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# HW8 - Mathematics Department UCLA T Richthammer winter 09...

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Mathematics Department, UCLA T. Richthammer winter 09, sheet 8 Feb 20, 2009 Homework assignments: Math 170A Probability, Sec. 1 093. The following functions are CDFs of RVs. Determine the corresponding PDF: (a) F ( x ) = 1 x 2 +1 1 ( -∞ , 0) ( x ) + 1 [0 , ) ( x ) (b) F ( x ) = e x 2 1 ( -∞ , 0) ( x ) + (1 - e - x 2 )1 [0 , ) ( x ) Answer: We use f ( x ) = F X ( x ), so (a) f ( x ) = - 2 x ( x 2 +1) 2 1 ( -∞ , 0) ( x ) (b) f ( x ) = 1 2 e -| x | . 094. Let X be a RV with PDF f ( x ) = 1 18 x 2 1 [ - 3 , 3] ( x ). Calculate R ( Y ) and the PDF of Y , where (a) Y = X + 2 (b) Y = 3 X (c) Y = e X (d) Y = X 2 (e) Y = - X Answer: R ( X ) = [ - 3 , 3] and for c R ( X ) we have F X ( c ) = 1 18 R c - 3 x 2 = c 3 +27 54 . (a) R ( Y ) = [ - 1 , 5] and for c [ - 1 , 5] we have F Y ( c ) = P ( X + 2 c ) = P ( X c - 2) = F X ( c - 2), so f Y ( c ) = f X ( c - 2) = 1 18 ( c - 2) 2 1 [ - 1 , 5] ( c ) (b) R ( Y ) = [ - 9 , 9] and for c [ - 9 , 9] we have F Y ( c ) = P (3 X c ) = P ( X c/ 3) = F X ( c/ 3), so f Y ( c ) = f X ( c/ 3) / 3 = 1 486 c 2 1 [ - 9 , 9] ( c ) (c) R ( Y ) = [ e - 3 , e 3 ] and for c R ( Y ) we have F Y ( c ) = P ( e X c ) = P ( X ln c ) = F X (ln c ), so f Y ( c ) = f X (ln c ) /c = 1 18 (ln c )2 c 1 [ e - 3 ,e 3 ] ( c ) (d) R ( Y ) = [0 , 9] and for c R ( Y ) we have F Y ( c ) = P ( X 2 c ) = P ( | X | ≤ c ) = F X ( c ) - F X ( - c ), so f Y ( c ) = f X ( c ) 1 2 c + f X ( - c ) 1 2 c = c 18 1 [0 , 9] ( c ) (e) R ( Y ) = [ - 3 , 3] and for c R ( Y ) we have F Y ( c ) = P ( - X c ) = P ( X ≥ - c ) = 1 - F X ( - c ), so f Y ( c ) = f X ( - c ) = 1 18 x 2 1 [ - 3 , 3] ( c ) = f ( c ) 095. For a > 0, b R express the PDF of Y = aX + b in terms of the PDF f X of the RV X . Answer: F Y ( c ) = P ( aX + b c ) = P ( X c - b a ) = F X ( c - b a ), so f Y ( c ) = 1 a f X ( c - b a ). 096. Compute P (0 < X < 3), E ( X ) and V ( X ) if X has a PDF given by (a) f ( x ) = 5 x 4 1 [0 , 1] ( x ) (b) f ( x ) = 8 x 3 1 [2 , ) ( x ) (c) f ( x ) = 3 4 (1 - x 2 )1 ( - 1 , 1) ( x ) Answer: (a) R 3 0 f ( x ) dx = 1, E ( X ) = 5 6 , E ( X 2 ) = 5 7 , so V ( X ) = 5 252 (b) R 3 0 f ( x ) dx = 5 9 , E ( X ) = 4, E ( X 2 ) = , so V ( X ) = (c) R 3 0 f ( x ) dx = 1 2 , E ( X ) = 0, so V ( X ) = E ( X 2 ) = 1 5 . 097. If the density function of X is f ( x ) = ( a + bx 2 )1 [0 , 1] ( x ) and E ( X ) = 3 5 , determine a and b .

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