一维随机变量的常见分布、期望、方差及其性质与推导过程
文章目录
- 必须知道的概率论知识
- 一维变量
- 离散随机变量
- def
- 常见分布
- 几何分布
- 期望
- 方差
- 二项分布——b(n,p)
- 期望
- 方差
- 泊松分布—— P ( λ ) P(\lambda) P(λ)
- 期望
- 方差
- 超几何分布——h(n,N,M)
- 期望
- 方差
- 连续型随机变量
- def
- 常见分布
- 均匀分布——U(a,b)
- 密度函数
- 分布函数
- 期望
- 方差
- 指数分布—— E x p ( λ ) Exp(\lambda) Exp(λ)
- 密度函数
- 分布函数
- 期望
- 方差
- 柯西分布
- 密度函数
- 正态分布—— N ( μ , σ 2 ) N(\mu, \sigma ^ 2) N(μ,σ2)
- 密度函数
- 分布函数
- 期望
- 方差
- 标准正态分布——N(0,1）
- 密度函数
- 分布函数
- 伽玛分布—— G a ( α , λ ) Ga(\alpha, \lambda) Ga(α,λ)
- 伽玛函数
- 密度函数
- 期望
- 方差
- 自由度为n的 χ 2 \chi ^2 χ2分布—— χ 2 ( n ) \chi ^2 (n) χ2(n)
- 密度函数
- 期望
- 方差
- 贝塔分布——Be(a,b)
- 贝塔函数
- 密度函数
- 期望
- 与均匀分布的关系
- 期望的性质
- 方差与方差的性质
必须知道的概率论知识
- 常见分布
- 期望
- 方差
- 性质
一维变量
离散随机变量
def
离散随机变量X的分布列
P = ( X = x i ) = p ( x i ) i = 1 , 2 , . . . , n P = (X = x_i) = p(x_i) \qquad i = 1,2,...,n P=(X=xi)=p(xi)i=1,2,...,n
则X的数学期望为： ∑ i = 1 n x i p ( x i ) \sum_{i=1}^{n} x_i p(x_i) ∑i=1nxip(xi)，记为;
E ( X ) = ∑ i = 1 n x i p ( x i ) E(X) = \sum_{i=1}^{n} x_i p(x_i) E(X)=i=1∑nxip(xi)
若X的取值可列，且无穷级数 ∑ i = 1 ∞ x i p ( x i ) \sum_{i=1}^{\infty} x_i p(x_i) ∑i=1∞xip(xi)收敛，则 E ( X ) = ∑ i = 1 ∞ x i p ( x i ) E(X) = \sum_{i=1}^{\infty} x_i p(x_i) E(X)=∑i=1∞xip(xi).
常见分布
几何分布
p ( x ) = P ( X = x ) = p ( 1 − p ) x − 1 , x = 1 , 2 , . . . . p(x) = P(X = x) = p(1 - p)^{x - 1},x = 1,2,.... p(x)=P(X=x)=p(1−p)x−1,x=1,2,.....
期望
E ( X ) = 1 p E(X) = \frac{1}{p} E(X)=p1.
- 推导
令 q = 1 − p q = 1 -p q=1−p，则有;
E ( X ) = ∑ x = 1 ∞ x p q x − 1 = p ∑ x = 1 ∞ x q x − 1 = p ∑ x = 1 ∞ d ( q x ) d q = p d d q ∑ x = 0 ∞ q x = p d d q ( 1 1 − q ) = 1 p E(X) = \sum_{x=1}^{\infty} x p q^{x -1} \\ = p \sum_{x=1}^{\infty} x q^{x -1} \\ = p \sum_{x=1}^{\infty} \frac{d(q^x)}{dq} \\ = p \frac{d}{dq} \sum_{x=0}^{\infty} q^x \\ = p \frac{d}{dq}(\frac{1}{1-q}) = \frac{1}{p} E(X)=x=1∑∞xpqx−1=px=1∑∞xqx−1=px=1∑∞dqd(qx)=pdqdx=0∑∞qx=pdqd(1−q1)=p1
方差
V a r ( X ) = 1 − p p 2 Var(X) = \frac{1 - p}{p^2} Var(X)=p21−p
- 推导
E ( X 2 ) = ∑ x = 1 ∞ x 2 p q x − 1 = p ∑ x = 1 ∞ x 2 q x − 1 = p ∑ x = 1 ∞ x d ( q x ) d q = p d d q ∑ x = 0 ∞ x q x E(X^2) =\sum_{x=1}^{\infty} x^2 p q^{x -1} \\ = p \sum_{x=1}^{\infty} x^2 q^{x -1} \\ = p \sum_{x=1}^{\infty} x \frac{d(q^x)}{dq} \\ = p \frac{d}{dq} \sum_{x=0}^{\infty} x q^x E(X2)=x=1∑∞x2pqx−1=px=1∑∞x2qx−1=px=1∑∞xdqd(qx)=pdqdx=0∑∞xqx
