各种分布

①:01分布 B(Binary)

二项分布

X B ( n , p ) X \sim B(n,p) XB(n,p)
E ( X ) = n p E(X)=np E(X)=np
D ( X ) = n p ( 1 p ) D(X)=np(1-p) D(X)=np(1p)

②:泊松分布 P(Poisson)

X P ( λ ) X\sim P(\lambda) XP(λ)
E ( X ) = D ( X ) = λ E(X)=D(X)=\lambda E(X)=D(X)=λ
p { x = k } = λ k k ! e λ p\{x=k\}=\frac{\lambda^k}{k!}e^{-\lambda} p{x=k}=k!λkeλ

理解

因为概率和为1
<munderover> k = 0 </munderover> λ k k ! e λ = 1 \sum_{k=0}^{\infty}\frac{\lambda^k}{k!}e^{-\lambda}=1 k=0k!λkeλ=1
所以:
<munderover> k = 0 </munderover> λ k k ! = e λ \sum_{k=0}^{\infty}\frac{\lambda^k}{k!}=e^{\lambda} k=0k!λk=eλ
其实是 e λ e^{\lambda} eλ的泰勒展开变形

③:均匀分布 U(Uniform)

X U ( a , b ) X\sim U(a,b) XU(a,b)
E ( X ) = a + b 2 E(X)=\frac{a+b}{2} E(X)=2a+b
D ( X ) = ( b a ) 2 12 D(X)=\frac{(b-a)^2}{12} D(X)=12(ba)2

④:指数分布 E(Exponential)

X E ( λ ) X\sim E(\lambda) XE(λ)
E ( X ) = 1 λ E(X)=\frac{1}{\lambda} E(X)=λ1
D ( X ) = 1 λ 2 D(X)=\frac{1}{\lambda^2} D(X)=λ21
X f ( x ) = { <mstyle displaystyle="false" scriptlevel="0"> λ e λ x , x > 0 </mstyle> <mstyle displaystyle="false" scriptlevel="0"> 0 , x 0 </mstyle> X\sim f(x)=\left\{\begin{matrix} \lambda e^{-\lambda x},x>0 \\ 0,x\leq0 \end{matrix}\right. Xf(x)={λeλx,x>00,x0

要背一哈积分

p x > t = t + λ e λ t d t = e λ t p{x>t}=\int_t^{+\infty}\lambda e^{-\lambda t}dt=e^{-\lambda t} px>t=t+λeλtdt=eλt

无记忆性

p x > t + s x > s = p ( x > t + s <mtext>    </mtext> , <mtext>    </mtext> x > s ) p ( x > s ) = p ( x > t + s ) p ( x > s ) = e λ ( t + s ) e λ s = e λ t p{x>t+s|x>s}=\frac{p(x>t+s\ \ ,\ \ x>s)}{p(x>s)}=\frac{p(x>t+s)}{p(x>s)}=\frac{e^{-\lambda(t+s)}}{e^{-\lambda s}}=e^{-\lambda t} px>t+sx>s=p(x>s)p(x>t+s  ,  x>s)=p(x>s)p(x>t+s)=eλseλ(t+s)=eλt

⑤: 正态分布 N(Normal)

X N ( μ , σ 2 ) X\sim N(\mu,\sigma^2) XN(μ,σ2)
E ( X ) = μ E(X)=\mu E(X)=μ
D ( X ) = σ 2 D(X)=\sigma^2 D(X)=σ2
X f ( x ) = 1 2 π σ e ( x μ ) 2 2 σ 2 X\sim f(x)=\frac{1}{\sqrt{2\pi}\sigma}e^{-\frac{(x-\mu)^2}{2\sigma^2}} Xf(x)=2π σ1e2σ2(xμ)2

标准正态

X N ( 0 , 1 ) X\sim N(0,1) XN(0,1)
φ ( x ) \varphi(x) φ(x)来表示
φ ( x ) 1 2 π e x 2 2 \varphi(x)\frac{1}{\sqrt{2\pi}}e^{-\frac{x^2}{2}} φ(x)2π 1e2x2

