diff --git a/probability-theory-and-mathematical-statistics/exercise/2-random-variables-and-distribution/random-variables-and-distribution.pdf b/probability-theory-and-mathematical-statistics/exercise/2-random-variables-and-distribution/random-variables-and-distribution.pdf index 0fabef9..85a122e 100644 Binary files a/probability-theory-and-mathematical-statistics/exercise/2-random-variables-and-distribution/random-variables-and-distribution.pdf and b/probability-theory-and-mathematical-statistics/exercise/2-random-variables-and-distribution/random-variables-and-distribution.pdf differ diff --git a/probability-theory-and-mathematical-statistics/exercise/2-random-variables-and-distribution/random-variables-and-distribution.tex b/probability-theory-and-mathematical-statistics/exercise/2-random-variables-and-distribution/random-variables-and-distribution.tex index c488d70..45fa3a7 100644 --- a/probability-theory-and-mathematical-statistics/exercise/2-random-variables-and-distribution/random-variables-and-distribution.tex +++ b/probability-theory-and-mathematical-statistics/exercise/2-random-variables-and-distribution/random-variables-and-distribution.tex @@ -198,7 +198,7 @@ $f(x)=\left\{\begin{array}{ll} \subsubsection{正态分布} -$f(x)=\dfrac{1}{\sqrt{2\pi}\sigma}\exp\left(-\dfrac{(x-\mu)^2}{2\sigma^2}\right)$（$-\infty0$），$X\sim N(\mu,\sigma^2)$。 +$f(x)=\dfrac{1}{\sqrt{2\pi}\sigma}e^{-\frac{(x-\mu)^2}{2\sigma^2}}$（$-\infty0$），$X\sim N(\mu,\sigma^2)$。 \textbf{例题：}已知随机变量$X\sim N(0,1)$，对给定的$\alpha$（$0<\alpha>1$），数$\mu_\alpha$满足$P\{X>\mu_\alpha\}=\alpha$，若$P\{\vert X\vert0$，$-1<\rho<1$。 @@ -236,7 +236,7 @@ $\because$面积为$\alpha$的下标为$\alpha$，$\therefore$面积为$\dfrac{1 \textbf{例题：}$(X,Y)\sim N(\mu_1,\mu_2;\sigma_1^2,\sigma_2^2;0)$，分布函数为$F(x,y)$，已知$F(\mu_1,y)=\dfrac{1}{4}$，求$y$。 -解：当$\rho=0$时，$F(X,Y)=\dfrac{1}{2\pi\sigma_1\sigma_2}\exp\left(-\dfrac{1}{2}\left(\left(\dfrac{x-\mu_1}{\sigma_1}\right)^2+\left(\dfrac{y-\mu_2}{\sigma_2}\right)^2\right)\right)\\=F_X(x)F_Y(y)$，即$XY$相互独立。 +解：当$\rho=0$时，$F(X,Y)=\dfrac{1}{2\pi\sigma_1\sigma_2}e^{-\frac{1}{2}\left[\left(\frac{x-\mu_1}{\sigma_1}\right)^2+\left(\frac{y-\mu_2}{\sigma_2}\right)^2\right]}\\=F_X(x)F_Y(y)$，即$XY$相互独立。 $X\sim N(\mu_1,\sigma_1)$，$Y\sim N(\mu_2,\sigma_2)$，$F_X(\mu_1)=P\{X\leqslant\mu_1\}=\dfrac{1}{2}$。 