1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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:
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook