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Thread: Hard prediction

  1. #1
    Actuary.com - Level II Poster
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    Unhappy Hard prediction

    Hi everyone,
    I look through this question and hardly figure out how to do. The solution seem to be doesnt make sense to me. I dont think anyone will predict it will be done that way. Any comment???

    Question:
    As part of the underwritting process for insurance, each prospective policyholder is tested for high blood pressure. Let X represent the number of tests completed when the first person with high blood pressure is found. The expected value X is 12.5. Calculate the probability that the sixth person tested is the first one with high blood pressure.

    The solution goes like this...
    we assume that each test for high blood pressure is a bernoulli trial with probability of success p. Then X has the geometric dist. and 12.5=E(x)=1/p.
    Therefore p=0.08. using this, we find
    p(first 5 do not have high blood pressure)*p(the 6th does have high blood pressure)=(1-0.08)^5.0.08=0.053.
    :goofy:

  2. #2
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    Quote Originally Posted by Jolin View Post
    Hi everyone,
    I look through this question and hardly figure out how to do. The solution seem to be doesnt make sense to me. I dont think anyone will predict it will be done that way. Any comment???

    Question:
    As part of the underwritting process for insurance, each prospective policyholder is tested for high blood pressure. Let X represent the number of tests completed when the first person with high blood pressure is found. The expected value X is 12.5. Calculate the probability that the sixth person tested is the first one with high blood pressure.

    The solution goes like this...
    we assume that each test for high blood pressure is a bernoulli trial with probability of success p. Then X has the geometric dist. and 12.5=E(x)=1/p.
    Therefore p=0.08. using this, we find
    p(first 5 do not have high blood pressure)*p(the 6th does have high blood pressure)=(1-0.08)^5.0.08=0.053.
    :goofy:
    This is simply a question about the geometric distribution (i.e. the probability of having a number x of trials before a first success).

    Let X--Geometric distribution with parameter p. Then P(X=x) = p*q^(x-1)

    In this case, let success: event that person has high blood pressure

    I strongly suggest that you know by heart that E(x) of a geo distr = 1/p. (The proof requires series expansion, which I will not show here) Therefore, 12.5=1/p --> p=0.08 ---> q = 1-p = 0.92.

    Then it is simply a matter of replacing p=0.08, q=0.92 and x=6 in the formula above.

    (edited a typo..)
    Last edited by Jo_M.; August 8th 2007 at 05:22 AM.

  3. #3
    Actuary.com - Level VI Poster jthias's Avatar
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    Quote Originally Posted by Jo_M. View Post
    This is simply a question about the geometric distribution (i.e. the probability of having a number x of failures before a first success).

    Let X--Geometric distribution with parameter p. Then P(X=x) = p*q^(x-1)

    In this case, let success: event that person has high blood pressure

    I strongly suggest that you know by heart that E(x) of a geo distr = 1/p. (The proof requires series expansion, which I will not show here) Therefore, 12.5=1/p --> p=0.08 ---> q = 1-p = 0.92.

    Then it is simply a metter of replacing p=0.08, q=0.92 and x=6 in the formula above.
    If I can recall correctly there are basically two different geometric distributions.

    Let X and Y represent these..

    If we define X as the number of failures (out of n independent trials) until the first success, where each trial has probability p of success, then

    P(X=0) = p, P(X=1) = qp, P(X=2) =q^2*p,...

    so the probability function looks like px(x) = q^x*p, x=0,1,2,3,...

    and has E(X) = q/p.

    If, however, we define Y to be the the trial number on which the 1st success occurs, then P(Y=0) = 0 which differs from the previous case. Also we have...

    P(X=1) = p, P(X=2) =qp, P(X=3) =q^2*p,...

    and the corresponding probability function is py(y) = q^(y-1)*p, y=1,2,3,...

    and has E(Y) = 1/p.

    In this problem we're dealing with the trial number on which the first success occurs.

    So the random variable is of the form py(y) = q^(y-1)*p, y=1,2,3,...

    and has expected value 1/p.

    Y starts at a value of 1. If we were to let U = Y-1

    then pu(u) = q^u*p, u=0,1,2,3,...

    Then U is of the same form as X given earlier, and so

    U is the number of failures until the 1st success. If we calculate the expected value of Y we will get...

    E(Y) = E(U+1) = E(U) + 1 = q/p + 1 = (1-p)/p + 1 = 1/p so that we still get the same expected value.
    Last edited by jthias; August 7th 2007 at 11:42 PM. Reason: corrections highlighted

  4. #4
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    Well, i understand the solution...but i am just wondering how you know that you will have to use geometric distribution???

  5. #5
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    Based on this line you should be able surmise its a geometric distribution.

    " Let X represent the number of tests completed when the first person with high blood pressure is found."

    JGET

  6. #6
    Actuary.com - Level VI Poster jthias's Avatar
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    Quote Originally Posted by Jolin View Post
    Well, i understand the solution...but i am just wondering how you know that you will have to use geometric distribution???
    You use a geometric distribution (one of two) whenever you have n independent trials performed in succession each with the same probability of success p (and prob. of failure 1-p), and you want to determine the probability of either "the number of trials that end in failure until you get the first success" or the probability that "kth trial will be the first that yields a successful outcome".

    Let X be the number of failure until the 1st success.

    px(x) = q^x*p, x=0,1,2,3,...

    Let Y be number of the trial (1st, 2nd, 3rd,..) on which the first successful outcome occurs.

    py(y) = q^(y-1)*p, y=1,2,3,...

    Notice P(X=0) = P(Y=1), P(X=1) = P(Y=2),...

    X and Y are diffenrent random variables but can be equated via the transformation

    X = Y-1

    p = P(X=0) = P(Y-1=0) = P(Y=1) = p

    qp = P(X=1) = P(Y-1=1) = P(Y=2) = qp, and so on.
    Last edited by jthias; August 8th 2007 at 09:30 PM.

  7. #7
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    Thanks guys...i deeply understand now...

  8. #8
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    the geometric distribution is just a variation of the binomial, and if you know how to do binomial, you should be able to do the problem anyways - IF you know to get the probability from the expected value

    that is pretty intuitive: for example, on average, how many tosses do you think it will take of a fair coin to get a "heads"? it would seem logical to figure 2, and the probability of a "heads" is 1/2

    on average, how many rolls of a die would it take to get a 3? since all numbers 1-6 are equally probable, it would on average take 6, so the probability is 1/6, and in this way you can see that the probability will end up 1/E(X)

    then if you have the probability (of high b.p. in this case), you can do the problem using the binomial distribution - it is the probability of getting 5 with low b.p., THEN 1 with high b.p., and there is only one possible way to do that: LLLLLH = (0.92)(0.92)(0.92)(0.92)(0.92)(0.08) = 0.053

  9. #9
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    That is a very good intuitive...i am appreciated with ur detail explaination. I wish i can think like you are...Thanks!

  10. #10
    Actuary.com - Level VI Poster Ken's Avatar
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    Quote Originally Posted by Jolin View Post
    Well, i understand the solution...but i am just wondering how you know that you will have to use geometric distribution???
    After doing many similar problems, you'll think this is a brainless question.
    Whether you are the lion or the gazelle, when the sun comes up, you better be running.

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