I am testing the variance of a negative binomial distribution, but I have problems in the derivative.
$$ V(x) = E(x(x-1))+E(x)-E(x)^{2} $$ $$ E(x(x-1)) = \displaystyle\sum^{\infty}_{\substack{x=0}} x (x-1) \; P(X=x) $$ $$ E(x(x-1)) = \displaystyle\sum^{\infty}_{\substack{x=0}} x (x-1) \displaystyle\binom{x-1}{r-1} \; p^{r} \; (1-p)^{x-r} $$ $$ E(x(x-1)) = \displaystyle\sum^{\infty}_{\substack{x=0}} x (x-1) \dfrac{(x-1)!}{(x-r)!\;(r-1)!} \; p^{r} \; (1-p)^{x-r} $$ $$ E(x(x-1)) = \displaystyle\sum^{\infty}_{\substack{x=0}} (x-1)r \dfrac{x!}{(x-r)!\;(r)!} \; p^{r} \; (1-p)^{x-r} $$ $$ E(x(x-1)) = r\;(\dfrac{p}{1-p})^{r} \displaystyle\sum^{\infty}_{\substack{x=0}} \displaystyle\binom{x}{r} \; (x-1) (1-p)^{x}$$ $$ E(x(x-1)) = r\left(\frac{p}{1-p}\right)^r\left[\sum^{\infty}_{x=r} \binom{x}{r} (x-1)y^{x}\right]_{y=1-p}$$ $$ E(x(x-1)) = r\left(\frac{p}{1-p}\right)^r\left[\sum^{\infty}_{x=r} \binom{x}{r} (xy^x-y^x)\right]_{y=1-p}$$
$$ E(x(x-1)) = r\left(\frac{p}{1-p}\right)^r\left[\sum^{\infty}_{x=r}\binom{x}{r}xy^x-\sum_{x=r}^{\infty }\binom{x}{r}y^x\right]_{y=1-p} $$
$$E(x(x-1)) =r\left(\frac{p}{1-p}\right)^r\left[y\frac{d}{dy}\sum^{\infty}_{x=r}\binom{x}{r} y^x-\sum_{x=r}^{\infty }\binom{x}{r}y^x\right]_{y=1-p} $$
Help me, please, Thanks