由无穷级数的理论可知 ∑ x = 0 ∞ x q x = q ( 1 − q ) 2 \sum_{x=0}^{\infty} x q^x = \frac{q}{(1-q)^2} ∑x=0∞xqx=(1−q)2q
从而：
接上式: E ( X 2 ) = p d d q ( q ( 1 − q ) 2 ) = 2 p − p 2 p 3 = 2 − p p 2 E(X^2) = p \frac{d}{dq} (\frac{q}{(1-q)^2)} = \frac{2p-p^2}{p^3} = \frac{2-p}{p^2} E(X2)=pdqd((1−q)2)q=p32p−p2=p22−p.从而得到方差。
二项分布——b(n,p)
p ( x ) = P ( X = x ) = ( n x ) p x ( 1 − p ) n − x x = 0 , 1 , . . . . , n p(x) = P(X = x) = \begin{pmatrix} n \\ x \end{pmatrix} p^x (1 - p)^{n -x} \qquad x = 0,1,....,n p(x)=P(X=x)=(nx)px(1−p)n−xx=0,1,....,n
n=1,时的二项分布b(1,p),又称为两点分布或0-1分布。
期望
E ( X ) = n p E(X) = np E(X)=np
- 推导
E ( X ) = ∑ x = 0 n x ( n x ) p x ( 1 − p ) n − x = n p ∑ x = 0 n ( n − 1 x − 1 ) p x − 1 ( 1 − p ) n − x = n p ∑ x = 0 n ( n − 1 x ) p x ( 1 − p ) n − 1 − x = n p [ p + ( 1 − p ) ] n − 1 = n p E(X) = \sum_{x = 0}^{n} x \begin{pmatrix} n \\ x \end{pmatrix} p^x (1 - p)^{n -x} \\ =np \sum_{x = 0}^{n} \begin{pmatrix} n - 1 \\ x - 1 \end{pmatrix} p^{x -1} (1 - p)^{n -x} \\ = np \sum_{x = 0}^{n} \begin{pmatrix} n - 1 \\ x \end{pmatrix} p^{x} (1 - p)^{n - 1 - x} \\ =np[p + (1 - p)]^{n -1} = np E(X)=x=0∑nx(nx)px(1−p)n−x=npx=0∑n(n−1x−1)px−1(1−p)n−x=npx=0∑n(n−1x)px(1−p)n−1−x=np[p+(1−p)]n−1=np
方差
V a r ( X ) = n p ( 1 − p ) Var(X) = np(1-p) Var(X)=np(1−p)
- 推导
E ( X 2 ) = ∑ x = 0 n x 2 ( n x ) p x ( 1 − p ) n − x = ∑ x = 2 n x ( x − 1 ) ( n x ) p x ( 1 − p ) n − x + ∑ x = 1 n x ( n x ) p x ( 1 − p ) n − x = n ( n − 1 ) p 2 ∑ x = 2 n ( n − 2 x − 2 ) p x − 2 ( 1 − p ) n − x + n p = n ( n − 1 ) p 2 + n p = n 2 p 2 + n p ( 1 − p ) E(X^2) = \sum_{x = 0} ^ n x ^2 \begin{pmatrix} n \\ x \end{pmatrix} p ^ x (1-p)^{n-x} \\ = \sum_{x = 2} ^ n x (x-1) \begin{pmatrix} n \\ x \end{pmatrix} p ^ x (1-p)^{n-x} + \sum_{x = 1} ^ n x \begin{pmatrix} n \\ x \end{pmatrix} p ^ x (1-p)^{n-x} \\ = n (n-1) p^2 \sum_{x = 2} ^ n \begin{pmatrix} n-2 \\ x-2 \end{pmatrix} p^{x-2} (1-p)^{n-x} + np \\ = n(n-1) p^2 + np = n^2p^2 + np(1-p) E(X2)=x=0∑nx2(nx)px(1−p)n−x=x=2∑nx(x−1)(nx)px(1−p)n−x+x=1∑nx(nx)px(1−p)n−x=n(n−1)p2x=2∑n(n−2x−2)px−2(1−p)n−x+np=n(n−1)p2+np=n2p2+np(1−p)
故:
V a r ( X ) = E ( X 2 ) − E ( X ) 2 = n p ( 1 − p ) Var(X) = E(X^2) - E(X)^2 = np (1-p) Var(X)=E(X2)−E(X)2=np(1−p)
泊松分布—— P ( λ ) P(\lambda) P(λ)
P ( X = x ) = λ x x ! e − λ x = 0 , 1 , . . . . λ > 0 P(X = x) = \frac{\lambda^x}{x!} e ^ {- \lambda} \qquad x = 0,1,.... \qquad \lambda > 0 P(X=x)=x!λxe−λx=0,1,....λ>0
