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  • 概率论各种基础分布期望和方差推导过程汇总

    万次阅读 多人点赞 2019-06-24 00:08:16
    0-1分布 XXX~b(1,p)b(1,p)b(1,p) E(X)=p∗1+(1−p)∗0=pE(X)=p*1+(1-p)*0=pE(X)=p∗1+(1−p)∗0=p E(X2)=02∗(1−p)+12∗p=pE(X^2)=0^2*(1-p)+1^2*p=pE(X2)=02∗(1−p)+12∗p=p D(X)=E(X2)−E(X)2=p−p2=p(1−p)D(X)...

    一、0-1分布 X X X~ B ( 1 , p ) B(1,p) B(1,p)

    E ( X ) = p ∗ 1 + ( 1 − p ) ∗ 0 = p E(X)=p*1+(1-p)*0=p E(X)=p1+(1p)0=p

    显然 E ( X 2 ) = 0 2 ∗ ( 1 − p ) + 1 2 ∗ p = p E(X^2)=0^2*(1-p)+1^2*p=p E(X2)=02(1p)+12p=p

    D ( X ) = E ( X 2 ) − E ( X ) 2 = p − p 2 = p ( 1 − p ) D(X)=E(X^2)-E(X)^2=p-p^2=p(1-p) D(X)=E(X2)E(X)2=pp2=p(1p)

    二、二项分布 X X X~ B ( n , p ) B(n,p) B(n,p)

    可认为 X = X 1 + X 2 + ⋯ + X n X=X_1+X_2+\cdots +X_n X=X1+X2++Xn X i X_i Xi~ B ( 1 , p ) B(1,p) B(1,p)
    E ( X ) = E ( X 1 + X 2 + ⋯ + X n ) = ∑ i = 1 n E ( X i ) = n p E(X)=E(X_1+X_2+\cdots +X_n)=\sum\limits^n_{i=1}E(X_i)=np E(X)=E(X1+X2++Xn)=i=1nE(Xi)=np
    D ( X ) = D ( ∑ i = 1 n X i ) = ∑ i = 1 n D ( X i ) = n p ( 1 − p ) D(X)=D(\sum\limits^n_{i=1}X_i)=\sum\limits^n_{i=1}D(X_i)=np(1-p) D(X)=D(i=1nXi)=i=1nD(Xi)=np(1p)

    三、均匀分布 X X X~ U ( a , b ) U(a,b) U(a,b)

    X X X的概率密度为 f ( x ) = { 1 b − a , a < x < b 0 其 他 f(x)=\begin{cases}\dfrac{1}{b-a} ,&a<x<b\\0 &其他\end{cases} f(x)=ba1,0a<x<b
    E ( X ) = ∫ − ∞ ∞ x f ( x ) d x = ∫ a b x b − a d x = a + b 2 E(X)=\int\limits^{\infty}_{-\infty}xf(x)dx=\int\limits^b_a\dfrac{x}{b-a}dx=\dfrac{a+b}{2} E(X)=xf(x)dx=abbaxdx=2a+b
    D ( X ) = E ( X 2 ) − E ( X ) 2 = ∫ − ∞ ∞ x 2 f ( x ) d x − ( a + b 2 ) 2 = ( b − a ) 2 12 D(X)=E(X^2)-E(X)^2=\int\limits^{\infty}_{-\infty}x^2f(x)dx-(\dfrac{a+b}{2})^2=\dfrac{(b-a)^2}{12} D(X)=E(X2)E(X)2=x2f(x)dx(2a+b)2=12(ba)2

    四、泊松分布 X X X~ π ( λ ) , P ( λ ) \pi(\lambda),P(\lambda) π(λ),P(λ)

    P ( X = k ) = λ k k ! e − λ P(X=k)=\dfrac{\lambda ^k}{k!}e^{-\lambda} P(X=k)=k!λkeλ
    E ( X ) = ∑ k = 0 ∞ k λ k k ! e − λ = 0 λ 0 0 ! e − λ + λ e − λ ∑ k − 1 = 0 ∞ λ k − 1 ( k − 1 ) ! = λ e − λ e λ = λ E(X)=\sum\limits^{\infty}_{k=0}k\dfrac{\lambda ^k}{k!}e^{-\lambda}=0\dfrac{\lambda ^0}{0!}e^{-\lambda}+\lambda e^{-\lambda}\sum\limits^{\infty}_{k-1=0}\dfrac{\lambda ^{k-1}}{(k-1)!}=\lambda e^{-\lambda}e^{\lambda}=\lambda E(X)=k=0kk!λkeλ=00!λ0eλ+λeλk1=0(k1)!λk1=λeλeλ=λ

    注释

    e x e^x ex x = 0 x=0 x=0处展开 e x = 1 + x + x 2 / 2 ! + … + x n / n ! + … = ∑ n = 0 ∞ x n n ! e^x=1+x+x^2/2!+…+x^n/n!+…=\sum\limits^\infty_{n=0}\dfrac{x^n}{n!} ex=1+x+x2/2!++xn/n!+=n=0n!xn

