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Maths

和差化积与积化和差

\[ \begin{aligned} \sin\alpha+\sin\beta &= 2\sin\frac{\alpha+\beta}{2}\cos\frac{\alpha-\beta}{2}\\ \cos\alpha+\cos\beta &= 2\cos\frac{\alpha+\beta}{2}\cos\frac{\alpha-\beta}{2}\\ \sin\alpha-\sin\beta &= 2\cos\frac{\alpha+\beta}{2}\sin\frac{\alpha-\beta}{2}\\ \cos\alpha-\cos\beta &= -2\sin\frac{\alpha+\beta}{2}\sin\frac{\alpha-\beta}{2}\\ \end{aligned} \]
\[ \begin{aligned} \sin\alpha\cos\beta &= \frac{1}{2}[\sin(\alpha+\beta)+\sin(\alpha-\beta)]\\ \cos\alpha\cos\beta &= \frac{1}{2}[\cos(\alpha+\beta)+\cos(\alpha-\beta)]\\ \cos\alpha\sin\beta &= \frac{1}{2}[\sin(\alpha+\beta)-\sin(\alpha-\beta)]\\ \sin\alpha\sin\beta &= -\frac{1}{2}[\cos(\alpha+\beta)-\cos(\alpha-\beta)]\\ \end{aligned} \]

积分

\[ \lim_{n\to\infty}\sum_{k=1}^n{\dfrac{1}{n+k}} = \lim_{n\to\infty}\sum_{k=1}^n{\dfrac{1}{1+\frac{k}{n}}}=\int_0^1{\dfrac{1}{1+x}}dx=\ln2. \]

常微分方程

  • \(\dfrac{dy}{dx}=f(x)g(y)\)
\[ \int{\dfrac{dy}{g(y)}=\int{f(x)dx}+C} \]
  • \(\dfrac{dy}{dx}=f(x,y)\)
\[ \begin{aligned} &\text{let}\ y = ux\\ f(x,y) &= f(x,ux)=\varphi(u)\\ \varphi(u) &= u+x\dfrac{du}{dx}\\ \int \dfrac{du}{\varphi(u)-u}&=\int \dfrac{dx}{x} + C=\ln|x| + C\\ \Phi(u)&=\Phi(\dfrac{y}{x}) = \ln|x| + C\\ \end{aligned} \]
  • \(y'+p(x)y=q(x)\)
\[ y = \dfrac{\displaystyle\int e^{\int p(x)dx} q(x)dx+C}{e^{\int p(x)dx}} \]
  • \(y'+p(x)y=q(x)y^n\)
\[ \begin{aligned} &y'y^{-n}+p(x)y^{1-n} = q(x)\\ \text{let}\ &z = y^{1-n},\dfrac{dz}{dx}=(1-n)\dfrac{dy}{dx}y^{-n}\\ &\dfrac{1}{1-n}\dfrac{dz}{dx}+p(x)z=q(x) \end{aligned} \]

Extended Euclidean Algorithm

\(q\) \(a\) \(b\) \(s_0\) \(s_1\) \(t_0\) \(t_1\)
\(-\) \(25\) \(21\) \(1\) \(0\) \(0\) \(1\)
\(1\) \(21\) \(4\) \(0\) \(1\) \(1\) \(-1\)
\(5\) \(4\) \(1\) \(1\) \(-5\) \(-1\) \(6\)
\(4\) \(1\) \(0\) \(-5\) \(21\) \(6\) \(-25\)
\[ (-5) * 25 + 6 * 21 = 1 \]

渐进无偏估计

设\(X_1, X_2, \cdots, X_n\)是来自总体\(X\)的样本，且\(E(X) = \mu\)已知，\(D(X)=\sigma^2\)未知，则\(\dfrac{1}{n}\displaystyle\sum_{i=1}^n(X_i-\mu)^2\)是\(\sigma^2\)的无偏估计。

\(\forall i=1,2,,\cdots,n\)，有\(E[(X_i-\mu)^2]=\sigma^2\)，则 $$ E\left(\dfrac{1}{n}\sum_{i=1}^n(X_i-\mu)^2\right) = \dfrac{1}{n}E\left(\sum_{i=1}^n(X_i-\mu)^2\right)=\dfrac{1}{n}n\sigma^2 $$

