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hw1_sol

# hw1_sol - ORIE 4580/5580/5581 Solutions for Homework...

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Unformatted text preview: ORIE 4580/5580/5581 Solutions for Homework Assignment 1 Question 1 a) Note that the largest value Y can take is 4 and the smallest value Y can take is 0. Therefore, F Y ( y ) = P { Y ≤ y } = if y < P { X 2 ≤ y } = P {- √ y ≤ X ≤ √ y } if 0 ≤ y ≤ 4 1 if y > 4 To compute the expression in the middle, P {- √ y ≤ X ≤ √ y } = ∫ √ y- √ y ( 3 28 x 2- 3 14 x + 3 28 ) dx = 1 28 ∫ √ y- √ y ( 3 x 2- 6 x + 3 ) dx = 1 28 ( x 3- 3 x 2 + 3 x ) √ y- √ y = 1 28 [2 y 3 / 2 + 6 y 1 / 2 ] . b) The probability density function of Y is f Y ( y ) = dF Y ( y ) dy = 1 28 [3 y 1 / 2 + 3 y- 1 / 2 ] if 0 ≤ y ≤ 4 , and 0 otherwise. c) E { Y } = ∫ 4 y 1 28 [3 y 1 / 2 + 3 y- 1 / 2 ] dy = 3 28 ∫ 4 [ y 3 / 2 + y 1 / 2 ] dy = 3 28 ( 2 5 y 5 / 2 + 2 3 y 3 / 2 ) 4 = 272 / 140 . E { Y 2 } = ∫ 4 y 2 1 28 [3 y 1 / 2 + 3 y- 1 / 2 ] dy = 3 28 ∫ 4 [ y 5 / 2 + y 3 / 2 ] dy = 3 28 ( 2 7 y 7 / 2 + 2 5 y 5 / 2 ) 4 = 1296 / 245 ....
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hw1_sol - ORIE 4580/5580/5581 Solutions for Homework...

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