§3 泊松过程
C1 指数分布
1)无记忆性
:∀s,t>0:P{X>s+t∣X>s}=P{X>t}\forall s,t\gt 0:P\{X\gt s+t|X\gt s\}=P\{X\gt t\}∀s,t>0:P{X>s+t∣X>s}=P{X>t}
X∼e(λ):E[X−t∣X>t]=EXX\sim e(\lambda): E[X-t|X\gt t]=EXX∼e(λ):E[X−t∣X>t]=EX
指数分布是唯一满足无记忆性的分布
证明:Fˉ(x)=1−F(x),∀s,t>0:Fˉ(s+t)=Fˉ(s)Fˉ(t)⟹F(x)=1−e−λt\bar F(x) = 1-F(x),\forall s,t\gt 0:\bar F(s+t) = \bar F(s)\bar F(t)\implies F(x)=1-e^{-\lambda t}Fˉ(x)=1−F(x),∀s,t>0:Fˉ(s+t)=Fˉ(s)Fˉ(t)⟹F(x)=1−e−λt
2)$T\sim e(\lambda),\forall f:E[\int\infin_0e{-\lambda x}f(x)\mathrm{d}x] =E[\int^T_0f(x)\mathrm{d}x] $
3)Xi∼e(λi):mini{Xi}∼e(∑iλi)X_i\sim e(\lambda_i):\min\limits_i\{X_i\}\sim e(\sum\limits_i\lambda_i)Xi∼e(λi):imin{Xi}∼e(i∑λi),且与排序无关
P{Xk=mini{Xi}}=λi∑iλiP\{X_k = \min\limits_i \{X_i\}\}= \frac{\lambda_i}{\sum\limits_i\lambda_i}P{Xk=imin{Xi}}=i∑λiλi
双服务台等待时间:
ET=E[T∣R1<R2]λ1λ1+λ2+E[T∣R2≤R1]λ2λ1+λ2ET = E[T|R_1\lt R_2]\frac{\lambda_1}{\lambda_1+\lambda_2}+E[T|R_2\le R_1]\frac{\lambda_2}{\lambda_1+\lambda_2}ET=E[T∣R1<R2]λ1+λ2λ1+E[T∣R2≤R1]λ1+λ2λ2
ET=1λ1+λ2ET = \frac{1}{\lambda_1+\lambda_2}ET=λ1+λ21
单服务台单队列,服务时间∼e(μ)\sim e(\mu)∼e(μ),每个人的等待极限∼e(θ)\sim e(\theta)∼e(θ)
第n个人被服务概率Pn=(1−θnθ+μ)Pn−1=μnθ+μP_n = (1-\frac{\theta}{n\theta+\mu})P_{n-1}=\frac{\mu}{n\theta + \mu}Pn=(1−nθ+μθ)Pn−1=nθ+μμ
第n个人等待时间:Wn=1nθ+μ+Wn−1=∑k=1n1kθ+μW_n = \frac{1}{n\theta+\mu}+W_{n-1}=\sum_{k=1}^n\frac{1}{k\theta+\mu}Wn=nθ+μ1+Wn−1=∑k=1nkθ+μ1
4)独立同分布Xi∼e(λ):S=∑k=1nXk∼Γ(n,λ),pS(s)=λe−λs(λs)n−1(n−1)!X_i\sim e(\lambda):S = \sum\limits_{k=1}^nX_k\sim \Gamma(n,\lambda),p_S(s) = \lambda e^{-\lambda s}\frac{(\lambda s)^{n-1}}{(n-1)!}Xi∼e(λ):S=k=1∑nXk∼Γ(n,λ),pS(s)=λe−λs(n−1)!(λs)n−1
5)独立Xi∼e(λi):S=∑k=1nXk,pS(s)=∑k=1n[λke−ks∏j≠iλjλj−λi]X_i\sim e(\lambda_i):S = \sum\limits_{k=1}^nX_k,p_S(s) = \sum\limits_{k=1}^n[\lambda_ke^{-k s}\prod\limits_{j\ne i}\frac{\lambda_j}{\lambda_j-\lambda_i}]Xi∼e(λi):S=k=1∑nXk,pS(s)=k=1∑n[λke−ksj=i∏λj−λiλj],称为亚指数分布
