Expectation a RV with negative binomial distribution

Hi,

In a derivation for the expectation of an RV with a geometric distribution I have the following:

$\displaystyle E(X) = \sum^{\infty}_{k=r} k \binom{k-1}{r-1}p^{r} (1-p)^{k-r} = \frac{r}{p} \sum^{\infty}_{k=r} \binom{k}{r} p^{r+1} (1-p)^{k-r} = \frac{r}{p} $

I have several questions.

1. What sorcery took place with the binomial coefficient and k, such that k disappeared, r appeared as a factor and its now $\displaystyle \binom{k}{r}$ rather than$\displaystyle \binom{k-1}{r-1}$ ?

2. Presumably the logic is that $\displaystyle \sum^{\infty}_{k=r} \binom{k}{r} p^{r+1} (1-p)^{k-r} = 1$. Can anyone show me how to prove this and what class of distribution is this, not binomial right?

Many thanks in advance. MD.

Re: Expectation a RV with negative binomial distribution

For your first question..

$\displaystyle k\binom{k-1}{r-1}p^r$

$\displaystyle =\frac{r}{p}\frac{k}{r}\binom{k-1}{r-1}p^{r+1}$

$\displaystyle =\frac{r}{p}\frac{k}{r}\frac{(k-1)!}{(k-r)!(r-1)!}p^{r+1}$

$\displaystyle =\frac{r}{p}\frac{k!}{(k-r)!r!}p^{r+1}$

$\displaystyle =\frac{r}{p}\binom{k}{r}p^{r+1}$

I haven't answered your second question but if you're trying to find E(X) for a geometric distribution I think there is an easier way.

$\displaystyle \displaystyle E(X)=\sum_{i=1}^\infty i p q ^{i-1}$

$\displaystyle \displaystyle =p\sum_{i=1}^\infty \frac{d}{dq}q^i$

$\displaystyle \displaystyle =p \frac{d}{dq} \sum_{i=1}^\infty q^i$

etc.