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Exam FM Solutions - November 2005 provided courtesy of Sam Broverman

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  • Exam FM Solutions - November 2005 provided courtesy of Sam Broverman

    Thank you Sam for providing your solutions for Exam M to users of our forums. Users may view the problems along with Sam Broverman's soutions at:

    http://www.sambroverman.com/Nov2005-...-Solutions.pdf

    Please contact Sam Broverman if you notice any corrections that need to be made.

    Samuel Broverman, Ph D. ASA, of The University of Toronto - http://www.sambroverman.com
    Actuary.com - www.actuary.com
    Email - [email protected]

  • #2
    PLease help

    Could you help me prove this?

    E[z(K)] = z(0) + summation([1 − G(k)] delta z(k)) evaluated from k=0 to positive infinity.

    Comment


    • #3
      If we are assuming that Z is a non-negative integer
      with probability function p(k) and cdf G(k) ,
      then p(k) = G(k) - G(k-1) = [1 - G(k-1)] - [1 - G(k)] .
      Then
      E[z(K)] = z(0)p(0) + z(1)p(1) + z(2)p(2) + z(3)p(3) + . . .
      = z(0)([1 - G(-1)] - [1 - G(0)])
      + z(1)([1 - G(0)] - [1 - G(1)])
      + z(2)([1 - G(1)] - [1 - G(2)])
      + z(3)([1 - G(2)] - [1 - G(3)]) + . . .

      If we regroup the terms, the sum becomes
      [1 - G(-1)]z(0)
      + [1 - G(0)][z(1) - z(0)]
      + [1 - G(1)][z(2) - z(1)]
      + [1 - G(2)][z(3) - z(2)]
      + . . .

      Since G(-1) = 0 (we are assuming that Z >= 0) ,
      this becomes the right side of the equation you wrote.

      Sam Broverman
      Sam Broverman

      [email protected]
      www.sambroverman.com

      Comment

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