随机变量的数字特征
本文最后更新于 2024年12月22日 凌晨
\(X\sim U(-\frac{1}{2},\frac{1}{2}),Y=g(X)=\begin{cases}lnX,X>0,\\0,X\leq 0\end{cases}\)
\(\begin{align} E(Y)=&\int^{+\infty}_{\infty}g(x)f_X(x)dx\\ =&\int^{\frac{1}{2} }_{-\frac{1}{2} }g(x)\cdot1dx\\ =&\int^{\frac{1}{2} }_{0}\ln xdx\\ =&-\frac{1}{2}\ln 2-\frac{1}{2}\\ \\ E(Y^2)=&\int^{+\infty}_{\infty}g^2(x)f_X(x)dx\\ =&\int^{\frac{1}{2} }_{0}\ln xdx\\ =&\frac{1}{2}\ln^2 2+1+\ln 2\\ D(Y)=&E(Y^2)-(E(Y))^2 \end{align}\)
| \(X_i\) | 1 | 0 |
|---|---|---|
| P | \(\frac{1}{n}\) | \(1-\frac{1}{n}\) |
\(E(X_i)=1\cdot \frac{1}{n}+0\cdot (1-\frac{1}{n})=\frac{1}{n}\)
\(E(X)=\mathop{\Sigma} _{ i=1}^{n}E(X_i)=1\)
\(E(X^2)=E(\Sigma_{i=1}^{n}X_i)^2=\Sigma_{i=1}^n E(X_i^2)+2\Sigma_{1\leq i\leq j\leq n}^nE(X_iX_j)\)
| \(X_iX_j\) | 1 | 0 |
|---|---|---|
| P | \(\frac{1}{n(n-1)}\) | \(1-\frac{1}{n(n-1)}\) |
\(E(X_iX_j)=\frac{1}{n(n-1)},i,j=1,2,...,n\)
\(E(X_i^2)=\frac{1}{n},n=1,2,...,n\)
\(E(X^2)=n\cdot \frac{1}{n}+2\cdot C^2_n\cdot \frac{1}{n(n-1)}=2\)
\(D(x)=2-1=1\)
标准化随机变量
设随机变量X的期望E(X),方差D(X)都存在,且D(X)$\(0,称\)X^=$为X的标准化随机变量
\(E(X^\star)=0,D(X^\star)=1\)
若已知分布的类型,及期望和方差,常能确定分布
\(\begin{align} &\int^1_0(Ax^2+Bx)dx=\frac{A}{3}+\frac{B}{2}=1\\ &E(X)=\int^1_0x(Ax^2+Bx)dx=\frac{A}{4}+\frac{B}{3}=0.5\\ \end{align}\)
\(得到:A=-6,B=6\)
协方差
\(E((X-E(X))(Y-E(Y))为随机变量X,Y的协方差\)
\(记作\text{cov}(X,Y)=E(XY)-E(X)E(Y)\)
称\(\begin{bmatrix}D(X)&\text{cov}(X,Y)\\\text{cov}(X,Y)&D(Y)\end{bmatrix}\)
为(X,Y)的协方差矩阵
\(\rho_{XY}=\frac{\text{cov}(X,Y)}{\sqrt{D(X)}\sqrt{D(Y)} }\)
事实上:\(\rho_{XY}=\text{cov}(X^\star,Y^\star)\)
若\(\rho_{XY}=0\),则称X,Y不相关
\(\text{cov}(X,Y)=E((X-E(X))(Y-E(Y)))\\=E(XY)-E(X)E(Y)\)
X,Y相互独立$$X,Y不相关
