sol6 - STA3007 Applied probability(08-09 Chapter 4 Solution...

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STA3007 Applied probability (08-09) Solution to Assignment 6 Chapter 4 Problem 3.1 (a) By definition, f (0) 00 = 0, and f (1) 00 = Pr( X 1 = 0 | X 0 = 0) = P 00 = 1 - a, f ( n ) 00 = Pr { X 1 6 = 0 , ··· , X n - 1 6 = 0 , X n = 0 | X 0 = 0 } = Pr { X 1 = 1 , ··· , X n - 1 = 1 , X n = 0 | X 0 = 0 } = Pr( X 1 = 1 | X 0 = 0)Pr( X n = 0 | X n - 1 = 1) n - 1 Y k =2 Pr( X k = 1 | X k - 1 = 1) = ab (1 - b ) n - 2 , ( n 2) . where Q 1 k =2 Pr( X k = 1 | X k - 1 = 1) = 1 is defined as 1. (b) We are asked to show b a + b + a a + b (1 - a - b ) n = P ( n ) 00 = n X k =0 f ( k ) 00 P ( n - k ) 00 for n 1. Since b a + b n X k =2 f ( k ) 00 = b a + b n X k =2 ab (1 - b ) k - 2 = ab 2 a + b 1 - (1 - b ) n - 1 1 - (1 - b ) = ab a + b (1 - (1 - b ) n - 1 ) . and a a + b n X k =2 f ( k ) 00 (1 - a - b ) n - k = a 2 b (1 - a - b ) n ( a + b )(1 - b ) 2 n X k =2 ( 1 - b 1 - a - b ) k = a 2 b (1 - a - b ) n ( a + b )(1 - b ) 2 ( 1 - b 1 - a - b ) 2 - ( 1 - b 1 - a - b ) n +1 1 - 1 - b 1 - a - b = ab a + b [(1 - b ) n - 1 - (1 - a - b ) n - 1 ] . 1
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Thus, n X k =0 f ( k ) 00 P ( n - k ) 00 = f (1) 00 P ( n - 1) 00 + n X k =2 f ( k ) 00 P ( n - k ) 00 = (1 - a )[ b a + b + a a + b (1 - a - b ) n - 1 ] + ab a + b [1 - (1 - a - b ) n - 1 ] = (1 - a ) b + ab a + b + a a + b (1 - a - b ) n - 1 (1 - a - b ) = b a + b + a a + b (1 - a - b ) n = P ( n ) 00 as was to be shown. Problem 3.2
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sol6 - STA3007 Applied probability(08-09 Chapter 4 Solution...

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