UA MATH564 概率分布总结
UA MATH564 概率分布總結
Part zero Continuous Uniform Distribution
Tip 1: X~U(a,b)X \sim U(a,b)X~U(a,b), F(x)=x?ab?a,f(x)=1b?aF(x) = \frac{x-a}{b-a},\ f(x) = \frac{1}{b-a}F(x)=b?ax?a?,?f(x)=b?a1?
Tip 2: First four moments
| b?a2\frac{b-a}{2}2b?a? | (b?a)212\frac{(b-a)^2}{12}12(b?a)2? | b3?a33(b?a)\frac{b^3-a^3}{3(b-a)}3(b?a)b3?a3? | b4?a44(b?a)\frac{b^4 - a^4}{4(b-a)}4(b?a)b4?a4? | b5?a55(b?a)\frac{b^5 - a^5}{5(b-a)}5(b?a)b5?a5? |
Tip 3: X(i)~Beta(i,n?i+1)X_{(i)} \sim Beta(i,n-i+1)X(i)?~Beta(i,n?i+1), (X(i),X(j))~Dirichlet(i,j?i,n?j+1),fori<j(X_{(i)},X_{(j)}) \sim Dirichlet(i,j-i,n-j+1),\ for\ i<j(X(i)?,X(j)?)~Dirichlet(i,j?i,n?j+1),?for?i<j
Tip 4: X(1),X(n)X_{(1)},X_{(n)}X(1)?,X(n)? are complete minimal sufficient statistics.
Tip 5: X(1),X(n)X_{(1)},X_{(n)}X(1)?,X(n)? are MLE of a,ba,ba,b.
Tip 6: A special case Y~U(0,θ)Y \sim U(0,\theta)Y~U(0,θ),
Part a Poisson Distribution
Tip 1: X~Pois(λ),f(x)=λxe?λx!,x=0,1,?X \sim Pois(\lambda), f(x) = \frac{\lambda^xe^{-\lambda}}{x!},x=0,1,\cdotsX~Pois(λ),f(x)=x!λxe?λ?,x=0,1,?
Tip 2: MX(t)=exp?(λ(et?1))M_X(t) = \exp(\lambda(e^t - 1))MX?(t)=exp(λ(et?1)) and first four moments
| λ\lambdaλ | λ\lambdaλ | λ?1/2\lambda^{-1/2}λ?1/2 | λ?1\lambda^{-1}λ?1 |
Tip 3: Additivity, Xi~Pois(λi),i=1,2,?,nX_i \sim Pois(\lambda_i),i=1,2,\cdots,nXi?~Pois(λi?),i=1,2,?,n and all XiX_iXi?s are independent, and then ∑i=1nXi~Pois(∑i=1nλi)\sum_{i=1}^n X_i \sim Pois(\sum_{i=1}^n \lambda_i)∑i=1n?Xi?~Pois(∑i=1n?λi?)
Tip 4: X1~Pois(λ1)X_1 \sim Pois(\lambda_1)X1?~Pois(λ1?) and X2~Pois(λ2)X_2 \sim Pois(\lambda_2)X2?~Pois(λ2?) are independent, and then X1∣X1+X2=x~Binom(x,λ1λ1+λ2)X_1|X_1+X_2 = x \sim Binom(x,\frac{\lambda_1}{\lambda_1+\lambda_2})X1?∣X1?+X2?=x~Binom(x,λ1?+λ2?λ1??)
Tip 5: MLE and MME are both Xˉ\bar{X}Xˉ, nXˉ~Pois(nλ)n\bar{X} \sim Pois(n\lambda)nXˉ~Pois(nλ)
Tip 6: Xˉ\bar{X}Xˉ is UMVUE, and complete minimal sufficient statistics.
