概率论期末复习要点

说明：此复习要点主要为关于各种疑难公式的推导过程，如果希望速成请直接背公式

随机变量的数字特征

1. $D(X\pm Y)=D(X)+D(Y)$

   D ( X ± Y ) = E ( [ ( X ± Y ) − E ( X ± Y ) ] 2 ) = E ( [   ( X − E ( X ) )   ±   ( Y − E ( Y )   )   ] 2 ) = E [ ( X − E ( X ) ) 2 ] + E [ ( Y − E ( Y ) ) 2 ] ± E [ ( X − E ( X ) ) ( Y − E ( Y ) ) ] = D ( X ) + D ( Y ) ± E [ ( X − E ( X ) ) ( Y − E ( Y ) ) ]

   D(X±Y)=E([(X±Y)−E(X±Y)]2)=E([ (X−E(X)) ± (Y−E(Y) ) ]2)=E[(X−E(X))2]+E[(Y−E(Y))2]±E[(X−E(X))(Y−E(Y))]=D(X)+D(Y)±E[(X−E(X))(Y−E(Y))]$$\begin{array}{rl}D(X\pm Y)& =E([(X\pm Y)-E(X\pm Y){]}^2)\\ & =E([(X-E(X))\pm (Y-E(Y)){]}^2)\\ & =E[(X-E(X){)}^2]+E[(Y-E(Y){)}^2]\pm E[(X-E(X))(Y-E(Y))]\\ & =D(X)+D(Y)\pm E[(X-E(X))(Y-E(Y))]\end{array}$$
   D(X±Y)​=E([(X±Y)−E(X±Y)]2)=E([ (X−E(X)) ± (Y−E(Y) ) ]2)=E[(X−E(X))2]+E[(Y−E(Y))2]±E[(X−E(X))(Y−E(Y))]=D(X)+D(Y)±E[(X−E(X))(Y−E(Y))]​

2. 协方差$cov(X,Y)=E(XY)-E(X)E(Y)$

   c o v ( X , Y ) = E { [ X − E ( X ) ] [ Y − E ( Y ) ] } = E [ X Y − X E ( Y ) − Y E ( X ) + E ( X ) E ( Y ) ] = E ( X Y ) − E ( X ) E ( Y ) − E ( Y ) E ( X ) + E ( X ) E ( Y ) = E ( X Y ) − E ( X ) E ( Y )

   cov(X,Y)=E{[X−E(X)][Y−E(Y)]}=E[XY−XE(Y)−YE(X)+E(X)E(Y)]=E(XY)−E(X)E(Y)−E(Y)E(X)+E(X)E(Y)=E(XY)−E(X)E(Y)$$\begin{array}{rl}cov(X,Y)& =E\{[X-E(X)][Y-E(Y)]\}\\ & =E[XY-XE(Y)-YE(X)+E(X)E(Y)]\\ & =E(XY)-E(X)E(Y)-E(Y)E(X)+E(X)E(Y)\\ & =E(XY)-E(X)E(Y)\end{array}$$
   cov(X,Y)​=E{[X−E(X)][Y−E(Y)]}=E[XY−XE(Y)−YE(X)+E(X)E(Y)]=E(XY)−E(X)E(Y)−E(Y)E(X)+E(X)E(Y)=E(XY)−E(X)E(Y)​

3. 相关系数$\rho (X,Y)=\frac{cov(X,Y)}{\sqrt{D(X)}\sqrt{D(Y)}}$

由于协方差的值会受到X和Y的量纲大小影响，为了克服这个缺点，我们引入相关系数的概念

定义1

设(X,Y)为一个二位随机变量，若cov(X,Y)存在，且$D(X)>0,D(Y)>0$,则称$\frac{cov(X,Y)}{\sqrt{D(X)}\sqrt{D(Y)}}$为X与Y的相关系数，记作$\rho (X,Y)或{\rho }_{xy}$
$$\rho (X,Y)=\frac{cov(X,Y)}{\sqrt{D(X)}\sqrt{D(Y)}}$$
相关系数$\rho (X,Y)$是一个无量纲的量，具有如下特性:

1. $|\rho (X,Y)|\le 1$
2. $|\rho (X,Y)|=1$的充要条件是$P(Y=aX+b)=1$，其中a，b为常数

以上的性质说明相关系数${\rho }_{xy}$表示的是X,Y取值之间的线性关系的密切程度。

当$|\rho (X,Y)|=1$时，X与Y的取值也许会有$Y\ne aX+b$的情况发生，不过这种情况发生的概率为0，可以不计（就好比连续函数可能取到某个值，但是取到其的概率为0）

定义2

若X与Y的相关系数$\rho (X,Y)=0$,则称X与Y不相关

(X，Y如果相互独立那么X，Y一定不相关，但是X，Y不相关不一定相互独立)

例题1：

设随机变量$X\sim N(0,1)$,$Y=X^2$,则$\rho (X,Y)=?$
∵ X ∼ N ( 0 , 1 ) ∴ E ( X 2 ) = D ( X ) + [ E ( X ) ] 2 = 1 ∴ E ( Y ) = E ( X 2 ) = 1 ∵ E ( X ) = 0 , E ( Y ) = 0 , E ( X Y ) = 0 ∴ c o v ( X , Y ) = E ( X Y ) − E ( X ) E ( Y ) = 0 − 0 = 0 ∴ ρ ( X , Y ) = c o v ( X , Y ) D ( X ) D ( Y ) = 0

