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Statistics Question: The 3rd Moment of Poisson Distribution

  • Thread starter Legendre
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Homework Statement



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)

Homework Equations



N.A.

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! :)
 

Answers and Replies

  • #2
1,838
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You can write:

x^3 = x (x-1)(x-2) + second degree polynomial in x
 
  • #3
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You can write:

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

Thanks a lot! :)
 
  • #4
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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.
 
  • #5
62
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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.
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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