Prove that the expected value of a negative binomial distribution is r/p.

Question

I've done most of the work but I am now stuck. I started by saying I've gotten too $$E[X] = \frac{r}{p} \sum_{k=r}^{\infty} p \binom{k}{r} p^r(1-p)^{k-r}.$$ I know I need to show that the summation is equal to 1 but I'm not sure how to do that.

Answer 1

I suggest totally probabilistic approach to problem. We know that the Negative binomial distribution is actually sum of several (in this case $r$) independent Geometrical distributions.

Then let $X$~$NB(r,p)$ and $Y_1,.., Y_r$~$Geometric(p)$.

$$\mathbb{E}[X]=\mathbb{E}[\sum_{k=1}^rY_k]=\sum_{k=1}^r\mathbb{E}[Y_k]=\sum_{k=1}^r \frac{1}{p}=\frac{r}{p}. $$

Answer 2

Hint: consider that $$\forall x\in(-1,1),\qquad \sum_{k\geq r}\binom{k}{r}x^{k-r} = \frac{1}{(1-x)^{r+1}}\tag{1}$$ since we have $\sum_{k=r}^{R}\binom{k}{r}=\binom{R+1}{k+1}$.

What happens if we differentiate both sides of $(1)$ with respect to $x$?