分布函数

ϕ \phi ϕ来表示
有个对称性的性质:
ϕ ( x ) = 1 ϕ ( x ) \phi(x)=1-\phi(-x) ϕ(x)=1ϕ(x)

一.独立事件

1

①:
<mover accent="true"> A B </mover> = <mover accent="true"> A ˉ </mover> <mover accent="true"> B ˉ </mover> \overline{A\cup B}=\bar{A}\cap \bar{B} AB=AˉBˉ
②:
<mover accent="true"> A B </mover> = <mover accent="true"> A ˉ </mover> <mover accent="true"> B ˉ </mover> \overline{A\cap B}=\bar{A}\cup \bar{B} AB=AˉBˉ
③:
<mover accent="true"> A B </mover> = <mover accent="true"> A ˉ </mover> B \overline{A- B}=\bar{A}\cup B AB=AˉB
这个感觉有点少见

2

①:
p ( B A ) = p ( B <mover accent="true"> A ˉ </mover> ) = p ( B ) p(B|A)=p(B|\bar A)=p(B) p(BA)=p(BAˉ)=p(B)

证明:

p ( B A ) = p ( A B ) p ( A ) = p ( A ) p ( B ) p ( A ) = p ( B ) p(B|A)=\frac{p(AB)}{p(A)}=\frac{p(A)p(B)}{p(A)}=p(B) p(BA)=p(A)p(AB)=p(A)p(A)p(B)=p(B)
p ( B <mover accent="true"> A ˉ </mover> ) p(B|\bar A) p(BAˉ)同理

②:
p ( A B ) = 1 p ( <mover accent="true"> A ˉ </mover> <mover accent="true"> B ˉ </mover> ) p(A|B)=1-p(\bar A|\bar B) p(AB)=1p(AˉBˉ)
这个怎么来的???

二.复合概率密度函数

X f ( x ) , Y g ( f ( x ) ) X\sim f(x),Y\sim g(f(x)) Xf(x),Yg(f(x))

定义法

f Y ( y ) = F Y ( y ) f_Y(y)=F'_Y(y) fY(y)=FY(y)

F Y ( y ) = p ( Y y ) = p ( g ( x ) y ) = g ( x ) y f x ( x ) d y F_Y(y)=p(Y\leq y)=p(g(x)\leq y)=\int_{g(x)\leq y}f_x(x)dy FY(y)=p(Yy)=p(g(x)y)=g(x)yfx(x)dy

一个结论

根据王式安老师说的,好像是个定理,要研究生的课才上
如果:
X f ( x ) , F ( x ) X\sim f(x),F(x) Xf(x),F(x)
并且有 Y = F ( X ) Y=F(X) Y=F(X)这个代换,那么
Y U ( 0 , 1 ) Y\sim U(0,1) YU(0,1)

简略理解证明

Y F Y ( y ) = p ( Y y ) = p ( F ( X ) y ) = p ( X F 1 ( y ) ) = F ( F 1 ( y ) ) = y Y\sim F_Y(y)=p(Y\leq y)=p(F(X)\leq y)=p(X\leq F^{-1}(y))=F(F^{-1}(y))=y YFY(y)=p(Yy)=p(F(X)y)=p(XF1(y))=F(F1(y))=y

f x ( x ) , f X ( x ) , f x ( X ) , f X ( X ) f_x(x),f_X(x),f_x(X),f_X(X) fx(x),fX(x),fx(X),fX(X)

协方差

C o v ( X , Y ) = E { [ X E ( X ) ] [ Y E ( Y ) ] } = E ( X Y ) E ( X ) E ( Y ) Cov(X,Y)=E\{[X-E(X)][Y-E(Y)]\}=E(XY)-E(X)E(Y) Cov(X,Y)=E{[XE(X)][YE(Y)]}=E(XY)E(X)E(Y)