\(\begin{align} cov(U,V)=&E(UV)-E(U)E(V)\\ =&E(a^2X^2-b^2Y^2)-[E(aX+bY)][E(aX-bY)]\\ =&a^2E(X^2)-b^2E(Y^2)-[a^2E^2(X)-b^2E^2(Y)]\\ =&a^2[E(X^2)-E^2(X)]-b^2[E(Y^2)-E^2(Y)]\\ =&a^2D(X)-b^2D(Y)\\ =&(a^2-b^2)\sigma^2\\ D(U)=&D(aX+bY)\\ =&a^2D(X)+b^2D(Y)=(a^2+b^2)\sigma^2\\ D(Y)=&(a^2+b^2)\sigma\\ \rho_{UV}=&\frac{cov(U,V)}{\sqrt {D(U)}\sqrt{D(V)} }\\ =&\frac{a^2-b^2}{a^2+b^2}\\ \end{align}\)
协方差的性质
- \(cov(X,Y)=cov(Y,X)=E(XY)-E(X)E(Y)\)
- \(\text{cov}(aX,bY)=ab\text{cov}(X,Y)\)
- \(\text{cov}(X,X)=D(X)\)
- \(|\text{cov}(X,Y)|^2\leq D(X)D(Y)\)当且仅当\(P(Y-E(Y)=t_0(X-E(X)))=1\)
Cauchy-Schwarz不等式
\(|E(XY)|^2\leq E(X^2)E(Y^2)\)
\(\begin{align} g(t)=&E[(X-E(X))\cdot t-(Y-E(Y))]^2\\ =&E[(X-E(X))^2t^2-2(X-E(X))t(Y-E(Y))+(Y-E(Y))^2]\\ =&D(X)t^2-2\text{cov}(X,Y)+D(Y)\\ \end{align}\)
对于任何实数\(t\),\(g(t)\geq 0\rightarrow 4\text{cov}^2(X,Y)-4D(X)D(Y)\leq 0\)
即\(|cov(X,Y)|^2\leq D(X)D(Y)\)
\(g(t_0)=0\)时,有两个相等的实零点
等号成立,\(|cov(x,Y)|^2=D(X)D(Y)\)
\(t_0=\pm\sqrt{\frac{D(Y)}{D(X)} }\)
\(\begin{cases}E[(Y-E(Y))-t_0(X-E(X))]^2=0\\E[(Y-E(Y))-t_0(X-E(X))]=0\end{cases}\leftrightarrow \begin{cases}D[(Y-E(Y))-t_0(X-E(X))]=0\\E[(Y-E(Y))-t_0(X-E(X))=0]=1\end{cases}\)
若X,Y服从二维正态分布,\(X,Y相互独立\Leftrightarrow X,Y不相关\)
\(\begin{align} &E(X)=E(Y)=1,D(X)=D(Y)=4,\rho_{XY}=\frac{1}{2}\\ &cov(X,Y)=\rho_{XY}\sqrt{D(X)}\sqrt{D(Y)}=2\\ &cov(X,Z)=cov(X,X)+cov(X,Y)=D(X)+\rho_{XY}\sqrt{D(X)}\sqrt{D(Y)}\\ &=4+\frac{1}{2}\cdot\sqrt{4}\cdot \sqrt{4}=6\\ &D(Z)=D(X)+D(Y)+2cov(X,Y)=12\\ &\rho_{XZ}=\frac{\sqrt {3} }{2}\end{align}\)
重要数学期望
\(E(X^k)\)X的k阶原点矩
\(E(|X|^k)\)X的k阶绝对原点矩
\(E((X-E(X))^k)\)X的k阶中心矩
马尔可夫不等式
\(P(X\geq \varepsilon)\leq \frac{E(x)}{\varepsilon}\)
\(证:P(X\geq\varepsilon)=\int^{+\infty}_{\varepsilon}f(x)dx\leq\int^{+\infty}_{\varepsilon}\frac{x}{\varepsilon}f(x)dx\\\leq\frac{1}{\varepsilon}\int^{+\infty}_0 xf(x)dx=\frac{E(X)}{\varepsilon}\)
推论1
\(P(|X|\geq \varepsilon)\leq\frac{E(|X|^k)}{\varepsilon^k}\) \[ \begin{aligned} &P(|X|\geq \varepsilon)=P(|X|^k\geq \varepsilon^k) \leq \dfrac{E(|X|^k)}{\varepsilon_k} \end{aligned} \]