∵X∼N(0,1)∴E(X2)=D(X)+[E(X)]2=1∴E(Y)=E(X2)=1∵E(X)=0,E(Y)=0,E(XY)=0∴cov(X,Y)=E(XY)−E(X)E(Y)=0−0=0∴ρ(X,Y)=cov(X,Y)D(X)−−−−−√D(Y)−−−−−√=0$$\begin{array}{rl}& \because X\sim N(0,1)\\ & \therefore E(X^2)=D(X)+[E(X){]}^2=1\\ & \therefore E(Y)=E(X^2)=1\\ & \because E(X)=0,E(Y)=0,E(XY)=0\\ & \therefore cov(X,Y)=E(XY)-E(X)E(Y)=0-0=0\\ & \therefore \rho (X,Y)=\frac{cov(X,Y)}{\sqrt{D(X)}\sqrt{D(Y)}}=0\end{array}$$

​∵X∼N(0,1)∴E(X2)=D(X)+[E(X)]2=1∴E(Y)=E(X2)=1∵E(X)=0,E(Y)=0,E(XY)=0∴cov(X,Y)=E(XY)−E(X)E(Y)=0−0=0∴ρ(X,Y)=D(X) ​D(Y) ​cov(X,Y)​=0​
所以为什么，E(XY)=0呢？

这里我们要知道，这道题中$E(XY)=E(X^3)$

于是，我们不妨开始递推$E(X^k)$的一般规律：
E ( X k ) = ∫ − ∞ + ∞ x k 1 2 π e − x 2 2 d x = ∫ − ∞ + ∞ 1 k + 1 1 2 π e − x 2 2 d x k + 1 = 1 2 π ( k + 1 ) x k + 1 e − x 2 2 ∣ − ∞ + ∞ + 1 ( k + 1 ) ∫ − ∞ + ∞ 1 2 π x k + 2 e − x 2 2 d x = 1 2 π ( k + 1 ) x k + 1 ∣ − ∞ + ∞ + 1 k + 1 E ( X k + 2 ) = 0 + 1 k + 1 E ( X k + 2 )

E(Xk)=∫+∞−∞xk12π−−√e−x22dx=∫+∞−∞1k+112π−−√e−x22dxk+1=12π−−√(k+1)xk+1e−x22|+∞−∞+1(k+1)∫+∞−∞12π−−√xk+2e−x22dx=12π−−√(k+1)xk+1|+∞−∞+1k+1E(Xk+2)=0+1k+1E(Xk+2)$$\begin{array}{rl}E(X^k)& ={\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}{x}^k\frac{1}{\sqrt{2\pi }}{e}^{-\frac{x^2}{2}}dx\\ & ={\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{1}{k+1}\frac{1}{\sqrt{2\pi }}{e}^{-\frac{x^2}{2}}d{x}^{k+1}\\ & =\frac{1}{\sqrt{2\pi }(k+1)}{x}^{k+1}{e}^{-\frac{x^2}{2}}{|}_{-\mathrm{\infty }}^{+\mathrm{\infty }}+\frac{1}{(k+1)}{\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{1}{\sqrt{2\pi }}{x}^{k+2}{e}^{-\frac{x^2}{2}}dx\\ & =\frac{1}{\sqrt{2\pi }(k+1)}{x}^{k+1}{|}_{-\mathrm{\infty }}^{+\mathrm{\infty }}+\frac{1}{k+1}E(X^{k+2})\\ & =0+\frac{1}{k+1}E(X^{k+2})\end{array}$$

E(Xk)​=∫−∞+∞​xk2π ​1​e−2x2​dx=∫−∞+∞​k+11​2π ​1​e−2x2​dxk+1=2π ​(k+1)1​xk+1e−2x2​∣−∞+∞​+(k+1)1​∫−∞+∞​2π ​1​xk+2e−2x2​dx=2π ​(k+1)1​xk+1∣−∞+∞​+k+11​E(Xk+2)=0+k+11​E(Xk+2)​
由此可得公式如下
E ( X k ) = { ( k − 1 ) E ( X k − 2 ) , k   M O D   2 = 0 1 , k = 2 0 , k   M O D   2 = 1 E(X^k)=

⎧⎩⎨⎪⎪(k−1)E(Xk−2)10,k MOD 2=0,k=2,k MOD 2=1$$\{\begin{array}{l}\begin{array}{rl}(k-1)E(X^{k-2})& ,kMOD2=0\\ 1& ,k=2\\ 0& ,kMOD2=1\end{array}\end{array}$$

E(Xk)=⎩⎪⎨⎪⎧​(k−1)E(Xk−2)10​,k MOD 2=0,k=2,k MOD 2=1​​

4. $E(X)={\int }_{-\infty }^{+\infty }xf(x)dx$

已知X的密度函数为 p ( x ) = { 0.5 x ∈ [ 0 , 2 ] 0 其 他 p(x)=

{0.50x∈[0,2]其他$$\{\begin{array}{l}\begin{array}{rlr}& 0.5& x\in [0,2]\\ & 0& 其他\end{array}\end{array}$$

p(x)={​0.50​x∈[0,2]其他​​，则X的数学期望$E(X)=?$
E ( X ) = ∫ − ∞ + ∞ x f ( x ) d x = ∫ 0 2 x f ( x ) d x = ∫ 0 2 x 2 d x = x 2 4 ∣ 0 2 = 1

