oldfinal1sol

# oldfinal1sol - UCSD ECE153 Handout#38 Prof Young-Han Kim...

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Unformatted text preview: UCSD ECE153 Handout #38 Prof. Young-Han Kim Thursday, June 2, 2011 Solutions to Final (Fall 2008) 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 . For 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) Find the pdf of Y 1 . (c) Find the pdf of Y 3 . (d) (Difficult.) Find 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 negationslash = j negationslash = 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 find f Y 3 ( y ) = 3 y 2 , ≤ y ≤ 1. Alternatively, we can see that by symmetry 1 − Y 3 and Y 1 should have the same pdf, which gives the same answer. (d) The event { Y 2 ≤ y } can be expressed as the union of following mutually exclusive events { Y 2 ≤ y } = { X 1 ,X 2 ≤ y,X 3 > y } ∪ { X 2 ,X 3 ≤ y,X 1 > y } ∪ { X 3 ,X 1 ≤ y,X 2 > y } ∪ { X 1 ,X 2 ,X 3 ≤ y } ....
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oldfinal1sol - UCSD ECE153 Handout#38 Prof Young-Han Kim...

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