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Unformatted text preview: Solutions to Homework 06 1. a. From the figure, f X ( x ) = (  x  > a c 1  x  a  x  a Now, 1 = Z  f X ( x ) dx = Z a a c 1  x  a ! dx = 2 Z a c 1 x a dx = 2 c " x x 2 2 a # a = ac c = 1 a b. F X ( x ) = 0 for x < a and F X ( x ) = 1 for x > a . For a x 0,  x  = x . Then F X ( x ) = 1 a Z x a 1 + x a ! dx = 1 2 + 1 a x + x 2 2 a ! For 0 x a ,  x  = x . Then F X ( x ) = Z a f X ( x ) dx + Z x f X ( x ) dx = 1 2 + 1 a Z x a 1 + x a ! dx = 1 2 + 1 a x x 2 2 a ! c. From part b , we know 0 < b < a . 1 2 = P [  X  < b ] = F X ( b ) F X ( b ) = 2 a b b 2 2 a ! b = a 1 1 2 ! 1 2. For a geometric random variable, P [ X = k ] = (1 p ) k 1 p , k = 1 , 2 , . a. A = { X > k } . F X ( x  A ) = P [ { X x } A ] P [ A ] = P [ { X x } { X > k } ] P [ { X > k } ] = ( F X ( x ) F X ( k ) 1 F X ( k ) for x > k else = ( (1 p ) k (1 p ) x (1 p ) k for x > k else b. A = { X < k } ....
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This note was uploaded on 09/03/2008 for the course ECSE 4510 taught by Professor Woods during the Spring '03 term at Rensselaer Polytechnic Institute.
 Spring '03
 WOODS

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