    同时 E [ X ( X − 1 ) ] = ∑ k = 0 ∞ k ( k − 1 ) λ k k ! e − λ = λ 2 E[X(X-1)]=\sum\limits^{\infty}_{k=0}k(k-1)\dfrac{\lambda ^k}{k!}e^{-\lambda}=\lambda^2 E[X(X1)]=k=0k(k1)k!λkeλ=λ2
    所以 E ( X 2 ) = E ( X ( X − 1 ) + X ) = E [ X ( X − 1 ) ] + E ( X ) = λ 2 + λ E(X^2)=E(X(X-1)+X)=E[X(X-1)]+E(X)=\lambda^2+\lambda E(X2)=E(X(X1)+X)=E[X(X1)]+E(X)=λ2+λ

    D ( X ) = E ( X 2 ) − E ( X ) 2 = λ 2 + λ − λ 2 = λ D(X)=E(X^2)-E(X)^2=\lambda^2+\lambda-\lambda^2=\lambda D(X)=E(X2)E(X)2=λ2+λλ2=λ

    五、 指数分布 X X X~ E ( θ ) E(\theta) E(θ)

    f ( x ) = { 1 θ e − x / θ , x > 0 0 x ≤ 0 , ( θ > 0 ) f(x)=\begin{cases}\dfrac{1}{\theta } e^{-x/\theta },&x> 0\\0 &x\le0\end{cases},(\theta>0) f(x)=θ1ex/θ,0x>0x0,(θ>0)
    (这是其中一种形式,还有形式有 λ = 1 / θ \lambda=1/\theta λ=1/θ作为参数的等等)
    E ( X ) = ∫ − ∞ ∞ x f ( x ) d x E(X)=\int\limits^{\infty}_{-\infty}xf(x)dx E(X)=xf(x)dx
    = ∫ 0 ∞ x 1 θ e − x / θ d x =\int\limits^{\infty}_{0}x\dfrac{1}{\theta } e^{-x/\theta }dx =0xθ1ex/θdx
    = [ 1 θ x ( − θ ) e − x / θ − ∫ 1 θ ( − θ ) e − x / θ d x ] 0 ∞ =\Big[\dfrac{1}{\theta }x(-\theta)e^{-x/\theta }-\int \dfrac{1}{\theta } (-\theta)e^{-x/\theta }dx\Big]^{\infty}_{0} =[θ1x(θ)ex/θθ1(θ)ex/θdx]0
    = [ − x e − x / θ − θ e − x / θ ] 0 ∞ =\Big[ -xe^{-x/\theta } -\theta e^{-x/\theta } \Big]^{\infty}_{0} =[xex/θθex/θ]0
    = θ =\theta =θ

    同理 E ( X 2 ) = ∫ − ∞ ∞ x 2 f ( x ) d x = 2 θ 2 E(X^2)=\int\limits^{\infty}_{-\infty}x^2f(x)dx=2\theta^2 E(X2)=x2f(x)dx=2θ2

    D ( X ) = E ( X 2 ) − E ( X ) 2 = 2 θ 2 − θ 2 = θ 2 D(X)=E(X^2)-E(X)^2=2\theta^2-\theta^2=\theta^2 D(X)=E(X2)E(X)2=2θ2θ2=θ2

    六、正态/高斯分布 X X X~ N ( μ , σ 2 ) N(\mu,\sigma^2) N(μ,σ2)

    f ( x ) = 1 2 π σ e − ( x − μ ) 2 2 σ 2 f(x)=\dfrac{1}{\sqrt{2\pi}\sigma}e^{-\dfrac{(x-\mu)^2}{2\sigma^2}} f(x)=2π σ1e2σ2(xμ)2
    Z = X − μ σ ∴ Z ∼ N ( 0 , 1 ) Z=\dfrac{X-\mu}{\sigma}\therefore Z\sim N(0,1) Z=σXμZN(0,1)
    容易知道 E ( Z ) = 0 E(Z)=0 E(Z)=0, D ( Z ) = 1 D(Z)=1 D(Z)=1
    X = μ + σ Z X=\mu+\sigma Z X=μ+σZ
    E ( X ) = E ( μ + σ Z ) = μ E(X)=E(\mu+\sigma Z)=\mu E(X)=E(μ+σZ)=μ
    D ( X ) = D ( μ + σ Z ) = σ 2 D(X)=D(\mu+\sigma Z)=\sigma^2 D(X)=D(μ+σZ)=σ2
    但是事实上用定义来做也能得出这个结果

    常用结论

    X ‾ − μ σ / n ∼ N ( 0 , 1 ) \dfrac{\overline X-\mu}{\sigma/\sqrt{n}} \sim N(0,1) σ/n XμN(0,1) 证明在下面样本均值模块

    七、卡方分布 χ 2 ∼ χ 2 ( n ) \chi^2\sim \chi^2(n) χ2χ2(n)

    χ 2 ( n ) = ∑ i = 1 n X i 2 , X i ∼ N ( 0 , 1 ) \chi^2(n)=\sum\limits^n_{i=1}X_i^2,X_i\sim N(0,1) χ2(n)=i=1nXi2,XiN(0,1)

    E ( X i 2 ) = E ( ( X i − 0 ) 2 ) = D ( X i ) = 1 E(X_i^2)=E((X_i-0)^2)=D(X_i)=1 E(Xi2)=E((Xi0)2)=D(Xi)=1

    E ( χ 2 ) = E ( ∑ i = 1 n X i 2 ) = n E(\chi^2)=E(\sum\limits^n_{i=1}X_i^2)=n E(χ2)=E(i=1nXi2)=n

    D ( X i 2 ) = E ( X i 4 ) − E ( X i 2 ) 2 = 3 − 1 = 2 D(X_i^2)=E(X^4_i)-E(X_i^2)^2=3-1=2 D(Xi2)=E(Xi4)E(Xi2)2=31=2