设\(X_1, X_2, \cdots, X_n\)是来自总体\(X\)的样本，\(E(X) = \mu\)未知，\(D(X)=\sigma^2\)未知，则\(\dfrac{1}{n}\displaystyle\sum_{i=1}^n(X_i-\overline{X})^2\)不是\(\sigma^2\)的无偏估计。

\[ \begin{aligned} \dfrac{1}{n}\displaystyle\sum_{i=1}^n(X_i-\overline{X})^2 &= \dfrac{1}{n}\sum_{i=1}^n\left((X_i-\mu) + (\mu - \overline{X})\right)^2\\ &= \dfrac{1}{n}\displaystyle\sum_{i=1}^n(X_i-\mu)^2 + 2(\overline{X}-\mu)(\mu-\overline{X}) + (\mu-\overline{X})^2\\ &= \dfrac{1}{n}\displaystyle\sum_{i=1}^n(X_i-\mu)^2 -(\mu-\overline{X})^2\\ &\leq \sigma^2 \end{aligned} \]

当\(\overline{X}\not=\mu\)时，\(\dfrac{1}{n}\displaystyle\sum_{i=1}^n(X_i-\overline{X})^2\)会造成对方差的低估，因此需要缩小分母。

已知

\[ \begin{aligned} E(\overline{X}) &= E(\dfrac{1}{n}\sum_{i=1}^nX_i) = \dfrac{1}{n}\sum_{i=1}^nE(X_i) = \dfrac{1}{n}n\mu = \mu\\ D(\overline{X}) &= D(\dfrac{1}{n}\sum_{i=1}^nX_i) = \dfrac{1}{n^2}D(\sum_{i=1}^nX_i) = \dfrac{1}{n^2}n\sigma^2 = \dfrac{1}{n}\sigma^2\\ E(\overline{X}^2) &= E(\overline{X})^2 + D(\overline{X}) = \mu^2 + \dfrac{1}{n}\sigma^2\\ \end{aligned} \]

于是有

\[ \begin{aligned} E\left(\sum_{i=1}^n(X_i-\overline{X})^2\right) &= E(\sum_{i=1}^nX_i^2-n\overline{X}^2)\\ &= \sum_{i=1}^nE(X_i^2) - nE(\overline{X}^2)\\ &= n(\mu^2 + \sigma^2) - n(\mu^2 + \dfrac{\sigma^2}{n})\\ &= (n - 1)\sigma^2\\ \end{aligned} \]

最终得到\(\dfrac{1}{n-1}\displaystyle\sum_{i=1}^n(X_i-\overline{X})^2\)是\(\sigma^2\)的无偏估计。

由于\(\dfrac{1}{n}\displaystyle\sum_{i=1}^n(X_i-\overline{X})^2=\dfrac{n}{n-1}\sigma^2\)，且\(\displaystyle\lim_{n\to\infty}\dfrac{n-1}{n}\sigma^2=\sigma^2\)，则称\(\dfrac{1}{n}\displaystyle\sum_{i=1}^n(X_i-\overline{X})^2\)为\(\sigma^2\)的渐进无偏估计。

概率论与数理统计

基本概念

常用统计量

  • 样本均值\(\overline{X} = \dfrac{1}{n}\displaystyle\sum^{n}_{i=1}\)，观测值\(\overline{x} = \dfrac{1}{n}\displaystyle\sum_{i=1}^{n}x_i\).

  • 样本方差\(S^2 = \dfrac{1}{n-1}\displaystyle\sum_{i=1}^{n}(X_i - \overline{X})^2 = \dfrac{1}{n-1}(\displaystyle\sum_{i=1}^nX_i^2 - n \overline{X}^2)\)

  • 样本标准差\(S = \sqrt{S^2}\)

  • 样本\(k\)阶原点矩\(A_k = \dfrac{1}{n}\displaystyle\sum_{i=1}^nX_i^k\)

  • 样本\(k\)阶中心距\(B_k = \dfrac{1}{n}\displaystyle\sum_{i=1}^n(X_i - \overline{X})^k\)

定理：对总体\(X\)有\(E(X) = \mu, D(X) = \sigma^2\)；从总体中取出样本，\(E(\overline{X}) = \mu, D(\overline{X}) = \dfrac{\sigma^2}{n},E(S^2) = \sigma^2\)

三种分布

\(\chi^2\)分布

若\(X_1,X_2\cdots X_n\)独立同分布，且满足\(X\sim N(0,1)\)，则随机变量\(X^2\)满足\(X\sim \chi^2(n)\)，其中\(n\)为自由度。

  • 若\(X\sim N(\mu, \sigma)\)，则\(\dfrac{1}{\sigma^2}\displaystyle\sum_{i=1}^n(X_i-\mu)\sim \chi^2(n)\)

  • \(E(x) = n, D(X) = 2n\)

