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Homework Help: Statistics Question: The 3rd Moment of Poisson Distribution

  1. Mar 21, 2010 #1
    1. The problem statement, all variables and given/known data

    X is a discrete random variable that has a Poisson Distribution with parameter L. Hence, the discrete mass function is [tex]f(x)[/tex] = [tex]L^{x} e^{-L} / x![/tex].

    Where L is a real constant, e is the exponential symbol and x! is x factorial.

    Without using generating functions, what is [tex]E(X^{3})[/tex]? (the 3rd moment)

    2. Relevant equations


    3. The attempt at a solution

    [tex]E(X^{3})[/tex] = [tex]\Sigma x^{3} L^{x} e^{-L} / x![/tex] from the definition of Expectation. Sigma is summation over all x values.

    I think I am suppose to rearrange all the terms in order to get something of the form "{summation that sums to 1} times {answer}" but I totally lost regarding what I should be manipulating the terms into. Or maybe this is the wrong approach?

    A little nudging would go a long way...and please don't give me the answer outright! Thanks in advance! :)
  2. jcsd
  3. Mar 21, 2010 #2
    You can write:

    x^3 = x (x-1)(x-2) + second degree polynomial in x
  4. Mar 22, 2010 #3
    This looks so trivial on hindsight! How did you arrive at this insight? Or did you come across this from somewhere?

    Thanks a lot! :)
  5. Mar 22, 2010 #4
    I think it's a pretty common trick found in intro statistics courses. I mean the whole idea is that you don't have to compute any sums or integrals, but just recognize how to turn a particular sum or integral into something that is well-known. Since the pmf of the Poisson distribution has the factorial term in the denominator, it's pretty clear why finding E[X(X-1)*...*(X-k+1)] is a good step towards determining E[X^k]. Of course, to actually find E[X^k], you still have to determine E[x], E[x^2], ..., E[X^(k-1)], which still takes time even if you can determine a general formula for E[X(X-1)*...*(X-k+1)]. This is probably one hassle that makes the moment-generating method so attractive. Anyways the same trick applies to other distributions, e.g., the binomial distribution.
  6. Mar 22, 2010 #5
    Thanks for the explanation. Sometimes I wonder if my brain is too slow to do mathematics.

    But I am oh so addicted. :rofl:
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