    D ( χ 2 ) = D ( ∑ i = 1 n X i 2 ) = 2 n D(\chi^2)=D(\sum\limits^n_{i=1}X_i^2)=2n D(χ2)=D(i=1nXi2)=2n

    注释

    E ( X 4 ) = 3 , X ∼ N ( 0 , 1 ) E(X^4)=3,X\sim N(0,1) E(X4)=3,XN(0,1) 证明用 E ( g ( X ) ) = ∫ − ∞ ∞ g ( x ) f ( x ) d x E(g(X))=\int\limits^{\infty}_{-\infty}g(x)f(x)dx E(g(X))=g(x)f(x)dx方法,也可以看最下面的解释

    常用结论

    X i ∼ N ( μ , σ 2 ) X_i\sim N(\mu,\sigma^2) XiN(μ,σ2)
    1. ∑ i = 1 n ( X i − μ σ ) 2 ∼ χ 2 ( n ) \sum\limits^n_{i=1}(\dfrac{X_i-\mu}{\sigma})^2 \sim \chi^2(n) i=1n(σXiμ)2χ2(n)
    2. ( n − 1 ) S 2 σ 2 = ∑ i = 1 n ( X i − X ‾ σ ) 2 ∼ χ 2 ( n − 1 ) \dfrac{(n-1)S^2}{\sigma^2}=\sum\limits^n_{i=1}(\dfrac{X_i-\overline X}{\sigma})^2 \sim \chi^2(n-1) σ2(n1)S2=i=1n(σXiX)2χ2(n1)(在1式情况下,当 μ \mu μ未知时,用 X ‾ \overline X X来代替 μ \mu μ)
    3. X ‾ , S 2 \overline X,S^2 XS2相互独立,且 X ‾ − μ S / n ∼ t ( n − 1 ) \dfrac{\overline X-\mu}{S/\sqrt{n}} \sim t(n-1) S/n Xμt(n1)(在把 X ‾ \overline X X标准化的时候,当 σ \sigma σ未知时,用 S S S来代替 σ \sigma σ t t t t t t-分布)

    八、抽样分布(1)样本均值 X ‾ ∼ N ( μ , σ 2 / n ) \overline X \sim N(\mu,\sigma^2/n) XN(μ,σ2/n)

    我们通常定义 X ‾ = 1 n ∑ i = 1 n X i , X i ∼ N ( μ , σ 2 ) \overline X=\dfrac{1}{n}\sum\limits_{i=1}^{n}X_i,X_i\sim N(\mu,\sigma^2) X=n1i=1nXiXiN(μ,σ2)
    E ( X ‾ ) = E ( 1 n ∑ i = 1 n X i ) = 1 n ∑ i = 1 n E ( X i ) = n μ n = μ E(\overline X)=E(\dfrac{1}{n}\sum\limits_{i=1}^{n}X_i)=\dfrac{1}{n}\sum\limits_{i=1}^{n}E(X_i)=\frac{n\mu}{n}=\mu E(X)=E(n1i=1nXi)=n1i=1nE(Xi)=nnμ=μ
    D ( X ‾ ) = D ( 1 n ∑ i = 1 n X i ) = ( 1 n ) 2 ∑ i = 1 n D ( X i ) = n σ 2 n 2 = σ 2 / n D(\overline X)=D(\dfrac{1}{n}\sum\limits_{i=1}^{n}X_i)=(\dfrac{1}{n})^2\sum\limits_{i=1}^{n}D(X_i)=\dfrac{n\sigma^2}{n^2}=\sigma^2/n D(X)=D(n1i=1nXi)=(n1)2i=1nD(Xi)=n2nσ2=σ2/n

    注释

    以下证明 X ‾ ∼ N ( μ , σ 2 / n ) \overline X \sim N(\mu,\sigma^2/n) XN(μ,σ2/n)
    X ∼ N ( μ x , σ x 2 ) , Y ∼ N ( μ y , σ y 2 ) X\sim N(\mu_x,\sigma^2_x),Y\sim N(\mu_y,\sigma^2_y) XN(μx,σx2),YN(μy,σy2) X , Y X,Y X,Y独立时,则 a X + b Y aX+bY aX+bY(a,b为不全为0的系数)也遵循正态分布, a X + b Y ∼ N ( a μ x + b μ y , a 2 σ x 2 + b 2 σ y 2 ) aX+bY \sim N(a\mu_x+b\mu_y,a^2\sigma_x^2+b^2\sigma^2_y) aX+bYN(aμx+bμy,a2σx2+b2σy2)
    所以 X ‾ = 1 n ∑ i = 1 n X i ∼ N ( 1 n ∑ i = 1 n μ i , 1 n 2 ∑ i = 1 n σ i 2 ) = N ( μ , σ 2 / n ) \overline X=\dfrac{1}{n}\sum\limits_{i=1}^{n}X_i \sim N(\dfrac{1}{n}\sum\limits_{i=1}^{n}\mu_i,\dfrac{1}{n^2}\sum\limits_{i=1}^{n}\sigma^2_i)=N(\mu,\sigma^2/n) X=n1i=1nXiN(n1i=1nμi,n21i=1nσi2)=N(μ,σ2/n)

