二项分布均值与方差计算

对于 $X∼B(n,p)$ 的第k项，其概率为 $P(X=k) = C_n^k p^k(1-p)^(n-k)$。

下面的计算中我用了几个变形技巧：

• 易证当存在 $n, k \in N, 且 n \ge k 时，有 kC_n^k = nC_{n-1}^{k-1}$
• 二项展开式 $(a + b)^n = \sum_{k=0}^{n} C_n^k a^{n-k} b^k$

则

$$ \begin{aligned} EX & = \sum_{k=0}^{n}kC_{n}^{k}p^{k}(1-p)^{n-k} \\ & = \sum_{k=1}^{n}kC_{n}^{k}p^{k}(1-p)^{n-k} \\ & = \sum_{k=1}^{n}nC_{n-1}^{k-1}p^{k}(1-p)^{n-k} \\ & = np\sum_{k=1}^{n}C_{n-1}^{k-1}p^{k-1}(1-p)^{n-k} \\ & = np(p+1-p)^{n-1} \\ & = np \end{aligned} $$

$$ \begin{aligned} DX & = \sum_{k=0}^{n}(k-EX)^2P(X=k) \\ & = \sum_{k=0}^{n}(k-np)^2C_{n}^{k}p^{k}(1-p)^{n-k} \\ & = \sum_{k=0}^{n}(k^2-2knp+n^2p^2)C_{n}^{k}p^{k}(1-p)^{n-k} \\ & = \sum_{k=0}^{n}k^2C_{n}^{k}p^{k}(1-p)^{n-k} -2np\sum_{k=0}^{n}kC_{n}^{k}p^{k}(1-p)^{n-k} + n^2p^2\sum_{k=0}^{n}C_{n}^{k}p^{k}(1-p)^{n-k} \\ & = \sum_{k=0}^{n}k^2C_{n}^{k}p^{k}(1-p)^{n-k} - 2np \cdot np + n^2p^2(p+1-p)^n \\ & = \sum_{k=1}^{n}k \cdot kC_{n}^{k}p^{k}(1-p)^{n-k} - n^2p^2 \\ & = \sum_{k=1}^{n}k \cdot nC_{n-1}^{k-1}p^{k}(1-p)^{n-k} - n^2p^2 \\ & = np\sum_{k=1}^{n}kC_{n-1}^{k-1}p^{k-1}(1-p)^{n-k} - n^2p^2 \\ & = np[\sum_{k=1}^{n}(k-1)C_{n-1}^{k-1}p^{k-1}(1-p)^{n-k} + \sum_{k=1}^{n}C_{n-1}^{k-1}p^{k-1}(1-p)^{n-k}] - n^2p^2 \\ & = np[(n-1)p + (p+1-p)^{n-1}] - n^2p^2 \\ & = np(1-p) \end{aligned} $$