随机过程学习笔记1:泊松过程

2020年06月05日 阅读数:447
这篇文章主要向大家介绍随机过程学习笔记1:泊松过程,主要内容包括基础应用、实用技巧、原理机制等方面,希望对大家有所帮助。

随机过程的通常概念

随机过程 T T ( , ) (-\infty,\infty) 的子集,对于每一个 t T t\in T X t X_t 是随机变量,则称随机变量的集合 { X t t T } \{X_t|t\in T\} 是随机过程, T T 是随机过程的 \underline{指标集}
随机过程的观测或实现:对每一个 t T t\in T ,都获得了 X t X_t 的观测值 x t x_t ,称 { x t t T } \{x_t|t\in T\} { X t t T } \{X_t|t\in T\} 是随机过程的一次实现或一次观测
轨迹或轨道:当 T T [ 0 , ) [0,\infty) ( , ) (-\infty,\infty) 时, { X t t T } \{X_t|t\in T\} 的一次观测又称为是一条轨道或一条轨迹
时间序列 T = N T=N T = Z T=Z 时,随机过程又称为时间序列html

计数过程的通常概念

计数过程:用来描述一段时间内发生的某类事件的次数的随机过程,计数过程 { N ( t ) } \{N(t)\} 定义为: N ( t ) N(t) ( 0 , t ] (0,t] 时间内某事件发生的次数,知足:
(1)对 t 0 t\ge 0 N ( t ) N(t) 是取非负整数值的随机变量
(2) t s t\ge s 时, N ( t ) N ( s ) 0 N(t)\ge N(s)\ge 0
(3) N ( t ) N ( s ) N(t)-N(s) 表示 ( s , t ] (s,t] 时间段内该事件发生的次数
(4) N ( t ) N(t) 的轨迹是单调不减右连续的阶梯函数
举例:计数过程能够表示某段时间内收到的手机短信数,某段时间内到达某个车站的乘客数目,某段时间内商场经理收到商品质量的投诉数,某段时间内股票市场股票成交的次数web

独立增量性:对于计数过程 { N ( t ) } \{N(t)\} ,若是任取 0 < t 1 < t 2 < < t n 0<t_1<t_2<\cdots<t_n ,随机变量 N ( t 1 ) , N ( t 2 ) N ( t 1 ) , , N ( t n ) N ( t n 1 ) N(t_1),N(t_2)-N(t_1),\cdots,N(t_n)-N(t_{n-1}) 是相互独立的,则称计数过程 { N ( t ) } \{N(t)\} 具备独立增量性,该计数过程称为独立增量过程app

平稳增量性:若是对计数过程 { N ( t ) } \{N(t)\} ,对任何的非负实数 s s ,随机变量
N ( t 2 ) N ( t 1 ) , N ( t 2 + s ) N ( t 1 + s ) N(t_2)-N(t_1),N(t_2+s)-N(t_1+s) 同分布,则称该计数过程具备平稳增量性,该计数过程称为平稳增量过程svg

泊松过程

泊松过程的定义

泊松过程 称知足如下条件的计数过程 { N ( t ) } \{N(t)\} 是强度为 λ \lambda 的泊松过程:
(1) { N ( t ) } \{N(t)\} 是知足 N ( 0 ) = 0 N(0)=0 的独立增量过程和平稳增量过程
(2) N ( t ) N(t) 知足参数为 λ t \lambda t 的泊松分布函数

强度的含义 E N ( t ) = λ t EN(t)=\lambda t ,故单位时间内发生的平均事件数为 E N ( t ) t = λ \frac{EN(t)}{t}=\lambda 所以,强度的实际含义是单位时间内发生时间的平均次数spa

等价定义:若是计数过程 { N ( t ) } \{N(t)\} 知足:
(1) N ( 0 ) = 0 N(0)=0
(2) { N ( t ) } \{N(t)\} 是独立增量过程和平稳增量过程
(3) { N ( t ) } \{N(t)\} 知足普通性,即: { P ( N ( h ) = 1 ) = λ h + o ( h ) P ( N ( h ) 2 ) = o ( h ) \begin{cases} P(N(h)=1)=\lambda h+o(h)\\ P(N(h)\ge 2)=o(h) \end{cases} { N ( t ) } \{N(t)\} 是强度为 λ \lambda 的泊松过程(通俗地讲是,在极短的时间内,发生一次时间的几率同时间几乎成正比,几乎不可能发生两次或以上事件)orm