    九、抽样分布(2) 样本方差 S 2 S^2 S2

    我们通常定义 S 2 = 1 n − 1 ∑ i = 1 n ( X i − X ‾ ) 2 , X i ∼ N ( μ , σ 2 ) S^2=\dfrac{1}{n-1}\sum\limits_{i=1}^{n}(X_i-\overline X)^2,X_i\sim N(\mu,\sigma^2) S2=n11i=1n(XiX)2XiN(μ,σ2)
    E ( S 2 ) = E [ 1 n − 1 ( ( ∑ i = 1 n X i 2 ) − n X ‾ 2 ) ] E(S^2)=E[\dfrac{1}{n-1}((\sum\limits_{i=1}^{n}X_i^2)-n\overline X^2)] E(S2)=E[n11((i=1nXi2)nX2)]
    = 1 n − 1 [ ( ∑ i = 1 n E ( X i 2 ) ) − n E ( X ‾ 2 ) ] =\dfrac{1}{n-1}[(\sum\limits_{i=1}^{n}E(X_i^2))-nE(\overline X^2)] =n11[(i=1nE(Xi2))nE(X2)]
    = 1 n − 1 [ ( ∑ i = 1 n σ 2 + μ 2 ) − n ( σ 2 / n + μ 2 ) ] =\dfrac{1}{n-1}[(\sum\limits_{i=1}^{n}\sigma^2+\mu^2)-n(\sigma^2/n+\mu^2)] =n11[(i=1nσ2+μ2)n(σ2/n+μ2)]
    = σ 2 =\sigma^2 =σ2

    实际上在上面提到对 S 2 S^2 S2 ( n − 1 ) S 2 σ 2 = ∑ i = 1 n ( X i − X ‾ σ ) 2 ∼ χ 2 ( n − 1 ) \dfrac{(n-1)S^2}{\sigma^2}=\sum\limits_{i=1}^{n}(\dfrac{X_i-\overline X}{\sigma})^2 \sim \chi^2(n-1) σ2(n1)S2=i=1n(σXiX)2χ2(n1)
    所以 2 ( n − 1 ) = D ( ( n − 1 ) S 2 σ 2 ) = ( n − 1 ) 2 σ 4 D ( S 2 ) 2(n-1)=D(\dfrac{(n-1)S^2}{\sigma^2})=\dfrac{(n-1)^2}{\sigma^4}D(S^2) 2(n1)=D(σ2(n1)S2)=σ4(n1)2D(S2),即

    D ( S 2 ) = 2 σ 4 n − 1 D(S^2)=\dfrac{2\sigma^4}{n-1} D(S2)=n12σ4

    反过来想,有 E ( ∑ i = 1 n ( X i − X ‾ ) 2 ) E(\sum\limits_{i=1}^{n}(X_i-\overline X)^2) E(i=1n(XiX)2)
    = ( n − 1 ) E ( 1 n − 1 ∑ i = 1 n ( X i − X ‾ ) 2 ) =(n-1)E(\dfrac{1}{n-1}\sum\limits_{i=1}^{n}(X_i-\overline X)^2) =(n1)E(n11i=1n(XiX)2)
    = ( n − 1 ) E ( S 2 ) =(n-1)E(S^2) =(n1)E(S2) = ( n − 1 ) σ 2 =(n-1)\sigma^2 =(n1)σ2

    十、正态变量的幂的统计量

    假设 X ∼ N ( μ , σ 2 ) X\sim N(\mu,\sigma^2) XN(μ,σ2)

    仍然有 Z = X − μ σ Z=\dfrac{X-\mu}{\sigma} Z=σXμ Z ∼ N ( 0 , 1 ) Z\sim N(0,1) ZN(0,1)

    E ( X 2 ) = D ( X ) + E ( X ) 2 = σ 2 + μ 2 E(X^2)=D(X)+E(X)^2=\sigma^2+\mu^2 E(X2)=D(X)+E(X)2=σ2+μ2.

    由于 E ( Z 3 ) = ∫ − ∞ ∞ z 3 φ ( z ) d z E(Z^3)=\int\limits^\infty_{-\infty}z^3\varphi(z)dz E(Z3)=z3φ(z)dz,且 φ ( z ) \varphi(z) φ(z)是偶函数, z 3 z^3 z3是奇函数,
    ∫ 0 ∞ z 3 φ ( z ) d z \int\limits^\infty_{0}z^3\varphi(z)dz 0z3φ(z)dz
    = ∫ 0 ∞ z 3 1 2 π e − z 2 / 2 d z =\int\limits^\infty_{0}z^3\dfrac{1}{\sqrt{2\pi}}e^{-z^2/2}dz =0z32π 1ez2/2dz
    = ∫ 0 ∞ z 3 2 3 / 2 2 3 / 2 2 π 2 e − z 2 / 2 d z 2 =\int\limits^\infty_{0} \dfrac{z^3}{2^{3/2}} \dfrac{2^{3/2}}{\sqrt{2\pi}} \sqrt{2}e^{-z^2/2}d\dfrac{z}{\sqrt{2}} =023/2z32π 23/22 ez2/2d2 z
    = 2 2 π ∫ 0 ∞ t 3 e − t 2 d t =\dfrac{2\sqrt{2}}{\sqrt{\pi}}\int\limits^\infty_{0}t^3e^{-t^2}dt =π 22 0t3et2dt, ( t = z / 2 ) (t=z/\sqrt{2}) (t=z/2 )
    = 2 2 π 1 2 Γ ( 2 ) = 2 π =\dfrac{2\sqrt{2}}{\sqrt{\pi}}\dfrac{1}{2}\Gamma(2)=\dfrac{\sqrt{2}}{\sqrt{\pi}} =π 22 21Γ(2)=π 2
    所以此积分收敛, E ( Z 3 ) = 0 E(Z^3)=0 E(Z3)=0
    同理可得 E ( Z 4 ) = 4 π Γ ( 5 2 ) = 3 E(Z^4)=\dfrac{4}{\sqrt{\pi}}\Gamma(\dfrac{5}{2})=3 E(Z4)=π 4Γ(25)=3等等