协方差的性质

①:
C o v ( a X , b Y ) = a b C o v ( X , Y ) Cov(aX,bY)=abCov(X,Y) Cov(aX,bY)=abCov(X,Y)
②:
C o v ( X 1 + X 2 , Y ) = C o v ( X 1 , Y ) + C o v ( X 2 , Y ) Cov(X_1+X_2,Y)=Cov(X_1,Y)+Cov(X_2,Y) Cov(X1+X2,Y)=Cov(X1,Y)+Cov(X2,Y)
用协方差来计算和的方差
D ( X ± Y ) = D ( X ) ± 2 C o v ( X , Y ) + D ( Y ) D(X\pm Y)=D(X)\pm2Cov(X,Y)+D(Y) D(X±Y)=D(X)±2Cov(X,Y)+D(Y)

相关系数

ρ X Y = C o v ( X , Y ) D ( X ) D ( Y ) \rho_{XY}=\frac{Cov(X,Y)}{\sqrt{D(X)D(Y)}} ρXY=D(X)D(Y) Cov(X,Y)

大数定理 中心定理

切比雪夫不等式

p ( X E ( X ) ε ) D ( X ) ε 2 p(|X-E(X)|\geq \varepsilon)\leq\frac{D(X)}{\varepsilon^2} p(XE(X)ε)ε2D(X)

切比雪夫大数定理 辛钦大数定理

大数定理这一节,截个王式安老师的图:

伯努利大数定理可以看成是上面两个的特殊情况
反正就是求期望就完事了~

中心定理


意思就是说加起来近似正态分布

样本及抽样分布

样本均值

<mover accent="true"> X </mover> = 1 n <munderover> i = 1 n </munderover> X i \overline X=\frac{1}{n}\sum_{i=1}^nX_i X=n1i=1nXi