diff --git a/probability-theory-and-mathematical-statistics/exercise/3-digital-features/digital-features.pdf b/probability-theory-and-mathematical-statistics/exercise/3-digital-features/digital-features.pdf index 00160e6..97c2416 100644 Binary files a/probability-theory-and-mathematical-statistics/exercise/3-digital-features/digital-features.pdf and b/probability-theory-and-mathematical-statistics/exercise/3-digital-features/digital-features.pdf differ diff --git a/probability-theory-and-mathematical-statistics/exercise/3-digital-features/digital-features.tex b/probability-theory-and-mathematical-statistics/exercise/3-digital-features/digital-features.tex index 70e68f2..7e67c53 100644 --- a/probability-theory-and-mathematical-statistics/exercise/3-digital-features/digital-features.tex +++ b/probability-theory-and-mathematical-statistics/exercise/3-digital-features/digital-features.tex @@ -40,6 +40,16 @@ \subsubsection{离散型随机变量} +可以根据随机变量分布律的形式拟合出已知的离散型随机变量分布，从而得到已知的期望。 + +\textbf{例题：}设随机变量$X$的分布律为$P\{X=k\}=\dfrac{1}{2^kk!(\sqrt{e}-1)}$，$k=1,2,\cdots$，求$EX$。 + +解：查看分布律中含有$k!$的形式，所以可以考虑转换为泊松分布。泊松分布的标准形式是$\dfrac{\lambda^k}{k!}e^{-\lambda}$。 + +$P\{X=k\}=\dfrac{1}{2^kk!(\sqrt{e}-1)}=\dfrac{\sqrt{e}}{\sqrt{e}-1}\dfrac{\left(\frac{1}{2}\right)^k}{k!}e^{-\frac{1}{2}}$，$X\sim\dfrac{\sqrt{e}}{\sqrt{e}-1}P\left(\dfrac{1}{2}\right)$。 + +$\therefore EX=\dfrac{\sqrt{e}}{2\sqrt{e}-2}$。 + \subsubsection{连续型随机变量} \textbf{例题：}连续型随机变量$X$的概率密度为$f(x)=\dfrac{1}{\pi(1+x^2)}$（$-\infty0$，设$\overline{X}=\dfrac{1}{n}\sum\limits_{i=1}^nX_i$，求$D(X_1-\overline{X})$。 + +解：由题已知$DX_i=\sigma_2$。 + +$D(X_1-\overline{X})=D\left(X_1-\dfrac{1}{n}\sum\limits_{i=1}^nX_i\right)=D\left(\dfrac{n-1}{n}X_1-\dfrac{1}{n}\sum\limits_{i=2}^nX_i\right)=\left(\dfrac{n-1}{n}\right)^2\\DX_1+\dfrac{1}{n^2}\sum\limits_{i=2}^nDX_i=\dfrac{n^2-2n+1}{n^2}\sigma^2+\dfrac{n-1}{n^2}\sigma^2=\dfrac{n-1}{n}\sigma^2$。 + +\subsubsection{期望关系} + +\textbf{例题：}已知随机变量$X_1$，$X_2$相互独立，且都服从正态分布$N(\mu,\sigma^2)$（$\sigma>0$），求$D(X_1X_2)$。 + +解：$X_1$，$X_2$服从$N(\mu,\sigma^2)$，则$EX_1=EX_2=\mu$。 + +$D(X_1X_2)=E[(X_1X_2)^2]-[E(X_1X_2)]^2=E(X_1^2X_2^2)-(EX_1EX_2)^2$。 + +若$X_1$，$X_2$相互独立则$X_1^2$，$X_2^2$相互独立，则$=EX_1^2EX_2^2-\mu^4$。 + +又$EX_1^2=EX_2^2=DX_1+(EX_1)^2=DX_2+(EX_2)^2=\sigma^2+\mu^2$。 + +$(\sigma^2+\mu^2)^2-\mu^4=\sigma^4+2\sigma^2\mu^2$。 + +\subsection{切比雪夫不等式} + +$P\{\vert X-EX\vert\leqslant\epsilon\}\leqslant\dfrac{DX}{\epsilon^2}$或$P\{\vert X-EX\vert<\epsilon\}\geqslant1-\dfrac{DX}{\epsilon^2}$。 + +\section{二维随机变量数字特征} + +\subsection{协方差} + +$Cov(X,Y)=E(XY)-E(X)E(Y)$。 + +\textbf{例题：}设随机变量$X_1,X_2,\cdots,X_n$独立同分布，且方差$\sigma^2>0$，$Y_1=\sum\limits_{i=2}^nX_i$和$Y_2=\sum\limits_{j=1}^{n-1}X_j$，求$Y_1$和$Y_n$的协方差$Cov(Y_1,Y_n)$。 + +解：$\because Y_1=\sum\limits_{i=2}^nX_i$，$Y_2=\sum\limits_{j=1}^{n-1}X_j$，$DX_i=\sigma^2$。 + \end{document} diff --git a/probability-theory-and-mathematical-statistics/exercise/4-law-of-large-numbers-and-central-limit-theorem/law-of-large-numbers-and-central-limit-theorem.pdf b/probability-theory-and-mathematical-statistics/exercise/4-law-of-large-numbers-and-central-limit-theorem/law-of-large-numbers-and-central-limit-theorem.pdf new file mode 100644 index 0000000..219da90 Binary files /dev/null and b/probability-theory-and-mathematical-statistics/exercise/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/exercise/4-law-of-large-numbers-and-central-limit-theorem/law-of-large-numbers-and-central-limit-theorem.tex b/probability-theory-and-mathematical-statistics/exercise/4-law-of-large-numbers-and-central-limit-theorem/law-of-large-numbers-and-central-limit-theorem.tex index 3052c89..cde8443 100644 --- a/probability-theory-and-mathematical-statistics/exercise/4-law-of-large-numbers-and-central-limit-theorem/law-of-large-numbers-and-central-limit-theorem.tex +++ b/probability-theory-and-mathematical-statistics/exercise/4-law-of-large-numbers-and-central-limit-theorem/law-of-large-numbers-and-central-limit-theorem.tex @@ -23,7 +23,7 @@ \usepackage[colorlinks,linkcolor=black,urlcolor=blue]{hyperref} % 超链接 \author{Didnelpsun} -\title{标题} +\title{大数定律与中心极限定理} \date{} \begin{document} \maketitle @@ -34,5 +34,8 @@ \newpage \pagestyle{plain} \setcounter{page}{1} -\section{} + +\section{大数定律} + +\section{中心极限定理} \end{document} diff --git a/probability-theory-and-mathematical-statistics/knowledge/2-random-variables-and-distribution/random-variables-and-distribution.pdf b/probability-theory-and-mathematical-statistics/knowledge/2-random-variables-and-distribution/random-variables-and-distribution.pdf