lims→∞rS(s)=mini{λi}\lim\limits_{s\to \infin} r_S(s) = \min\limits_i\{\lambda_i\}s→∞limrS(s)=imin{λi}
6)失败率函数
:r(t)=f(t)1−F(t),r(0)=0r(t) = \frac{f(t)}{1-F(t)},r(0)=0r(t)=1−F(t)f(t),r(0)=0,即时刻t发生损坏的概率
对于指数分布:r(t)≡λr(t)\equiv \lambdar(t)≡λ失败率函数可唯一确定分布:F(t)=1−e−∫0tr(x)dxF(t) = 1-e^{-\int^t_0r(x)\mathrm{d}x}F(t)=1−e−∫0tr(x)dx
C2 泊松过程
1)泊松过程
:满足以下条件的计数过程{N(t)∣t≥0}\{N(t)|t\ge 0\}{N(t)∣t≥0}:
N(0)=0,N(t)∈ZN(0) = 0,N(t)\in ZN(0)=0,N(t)∈Z具有独立增量
和平稳增量
:不重叠时间段内增量独立,且只与时段长度相关P{N(t+h)−N(t)=1}=λh+o(h)P\{N(t+h)-N(t) = 1\} = \lambda h+o(h)P{N(t+h)−N(t)=1}=λh+o(h),λ\lambdaλ称为到达率P{N(t+h)−N(t)=2}=h+o(h)P\{N(t+h)-N(t) = 2\} = h+o(h)P{N(t+h)−N(t)=2}=h+o(h)
2)增量分布:N(t+s)−N(s)∼p(λt)N(t+s)-N(s) \sim p(\lambda t)N(t+s)−N(s)∼p(λt)
证明:
对N(t)N(t)N(t)进行拉普拉斯变换:g(s,t)=Ee−sN(t)g(s,t) = Ee^{-sN(t)}g(s,t)=Ee−sN(t)
则∂g∂t(s,t)=limh→0h−1[Ee−sN(t+h)−Ee−sN(t)]\frac{\partial g}{\partial t}(s,t) = \lim\limits_{h\to 0}h^{-1}[Ee^{-sN(t+h)}-Ee^{-sN(t)}]∂t∂g(s,t)=h→0limh−1[Ee−sN(t+h)−Ee−sN(t)]
Ee−sN(t+h)=E[e−sN(t)e−s(N(t+h)−N(t))]Ee^{-sN(t+h)}=E[e^{-sN(t)}e^{-s(N(t+h)-N(t))}]Ee−sN(t+h)=E[e−sN(t)e−s(N(t+h)−N(t))],由平稳增量性Ee−sN(t+h)=Ee−sN(t)Ee−s(N(t+h)−N(t))Ee^{-sN(t+h)}=Ee^{-sN(t)}Ee^{-s(N(t+h)-N(t))}Ee−sN(t+h)=Ee−sN(t)Ee−s(N(t+h)−N(t))
Ee−s(N(t+h)−N(t))=Ee−sN(h)Ee^{-s(N(t+h)-N(t))}=Ee^{-sN(h)}Ee−s(N(t+h)−N(t))=Ee−sN(h),取条件于N(h)N(h)N(h),结合定义3,4得Ee−sN(h)=1−λh+λhe−s+o(h)Ee^{-sN(h)}=1-\lambda h+\lambda he^{-s} +o(h)Ee−sN(h)=1−λh+λhe−s+o(h)
得∂g∂t(s,t)=g(s,t)λ(e−s−1),g(s,t)=eλt(e−s−1)\frac{\partial g}{\partial t}(s,t) = g(s,t)\lambda(e^{-s}-1),g(s,t) = e^{\lambda t(e^{-s}-1)}∂t∂g(s,t)=g(s,t)λ(e−s−1),g(s,t)=eλt(e−s−1)
