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Gore Tree Lover
10-18-2001, 03:02 PM
Can anyone explain how to do this problem? These posterior distributionns confuse me? Seems like it should be easy but I just can't get it.

New at pd
10-18-2001, 03:04 PM
What is the question? I don't have the exams in front of me.

Dr T Non-Fan
10-18-2001, 03:05 PM
I think more people are confused about which course's exam you're discussing. If only...
(Yes, I'm hollow-victory-dancing.)

Gore Tree Lover
10-18-2001, 03:20 PM
Sorry, Exam 4 (spring 2001). If only.

The question asked to calculate the probability of a person earning more than 50,000 if they had no claims in the previous year.

I don't have the exam in front of me right now either.

Dr T Non-Fan
10-18-2001, 03:31 PM
If you're online, then you could at least have it minimized, from the SOA site.

Posterior probability question.
Number of claims for an employee follows Poisson with mean (100-p)/100;
p = salary in thousands.
Distribution of p is uniform along (0,100].

How to solve: I don't know. Already have credit for Exam 4. Don't have a text in front of me.

New at pd
10-18-2001, 03:31 PM
This is a straight-forward application of Bayesian credibility:

First, not that for prior distributions that are uniform distributions are negligible when calculating the posterior distribution.

A Poisson dist of mean lambda has
Pr(N=0 | lambda) = exp(-lambda) =
exp(-(1 - p/100))

The post. of p is then exp(-(1 - p/100)), divided by the integral of the above over p.

Then, just integrate the post. dist from 50K to 100K to get the probability that the employees salary is over 50K.

Gore Tree Lover
10-18-2001, 03:46 PM
The Soa solution talked about a normalizing Constant. What is that? I have been doing calculus for many many years and do not recall that phrase or maybe I just have a bad memory.

New at pd
10-18-2001, 03:54 PM
The normalizing constant that they are talking about is the integral of

Pr(p | no claims) * Pr(p), over all p

This integral is usually not 1, and is used to divide Pr(p | no claims) * Pr(p) so that the cumulative probablity of the posterior distribution is 1.