S has a compound Poisson distribution with = 3 and probability of claims of size
ID: 2958052 • Letter: S
Question
S has a compound Poisson distribution with = 3 and probability of claims of size 1, 2, and 3 being 0.5, 0.25, and 0.25, respectively. Compute fs(x) for x = 0, 1, 2, 3 and 4.
Below, at the very bottom of the page is a sample problem and how it is solved...
I know how to get the very bottom which is the probabilities by using the formula:
e-.8(.8)n/n!... this is of course based on a problem where =.8 and probabilities .25, .375 and .375 (try to read the problem-- you need to squint your eyes a bit) But how do you get the rest of the data in the table.
Pls show the solution.
Explanation / Answer
First, let's talk about the table, i.e. the numbers in the table and not the margins. What's given are the conditional probabilities that S =x given that N=n. That is, the probability that the total value of the claims is x, given that there are n claims. 1st column: the prob. that S=0 given that there are no claims must be 1 -- hence there is a 1 at x=0 and the other x's are blanked out because they cannot arise. 2nd column: Here n=1. There is one claim. That claim could have 1, 2 or 3, with probs. 0.25, 0.375, 0.375, respectively. So P(S=1|N=1) = 0.25, P(S=2|N=1) = 0.375 and P(S=3|N=1) = 0.375. These conditional probs. are entered against x=1, x=2, x=3, resp. 3rd column: Here n=2. The total value of the claims could be 2 or 3 or ... or 6. How could x=2 arise? Well the values for both claims could be 1. So the prob. of this (assuming independence) is 0.25 x 0.25 = 0.0625. This is entered against x=2. How could x=3 arise? We could have (1,2) or (2,1). The probability that 1st claim have value 1 and second has value 2 is 0.25 x 0.375 = .09375. The probability that the 1st claim has value 2 and the second value 1 is also 0.09375. So the prob. that S=3 is 0.09375 + 0.09375 = 0.1875. This is entered against x=3. How could x=4 arise? We could have (1,3) (2,2) or (3,1), with probs. 0.25 x 0.375 + 0.375 x 0.375 + 0.25 x 0.375 = 0.328125. How could x=5 arise? We could have (2,3) or (3,2), with probs. 0.375 x 0.375 + 0.375 x 0.375 = 0.28125. How could x=6 arise? This can only happen if we get (3,3), with prob. .375 x .375 = 0.140625. So, does this make sense for the second column? 4th column: Here n=3 and so the total value of the claims could be 3 or 4 or ... or 9. How could x=3 arise? Well, only when we get (1,1,1). The probability of this is 0.25 x 0.25 x 0.25 = 0.015625. How could x=4 arise? Only if we get (1,1,2) or (1,2,1) or (2,1,1), with prob. 0.25 x 0.25 x 0.375 + 0.25 x 0.375 x 0.25 + 0.375 x 0.25 x 0.25 = 0.0234375 (the table is rounded here) How could x= 5 arise? Only if we get (1,1,3), (1,3,1), (3,1,1), (1,2,2), (2,1,2), (2,2,1). The probability of this is 3 x [ 0.25 x 0.25 x 0.375] + 3 x [ 0.25 x 0.375 x 0.375] = 0.17578125. How could x=6 arise? Only if we get (2,2,2) or (1,2,3), (2,1,3) or (2,3,1) or (3,1,2) or (3,2,1). The prob. is 0.375 x 0.375 x 0.375 + 6 x [ 0.25 x 0.375 x 0.375] = 0.263671875. OK? I hope you can see now how to understand these entries in the table and so do it yourself. Now, we come to the final column. These give the unconditional probabilities of the values taken by S. The basic formula here is P(S =x) = SUM[over n] P(S =x | N=n) x P(N=n). Now look at the table along rows. 1st row: So, for x=0 there is only one way this can happen -- when n=0. So P(S=0) = 1 x P(N=0) = 0.449329. That is, multiply the numbers in row 1 by the corresponding probs in the row margin and add up. 4th row: Let's consider x=3. This can arise if we have 1 or 2 or 3 claims. Hence, P(S=3) = P(S=3|N=1)x P(N=1) + P(S=3|N=2) x P(N=2) + P(S=3|N=3) xP(N=3) This is 0.375 x 0.359463 + 0.1875 x 0.143785 + 0.015625 x 0.038343 = 0.162358 (6 dp). Let's consider x=5 on 6th row P(S=5) = P(S=5|N=2) xP(N=2) + P(S=5|N=3) x P(N=3) + P(S=5|N=4)x P(N=4) + P(S=5|N=5) xP(N=5) Let's use the table to work this out. P(S=5) = 0.28125 x 0.143785 + 0.175781 x 0.038343 + 0.023438 x 0.007669 + 0.000977 x 0.001227 = 0.047360 (6 dp). OK? Is this enough? Can you see how this is working? I hope so.
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