逆变换得:N(t)∼p(λt)N(t) \sim p(\lambda t)N(t)∼p(λt)
3)每个事件到达时间独立,到达间隔时间T∼e(λ)T\sim e(\lambda)T∼e(λ)
第n次事件到达时间Sn=∑k=1nTk∼Γ(n,λ)S_n = \sum\limits_{k=1}^n T_k \sim \Gamma(n,\lambda)Sn=k=1∑nTk∼Γ(n,λ)P{T>t}=P{N(t)=0}=e−λtP\{T\gt t\} = P\{N(t)=0\}=e^{-\lambda t}P{T>t}=P{N(t)=0}=e−λt
收集k类奖券需要的抽奖次数ENENEN:
将抽奖过程描述为泊松分布p(1)p(1)p(1)
记Xi∼e(piλ)X_i\sim e(p_i\lambda)Xi∼e(piλ)为第iii类奖券首次出现的时间,X=maxiXiX = \max\limits_i X_iX=imaxXi即集齐全套的时间
P(X<t)=∏i(1−e−piλt)P(X\lt t)=\prod\limits_i (1-e^{-p_i\lambda t})P(X<t)=i∏(1−e−piλt)
EX=∫0∞P(X>t)dt=∫0∞(1−∏(1−e−piλt))dtEX = \int_0^\infin P(X\gt t)\mathrm{d}t=\int_0^\infin (1-\prod(1-e^{-p_i\lambda t}))\mathrm{d}tEX=∫0∞P(X>t)dt=∫0∞(1−∏(1−e−piλt))dt
又X=∑i=1NTi,TiX = \sum\limits_{i=1}^N T_i,T_iX=i=1∑NTi,Ti为第iii次到达间隔
得EX=ENETi=ENEX = ENET_i=ENEX=ENETi=EN
发生多于nnn次时间的概率∑k=n+1∞λkk!e−λ=∫0tλxe−λx(λx)nn!dx\sum\limits_{k=n+1}^\infin \frac{\lambda^k}{k!}e^{-\lambda}=\int_0^t\lambda xe^{-\lambda x}\frac{(\lambda x)^n}{n!}\mathrm{d} xk=n+1∑∞k!λke−λ=∫0tλxe−λxn!(λx)ndx第n+1n+1n+1次事件发生早于ttt的概率两独立泊松过程N1(t),N2(t)N_1(t),N_2(t)N1(t),N2(t),P{S1n<S2m}P\{S_{1n}<S_{2m}\}P{S1n<S2m}等价于投掷硬币nnn次正面先于mmm次反面的概率,其中出现正面的概率为λ1λ1+λ2\frac{\lambda_1}{\lambda_1+\lambda_2}λ1+λ2λ1,即∑k=nn+m−1Cn+m−1k(λ1λ1+λ2)k(λ2λ1+λ2)n+m−1−k\sum_{k=n}^{n+m-1}C_{n+m-1}^k(\frac{\lambda_1}{\lambda_1+\lambda_2})^k(\frac{\lambda_2}{\lambda_1+\lambda_2})^{n+m-1-k}∑k=nn+m−1Cn+m−1k(λ1+λ2λ1)k(λ1+λ2λ2)n+m−1−k
4)若事件以概率p1,…,pkp_1,\dots,p_kp1,…,pk分为kkk类,则Nk(t)∼p(λpit)N_k(t)\sim p(\lambda p_i t)Nk(t)∼p(λpit)且相独立
5)均匀到达性:
记次序统计量Y(i)Y_{(i)}Y(i)表示第iii小元素,则联合分布为:f(y1,…,yn)=n!∏k=1nf(yi)f(y_1,\dots,y_n)=n!\prod\limits_{k=1}^n f(y_i)f(y1,…,yn)=n!k=1∏nf(yi)。若Y(i)∼U[0,t]Y_{(i)}\sim U[0,t]Y(i)∼U[0,t],则f=n!tnf = \frac{n!}{t^n}f=tnn!