E(X)=∫+∞−∞xf(x)dx=∫20xf(x)dx=∫20x2dx=x24|20=1$$\begin{array}{rl}E(X)& ={\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}xf(x)dx={\int }_0^{2}xf(x)dx\\ & ={\int }_0^2\frac{x}{2}dx=\frac{x^2}{4}{|}_0^2=1\end{array}$$

E(X)​=∫−∞+∞​xf(x)dx=∫02​xf(x)dx=∫02​2x​dx=4x2​∣02​=1​

5. $E(C)=C$

6. $E(XY)={\int }_{-\infty }^{+\infty }{\int }_{-\infty }^{+\infty }xyf(x,y)dxdy$

例题：设随机变量(X,Y)具有概率密度， f ( x , y ) = { 24 x y 0 ≤ x ≤ 1 , 0 ≤ y ≤ 1 , x + y ≤ 1 0 其 他 f(x,y)=

{24xy00≤x≤1,0≤y≤1,x+y≤1其他$$\{\begin{array}{l}\begin{array}{rlr}& 24xy& 0\le x\le 1,0\le y\le 1,x+y\le 1\\ & 0& 其他\end{array}\end{array}$$

f(x,y)={​24xy0​0≤x≤1,0≤y≤1,x+y≤1其他​​

求$E(X),E(Y),E(XY)$

1. 求$E(X)$
   E ( X ) = ∫ − ∞ + ∞ ∫ − ∞ + ∞ 24 x 2 y d x d y = ∫ 0 1 24 x 2 d x ∫ 0 1 − x y d y = ∫ 0 1 ( 1 − x ) 2 2 24 x 2 d x = ∫ 0 1 12 x 4 − 24 x 3 + 12 x 2 d x = 12 5 x 5 − 6 x 4 + 4 x 3 ∣ 0 1 = 0.4

   E(X)=∫+∞−∞∫+∞−∞24x2ydxdy=∫1024x2dx∫1−x0ydy=∫10(1−x)2224x2dx=∫1012x4−24x3+12x2dx=125x5−6x4+4x3|10=0.4$$\begin{array}{rl}E(X)& ={\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}{\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}24x^{2}ydxdy\\ & ={\int }_0^{1}24x^{2}dx{\int }_0^{1-x}ydy\\ & ={\int }_0^1\frac{(1-x{)}^2}{2}24x^{2}dx\\ & ={\int }_0^{1}12x^4-24x^3+12x^{2}dx\\ & =\frac{12}{5}{x}^5-6x^4+4x^3{|}_0^1\\ & =0.4\end{array}$$
   E(X)​=∫−∞+∞​∫−∞+∞​24x2ydxdy=∫01​24x2dx∫01−x​ydy=∫01​2(1−x)2​24x2dx=∫01​12x4−24x3+12x2dx=512​x5−6x4+4x3∣01​=0.4​

2. 求$E(Y)$
   E ( Y ) = ∫ − ∞ + ∞ ∫ − ∞ + ∞ 24 x y 2 d x d y = ∫ 0 1 24 y 2 d y ∫ 0 1 − y x d x = ∫ 0 1 ( 1 − y ) 2 2 24 y 2 d x = ∫ 0 1 12 y 4 − 24 y 3 + 12 y 2 d y = 12 5 y 5 − 6 y 4 + 4 y 3 ∣ 0 1 = 0.4

   E(Y)=∫+∞−∞∫+∞−∞24xy2dxdy=∫1024y2dy∫1−y0xdx=∫10(1−y)2224y2dx=∫1012y4−24y3+12y2dy=125y5−6y4+4y3|10=0.4$$\begin{array}{rl}E(Y)& ={\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}{\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}24x{y}^{2}dxdy\\ & ={\int }_0^{1}24y^{2}dy{\int }_0^{1-y}xdx\\ & ={\int }_0^1\frac{(1-y{)}^2}{2}24y^{2}dx\\ & ={\int }_0^{1}12y^4-24y^3+12y^{2}dy\\ & =\frac{12}{5}{y}^5-6y^4+4y^3{|}_0^1\\ & =0.4\end{array}$$
   E(Y)​=∫−∞+∞​∫−∞+∞​24xy2dxdy=∫01​24y2dy∫01−y​xdx=∫01​2(1−y)2​24y2dx=∫01​12y4−24y3+12y2dy=512​y5−6y4+4y3∣01​=0.4​
   (其实对称性可以直接求)

3. 求$E(XY)$
   E ( X Y ) = ∫ − ∞ + ∞ ∫ − ∞ + ∞ 24 x 2 y 2 d x d y = ∫ 0 1 24 y 2 d y ∫ 0 1 − y x 2 d x = ∫ 0 1 ( 1 − y ) 3 3 24 y 2 d x = 8 ∫ 0 1 ( − y 5 + 3 y 4 − 3 y 3 + y 2 ) d y = 1 60

