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# s7 - Math 4653 Elementary Probability Spring 2007...

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Math 4653: Elementary Probability: Spring 2007 Homework #7. Problems and Solutions 1. Ch. 4, Review: #21: Suppose R 1 and R 2 are two independent random variables with the same density function f ( x ) = x exp( - 1 2 x 2 ) for x 0. Find a) the density of Y = min { R 1 , R 2 } ; b) the density of Y 2 ; c) E ( Y 2 ). Solution. a) For each y > 0, we have P ( R k > y ) = y f ( x ) dx = exp - 1 2 y 2 , k = 1 , 2; P ( Y > y ) = P (min { R 1 , R 2 } > y ) = P ( R 1 > y, R 2 > y ) = P ( R 1 > y ) · P ( R 2 > y ) = exp( - y 2 ) . Therefore, f Y ( y ) = - d dy P ( Y > y ) = 2 y exp( - y 2 ) for y > 0 , and f Y ( y ) = 0 for y 0. b) The density of Z = Y 2 is f Z ( z ) = f Y ( y ) dz/dy = 2 y exp( - y 2 ) 2 y = e - z for z > 0 , and f Z ( z ) = 0 for z 0. c) E ( Y 2 ) = E ( Z ) = λ = 1. 2. Sec. 5.2: #16: Suppose X 1 , X 2 , X 3 are independent exponential random variables with parameters λ 1 , λ 2 , λ 3 respectively. Evaluate P ( X 1 < X 2 < X 3 ). Solution. P ( X 1 < X 2 < X 3 ) = x 1 <x 2 <x 3 λ 1 e - λ 1 x 1 λ 2 e - λ 2 x 2 λ 3 e - λ 3 x 3 dx 1 dx 2 dx 3 = 0 λ 1 e - λ 1 x 1 x 1 λ 2 e - λ 2 x 2 x 2 λ 3 e - λ 3 x 3 dx 3 dx 2 dx 1 = 0 λ 1 e - λ 1 x 1 x 1 λ 2 e - ( λ 2 + λ 3 ) x 2 dx 2 dx 1 = λ 1 λ 2 λ 2 + λ 3 0 e - ( λ 1 + λ 2 + λ 3 ) x 1 dx 1 = λ 1 λ 2 ( λ 2 + λ 3 )( λ 1 + λ 2 + λ 3 ) . 3. Sec. 5.3: #2: Let X and Y be independent random variables, with E ( X ) = 1, E ( Y ) = 2, Var ( X ) = 3, and Var ( Y ) = 4. a) Find E (10 X 2 + 8 Y 2 - XY + 8 X + 5 Y - 1). b) Assuming all variables are normally distributed, find P (2 X > 3 Y - 5). Solution. a) We have E ( X 2 ) = Var ( X ) + [ E ( X )] 2 = 3 + 1 2 = 4 , E ( Y 2 ) = Var ( Y ) + [ E ( Y )] 2 = 4 + 2 2 = 8 , 1

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and, since X and Y and independent, E ( XY ) = E ( X ) · E ( Y ) = 1 · 2 = 2. Therefore, E (10 X 2 + 8 Y 2 - XY + 8 X + 5 Y - 1) = 10 · 4 + 8 · 8 - 2 + 8 · 1 + 5 · 2 - 1 = 119 .
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