给定N(t)=nN(t)=nN(t)=n,则事件均匀独立到达
f(s1,…,sn)=λe−λs1⋅λe−λ(s2−s1)⋯λe−λ(sn−sn−1)⋅e−λ(t−sn)(λt)nn!e−λt=n!tnf(s_1,\dots,s_n)=\frac{\lambda e^{-\lambda s_1}\cdot\lambda e^{-\lambda (s_2-s_1)}\cdots\lambda e^{-\lambda (s_n-s_{n-1})}\cdot e^{-\lambda (t-s_n)} }{\frac{(\lambda t)^n}{n!}e^{-\lambda t}}=\frac{n!}{t^n}f(s1,…,sn)=n!(λt)ne−λtλe−λs1⋅λe−λ(s2−s1)⋯λe−λ(sn−sn−1)⋅e−λ(t−sn)=tnn!
用例:Si∼p(λ)S_i\sim p(\lambda)Si∼p(λ)表示理赔时间,CiC_iCi表示均值为μ\muμ,分布为GGG的理赔金额,计算D(t)=∑i=1N(t)e−αSiCiD(t)=\sum\limits_{i=1}^{N(t)}e^{-{\alpha S_i}}C_iD(t)=i=1∑N(t)e−αSiCi
ED(t)=∑n=0∞E[D(t)∣N(t)=n](λt)nn!e−λtED(t) =\sum\limits_{n=0}^\infin E[D(t)|N(t)=n]\frac{(\lambda t)^n}{n!}e^{-\lambda t}ED(t)=n=0∑∞E[D(t)∣N(t)=n]n!(λt)ne−λt
由于S1…SnS_1\dots S_nS1…Sn与次序统计量U1…UnU_{1}\dots U_nU1…Un同分布:
E[D(t)∣N(t)=n]=E∑i=1ne−αUiCi=μn∫0t1te−αxdxE[D(t)|N(t)=n] = E\sum\limits_{i=1}^ne^{-\alpha U_i}C_i=\mu n\int_0^t\frac{1}{t}e^{-\alpha x}\mathrm{d}xE[D(t)∣N(t)=n]=Ei=1∑ne−αUiCi=μn∫0tt1e−αxdx
得ED(t)=λμα(1−e−αt)ED(t) = \frac{\lambda \mu}{\alpha}(1-e^{-\alpha t})ED(t)=αλμ(1−e−αt)
推论:给定第nnn个事件时间tnt_ntn的条件下,前n−1n-1n−1个事件分布同n−1n-1n−1个U(0,tn)U(0,t_n)U(0,tn)随机量的分布相同
6)时间抽样
:若事件在时刻ttt以Pi(t)P_i(t)Pi(t)被分入iii类,则Ni(t)∼p(λ∫0tPi(x)dx)N_i(t)\sim p(\lambda\int^t_0P_i(x)\mathrm{d}x)Ni(t)∼p(λ∫0tPi(x)dx)
证明:
0−t0-t0−t时段内任意事件均匀到达,并以Pi(s)P_i(s)Pi(s)被分类,故Pi=1t∫0tPi(s)dsP_i=\frac{1}{t}\int^t_0P_i(s)\mathrm{d}sPi=t1∫0tPi(s)ds
故联合分布满足多项分布n!n1!⋯nk!P1n1⋯Pknk⋅(λt)nn!e−λt=∏k=1n(λtPi)nini!e−λtPi\frac{n!}{n_1!\cdots n_k!}P_1^{n_1}\cdots P_k^{n_k}\cdot \frac{(\lambda t)^n}{n!}e^{-\lambda t}=\prod\limits_{k=1}^n\frac{(\lambda tP_i)^{n_i}}{n_i!}e^{-\lambda tP_i}n1!⋯nk!n!P1n1⋯Pknk⋅n!(λt)ne−λt=k=1∏nni!(λtPi)nie−λtPi,即可分离形式