   E(XY)=∫+∞−∞∫+∞−∞24x2y2dxdy=∫1024y2dy∫1−y0x2dx=∫10(1−y)3324y2dx=8∫10(−y5+3y4−3y3+y2)dy=160$$\begin{array}{rl}E(XY)& ={\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}{\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}24x^2{y}^{2}dxdy\\ & ={\int }_0^{1}24y^{2}dy{\int }_0^{1-y}{x}^{2}dx\\ & ={\int }_0^1\frac{(1-y{)}^3}{3}24y^{2}dx\\ & =8{\int }_0^1(-y^5+3y^4-3y^3+y^2)dy\\ & =\frac{1}{60}\end{array}$$
   E(XY)​=∫−∞+∞​∫−∞+∞​24x2y2dxdy=∫01​24y2dy∫01−y​x2dx=∫01​3(1−y)3​24y2dx=8∫01​(−y5+3y4−3y3+y2)dy=601​​

大数定理和中心极限定理

切比雪夫不等式

定理1

设随机变量$X$的数学期望$E(X)$与方差$D(X)$都存在，则对任意的常数$ϵ>0$,都有：
$$P(|X-E(X)|\ge ϵ)\le \frac{D(X)}{{ϵ}^2}$$
含义：

对于任意一个分布，其随机变量$X$落在其期望$E(X)$的长度为$ϵ$的邻域的范围外的概率小于一定的值

例题:已知正常成人男性的血液中，单位白细胞数平均是$7300个/ml$,均方差是700，利用切比雪夫不等式估计单位白细胞数在5200~9400之间的概率

**解:**设$X$表示成年男性血液中白细胞数，由题知:

$E(X)=7300$，$D(X)=700^2$,由切比雪夫不等式得:
P ( 5200 < X < 9400 ) = P ( − 2100 < X − 7300 < 2100 ) = P ( ∣ X − 7300 ∣ < 2100 ) = 1 − P ( ∣ X − 7300 ∣ ≥ 2100 ) ≥ 1 − D ( X ) ϵ 2 = 1 − 70 0 2 210 0 2 = 8 9

P(5200<X<9400)=P(−2100<X−7300<2100)=P(|X−7300|<2100)=1−P(|X−7300|≥2100)≥1−D(X)ϵ2=1−700221002=89$$\begin{array}{rl}P(5200<X<9400)& =P(-2100<X-7300<2100)\\ & =P(|X-7300|<2100)\\ & =1-P(|X-7300|\ge 2100)\ge 1-\frac{D(X)}{{ϵ}^2}\\ & =1-\frac{{700}^2}{{2100}^2}=\frac{8}{9}\end{array}$$

P(5200<X<9400)​=P(−2100<X−7300<2100)=P(∣X−7300∣<2100)=1−P(∣X−7300∣≥2100)≥1−ϵ2D(X)​=1−210027002​=98​​
由此可知，切比雪夫不等式在计算连续随机变量的区间概率是很方便的

大数定理

设$X_1,X_2,...,X_n$是相互独立的，服从同一分布的随机变量序列，且具有数学期望$E(X_k)=\mu (k=1,2,...,n)$，则对于任意$ϵ>0$，有：
lim ⁡ n − > ∞ P { ∣ 1 n ∑ k = 1 n X k − μ ∣ < ϵ } = 1 \lim_{n->\infty}P\{|\frac{1}{n}\sum_{k=1}^nX_k-\mu|<\epsilon\}=1 n−>∞lim​P{∣n1​k=1∑n​Xk​−μ∣<ϵ}=1

含义:

当样本空间足够大的时候，样本的期望就等于总体的期望

证明:
E ( 1 n ∑ k = 1 n X k ) = 1 n ∑ k = 1 n E ( X k ) = μ D ( 1 n ∑ k = 1 n X k ) = 1 n 2 ∑ k = 1 n D ( X k ) = σ 2 n 由 切 比 雪 夫 不 等 式 可 知 : P ( ∣ 1 n ∑ k = 1 n X k − μ ∣ ≥ ϵ ) ≤ D ( x ) ϵ 2 ∴ 1 − D ( X ) ϵ 2 ≤ P ( ∣ 1 n ∑ k = 1 n X k − μ ∣ < ϵ ) ≤ 1 即 1 − σ 2 n ϵ 2 ≤ P ( ∣ 1 n ∑ k = 1 n X k − μ ∣ < ϵ ) ≤ 1 ∴ 1 ≤ lim ⁡ n − > ∞ P ( ∣ 1 n ∑ k = 1 n X k − μ ∣ < ϵ ) ≤ 1 ∴ lim ⁡ n − > ∞ P ( ∣ 1 n ∑ k = 1 n X k − μ ∣ < ϵ ) = 1

