4
$\begingroup$

I am trying to caclulate the Factorial Moment of the Geometric Distribution #2 with parameter $p$. Therefore I set $\Omega = \mathbb{N}_0$ and have by using the pochhammer symbol and setting $q=1-q$ that

$E((k)_l)= \sum _{k=0}^{\infty } (k)_l p q^k = p^{-l} q \cdot l! \sum _{k=0}^{\infty } (\frac{(k+l-1)!}{(k-1)! \cdot l!}\cdot p^{l+1} q^{k-1}) $

Now Mathematica tells me that $\sum _{k=0}^{\infty } (\frac{(k+l-1)!}{(k-1)! \cdot l!}\cdot p^{l+1} q^{k-1})=1$, but I cannot see why this identity is true. Also when using

FactorialMoment[GeometricDistribution[p], l] 

Mathematica suggests that $E((k)_l)=(\frac{q}{p})^l l!$. Thank you in Advance for your help.

2 Answers 2

5

Let $X$ have geometric distribution, where $X$ is the number of failures before the first success.

The easiest approach to the factorial moments in this case is to find the factorial moment generating function, which is $E(t^X)$ Suppose the probability of success is $p$. We want $\sum_{n=0}^\infty pq^n t^n$ where as usual $q=1-p$. So we want $\sum_{n=0}^\infty p(qt)^n$ Sum this infinite geometric series. We get $\frac{p}{1-qt}$ To find the $k$-th factorial moment, find the $k$-th derivative of the factorial moment generating function (with respect to $t$) at $t=1$. In our particular case, finding the $k$-th derivative is easy.

If by geometric distribution you mean total number of trials until first success (so values are $1$, $2$, and so on) a small modification of the above calculation will give the answer.

Addendum: The easiest way to find the sum $\sum_{k=1}^\infty (k)(k-1)\cdots(n-\ell+1)x^k$ that was asked about is to express this as $x^{\ell}\sum_{k=1}^\infty (k)(k-1)\cdots(n-\ell+1)x^{k-\ell}$ and observe that $(k)(k-1)\cdots(n-\ell+1)x^{k-\ell}$ is the $\ell$-th derivative of $x^k$. So the desired sum is the $\ell$-th derivative of $1+x+x^2+ x^3+\cdots$, that is, of $1/(1-x)$.

  • 0
    Ok I did this and now got the same result as Mathematica. Thanks!2011-05-10
4

The factorial moments of an integer valued random variable $X$ are linked to the successive derivatives of the generating function $g_X$ of $X$, defined by $ g_X(s)=E(s^X)=\sum_{n=0}^{+\infty}P(X=n)s^n. $ For every $k\ge0$, the $k$th derivative is $ g_X^{(k)}(s)=E((X)_ks^{X-k}), $ hence the value at $s=1$ yields the factorial moment.

Now, what is $g_X$ for $X$ geometric?

  • 0
    Yes I got the result now, thanks!2011-05-10