index 56ef743..9a9db6d 100644 Binary files a/probability-theory-and-mathematical-statistics/knowledge/2-random-variables-and-distribution/random-variables-and-distribution.pdf and b/probability-theory-and-mathematical-statistics/knowledge/2-random-variables-and-distribution/random-variables-and-distribution.pdf differ diff --git a/probability-theory-and-mathematical-statistics/knowledge/2-random-variables-and-distribution/random-variables-and-distribution.tex b/probability-theory-and-mathematical-statistics/knowledge/2-random-variables-and-distribution/random-variables-and-distribution.tex index f6d5fc2..894641b 100644 --- a/probability-theory-and-mathematical-statistics/knowledge/2-random-variables-and-distribution/random-variables-and-distribution.tex +++ b/probability-theory-and-mathematical-statistics/knowledge/2-random-variables-and-distribution/random-variables-and-distribution.tex @@ -275,7 +275,7 @@ $=1-F(t)=1-P\{X\leqslant t\}=P\{X>t\}$。 \subsubsection{正态分布} -\textcolor{violet}{\textbf{定义：}}如果$X$的概率密度为$f(x)=\dfrac{1}{\sqrt{2\pi}\sigma}\exp\left(-\dfrac{(x-\mu)^2}{2\sigma^2}\right)$（$-\infty0$），则称$X$服从参数为$(\mu,\sigma^2)$的\textbf{正态分布}，称$X$为\textbf{正态变量}，记为$X\sim N(\mu,\sigma^2)$。 +\textcolor{violet}{\textbf{定义：}}如果$X$的概率密度为$f(x)=\dfrac{1}{\sqrt{2\pi}\sigma}e^{-\frac{(x-\mu)^2}{2\sigma^2}}$（$-\infty0$），则称$X$服从参数为$(\mu,\sigma^2)$的\textbf{正态分布}，称$X$为\textbf{正态变量}，记为$X\sim N(\mu,\sigma^2)$。 $f(x)$的图形关于$x=\mu$对称，即$f(\mu-x)=f(\mu+x)$，并在$x=\mu$处有唯一最大值$f(\mu)=\dfrac{1}{\sqrt{2\pi}\sigma}$。$\mu-\sigma$和$\mu+\sigma$为拐点。 @@ -291,7 +291,7 @@ $f(x)$的图形关于$x=\mu$对称，即$f(\mu-x)=f(\mu+x)$，并在$x=\mu$处 \filldraw[black] (1,1) node{$f(x)$}; \end{tikzpicture} -当$\mu=0$，$\sigma=1$时的正态分布$N(0,1)=\dfrac{1}{\sqrt{2\pi}}\exp\left(-\dfrac{x^2}{2}\right)$为\textbf{标准正态分布}，记为$\varPhi(x)$，$\varPhi(x)$为偶函数，$\varPhi(0)=\dfrac{1}{2}$，$\varPhi(-x)=1-\varPhi(x)$。 +当$\mu=0$，$\sigma=1$时的正态分布$N(0,1)=\dfrac{1}{\sqrt{2\pi}}e^{-\frac{x^2}{2}}$为\textbf{标准正态分布}，记为$\varPhi(x)$，$\varPhi(x)$为偶函数，$\varPhi(0)=\dfrac{1}{2}$，$\varPhi(-x)=1-\varPhi(x)$。 若$X\sim N(0,1)$，$P\{X>\mu_\alpha\}=\alpha$，则称$\mu_\alpha$为标准正态分布的\textbf{上侧$\alpha$分位数/上$\alpha$分位点}。 @@ -524,7 +524,7 @@ $p_{\cdot j}=P\{Y=y_i\}=\sum\limits_{i=1}^\infty P\{X=x_i,Y=y_j\}=\sum\limits_{i \textcolor{violet}{\textbf{定义：}}若$(X,Y)$的概率密度为： -{\fontsize{8.2pt}{10pt}$f(x,y)=\dfrac{1}{2\pi\sigma_1\sigma_2\sqrt{1-\rho^2}}\exp\left(-\dfrac{1}{2(1-\rho^2)}\left(\left(\dfrac{x-\mu_1}{\sigma_1}\right)^2-2\rho\left(\dfrac{x-\mu_1}{\sigma_1}\right)\left(\dfrac{y-\mu_2}{\sigma_2}\right)+\left(\dfrac{y-\mu_2}{\sigma_2}\right)^2\right)\right)$} +{\fontsize{8.2pt}{10pt}$f(x,y)=\dfrac{1}{2\pi\sigma_1\sigma_2\sqrt{1-\rho^2}}\exp\left\{-\dfrac{1}{2(1-\rho^2)}\left[\left(\dfrac{x-\mu_1}{\sigma_1}\right)^2-2\rho\left(\dfrac{x-\mu_1}{\sigma_1}\right)\left(\dfrac{y-\mu_2}{\sigma_2}\right)+\left(\dfrac{y-\mu_2}{\sigma_2}\right)^2\right]\right\}$} 其中$\mu_1,\mu_2\in R$，$\sigma_1,\sigma_2>0$，$-1<\rho<1$，则称$(X,Y)$服从参数为$\mu_1,\mu_2,\sigma_1^2,\sigma_2^2,\rho$的\textbf{二维正态分布}，记为$(X,Y)\sim N(\mu_1,\mu_2;\sigma_1^2,\sigma_2^2;\rho)$。此时： diff --git a/probability-theory-and-mathematical-statistics/knowledge/3-digital-features/digital-features.pdf b/probability-theory-and-mathematical-statistics/knowledge/3-digital-features/digital-features.pdf index b3c7e5b..3eca5a3 100644 Binary files a/probability-theory-and-mathematical-statistics/knowledge/3-digital-features/digital-features.pdf and b/probability-theory-and-mathematical-statistics/knowledge/3-digital-features/digital-features.pdf differ diff --git a/probability-theory-and-mathematical-statistics/knowledge/3-digital-features/digital-features.tex b/probability-theory-and-mathematical-statistics/knowledge/3-digital-features/digital-features.tex index df49085..1a56bef 100644 --- a/probability-theory-and-mathematical-statistics/knowledge/3-digital-features/digital-features.tex +++ b/probability-theory-and-mathematical-statistics/knowledge/3-digital-features/digital-features.tex @@ -71,7 +71,9 @@ \subsubsection{概念} -\textcolor{violet}{\textbf{定义：}}设$X$是随机变量，若$E[(X-EX)^2]$存在，则称$E[(X-EX)^2]$为$X$的\textbf{方差}，记为$DX$，即$DX=E[(X-EX)^2]=E(X^2)-(EX)^2$。称$\sqrt{DX}$为$X$的\textbf{标准差}或\textbf{均方差}，记为$\sigma(X)$，称随机变量$X^*=\dfrac{X-EX}{\sqrt{DX}}$为$X$的\textbf{标准化随机变量}，此时$EX^*=0$，$DX^*=1$。 +\textcolor{violet}{\textbf{定义：}}设$X$是随机变量，若$E[(X-EX)^2]$存在，则称$E[(X-EX)^2]$为$X$的\textbf{方差}，记为$D(X)$或$DX$，即$DX=E[(X-EX)^2]=E(X^2)-(EX)^2=EX^2-E^2X$。称$\sqrt{DX}$为$X$的\textbf{标准差}或\textbf{均方差}，记为$\sigma(X)$，称随机变量$X^*=\dfrac{X-EX}{\sqrt{DX}}$为$X$的\textbf{标准化随机变量}，此时$EX^*=0$，$DX^*=1$。 + +当$X$为离散型随机变量时$D(X)=\sum\limits_{i=1}^\infty[x_i-E(X)]^2p_i$，当$X$为连续型随机变量时$D(X)=\int_{-\infty}^{+\infty}[x-E(X)]^2f(x)\,\textrm{d}x$。 \subsubsection{性质} @@ -79,15 +81,21 @@ \item $DX\geqslant0$，$E(X^2)=DX+(EX)^2\geqslant(EX)^2$。 \item $Dc=0$。 \item $D(aX+b)=a^2DX$。 - \item $D(X\pm Y)=DX+DY\pm2Cov(X,Y)$。 - \item 若$XY$相互独立，则$D(aX+bY)=a^2DX+b^2DY$，一般若$X_1,X_2,\cdots,X_n$相互独立，$g_i(x)$为关于$x$的连续函数，则$D\left(\sum\limits_{i=1}^na_iX_i\right)=\sum\limits_{i=1}^na_i^2DX_i$，$D\left[\sum\limits_{i=1}^ng_i(X_i)\right]=\sum\limits_{i=1}^nD[g_i(X_i)]$。 + \item $D(X\pm Y)=DX+DY\pm2Cov(X,Y)=DX+DY\pm2E[(X-EX)(Y-EY)]$。 + \item 若$XY$相互独立，则$D(aX\pm bY)=a^2DX+b^2DY$，一般若$X_1,X_2,\cdots,X_n$相互独立，$g_i(x)$为关于$x$的连续函数，则$D\left(\sum\limits_{i=1}^na_iX_i\right)=\sum\limits_{i=1}^na_i^2DX_i$，$D\left[\sum\limits_{i=1}^ng_i(X_i)\right]=\sum\limits_{i=1}^nD[g_i(X_i)]$。 \end{itemize} \subsection{切比雪夫不等式} \textcolor{violet}{\textbf{定义：}}若随机变量$X$的方差$DX$存在，则对任意$\epsilon>0$，有$P\{\vert X-EX\vert\leqslant\epsilon\}\leqslant\dfrac{DX}{\epsilon^2}$或$P\{\vert X-EX\vert<\epsilon\}\geqslant1-\dfrac{DX}{\epsilon^2}$。 -即一个变量不会离标准差太大距离。 +$P\{\vert X-EX\vert\geqslant\epsilon\}$即代表变量与期望的差距大于某个值的概率，$DX$就是方差，$DX$越小证明波动越小，波动在$\epsilon$外的概率就越小，反之同理，而$\epsilon$越小，则$\dfrac{DX}{\epsilon^2}$越大，则代表$X$靠近期望$EX$的概率越大，反之同理。 + +证明：若$X$是连续型随机变量，令$\vert X-EX\vert\geqslant\epsilon=D$，则$P\{\vert X-EX\vert\geqslant\epsilon\}=\int\limits_Df(x)\,\textrm{d}x$，又该区间上$\vert X-EX\vert\geqslant\epsilon$，$\therefore\dfrac{(X-EX)^2}{\epsilon^2}\geqslant 1$。 + +$\displaystyle{\int\limits_Df(x)\,\textrm{d}x\leqslant\int\limits_D\dfrac{(X-EX)^2}{\epsilon^2}f(x)\,\textrm{d}x\leqslant\int_{-\infty}^{+\infty}\dfrac{(X-EX)^2}{\epsilon^2}f(x)\,\textrm{d}x}$ + +$=\displaystyle{\dfrac{1}{\epsilon^2}\int_{-\infty}^{+\infty}(X-EX)^2f(x)\,\textrm{d}x}=\dfrac{DX}{\epsilon^2}$。 可以用于估算随机变量在某范围中取值的概率，也可以证明某些收敛性问题（如数学统计章节中的一致性）。 @@ -105,7 +113,7 @@ 二项分布$B(n,p)$ & $P\{X=k\}=C_n^kp^k(1-p)^{n-k}$，$k=0,\cdots,n$ & $np$ & $np(1-p)$ \\ \hline 泊松分布$P(\lambda)$ & $P\{X=k\}=\dfrac{\lambda^k}{k!