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UA MATH564 概率分布总结

發布時間：2025/4/14 编程问答 26 豆豆

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UA MATH564 概率分布總結

Part zero Continuous Uniform Distribution
Tip 1: $\sim U(a,b)$ , $\frac{x-a}{b-a},\ f(x) = \frac{1}{b-a}$
Tip 2: First four moments

E X

V a r (X)

EX^2

EX^3

EX^4

b?a2\frac{b-a}{2}

(b?a)212\frac{(b-a)^2}{12}

b3?a33(b?a)\frac{b^3-a^3}{3(b-a)}

b4?a44(b?a)\frac{b^4 - a^4}{4(b-a)}

b5?a55(b?a)\frac{b^5 - a^5}{5(b-a)}

Tip 3: $X(i)～Beta(i,n?i+1)X_{(i)} \sim Beta(i,n-i+1)$ , $(X(i),X(j))～Dirichlet(i,j?i,n?j+1),fori<j(X_{(i)},X_{(j)}) \sim Dirichlet(i,j-i,n-j+1),\ for\ i<j$
Tip 4: $X_{(1)},X_{(n)}$ are complete minimal sufficient statistics.
Tip 5: $X_{(1)},X_{(n)}$ are MLE of $a, b$ .

Tip 6: A special case $\sim U(0,\theta)$ ,

U(0,θ)U(0,\theta)

is Complete.

X_{(n)}

is complete minimal sufficient statistics.

MLE is

X(n)～θ?Beta(n,1)X_{(n)} \sim \theta \cdot Beta(n,1)

MME is

2Xˉ2\bar{X}

(approximately

N(θ,θ23n)N(\theta,\frac{\theta^2}{3n})

)

UMVUE is

n+1nX(n)\frac{n+1}{n}X_{(n)}

Part a Poisson Distribution
Tip 1: $\sim Pois(\lambda), f(x) = \frac{\lambda^xe^{-\lambda}}{x!},x=0,1,\cdots$
Tip 2: $MX(t)=exp?(λ(et?1))M_X(t) = \exp(\lambda(e^t - 1))$ and first four moments

E X

V a r (X)

skewnesskurtosis

λ\lambda

λ\lambda

λ?1/2\lambda^{-1/2}

λ?1\lambda^{-1}

Tip 3: Additivity, $Xi～Pois(λi),i=1,2,?,nX_i \sim Pois(\lambda_i),i=1,2,\cdots,n$ and all $X_i$ s are independent, and then $∑i=1nXi～Pois(∑i=1nλi)\sum_{i=1}^n X_i \sim Pois(\sum_{i=1}^n \lambda_i)$
Tip 4: $X1～Pois(λ1)X_1 \sim Pois(\lambda_1)$ and $X2～Pois(λ2)X_2 \sim Pois(\lambda_2)$ are independent, and then $X1∣X1+X2=x～Binom(x,λ1λ1+λ2)X_1|X_1+X_2 = x \sim Binom(x,\frac{\lambda_1}{\lambda_1+\lambda_2})$
Tip 5: MLE and MME are both $Xˉ\bar{X}$ , $nXˉ～Pois(nλ)n\bar{X} \sim Pois(n\lambda)$
Tip 6: $Xˉ\bar{X}$ is UMVUE, and complete minimal sufficient statistics.

Part b Bernoulli Distribution
Tip 1: $\sim Ber(p),\ f(x) = p^x(1-p)^{1-x},x=0,1$

Tip 2: $MX(t)=exp?(λ(et?1))M_X(t) = \exp(\lambda(e^t - 1))$ and first four moments

E X

V a r (X)

skewnesskurtosis

p

p (1 ? p)

1?2pp(1?p)\frac{1-2p}{\sqrt{p(1-p)}}

6p2?6p+1p(1?p)\frac{6p^2-6p+1}{p(1-p)}

Tip 3: Additivity, $Xi～iidBer(p),i=1,?,nX_i \sim_{iid} Ber(p),i=1,\cdots,n$ , $∑i=1nXi～Binom(n,p)\sum_{i=1}^n X_i \sim Binom(n,p)$
Tip 4: MME and MLE are both $Xˉ\bar{X}$ , $nXˉ～Binom(n,p)n\bar{X} \sim Binom(n,p)$
Tip 5: $Xˉ\bar{X}$ is UMVUE

