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∼B(n,p)X\sim \mathrm{B}(n,p)

则有期望

μ=∑k=0nP(X=k)k=∑k=0nCnkpk(1−p)n−kk=∑k=1nCn−1k−1pk−1(1−p)(n−1)−(k−1)np=[p+(1−p)]n−1np=np\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

σ2=∑k=0nP(X=k)(k−μ)2=∑k=0nP(X=k)k2−2μ∑k=0nP(X=k)k+μ2=∑k=0nP(X=k)k2−2μ2+μ2=∑k=0nCnkpk(1−p)n−kk2−μ2=np∑k=1nCn−1k−1pk−1(1−p)(n−1)−(k−1)k−μ2=np∑k=1nCn−1k−1pk−1(1−p)(n−1)−(k−1)[(k−1)+1]−μ2=np∑k=2nCn−1k−1pk−1(1−p)(n−1)−(k−1)(k−1)+∑k=1nCn−1k−1pk−1(1−p)(n−1)−(k−1)−μ2=n(n−1)p2∑k=2nCn−2k−2pk−2(1−p)(n−2)−(k−2)+∑k=1nCn−1k−1pk−1(1−p)(n−1)−(k−1)−μ2=n(n−1)p2[p+(1−p)]n−2+[p+(1−p)]n−1−μ2=n(n−1)p2+1−n2p2=np(1−p)\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 nn，均值 μ\mu，方差 σ2\sigma^2
共 kk 层，第 ii 层样本数量为 nin_i，均值为 μi\mu_i，方差为 σi2\sigma_i^2
第 ii 层第 jj 个样本为 xijx_{ij}
则有方差

σ2=1n∑i=1k∑j=1ni(xij−μ)2=1n∑i=1k∑j=1ni[(xij−μi)+(μi−μ)]2=1n∑i=1k[∑j=1ni(xij−μi)2+2(μi−μ)∑j=1ni(xij−μi)+∑j=1ni(μi−μ)2]=1n∑i=1k[niσi2+2(μi−μ)(∑j=1nixij−niμi)+ni(μi−μ)2]=1n∑i=1k[niσi2+ni(μi−μ)2]=1n∑i=1kni[σi2+(μi−μ)2]\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}