    注释

    事实上,有 ∫ 0 ∞ z n φ ( z ) d z = 2 n 2 π Γ ( n + 1 2 ) ( n ∈ Z + ) \int\limits^\infty_{0}z^n\varphi(z)dz=\dfrac{\sqrt{2}^n}{2\sqrt{\pi}}\Gamma(\dfrac{n+1}{2})(n\in Z^+) 0znφ(z)dz=2π 2 nΓ(2n+1)(nZ+)
    所以
    ∫ − ∞ ∞ z n φ ( z ) d z = { 2 n π Γ ( n + 1 2 ) ( n = 2 k ) 0 ( n = 2 k + 1 ) ( k ∈ Z + ) \int\limits^\infty_{-\infty}z^n\varphi(z)dz=\left\{ \begin{aligned} \dfrac{\sqrt{2}^n}{\sqrt{\pi}}\Gamma(\dfrac{n+1}{2}) & & (n=2k) \\ 0 & & (n=2k+1) \\ \end{aligned} \right.(k\in Z^+) znφ(z)dz=π 2 nΓ(2n+1)0(n=2k)(n=2k+1)(kZ+)
    其中 Γ ( x ) \Gamma(x) Γ(x) Γ \Gamma Γ函数。 Γ ( a + 1 ) = a Γ ( a ) = a ! , Γ ( 1 / 2 ) = π \Gamma(a+1)=a\Gamma(a)=a!,\Gamma(1/2)=\sqrt{\pi} Γ(a+1)=aΓ(a)=a!,Γ(1/2)=π

    另一方面
    E ( Z 3 ) = E ( X − μ σ ) 3 = E ( 1 σ 3 ( X 3 − 3 μ X 2 + 3 μ 2 X − μ 3 ) ) E(Z^3)=E(\dfrac{X-\mu}{\sigma})^3=E(\dfrac{1}{\sigma^3}(X^3-3\mu X^2+3\mu^2X-\mu^3)) E(Z3)=E(σXμ)3=E(σ31(X33μX2+3μ2Xμ3))
    = 1 σ 3 ( E ( X 3 ) − 3 μ E ( X 2 ) + 3 μ 2 E ( X ) − μ 3 ) =\dfrac{1}{\sigma^3}(E(X^3)-3\mu E(X^2)+3\mu^2E(X)-\mu^3) =σ31(E(X3)3μE(X2)+3μ2E(X)μ3)
    = 1 σ 3 ( E ( X 3 ) − 3 μ ( σ 2 + μ 2 ) + 3 μ 2 μ − μ 3 ) =\dfrac{1}{\sigma^3}(E(X^3)-3\mu(\sigma^2+\mu^2)+3\mu^2\mu-\mu^3) =σ31(E(X3)3μ(σ2+μ2)+3μ2μμ3)
    可得

    E ( X 3 ) = 3 μ σ 2 + μ 3 E(X^3)=3\mu\sigma^2+\mu^3 E(X3)=3μσ2+μ3

    以这种方法可以算出高次幂的期望,进而根据 D ( X n ) = E ( X 2 n ) − E ( X n ) 2 D(X^n)=E(X^{2n})-E(X^n)^2 D(Xn)=E(X2n)E(Xn)2可以算出高次幂的方差。

    展开全文
  • 常见分布期望和方差的推导

    千次阅读 2020-06-01 09:31:09
    设有一个随机变量X, 其期望存在为E(X),方差存在为D(X) 有结论D(X)=E(X2)−[E(X)]2D(X) = E(X^2) - [E(X)]^2D(X)=E(X2)−[E(X)]2 1.二项分布 X~b(n, p) P{X=k}=(nk)pk(1−p)n−k, k=0,1,2,...,n P\{X=k\} = {n...

    设有一个随机变量X, 其期望存在为E(X),方差存在为D(X)
    有结论 D ( X ) = E ( X 2 ) − [ E ( X ) ] 2 D(X) = E(X^2) - [E(X)]^2 D(X)=E(X2)[E(X)]2

    1.二项分布

    表达式:
    X ~ b ( n , p ) P { X = k } = ( n k ) p k ( 1 − p ) n − k ,   k = 0 , 1 , 2 , . . . , n X \text{\textasciitilde} b(n, p)\\ P\{X=k\} = {n \choose k}p^k(1-p)^{n-k}, \space k=0,1,2,...,n X~b(n,p)P{X=k}=(kn)pk(1p)nk, k=0,1,2,...,n

