t + 90 120 e 3 t + 25 120 e 5 t ± ± 0 = 240 120 = 2. Var( Y ) = E ( Y 2 ) - [ E ( Y )] 2 = 2 - 1 2 = 1. (b) The probability function is: y 0 1 2 3 4 5 p ( y ) 44 120 45 120 20 120 10 120 0 1 120 Therefire, P ( Y 3) = P ( Y = 3)+ P ( Y = 4)+ P ( Y = 5) = 10 120 +0+ 1 120 = 11 120 0 . 092. (5) Note that m Y ( t ) = ( 0 . 6 e t + 0 . 4 ) 8 and m W ( t ) = e 4 t m Y (3 t ) = e 4 t ( 0 . 6 e 3 t + 0 . 4 ) 8 . (6) (a) (i) If y 6 = r,r + 1 ,r + 2 ,... , then p ( y ) = 0 is nonnegative. If y = r,r + 1 ,r + 2 ,... then y r , so y - r 0 and since p and q are positive we have that p r > 0 and q y - r > 0. Moreover, since y - 1 r - 1 0 we have that ( y - 1)! 1
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2 SOLUTIONS (
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Unformatted text preview: r-1)! 1 and ( y-r )! 1, so y-1 r-1 = ( y-1)! ( r-1)!( y-r )! 0. Therefore, y-1 r-1 p r q y-r 0. (ii) X y = r p ( y ) = X y = r y-1 r-1 p r q y-r = X k =0 k + r-1 r-1 p r q k = p r X k =0 k + r-1 r-1 q k = p r (1-q )-r = 1 (b) m ( t ) = X y = r e ty p ( y ) = X y = r e ty y-1 r-1 p r q y-r = X k =0 e t ( k + r ) k + r-1 r-1 p r q k = e tr p r X k =0 k + r-1 r-1 ( e t q ) k = e tr p r (1-e t q )-r = pe t 1-qe t r...
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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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Assignment3 - r-1)! 1 and ( y-r )! 1, so y-1 r-1 = ( y-1)!...

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