期望
E ( X ) = λ E(X) = \lambda E(X)=λ
- 推导
E ( X ) = ∑ x = 0 ∞ x ⋅ λ x x ! e − λ = λ e − λ ∑ x = 1 ∞ λ x − 1 ( x − 1 ) ! = λ E(X) = \sum_{x = 0}^{\infty} x \cdot \frac{\lambda^x}{x!} e ^ {- \lambda} = \lambda e ^{- \lambda} \sum_{x = 1}^{\infty} \frac{\lambda ^{x - 1}}{(x - 1)!} = \lambda E(X)=x=0∑∞x⋅x!λxe−λ=λe−λx=1∑∞(x−1)!λx−1=λ
方差
V a r ( X ) = λ Var(X) = \lambda Var(X)=λ
- 推导
E ( X − λ ) 2 = E [ X 2 − 2 λ X + λ 2 ] = E ( X 2 ) − 2 λ E ( X ) + λ 2 E(X - \lambda)^2 = E[X^2 - 2\lambda X + \lambda ^2]= E(X^2) - 2 \lambda E(X) + \lambda ^2 E(X−λ)2=E[X2−2λX+λ2]=E(X2)−2λE(X)+λ2
又
E ( X 2 ) = ∑ x = 0 ∞ x 2 ⋅ λ x x ! e − λ = ∑ x = 1 ∞ x ⋅ λ x ( x − 1 ) ! e − λ = ∑ x = 1 ∞ [ ( x − 1 ) + 1 ] ⋅ λ x ( x − 1 ) ! e − λ = λ 2 e − λ ∑ x = 2 ∞ λ x − 2 ( x − 2 ) ! + λ e − λ ∑ x = 1 ∞ λ x − 1 ( x − 1 ) ! = λ 2 + λ E(X^2) = \sum_{x =0} ^ {\infty} x ^ 2 \cdot \frac{\lambda ^ x}{x!} e ^ {- \lambda} \\ = \sum_{x = 1} ^ {\infty} x \cdot \frac{\lambda ^ x}{(x-1)!} e ^ {- \lambda} \\ = \sum_{x = 1} ^ {\infty} [(x -1) + 1] \cdot \frac{\lambda ^ x}{(x-1)!} e ^ {- \lambda} \\ = \lambda ^2 e ^ {- \lambda} \sum_{x = 2} ^ {\infty} \frac{\lambda ^ {x - 2}}{(x-2)!} + \lambda e ^ {- \lambda} \sum_{x = 1} ^ {\infty} \frac{\lambda ^ {x -1}}{(x-1)!} \\ = \lambda ^2 + \lambda E(X2)=x=0∑∞x2⋅x!λxe−λ=x=1∑∞x⋅(x−1)!λxe−λ=x=1∑∞[(x−1)+1]⋅(x−1)!λxe−λ=λ2e−λx=2∑∞(x−2)!λx−2+λe−λx=1∑∞(x−1)!λx−1=λ2+λ
故： E ( X − λ ) 2 = λ 2 + λ − 2 λ 2 + λ 2 = λ E(X - \lambda)^2 = \lambda ^2 + \lambda - 2 \lambda ^2 + \lambda ^2 = \lambda E(X−λ)2=λ2+λ−2λ2+λ2=λ
超几何分布——h(n,N,M)
P ( X = x ) = ( M x ) ( N − M n − x ) ( N n ) x = 0 , 1 , … , r , r = m i n ( n , M ) P(X = x) = \frac{\begin{pmatrix} M \\ x \end{pmatrix} \begin{pmatrix} N-M \\ n-x \end{pmatrix}}{\begin{pmatrix} N \\ n \end{pmatrix}} \qquad x=0,1,\dots,r,\ r=min(n,M) P(X=x)=(Nn)(Mx)(N−Mn−x)x=0,1,…,r, r=min(n,M)
期望
E ( X ) = n M N E(X) = \frac{nM}{N} E(X)=NnM
- 推导
E ( X ) = ∑ x = 0 r x ( M x ) ( N − M n − x ) ( N n ) = n M N ∑ x = 1 r ( M − 1 x − 1 ) ( N − M n − x ) ( N − 1 n − 1 ) = n M N \begin{aligned} E(X) & = \sum_{x = 0}^{r} x \frac{\begin{pmatrix} M \\ x \end{pmatrix} \begin{pmatrix} N-M \\ n-x \end{pmatrix}}{\begin{pmatrix} N \\ n \end{pmatrix}} \\ & = \frac{nM}{N} \sum_{x = 1}^{r} \frac{\begin{pmatrix} M - 1 \\ x - 1 \end{pmatrix} \begin{pmatrix} N-M \\ n-x \end{pmatrix}}{\begin{pmatrix} N - 1 \\ n - 1 \end{pmatrix}}\\ & = \frac{nM}{N} \end{aligned} E(X)=x=0∑rx(Nn)(Mx)(N−Mn−x)=NnMx=1∑r(N−1n−1)(M−1x−1)(N−Mn−x)=NnM