E(1n∑k=1nXk)D(1n∑k=1nXk)由切比雪夫不等P(|1n∑k=1nXk∴1−D(X)ϵ2≤即1−σ2nϵ2≤P∴1≤limn−>∞P(|∴limn−>∞P(|1n∑k=1n=1n∑k=1nE(Xk)=μ=1n2∑k=1nD(Xk)=σ2n式可知:−μ|≥ϵ)≤D(x)ϵ2P(|1n∑k=1nXk−μ|<ϵ)≤1(|1n∑k=1nXk−μ|<ϵ)≤11n∑k=1nXk−μ|<ϵ)≤1Xk−μ|<ϵ)=1$$\begin{array}{rl}E(\frac{1}{n}\sum _{k=1}^n{X}_k)& =\frac{1}{n}\sum _{k=1}^{n}E(X_k)=\mu \\ D(\frac{1}{n}\sum _{k=1}^n{X}_k)& =\frac{1}{n^2}\sum _{k=1}^{n}D(X_k)=\frac{{\sigma }^2}{n}\\ 由切比雪夫不等& 式可知:\\ P(|\frac{1}{n}\sum _{k=1}^n{X}_k& -\mu |\ge ϵ)\le \frac{D(x)}{{ϵ}^2}\\ \therefore 1-\frac{D(X)}{{ϵ}^2}\le & P(|\frac{1}{n}\sum _{k=1}^n{X}_k-\mu |<ϵ)\le 1\\ 即1-\frac{{\sigma }^2}{n{ϵ}^2}\le P& (|\frac{1}{n}\sum _{k=1}^n{X}_k-\mu |<ϵ)\le 1\\ \therefore 1\le \underset{n->\mathrm{\infty }}{lim}P(|& \frac{1}{n}\sum _{k=1}^n{X}_k-\mu |<ϵ)\le 1\\ \therefore \underset{n->\mathrm{\infty }}{lim}P(|\frac{1}{n}\sum _{k=1}^n& X_k-\mu |<ϵ)=1\end{array}$$

E(n1​k=1∑n​Xk​)D(n1​k=1∑n​Xk​)由切比雪夫不等P(∣n1​k=1∑n​Xk​∴1−ϵ2D(X)​≤即1−nϵ2σ2​≤P∴1≤n−>∞lim​P(∣∴n−>∞lim​P(∣n1​k=1∑n​​=n1​k=1∑n​E(Xk​)=μ=n21​k=1∑n​D(Xk​)=nσ2​式可知:−μ∣≥ϵ)≤ϵ2D(x)​P(∣n1​k=1∑n​Xk​−μ∣<ϵ)≤1(∣n1​k=1∑n​Xk​−μ∣<ϵ)≤1n1​k=1∑n​Xk​−μ∣<ϵ)≤1Xk​−μ∣<ϵ)=1​

抽样分布

1. 样本均值:$\bar{X}=\frac{1}{n}{\sum }_{i=1}^n{X}_i$

2. 未修正的样本方差:$S_0^2=\frac{1}{n}{\sum }_{i=1}^n(X_i-\bar{X}{)}^2$

3. 修正的样本方差:$S^2=\frac{1}{n-1}{\sum }_{i=1}^n(X_i-\bar{X}{)}^2$

为社么要修正？

可以观察第五个式子，如果我们没有修正，那么未修正的样本方差我们会发现不等于${\sigma }^2$,那么为了能无偏估计，我们只能使用修正后的样本方差，来将结果“纠正”

4. $E(\bar{X})=\mu$

推导:
$$E(\bar{X})=E(\frac{1}{n}\sum _{i=1}^n{X}_i)=\frac{1}{n}\sum _{i=1}^{n}E(X_i)=\frac{1}{n}n\mu =\mu$$

5. $D(\bar{X})=\frac{{\sigma }^2}{n}$

推导:
$$D(\bar{X})=D(\frac{1}{n}\sum _{i=1}^n{X}_i)=\frac{1}{n^2}\sum _{i=1}^{n}D(X_i)=\frac{1}{n^2}n{\sigma }^2=\frac{{\sigma }^2}{n}$$

6. $E(S^2)={\sigma }^2$

推导:
E ( S 2 ) = E ( 1 n − 1 ∑ i = 1 n ( X i − X ˉ ) 2 ) = 1 n − 1 E ( ∑ i = 1 n [ ( X i − μ ) − ( X ˉ − μ ) ] 2 ) = 1 n − 1 E [ ∑ i = 1 n ( X i − μ ) 2 − 2 ∑ i = 1 n ( X i − μ ) ( X ˉ − μ ) + ∑ i = 1 n ( X ˉ − μ ) 2 ] = 1 n − 1 { E [ ∑ i = 1 n ( X i − μ ) 2 ] − 2 E [ ∑ i = 1 n ( X i − μ ) ( X ˉ − μ ) ] + E [ ∑ i = 1 n ( X ˉ − μ ) 2 ] } = 1 n − 1 [ n σ 2 − 2 n E ( X ˉ − μ ) 2 + n E ( X ˉ − μ ) 2 ] = 1 n − 1 [ n σ 2 − n E ( X ˉ − μ ) 2 ] = 1 n − 1 [ n σ 2 − n D ( X ˉ ) ] = 1 n − 1 [ n σ 2 − n σ 2 n ] = σ 2

