The equation below indicates expected value of negative binomial distribution. I need a derivation for this formula. I have searched a lot but can't find any solution. Thanks for helping :)
$ E(X)=\sum _{x=r}^{}x\times \left(\begin{array}{c}x-1\\ r-1\end{array}\right)\times {p}^{r}\times (1-p{)}^{x-r}=\frac{r}{p} $
I have tried: $ E(X)=\sum _{x=r}^{}x\times \left(\begin{array}{c}x-1\\ r-1\end{array}\right)\times {p}^{r}\times (1-p{)}^{x-r}=\sum _{x=r}^{}x\times \frac{(x-1)!}{(r-1)!\times ((x-1-(r-1))!}\times {p}^{r}\times (1-p{)}^{x-r}=\sum _{x=r}^{}\frac{x!}{(r-1)!\times ((x-r)!}\times {p}^{r}\times (1-p{)}^{x-r}\Rightarrow \phantom{\rule{0ex}{0ex}}\sum _{x=r}^{}r\times \frac{x!}{r!\times (x-r)!}\times {p}^{r}\times (1-p{)}^{x-r}=\frac{r}{p}\times \sum _{x=r}^{}\frac{x!}{r!\times (x-r)!}\times {p}^{r+1}\times (1-p{)}^{x-r} $\begin{align} E(X)&=\sum _{x=r}^{}x\times \left(\begin{array}{c}x-1\\ r-1\end{array}\right)\times {p}^{r}\times (1-p{)}^{x-r}\\&=\sum _{x=r}^{}x\times \frac{(x-1)!}{(r-1)!\times ((x-1-(r-1))!}\times {p}^{r}\times (1-p{)}^{x-r}\\ &=\sum _{x=r}^{}\frac{x!}{(r-1)!\times ((x-r)!}\times {p}^{r}\times (1-p{)}^{x-r}\\ &\Rightarrow \phantom{\rule{0ex}{0ex}}\sum _{x=r}^{}r\times \frac{x!}{r!\times (x-r)!}\times {p}^{r}\times (1-p{)}^{x-r}\\ &=\frac{r}{p}\times \sum _{x=r}^{}\frac{x!}{r!\times (x-r)!}\times {p}^{r+1}\times (1-p{)}^{x-r} \end{align}
If the power of p in the last equation were not r + 1, I can implement Newton Binomial. So It will be true. But I am stuck here.
