I am trying (and failing) to find a recursive formula for the $nth$ moment of a geometric distribution. I have arrived at bogus results, and I think it has something to do with the convergence of the power series in question, and my own mathematically dubious operator-interchange. The problem:

Suppose we have a random variable $X$ that follows a geometric distribution (with parameter $p$, and let $q=1-p$), over the support $x \in \mathbb{N}$. The p.m.f is given by $$ f(x) = pq^{x-1} $$

I want to be able to find the $nth$ moment, that is, $E[X^n]$. I've seen the following clever trick used to find $E[X]$: $$ E[X] = \sum_{x=1}^\infty xpq^{x-1} \\ = p\sum_{x=1}^\infty \frac{d}{dq}q^x \\ = p\frac{d}{dq}\sum_{x=1}^\infty q^x \\ = p\frac{d}{dq}(\frac{q}{1-q}) \\ = p\frac{(1-q) - (-q)}{p^2} = \frac{p}{p^2} = \frac{1}{p} = \frac{1}{1-q} $$ This relies on interchanging the summation and differentiation operators (and happily factoring out the $p$, even though it is actually a function of $q$). I am trying to extend the idea to calculating the $nth$ moment (in terms of the $(n-1)$th moment):

First, notice that $$ E[X^{n-1}] = \sum_{x=1}^\infty x^{n-1}pq^{x-1} $$


$$ E[X^{n}] = \sum_{x=1}^\infty x^{n}pq^{x-1}= \sum_{x=1}^\infty x^{n-1}pxq^{x-1} \\ = \sum_{x=1}^\infty x^{n-1}p\frac{d}{dq}q^{x} $$

I imagine that the following step is the source of error: $$ = \frac{d}{dq}\sum_{x=1}^\infty x^{n-1}pq^{x} \\ = \frac{d}{dq}q\sum_{x=1}^\infty x^{n-1}pq^{x-1} \\ = \frac{d}{dq}qE[X^{n-1}] $$

This is obviously incorrect; setting $n=2$: $$ E[X^2] = \frac{d}{dq}qE[X^{n-1}] = \frac{d}{dq}q\frac{1}{1-q} \\ = \frac{1}{(1-q)^2} = \frac{1}{p^2} $$ and substituting into the formula for variance:

$$ \sigma^2 = E[X^2] - (E[X])^2 $$ yields: $$ \sigma^2 = \frac{1}{p^2} - \frac{1}{p^2} = 0 $$

which I know is not true. Clearly I have made an error, and I don't know where, or more importantly, why. Can someone point me in the right direction please?

  • 3
    $\begingroup$ The problem is that $p$ and $q$ are not independent variables :) Whereas it was correct to pull out the $p$ for purposes of evaluating the series, you can't leave it inside the $d/dq$ with impunity. $\endgroup$ – Ted Shifrin Jun 11 '14 at 13:33
  • $\begingroup$ @TedShifrin I don't I understand your comment; if $p$ and $q$ are not independent, then how come you can take out the $p$ in the first place? $\endgroup$ – Daniel Levin Jun 11 '14 at 13:49
  • 1
    $\begingroup$ he means that, since $p=1-q$, $\frac{d}{dq}(pq^k) \ne p \frac{d}{dq}(q^k)$. For the moments, you might try differentiating the moment-generating function. $\endgroup$ – PA6OTA Jun 11 '14 at 14:20

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