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Let X be a geometric random variable with parameter $p$ compute $E[X^3]$.

How would I approach this and how would I simplify the series? Can I use a moment generating function?
I am able to write out a formula for expectation, I believe it is the sum from 1 to n of $k^3p(1-p)^{k-1}$, I apologize for the terrible notation but I am not sure how to proceed from here.

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Thank you! that was very helpful! –  Olivia Irving Dec 14 '12 at 2:04
    
Just enclosed it in $ signs, and the exponent in {}. –  Berci Dec 14 '12 at 2:04

1 Answer 1

up vote 0 down vote accepted

I would write $k^3=k(k^2-1)+k$, then $$E[X^3]=p(1-p)\sum_{k=1}^\infty (k+1)k(k-1)\cdot (1-p)^{k-2} + p\sum_{k=1}^\infty k\cdot (1-p)^{k-1}, $$ and use the famous $\displaystyle\sum_{n=0}^\infty q^n= \frac 1{1-q} $, and its derivatives.

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