E(S2)=E(1n−1∑i=1n(Xi−X¯)2)=1n−1E(∑i=1n[(Xi−μ)−(X¯−μ)]2)=1n−1E[∑i=1n(Xi−μ)2−2∑i=1n(Xi−μ)(X¯−μ)+∑i=1n(X¯−μ)2]=1n−1{E[∑i=1n(Xi−μ)2]−2E[∑i=1n(Xi−μ)(X¯−μ)]+E[∑i=1n(X¯−μ)2]}=1n−1[nσ2−2nE(X¯−μ)2+nE(X¯−μ)2]=1n−1[nσ2−nE(X¯−μ)2]=1n−1[nσ2−nD(X¯)]=1n−1[nσ2−nσ2n]=σ2$$\begin{array}{rl}E(S^2)& =E(\frac{1}{n-1}\sum _{i=1}^n(X_i-\overline{X}{)}^2)\\ & =\frac{1}{n-1}E(\sum _{i=1}^n[(X_i-\mu )-(\overline{X}-\mu ){]}^2)\\ & =\frac{1}{n-1}E[\sum _{i=1}^n(X_i-\mu {)}^2-2\sum _{i=1}^n(X_i-\mu )(\overline{X}-\mu )+\sum _{i=1}^n(\overline{X}-\mu {)}^2]\\ & =\frac{1}{n-1}\{E[\sum _{i=1}^n(X_i-\mu {)}^2]-2E[\sum _{i=1}^n(X_i-\mu )(\overline{X}-\mu )]+E[\sum _{i=1}^n(\overline{X}-\mu {)}^2]\}\\ & =\frac{1}{n-1}[n{\sigma }^2-2nE(\overline{X}-\mu {)}^2+nE(\overline{X}-\mu {)}^2]\\ & =\frac{1}{n-1}[n{\sigma }^2-nE(\overline{X}-\mu {)}^2]\\ & =\frac{1}{n-1}[n{\sigma }^2-nD(\overline{X})]\\ & =\frac{1}{n-1}[n{\sigma }^2-n\frac{{\sigma }^2}{n}]\\ & ={\sigma }^2\end{array}$$

E(S2)​=E(n−11​i=1∑n​(Xi​−Xˉ)2)=n−11​E(i=1∑n​[(Xi​−μ)−(Xˉ−μ)]2)=n−11​E[i=1∑n​(Xi​−μ)2−2i=1∑n​(Xi​−μ)(Xˉ−μ)+i=1∑n​(Xˉ−μ)2]=n−11​{E[i=1∑n​(Xi​−μ)2]−2E[i=1∑n​(Xi​−μ)(Xˉ−μ)]+E[i=1∑n​(Xˉ−μ)2]}=n−11​[nσ2−2nE(Xˉ−μ)2+nE(Xˉ−μ)2]=n−11​[nσ2−nE(Xˉ−μ)2]=n−11​[nσ2−nD(Xˉ)]=n−11​[nσ2−nnσ2​]=σ2​

练习题:

样本$(X_1,...,X_n)$取自标准正态总体$N(0,1)$,$\bar{X},S$分别为样本均值及样本标准差，则$n\bar{X}\sim ?$

∵ E ( n X ˉ ) = E ( ∑ i = 1 n X i ) = ∑ i = 1 n E ( X i ) = n μ = 0 D ( n X ˉ ) = n 2 D ( X ˉ ) = n 2 σ 2 n = n σ 2 = n ∴ n X ˉ ∼ N ( 0 , n )

∵∴E(nX¯)=E(∑i=1nXi)=∑i=1nE(Xi)=nμ=0D(nX¯)=n2D(X¯)=n2σ2n=nσ2=nnX¯∼N(0,n)$$\begin{array}{rl}\because & E(n\overline{X})=E(\sum _{i=1}^n{X}_i)=\sum _{i=1}^{n}E(X_i)=n\mu =0\\ & D(n\overline{X})=n^{2}D(\overline{X})=n^2\frac{{\sigma }^2}{n}=n{\sigma }^2=n\\ \therefore & n\overline{X}\sim N(0,n)\end{array}$$

∵∴​E(nXˉ)=E(i=1∑n​Xi​)=i=1∑n​E(Xi​)=nμ=0D(nXˉ)=n2D(Xˉ)=n2nσ2​=nσ2=nnXˉ∼N(0,n)​

7. 卡方分布

定义：

设样本$(X_1,X_2,...,X_{n-1},X_n)$为$N(0,1)$,即标准正态分布的样本，则称随机变量$X^2={\sum }_{i=1}^n{X}_i^2$所服从的分布是自由度为n的${\chi }^2$分布，记为$X^2\sim {\chi }^2(n)$

性质:

(1)当$X^2\sim {\chi }^2(n)$时，$E(X^2)=n,D(X^2)=2n$
E ( X 2 ) = E ( ∑ i = 1 n X i 2 ) = ∑ i = 1 n E ( X i 2 ) = ∑ i = 1 n { D ( X i ) + [ E ( X i ) ] 2 } = ∑ i = 1 n 1 = n

E(X2)=E(∑i=1nX2i)=∑i=1nE(X2i)=∑i=1n{D(Xi)+[E(Xi)]2}=∑i=1n1=n$$\begin{array}{rl}E(X^2)& =E(\sum _{i=1}^n{X}_i^2)=\sum _{i=1}^{n}E(X_i^2)\\ & =\sum _{i=1}^n\{D(X_i)+[E(X_i){]}^2\}\\ & =\sum _{i=1}^{n}1=n\end{array}$$

E(X2)​=E(i=1∑n​Xi2​)=i=1∑n​E(Xi2​)=i=1∑n​{D(Xi​)+[E(Xi​)]2}=i=1∑n​1=n​

D ( X 2 ) = D ( ∑ i = 1 n X i 2 ) = ∑ i = 1 n D ( X i 2 ) = ∑ i = 1 n [ E ( X i 4 ) − E 2 ( X i 2 ) ] = ∑ i = 1 n [ E ( X i 4 ) − 1 ]