Part c Negative Binomial Distribution
Tip 1: Number of failed trials to get $r$ successfal trials. $\sim NB(r,p), f(x) = C_{r+x-1}^x p^r(1-p)^{x},x=0,1,\cdots$
Tip 2: $MX(t)=(pet1?(1?p)et)rM_X(t) = \left( \frac{pe^t}{1-(1-p)e^t}\right)^r$ and first four moments

E X

V a r (X)

skewnesskurtosis

r(1?p)p\frac{r(1-p)}{p}

r(1?p)p2\frac{r(1-p)}{p^2}

2?pr(1?p)\frac{2-p}{\sqrt{r(1-p)}}

6r+p2r(1?p)\frac{6}{r} + \frac{p^2}{r(1-p)}

Tip 3: Additivity, $Xi～iidGeo(p),i=1,?,nX_i \sim_{iid} Geo(p),i=1,\cdots,n$ , $∑i=1nXi～NB(n,p)\sum_{i=1}^n X_i \sim NB(n,p)$
Tip 4: $N B (1, p)$ is $G e o (p)$ (Number of failed trials before the first success)
Tip 5: A special case $\sim NB(r,p)$ where $r$ is known constant. MLE and MME are both $r/(r+Xˉ)r/(r+\bar{X})$ , $nXˉ～NB(nr,p)n\bar{X} \sim NB(nr,p)$ .

Part d Exponential Distribution
Tip 1: $\sim EXP(\lambda), f(x) = \lambda e^{-\lambda x},x>0, F(x) = e^{-\lambda x},x>0$
Tip 2: $MX(t)=(1?tλ)?1M_X(t) = \left(1- \frac{t}{\lambda}\right)^{-1}$ and first four moments

E X

V a r (X)

skewnesskurtosis

1λ\frac{1}{\lambda}

1λ2\frac{1}{\lambda^2}

2

6

Tip 3: $EXP(λ)=dΓ(1,λ)EXP(\lambda) =_d \Gamma(1,\lambda)$
Tip 4: Additivity, $Xi～iidEXP(λ),i=1,?,nX_i \sim_{iid} EXP(\lambda),i=1,\cdots,n$ , $∑i=1nXi～Γ(n,λ)\sum_{i=1}^n X_i \sim \Gamma(n,\lambda)$
Tip 5: Scale family, $\sim EXP(\lambda/c)$
Tip 6: $Xˉ\bar{X}$ is MLE and MME of $1/λ1/\lambda$ and also $1/Xˉ1/\bar{X}$ is MLE and MME of $λ\lambda$ , $nXˉ～Γ(n,λ)n\bar{X} \sim \Gamma(n,\lambda)$

Part e Gamma Distribution
Tip 1: $\sim \Gamma(\alpha,\lambda), f(x) = \frac{\lambda^{\alpha}x^{\alpha-1}}{\Gamma(\alpha)} e^{-\lambda x},x>0$ , $Γ(α)=∫0+∞xα?1e?xdx\Gamma(\alpha) = \int_0^{+\infty} x^{\alpha-1}e^{-x}dx$ , $Γ(n)=(n?1)!,Γ(1)=1,Γ(1/2)=π\Gamma(n) = (n-1)!, \Gamma(1) =1, \Gamma(1/2) = \sqrt{\pi}$
Tip 2: $MX(t)=(1?tλ)?αM_X(t) = \left(1- \frac{t}{\lambda}\right)^{-\alpha}$ and first four moments

E X

V a r (X)

skewnesskurtosis

αλ\frac{\alpha}{\lambda}

αλ2\frac{\alpha}{\lambda^2}

2α\frac{2}{\sqrt{\alpha}}

6α\frac{6}{\alpha}

Tip 3: Additivity, $Γ(α1,λ)+Γ(α2,λ)=Γ(α1+α2,λ)\Gamma(\alpha_1,\lambda)+\Gamma(\alpha_2,\lambda) = \Gamma(\alpha_1+\alpha_2,\lambda)$ (Independence)
Tip 4: $Γ(1,λ)=EXP(λ),Γ(n2,12)=χn2\Gamma(1,\lambda) = EXP(\lambda),\ \Gamma(\frac{n}{2},\frac{1}{2}) = \chi^2_n$
Tip 5: Scale family, $\sim \Gamma(\alpha,\lambda/c)$