  • 若\(X\sim \chi^2(m),Y\sim \chi^x(n)\)，则\(X+Y\sim \chi^2(m+n)\)

  • 对于\(0 < \alpha < 1\)，当\(P(X^2 \geq \chi^2_{\alpha}(n)) = \alpha\)称为\(\chi_{\alpha}^2(n)\)的\(\alpha\)的上侧分位数

\(t\)分布

若\(X,Y\)独立分布，\(X\sim N(0,1), Y\sim \chi^(n)\)，则\(T = \dfrac{X}{\sqrt{Y/n}}\)称为\(t(n)\)分布（密度函数关于\(y\)轴对称）

  • 对于\(0 < \alpha < 1\)，当\(P(t \geq t_{\alpha}(n)) = \alpha\)称为\(t_{\alpha}(n)\)的\(\alpha\)的上侧分位数

\(F\)分布

若\(X\sim \chi^2(m), Y\sim \chi^2(n)\)相互独立，则\(F = \dfrac{X/m}{Y/n}\)称为\(F(m,n)\)分布

  • 设\(X\sim F(m,n)\)，则\(\dfrac{1}{X}\sim F(n,m)\)

  • 对于\(0< \alpha < 1\)，当\(P(F \geq F_{\alpha}(m,n)) = \alpha\)称为\(F_{\alpha}(m,n)\)的上侧分位数

正态总体统计量的分布

单变量

若\(X\sim N(\mu, \sigma^2)\)，有

  • \(\overline{X} \sim N(\mu, \dfrac{\sigma^2}{n})\)，\(\dfrac{\overline{X} - \mu}{\sigma / \sqrt{n}}\sim N(0,1)\)
  • \(\dfrac{1}{\sigma^2}\displaystyle\sum_{i=1}^n(X_i - \mu)^2\sim \chi^2(n)\)
  • \(\overline{X}\)与\(S^2\)相互独立，且\(\dfrac{(n-1)S^2}{\sigma^2}\sim \chi^2(n-1)\)

\(X^2 = \displaystyle\sum_{i=1}^n (\dfrac{X_i-\overline{X}}{\sigma})^2\)，但是\(\displaystyle\sum _{i=1}^n(\dfrac{X_i-\overline{X}}{\sigma}) = 0\)是一个约束，故自由度从\(n\)减为\(n-1\).

  • \(\dfrac{\overline{X} - \mu}{S/\sqrt{n}} \sim t(n-1)\)

\(z = \dfrac{\overline{X} - \mu}{\sigma / \sqrt{n}}\sim N(0,1)\)，\(\chi^2 = \dfrac{(n-1)S^2}{\sigma^2}\sim \chi^2(n-1)\)满足相互独立，则\(t = \dfrac{z}{\sqrt{\frac{\chi^2}{n-1}}}\)是自由度为\(n-1\)的\(t\)分布

双变量

若\(X\sim N(\mu, \sigma^2),Y\sim N(0,1)\)，有

  • \(\dfrac{\overline{X} - \overline{Y} - (\mu_1 - \mu_2)}{\sqrt{\frac{\sigma_1^2}{m}+\frac{\sigma_2^2}{n}}}\sim N(0,1)\)

  • \(\dfrac{\overline{X} - \overline{Y} - (\mu_1 - \mu_2)}{S_w\sqrt{\frac{1}{m} + \frac{1}{n}}}\sim t(m+n-2),其中S_w^2 = \sqrt{\dfrac{(m-1)S_1^2+(n-1)S_2^2}{m+n-2}}\)

  • \(\dfrac{S_1^2/S^2_2}{\sigma_1^2/\sigma_2^2}\sim F(m-1,n-1)\)

\(\dfrac{(n-1)S_1^2}{\sigma_1^2}\sim \chi^2(m-1)\),\(\dfrac{(n-1)S_2^2}{\sigma_2^2}\sim \chi^2(n-1)\)两者相互独立

参数估计

点估计

矩估计

用样本均值\(\overline{X}\)代替总体均值\(\mu\)，用样本的二阶中心距代替总体方差\(\sigma^2\).

