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An insurance company has two insurance portfolios. Claims in Portfolio Poccur inaccordance with a Poisson process with mean 3 per year. Claims in portfolio Qoccur inaccordance with a Poisson process with mean 5 per year. The two processes areindependent.Calculate the probability that 3 claims occur in Portfolio Pbefore 3 claims occur inPortfolio Q. I proceeded with this as you received three claims out of five from p and the sixth claim will be from q, like a normal negative binomial. Solution however assumes as min 3 claims out of first five from p. I didn’t quite get the logic

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Consider the process X = P + Q. This is also a Poisson process with mean 8 (=3+5). Probability of a claim in this process belonging to P is 3/8 and to Q is 5/8. If we think of a claim belonging to P as a success, then the random variable Y denoting the number of successes (claims from P) out of the first 5 claims is a Binomial random variable with success probability p = 3/8. Now we can ask what is the probability of Y=3,4 or 5.

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  • $\begingroup$ But it says only 3from p. , why do we take 4 and 5 $\endgroup$ Jun 15 '20 at 17:45
  • $\begingroup$ Because we are interested in the event that 3 from P occur before 3 from Q. So when looking at the first 5 claims, we must account for all the paths of the Poisson process where that happens. If 4 of the first 5 are from P, then that path is also included in the event "3 from P occur before 3 from Q". And similarly for 5 out of 5. Hence we must take at least 3, instead of exactly 3. $\endgroup$
    – dshirodkar
    Jun 16 '20 at 4:51

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