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When to use bayes theorem, and when to not.

  1. Jun 10, 2013 #1
    Just need the 'auto' portion answered. this is more of a general question.

    1. The problem statement, all variables and given/known data
    (This is problem #7 on the sample actuarial exam P, soa 153)

    An insurance company estimates that 40% of policyholders who have only an auto policy will renew next year and 60% of policyholders who have only a homeowners policy will renew next year. The company estimates that 80% of policyholders who have both an auto and a homeowners policy will renew at least one of those policies next year. Company records show that 65% of policyholders have an auto policy, 50% of policyholders have a homeowners policy, and 15% of policyholders have both an auto and a homeowners policy.

    Using the company’s estimates, calculate the percentage of policyholders that will
    renew at least one policy next year.


    3. The attempt at a solution


    My first time looking at this, my thought process was that I will break this up in to cases and sum it all up:
    Probability they will renew, given that they have Auto only (.5)
    Probability they will renew, given that they have Homeowners only (.35)
    Probability they will renew, given that they have Auto and Homeowners (.15)

    And I immediately though "Bayes theorem." P[A given B] = P [AB]/ P

    So for Auto, it would be P [ they renewed, and they have auto] / P [they have auto]. which would be .5/.4.

    When I looked at the solution, it was not .5/.4, but it was .4 * .5. This actually makes sense intuitively (.4 of the .5), but I'm just in general, I dont quite understand when to apply Bayes theorem, and when to not.
     
  2. jcsd
  3. Jun 10, 2013 #2

    D H

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    Bayes' theorem isn't applicable here for a number reasons.

    First off, P(A|B) = P(A ∩ B) / P(B) -- that's not Bayes' theorem. That is Kolmogorov's definition of conditional probability. Bayes' theorem is P(A|B) = P(B|A) P(A) / P(B). Bayes' theorem tells us how to revert the sense of the conditionality, that is, how to go from P(B|A) to P(A|B). While Bayes' theorem is a direct consequence of that definition, you shouldn't conflate the definition with Bayes' theorem. They are distinct things.

    Secondly, the problem isn't to solve for the conditional probabilities. Those conditional probabilities are givens in this problem. For example "an insurance company estimates that 40% of policyholders who have only an auto policy will renew next year" is, in math, P(renewal | auto-only)=0.4.

    Thirdly, the problem doesn't provide any information on the joint probabilities P(A ∩ B) or on the reverse sense conditional probabilities P(B|A). You can't use either the definition of conditional probability or Bayes' theorem here for the simple reason that both formulae require missing information.


    What this problem is asking you to solve is the total probability P(renewal). What's the total probability P(B) given some disjoint partitioning of the sample space {Ei}, and given probabilities P(B|Ei) and P(Ei) for each partition Ei?
     
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