diff --git a/advanced-math/knowledge/5-vector-algebra-and-space-analytic-geometry/vector-algebra-and-space-analytic-geometry.pdf b/advanced-math/knowledge/5-vector-algebra-and-space-analytic-geometry/vector-algebra-and-space-analytic-geometry.pdf index be0e798..d286c52 100644 Binary files a/advanced-math/knowledge/5-vector-algebra-and-space-analytic-geometry/vector-algebra-and-space-analytic-geometry.pdf and b/advanced-math/knowledge/5-vector-algebra-and-space-analytic-geometry/vector-algebra-and-space-analytic-geometry.pdf differ diff --git a/advanced-math/knowledge/5-vector-algebra-and-space-analytic-geometry/vector-algebra-and-space-analytic-geometry.tex b/advanced-math/knowledge/5-vector-algebra-and-space-analytic-geometry/vector-algebra-and-space-analytic-geometry.tex index 35f6ba7..e856597 100644 --- a/advanced-math/knowledge/5-vector-algebra-and-space-analytic-geometry/vector-algebra-and-space-analytic-geometry.tex +++ b/advanced-math/knowledge/5-vector-algebra-and-space-analytic-geometry/vector-algebra-and-space-analytic-geometry.tex @@ -60,7 +60,7 @@ \begin{itemize} \item $\vec{a}\cdot\vec{b}=(a_x,a_y,a_z)\cdot(b_x,b_y,b_z)=a_xb_x+a_yb_y+a_zb_z$。 - \item $\vec{a}\cdot\vec{b}=\vert a\vert\vert b\vert\cos\theta$，则$\cos\theta=\dfrac{a\cdot b}{\vert a\vert\vert b\vert}=\dfrac{a_xb_x+a_yb_y+a_zb_z}{\sqrt{a_x^2+a_y^2+a_z^2}\cdot\sqrt{b_x^2+b_y^2+b_z^2}}$，其中$\theta$为$\vec{a},\vec{b}$夹角。 + \item $\vec{a}\cdot\vec{b}=\vert\vec{a}\vert\vert\vec{b}\vert\cos\theta$，则$\cos\theta=\dfrac{\vec{a}\cdot\vec{b}}{\vert a\vert\vert b\vert}=\dfrac{a_xb_x+a_yb_y+a_zb_z}{\sqrt{a_x^2+a_y^2+a_z^2}\cdot\sqrt{b_x^2+b_y^2+b_z^2}}$，其中$\theta$为$\vec{a},\vec{b}$夹角。 \item $a\bot b\Leftrightarrow\theta=\dfrac{\pi}{2}\Leftrightarrow a\cdot b=\vert a\vert\vert b\vert\cos\theta=0\Leftrightarrow a_xb_x+a_yb_y+a_zb_z=0$。 \item $Prj_ba=\dfrac{a\cdot b}{\vert b\vert}=\dfrac{a_xb_x+a_yb_y+a_zb_z}{\sqrt{b_x^2+b_y^2+b_z^2}}$称$a$在$b$上的\textbf{投影}。 \end{itemize} diff --git a/advanced-math/knowledge/8-infinite-series/infinite-series.pdf b/advanced-math/knowledge/8-infinite-series/infinite-series.pdf index 184a882..e592021 100644 Binary files a/advanced-math/knowledge/8-infinite-series/infinite-series.pdf and b/advanced-math/knowledge/8-infinite-series/infinite-series.pdf differ diff --git a/advanced-math/knowledge/8-infinite-series/infinite-series.tex