方差
V a r ( X ) = n M N ( 1 − M N ) ( N − n N − 1 ) Var(X) = \frac{nM}{N}(1-\frac{M}{N})(\frac{N-n}{N-1}) Var(X)=NnM(1−NM)(N−1N−n)
- 推导
E ( X 2 ) = ∑ x = 0 r x 2 ( M x ) ( N − M n − x ) ( N n ) = n M N ∑ x = 1 r x ( M − 1 x − 1 ) ( N − M n − x ) ( N − 1 n − 1 ) = n M N ∑ x = 1 r ( x − 1 ) ( M − 1 x − 1 ) ( N − M n − x ) ( N − 1 n − 1 ) + n M N ∑ x = 1 r ( M − 1 x − 1 ) ( N − M n − x ) ( N − 1 n − 1 ) = n M N ( M − 1 ) ∑ x = 1 r ( M − 2 x − 2 ) ( N − M n − x ) ( N − 1 n − 1 ) + n M N = n M N [ ( M − 1 ) ( n − 1 ) N − 1 + 1 ] \begin{aligned} E(X^2) & = \sum_{x = 0}^{r} x^2 \frac{\begin{pmatrix} M \\ x \end{pmatrix} \begin{pmatrix} N-M \\ n-x \end{pmatrix}}{\begin{pmatrix} N \\ n \end{pmatrix}} \\ & = \frac{nM}{N} \sum_{x = 1}^{r} x \frac{\begin{pmatrix} M - 1 \\ x - 1 \end{pmatrix} \begin{pmatrix} N-M \\ n-x \end{pmatrix}}{\begin{pmatrix} N - 1 \\ n - 1 \end{pmatrix}}\\ & = \frac{nM}{N} \sum_{x = 1}^{r} (x -1) \frac{\begin{pmatrix} M - 1 \\ x - 1 \end{pmatrix} \begin{pmatrix} N-M \\ n-x \end{pmatrix}}{\begin{pmatrix} N - 1 \\ n - 1 \end{pmatrix}} + \frac{nM}{N} \sum_{x = 1}^{r} \frac{\begin{pmatrix} M - 1 \\ x - 1 \end{pmatrix} \begin{pmatrix} N-M \\ n-x \end{pmatrix}}{\begin{pmatrix} N - 1 \\ n - 1 \end{pmatrix}} \\ & = \frac{nM}{N} (M-1) \sum_{x = 1}^{r} \frac{\begin{pmatrix} M - 2 \\ x - 2 \end{pmatrix} \begin{pmatrix} N-M \\ n-x \end{pmatrix}}{\begin{pmatrix} N - 1 \\ n - 1 \end{pmatrix}} + \frac{nM}{N} \\ & = \frac{nM}{N}[\frac{(M-1)(n-1)}{N-1} + 1] \end{aligned} E(X2)=x=0∑rx2(Nn)(Mx)(N−Mn−x)=NnMx=1∑rx(N−1n−1)(M−1x−1)(N−Mn−x)=NnMx=1∑r(x−1)(N−1n−1)(M−1x−1)(N−Mn−x)+NnMx=1∑r(N−1n−1)(M−1x−1)(N−Mn−x)=NnM(M−1)x=1∑r(N−1n−1)(M−2x−2)(N−Mn−x)+NnM=NnM[N−1(M−1)(n−1)+1]
再由方差公式即可求得。
连续型随机变量
def
p ( x ) p(x) p(x)是实数轴上的一个函数，满足：
（1） p ( x ) ≥ 0 p(x) \geq 0 p(x)≥0.（非负）
（2） ∫ − ∞ ∞ = 1 \int_{- \infty}^{\infty} = 1 ∫−∞∞=1
则称 p ( x ) p(x) p(x)为概率密度函数。
期望： E ( X ) = ∫ − ∞ ∞ x p ( x ) d x E(X) = \int_{- \infty } ^ {\infty} x p(x) d x E(X)=∫−∞∞xp(x)dx.
分布函数：
F ( x ) = P ( X ≤ x ) = ∫ − ∞ x p ( x ) d x F(x) = P(X \leq x) = \int_{- \infty}^{x} p(x)dx F(x)=P(X≤x)=∫−∞xp(x)dx.