极大似然估计

选择能够使概率取到极值的参数 $$ \begin{aligned} &P(X =x) = f(x;\theta)\ 极大似然函数&L_n(\theta) = \displaystyle\prod_{i=1}^{n}f(x_i;\theta)\ 解方程&\dfrac{d}{d\theta}\ln {L_n(\theta)} = 0\ \end{aligned} $$

衡量点估计的标准

  • 无偏性：\(E(\hat{\theta}) = \theta\)
  • 有效性：若\(D(\hat{\theta_1}) < D(\hat{\theta_2})\)，则\(\theta_1\)比\(\theta_2\)更有效
  • 一致性：\(\displaystyle\lim_{n \to \infty}P(\lvert \hat{\theta_n} - \theta\rvert < \varepsilon) = 1\)

样本均值是总体均值的无偏且一致的估计量；

样本方差是总体方差的无偏且一致估计量

区间估计

单正态总体下的参数区间估计

用枢轴量法求参数参数\(\mu\)的置信水平为\(1-\alpha\)的置信区间

  • \(\sigma\)已知

构造\(\overline{X}\)和\(\mu\)的函数 $$ \begin{aligned} &Z = \dfrac{\overline{X}-\mu}{\sigma/\sqrt{n}}\sim N(0,1)\ 对给定\alpha，取a<b，满足&P(a\leq \dfrac{\overline{X}-\mu}{\sigma/\sqrt{n}} \leq b) = 1-\alpha\ 为使置信区间最短，取&a = -z_{\frac{\alpha}{2}},b = z_{\frac{\alpha}{2}}\ 则\mu的置信水平为1-\alpha 的置信区间为&[\overline{X}-\dfrac{\sigma}{\sqrt{n}}z_{\frac{\alpha}{2}},\overline{X}+\dfrac{\sigma}{\sqrt{n}}z_{\frac{\alpha}{2}}]\ \end{aligned} $$

  • \(\sigma\)未知

用样本标准差代替总体标准差，构造\(\overline{X}\)和\(\mu\)的函数 $$ \begin{aligned} &T = \dfrac{\overline{X}-\mu}{S/\sqrt{n}} \sim t(n-1)\ 对给定\alpha，取a<b，满足&P(a\leq \dfrac{\overline{X}-\mu}{S/\sqrt{n}} \leq b) = 1-\alpha\ t分布也是对称分布，为使区间长度最短，取&a = -t_{\frac{\alpha}{2}}(n-1), b = t_{\frac{\alpha}{2}}(n-1)\ 则\mu的置信水平为1-\alpha的置信区间为&[\overline{X} - \dfrac{S}{\sqrt{n}}t_{\frac{\alpha}{2}}(n-1),\overline{X} + \dfrac{S}{\sqrt{n}}t_{\frac{\alpha}{2}}(n-1)]\ \end{aligned} $$ 用枢轴量法求参数参数\(\sigma^2\)的置信水平为\(1-\alpha\)的置信区间

  • \(\mu\)已知

参数\(\sigma^2\)的无偏估计为\(\hat{\sigma^2} = \dfrac{1}{n}\displaystyle\sum_{i = 1}^n(X_i - \mu)^2\)构造\(\dfrac{1}{n}\displaystyle\sum_{i=1}^n(X_i - \mu)^2\)和参数\(\sigma^2\)的函数 $$ \begin{aligned} &G = \dfrac{n\hat{\sigma^2}}{\sigma^2} = \dfrac{\displaystyle\sum_{i=1}^n (X_i - \mu)^2}{\sigma^2}\sim \chi^2(n)\ 对给定\alpha，取a<b，满足&P(a\leq \dfrac{\displaystyle\sum_{i=1}^n (X_i - \mu)^2}{\sigma^2} \leq b) = 1-\alpha\ 通常选取&P(G < a) = P(G > b) = \dfrac{\alpha}{2},即a = \chi_{1-\alpha /2}^2(n),b = \chi_{\alpha / 2}^2(n)\ 由&\chi_{1-\alpha /2}^2(n) \leq \dfrac{\displaystyle\sum_{i=1}^n (X_i - \mu)^2}{\sigma^2} \leq \chi_{\alpha / 2}^2(n)\ 求得置信区间为&\bigg[\dfrac{\displaystyle\sum_{i=1}^n (X_i-\mu)^2}{\chi_{\alpha / 2}^2(n)}, \dfrac{\displaystyle\sum_{i=1}^n (X_i-\mu)^2}{\chi^2_{1-\alpha/ 2}(n)}\bigg]\ \end{aligned} $$

  • \(\mu\)未知

由于\(\mu\)未知，参数\(\sigma^2\)的无偏估计为\(S^2 = \dfrac{1}{n-1}\displaystyle\sum_{i = 1}^n(X_i - \overline{X})^2\)，构造\(S^2\)和\(\sigma^2\)的函数 $$ \begin{aligned} &\chi^2 = \dfrac{(n-1)S^2}{\sigma^2}\sim \chi^2(n-1)\ 类似地，可以得到置信区间为&\bigg[\dfrac{\displaystyle\sum_{i=1}^n (X_i-\mu)^2}{\chi_{\alpha / 2}^2(n-1)}, \dfrac{\displaystyle\sum_{i=1}^n (X_i-\mu)^2}{\chi^2_{1-\alpha/ 2}(n-1)}\bigg] \end{aligned} $$