证:
f 0 ( h ) = P { N ( h ) = 0 } = 1 P { N ( h ) = 1 } P { N ( h ) 2 } = 1 λ h + o ( h ) f_0(h)=P\{N(h)=0\}=1-P\{N(h)=1\}-P\{N(h)\ge 2\}=1-\lambda h+o(h) ,则当 Δ h > 0 \Delta h>0 f 0 ( h + Δ h ) = P { N ( h + Δ h ) = 0 } = P { N ( h ) = 0 , N ( h + Δ h ) N ( h ) = 0 } = P { N ( h ) = 0 } P { N ( Δ h ) = 0 } = f 0 ( h ) f 0 ( Δ h ) \begin{aligned} &f_0(h+\Delta h)=P\{N(h+\Delta h)=0\}\\=&P\{N(h)=0,N(h+\Delta h)-N(h)=0\}\\ =&P\{N(h)=0\}P\{N(\Delta h)=0\}\\ =&f_0(h)f_0(\Delta h) \end{aligned} 上式的第二个等号,能够从两个角度理解
第一个角度,计数过程的取值都是非负整数,且知足单调性, Δ h > 0 \Delta h>0 ,而 N ( h + Δ h ) = N ( h ) + N ( h + Δ h ) N ( h ) N(h+\Delta h)=N(h)+N(h+\Delta h)-N(h) ,因为 N ( h ) , N ( h + Δ h ) N ( h ) N(h),N(h+\Delta h)-N(h) 都取非负整数值,二者之和为零的充要条件是二者均为0,若是 Δ h < 0 \Delta h<0
第二个角度,从计数过程的定义来看, ( 0 , h + Δ h ] (0,h+\Delta h] 这段时间内没有任何事件发生的充要条件固然是 ( 0 , h ] , ( h , h + Δ h ] (0,h],(h,h+\Delta h] 时间段内都没有任何事件发生,这两段时间内只要有任何一段发生了一次事件,那么 ( 0 , h + Δ h ] (0,h+\Delta h] 这段时间内都至少会发生1次事件
因而,当 Δ h > 0 \Delta h>0 时,就有 f 0 ( h + Δ h ) f 0 ( h ) = [ f 0 ( Δ h ) 1 ] f ( h ) = [ λ Δ h + o ( Δ h ) ] f 0 ( h ) f_0(h+\Delta h)-f_0(h)=[f_0(\Delta h)-1]f(h)=[-\lambda \Delta h+o(\Delta h)]f_0(h) 两边除以 Δ h \Delta h ,再令 Δ h 0 + \Delta h\to 0^+ ,便可证得 f ( h ) f(h) 的右导数存在,而且 f 0 + ( h ) = λ f 0 ( h ) f_0^{+}(h)=-\lambda f_0(h) 若是 Δ h < 0 \Delta h<0 ,那么 f 0 ( h ) = f 0 ( h + Δ h ) f 0 ( Δ h ) f_0(h)=f_0(h+\Delta h)f_0(-\Delta h) 所以 f 0 ( h + Δ h ) f 0 ( h ) = f 0 ( h + Δ h ) [ 1 f 0 ( Δ h ) ] = f 0 ( h + Δ h ) [ λ Δ h + o ( Δ h ) ] f_0(h+\Delta h)-f_0(h)=f_0(h+\Delta h)[1-f_0(-\Delta h)]=f_0(h+\Delta h)[-\lambda \Delta h+o(\Delta h)] 由上式不可贵出 lim Δ 0 f 0 ( h + Δ h ) = f ( h ) \displaystyle \lim_{\Delta \to 0^-}f_0(h+\Delta h)=f(h) ,两边除以 Δ h \Delta h ,再令 Δ h 0 \Delta h\to 0^- ,便可证得 f ( h ) f(h) 的左导数也存在,而且 f 0 ( h ) = λ f 0 ( h ) f^{-}_0(h)=-\lambda f_0(h) f ( h ) f(h) h = 0 h=0 处可导,且知足微分方程 f 0 ( h ) = λ f 0 ( h ) f_0^\prime(h)=-\lambda f_0(h) 方程的通解为 f 0 ( h ) = C e λ h f_0(h)=Ce^{-\lambda h} h = 0 h=0 f 0 ( h ) = 1 f_0(h)=1 ,故 C = 1 C=1 ,从而获得 P { N ( h ) = 0 } = e λ h P\{N(h)=0\}=e^{-\lambda h} 。接下来就能够求解 P { N ( h ) = 1 } P\{N(h)=1\} ,一样地,咱们令 f 1 ( h ) = P { N ( h ) = 1 } f_1(h)=P\{N(h)=1\} ,当 Δ h > 0 \Delta h>0 f 1 ( h + Δ h ) = f 1 ( h ) f 0 ( Δ h ) + f 1 ( Δ h ) f 0 ( h ) f_1(h+\Delta h)=f_1(h)f_0(\Delta h)+f_1(\Delta h)f_0(h) 所以 f 1 ( h + Δ h ) f 1 ( h ) = f 1 ( h ) [ f 0 ( Δ h ) 1 ] + f 1 ( Δ h ) f 0 ( h ) = f 1 ( h ) [ λ Δ h + o ( Δ h ) ] + [ λ Δ h + o ( Δ h ) ] e λ h \begin{aligned} &f_1(h+\Delta h)-f_1(h)=f_1(h)[f_0(\Delta h)-1]+f_1(\Delta h)f_0(h)\\ =&f_1(h)[-\lambda \Delta h +o(\Delta h)]+[\lambda \Delta h+o(\Delta h)]e^{-\lambda