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Unformatted text preview: Mathematics Department, UCLA T. Richthammer spring 09, sheet 2 Apr 03, 2009 Homework assignments: Math 170B Probability, Sec. 1 10. X, Y have the joint PDF f ( x, y ) = xe x ( y +1) 1 { x,y> } . Calculate (a) E ( e XY (1+ Y ) 2 ) (b) E ( X ) (c) E ( XY ) (d*) E ( Y ) Would you expect that E ( XY ) = E ( X ) E ( Y ) without doing the calculations? Answer: (a) E ( e XY (1+ Y ) 2 ) = R dxdy e xy (1+ y ) 2 f ( x, y ) = R dx R dy x (1+ y ) 2 e x = R dxxe x R dy 1 (1+ y ) 2 = 1. (b) E ( X ) = R dxx 2 R dye x ( y +1) = R dxx 2 1 x e x = 1. (c) E ( XY ) = R dxx 2 R dyye x ( y +1) = R dxx 2 1 x 2 e x = 1. (d) E ( Y ) = R dxx R dyye x ( y +1) = R dxx 1 x 2 e x = R dx 1 x e x = since 1 x e x 1 2 x near 0 and R dx 1 x = . You should not expect E ( XY ) = E ( X ) E ( Y ) since X, Y are not independent. 11. In the situation of problem 09, calculate the expected length of the middle interval and the expected length of the interval containing 1/2 without calculating the PDF first, but using the formula for the expectation of a function of several RVs. Answer: As in the solution of problem 09, the length of the middle interval is L =  X 1 X 2  , so E ( L ) = R dx 1 dx 2  x 1 x 2  1 [0 , 1] 2 ( x 1 , x 2 ) = R 1 dx 1 R 1 dx 2 ( x 1 x 2 )1 { x 1 >x 2 } + R 1 dx 1 R 1 dx 2 ( x 2 x 1 )1 { x 2 >x 1 } = 2 R 1 dx 1 R x 1 dx 2 ( x 1 x 2 ) = 2 R 1 dx 1 1 2 x 2 1 = 1 3 . As in the solution of problem 09, the length of the interval containing 1 2 is K = (1 max { X 1 , X 2 } )1 { X 1 ,X 2 < 1 2 } +min { X 1 , X 2 } 1 { X 1 ,X 2 > 1 2 } +  X 1 X 2  1 { X 1 < 1 2 <X 2 or X 2 < 1 2 <X 1 } , and distinguishing the cases which of X 1 , X 2 is smaller we get E ( K ) = E ((1 X 2 )1 { X 1 <X 2 < 1 2 } )+ E ((1 X 1 )1 { X 2 <X 1 < 1 2 } )+ E ( X 2 1 { X 1 >X 2 > 1 2 } )+ E ( X 1 1 { X 2 >X 1 > 1 2 } )+ E (( X 1 X 2 )1 { X 2 < 1 2 <X 1 } )+ E (( X 2 X 1 )1 { X 1 < 1 2 <X 2 } ). Using symmetry and the formula for the expectation we thus get E ( K ) = 2 R 1 / 2 dx 2 (1 x 2 ) R x 2 dx 1 +2 R 1 1 / 2 dx 2 x 2 R 1 x 2 dx 1 +2 R 1 1 / 2 dx 1 R 1 / 2 dx 2 ( x 1 x 2 ) = 2 R 1 / 2 dx 2 ( x 2 x 2 2 ) + 2 R 1 1 / 2 dx 2 ( x 2 x 2 2 ) + 2 R 1 1 / 2 dx 1 ( 1 2 x 1 1 8 ) = 2( 1 2 1 3 ) + 1 4 = 7 12 ....
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This note was uploaded on 07/07/2010 for the course MATH 170B 170B taught by Professor Richthammer during the Winter '10 term at UCLA.
 Winter '10
 RICHTHAMMER

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