用例:在一段道路上车辆服从p(λ)p(\lambda)p(λ),通过的时间服从分布GGG,则t时刻仍在道路上的车辆数期望:
在s时进入的车辆有概率1−G(t−s)1-G(t-s)1−G(t−s)未能完成通行,G(t−s)G(t-s)G(t−s)完成,故所求为λ∫0t(1−G(t−s))ds\lambda\int^t_0 (1-G(t-s))\mathrm{d} sλ∫0t(1−G(t−s))ds
C3 广义泊松过程
1)非时齐泊松过程
:
定义:
N(0)=0N(0)=0N(0)=0具有独立增量非平稳增量:P{N(t+h)−N(t)=1}=λ(t)h+o(h)P\{N(t+h)-N(t)=1\}=\lambda(t)h+o(h)P{N(t+h)−N(t)=1}=λ(t)h+o(h)λ(t)\lambda(t)λ(t)称为强度函数
,m(t)=∫0tλ(x)dxm(t)=\int_0^t\lambda(x)\mathrm{d}xm(t)=∫0tλ(x)dx称均值函数
P{N(t+h)−N(t)≥2}=o(h)P\{N(t+h)-N(t)\ge 2\} = o(h)P{N(t+h)−N(t)≥2}=o(h)
增量分布:N(t+s)−N(s)∼p(m(t+s)−m(s))N(t+s)-N(s)\sim p(m(t+s)-m(s))N(t+s)−N(s)∼p(m(t+s)−m(s))
证明:仿照泊松过程证明有∂g∂t(s,t)=g(s,t)λ(t)(e−s−1)\frac{\partial g}{\partial t}(s,t) =g(s,t)\lambda(t)(e^{-s}-1)∂t∂g(s,t)=g(s,t)λ(t)(e−s−1)
泊松过程的时间抽样为非泊松过程
第nnn个事件到达时间:∼λ(t)e−m(t)m(t)n−1(n−1)!\sim \lambda(t)e^{-m(t)}\frac{m(t)^{n-1}}{(n-1)!}∼λ(t)e−m(t)(n−1)!m(t)n−1
2)复合泊松过程
:X(t)=∑i=1N(t)Yi,N(t)∼p(λ),YiX(t) = \sum\limits_{i=1}^{N(t)}Y_i,N(t)\sim p(\lambda),Y_iX(t)=i=1∑N(t)Yi,N(t)∼p(λ),Yi独立同分布
EX(t)=λtEY,DX(t)=λtEY2EX(t)=\lambda tEY,DX(t)=\lambda tEY^2EX(t)=λtEY,DX(t)=λtEY2
单服务台单队列不离队队列忙期,泊松到达,首个顾客服务时间为SSS
记N(S)N(S)N(S)为首个顾客服务期间的到达数
N(S)=1N(S) = 1N(S)=1时,易见T=S+T1,T1T = S+T_1,T_1T=S+T1,T1与TTT同分布
N(S)≠1N(S)\neq 1N(S)=1可认为在某顾客服务期间到达的人全部插队到其后一个位置,
T=S+∑i=1N(S)Ti,TiT = S + \sum\limits_{i=1}^{N(S)} T_i,T_iT=S+i=1∑N(S)Ti,Ti与TTT同分布
ET=ES=+λES⋅ET=ES1−λESET = ES =+ \lambda ES\cdot ET = \frac{ES}{1-\lambda ES}ET=ES=+λES⋅ET=1−λESES
若YYY为可数集,则当ttt增大时X(t)X(t)X(t)趋近于正态分布N(λtEY,λtEY2)N(\lambda tEY,\lambda tEY^2)N(λtEY,λtEY2)