推论2 切比雪夫不等式
\(P(|X-E(X)|\geq \varepsilon)\leq \frac{D(X)}{\varepsilon^2}\)
\(proof:P(|X-E(x)|\geq \varepsilon)=P(|X-E(x)|^2\geq \varepsilon^2)\\\leq \frac{E(|X-E(X)|^2)}{\varepsilon^2}=\frac{D(X)}{\varepsilon^2}\)
\(X\sim B(6000,\frac{1}{6})\\E(X)=1000\\D(X)=6000\times\frac{1}{6}\times\frac{5}{6}=\frac{5000}{6}\)
\(P(|\frac{X}{6000}-\frac{1}{6}|<0.01)=P(|X-6000|<60)\\\geq 1-\frac{\frac{5000}{6} }{60^2}=0.7685\)
切比雪夫大数定律
\(D(X_k)=\sigma^2_k\leq c,k=1,2,...\)
则有
\(\lim\limits_{n\rightarrow \infty}D(\frac{1}{n}\mathop\Sigma\limits_{k=1}^nE(X_k))=0\)
\(proof:E(\frac{1}{n}\mathop\Sigma\limits^n_{i=1}X_i)=\frac{1}{n}\mathop\Sigma\limits^n_{i=1}E(X_i)\\D(\frac{1}{n}\mathop\Sigma^n_{i=1}X_i)=\frac{1}{n^2}\mathop\Sigma^n_{i=1}D(X_i)=\frac{1}{n^2}\cdot nc\\=\frac{c}{n}\)
定义
\(\lim\limits_{n\rightarrow \infty}P(|Y_n-a|\geq \varepsilon)=0\\\lim\limits_{n\rightarrow \infty}P(|Y_n-a|< \varepsilon)=1\)
记作\(Y_n\mathop\rightarrow\limits^P_{n\rightarrow \infty}a\)
辛钦大数定律
\(\lim\limits_{n\rightarrow \infty}(|\frac{1}{n}\mathop\Sigma\limits_{k=1}^nX_k-\mu|\geq\varepsilon)=0\)
伯努利大数定律
\(\lim\limits_{n\rightarrow \infty} P(|\frac{n_A}{n}-p|\geq \varepsilon)=0\)
意义
回答了:为何能以某事件发生的概率作为事件的概率的估计
独立同分布的中心极限定理
\(随机变量序列X_1,X_2,...,X_n,...相互独立,\\服从同一分布,且有期望和方差:\\E(X_k)=\mu,D(X_k)=\sigma^2,k=1,2...\)
\(\lim\limits_{n\rightarrow \infty}P(\frac{\mathop\Sigma\limits^{n}_{k=1}X_k-n\mu}{\sqrt{n}\sigma}\leq x)=\frac{1}{\sqrt{2\pi} }\int^x_{-\infty}e^{-\frac{t^2}{2} }dt\)
德莫佛-拉普拉斯中心极限定理
\(Y_n\sim B(n,p),0<p<1,n=1,2,...\)
\(\lim\limits_{n\rightarrow \infty}P(\frac{Y_n-n\mu}{\sqrt{np(1-p)} }\leq x)=\frac{1}{\sqrt{2\pi} }\int^x_{-\infty}e^{-\frac{t^2}{2} }dt\)
即对任意的a<b,
\(\lim\limits_{n\rightarrow \infty}P(a<\frac{Y_n-np}{\sqrt{np(1-p)} }\leq b)=\frac{1}{\sqrt{2\pi} }\int^b_ae^{-\frac{t^2}{2} }dt\)
\(Y_n\sim N(np,np(1-p))(近似)\)