样本方差

S 2 = 1 n 1 <munderover> i 1 n </munderover> ( X i <mover accent="true"> X </mover> ) 2 S^2=\frac{1}{n-1}\sum_{i-1}^n(X_i-\overline X)^2 S2=n11i1n(XiX)2
样本方差阔以化为两种形状:
①:
1 n 1 <munderover> i 1 n </munderover> ( X i <mover accent="true"> X </mover> ) 2 = 1 n 1 [ <munderover> i = 1 n </munderover> X i 2 n <mover accent="true"> X </mover> 2 ] \frac{1}{n-1}\sum_{i-1}^n(X_i-\overline X)^2=\frac{1}{n-1}[\sum_{i=1}^nX_i^2-n\overline X^2] n11i1n(XiX)2=n11[i=1nXi2nX2]
过程:
S 2 = 1 n 1 <munderover> i 1 n </munderover> ( X i <mover accent="true"> X </mover> ) 2 = 1 n 1 <munderover> i 1 n </munderover> ( X i 2 2 X i <mover accent="true"> X </mover> + <mover accent="true"> X </mover> 2 ) = 1 n 1 [ <munderover> i 1 n </munderover> X i 2 2 <mover accent="true"> X </mover> <munderover> i = 1 n </munderover> X i + <munderover> i = 1 n </munderover> <mover accent="true"> X </mover> 2 ] = 1 n 1 [ <munderover> i 1 n </munderover> X i 2 2 <mover accent="true"> X </mover> n <mover accent="true"> X </mover> + <munderover> i = 1 n </munderover> <mover accent="true"> X </mover> 2 ] = 1 n 1 ( <munderover> i 1 n </munderover> X i 2 n <mover accent="true"> X </mover> 2 ) S^2=\frac{1}{n-1}\sum_{i-1}^n(X_i-\overline X)^2=\frac{1}{n-1}\sum_{i-1}^n(X_i^2-2X_i\overline X+\overline X^2)=\frac{1}{n-1}[\sum_{i-1}^nX_i^2-2\overline X\sum_{i=1}^nX_i+\sum_{i=1}^n\overline X^2]=\frac{1}{n-1}[\sum_{i-1}^nX_i^2-2\overline X\cdot n\overline{X}+\sum_{i=1}^n\overline X^2]=\frac{1}{n-1}(\sum_{i-1}^nX_i^2-n\overline X^2) S2=n11i1n(XiX)2=n11i1n(Xi22XiX+X2)=n11[i1nXi22Xi=1nXi+i=1nX2]=n11[i1nXi22XnX+i=1nX2]=n11(i1nXi2nX2)
②:
1 n 1 <munderover> i 1 n </munderover> ( X i <mover accent="true"> X </mover> ) 2 = <munderover> i = 1 n </munderover> ( X i μ ) 2 n ( <mover accent="true"> X </mover> μ ) 2 \frac{1}{n-1}\sum_{i-1}^n(X_i-\overline X)^2=\sum_{i=1}^n(X_i-\mu)^2-n(\overline X-\mu)^2 n11i1n(XiX)2=i=1n(Xiμ)2n(Xμ)2
过程:
1 n 1 <munderover> i 1 n </munderover> ( X i <mover accent="true"> X </mover> ) 2 = 1 n 1 <munderover> i 1 n </munderover> [ ( X i μ ) ( <mover accent="true"> X </mover> μ ) ] 2 = 1 n 1 <munderover> i 1 n </munderover> [ ( X i μ ) 2 2 ( X i μ ) ( <mover accent="true"> X </mover> μ ) + ( <mover accent="true"> X </mover> μ ) 2 ] = <munderover> i = 1 n </munderover> [ ( X i μ ) 2 ( <mover accent="true"> X </mover> μ ) 2 ] = <munderover> i = 1 n </munderover> ( X i μ ) 2 n ( <mover accent="true"> X </mover> μ ) 2 \frac{1}{n-1}\sum_{i-1}^n(X_i-\overline X)^2=\frac{1}{n-1}\sum_{i-1}^n[(X_i-\mu)-(\overline X-\mu)]^2=\frac{1}{n-1}\sum_{i-1}^n[(X_i-\mu)^2-2(X_i-\mu)(\overline X-\mu)+(\overline X-\mu)^2]=\sum_{i=1}^n[(X_i-\mu)^2-(\overline X-\mu)^2]=\sum_{i=1}^n(X_i-\mu)^2-n(\overline X-\mu)^2 n11i1n(XiX)2=n11i1n[(Xiμ)(Xμ)]2=n11i1n[(Xiμ)22(Xiμ)(Xμ)+(Xμ)2]=i=1n[(Xiμ)2(Xμ)2]=i=1n(Xiμ)2n(Xμ)2