b/advanced-math/knowledge/8-infinite-series/infinite-series.tex index f8bdb03..ce04375 100644 --- a/advanced-math/knowledge/8-infinite-series/infinite-series.tex +++ b/advanced-math/knowledge/8-infinite-series/infinite-series.tex @@ -64,7 +64,7 @@ $S_n=u_1+u_2+\cdots+u_n$称为级数的\textbf{部分和}，$\{S_n\}$是级数 \begin{enumerate} \item 线性性质：若级数$\sum\limits_{n=1}^\infty u_n$，$\sum\limits_{n=1}^\infty v_n$均收敛，且其和分别为$S$，$T$，则任给常数$a,b$，有$\sum\limits_{n=1}^\infty(au_n+bv_n)$也收敛，且其和为$aS+bT$，即$\sum\limits_{n=1}^\infty(au_n+bv_n)=a\sum\limits_{n=1}^\infty u_n+b\sum\limits_{n=1}^\infty v_n$。 \item 若级数$\sum\limits_{n=1}^\infty u_n$收敛，则其任意$m$项后余项$\sum\limits_{n=m+1}^\infty u_n$也收敛；若存在$m$项后余项$\sum\limits_{n=m+1}^\infty u_n$收敛，则$\sum\limits_{n=1}^\infty u_n$也收敛。 - \item 级数收敛必要条件：若级数$\sum\limits_{n=1}^\infty u_n$首先，则$\lim\limits_{n\to\infty}u_n=0$。 + \item 级数收敛必要条件：若级数$\sum\limits_{n=1}^\infty u_n$收敛，则$\lim\limits_{n\to\infty}u_n=0$。 \end{enumerate} 证明性质三：$u_n=S_n-S_{n-1}$，所以$\lim\limits_{n\to\infty}=\lim\limits_{n\to\infty}(S_n-S_{n-1})=\lim\limits_{n\to\infty}S_n-\lim\limits_{n\to\infty}S_{n-1}=S-S=0$。极限为0不一定收敛。 @@ -87,7 +87,7 @@ $S_n=u_1+u_2+\cdots+u_n$称为级数的\textbf{部分和}，$\{S_n\}$是级数 基本就是使用放缩法判断是否有界。 -\textcolor{aqua}{\textbf{定理：}}等比级数（几何级数）：$\sum\limits_{n=21}^\infty\dfrac{1}aq^{n-1}\left\{\begin{array}{l} +\textcolor{aqua}{\textbf{定理：}}等比级数（几何级数）：$\sum\limits_{n=1}^\infty\dfrac{1}aq^{n-1}\left\{\begin{array}{l} \vert q\vert<1, \text{收敛} \\ \vert q\vert\geqslant 1, \text{发散} \end{array}\right.$。 diff --git a/probability-theory-and-mathematical-statistics/knowledge/4-law-of-large-numbers-and-central-limit-theorem/law-of-large-numbers-and-central-limit-theorem.pdf b/probability-theory-and-mathematical-statistics/knowledge/4-law-of-large-numbers-and-central-limit-theorem/law-of-large-numbers-and-central-limit-theorem.pdf index e64edb8..6839b3e 100644 Binary files a/probability-theory-and-mathematical-statistics/knowledge/4-law-of-large-numbers-and-central-limit-theorem/law-of-large-numbers-and-central-limit-theorem.pdf and b/probability-theory-and-mathematical-statistics/knowledge/4-law-of-large-numbers-and-central-limit-theorem/law-of-large-numbers-and-central-limit-theorem.pdf differ diff --git a/probability-theory-and-mathematical-statistics/knowledge/4-law-of-large-numbers-and-central-limit-theorem/law-of-large-numbers-and-central-limit-theorem.tex