分布函数的性质
-
F ( x ) F(x) F(x)是直线上的连续函数
-
P ( X = x ) = 0 P(X = x) = 0 P(X=x)=0
-
P ( a ≤ X ≤ b ) = P ( a ≤ X < b ) = P ( a < X ≤ b ) = P ( a < X < b ) P(a \leq X \leq b) = P(a \leq X < b) = P(a < X \leq b) = P(a < X < b) P(a≤X≤b)=P(a≤X<b)=P(a<X≤b)=P(a<X<b)
-
在 F ( x ) F(x) F(x)导数存在的点x 上有：
F ′ ( x ) = p ( x ) F'(x) = p(x) F′(x)=p(x)
常见分布
均匀分布——U(a,b)
密度函数
p ( x ) = { 1 b − a a ≤ x ≤ b 0 其 他 p(x)= \begin{cases} \frac{1}{b - a} & a \leq x \leq b \\ 0 & 其他 \end{cases} p(x)={b−a10a≤x≤b其他
分布函数
F ( x ) = { 0 x < a x − a b − a a ≤ x ≤ b 1 x > b F(x)= \begin{cases} 0 & x < a \\ \frac{x - a}{b - a} & a \leq x \leq b \\ 1 & x > b \end{cases} F(x)=⎩⎪⎨⎪⎧0b−ax−a1x<aa≤x≤bx>b
期望
E ( X ) = ∫ − ∞ ∞ x p ( x ) d x = ∫ a b x ⋅ 1 b − a d x = 1 b − a x 2 2 ∣ a b = b 2 − a 2 2 ( b − a ) = a + b 2 E(X) = \int_{- \infty } ^ {\infty} x p(x) d x = \int_a^b x \cdot \frac{1}{b -a}dx = \frac{1}{b -a} \frac{x^2}{2}|_a^b = \frac{b^2 - a^2}{2 (b -a)} = \frac{a + b}{2} E(X)=∫−∞∞xp(x)dx=∫abx⋅b−a1dx=b−a12x2∣ab=2(b−a)b2−a2=2a+b
方差
E ( X 2 ) = ∫ a b x 2 b − a d x = 1 b − a x 3 3 ∣ a b = 1 3 ( b 2 + a b + a 2 ) E(X^2) = \int_a^b \frac{x^2}{b -a}dx \\ = \frac{1}{b -a} \frac{x^3}{3}|_a^b \\ = \frac{1}{3}(b^2 + ab + a^2) E(X2)=∫abb−ax2dx=b−a13x3∣ab=31(b2+ab+a2)
故:
V a r ( X ) = ( b − a ) 2 12 Var(X) = \frac{(b-a)^2}{12} Var(X)=12(b−a)2
指数分布—— E x p ( λ ) Exp(\lambda) Exp(λ)
密度函数
p ( x ) = { λ e − λ x x ≥ 0 0 x < 0 p(x)= \begin{cases} \lambda e ^{- \lambda x} & x \geq 0 \\ 0 & x < 0 \end{cases} p(x)={λe−λx0x≥0x<0
分布函数
F ( x ) = { 1 − e − λ x x ≥ 0 0 x < 0 F(x)= \begin{cases} 1 - e ^{- \lambda x} & x \geq 0 \\ 0 & x < 0 \end{cases} F(x)={1−e−λx0x≥0x<0
期望
E ( X ) = ∫ − ∞ ∞ x p ( x ) d x = ∫ 0 ∞ λ x e − λ x d x = 1 λ E(X) = \int_{- \infty } ^ {\infty} x p(x) d x = \int_0^{\infty} \lambda x e ^ {- \lambda x}dx = \frac{1}{\lambda} E(X)=∫−∞∞xp(x)dx=∫0∞λxe−λxdx=λ1
方差
V a r ( X ) = 1 λ 2 Var(X) = \frac{1}{\lambda ^ 2} Var(X)=λ21
推导 见伽马分布。
柯西分布
密度函数
p ( x ) = 1 π ( 1 + x 2 ) p(x) = \frac{1}{\pi (1 + x^2)} p(x)=π(1+x2)1
数学期望不存在
正态分布—— N ( μ , σ 2 ) N(\mu, \sigma ^ 2) N(μ,σ2)
密度函数
p ( x ) = 1 2 π σ e − ( x − μ ) 2 2 σ 2 − ∞ < x < ∞ p(x) = \frac{1}{\sqrt{2 \pi} \sigma} e ^ {- \frac{(x - \mu) ^ 2}{2 \sigma ^ 2}} \qquad - \infty < x < \infty p(x)=2πσ1e−2σ2(x−μ)2−∞<x<∞
分布函数
F ( x ) = 1 2 π σ ∫ − ∞ x e − ( x − μ ) 2 2 σ 2 d x − ∞ < x < ∞ F(x) = \frac{1}{\sqrt{2 \pi} \sigma} \int _{- \infty}^{x} e ^ {- \frac{(x - \mu) ^ 2}{2 \sigma ^ 2}} dx \qquad - \infty < x < \infty F(x)=2πσ1∫−∞xe−2σ2(x−μ)2dx−∞<x<∞
期望
E ( X ) = μ E(X) = \mu E(X)=μ.
