# Negative Binomial mean

Let $$X\sim BN(r,p)$$.

If I try to compute E[X] through the Moment Generating Function I get the following:

\begin{aligned} E[X] &=\left.\frac{d}{d t} M_{X}(t)\right|_{t=0} \ &=\left.\frac{d}{d t} p^{r}\left(1-p e^{t}(1-p)\right)^{-r}\right|_{t=0} \ &=p^{r+1} r(1-p)(1-p(1-p))^{-r-1} \end{aligned}

But if I find it through the first factorial moment I get
$$E\left[X^{(h)}\right]$$ $$=\sum_{x=0}^{\infty} \frac{x !}{(x-h) !} \frac{(x+r-1) !}{x !(r-1) !} p^{r}(1-p)^{x}$$ $$=\sum_{x=h}^{\infty} \frac{(x+r-1) !}{(x-h) !(r-1) !} p^{r}(1-p)^{x},\left(x^{*}=x-h\right)$$ $$=\sum_{x^{*}=0}^{\infty} \frac{\left(x^{*}+r+h-1\right) !}{x^{*} !(r-1) !} p^{r}(1-p)^{x^{*}+h}$$ $$=\frac{(r+h-1) !}{(r-1) !} \frac{(1-p)^{h}}{p^{h}} \sum_{x^{*}=0}^{\infty} \frac{\left(x^{*}+h+r+1\right) !}{x^{*} !(r+h-1) !} p^{r+h}(1-p)^{x^{*}}$$ $$=\frac{(r+h-1) !}{(r-1) !} \frac{(1-p)^{h}}{p^{h}}$$
Therefore,
$$E[X]=\frac{r(1-p)}{p}$$
Where did I do wrong?

• You have an extra $p$ in $p^{r}\left(1-p e^{t}(1-p)\right)^{-r}$ (it should be $p^{r}\left(1- e^{t}(1-p)\right)^{-r}$). Remove it and you should get the answer. Jun 28 at 8:06

Your second derivation seems right. I can only guess that you made some minor error taking derivative.

Maybe it is better to compute abstract: Let $$M_Y(t)$$ be the moment generating function of a geometrically distributed random variable $$Y$$ with parameter $$p\in(0,1)$$, then $$M_X(t)=(M_Y(t))^r$$.

Therefore by chain rule $$M_X(t)'=r(M_Y(t))^{r-1}\cdot M_Y'(t)$$

Taking $$t$$ to zero, the result is just $$\mathbb E[X]=r\cdot\mathbb E[Y]=r\cdot\frac{1-p}{p}$$, as you correctly derived using the second method.

The MGF is $$\left(\dfrac{1-p}{1-p\mathrm{e}^x}\right)^r$$ So taking derivatives we get

$$\frac{p\,r\,\mathrm{e}^x\cdot\left(\frac{1-p}{1-p\,\mathrm{e}^x}\right)^{r-1}}{1-p\,\mathrm{e}^x}$$ Evaluating at $$x=0$$ gives us $$\frac{pr}{1-p}$$

• He is talking about negativ binomial... Jun 28 at 7:48
• My bad. Corrected it Jun 28 at 7:55