b/probability-theory-and-mathematical-statistics/knowledge/4-law-of-large-numbers-and-central-limit-theorem/law-of-large-numbers-and-central-limit-theorem.tex index 1569a4d..8006b5f 100644 --- a/probability-theory-and-mathematical-statistics/knowledge/4-law-of-large-numbers-and-central-limit-theorem/law-of-large-numbers-and-central-limit-theorem.tex +++ b/probability-theory-and-mathematical-statistics/knowledge/4-law-of-large-numbers-and-central-limit-theorem/law-of-large-numbers-and-central-limit-theorem.tex @@ -52,13 +52,21 @@ 在满足一定的条件下，大数定律均为$\dfrac{1}{n}\sum\limits_{i=1}^nX_i\overset{P}{\rightarrow}E\left(\dfrac{1}{n}\sum\limits_{i=1}^nX_i\right)$。 -\textcolor{violet}{\textbf{定义：}}假设随机变量序列$\{X_n\}$（$n=1,2,3\cdots$）是\textbf{相互独立}的，若方差$DX_i$（$i\geqslant1$）\textbf{存在且一致有上界}，即存在常数$C$，使得$DX_i\leqslant C$对一切$i\geqslant1$均成立，则$\{X_n\}$服从大数定律：$\dfrac{1}{n}\sum\limits_{i=1}^nX_i\overset{P}{\longrightarrow}\dfrac{1}{n}\sum\limits_{i=1}^nEX_i$。即$\overline{X}\overset{P}{\rightarrow}E\overline{X}$。 - 所以大数定律一般是考定律成立条件与结论正确性。 \subsection{切比雪夫大数定律} -\textcolor{violet}{\textbf{定义：}}假设随机变量序列$\{X_n\}$（$n=1,2,3\cdots$）是\textbf{相互独立}的，若方差$DX_i$（$i\geqslant1$）\textbf{存在且一致有上界}，即存在常数$C$，使得$DX_i\leqslant C$对一切$i\geqslant1$均成立，则$\{X_n\}$服从大数定律：$\dfrac{1}{n}\sum\limits_{i=1}^nX_i\overset{P}{\longrightarrow}\dfrac{1}{n}\sum\limits_{i=1}^nEX_i$。即$\overline{X}\overset{P}{\rightarrow}E\overline{X}$。 +\textcolor{violet}{\textbf{定义：}}假设随机变量序列$\{X_n\}$（$n=1,2,3\cdots$）是\textbf{相互独立}的（不一定同分布），若期望$EX$存在，且方差$DX_i$（$i\geqslant1$）\textbf{存在且一致有上界}，即存在常数$C$，使得$DX_i\leqslant C$对一切$i\geqslant1$均成立，则$\forall\epsilon>0$，$\lim\limits_{n\to\infty}P\left\{\left\vert\dfrac{1}{n}\sum\limits_{i=1}^nX_i-\dfrac{1}{n}\sum\limits_{i=1}^nEX_i\right\vert<\epsilon\right\}=1$，$\{X_n\}$服从大数定律：$\dfrac{1}{n}\sum\limits_{i=1}^nX_i\overset{P}{\longrightarrow}\dfrac{1}{n}\sum\limits_{i=1}^nEX_i$。即$\overline{X}\overset{P}{\rightarrow}E\overline{X}$。 + +即均值收敛于期望的均值。 + +证明：$E\left(\dfrac{1}{n}\sum\limits_{i=1}^nX_i\right)=\dfrac{1}{n}\sum\limits_{i=1}^nEX_i$，又$X_i$不相关，则$Cov(X_i,X_j)=0$。 + +则$D\left(\dfrac{1}{n}\sum\limits_{i=1}^nX_i\right)=\dfrac{1}{n^2}\sum\limits_{i=1}^nDX_i\leqslant\dfrac{nC}{n^2}=\dfrac{C}{n}$。 + +又根据切比雪夫不等式$1\geqslant\lim\limits_{n\to\infty}P\left\{\left\vert\dfrac{1}{n}\sum\limits_{i=1}^nX_i-\dfrac{1}{n}\sum\limits_{i=1}^nEX_i\right\vert<\epsilon\right\}\geqslant1-\dfrac{D\left(\dfrac{1}{n}\sum_{i=1}^nX_i\right)}{\epsilon^2}\geqslant1-\dfrac{C}{n\epsilon^2}\xrightarrow{n\to+\infty}1$。 + +根据夹逼定理$\lim\limits_{n\to\infty}P\left\{\left\vert\dfrac{1}{n}\sum\limits_{i=1}^nX_i-\dfrac{1}{n}\sum\limits_{i=1}^nEX_i\right\vert<\epsilon\right\}$=1。 \textbf{例题：}设$X_1,X_2,\cdots,X_n,\cdots$为相互独立的随机变量序列，$X_n$服从参数为$n$的指数分布（$n\leqslant1$），则下列随机变量序列中不服从切比雪夫大数定律的是()。 @@ -78,7 +86,7 @@ $\because X_n\sim E(n)$，$\therefore EX_n=\dfrac{1}{n}$，$DX_n=\dfrac{1}{n^2}$ \subsection{伯努利大数定律} -\textcolor{violet}{\textbf{定义：}}假设$\mu_n$是$n$重伯努利试验中事件$A$发生的次数，在每次试验中事件$A$发生的概率为$p$（$00$，有$\lim\limits_{n\to\infty}P\left\{\left\vert\dfrac{\mu_n}{n}-p\right\vert<\epsilon\right\}=1$。 +\textcolor{violet}{\textbf{定义：}}假设$\mu_n$是$n$重伯努利试验中事件$A$发生的次数，在每次试验中事件$A$发生的概率为$p$（$00$，有$\lim\limits_{n\to\infty}P\left\{\left\vert\dfrac{\mu_n}{n}-p\right\vert<\epsilon\right\}=\lim\limits_{n\to\infty}P\left\{\left\vert\dfrac{1}{n}\sum\limits_{i=1}^nX_i-\dfrac{1}{n}\sum\limits_{i=1}^nEX_i\right\vert<\epsilon\right\}=\lim\limits_{n\to\infty}P\\\left\{\left\vert\dfrac{1}{n}\sum\limits_{i=1}^nX_i-\dfrac{1}{n}\sum\limits_{i=1}^nEX_i\right\vert<\epsilon\right\}=1$。 可以看作通过$n$重伯努利试验，一个事件的试验概率$\dfrac{\mu_n}{n}$会逼近一个固定的事件概率$p$。 @@ -92,8 +100,12 @@ $\because X_n\sim E(n)$，$\therefore EX_n=\dfrac{1}{n}$，$DX_n=\dfrac{1}{n^2}$ \subsection{辛钦大数定律} +辛钦大数定律类似切比雪夫大数定律的特殊化，将序列约束为同分布。（但是对方差没有要求，所以不能按切比雪夫定律的证明来做） + \textcolor{violet}{\textbf{定义：}}假设随机变量序列$\{X_n\}$（$n=1,2,3\cdots$）是\textbf{相互独立}的\textbf{同分布}的，如果$EX_i=\mu$（$i=1,2,\cdots$）\textbf{存在}，则$\dfrac{1}{n}\sum\limits_{i=1}^nX_i\overset{P}{\longrightarrow}\mu$，即对任意的$\epsilon>0$有$\lim\limits_{n\to\infty}P\left\{\left\vert\dfrac{1}{n}\sum\limits_{i=1}^nX_i-\mu\right\vert<\epsilon\right\}=1$。也可以转换为即$\overline{X}\overset{P}{\rightarrow}E\overline{X}$。\medskip +即能用平均数可以来逼近期望。 + \textbf{例题：}假设随机变量序列$X_1,X_2,\cdots,X_n,\cdots$相互独立，根据辛钦大数定律，当$n\to\infty$时，$\dfrac{1}{n}\sum\limits_{i=1}^nX_i$依概率收敛于数学期望，只要$\{X_n\}$()。 $A.$有相同的数学期望\qquad$B.