这儿有篇苏剑林写的关于无偏估计的:为啥是n-1

E ( <mover accent="true"> X </mover> ) = E ( 1 n <munderover> i = 1 n </munderover> X i ) = <munderover> i = 1 n </munderover> E ( 1 n X i ) = <munderover> i = 1 n </munderover> 1 n μ = μ E(\overline X)=E(\frac{1}{n}\sum_{i=1}^nX_i)=\sum_{i=1}^nE(\frac{1}{n}X_i)=\sum_{i=1}^n\frac{1}{n}\mu=\mu E(X)=E(n1i=1nXi)=i=1nE(n1Xi)=i=1nn1μ=μ
D ( <mover accent="true"> X </mover> ) = D ( 1 n <munderover> i = 1 n </munderover> X i ) = <munderover> i = 1 n </munderover> D ( 1 n X i ) = <munderover> i = 1 n </munderover> 1 n 2 D ( X i ) = <munderover> i = 1 n </munderover> 1 n 2 σ 2 = σ 2 n D(\overline X)=D(\frac{1}{n}\sum_{i=1}^nX_i)=\sum_{i=1}^nD(\frac{1}{n}X_i)=\sum_{i=1}^n\frac{1}{n^2}D(X_i)=\sum_{i=1}^n\frac{1}{n^2}\sigma^2=\frac{\sigma^2}{n} D(X)=D(n1i=1nXi)=i=1nD(n1Xi)=i=1nn21D(Xi)=i=1nn21σ2=nσ2
E ( S 2 ) = 1 n 1 [ <munderover> i = 1 n </munderover> E ( X i 2 ) n E ( <mover accent="true"> X </mover> 2 ) ] = 1 n 1 [ <munderover> i = 1 n </munderover> ( D ( X i ) + E 2 ( X i ) ) n ( D ( <mover accent="true"> X </mover> ) + E 2 ( <mover accent="true"> X </mover> ) ) ] = 1 n 1 [ <munderover> i = 1 n </munderover> ( σ 2 + μ 2 ) n ( σ 2 n + μ 2 ) ] = 1 n 1 [ n σ 2 + n μ 2 σ 2 n μ 2 ] = 1 n 1 [ ( n 1 ) σ 2 ] = σ 2 E(S^2)=\frac{1}{n-1}[\sum_{i=1}^nE(X_i^2)-nE(\overline X^2)]=\frac{1}{n-1}[\sum_{i=1}^n(D(X_i)+E^2(X_i))-n(D(\overline X)+E^2(\overline X))]=\frac{1}{n-1}[\sum_{i=1}^n(\sigma^2+\mu^2)-n(\frac{\sigma^2}{n}+\mu^2)]=\frac{1}{n-1}[n\sigma^2+n\mu^2-\sigma^2-n\mu^2]=\frac{1}{n-1}[(n-1)\sigma^2]=\sigma^2 E(S2)=n11[i=1nE(Xi2)nE(X2)]=n11[i=1n(D(Xi)+E2(Xi))n(D(X)+E2(X))]=n11[i=1n(σ2+μ2)n(nσ2+μ2)]=n11[nσ2+nμ2σ2nμ2]=n11[(n1)σ2]=σ2

开方分布

X χ 2 ( n ) X\sim \chi^2(n) Xχ2(n)
E ( X ) = n E(X)=n E(X)=n
D ( X ) = 2 n D(X)=2n D(X)=2n
还有一个关于 开方分布 与 样本方差 的一个定理,感觉经常用,但是证明很麻烦,是书上P143,证明在章末附录
( n 1 ) S 2 σ 2 χ 2 ( n 1 ) \frac{(n-1)S^2}{\sigma^2}\sim\chi^2(n-1) σ2(n1)S2χ2(n1)

t分布

X N ( 0 , 1 ) Y χ 2 ( n ) X\sim N(0,1)\\Y\sim \chi^2(n) XN(0,1)Yχ2(n)
T = X Y / n T=\frac{X}{\sqrt{Y/n}} T=Y/n X
t 1 α ( n ) = t α ( n ) t_{1-\alpha}(n)=-t_{\alpha}(n) t1α(n)=tα(n)
关于 t分布 与 样本方差 的一个定理
T = <mover accent="true"> X </mover> μ S 2 / n t ( n 1 ) T=\frac{\overline X-\mu}{\sqrt{S^2/n}}\sim t(n-1) T=S2/n Xμt(n1)

正态总体的样本均值与样本方差的分布

①:
<mover accent="true"> X </mover> N ( μ , σ 2 n ) , <mover accent="true"> X </mover> μ σ 2 / n N ( 0 , 1 ) \overline X\sim N(\mu,\frac{\sigma^2}{n}),\frac{\overline X-\mu}{\sqrt{\sigma^2/n}}\sim N(0,1) XN(μ,nσ2),σ2/n XμN(0,1)
②:
<mover accent="true"> X </mover> S 2 ( n 1 ) S 2 σ 2 χ 2 ( n 1 ) \overline X与S^2相互独立,且\frac{(n-1)S^2}{\sigma^2} \sim \chi^2(n-1) XS2σ2(n1)S2χ2(n1)
③:
T = <mover accent="true"> X </mover> μ S 2 / n t ( n 1 ) T=\frac{\overline X-\mu}{\sqrt{S^2/n}}\sim t(n-1) T=S2/n Xμt(n1)
背一个积分
0 x n e x d x = n ! \int_0^\infty x^ne^{-x}dx=n! 0xnexdx=n!