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Soa #50

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  • Soa #50

    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. What is the expected amount paid to the company under this policy during a one-year
    period?

    The answer says... Let N be the number of major snowstorms per year, and let P be the amount paid to the company under the policy. Then Pr[N = n] =
    [(3/2)^n*e-(3/2)]/(n!) , n = 0, 1, 2, . . . P=0 for N=0 and P=10,000(N-1) for N ≥1.

    Now observe that E[P] = Summation(n=1 to inifinity) 10,000(n-1)[(3/2)^n*e-(3/2)]/(n!)

    Everything makes sense so far. The next step is what gets me. The next step says

    = 10,000e^(–3/2) + Summation(n=0 to inifinity) 10,000(n-1)[(3/2)^n*e-(3/2)]/(n!)
    = 10,000 e–3/2 + E[10,000 (N – 1)]
    = 10,000 e–3/2 + E[10,000N] – E[10,000] = 10,000 e–3/2 + 10,000 (3/2) – 10,000 = 7,231 .

    It seems that they added in the 10,000e^(–3/2) to balance out what would happen if you plug 0 into the summation, but why do we have to change the n from starting with 1 to starting with 0?
    Last edited by ohmymath; May 16 2010, 10:39 PM.

  • #2
    There's a few threads on this already.

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    • #3
      http://www.actuary.com/actuarial-dis...hlight=poisson for one

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