$服从同一离散型分布 diff --git a/probability-theory-and-mathematical-statistics/knowledge/5-mathematical-statistics/mathematical-statistics.pdf b/probability-theory-and-mathematical-statistics/knowledge/5-mathematical-statistics/mathematical-statistics.pdf index ec20f51..e6ac91b 100644 Binary files a/probability-theory-and-mathematical-statistics/knowledge/5-mathematical-statistics/mathematical-statistics.pdf and b/probability-theory-and-mathematical-statistics/knowledge/5-mathematical-statistics/mathematical-statistics.pdf differ diff --git a/probability-theory-and-mathematical-statistics/knowledge/5-mathematical-statistics/mathematical-statistics.tex b/probability-theory-and-mathematical-statistics/knowledge/5-mathematical-statistics/mathematical-statistics.tex index 07fd377..d7819af 100644 --- a/probability-theory-and-mathematical-statistics/knowledge/5-mathematical-statistics/mathematical-statistics.tex +++ b/probability-theory-and-mathematical-statistics/knowledge/5-mathematical-statistics/mathematical-statistics.tex @@ -217,7 +217,7 @@ $\therefore P\{F>F_\alpha(n_2,n_1)\}=\alpha$，$P\{F\leqslant F_\alpha(n_2,n_1)\ \begin{enumerate} \item $\overline{X}\sim N\left(\mu,\dfrac{\sigma^2}{n}\right)$，即$\dfrac{\overline{X}-\mu}{\sigma/\sqrt{n}}=\dfrac{\sqrt{n}(\overline{X}-\mu)}{\sigma}\sim N(0,1)$。 - \item $\dfrac{1}{\sigma^2}\sum\limits_{i=1}^n(X_i-\mu)^2\sim\chi^2(n)$。 + \item $\dfrac{1}{\sigma^2}\sum\limits_{i=1}^n(X_i-\mu)^2\sim\chi^2(n)$。$\dfrac{(n-1)S^2}{\sigma^2}\sim\chi^2(n-1)$。 \item $\dfrac{(n-1)S^2}{\sigma^2}=\sum\limits_{i=1}^n\left(\dfrac{X_i-\overline{X}}{\sigma}\right)^2\sim\chi^2(n-1)$（$\mu$未知时，在2中用$\overline{X}$代替$\mu$）。 \item $\overline{X}$与$S^2$相互独立，$\dfrac{\sqrt{n}(\overline{X}-\mu)}{S}\sim t(n-1)$（$\sigma$未知时在1中用$S$代替$\sigma$）。进一步有$\dfrac{n(\overline{X}-\mu)^2}{S^2}\sim F(1,n-1)$。 \end{enumerate} @@ -236,6 +236,14 @@ $\therefore P\{F>F_\alpha(n_2,n_1)\}=\alpha$，$P\{F\leqslant F_\alpha(n_2,n_1)\ \subsubsection{矩估计法} +使用替换思想，用样本距来估计总体距。 + +\begin{enumerate} + \item 写出总体$k$阶矩$\mu_k=E(X^k)$，其中$\mu$含有参数$\theta$。 + \item 写出样本$k$阶矩$A_k=\dfrac{1}{n}\sum\limits_{i=1}^nX^k$，$A_k$只与样本有关。 + \item 令总体$k$阶矩=样本$k$阶矩，即$E(X^k)=\dfrac{1}{n}\sum\limits_{i=1}^nX^k$，就得到了$\theta$的方程。 +\end{enumerate} + \textbf{例题：}来自总体的$X$的简单随机样本$X_1,X_2,\cdots,X_n$，总体$X$的概率分布为$X\sim\left(\begin{array}{ccc} -1 & 0 & 2 \\ 2\theta & \theta & 1-3\theta @@ -329,7 +337,7 @@ $ED=\sigma^2=kE(\sum\lim\limits_{i=1}^{n-1}(X_{i+1}-X_i)^2)=kE(\sum\limits_{i=1} \subsubsection{有效性} -也称为最小方差性。 +也称为最小方差性。只有同样的无偏性才能比较有效性。 \textcolor{violet}{\textbf{定义：}}设$\hat{\theta_1}=\hat{\theta_1}(X_1,X_2,\cdots,X_n)$与$\hat{\theta_2}=\hat{\theta_2}(X_1,X_2,\cdots,X_n)$都是$\theta$的无偏估计量，若$D(\hat{\theta_1})