Part b Bernoulli Distribution
Tip 1: X~Ber(p),f(x)=px(1?p)1?x,x=0,1X \sim Ber(p),\ f(x) = p^x(1-p)^{1-x},x=0,1X~Ber(p),?f(x)=px(1?p)1?x,x=0,1
Tip 2: MX(t)=exp?(λ(et?1))M_X(t) = \exp(\lambda(e^t - 1))MX?(t)=exp(λ(et?1)) and first four moments
| ppp | p(1?p)p(1-p)p(1?p) | 1?2pp(1?p)\frac{1-2p}{\sqrt{p(1-p)}}p(1?p)?1?2p? | 6p2?6p+1p(1?p)\frac{6p^2-6p+1}{p(1-p)}p(1?p)6p2?6p+1? |
Tip 3: Additivity, Xi~iidBer(p),i=1,?,nX_i \sim_{iid} Ber(p),i=1,\cdots,nXi?~iid?Ber(p),i=1,?,n, ∑i=1nXi~Binom(n,p)\sum_{i=1}^n X_i \sim Binom(n,p)∑i=1n?Xi?~Binom(n,p)
Tip 4: MME and MLE are both Xˉ\bar{X}Xˉ, nXˉ~Binom(n,p)n\bar{X} \sim Binom(n,p)nXˉ~Binom(n,p)
Tip 5: Xˉ\bar{X}Xˉ is UMVUE
Part c Negative Binomial Distribution
Tip 1: Number of failed trials to get rrr successfal trials. X~NB(r,p),f(x)=Cr+x?1xpr(1?p)x,x=0,1,?X \sim NB(r,p), f(x) = C_{r+x-1}^x p^r(1-p)^{x},x=0,1,\cdotsX~NB(r,p),f(x)=Cr+x?1x?pr(1?p)x,x=0,1,?
Tip 2: MX(t)=(pet1?(1?p)et)rM_X(t) = \left( \frac{pe^t}{1-(1-p)e^t}\right)^rMX?(t)=(1?(1?p)etpet?)r and first four moments
| r(1?p)p\frac{r(1-p)}{p}pr(1?p)? | r(1?p)p2\frac{r(1-p)}{p^2}p2r(1?p)? | 2?pr(1?p)\frac{2-p}{\sqrt{r(1-p)}}r(1?p)?2?p? | 6r+p2r(1?p)\frac{6}{r} + \frac{p^2}{r(1-p)}r6?+r(1?p)p2? |
Tip 3: Additivity, Xi~iidGeo(p),i=1,?,nX_i \sim_{iid} Geo(p),i=1,\cdots,nXi?~iid?Geo(p),i=1,?,n, ∑i=1nXi~NB(n,p)\sum_{i=1}^n X_i \sim NB(n,p)∑i=1n?Xi?~NB(n,p)
Tip 4: NB(1,p)NB(1,p)NB(1,p) is Geo(p)Geo(p)Geo(p) (Number of failed trials before the first success)
Tip 5: A special case Y~NB(r,p)Y \sim NB(r,p)Y~NB(r,p) where rrr is known constant. MLE and MME are both r/(r+Xˉ)r/(r+\bar{X})r/(r+Xˉ), nXˉ~NB(nr,p)n\bar{X} \sim NB(nr,p)nXˉ~NB(nr,p).
Part d Exponential Distribution
Tip 1: X~EXP(λ),f(x)=λe?λx,x>0,F(x)=e?λx,x>0X \sim EXP(\lambda), f(x) = \lambda e^{-\lambda x},x>0, F(x) = e^{-\lambda x},x>0X~EXP(λ),f(x)=λe?λx,x>0,F(x)=e?λx,x>0
Tip 2: MX(t)=(1?tλ)?1M_X(t) = \left(1- \frac{t}{\lambda}\right)^{-1}MX?(t)=(1?λt?)?1 and first four moments
| 1λ\frac{1}{\lambda}λ1? | 1λ2\frac{1}{\lambda^2}λ21? | 222 | 666 |
Tip 3: EXP(λ)=dΓ(1,λ)EXP(\lambda) =_d \Gamma(1,\lambda)EXP(λ)=d?Γ(1,λ)
Tip 4: Additivity, Xi~iidEXP(λ),i=1,?,nX_i \sim_{iid} EXP(\lambda),i=1,\cdots,nXi?~iid?EXP(λ),i=1,?,n, ∑i=1nXi~Γ(n,λ)\sum_{i=1}^n X_i \sim \Gamma(n,\lambda)∑i=1n?Xi?~Γ(n,λ)
Tip 5: Scale family, cX~EXP(λ/c)cX \sim EXP(\lambda/c)cX~EXP(λ/c)
Tip 6: Xˉ\bar{X}Xˉ is MLE and MME of 1/λ1/\lambda1/λ and also 1/Xˉ1/\bar{X}1/Xˉ is MLE and MME of λ\lambdaλ, nXˉ~Γ(n,λ)n\bar{X} \sim \Gamma(n,\lambda)nXˉ~Γ(n,λ)
Part e Gamma Distribution
Tip 1: X~Γ(α,λ),f(x)=λαxα?1Γ(α)e?λx,x>0X \sim \Gamma(\alpha,\lambda), f(x) = \frac{\lambda^{\alpha}x^{\alpha-1}}{\Gamma(\alpha)} e^{-\lambda x},x>0X~Γ(α,λ),f(x)=Γ(α)λαxα?1?e?λx,x>0, Γ(α)=∫0+∞xα?1e?xdx\Gamma(\alpha) = \int_0^{+\infty} x^{\alpha-1}e^{-x}dxΓ(α)=∫0+∞?xα?1e?xdx, Γ(n)=(n?1)!,Γ(1)=1,Γ(1/2)=π\Gamma(n) = (n-1)!, \Gamma(1) =1, \Gamma(1/2) = \sqrt{\pi}Γ(n)=(n?1)!,Γ(1)=1,Γ(1/2)=π?