若X(t),Y(t)X(t),Y(t)X(t),Y(t)是具有到达率λ1,λ2\lambda_1,\lambda_2λ1,λ2和分布F1,F2F_1,F_2F1,F2的独立复合泊松过程,则X(t)+Y(t)X(t)+Y(t)X(t)+Y(t)是具有到达率λ1+λ2\lambda_1+\lambda_2λ1+λ2和分布λ1λ1+λ2F1+λ2λ1+λ2F2\frac{\lambda_1}{\lambda_1+\lambda_2}F_1+\frac{\lambda_2}{\lambda_1+\lambda_2}F_2λ1+λ2λ1F1+λ1+λ2λ2F2的泊松过程(事件以概率λiλ1+λ2\frac{\lambda_i}{\lambda_1+\lambda_2}λ1+λ2λi来自FiF_iFi)
3)条件(混合)泊松过程
:λ\lambdaλ为一个正随机量LLL的取值
P{N(t+s)−N(s)=n}=∫0∞P{N(t+s)−N(s)=n∣L=l}P{L=l}dl=∫0∞lnn!e−lP{L=l}dlP\{N(t+s)-N(s)=n\}=\int_0^\infin P\{N(t+s)-N(s)=n|L=l\}P\{L=l\}\mathrm{d}l=\int_0^\infin \frac{l^n}{n!}e^{-l}P\{L=l\}\mathrm{d}lP{N(t+s)−N(s)=n}=∫0∞P{N(t+s)−N(s)=n∣L=l}P{L=l}dl=∫0∞n!lne−lP{L=l}dl由于一个时间段内计数可能反映LLL,一般不具有独立增量EN(t)=tEL;DN(t)=tEL+t2DLEN(t) = tEL;DN(t) = tEL+t^2DLEN(t)=tEL;DN(t)=tEL+t2DL
C4 霍克斯过程
1)随机强度
:到达率是依赖于ttt前历史的随机变量R(t)R(t)R(t)
2)霍克斯过程
:R(t)=λ+∑i=1N(t)Miλe−α(t−Si)R(t) = \lambda + \sum\limits_{i=1}^{N(t)}M_i\lambda e^{-\alpha(t-S_i)}R(t)=λ+i=1∑N(t)Miλe−α(t−Si)
R(0)=λR(0) = \lambdaR(0)=λ,基础强度每个事件到达将增强随机强度,增强量MiM_iMi取决于事件增强量随时间指数衰减
3)EN(t)=∫0tER(s)dsEN(t) = \int_0^t ER(s)\mathrm{d}sEN(t)=∫0tER(s)ds
若EM=μEM=\muEM=μ,则EN(t)=λt+λμ(μ−α)2(e(μ−α)t−1−(μ−α)t);ER(t)=λ+λμμ−α(e(μ−α)t−1)EN(t)=\lambda t+\frac{\lambda\mu}{(\mu-\alpha)^2}(e^{(\mu-\alpha)t}-1-(\mu-\alpha)t);ER(t) =\lambda+\frac{\lambda\mu}{\mu-\alpha}(e^{(\mu-\alpha)t}-1)EN(t)=λt+(μ−α)2λμ(e(μ−α)t−1−(μ−α)t);ER(t)=λ+μ−αλμ(e(μ−α)t−1)证明:
E[N(t+h)∣N(t),R(t)]=N(t)+R(t)h+o(h)E[N(t+h)|N(t),R(t)] = N(t) + R(t)h + o(h)E[N(t+h)∣N(t),R(t)]=N(t)+R(t)h+o(h)
两边取期望得EN(t+h)=EN(t)+ER(t)h+o(h)EN(t+h) = EN(t)+ER(t)h+o(h)EN(t+h)=EN(t)+ER(t)h+o(h)
(EN(t))′=ER(t)(EN(t))'=ER(t)(EN(t))′=ER(t)