finalsol - UCSD ECE 153 Prof Young-Han Kim Handout#28...

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UCSD ECE 153 Handout #28 Prof. Young-Han Kim Wednesday, December 10, 2008 Solutions to Final Exam (Total: 200 points) There are 4 problems. The Frst 3 problems have 4 parts each, while the last problem has 8 parts. Each part is uniformly worth 10 points. Your answer should be as clear and readable as possible. In particular, if the answer involves a pmf or pdf, make sure to identify the values or intervals for which the pmf or pdf is nonzero. 1. Order statistics. Let X 1 ,X 2 ,X 3 be independent and uniformly drawn from the interval [0 , 1]. Let Y 1 be the smallest of X 1 ,X 2 ,X 3 , let Y 2 be the median (second smallest) of X 1 ,X 2 ,X 3 , and let Y 3 be the largest of X 1 ,X 2 ,X 3 . ±or example, if X 1 = . 3 ,X 2 = . 1 ,X 3 = . 7, then Y 1 = . 1 ,Y 2 = . 3 ,Y 3 = . 7. The random variables Y 1 ,Y 2 ,Y 3 are called the order statistics of X 1 ,X 2 ,X 3 . (a) What is the probability P { X 1 X 2 X 3 } ? (b) ±ind the pdf of Y 1 . (c) ±ind the pdf of Y 3 . (d) (Di²cult.) ±ind the pdf of Y 2 . (Hint: Y 2 y if and only if at least two among X 1 ,X 2 ,X 3 are y .) Solution: (a) By symmetry, P { X i X j X k } should be identical for all i n = j n = k . Since there are 3! = 6 such ( i,j,k ), the probability should be 1 / 6. (b) We have P { Y 1 > y } = P { X 1 ,X 2 ,X 3 > y } = P { X 1 > y } P { X 2 > y } P { X 3 > y } = (1 y ) 3 . Hence, f Y 1 ( y ) = d dy (1 (1 y ) 3 ) = 3(1 y ) 2 for 0 y 1. (c) We can use the similar steps to part (b) to Fnd f Y 3 ( y ) = 3 y 2 , 0 y 1. Alternatively, we can see that by symmetry 1
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finalsol - UCSD ECE 153 Prof Young-Han Kim Handout#28...

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