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bv12
May 8th 2006, 03:03 PM
the probability generating function. I haven't seen any and I am wondering how they are used on the exam since they are given for each distribution. If this is a dumb question I would not be surprised but thanks for any responses anyway.

ACDW
May 10th 2006, 03:01 PM
In the Nov 2005 there is one question -#27

.Godspeed.
May 10th 2006, 03:10 PM
the probability generating function. I haven't seen any and I am wondering how they are used on the exam since they are given for each distribution. If this is a dumb question I would not be surprised but thanks for any responses anyway.
Also, see Problem #40 of Spring 2004 CAS 3, which is indeed fair game for M. You can solve this problem rather quickly by using the probability generating function of the Poisson distribution to find the probability that aggregate losses are 0.

What we are really asked is P(Aggregate Losses - Aggregate Deductible > Premium Received) = P(S - 100 > 150) = P(S > 250) = 1- P(S<=250).

After applying the \$500 per-event deductible, we see that the probability of a loss, X, of \$0 is .10+.25=.35.

So, P(S<=250) really just equals the probability that aggregate losses are 0. Pr(S=0)=PGF(P(X=0))=PGF(.35), where PGF() denotes the probability generating function of the frequency distribution. Our PGF, P(z), is Poisson: e^[lambda(z-1)]; so, P(S=0)=PGF(.35)=e^[.15(.35-1)]=.9071.

Thus, P(S > 250) = 1-.9071=.093 (E).

Is that at least somewhat helpful?