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Thread: expected payment, Poisson distribution.

  1. #1
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    expected payment, Poisson distribution.

    A company buys a policy to insure its revenue in the event of major snowstorms that shut down business. The policy pays nothing for the first such snowstorm of the year and 10,000 for each one thereafter, until the end of the year. The number of major snowstorms per year that shut down business is assumed to have a Poisson distribution with mean 1.5. Calculate the expected amount paid to the company under this policy during a one-year period.

    answer option:
    (A) 2,769 (B) 5,000 (C) 7,231 (D) 8,347 (E) 10,578



    I will denote X as the event when a snowstorm occurs. where x follows parameter of poisson 1.5 mean.

    I will now make the distribution payment of the snowstorm (y) = 0, x<=1, and y=10,000, x>1

    So E[X] = 0 * p(x<=1) + 10,000*p(x>1)


    so E[X]= 10,000 * p(x>1)

    p(x>1)= ....? * this would be an easy problem if the mean was an exponential *lol*. So since this is a discrete case:

    would I do: 10,000* p(x=1) + 10,000 * p(x<=1). <---- complement of p(x>1) = p(x<=1), and I know for discrete cases you have to evaluate these things at p(x=1),

    for some reason I believe I am missing something here, such as a series of some sort.
    Last edited by math951; Nov 12th 2018 at 09:18 AM.
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    Re: expected payment, Poisson distribution.

    $\sum \limits_{k=2}^\infty~10000 \cdot p_{\lambda}(k) = $

    $\lambda=1.5$

    $10000(1-p_{\lambda}(0)-p_{\lambda}(1)) = $

    $10000\left(1 - \dfrac{\lambda^0 e^{-\lambda}}{0!} - \dfrac{\lambda^1 e^{-\lambda}}{1!}\right) \approx $

    $10000 \cdot 0.442175 = \$4421.75$

    I see this doesn't match any of the options. The problem seems pretty straightforward. I'll double check things.
    Last edited by romsek; Nov 12th 2018 at 09:26 AM.
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    Re: expected payment, Poisson distribution.

    why don't you have the series set to k=0? Don't we have to account for the first two snowstorms, even if the pay is nothing..? Well I guess it would not matter then. Never mind.
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    Re: expected payment, Poisson distribution.

    Quote Originally Posted by math951 View Post
    why don't you have the series set to k=0? Don't we have to account for the first two snowstorms, even if the pay is nothing..? Well I guess it would not matter then. Never mind.
    just the first 1.

    the event k=0 means it never snowed but we've still bought the insurance.

    Did the problem mention what the cost of the insurance was? Maybe they want the net payment which would subtract off that cost.
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    Re: expected payment, Poisson distribution.

    Quote Originally Posted by romsek View Post
    just the first 1.

    the event k=0 means it never snowed but we've still bought the insurance.

    Did the problem mention what the cost of the insurance was? Maybe they want the net payment which would subtract off that cost.
    No. The question is all they gave us (and the answer choices).
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    Re: expected payment, Poisson distribution.

    Quote Originally Posted by math951 View Post
    No. The question is all they gave us (and the answer choices).
    I screwed this up.

    it's actually

    $E[X] = 10000 \sum \limits_{k=2}^\infty~(k-1) \dfrac{\lambda^k e^{-\lambda}}{k!}$

    $E[X] = $7231.30$

    i.e. choice (c)

    see if you can reproduce that result from the sum.
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    Re: expected payment, Poisson distribution.

    Why do we use (k-1)? Is it because we take in account for when it snowed, but no payment?
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    Re: expected payment, Poisson distribution.

    Quote Originally Posted by math951 View Post
    Why do we use (k-1)? Is it because we take in account for when it snowed, but no payment?
    if there are 2 crazy snows they pay out 1x10000 dollars

    if there are k crazy snows they pay out (k-1)x10000 dollars
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    Re: expected payment, Poisson distribution.

    Quote Originally Posted by romsek View Post
    I screwed this up.

    it's actually

    $E[X] = 10000 \sum \limits_{k=2}^\infty~(k-1) \dfrac{\lambda^k e^{-\lambda}}{k!}$

    $E[X] = $7231.30$

    i.e. choice (c)

    see if you can reproduce that result from the sum.
    $\lambda=1.5$

    $p_k = \dfrac{\lambda^k e^{-\lambda}}{k!}$

    $p_0 = \dfrac{\lambda^0 e^{-\lambda}}{0!} = e^{-\lambda}$

    $p_1= \lambda e^{-\lambda}$

    $E[X] = 10000 \sum \limits_{k=2}^\infty~(k-1) p_k$

    $E[X] = 10000\left(\sum \limits_{k=2}^\infty~k p_k - \sum \limits_{k=2}^\infty~p_k\right) = $

    $10000\left((1.5-(1)p_1 - (0)p_0) - (1 - p_1 - p_0)\right) = $

    $10000\left(0.5+p_0\right) = $

    $5000 + 10000e^{-1.5} = 7231.3$
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