# HW9sol - ORIE 3500/5500 Fall Term 2008 Assignment 9 Solution 1 The total rain T = T1 TN where N Binomial(30.25 and Ti exp(1/2(a E(Ti = 2 So E(T |N

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ORIE 3500/5500 Fall Term 2008 Assignment 9 Solution 1. The total rain T = T 1 + ··· T N , where N Binomial (30 , . 25) and T i exp(1 / 2). (a) E ( T i ) = 2. So E ( T | N ) = 2 N . Hence E ( T ) = E [ E ( T | N )] = E [2 N ] = 2 · 30 · 0 . 25 = 15 . (b) var ( T 1 ) = 2 2 = 4. So var ( T | N ) = Nvar ( T 1 ) = 4 N . Hence var ( T ) = var [ E ( T | N )] + E [ var ( T | N )] = var [2 N ] + E [4 N ] = 4 · 30 · 0 . 25 · (1 - 0 . 25) + 4 · 30 · 0 . 25 = 52 . 5 . 2. f S ( s ) = e - s for s 0 and o otherwise. f W ( w ) = 1 for 0 w 1 and 0 otherwise. So f S,Y ( s, y ) = f S ( s ) f Y | S ( y | s ) = ± e - s , 0 s y s + 1 0 , otherwise . So the marginal of Y f Y ( y ) = Z y max( y - 1 , 0) e - s ds = 1 - e - y , 0 y 1 e - ( y - 1) - e - y , y 1 0 , otherwise . So E ( S | Y = y ) equals Z y max( y - 1 , 0) s f S,Y ( s, y ) f Y ( y ) ds = ± [1 - (1 + y ) e - y ] / [1 - e - y ] , 0 y 1 [ ye - ( y - 1) - (1 + y ) e - y ] / [ e - ( y - 1) - (1 + y ) e - y ] , y 1 .

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## This note was uploaded on 12/04/2008 for the course STSCI 6000 taught by Professor Turnbull during the Fall '08 term at Cornell University (Engineering School).

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HW9sol - ORIE 3500/5500 Fall Term 2008 Assignment 9 Solution 1 The total rain T = T1 TN where N Binomial(30.25 and Ti exp(1/2(a E(Ti = 2 So E(T |N

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