Tip 2: MX(t)=(1?tλ)?αM_X(t) = \left(1- \frac{t}{\lambda}\right)^{-\alpha}MX?(t)=(1?λt?)?α and first four moments
| αλ\frac{\alpha}{\lambda}λα? | αλ2\frac{\alpha}{\lambda^2}λ2α? | 2α\frac{2}{\sqrt{\alpha}}α?2? | 6α\frac{6}{\alpha}α6? |
Tip 3: Additivity, Γ(α1,λ)+Γ(α2,λ)=Γ(α1+α2,λ)\Gamma(\alpha_1,\lambda)+\Gamma(\alpha_2,\lambda) = \Gamma(\alpha_1+\alpha_2,\lambda)Γ(α1?,λ)+Γ(α2?,λ)=Γ(α1?+α2?,λ) (Independence)
Tip 4: Γ(1,λ)=EXP(λ),Γ(n2,12)=χn2\Gamma(1,\lambda) = EXP(\lambda),\ \Gamma(\frac{n}{2},\frac{1}{2}) = \chi^2_nΓ(1,λ)=EXP(λ),?Γ(2n?,21?)=χn2?
Tip 5: Scale family, cX~Γ(α,λ/c)cX \sim \Gamma(\alpha,\lambda/c)cX~Γ(α,λ/c)
Part f Beta Distribution
Tip 1: X~B(α,β),f(x)=Γ(α+β)Γ(α)Γ(β)xα?1(1?x)β?1,x∈(0,1)X \sim \Beta(\alpha,\beta), f(x) = \frac{\Gamma(\alpha+\beta)}{\Gamma(\alpha)\Gamma(\beta)} x^{\alpha-1}(1-x)^{\beta-1},x \in (0,1)X~B(α,β),f(x)=Γ(α)Γ(β)Γ(α+β)?xα?1(1?x)β?1,x∈(0,1)
Tip 2: EXk=(α)k(α+β)kEX^k = \frac{(\alpha)_k}{(\alpha+\beta)_k}EXk=(α+β)k?(α)k??, first two moments
| αα+β\frac{\alpha}{\alpha+\beta}α+βα? | αβ(α+β)2(α+β+1)\frac{\alpha\beta}{(\alpha+\beta)^2(\alpha+\beta+1)}(α+β)2(α+β+1)αβ? |
Part g Normal Distribution
Tip 1: X~N(μ,σ2),f(x)=12πe?(x?μ)22σ2X \sim N(\mu,\sigma^2), f(x) = \frac{1}{\sqrt{2\pi}} e^{-\frac{(x-\mu)^2}{2\sigma^2}}X~N(μ,σ2),f(x)=2π?1?e?2σ2(x?μ)2?
Tip 2: MX(t)=exp?(μt+σ22t2)M_X(t) = \exp \left( \mu t + \frac{\sigma^2}{2}t^2 \right)MX?(t)=exp(μt+2σ2?t2) and first four moments
| μ\muμ | σ2\sigma^2σ2 | 0 | 0 |
Tip 3: Additivity\ Linearity
Tip 4: (Xˉ,S2)(\bar{X},S^2)(Xˉ,S2) are OLS, UMVUE, complete minimal sufficient statistics, Xˉ~N(μ,σ2/n)\bar{X} \sim N(\mu,\sigma^2/n)Xˉ~N(μ,σ2/n), (n?1)S2~σ2?χn?12(n-1)S^2 \sim \sigma^2 \cdot \chi^2_{n-1}(n?1)S2~σ2?χn?12? or S2~Γ(n?12,n?12σ2)S^2 \sim \Gamma(\frac{n-1}{2},\frac{n-1}{2\sigma^2})S2~Γ(2n?1?,2σ2n?1?)
