HW5 - Mathematics Department, UCLA T. Richthammer spring...

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Mathematics Department, UCLA T. Richthammer spring 09, sheet 5 Apr 24, 2009 Homework assignments: Math 170B Probability, Sec. 1 42. Choose a number n completely at random from { 1 , 2 , 3 , 4 } , then choose a number k com- pletely at random from { 1 , . . . , n } . Given the value of k , calculate the distribution and the expected value of the first number in all cases k = 1 , 2 , 3 , 4. Answer: Let X denote the first number and Y the second number. Given that Y = k , the possible values of X are k, . . . , 4 and we have p X ( n | Y = k ) = P ( X = n,Y = k ) P ( Y = k ) . P ( X = n, Y = k ) = 1 4 1 n and P ( Y = k ) = P ( X = k, Y = k ) + . . . + P ( X = 4 , Y = k ) = 1 4 ( 1 k + . . . + 1 4 ). Thus p X ( n | Y = k ) = 1 n 1 k + ... + 1 4 . Using that formula we get: Given Y = 1, X = 1 , 2 , 3 , 4 with probabilities 12 25 , 6 25 , 4 25 , 3 25 and E ( X | Y = 1) = 48 25 = 1 . 92. Given Y = 2, X = 2 , 3 , 4 with probabilities 6 13 , 4 13 , 3 13 and E ( X | Y = 2) = 36 13 2 . 77. Given Y = 3, X = 3 , 4 with probabilities 4 7 , 3 7 and E ( X | Y = 3) = 24 7 3 . 43. Given Y = 4, X = 4 with probability 1 and E ( X | Y = 4) = 4. 43. Let X, Y have the joint PDF f ( x, y ) = 24 xy 1 { x 0 ,y 0 ,x + y 1 } . (a) Calculate P ( Y 1 2 | X = 0), P ( Y 1 2 | X = 1 4 ), P ( Y 1 2 | X = 1 2 ). (b) Calculate E ( Y | X = 1 2 ) and E ( 1 Y | X = 1 2 ). Answer: (a) For 0 x 1 we have f X ( x ) = R 1 - x 0 24 xydy = 12 x (1 - x ) 2 , so f Y ( y | X = x ) = 2 y (1 - x ) 2 for 0 y 1 - x . We get P ( Y 1 2 | X = 0) = R 1 1 / 2 2 ydy = 3 4 , P ( Y 1 2 | X = 1 4 ) = R 3 / 4 1 / 2 32 9 ydy = 5 9 , P ( Y 1 2 | X = 1 2 ) = R 1 / 2 1 / 2 8 ydy = 0. (b) f Y ( y | X = 1 2 ) = 8 y 1 [0 , 1 / 2] ( y ) from (a). Thus E ( Y | X = 1 2 ) = R 1 / 2 0 8 y 2 dy = 1 3 and E ( 1 Y | X = 1 2 ) = R 1 / 2 0 8 dy = 4. 44. Let X, Y have the joint PDF f ( x, y ) = e - y y 1 { 0 <x<y } . Calculate E ( X 3 | Y = y ). Answer: f Y ( y ) = e - y , so f X ( x | Y = y ) = 1 y for 0 < x < y , i.e. X is a uniform RV. Using this density we get E ( X 3 | Y = y ) = R y 0 x 3 1 y = 1 y 1 4 y 4 = y 3 4 . 45. We choose a point X from [0 , 1], and after that we independently choose X 1 and X 2 from [0 , X ] and [ X, 1] respectively. (All choices are made completely at random.) (a) Calculate the expected distance of X 1 and X 2 , given that X = x . (b) Explain your result in (a) without any calculation. (c) Calculate the expected value of X , given that X 1 = x 1 and X 2 = x 2 . (d) Check your result of (c) in the special case x 1 = 1 2 - c , x 2 = 1 2 + c for arbitrary c [0 , 1 2 ]. Which value do you get for x 1 = 1 4 , x 2 = 1 2 ?
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Answer: (a) We are given that f X 1 ,X 2 ( x 1 , x 2 | X = x ) = 1 x 1 [0 ,x ] ( x 1 ) 1 1 - x 1 [ x, 1] ( x 2 ). Using this density we get E ( X 2 - X 1 | X = x ) = R x 2 - x 1 x - x 2 1 [0 ,x ] ( x 1 )1 [ x, 1] ( x 2 ) dx 1 dx 2 = 1 x - x 2 R x 0 dx 1 R 1 x dx 2 ( x 2 - x 1 ) = 1 x - x 2 R x 0 dx 1 ( 1 2 - 1 2 x 2 - x 1 (1 - x )) = 1 2 .
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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.

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HW5 - Mathematics Department, UCLA T. Richthammer spring...

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