D(X2)=D(∑i=1nX2i)=∑i=1nD(X2i)=∑i=1n[E(X4i)−E2(X2i)]=∑i=1n[E(X4i)−1]$$\begin{array}{rl}D(X^2)& =D(\sum _{i=1}^n{X}_i^2)=\sum _{i=1}^{n}D(X_i^2)\\ & =\sum _{i=1}^n[E(X_i^4)-E^2(X_i^2)]\\ & =\sum _{i=1}^n[E(X_i^4)-1]\end{array}$$

D(X2)​=D(i=1∑n​Xi2​)=i=1∑n​D(Xi2​)=i=1∑n​[E(Xi4​)−E2(Xi2​)]=i=1∑n​[E(Xi4​)−1]​

由公式6可知，$E(X_i^4)=3E(X_i^2)=3$
$$\therefore D(X^2)=2n$$
(2)${\chi }^2$分布的可加性：

设$X_1^2$与$X_2^2$相互独立，且$X_1^2\sim {\chi }^2(m),X_2^2\sim {\chi }^2(n)$,则
$$X_1^2+X_2^2\sim {\chi }^2(m+n)$$
(3)设$(X_1,X_2,...,X_n)$为正态分布$N(\mu ,{\sigma }^2)$的样本，则
$$X^2=\frac{1}{{\sigma }^2}\sum _{i=1}^n(X_i-\mu {)}^2\sim {\chi }^2(n)$$
其实这个不就是简单的标准化吗，标准化以后就是标准正态分布了，那么当然服从卡方分布了

参数估计

参数估计就是，使用样本计算得到的参数来估计总体的参数(均值、方差。。。)

1. 矩估计的定理：

   假设$(X_1,X_2,...,X_n)$是取自总体$X$的一个样本，$E(X)=\mu ,D(X)={\sigma }^2$

   (1)$\bar{X}$是未知参数$\mu$的矩估计

   (2)当$\mu$未知时,$S_n^2$是未知参数${\sigma }^2$的矩估计,$S_n$是未知参数$\sigma$的矩估计

   (3)当$\mu$已知时，$\frac{1}{n}{\sum }_{i=1}^n{X}_i^2-{\mu }^2$是未知参数${\sigma }^2$的矩估计,$\sqrt{\frac{1}{n}{\sum }_{i=1}^n{X}_i^2-{\mu }^2}$是未知参数$\sigma$的矩估计

   • 例题1

     设总体$X$在$[a,b]$上服从均匀分布，a,b未知.$X_1,X_2,...,X_n$是来自$X$的样本,试求$a,b$的矩估计量

     分析:要求矩估计量，其实就是让我们找出未知量与矩的关系，然后通过这些关系，用已知的矩估计量来代替关系式中的矩，以求得未知量的矩估计

     $解:$
     μ 1 = E ( X ) = a + b 2 μ 2 = E ( X 2 ) = D ( X ) + [ E ( X ) ] 2 = ( b − a ) 2 12 + μ 1 2 ∴ { a + b = 2 μ 1 b − a = 12 ( μ 2 − μ 1 2 ) ∴ { a = μ 1 − 3 ( μ 2 − μ 1 2 ) b = μ 1 + 3 ( μ 2 − μ 1 2 ) 使 用 A 1 = X ˉ 代 替 μ 1 , 使 用 A 2 = f v o 6 y 1 n ∑ i = 1 n X i 2 代 替 μ 2 a ^ = X ˉ − 3 ( 1 n ∑ i = 1 n X i 2 − X ˉ 2 ) b ^ = X ˉ + 3 ( 1 n ∑ i = 1 n X i 2 − X ˉ 2 ) 其 中 1 n ∑ i = 1 n X i 2 − X ˉ 2 = 1 n ∑ i = 1 n ( X i − X ˉ ) 2 ∴ a ^ = X ˉ − 3 ( 1 n ∑ i = 1 n ( X i − X ˉ ) 2 ) b ^ = X ˉ + 3 ( 1 n ∑ i = 1 n ( X i − X ˉ ) 2 )