Tip 5: Xi~iidN(0,1),∑i=1nXi2~χn2X_i\sim_{iid}N(0,1),\sum_{i=1}^n X_i^2 \sim \chi^2_nXi?~iid?N(0,1),∑i=1n?Xi2?~χn2?
Tip 6: X,Y~iidN(0,σ2),X2+Y2~Rayleigh(σ2),X/Y~Cauchy(0,1)X,Y \sim _{iid}N(0,\sigma^2),\sqrt{X^2+Y^2}\sim Rayleigh(\sigma^2), X/Y \sim Cauchy(0,1)X,Y~iid?N(0,σ2),X2+Y2?~Rayleigh(σ2),X/Y~Cauchy(0,1)
Part h Log Normal Distribution
Tip 1: X~LN(μ,σ2)X \sim LN(\mu,\sigma^2)X~LN(μ,σ2), f(x)=12πσxexp?(?(ln?x?μ)22σ2),x>0f(x) = \frac{1}{\sqrt{2\pi}\sigma x}\exp (-\frac{(\ln x-\mu)^2}{2\sigma^2}),x>0f(x)=2π?σx1?exp(?2σ2(lnx?μ)2?),x>0
Tip 2: EXk=exp?(μk+σ22k)EX^k = \exp (\mu k + \frac{\sigma^2}{2}k)EXk=exp(μk+2σ2?k)
Tip 3: X1~LN(μ1,σ12),X2~LN(μ2,σ22)X_1 \sim LN(\mu_1,\sigma^2_1),X_2 \sim LN(\mu_2,\sigma^2_2)X1?~LN(μ1?,σ12?),X2?~LN(μ2?,σ22?), X1X2~LN(μ1+μ2,σ12+σ22)X_1X_2 \sim LN(\mu_1+\mu_2,\sigma_1^2+\sigma^2_2)X1?X2?~LN(μ1?+μ2?,σ12?+σ22?) (Independence)
Tip 4: (1n∑i=1nln?Xi,1n?1∑i=1n(ln?Xi?ln?Xˉ)2)(\frac{1}{n}\sum_{i=1}^n \ln X_i, \frac{1}{n-1}\sum_{i=1}^n (\ln X_i-\bar{\ln X})^2)(n1?∑i=1n?lnXi?,n?11?∑i=1n?(lnXi??lnXˉ)2) are OLS, UMVUE, complete minimal sufficient statistics. The former follows N(μ,σ2/n)N(\mu,\sigma^2/n)N(μ,σ2/n) and the latter follows Γ(n?12,n?12σ2)\Gamma(\frac{n-1}{2},\frac{n-1}{2\sigma^2})Γ(2n?1?,2σ2n?1?)
Part i Cauchy Distribution
Tip 1: X~Cauchy(x0,γ),f(x)=1πγ(x?x0)2+γ2X \sim Cauchy(x_0,\gamma), f(x) = \frac{1}{\pi}\frac{\gamma}{(x-x_0)^2+\gamma^2}X~Cauchy(x0?,γ),f(x)=π1?(x?x0?)2+γ2γ?, F(x)=1πarctan?((x?x0)/γ)+1/2F(x) = \frac{1}{\pi}\arctan((x-x_0)/\gamma)+1/2F(x)=π1?arctan((x?x0?)/γ)+1/2
Tip 2: No moments
Tip 3: Cauchy(0,1)=dt(1)Cauchy(0,1) =_d t(1)Cauchy(0,1)=d?t(1)
Part j Rayleigh Distribution
Tip 1: X~Ray(σ),f(x)=xσ2e?x22σ2,x≥0X \sim Ray(\sigma), f(x) = \frac{x}{\sigma^2}e^{-\frac{x^2}{2\sigma^2}},x \ge 0X~Ray(σ),f(x)=σ2x?e?2σ2x2?,x≥0
Tip 2: first two moments
| π2σ\frac{\pi}{2}\sigma2π?σ | 4?π2σ2\frac{4-\pi}{2}\sigma^224?π?σ2 |
總結
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