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Unformatted text preview: ISyE 2027B Probability with Applications Fall 2010 Homework 7 Solutions 1. For a < 1, F X ( a ) = 0; for a 1, F X ( a ) = a 1 x +1 dx = 1 a Thus F Y ( y ) = P ( Y y ) = P (ln( X ) y ) = P ( X e y ) = F X ( e y ) . For y < 0 ( e y < 1), F Y ( y ) = 0; for y 0 ( e y 1), F Y ( y ) = 1 e y . Y has exponential distribution with parameter . 2. F Y ( y ) = P ( Y y ) = P ( X y ) = P ( X y ) = 1 P ( X y ) = 1 F X ( y ) Take derivative on both sides, f Y ( y ) = f X ( y ). 3. Since we know f Z ( z ) = nz n 1 and f V ( v ) = n (1 v ) n 1 for z,v [0 , 1]. (a) Let n = 2, f Z ( z ) = 2 z and f V ( v ) = 2(1 v ), where Z = max { X 1 ,X 2 } and V = min { X 1 ,X 2 } . E [ Z ] = 1 2 z 2 dz = 2 3 E [ V ] = 1 2(1 v ) vdv = 1 3 (b) Similarly, when Z = max { X 1 ,X 2 ,...,X n } and V = min { X 1 ,X 2 ,...,X n } . E [ Z ] = 1 znz n 1 dz = 1 nz n dz = n n + 1 E [ V ] = 1 vn (1 v ) n 1 dv = 1 n (1 u ) u n 1...
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This note was uploaded on 02/21/2011 for the course ISYE 2027 taught by Professor Zahrn during the Fall '08 term at Georgia Institute of Technology.
 Fall '08
 Zahrn

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