Part f Beta Distribution
Tip 1: $\sim \Beta(\alpha,\beta), f(x) = \frac{\Gamma(\alpha+\beta)}{\Gamma(\alpha)\Gamma(\beta)} x^{\alpha-1}(1-x)^{\beta-1},x \in (0,1)$
Tip 2: $EXk=(α)k(α+β)kEX^k = \frac{(\alpha)_k}{(\alpha+\beta)_k}$ , first two moments

E X

V a r (X)

αα+β\frac{\alpha}{\alpha+\beta}

αβ(α+β)2(α+β+1)\frac{\alpha\beta}{(\alpha+\beta)^2(\alpha+\beta+1)}

Part g Normal Distribution
Tip 1: $\sim N(\mu,\sigma^2), f(x) = \frac{1}{\sqrt{2\pi}} e^{-\frac{(x-\mu)^2}{2\sigma^2}}$
Tip 2: $MX(t)=exp?(μt+σ22t2)M_X(t) = \exp \left( \mu t + \frac{\sigma^2}{2}t^2 \right)$ and first four moments

E X

V a r (X)

skewnesskurtosis

μ\mu

σ2\sigma^2

Tip 3: Additivity\ Linearity
Tip 4: $(Xˉ,S2)(\bar{X},S^2)$ are OLS, UMVUE, complete minimal sufficient statistics, $Xˉ～N(μ,σ2/n)\bar{X} \sim N(\mu,\sigma^2/n)$ , $(n?1)S2～σ2?χn?12(n-1)S^2 \sim \sigma^2 \cdot \chi^2_{n-1}$ or $S2～Γ(n?12,n?12σ2)S^2 \sim \Gamma(\frac{n-1}{2},\frac{n-1}{2\sigma^2})$
Tip 5: $Xi～iidN(0,1),∑i=1nXi2～χn2X_i\sim_{iid}N(0,1),\sum_{i=1}^n X_i^2 \sim \chi^2_n$
Tip 6: $\sim _{iid}N(0,\sigma^2),\sqrt{X^2+Y^2}\sim Rayleigh(\sigma^2), X/Y \sim Cauchy(0,1)$

Part h Log Normal Distribution
Tip 1: $\sim LN(\mu,\sigma^2)$ , $\frac{1}{\sqrt{2\pi}\sigma x}\exp (-\frac{(\ln x-\mu)^2}{2\sigma^2}),x>0$
Tip 2: $EXk=exp?(μk+σ22k)EX^k = \exp (\mu k + \frac{\sigma^2}{2}k)$
Tip 3: $X1～LN(μ1,σ12),X2～LN(μ2,σ22)X_1 \sim LN(\mu_1,\sigma^2_1),X_2 \sim LN(\mu_2,\sigma^2_2)$ , $X1X2～LN(μ1+μ2,σ12+σ22)X_1X_2 \sim LN(\mu_1+\mu_2,\sigma_1^2+\sigma^2_2)$ (Independence)
Tip 4: $(1n∑i=1nln?Xi,1n?1∑i=1n(ln?Xi?ln?Xˉ)2)(\frac{1}{n}\sum_{i=1}^n \ln X_i, \frac{1}{n-1}\sum_{i=1}^n (\ln X_i-\bar{\ln X})^2)$ are OLS, UMVUE, complete minimal sufficient statistics. The former follows $N(μ,σ2/n)N(\mu,\sigma^2/n)$ and the latter follows $Γ(n?12,n?12σ2)\Gamma(\frac{n-1}{2},\frac{n-1}{2\sigma^2})$

Part i Cauchy Distribution
Tip 1: $\sim Cauchy(x_0,\gamma), f(x) = \frac{1}{\pi}\frac{\gamma}{(x-x_0)^2+\gamma^2}$ , $\frac{1}{\pi}\arctan((x-x_0)/\gamma)+1/2$
Tip 2: No moments
Tip 3: $Cauchy(0,1) =_d t(1)$

Part j Rayleigh Distribution
Tip 1: $\sim Ray(\sigma), f(x) = \frac{x}{\sigma^2}e^{-\frac{x^2}{2\sigma^2}},x \ge 0$
Tip 2: first two moments

E X

V a r (X)

π2σ\frac{\pi}{2}\sigma

4?π2σ2\frac{4-\pi}{2}\sigma^2

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