- 推导
令 z = x − μ σ , E ( X ) = 1 2 π σ ∫ − ∞ ∞ x e − ( x − μ ) 2 2 σ 2 d x = 1 2 π σ ∫ − ∞ ∞ ( σ z + μ ) e − z 2 2 d z = 1 2 π [ σ ∫ − ∞ ∞ z e − x 2 2 d z + μ ∫ − ∞ ∞ e − x 2 2 d z ] = 0 + μ = μ 令 z = \frac{x - \mu}{\sigma},\\ E(X) = \frac{1}{\sqrt{2 \pi} \sigma} \int _{- \infty}^{\infty} x e ^ {- \frac{(x - \mu) ^ 2}{2 \sigma ^ 2}} dx \\ =\frac{1}{\sqrt{2 \pi} \sigma} \int _{- \infty}^{\infty} (\sigma z + \mu) e ^ {- \frac{z ^ 2}{2}} dz \\ = \frac{1}{\sqrt{2 \pi}} [\sigma \int _{- \infty}^{\infty} z e ^ {- \frac{x ^ 2}{2}}dz + \mu \int _{- \infty}^{\infty} e ^ {- \frac{x ^ 2}{2}}dz] \\ = 0 + \mu = \mu 令z=σx−μ,E(X)=2πσ1∫−∞∞xe−2σ2(x−μ)2dx=2πσ1∫−∞∞(σz+μ)e−2z2dz=2π1[σ∫−∞∞ze−2x2dz+μ∫−∞∞e−2x2dz]=0+μ=μ
方差
V a r ( X ) = E ( X − E ( X ) ) 2 = E ( X − μ ) 2 = 1 2 π σ ∫ − ∞ ∞ ( x − μ ) 2 e − ( x − μ ) 2 2 σ 2 d x = u = x − μ σ σ 2 2 π ∫ − ∞ ∞ u 2 e − u 2 2 d u = 2 σ 2 2 π ∫ 0 ∞ u 2 e − u 2 2 d u = y = u 2 2 2 σ 2 2 π 2 ∫ 0 ∞ y 1 2 e − y d y = 2 σ 2 2 π 2 Γ ( 3 2 ) = 2 σ 2 2 π 2 π 2 = σ 2 Var(X) = E(X - E(X))^2 \\ = E(X - \mu) ^ 2 \\ = \frac{1}{\sqrt{2 \pi}} \sigma \int _{- \infty}^{\infty} (x - \mu) ^ 2 e ^ {- \frac{(x - \mu) ^ 2}{2 \sigma ^ 2}} dx \\ \overset{u = \frac{x - \mu}{\sigma}}{=} \frac{\sigma ^2}{\sqrt{2 \pi}}\int _{- \infty}^{\infty} u ^2 e ^ {- \frac{u ^ 2}{2}} d u \\ = \frac{2 \sigma ^2}{\sqrt{2 \pi}}\int _{0}^{\infty} u ^2 e ^ {- \frac{u ^ 2}{2}} d u \\ \overset{y = \frac{u^2}{2}}{=} \frac{2 \sigma ^2}{\sqrt{2 \pi}} \sqrt{2} \int _{0}^{\infty} y ^{\frac{1}{2}} e ^ {- y} d y \\ = \frac{2 \sigma ^2}{\sqrt{2 \pi}} \sqrt{2} \Gamma (\frac{3}{2}) = \frac{2 \sigma ^2}{\sqrt{2 \pi}} \frac{\sqrt{2 \pi}}{2} = \sigma ^2 Var(X)=E(X−E(X))2=E(X−μ)2=2π1σ∫−∞∞(x−μ)2e−2σ2(x−μ)2dx=u=σx−μ2πσ2∫−∞∞u2e−2u2du=2π2σ2∫0∞u2e−2u2du=y=2u22π2σ22∫0∞y21e−ydy=2π2σ22Γ(23)=2π2σ222π=σ2
标准正态分布——N(0,1）
密度函数
ϕ ( μ ) = 1 2 π e − u 2 2 − ∞ < u < ∞ \phi(\mu) = \frac{1}{\sqrt{2 \pi}} e ^ {- \frac{ u ^ 2}{2}} \qquad - \infty < u < \infty ϕ(μ)=2π1e−2u2−∞<u<∞
分布函数
Φ ( μ ) = 1 2 π ∫ − ∞ u e − x 2 2 d x − ∞ < u < ∞ \Phi(\mu) = \frac{1}{\sqrt{2 \pi}} \int _{- \infty}^{u} e ^ {- \frac{ x ^ 2}{2}}dx \qquad - \infty < u < \infty Φ(μ)=2π1∫−∞ue−2x2dx−∞<u<∞
- 对任意实数u，有：
Φ ( − u ) = 1 − Φ ( u ) \Phi (- u) = 1 - \Phi(u) Φ(−u)=1−Φ(u)
- 当 X ∼ N ( μ , σ 2 ) X \sim N(\mu, \sigma^2) X∼N(μ,σ2)时， U = X − μ σ U = \frac{X - \mu}{\sigma} U=σX−μ是标准正态变量，即 U ∼ N ( 0 , 1 ) U \sim N(0,1) U∼N(0,1).
