Standard Combinatorial Identity: \[\begin{align*} \binom{k+r-1}{k} & = \frac{(k+r-1)!}{k!(r-1)!} \\ \text{and: }k \binom{k+r-1}{k} & = k \cdot \frac{(k+r-1)!}{k!(r-1)!} \\ \text{and } \frac{k}{k!} &= \frac{1}{(k-1)!} \\ & = \frac{(k+r-1)!}{(k-1)!(r-1)!} \\ &= r \cdot \frac{(k+r-1)!}{(k-1)!r!} \\ &= r \binom{k+r-1}{k-1} \end{align*}\]
Normalization of a negative binomial type series
For \(|x| <1\):
\[\begin{align*} \sum_{j=0}^{\infty} \binom{j+r}{j} x^j &= \frac{1}{(1-x)^{r+1}} \\ \text{let: } x &= 1-p \\ \sum_{j=0}^{\infty} \binom{j+r}{j} (1-p)^j &= \frac{1}{p^{r+1}} \\ \text{Multiplying both sides by } p^r \\ \sum_{j=0}^{\infty} \binom{j+r}{j} p^r (1-p)^j &= p^r \cdot \frac{1}{p^{r+1}} \\ &= \frac{1}{p} \end{align*}\]
Standard Combinatorial Identity: \[\begin{align*} k(k-1) \binom{k+r-1}{k} &= k(k-1) \cdot \frac{(k+r-1)!}{k!(r-1)!} \\ \text{and } k! &= k(k-1)(k-2)! \\ & = \frac{(k+r-1)!}{(k-2)!(r-1)!} \\ \end{align*}\]
Now, expanding the RHS:
\[\begin{align*} \binom{k+r-1}{k-2} &= \frac{(k+r-1)!}{(k-2)!(r+1)!} \\ r(r+1)\binom{k+r-1}{k-2} &= r(r+1)\frac{(k+r-1)!}{(k-2)!(r+1)!} \\ \text{Since: } (r+1)! &= (r+1)r(r-1)! \\ &= \frac{(k+r-1)!}{(k-2)!(r-1)!} \\ \end{align*}\]
\[\begin{align*} k(k-1) \binom{k+r-1}{k} &= r(r+1)\binom{k+r-1}{k-2} \\ \end{align*}\]
Normalization of a negative binomial type series
For \(|x| <1\): \[\begin{align*} \sum_{j=0}^{\infty} \binom{j+r-1}{j} x^j &= \frac{1}{(1-x)^r} \\ \text{From the Generalized Binomial Theorem:} \\ (1-x)^{-r} &= \sum_{j=0}^\infty \binom{-r}{j}(-x)^j \\ \text{And: } \binom{-r}{j} &= \frac{(-r)(-r-1)\cdots(-r-j+1)}{j!} \\ &= (-1)^j \frac{r(r+1)\cdots(r+j-1)}{j!} \\ \text{But: } r(r+1)\cdots(r+j-1) &= \frac{(r+j-1)!}{(r-1)!} \\ \binom{-r}{j} &= (-1)^j \frac{(r+j-1)!}{j!(r-1)!} \\ &= (-1)^j \binom{r+j-1}{j} \\ \text{Then: } (1-x)^{-r} &= \sum_{j=0}^\infty (-1)^j \binom{r+j-1}{j}(-x)^j \\ &= \sum_{j=0}^\infty \binom{r+j-1}{j}x^j \\ \end{align*}\]
\[\begin{align*} E(X) &= \sum_{k=0}^{\infty} k \binom{k+r-1}{k}p^r(1-p)^k\\ &= \sum_{k=0}^\infty r\binom{k+r-1}{k-1} p^r (1-p)^{k}\\ &= r(1-p) \sum_{k=1}^\infty \binom{k+r-1}{k-1} p^r (1-p)^{k-1} \\ &= r(1-p) \sum_{j=0}^\infty \binom{j+r}{j} p^r (1-p)^j \\ &= \frac{r(1-p)}{p} \end{align*}\]
\[\begin{align*} E[X(X-1)] &= \sum_{k=0}^{\infty} k(k-1) \binom{k+r-1}{k}p^r(1-p)^k\\ &= \sum_{k=0}^{\infty} r(r+1) \binom{k+r-1}{k-2}p^r(1-p)^k\\ &= r(r+1)p^r \sum_{j=0}^\infty \binom{j+r-1}{j} (1-p)^{j+2} \\ &= r(r+1)p^r(1-p)^2 \frac{1}{p^{r+2}} \\ &= \frac{r(r+1)(1-p)^2}{p^2} \end{align*}\]
Now: \(X^2 = X(X-1)+X\)
\[\begin{align*} E[X^2] &= E[X(X-1)] + E[X] \\ &= \frac{r(r+1)(1-p)^2}{p^2}+\frac{r(1-p)}{p} \\ &= \frac{r(r+1)(1-p)^2 + rp(1-p)}{p^2}\\ &= \frac{r(1-p)[(r+1)(1-p) + p]}{p^2}\\ &= \frac{r(1-p)[r(1-p)-1]}{p^2} \\ Var(X) &= E[X^2] - (E[X])^2\\ &= \frac{r(1-p)[r(1-p)-1]}{p^2} - \left(\frac{r(1-p)}{p}\right)^2 \\ &= \frac{r(1-p)[r(1-p)-1]}{p^2} - \frac{r^2(1-p)^2}{p^2}\\ &= \frac{r(1-p)}{p^2}[r(1-p)+1-r(1-p)] \\ &= \frac{r(1-p)}{p^2} \end{align*}\]
\[\begin{align*} M_X(t) &= \sum_{k=0}^\infty e^{tk} \binom{k+r-1}{r-1} (1-p)^kp^r\\ &= p^r \sum_{k=0}^\infty \binom{y+r-1}{r-1} [e^t(1-p)]^k\\ &= p^r \sum_{k=0}^\infty \binom{-r}{k} (-1)^k[e^t(1-p)]^k \\ \text{Using Newton's Binomial Theorem: } (1+x)^r &= \sum_{i=0}^\infty \binom{r}{i}x^i \\ &= \left(\frac{p}{1-(1-p)e^t}\right)^r, t<-\log(1-p) \end{align*}\]