待估参数 待估参数 \(G(\hat{\theta}, \theta)\) 双侧置信区间
均值\(\mu\) \(\sigma\)已知 \(G = \dfrac{(\overline{X}-\mu)}{\sigma/\sqrt{n}}\sim N(0,1)\) \([\overline{X}-\dfrac{\sigma}{\sqrt{n}}z_{\frac{\alpha}{2}},\overline{X}+\dfrac{\sigma}{\sqrt{n}}z_{\frac{\alpha}{2}}]\)
均值\(\mu\) \(\sigma\)未知 \(G = \dfrac{(\overline{X}-\mu)}{S/\sqrt{n}}\sim t(n-1)\) \([\overline{X} - \dfrac{S}{\sqrt{n}}t_{\frac{\alpha}{2}}(n-1),\overline{X} + \dfrac{S}{\sqrt{n}}t_{\frac{\alpha}{2}}(n-1)]\)
方差\(\sigma\) \(\mu\)已知 \(G = \dfrac{1}{\sigma^2}\displaystyle\sum_{i=1}^n(X_i-\mu)^2\sim \chi^2(n)\) \(\bigg[\dfrac{\displaystyle\sum_{i=1}^n (X_i-\mu)^2}{\chi_{\alpha / 2}^2(n)}, \dfrac{\displaystyle\sum_{i=1}^n (X_i-\mu)^2}{\chi^2_{1-\alpha/ 2}(n)}\bigg]\)
方差\(\sigma\) \(\mu\)未知 \(G = \dfrac{(n-1)S^2}{\sigma^2}\sim \chi^2(n-1)\) \(\bigg[\dfrac{\displaystyle\sum_{i=1}^n (X_i-\mu)^2}{\chi_{\alpha / 2}^2(n-1)}, \dfrac{\displaystyle\sum_{i=1}^n (X_i-\mu)^2}{\chi^2_{1-\alpha/ 2}(n-1)}\bigg]\)

两个正态总体下的参数区间估计

待估参数 待估参数 \(G(\hat{\theta}, \theta)\) 双侧置信区间
均值差\(\mu_1-\mu_2\) \(\sigma_1,\sigma_2\)已知 \(G = \dfrac{\overline{X} - \overline{Y}-(\mu_1 - \mu_2)}{\sqrt{\frac{\sigma_1^2}{m} + \frac{\sigma_2^2}{n}}}\sim N(0,1)\) 不背， 现推:sleepy:
均值差\(\mu_1 - \mu_2\) \(\sigma_1=\sigma_2=\sigma^2\)，但\(\sigma^2\)未知 \(G = \dfrac{\overline{X} - \overline{Y}-(\mu_1 - \mu_2)}{S_w\sqrt{\frac{1}{m} + \frac{1}{n}}}\sim t(m+n-2)\) \(\cdots\)
方差比\(\dfrac{\sigma_1^2}{\sigma_2^2}\) \(\mu_1, \mu_2\)已知 \(\dfrac{\hat{\sigma^2_1}/\hat{\sigma_2^2}}{\sigma_1^2/\sigma_2^2}\sim F(m.n)\) \(\cdots\)
方差比\(\dfrac{\sigma_1^2}{\sigma_2^2}\) \(\mu_1, \mu_2\)未知 \(G = \dfrac{S_1^2/S_2^2}{\sigma_1^2/\sigma_2^2}\sim F(m-1,n-1)\) \(\cdots\)

假设检验

  1. 根据问题提出原假设\(H_0\)和备择假设\(H_1\)；
  2. 选择统计量，在原假设成立的条件下确定统计量的分布；
  3. 根据\(\alpha\)确定统计量对应的临界值；
  4. 根据样本观测值计算统计量的观测值与临界值比较，选择接受或拒绝原假设\(H_0\)；

两类错误

  • 弃真：当且仅当小概率事件\(A\)发生时才拒绝\(H_0\)，\(P(A|H_0)\leq \alpha\)
  • 取伪：当\(\alpha\)确定后，可以通过增加样本容量来减小取伪概率

显著性检验只对第一类错误的概率加以控制，而不考虑犯第二类错误的概率，称犯第一类错误的概率\(\alpha\)为显著性水平

后果严重的作为原假设

正态总体的假设检验

单边检验：拒绝域与备择假设的不等号方向一致