伽玛分布—— G a ( α , λ ) Ga(\alpha, \lambda) Ga(α,λ)
伽玛函数
含参数 α \alpha α的积分;
Γ ( α ) = ∫ 0 ∞ x α − 1 e − x d x α > 0 \Gamma (\alpha) = \int_0^{\infty} x^{\alpha - 1}e^{- x}dx \qquad \alpha > 0 Γ(α)=∫0∞xα−1e−xdxα>0
性质：
- Γ ( 1 ) = 1 , Γ ( 1 2 ) = π \Gamma(1) = 1, \Gamma (\frac{1}{2}) = \sqrt{\pi} Γ(1)=1,Γ(21)=π
- 递推公式： Γ ( α + 1 ) = α Γ ( α ) \Gamma (\alpha + 1) = \alpha \Gamma (\alpha) Γ(α+1)=αΓ(α).特别对于n， Γ ( n + 1 ) = n ! \Gamma (n + 1) = n! Γ(n+1)=n!
- ∫ 0 ∞ x α − 1 e − λ x d x = Γ ( α ) λ α \int _0 ^{\infty} x ^ {\alpha - 1} e ^ {-\lambda x} d x = \frac {\Gamma (\alpha)}{\lambda ^ {\alpha}} ∫0∞xα−1e−λxdx=λαΓ(α)
密度函数
p ( x ) = { λ α Γ ( α ) x α − 1 e − λ x x > 0 0 x ≤ 0 p(x) = \begin{cases} \frac{\lambda ^ {\alpha}}{\Gamma (\alpha)} x ^ {\alpha - 1} e ^ {- \lambda x} & x > 0 \\ 0 & x \leq 0 \end{cases} p(x)={Γ(α)λαxα−1e−λx0x>0x≤0
简记为;
p ( x ) = λ α Γ ( α ) x α − 1 e − λ x x > 0 p(x) = \frac{\lambda ^ {\alpha}}{\Gamma (\alpha)} x ^ {\alpha - 1} e ^ {- \lambda x} \qquad x > 0 p(x)=Γ(α)λαxα−1e−λxx>0
α = 1 \alpha = 1 α=1的时候伽马分布就是指数分布，密度函数
p ( x ) = λ e − λ x x > 0 p(x) = \lambda e ^{- \lambda x} \qquad x > 0 p(x)=λe−λxx>0
α = n 2 , λ = 1 2 \alpha = \frac{n}{2}, \lambda = \frac{1}{2} α=2n,λ=21时伽马分布为卡方分布。
期望
E ( X ) = λ α Γ ( α ) ∫ 0 ∞ x α e − λ x d x = Γ ( α + 1 ) Γ ( α ) 1 λ = α λ E(X) = \frac{\lambda ^ {\alpha}}{\Gamma (\alpha)} \int_0 ^{\infty} x ^ {\alpha} e ^ {- \lambda x} d x = \frac{\Gamma (\alpha + 1)}{\Gamma (\alpha)} \frac{1}{\lambda} = \frac{\alpha}{\lambda} E(X)=Γ(α)λα∫0∞xαe−λxdx=Γ(α)Γ(α+1)λ1=λα
方差
E ( X 2 ) = λ α Γ ( α ) ∫ 0 ∞ x 2 ⋅ x α e − λ x d x = λ α Γ ( α ) Γ ( α + 2 ) λ α + 2 = α ( α + 1 ) λ 2 E(X^2) = \frac{\lambda ^ {\alpha}}{\Gamma (\alpha)} \int_0 ^{\infty} x^2 \cdot x ^ {\alpha} e ^ {- \lambda x} d x \\ = \frac{\lambda ^ {\alpha}}{\Gamma (\alpha)} \frac{\Gamma (\alpha + 2)}{\lambda ^ {\alpha + 2}} \\ = \frac{\alpha(\alpha +1)}{\lambda ^2} E(X2)=Γ(α)λα∫0∞x2⋅xαe−λxdx=Γ(α)λαλα+2Γ(α+2)=λ2α(α+1)
自由度为n的 χ 2 \chi ^2 χ2分布—— χ 2 ( n ) \chi ^2 (n) χ2(n)