h} \end{aligned} 两边除以 Δ h \Delta h ,令 Δ 0 + \Delta \to 0^+ ,便可证得 f 1 ( h ) f_1(h) h h 处存在右导数,而且 f 1 + ( h ) = λ f 1 ( h ) + λ e λ h f^+_1(h)=-\lambda f_1(h)+\lambda e^{-\lambda h} 若是 Δ h < 0 \Delta h<0 ,有 f 1 ( h + Δ h ) f ( h ) = f 1 ( h + Δ h ) [ 1 f 0 ( Δ h ) ] f 1 ( Δ h ) f 0 ( h + Δ h ) = [ λ Δ h + o ( Δ h ) ] f 1 ( h + Δ h ) [ λ Δ h + o ( Δ h ) ] f 0 ( h + Δ h ) \begin{aligned} &f_1(h+\Delta h)-f(h)\\ =&f_1(h+\Delta h)[1-f_0(-\Delta h)]-f_1(-\Delta h)f_0(h+\Delta h)\\ =&[-\lambda \Delta h +o(\Delta h)]f_1(h+\Delta h)-[-\lambda \Delta h+o(\Delta h)]f_0(h+\Delta h) \end{aligned} 同理可得 f 1 ( h ) f_1^{-}(h) 存在,而且 f 1 ( h ) = λ f 1 ( h ) + λ e λ h f_1^{-}(h)=-\lambda f_1(h)+\lambda e^{-\lambda h} 因而 f 1 ( h ) f_1(h) h h 处可导,而且知足微分方程 f 1 ( h ) = λ f 1 ( h ) + λ e λ h f_1^\prime(h)=-\lambda f_1(h)+\lambda e^{-\lambda h} 同时 f 1 ( 0 ) = 0 f_1(0)=0 ,方程组的通解为 f 1 ( h ) = λ h e λ h + C e λ h f_1(h)=\lambda h e^{-\lambda h}+Ce^{-\lambda h} f 1 ( 0 ) = 0 f_1(0)=0 ,获得 C = 0 C=0 f 1 ( h ) = λ h e λ h f_1(h)=\lambda h e^{-\lambda h} ,接下来咱们用数学概括法完成接下来的证实,假设对 k n k\le n ,都有 P { N ( h ) = k } = ( λ h ) k k ! e λ h P\{N(h)=k\}=\frac{(\lambda h)^k}{k!}e^{-\lambda h} 咱们来求 f n + 1 ( h ) = P { N ( h ) = n + 1 } f_{n+1}(h)=P\{N(h)=n+1\} ,咱们这里设 f k ( h ) = ( λ h ) k k ! e λ h , k = 1 , , n f_k(h)=\frac{(\lambda h)^k}{k!}e^{-\lambda h},k=1,\cdots,n ,一样地,当 Δ h > 0 \Delta h>0 时,有 f n + 1 ( h + Δ h ) = k = 0 n + 1 f k ( h ) f n + 1 k ( Δ h ) f_{n+1}(h+\Delta h)=\sum_{k=0}^{n+1}f_k(h)f_{n+1-k}(\Delta h) f n + 1 ( h + Δ h ) f n + 1 ( h ) = f n + 1 ( h ) ( 1 f 0 ( Δ h ) ) + k = 0 n f k ( h ) f n + 1 k ( Δ h ) f_{n+1}(h+\Delta h)-f_{n+1}(h)=f_{n+1}(h)(1-f_0(\Delta h))+\sum_{k=0}^nf_k(h)f_{n+1-k}(\Delta h) 两边除以 Δ h \Delta h ,再令 Δ h 0 + \Delta h\to 0^+ ,可证得 f n + 1 ( h ) f_{n+1}(h) h h 处右导数存在,而且 f n + 1 + ( h ) = λ f n + 1 ( h ) + λ f n ( h ) f_{n+1}^{+}(h)=\lambda f_{n+1}(h)+\lambda f_n(h) 模仿上面的证实,能够证得 f n + 1 f_{n+1} h h 处左导数也存在,而且 f n + 1 ( h ) = λ f n + 1 ( h ) + λ f n ( h ) f_{n+1}^{-}(h)=\lambda f_{n+1}(h)+\lambda f_n(h) f n + 1 ( h ) = λ f n + 1 ( h ) + λ n + 1 h n n ! e λ h f_{n+1}^\prime(h)=\lambda f_{n+1}(h)+\frac{\lambda^{n+1}h^n}{n!}e^{-\lambda h} 该微分方程的通解为 f n + 1 ( h ) = ( λ h ) n + 1 ( n + 1 ) ! e λ h + C e λ h f_{n+1}(h)=\frac{(\lambda h)^{n+1}}{(n+1)!}e^{-\lambda h}+C e^{-\lambda h} 再由 f n + 1 ( 0 ) = 0 f_{n+1}(0)=0 ,得 C = 0 C=0 ,故 f n + 1 ( h ) = ( λ h ) n + 1 ( n + 1 ) ! e λ h f_{n+1}(h)=\frac{(\lambda h)^{n+1}}{(n+1)!}e^{-\lambda h} 由概括法,对任意的 n = 1 , 2 , n=1,2,\cdots ,就有 P { N ( h ) = n } = ( λ t ) n n ! e λ t P\{N(h)=n\}=\frac{(\lambda t)^n}{n!}e^{-\lambda t} { N ( t ) } \{N(t)\} 是泊松过程xml

