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Unformatted text preview: f X ( x ) = ½ 1 / 41 ≤ x ≤ 3 otherwise We recognize that X is a uniform random variable from [1,3]. (a) E [ X ] = 1 and Var [ X ] = ( 3 + 1 ) 2 12 = 4 / 3. (b) The new random variable Y is defined as Y = h ( X ) = X 2 . Therefore h ( E [ X ]) = h ( 1 ) = 1 and E [ h ( X )] = E £ X 2 ¤ = Var [ X ]+ E [ X ] 2 = 4 / 3 + 1 = 7 / 3 Finally E [ Y ] = E [ h ( X )] = E £ X 2 ¤ = 7 / 3 Var [ Y ] = E £ X 4 ¤E £ X 2 ¤ 2 = 31 x 4 4 dx49 9 = 61 549 9 Problem 4.3.4 (a) We can find the expected value of X by direct integration of the given PDF. f Y ( y ) = ½ y / 2 ≤ y ≤ 2 otherwise The expectation is E [ Y ] = 2 y 2 2 dy = 4 / 3 (b) E £ Y 2 ¤ = 2 y 3 2 dy = 2 Var [ Y ] = E £ Y 2 ¤E [ Y ] 2 = 2( 4 / 3 ) 2 = 2 / 9 2...
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 Spring '09
 MarshallSylvan
 Probability theory, probability density function, Roy D. Yates

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