其实是在高二刚学的时候推的 拖到现在才搞成电子稿（
二项分布
设有二项分布
X\sim \mathrm{B}(n,p)
则有期望
\begin{aligned}
\mu&=\sum_{k=0}^{n}{\mathrm{P}(X=k)k}\\
&=\sum_{k=0}^{n}{\mathrm{C}_n^k p^k (1-p)^{n-k} k}\\
&=\sum_{k=1}^{n}{\mathrm{C}_{n-1}^{k-1} p^{k-1} (1-p)^{(n-1)-(k-1)} np}\\
&=[p+(1-p)]^{n-1} np\\
&=np
\end{aligned}
方差
\begin{aligned}
\sigma^2&=\sum_{k=0}^{n}{\mathrm{P}(X=k)(k-\mu)^2}\\
&=\sum_{k=0}^{n}{\mathrm{P}(X=k)k^2} - 2\mu\sum_{k=0}^{n}{\mathrm{P}(X=k)k} + \mu^2\\
&=\sum_{k=0}^{n}{\mathrm{P}(X=k)k^2} - 2\mu^2 + \mu^2\\
&=\sum_{k=0}^{n}{\mathrm{C}_n^k} p^k (1-p)^{n-k} k^2 - \mu^2\\
&=np\sum_{k=1}^{n}{\mathrm{C}_{n-1}^{k-1} p^{k-1} (1-p)^{(n-1)-(k-1)}k} - \mu^2\\
&=np\sum_{k=1}^{n}{\mathrm{C}_{n-1}^{k-1} p^{k-1} (1-p)^{(n-1)-(k-1)}[(k-1)+1]} - \mu^2\\
&=np\sum_{k=2}^{n}{\mathrm{C}_{n-1}^{k-1} p^{k-1} (1-p)^{(n-1)-(k-1)}(k-1)} + \sum_{k=1}^{n}{\mathrm{C}_{n-1}^{k-1}} p^{k-1} (1-p)^{(n-1)-(k-1)} - \mu^2\\
&=n(n-1)p^2\sum_{k=2}^{n}{\mathrm{C}_{n-2}^{k-2} p^{k-2} (1-p)^{(n-2)-(k-2)}} + \sum_{k=1}^{n}{\mathrm{C}_{n-1}^{k-1} p^{k-1} (1-p)^{(n-1)-(k-1)}} - \mu^2\\
&=n(n-1)p^2 [p+(1-p)]^{n-2} + [p+(1-p)]^{n-1} - \mu^2\\
&=n(n-1)p^2 + 1 - n^2 p^2\\
&=np(1-p)
\end{aligned}
分层抽样
样本数量 n，均值 \mu，方差 \sigma^2
共 k 层，第 i 层样本数量为 n_i，均值为 \mu_i，方差为 \sigma_i^2
第 i 层第 j 个样本为 x_{ij}
则有方差
\begin{aligned}
\sigma^2&=\frac{1}{n}\sum_{i=1}^{k}{\sum_{j=1}^{n_i}{(x_{ij}-\mu)^2}}\\
&=\frac{1}{n}\sum_{i=1}^{k}{\sum_{j=1}^{n_i}{[(x_{ij}-\mu_i)+(\mu_i-\mu)]^2}}\\
&=\frac{1}{n}\sum_{i=1}^{k}{\left[\sum_{j=1}^{n_i}{(x_{ij}-\mu_i)^2} + 2(\mu_i-\mu)\sum_{j=1}^{n_i}{(x_{ij}-\mu_i)} + \sum_{j=1}^{n_i}{(\mu_i-\mu)^2}\right]}\\
&=\frac{1}{n}\sum_{i=1}^{k}{\left[n_i\sigma_i^2 + 2(\mu_i-\mu)\left(\sum_{j=1}^{n_i}{x_{ij}-n_i\mu_i}\right) + n_i(\mu_i-\mu)^2\right]}\\
&=\frac{1}{n}\sum_{i=1}^{k}{\left[n_i\sigma_i^2 + n_i(\mu_i-\mu)^2\right]}\\
&=\frac{1}{n}\sum_{i=1}^{k}{n_i\left[\sigma_i^2 + (\mu_i-\mu)^2\right]}
\end{aligned}