}e^{-\lambda}$，$k=0,\cdots$ & $\lambda$ & $\lambda$ \\ \hline 几何分布$G(p)$ & $P\{X=k\}=(1-p)^{k-1}$，$p,k=1,\cdots$ & $\dfrac{1}{p}$ & $\dfrac{1-p}{p^2}$ \\ \hline - 正态分布$N(\mu,\sigma^2)$ & $f(x)=\dfrac{1}{\sqrt{2\pi}\sigma}\exp\left\{-\dfrac{(x-\mu)^2}{2\sigma^2}\right\}$，$x\in R$ & $\mu$ & $\sigma^2$ \\ \hline + 正态分布$N(\mu,\sigma^2)$ & $f(x)=\dfrac{1}{\sqrt{2\pi}\sigma}e^{-\frac{(x-\mu)^2}{2\sigma^2}}$，$x\in R$ & $\mu$ & $\sigma^2$ \\ \hline 均匀分布$U(a,b)$ & $f(x)=\dfrac{1}{b-a}$，$a0$ & $\dfrac{1}{\lambda}$ & $\dfrac{1}{\lambda^2}$ \\ \hline \end{tabular} @@ -130,15 +138,19 @@ 从定义来看，方差$DX$就是自己的协方差$Cov(X,X)$。 +协方差也可以标准化，已知$X^*=\dfrac{X-EX}{\sqrt{DX}}$，$Y^*=\dfrac{Y-EY}{\sqrt{DY}}$，则$Cov(X^*,Y^*)\\=Cov(\dfrac{X-EX}{\sqrt{DX}},\dfrac{Y-EY}{\sqrt{DY}})=\dfrac{Cov(X-EX,Y-EY)}{\sqrt{DX}\sqrt{DY}}\\=\dfrac{Cov(X,Y)-Cov(X,EY)-Cov(EX,Y)+Cov(EX,EY)}{\sqrt{DX}\sqrt{DY}}=\dfrac{Cov(X,Y)}{\sqrt{DXDY}}$。 + \textcolor{violet}{\textbf{定义：}}$\rho_{XY}=\dfrac{Cov(X,Y)}{\sqrt{DX}\sqrt{DY}}$为随机变量$XY$的\textbf{相关系数}。若$\rho_{XY}=0$，则$XY$不相干，否则相关。 -相关系数是描述随机变量$XY$之间的线性关系。相关系数为0不代表没有其之间没有关系，也可能存在非线性关系。 +相关系数是描述随机变量$XY$之间的线性关系，绝对值越靠近1则越线性相关。相关系数为0不代表没有其之间没有关系，也可能存在非线性关系。 \subsubsection{性质} \begin{itemize} \item 对称性：$Cov(X,Y)=Cov(Y,X)$，$\rho_{XY}=\rho_{YX}$，$Cov(X,X)=DX$，$\rho_{XX}=1$。 \item 线性性：$Cov(X,c)=0$，$Cov(aX+b,Y)=aCov(X,Y)$，$Cov(X_1+X_2,Y)=Cov(X_1,Y)+Cov(X_2,Y)$。一般$Cov\left(\sum\limits_{i=1}^na_iX_i,Y\right)=\sum\limits_{i=1}^nCov(X_i,Y)$。 + \item 若$XY$相互独立，则$Cov(X,Y)=0$。$D(\sum X)=\sum DX$。 + \item $D(X\pm Y)=DX+DY\pm2Cov(X,Y)$。 \item 相关系数有界性：$\vert\rho_{XY}\vert\leqslant1$。 \item 线性关系下的相关系数：若$Y=aX+b$，则$\rho_{XY}=\left\{\begin{array}{ll} 1, & a>0 \\ @@ -203,6 +215,8 @@ $P\{Z=0\}=1-P\{Z=1\}-P\{Z=-1\}=\dfrac{1}{3}$。 \section{独立性与相关性} +相关性是线性相关性。 + \begin{itemize} \item 独立则一定不相关，但是不相关不一定独立。 \item 如果相关则一定不独立。 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 47c0be5..e64edb8 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 a72b38b..1569a4d 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 @@ -44,10 +44,16 @@ \textcolor{violet}{\textbf{定义：}}设随机变量$X$与随机变量序列$\{X_n\}$（$n=1,2,3\cdots$），如果对任意的$\epsilon>0$，有$\lim\limits_{n\to\infty}P\{\vert X_n-X\vert\geqslant\epsilon\}=0$或$\lim\limits_{n\to\infty}P\{\vert X_n-X\vert<\epsilon\}=1$，则称随机变量序列$\{X_n\}$\textbf{依概率收敛于随机变量$X$}，记为$\lim\limits_{n\to\infty}X_n=X(P)$或$X_n\overset{P}{\rightarrow}X(n\to\infty)$。 +即在某项后面的全部项全部落在区域内的概率为1。