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Thread: Discrete Probability Distribution

  1. #1
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    Discrete Probability Distribution

    I have been unable to solve the following exercise from my Probability and Statistics book.

    Under an insurance policy, a maximum of five claims may be filed per year by a policyholder. Let $\displaystyle p_n$ be the probability that a policyholder files n claims during a given year, where $\displaystyle n$ = 1,2,3,4,5. An actuary makes the following observations:

    (i) $\displaystyle p_n \geq p_{n+1}$ for $\displaystyle n$ = 1,2,3,4.
    (ii) The difference between $\displaystyle p_n$ and $\displaystyle p_{n+1}$ is the same for $\displaystyle n$ = 1,2,3,4.
    (iii) Exactly 40% of the policyholders file fewer than three claims during a given year.

    Calculate the probability that a random policyholder will file more than three claims during a given year.

    (A) 0.14 (B) 0.16 (C) 0.27 (D) 0.29 (E) 0.33
    I succeeded in expressing observations (ii) and (iii) as mathematical statements:

    (ii) $\displaystyle p_n - p_{n+1} = k$ for $\displaystyle n$ = 1,2,3,4.
    (iii) $\displaystyle p_1 + p_2 = 0.4$

    I also know the probabilities sum to 1, so

    $\displaystyle p_1 + p_2 + p_3 + p_4 + p_5 = 1$

    To solve the exercise, I need to find $\displaystyle p_4 + p_5$, but I have been unable to do so. I tried rewriting the sum of probabilities like so:

    $\displaystyle p_1 + (p_1 - k) + (p_1 - 2k) + (p_1 - 3k) + p_5 = 1 \iff 4p_1 - 6k + p_5 = 1$

    I have tried other approaches, but it seems I am aimlessly solving equations that are just restatements of each other. I believe if I could find $\displaystyle p_1$ and $\displaystyle k$, then I could find $\displaystyle p_4$ and $\displaystyle p_5$. If anyone could point me in the right direction, I would appreciate it.

    By the way, the solutions manual says the answer is (C) 0.27.
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  2. #2
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    Re: Discrete Probability Distribution

    Quote Originally Posted by NOX Andrew View Post
    I have been unable to solve the following exercise from my Probability and Statistics book.



    I succeeded in expressing observations (ii) and (iii) as mathematical statements:

    (ii) $\displaystyle p_n - p_{n+1} = k$ for $\displaystyle n$ = 1,2,3,4.
    (iii) $\displaystyle p_1 + p_2 = 0.4$

    I also know the probabilities sum to 1, so

    $\displaystyle p_1 + p_2 + p_3 + p_4 + p_5 = 1$

    To solve the exercise, I need to find $\displaystyle p_4 + p_5$, but I have been unable to do so. I tried rewriting the sum of probabilities like so:

    $\displaystyle p_1 + (p_1 - k) + (p_1 - 2k) + (p_1 - 3k) + p_5 = 1 \iff 4p_1 - 6k + p_5 = 1$

    I have tried other approaches, but it seems I am aimlessly solving equations that are just restatements of each other. I believe if I could find $\displaystyle p_1$ and $\displaystyle k$, then I could find $\displaystyle p_4$ and $\displaystyle p_5$. If anyone could point me in the right direction, I would appreciate it.

    By the way, the solutions manual says the answer is (C) 0.27.


    i dont understand the proposed solution.

    $\displaystyle p_1 = p_2 = p_3 = p_4 = p_5= 0.2$ satisfies all the conditions and would give an answer of 0.4.


    Moreover, the above solution is unique:

    (ii)
    $\displaystyle p_n - p_{n+1} = k$
    =>$\displaystyle p_n =a + kn$

    (iii)
    $\displaystyle p_1 + p_2 = 0.4$
    =>$\displaystyle (a+k) + (a + 2k) = 0.4$
    =>$\displaystyle 2a+3k = 0.4~~~~~~~~~~:$ [1]

    And as you already noticed:
    $\displaystyle p_1 + p_2 + p_3 + p_4 + p_5 = 1$
    =>$\displaystyle 5a + 15k = 1 ~~~~~~~~~:$ [2]

    The only solutions to [1] and [2] are a=0.2, k=0

    and so $\displaystyle p_n = 0.2$
    Last edited by SpringFan25; Jul 10th 2011 at 01:28 AM.
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    Re: Discrete Probability Distribution

    Thank you for your reply. In [2], it appears you made the substitution $\displaystyle p_5 = a + 5k$, but $\displaystyle p_n - p_{n + 1} = k$ only applies for $\displaystyle n = 1,2,3,4$. Does this change anything?

    Since it doesn't apply to $\displaystyle p_5$, a possible distribution satisfying the conditions is as follows $\displaystyle p_1 = 0.21$, $\displaystyle p_2 = 0.19$, $\displaystyle p_3 = 0.17$, $\displaystyle p_4 = 0.15$, $\displaystyle p_5 = 0.28$, in which case the answer is 0.43.
    Last edited by NOX Andrew; Jul 10th 2011 at 09:30 AM.
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  4. #4
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    Re: Discrete Probability Distribution

    the relationship $\displaystyle p_n = a + kn$ applies to n=1,2,3,4,5.

    You already noted that $\displaystyle p_n - p_{n+1} = k$ (n=1,2,3,4)

    use n=4, so:
    $\displaystyle p_4 - p_5 = k$

    and then use the same reasoning as before.


    Remark
    If you believe that the above isn't valid for $\displaystyle p_5$ then there are infinately many solutions to the problem(this isn't hard to show).
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  5. #5
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    Re: Discrete Probability Distribution

    Of course! It seems so obvious now. Thank you for understanding my question about applying it to p_5 and providing such a clear, concise explanation.
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  6. #6
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    Re: Discrete Probability Distribution

    FYI there is a mistake in my first post

    when i said
    $\displaystyle p_n - p_{n+1} = k$
    $\displaystyle => p_n =a + kn$
    I changed the sign of k. This doesn't affect the result
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