    期望推导过程:
    E ( X ) = ∑ k = 0 n P { X = k } ⋅ k = ∑ k = 1 n ( n k ) k p k ( 1 − p ) n − k = ∑ k = 1 n n p ( n − 1 k − 1 ) p k − 1 ( 1 − p ) n − k =   n p ∑ k = 1 n ( n − 1 k − 1 ) p k − 1 ( 1 − p ) n − k = n p \begin{aligned} E(X) = &\sum_{k=0}^nP\{X=k\}\sdot k = \sum_{k=1}^n{n \choose k}kp^k(1-p)^{n-k}={\sum_{k=1}^n}np{n-1 \choose k-1}p^{k-1}(1-p)^{n-k}\\ =&\space np\sum_{k=1}^n {n-1 \choose k-1}p^{k-1}(1-p)^{n-k} = np \end{aligned} E(X)==k=0nP{X=k}k=k=1n(kn)kpk(1p)nk=k=1nnp(k1n1)pk1(1p)nk npk=1n(k1n1)pk1(1p)nk=np

    这里面有2个变换:
    ( n k ) = n k ( n − 1 k − 1 ) ( n − 1 k − 1 ) p k − 1 ( 1 − p ) n − k = ( p + 1 − p ) n − 1 = 1 {n \choose k} = \frac n k {n-1 \choose k-1} \\ {n-1 \choose k-1}p^{k-1}(1-p)^{n-k} = (p+1-p)^{n-1} = 1 (kn)=kn(k1n1)(k1n1)pk1(1p)nk=(p+1p)n1=1

    方差的推导过程:
    E ( X 2 ) = ∑ k = 0 n k 2 ( n k ) p k ( 1 − p ) n − k 令 1 − p = q E ( X 2 ) = ∑ k = 1 n k 2 ( n k ) p k q n − k = ∑ k = 1 n n k ( n − 1 k − 1 ) p k q n − k = ∑ k = 1 n n ( k − 1 + 1 ) ( n − 1 k − 1 ) p k q n − k = ∑ k = 1 n n ( k − 1 ) ( n − 1 k − 1 ) p k q n − k   + ∑ k = 1 n n p ( n − 1 k − 1 ) p k − 1 q n − k = ∑ k = 2 n n ( n − 1 ) p 2 ( n − 2 k − 2 ) p k − 2 q n − k + n p = n ( n − 1 ) p 2 + n p D ( X ) = E ( X 2 ) − [ E ( X ) 2 ] = n ( n − 1 ) p 2 + n p − ( n p ) 2 = n p ( 1 − p ) \begin{aligned} E(X^2) &= \sum_{k=0}^nk^2{n \choose k}p^k(1-p)^{n-k} \quad 令 1-p = q\\ E(X^2) &= \sum_{k=1}^n k^2 {n \choose k} p^k q^{n-k} = \sum_{k=1}^n nk {n-1 \choose k-1} p^k q^{n-k}\\ &= \sum_{k=1}^n n(k -1 + 1){n-1 \choose k-1} p^k q^{n-k}\\ &= \sum_{k=1}^n n(k-1) {n-1 \choose k-1}p^k q^{n-k} \space + \sum_{k=1}^n np {n-1 \choose k-1} p^{k-1} q^{n-k}\\ &= \sum_{k=2}^n n(n-1) p^2 {n-2 \choose k-2} p^{k-2} q^{n-k} + np\\ &= n(n-1)p^2 + np\\ \\ D(X) &= E(X^2) - [E(X)^2]\\ &= n(n-1)p^2 + np -(np)^2\\ &= np(1-p) \end{aligned} E(X2)E(X2)D(X)=k=0nk2(kn)pk(1p)nk1p=q=k=1nk2(kn)pkqnk=k=1nnk(k1n1)pkqnk=k=1nn(k1+1)(k1n1)pkqnk=k=1nn(k1)(k1n1)pkqnk +k=1nnp(k1n1)pk1qnk=k=2nn(n1)p2(k2n2)pk2qnk+np=n(n1)p2+np=E(X2)[E(X)2]=n(n1)p2+np(np)2=np(1p)

    2.泊松分布

    表达式:
    X ~ π ( λ ) P { X = k } = λ k e − λ k ! , k = 0 , 1 , 2 , . . . X \text{\textasciitilde} \pi(\lambda)\\ P\{X=k\} = \frac {\lambda^k e^{-\lambda}} {k!}, k=0,1,2,... X~π(λ)P{X=k}=k!λkeλ,k=0,1,2,...

    期望推导过程:
    E ( X ) = ∑ k = 0 ∞ k λ k e − λ k ! = λ e − λ ∑ k = 1 ∞ λ k − 1 k ! = λ e − λ e λ = λ \begin{aligned} E(X) &= \sum_{k=0}^\infin k {\frac {\lambda^k e^{-\lambda} } {k!}} = \lambda e^{-\lambda} \sum_{k=1}^\infin {\frac {\lambda^{k-1}} {k!} } \\ &= \lambda e^{-\lambda} e^{\lambda}\\ &= \lambda\\ \end{aligned} E(X)=k=0kk!λkeλ=λeλk=1k!λk1=λeλeλ=λ

    这里也有一个变换(就是 e^x的泰勒展开式):
    ∑ k = 0 ∞ λ k k ! = e λ \sum_{k=0}^\infin {\frac {\lambda^{k}} {k!} } = e^{\lambda} k=0k!λk=eλ