（不是严格的极限，可能存在超过范围的点，但是不影响后面的点在区域内） + +通常使用伯努利试验类似的频率来估计概率，从而来极限逼近真实概率，但是进行试验时很可能不凑巧出现了概率很小的情况从而破坏了试验的概率随着频率变大而逼近真实概率，所以这种就是依概率收敛。比如抛硬币试验概率：$\dfrac{6}{10}$，$\dfrac{16}{20}$，$\dfrac{21}{40}$，其中$\dfrac{15}{20}$就是一个特殊的情况，但是不影响后面的收敛。 + \section{大数定律} 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} @@ -72,9 +78,17 @@ $\because X_n\sim E(n)$，$\therefore EX_n=\dfrac{1}{n}$，$DX_n=\dfrac{1}{n^2}$ \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$\lim\limits_{x\to\infty}P\left\{\left\vert\dfrac{\mu_n}{n}-p\right\vert<\epsilon\right\}=1$。 +\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$\lim\limits_{n\to\infty}P\left\{\left\vert\dfrac{\mu_n}{n}-p\right\vert<\epsilon\right\}=1$。 -可以看作通过$n$重伯努利试验，一个事件的概率会逼近一个固定的值。 +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$p$。 + +证明：$n$重伯努利试验，则$\mu_n\sim B(n,p)$，$E_{\mu_n}=np$，$D_{\mu_n}=np(1-p)$。 + +则$E\left(\dfrac{\mu_n}{n}\right)=p$，$D\left(\dfrac{\mu_n}{n}\right)=\dfrac{p(1-p)}{n}$。 + +切比雪夫不等式：$\forall\epsilon>0$，$1\geqslant P\{\vert\dfrac{\mu_n}{n}-p\vert<\epsilon\}\geqslant1-\dfrac{p(1-p)}{n\epsilon^2}\xrightarrow{n\to+\infty}1$。 + +根据夹逼定理$\lim\limits_{n\to\infty}P\left\{\left\vert\dfrac{\mu_n}{n}-p\right\vert<\epsilon\right\}=1$。 \subsection{辛钦大数定律} @@ -112,10 +126,24 @@ $C.$服从同一泊松分布\qquad$D.$服从同一连续型分布 定理要求：独立、同分布、期望方差存在。 +$\dfrac{\sum\limits_{i=1}^nX_i-n\mu}{\sqrt{n}\sigma}\sim N(0,1)$，$\sum\limits_{i=1}^nX_i\sim N(n\mu,n\sigma^2)$。 + +\textbf{例题：}已知手套是使用寿命服从指数分布，单位为小时，且平均寿命为20小时。若一个人需要带手套进行工作，发现手套坏了就立刻换新继续工作，为保证该工人有95\%的把握能工作2000小时，求应该为其准备手套的副数。 + +解：题目中需要为其准备手套，且所有使用寿命加在一起大于2000的概率为95\%。这个问题是概率分布的和的问题，所以使用列维-林德伯格定理。 + +假设第$i$副手套的使用寿命为$X_i$，则$X_i\sim E(\lambda)$，又平均寿命为$20$小时，则$E(X_i)=\dfrac{1}{\lambda}=20$，即$\lambda=\dfrac{1}{20}$，$D(X_i)=\dfrac{1}{\lambda^2}=400$。 + +又根据中心极限定理，$\sum\limits_{i=1}^nX_i\sim N(n\mu,n\sigma^2)=N(20n,400n)$， + +保证该工人有95\%的把握能工作2000小时，则$P\{\sum\limits_{i=1}^nX_i\geqslant2000\}\approx0.95$，则标准化$P\left\{\dfrac{\sum X_i-20n}{20\sqrt{n}}\geqslant\dfrac{2000-20n}{\sqrt{20\sqrt{n}}}\right\}\approx0.95$，$1-P\left\{Z<\dfrac{100-n}{\sqrt{n}}\right\}\approx0.95$，$Z=\dfrac{\sum X_i-20n}{20\sqrt{n}}\sim N(0,1)$，即$\varPhi(\dfrac{100-n}{\sqrt{n}})\approx0.05$，查表，$\dfrac{n-100}{\sqrt{n}}\approx1.64$，$n\approx118$。 + \subsection{棣莫弗-拉普拉斯定理} \textcolor{violet}{\textbf{定义：}}假设随机变量$Y_n\sim B(n,p)$（$0