即伽玛分布 G a ( n 2 , 1 2 ) Ga(\frac{n}{2}, \frac{1}{2}) Ga(2n,21),
密度函数
p ( x ) = 1 Γ ( n 2 ) 2 n 2 x n 2 − 1 e − x 2 x > 0 p(x) = \frac{1}{\Gamma (\frac{n}{2}) 2 ^{\frac{n}{2}}} x ^ {\frac{n}{2} -1} e ^{- \frac{x}{2}} \qquad x > 0 p(x)=Γ(2n)22n1x2n−1e−2xx>0
期望
E ( X ) = n E(X) = n E(X)=n
方差
V a r ( X ) = 2 n Var(X) = 2n Var(X)=2n
推导见伽马分布。
贝塔分布——Be(a,b)
贝塔函数
含参数a，b的积分：
β ( a , b ) = ∫ 0 1 x a − 1 ( 1 − x ) b − 1 d x a > 0 , b > 0 \beta (a,b) = \int _0 ^1 x ^{a - 1}(1 - x)^{b -1} d x \qquad a > 0, b > 0 β(a,b)=∫01xa−1(1−x)b−1dxa>0,b>0
性质：
-
β ( a , b ) = β ( b , a ) \beta(a,b) = \beta(b,a) β(a,b)=β(b,a)
-
贝塔函数与伽玛函数的关系：
β ( a , b ) = Γ ( a ) Γ ( b ) Γ ( a + b ) \beta(a,b) = \frac{\Gamma(a) \Gamma(b)}{\Gamma(a+b)} β(a,b)=Γ(a+b)Γ(a)Γ(b)
密度函数
p ( x ) = Γ ( a + b ) Γ ( a ) Γ ( b ) x a − 1 ( 1 − x ) b − 1 0 ≤ x ≤ 1 p(x) = \frac{\Gamma(a+b)}{\Gamma(a) \Gamma(b)} x ^ {a - 1} (1 -x)^ {b - 1} \qquad 0 \leq x \leq 1 p(x)=Γ(a)Γ(b)Γ(a+b)xa−1(1−x)b−10≤x≤1
期望
E ( X ) = Γ ( a + b ) Γ ( a ) Γ ( b ) ∫ 0 1 x a − 1 ( 1 − x ) b − 1 d x = Γ ( a + b ) Γ ( a ) Γ ( b ) ⋅ Γ ( a + 1 ) Γ ( b ) Γ ( a + b + 1 ) = a a + b E(X) =\frac{\Gamma(a+b)}{\Gamma(a) \Gamma(b)} \int _ 0 ^ 1 x ^ {a - 1} (1 -x)^ {b - 1} d x \\ =\frac{\Gamma(a+b)}{\Gamma(a) \Gamma(b)} \cdot \frac{\Gamma(a + 1) \Gamma(b)}{\Gamma(a+b + 1)} \\ = \frac{a}{a + b} E(X)=Γ(a)Γ(b)Γ(a+b)∫01xa−1(1−x)b−1dx=Γ(a)Γ(b)Γ(a+b)⋅Γ(a+b+1)Γ(a+1)Γ(b)=a+ba
与均匀分布的关系
Be（1,1）就是[0,1] 上的均匀分布。
期望的性质
- E [ c g ( X ) ] = c E [ g ( X ) ] E[cg(X)] = c E[g(X)] E[cg(X)]=cE[g(X)]
- E [ g ( X ) ± h ( X ) ] = E [ g ( X ) ] ± E [ h ( X ) ] E[g(X) \pm h(X)] = E[g(X)] \pm E[h(X)] E[g(X)±h(X)]=E[g(X)]±E[h(X)]
- E ( c ) = c E(c) =c E(c)=c,c为常数
方差与方差的性质
V a r ( X ) = E [ X − E ( X ) ] 2 Var(X) = E[X - E(X)]^2 Var(X)=E[X−E(X)]2
性质：
- V a r ( c ) = 0 Var(c) = 0 Var(c)=0,c为常数
- V a r ( a X + b ) = a 2 V a r ( X ) Var(a X + b) = a ^ 2 Var(X) Var(aX+b)=a2Var(X)
- V a r ( X ) = E ( X 2 ) − [ E ( X ) ] 2 Var(X) = E(X^2) - [E(X)]^ 2 Var(X)=E(X2)−[E(X)]2
多元情况后续更新。