     ∴∴∴μ1=E(X)=a+b2μ2=E(X2)=D(X)+[E(X)]2=(b−a)212+μ21{a+b=2μ1b−a=12(μ2−μ21)−−−−−−−−−√⎧⎩⎨⎪⎪a=μ1−3(μ2−μ21)−−−−−−−−−√b=μ1+3(μ2−μ21)−−−−−−−−−√使用A1=X¯代替μ1,使用A2=fvo6y1n∑i=1nX2i代替μ2a^=X¯−3(1n∑i=1nX2i−X¯2)−−−−−−−−−−−−−−√b^=X¯+3(1n∑i=1nX2i−X¯2)−−−−−−−−−−−−−−√其中1n∑i=1nX2i−X¯2=1n∑i=1n(Xi−X¯)2a^=X¯−3(1n∑i=1n(Xi−X¯)2)−−−−−−−−−−−−−−−√b^=X¯+3(1n∑i=1n(Xi−X¯)2)−−−−−−−−−−−−−−−√$$\begin{array}{rl}& {\mu }_1=E(X)=\frac{a+b}{2}\\ & {\mu }_2=E(X^2)=D(X)+[E(X){]}^2=\frac{(b-a{)}^2}{12}+{\mu }_1^2\\ \therefore & \{\begin{array}{l}a+b=2{\mu }_1\\ b-a=\sqrt{12({\mu }_2-{\mu }_1^2)}\end{array}\\ \therefore & \{\begin{array}{l}a={\mu }_1-\sqrt{3({\mu }_2-{\mu }_1^2)}\\ b={\mu }_1+\sqrt{3({\mu }_2-{\mu }_1^2)}\end{array}\\ & 使用A_1=\overline{X}代替{\mu }_1,使用A_2=fvo6y\frac{1}{n}\sum _{i=1}^n{X}_i^2代替{\mu }_2\\ & \hat{a}=\overline{X}-\sqrt{3(\frac{1}{n}\sum _{i=1}^n{X}_i^2-{\overline{X}}^2)}\\ & \hat{b}=\overline{X}+\sqrt{3(\frac{1}{n}\sum _{i=1}^n{X}_i^2-{\overline{X}}^2)}\\ & 其中\frac{1}{n}\sum _{i=1}^n{X}_i^2-{\overline{X}}^2=\frac{1}{n}\sum _{i=1}^n(X_i-\overline{X}{)}^2\\ \therefore & \hat{a}=\overline{X}-\sqrt{3(\frac{1}{n}\sum _{i=1}^n(X_i-\overline{X}{)}^2)}\\ & \hat{b}=\overline{X}+\sqrt{3(\frac{1}{n}\sum _{i=1}^n(X_i-\overline{X}{)}^2)}\end{array}$$

     ∴∴∴​μ1​=E(X)=2a+b​μ2​=E(X2)=D(X)+[E(X)]2=12(b−a)2​+μ12​{a+b=2μ1​b−a=12(μ2​−μ12​) ​​{a=μ1​−3(μ2​−μ12​) ​b=μ1​+3(μ2​−μ12​) ​​使用A1​=Xˉ代替μ1​,使用A2​=fvo6yn1​i=1∑n​Xi2​代替μ2​a^=Xˉ−3(n1​i=1∑n​Xi2​−Xˉ2) ​b^=Xˉ+3(n1​i=1∑n​Xi2​−Xˉ2) ​其中n1​i=1∑n​Xi2​−Xˉ2=n1​i=1∑n​(Xi​−Xˉ)2a^=Xˉ−3(n1​i=1∑n​(Xi​−Xˉ)2) ​b^=Xˉ+3(n1​i=1∑n​(Xi​−Xˉ)2) ​​

     这里列出$\frac{1}{n}\sum_{i=1}^nX_i2-\bar{X}^{2=\frac{1}{n}\sum_{i=1}}n(X_i-\bar{X})^2\$的证明
     1 n ∑ i = 1 n ( X i − X ˉ ) 2 = 1 n ∑ i = 1 n ( X i 2 − 2 X i X ˉ + X ˉ 2 ) = 1 n ∑ i = 1 n ( X i 2 ) − 2 1 n ∑ i = 1 n X i X ˉ + 1 n ∑ i = 1 n X ˉ 2 = 1 n ∑ i = 1 n X i 2 − 2 X ˉ 2 + X ˉ 2 = 1 n ∑ i = 1 n ( X i 2 − X ˉ 2 )

     1n∑i=1n(Xi−X¯)2=1n∑i=1n(X2i−2XiX¯+X¯2)=1n∑i=1n(X2i)−21n∑i=1nXiX¯+1n∑i=1nX¯2=1n∑i=1nX2i−2X¯2+X¯2=1n∑i=1n(X2i−X¯2)$$\begin{array}{rl}\frac{1}{n}\sum _{i=1}^n(X_i-\overline{X}{)}^2& =\frac{1}{n}\sum _{i=1}^n(X_i^2-2X_i\overline{X}+{\overline{X}}^2)\\ & =\frac{1}{n}\sum _{i=1}^n(X_i^2)-2\frac{1}{n}\sum _{i=1}^n{X}_i\overline{X}+\frac{1}{n}\sum _{i=1}^n{\overline{X}}^2\\ & =\frac{1}{n}\sum _{i=1}^n{X}_i^2-2{\overline{X}}^2+{\overline{X}}^2\\ & =\frac{1}{n}\sum _{i=1}^n(X_i^2-{\overline{X}}^2)\end{array}$$

     n1​i=1∑n​(Xi​−Xˉ)2​=n1​i=1∑n​(Xi2​−2Xi​Xˉ+Xˉ2)=n1​i=1∑n​(Xi2​)−2n1​i=1∑n​Xi​Xˉ+n1​i=1∑n​Xˉ2=n1​i=1∑n​Xi2​−2Xˉ2+Xˉ2=n1​i=1∑n​(Xi2​−Xˉ2)​

2. 设$\hat{\theta }$是未知参数$\theta$的一个估计量，若$E(\hat{\theta })=\theta$,则称$\hat{\theta }$为参数$\theta$的一个无偏估计量

3. 求最好值的一般方法：

   先求每个估计量的期望，如果期望与总体的期望不一样就$pass$

   再求每个估计量的方差，方差最小的那个最好

4. 求极大似然估计

   1. 将几个样本带入分布函数，将结果相乘
   2. 令相乘的结果为$L(\theta )$，对$L(\theta )$取对数
   3. 对两边同时求导，求出取最大值时${\theta }_0$的值
   4. 此时的${\theta }_0$就是极大似然估计值