    方差推导过程:
    E ( X 2 ) = ∑ k = 0 ∞ k 2 λ k e − λ k ! = ∑ k = 1 ∞ k e − λ λ   λ k − 1 ( k − 1 ) ! = ∑ k = 1 ∞ ( k − 1 + 1 ) λ e − λ λ k − 1 ( k − 1 ) ! = ∑ k = 1 ∞ ( k − 1 ) λ e − λ λ k − 1 ( k − 1 ) !   + λ e − λ ∑ k = 1 ∞ λ k − 1 ( k − 1 ) ! = λ 2 e − λ ∑ k = 2 ∞ λ k − 2 ( k − 2 ) !   + λ = λ 2 + λ D ( X ) = λ 2 + λ − λ 2 = λ \begin{aligned} E(X^2) &= \sum_{k=0}^\infin k^2 {\frac {\lambda^k e^{-\lambda}} {k!}} = \sum_{k=1}^\infin k e^{-\lambda} \lambda \space {\frac {\lambda^{k-1}} {(k-1)!}} \\ &= \sum_{k=1}^\infin (k-1+1) \lambda e^{-\lambda} {\frac {\lambda^{k-1}} {(k-1)!}} \\ &= \sum_{k=1}^\infin (k-1) \lambda e^{-\lambda} {\frac {\lambda ^{k-1}} {(k-1)!}} \space + \lambda e^{-\lambda} \sum_{k=1}^\infin {\frac {\lambda^{k-1}} {(k-1)!}} \\ &= \lambda^2 e^{-\lambda} \sum_{k=2}^\infin {\frac {\lambda^{k-2}} {(k-2)!}} \space + \lambda \\ &= \lambda^2 + \lambda \\ \\ D(X) &= \lambda^2 + \lambda - \lambda^2 = \lambda \end{aligned} E(X2)D(X)=k=0k2k!λkeλ=k=1keλλ (k1)!λk1=k=1(k1+1)λeλ(k1)!λk1=k=1(k1)λeλ(k1)!λk1 +λeλk=1(k1)!λk1=λ2eλk=2(k2)!λk2 +λ=λ2+λ=λ2+λλ2=λ

    3.几何分布

    表达式:
    X ~ G ( p ) P { X = k } = ( 1 − p ) k − 1 p ,   k = 1 , 2 , 3 , . . . X \text{\textasciitilde} G(p)\\ P\{X=k\} = (1-p)^{k-1}p, \space k= 1, 2, 3, ... X~G(p)P{X=k}=(1p)k1p, k=1,2,3,...

    期望的推导:
    E ( X ) = ∑ k = 1 ∞ k ( 1 − p ) k − 1 p = p ∑ k = 1 ∞ k ( 1 − p ) k − 1 令 q = 1 − p ,   S = ∑ k = 1 ∞ k q k − 1 S = 1 + 2 q + 3 q 2 + 4 q 3 + . . . + k q k − 1 , k = ∞ ① q S = q + 2 q 2 + 3 q 3 + 4 q 4 + . . . + k q k , k = ∞ ② ① − ② 得 ( 1 − q ) S = 1 + q + q 2 + q 3 + . . . + q k − 1 − k q k S = 1 − q k ( 1 − q ) 2 − k q k 1 − q = 1 ( 1 − q ) 2 = 1 p 2 , 这 里 因 为 k = ∞ E ( X ) = p 1 p 2 = 1 p \begin{aligned} E(X) &= \sum_{k=1}^\infin k (1-p)^{k-1} p = p \sum_{k=1}^\infin k (1-p)^{k-1} \\ 令q = &1 - p, \space S = \sum_{k=1}^\infin kq^{k-1} \\ &\\ S = 1 &+ 2q + 3q^2 + 4q^{3} + ... + kq^{k-1}, k=\infin \qquad ① \\ qS = q&+ 2q^2 + 3q^3 + 4q^4 + ... + kq^k, k=\infin \qquad ② \\ ① - ②得 \quad &(1-q)S = 1+ q +q^2 + q^3 + ... + q^{k-1} - kq^k \\ S = & {\frac {1-q^k} {(1-q)^2} } - {\frac {kq^k} {1-q}} = {\frac 1 {(1-q)^2} } = \frac 1 {p^2}, \quad 这里因为k=\infin \\ \\ E(X) &= p {\frac 1 {p^2}} = {\frac 1 p} \\ \end{aligned} E(X)q=S=1qS=qS=E(X)=k=1k(1p)k1p=pk=1k(1p)k11p, S=k=1kqk1+2q+3q2+4q3+...+kqk1,k=+2q2+3q3+4q4+...+kqk,k=(1q)S=1+q+q2+q3+...+qk1kqk(1q)21qk1qkqk=(1q)21=p21,k==pp21=p1

