tu4-solutions - ACTL2003&5103 Tutorial 4 Sample...

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Unformatted text preview: ACTL2003&5103 Tutorial 4- Sample Solutions 1. Define { N B ( t ) } as a Poisson process with rate λB . If we know what B is, then obviously for x = , 1 , 2 ,... Prob( N B ( t ) = x | B = b ) = e- λbt ( λbt ) x x ! . From standard conditional probability definitions, we have: Prob( B = b | N B ( t ) = x ) = Prob( B = b,N B ( t ) = x ) Prob( N B ( t ) = x ) The numerator represents the joint probability of B and { N B ( t ) } . For x = 0 , 1 , 2 ,... and b ∈ { 1 , 2 , 3 ,... } , we have Prob( B = b,N B ( t ) = x ) = Prob( N B ( t ) = x | B = b )Prob( B = b ) = e- λbt ( λbt ) x x ! π ( b ) , from which we can also derive the denominator: Prob( N B ( t ) = x ) = X b Prob( B = b,N B ( t ) = x ) = X b e- λbt ( λbt ) x x ! π ( b ) . And finally we have: Prob( B = b | N B ( t ) = x ) = e- λbt ( λbt ) x x ! π ( b ) ∑ a e- λat ( λat ) x x ! π ( a ) . To derive the probability that the specified individual will not have an accident in ( t,t + 1) , we can do the following: Prob(...
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This note was uploaded on 06/12/2011 for the course ASB 2003 taught by Professor Kim during the Three '11 term at University of New South Wales.

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tu4-solutions - ACTL2003&5103 Tutorial 4 Sample...

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