{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

tu4-solutions - ACTL2003&5103 Tutorial 4 Sample Solutions 1...

Info iconThis preview shows pages 1–2. Sign up to view the full content.

View Full Document Right Arrow Icon
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 = 0 , 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 ( N B ( t + 1) - N B ( t ) = 0 | N B ( t ) = x ) = X b Prob ( N B ( t + 1) - N B ( t ) = 0 | B = b, N B ( t ) = x ) Prob ( B = b | N B ( t ) = x ) = X b ( e - λb ) e - λbt ( λbt ) x x ! π ( b ) a e - λat ( λat ) x x ! π ( a ) = b e - λb ( t +1) b x π ( b ) a e - λat a x π ( a ) .
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Image of page 2
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}