음이항분포

\binom{x-1}{r-1} \ge 0, ~~ p^{r}(1-p)^{x-r} \ge 0, ~~ I(x \in \{r, r+1, ...\}) \ge 0

\begin{align*} \sum_{x}~P(X=x) &= ~\sum_{x} \binom{x-1}{r-1}~ p^{r}~(1-p)^{x-r} \\[20pt] &= p^r~\sum_{x=0}^{\infty} \binom{r+x-1}{r-1}~ (1-p)^{k} \\[20pt] &= p^r~ \{1-(1-p)^r\} \\[20pt] &= p^r \cdot p^{-r} = 1 \end{align*}

\therefore~\sum~P(X=x) = 1

E(X) = \dfrac{r}{p}

\begin{align*} E(X) &= \sum_{x}~x~P(X=x) \\[20pt] &= \sum_{x=r}^{\infty}~x\cdot \binom{x-1}{r-1}~p^r(1-p)^{x-r} \\[20pt] &= \sum_{x=r}^{\infty}~ \dfrac {x\cdot(x-1)!} {(r-1)!(x-r)!}~ p^r(1-p)^{x-r} \\[20pt] &= \sum_{x=r}^{\infty}~ \dfrac {x! \cdot r} {(r-1)!(x-r)!\cdot r} ~~p^r(1-p)^{x-r} \\[20pt] &= \sum_{x=r}^{\infty}~ r~\binom{x}{r}~p^r(1-p)^{x-r} \\[20pt] &= r~\sum_{x=r}^{\infty} \binom{x}{r}~p^r(1-p)^{x-r} \\[20pt] &=~rp^r \cdot \sum_{x=r}^{\infty} \binom{r+k}{r} (1-p)^k \\[20pt] &=~rp^r \cdot \sum_{x=r}^{\infty} \binom {(r+1)-1-k} {(r+1)-1} (1-p)^k \\[20pt] &=~rp^r \cdot \sum_{x=r}^{\infty} \binom {r_*-1+k} {r_*-1} (1-p)^k \\[20pt] &= rp^r \cdot \{1-(1-p)\}^{-r_n} = rp^r \cdot \{1-(1-p)\}^{-r+1} \\[20pt] &= rp^r\cdot\dfrac{1}{p^{r+1}} = \dfrac{r}{p} \end{align*}

\begin{align*} E(X^2) &= E(X(X-1)) - E(X) \\[10pt] &= \dfrac{r(r+1)}{p^2} - \dfrac{r}{p} \end{align*}

\begin{align*} E(X(X-1)) &=~ \sum_{x=r}^{\infty} ~x(x+1)~\binom{x-r}{r-1}~ p^r~(1-p)^{x-r} \\[15pt] &=~r(r+1)~p^r~\sum_{x=r}^{\infty} \binom{x+1}{r+1}~(1-p)^{x-r} \\[15pt] &=~r(r+1)~p^r~\sum_{k=0}^{\infty} \binom{k+r+1}{r+1}~(1-p)^k \\[15pt] &=~r(r+1)~p^r~\sum_{k=0}^{\infty} \binom{(r+2)-1+k}{(r+2)-1}~(1-p)^k \\[15pt] &=~r(r+1)~p^r~\{1-(1-p)\}^{-(r+2)} \\[15pt] &=\dfrac{r(r+1)}{p^2} \end{align*}

\begin{align*} Var(X) &= E(X^2)-E(X)^2 \\[15pt] &= \bigg\{ \dfrac{r(r+1)}{p^2} - \dfrac{r}{p} \bigg\} -\dfrac{r^2}{p^2} \\[15pt] &= \dfrac{r}{p^2} ~(r+1-p-r) \\[15pt] &= \dfrac{r(1-r)}{p^2} \end{align*}

M_X(t) = \Bigg\{ \dfrac{pe^t}{1-e^t(1-p)} \Bigg\}^r

\begin{align*} M_X(t) &= E(e^{tX}) \\[15pt] &=\sum_{x}~e^{tx}~ \binom{x-1}{r-1}~p^r~(1-p)^{x-r} \\[20pt] &=\sum_{x}~e^{t(x-r+r)}~ \binom{x-1}{r-1}~ p^r~(1-p)^{x-r} \\[20pt] &=e^{tr}~p^r~\sum_{x} \binom{x-1}{r-1} \Big\{e^t(1-p)\Big\}^{x-r} \\[20pt] &= (pe^t)^r~\sum_{x=0}^{\infty} \dfrac{r+k-1}{r-1} \big\{e^t(1-p)\big\}^n \\[20pt] &= \dfrac {(pe^t)^r} {\Big\{ 1-e^t(1-p) \Big\} ^r} = \bigg\{ \dfrac {pe^t} {1-e^t(1-p)} \bigg\}^r \end{align*}

\begin{align*} M_X(t) &= \Bigg\{ \dfrac{pe^t}{1-e^t(1-p)} \Bigg\}^r \\[20pt] \log M_X(t) &= r\log pe^t - r\log (1-e^t(1-p)) \\[10pt] &= r\log p + rt - r\log (1-e^t(1-p)) \\[35pt] \dfrac{d}{dt}~\log M_X(t) &= r + \dfrac {re^t(1-p)}{1-e^t(1-p)} = r \bigg( 1 + \dfrac {e^t(1-p)} {1-e^t(1-p)} \bigg) \\[20pt] &=\dfrac{r}{1-e^t(1-p)} \\[35pt] \dfrac{d^2}{dt^2}~\log M_X(t) &= \dfrac{d}{dt}~ \dfrac{r}{1-e^t(1-p)} = \dfrac{d}{dt}~ r\Big\{1-e^t(1-p)\Big\}^{-1} \\[20pt] &=r\cdot \Big\{1-e^t(1-p)\Big\}^{-2} \cdot(-1) \cdot\Big\{-e^t(1-p)\Big\} \\[20pt] &=r~(1-p)~e^t \Big\{1-e^t(1-p)\Big\}^{-2} \end{align*}