    方差推导过程:
    E ( X 2 ) = ∑ k = 1 ∞ k 2 ( 1 − p ) k − 1 p = p ∑ k − 1 ∞ k 2 ( 1 − p ) k − 1 令 1 − p = q S = ∑ k = 1 ∞ k 2 q k − 1 = ∑ k = 1 ∞ ( k q k ) ′ = [ ∑ k = 1 ∞ k q n ] ′ = [ q ( 1 − q ) 2 ] ′ = ( 1 − q ) 2 + 2 ( 1 − q ) q ( 1 − q ) 4 = 2 p − p 2 p 4 = 2 − p p 3 所 以 E ( X 2 ) = 2 − p p 2 D ( X ) = 2 − p p 2 − ( 1 p ) 2 = 1 − p p 2 \begin{aligned} E(X^2) &= \sum_{k=1}^\infin k^2 (1-p)^{k-1} p = p\sum_{k-1}^\infin k^2 {(1-p)}^{k-1} \\ 令1-p &= q \\ S &= \sum_{k=1}^\infin k^2 q^{k-1} = \sum_{k=1}^\infin (kq^k)' = {[\sum_{k=1}^\infin kq^n]}' \\ &= [\frac q {(1-q)^2}]' \\ &= \frac {(1-q)^2 + 2(1-q)q} {(1-q)^4} \\ &= \frac {2p - p^2} {p^4} \\ &= \frac {2-p} {p^3} \\ 所以 E(X^2) &= \frac {2-p} {p^2} \\ D(X) &= \frac {2-p} {p^2} - {(\frac 1 p)}^2 \\ &= \frac {1-p} {p^2}\\ \end{aligned} E(X2)1pSE(X2)D(X)=k=1k2(1p)k1p=pk1k2(1p)k1=q=k=1k2qk1=k=1(kqk)=[k=1kqn]=[(1q)2q]=(1q)4(1q)2+2(1q)q=p42pp2=p32p=p22p=p22p(p1)2=p21p

    4.超几何分布

    表达式:
    X ~ H ( n , K , N ) P { X = k } = ( K k ) ( N − K n − k ) ( N n ) X \text{\textasciitilde} H(n, K, N) \\ P\{X=k\} = {\frac {{K \choose k} {N-K \choose n-k} } {N \choose n} } \\ X~H(n,K,N)P{X=k}=(nN)(kK)(nkNK)

    期望推导过程:
    E ( X ) = ∑ k = 0 n k ( K k ) ( N − K n − k ) ( N n ) = K n N ∑ k = 1 n ( K − 1 k − 1 ) ( N − K n − k ) ( N − 1 n − 1 ) = K n N \begin{aligned} E(X) &= \sum_{k=0}^n k {\frac {{K \choose k} {N-K \choose n-k} } {N \choose n}} \\ &= {\frac {Kn} {N}} \sum_{k=1}^n {\frac {{K-1 \choose k-1} {N-K \choose n-k}} {N-1 \choose n-1}} \\ &= {\frac {Kn} {N}}\\ \end{aligned} E(X)=k=0nk(nN)(kK)(nkNK)=NKnk=1n(n1N1)(k1K1)(nkNK)=NKn

    方差的推导过程:
    E ( X 2 ) = ∑ k = 0 n k 2 ( K k ) ( N − K n − k ) ( N n ) = K n N ∑ k = 1 n ( k − 1 + 1 ) ( K − 1 k − 1 ) ( N − K n − k ) ( N − 1 n − 1 ) = K n N [ ∑ k = 1 n ( k − 1 ) ( K − 1 k − 1 ) ( N − K n − k ) ( N − 1 n − 1 ) + ∑ k = 1 n ( K − 1 k − 1 ) ( N − K n − k ) ( N − 1 n − 1 ) ] = K n N [ ( K − 1 ) ( n − 1 ) N − 1 ∑ k = 2 n ( K − 2 k − 2 ) ( N − K n − k ) ( N − 2 n − 2 ) + 1 ] = K n N [ ( K − 1 ) ( n − 1 ) N − 1 + 1 ] D ( X ) = K n N [ ( K − 1 ) ( n − 1 ) N − 1 + 1 ] − ( K n N ) 2 = K n N ( N − K ) ( N − n ) N ( N − 1 ) \begin{aligned} E(X^2) &= \sum_{k=0}^n k^2 {\frac {{K \choose k} {N-K \choose n-k}} {N \choose n} } \\ &= {\frac {Kn} {N}} \sum_{k=1}^n (k-1+1) {\frac { {K-1 \choose k-1} {N-K \choose n-k} } {N-1 \choose n-1} } \\ &={\frac {Kn} {N}} [ \sum_{k=1}^n (k-1){\frac { {K-1 \choose k-1} {N-K \choose n-k} } {N-1 \choose n-1}} + \sum_{k=1}^n{\frac { {K-1 \choose k-1} {N-K \choose n-k} } {N-1 \choose n-1}} ] \\ &= {\frac {Kn} {N}} [{\frac {(K-1)(n-1)} {N-1}}\sum_{k=2}^n {\frac { {K-2 \choose k-2} {N-K \choose n-k} } {N-2 \choose n-2}} + 1] \\ &= {\frac {Kn} {N}}[{\frac {(K-1)(n-1)} {N-1}} +1] \\ \\ D(X) &= {\frac {Kn} {N}}[{\frac {(K-1)(n-1)} {N-1}} +1] - ({\frac {Kn} N})^2 \\ &= {\frac {Kn} {N}} {\frac {(N-K)(N-n)} {N(N-1)}} \\ \end{aligned} E(X2)D(X)=k=0nk2(nN)(kK)(nkNK)=NKnk=1n(k1+1)(n1N1)(k1K1)(nkNK)=NKn[k=1n(k1)(n1N1)(k1K1)(nkNK)+k=1n(n1N1)(k1K1)(nkNK)]=NKn[N1(K1)(n1)k=2n(n2N2)(k2K2)(nkNK)+1]=NKn[N1(K1)(n1)+1]=NKn[N1(K1)(n1)+1](NKn)2=NKnN(N1)(NK)(Nn)

    5.均匀分布

    表达式:

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    做n次0-1试验,每次实验为1的概率为p,为0的概率为1-p;有k次为1,n-k次为0的概率,就是二项分布B(n,p,k)。

     

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