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flyless
July 29th 2008, 09:39 PM
I have one question about aggregate loss. It is in ASM manual:

14.6 You are given:
*An aggregate loss distribution has a compound Poisson distribution with expected number of claims equal to 1.25;
*Individual claim amounts can take only the value 1,2 or 3, with equal probability.

Determine the probability that aggregate losses exceed 3.
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The first step is to factor out the the probability of 0, which is e^(-1.25)=0.2865048, however, the answer in the manual just simply multiples that number. Did i miss sth here?

Any thoughts?:wacko:

Abraham Weishaus
August 2nd 2008, 10:59 PM
I have one question about aggregate loss. It is in ASM manual:

14.6 You are given:
*An aggregate loss distribution has a compound Poisson distribution with expected number of claims equal to 1.25;
*Individual claim amounts can take only the value 1,2 or 3, with equal probability.

Determine the probability that aggregate losses exceed 3.
-----------------------------------------------------------
The first step is to factor out the the probability of 0, which is e^(-1.25)=0.2865048, however, the answer in the manual just simply multiples that number. Did i miss sth here?

Any thoughts?:wacko:

What exactly don't you understand?

I calculate probability of 0, 1, 2. I then take 1 minus those 3 probabilities, which results in the probability of 3 or more.

The probabilities of 0, 1, and 2 are all e^{-1.25} times something else. In each of the 3 probabilities, I calculate the "something else" and then multiply the sum by e^{-1.25}. You may multiply each probability separately by e^{-1.25} if you wish - my method saves 2 operations.