二項分佈#
具有參數 \(\left(n,p\right)\) 的二項隨機變數可以描述為 \(n\) 個參數為 \(p\) 的獨立白努利隨機變數的總和；
\[Y=\sum_{i=1}^{n}X_{i}.\]
因此，此隨機變數計算在 \(n\) 次獨立隨機實驗中成功的次數，其中成功的機率為 \(p.\)
\begin{eqnarray*} p\left(k;n,p\right) & = & \left(\begin{array}{c} n\\ k\end{array}\right)p^{k}\left(1-p\right)^{n-k}\,\, k\in\left\{ 0,1,\ldots n\right\} ,\\ F\left(x;n,p\right) & = & \sum_{k\leq x}\left(\begin{array}{c} n\\ k\end{array}\right)p^{k}\left(1-p\right)^{n-k}=I_{1-p}\left(n-\left\lfloor x\right\rfloor ,\left\lfloor x\right\rfloor +1\right)\quad x\geq0\end{eqnarray*}
其中不完全貝塔積分為
\[I_{x}\left(a,b\right)=\frac{\Gamma\left(a+b\right)}{\Gamma\left(a\right)\Gamma\left(b\right)}\int_{0}^{x}t^{a-1}\left(1-t\right)^{b-1}dt.\]
現在
\begin{eqnarray*} \mu & = & np\\ \mu_{2} & = & np\left(1-p\right)\\ \gamma_{1} & = & \frac{1-2p}{\sqrt{np\left(1-p\right)}}\\ \gamma_{2} & = & \frac{1-6p\left(1-p\right)}{np\left(1-p\right)}.\end{eqnarray*}
\[M\left(t\right)=\left[1-p\left(1-e^{t}\right)\right]^{n}\]