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Homework Help: The probability of a member being part of a group

  1. Sep 27, 2012 #1
    1. The problem statement, all variables and given/known data
    There are two groups, group 1 and 2. Group 1 has a 0.5 chance of losing $36, while group 2 has a 0.1 chance of losing $36 dollars. The groups are of equal size. Now an insurance company is willing to cover the losses for a payment. However, the insurance company has an imperfect test for determining what group a person belongs to. The chance that a person actually belongs to the group the insurance thinks they are part of, is X. (0<x<1)
    If a person from group 2 is willing to pay a maximum of $4 to the insurance company, what value should X have?

    2. Relevant equations
    I can't really think of any relevant equations, it seems like a rather straightforward probability question.

    3. The attempt at a solution
    Assuming that the insurance company had no test to determine who was part of what group, they would assume that the probability of someone losing $36 would be equal to (0.1+0.5)/2 = 0.3; they would ask them to pay $10.8. This is more than the $4 that group 2 members are willing to pay.
    However, if the probability is p, then for p = 1/9 members of group 2 will pay the insurance money, as 1/9*36 = $4. However, I don't know how to go to a value of X from here. All I know is that the insurance probably has to think that a large portion of the combined group actually belongs to group 2, to lower the average probability. But whatever I try, I can't seem to come up with a good approach for calculation the value of X. Any push in the right direction would be much appreciated.

    Possibly, it could be solved like this:
    If the insurance selects someone from group 2, they will ask a price of 0.1X x 36 + 0.5(1-X) x 36. Solving this for $4:
    0.1X*36+0.5(1-X)*36 = 4 → 0.1X+0.5-0.5X = 1/9 → 0.4X = 0.5 - 1/9
    So for X = 0.972, group 2 will pay the insurance money

    Is that correct?
    Last edited: Sep 27, 2012
  2. jcsd
  3. Sep 27, 2012 #2

    Ray Vickson

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    There is crucial information missing from your problem description: what does the insurance company do with the information supplied by the test? Do they just give the test and then ignore the results? Or, do they sell insurance only if they think the person belongs to group 2?

    BTW: your writeup of your solution has some things backwards. The company does not ask a price of 0.1X x 36 + 0.5(1-X) x 36. It asks a price of $4. The company's payout is $0 if no loss occurs, and is $36 if a loss occurs. So, you need to compute the expected payout, which must be computed using *conditional* probabilities of belonging to group 1 or group 2, given the results of the test.

    Last edited: Sep 27, 2012
  4. Sep 27, 2012 #3
    Hm, alright, well I rephrased the question a bit. The actual question was

    There are two groups of equal size, each with a utility function given by U(M) = M^1/2
    where M = 100 is the initial wealth level for every individual. Each member of group 1
    faces a loss of 36 with probability 0.5. Each member of group 2 faces the same loss with
    probability 0.1.
    a. What is the most a member of each group would be willing to pay to insure against
    this loss?
    b. In part (a), if it is impossible for outsiders to discover which individuals belong to
    which group, will it be practical for members of group 2 to insure against this loss in
    a competitive insurance market? (For simplicity, you may assume that insurance
    companies charge only enough in premiums to cover their expected benefit payments.)
    c. Now suppose that the insurance companies in part (b) have an imperfect test for
    identifying which persons belong to which group. If the test says that a person belongs
    to a particular group, the probability that he really does belong to that group
    is x < 1.0. How large must x be in order to alter your answer to part (b)?

    I already solved question a, which gave me the $4 information, and b was just (I think) the 0.3x36=$10.8.

    I don't think that the information that is missing is in here, or I am not seeing it. I think that the insurance company does the test, but ignores the result; they don't know if their test has failed or not.

    Your comment does make sense, but I wouldn't know how to incorporate that. Could you give me a hint?
    Last edited: Sep 27, 2012
  5. Sep 27, 2012 #4

    Ray Vickson

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    Of course the company does not know if the test failed or not, but if they ignore the results they might as well not bother doing the test. The only reason to do the test is to use the results in some way, and the two most common forms of that would be: (i) sell insurance at a higher premium if you think the person is in group 1; or (ii) refuse to sell insurance to someone you think is in group 1. Both methods are used in practice.

    It is important to know whether the group1/2 losses are independent or not; that is, does every member of group 1 have a 0.5 chance of a loss, independent of what happens to other members of group 1, or is it the case that they all either win or lose together? If the first applies (which is, typically what happens in real insurance problems) then the actual number of $36 payouts the insurance company would make to group 1 is a binomially-distributed random variable, with parameters N1 and 0.1, where N1 = number of members of group 1 members the company actually insures. The corresponding number of payouts to group 2 would be binomial(N2,0.1), where N2 = number of group 2 members it insures. On the other hand, if all members win or lose together, the payout to group 1 would be 0 with probability 0.5 and 36N1 with probability 0.5, where N1 = number of group 1's insured, etc.
    Now: about the test. The same considerations apply here: are the individual test results independent among the different members of the group? If so, then when testing N group 1 members for membership in group 1, the number of correct results is binomial with parameters N and x, hence the number of group 1's that are said to be in group 2 is the complement of that, which will be binomial with parameters N and 1-x. That is the N1 referred to before. Similarly, for testing N members of group 2 we get N2 = binomial(N,x).

    Now, I would not expect all these things to be done properly in a first course in probability at the pre-calculus level, so I would expect simplifying approximations to be made. For example, we could ignore distributions and just look at mean values; and we could approximate the mean of a product of two random variables as the product of their means, etc.

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