泊松呼叫流、等待时间、年龄、寿命

先给出这几个概念,再给出他们的分布。假设对 { N ( t ) } \{N(t)\} 对应的事件流,第一次事件发生的时间点为 S 1 S_1 S 2 S_2 S 3 S_3 \cdots ,咱们将其画在时间轴上,即
在这里插入图片描述
泊松呼叫流:计数过程 { N ( t ) } \{N(t)\} 对应的各事件发生的时刻 { S n } \{S_n\} { S n } \{S_n\} 是单调递增的非负时间序列,由泊松呼叫流的一次实现能够获得 { N ( t ) } \{N(t)\} 的一次实现,因为 N ( t ) N(t) 的含义是在 ( 0 , t ] (0,t] 内发生的事件的次数,故咱们只须要数 ( 0 , t ] (0,t] 时间内有多少个 S i S_i ,若是 N ( t ) = n N(t)=n 说明: S n t S_n\le t S n + 1 > t S_{n+1}>t ,咱们获得如下的基本关系式
{ N ( t ) < n } = { S n > t } { N ( t ) = n } = { S n t < S n + 1 } \{N(t)< n\}=\{S_n>t\}\\ \{N(t)=n\}=\{S_n\le t < S_{n+1}\}
等待时间:即相邻两次事件发生的时间间隔,即 X i = S i S i 1 , i = 1 , 2 , X_i=S_{i}-S_{i-1},i=1,2,\cdots ,这里定义 S 0 = 0 S_0=0
年龄和寿命:咱们把 S 1 , S 2 , S_1,S_2,\cdots 想象成更换某种零件的时间点,具体地讲,在0时刻第一个零件开始运做,在 S 1 S_1 时刻第一个零件损坏,当即更换上第二个零件,在 S 2 S_2 时刻,第二个零件损坏,当即更换上第三个零件,以此类推。假设 N ( t ) = n N(t)=n ,此时已经更换了 n n 次零件,即此时运做的是机器的第 n + 1 n+1 个零件,那么 S n S_n 是这个零件开始运做的时刻, S n + 1 S_{n+1} 是这个零件损坏的时刻,那么,这个零件的年龄是 t S n t-S_n ,剩余寿命是 S n + 1 t S_{n+1}-t ,将 n n 换成 N ( t ) N(t) ,年龄就是 t S N ( t ) t-S_{N(t)} ,寿命就是 S N ( t ) + 1 t S_{N(t)+1}-t ,前者咱们记为 A ( t ) A(t) ,后者咱们记为 R ( t ) R(t) htm

呼叫流的联合分布

如今咱们来求呼叫流 ( S 1 , , S n ) (S_1,\cdots,S_n) 的分布,首先求单个 S n S_n 的分布:实际上 { S n t } = { N ( t ) n } \{S_n\le t\}=\{N(t)\ge n\} ,因而就获得 S n S_n 的分布函数 F n ( t ) = k = n ( λ t ) k k ! e λ t F_n(t)=\sum_{k=n}^\infty \frac{(\lambda t)^k}{k!}e^{-\lambda t} 求导便可获得其密度函数 p n ( t ) = k = n k λ k t k 1 k ! e λ t λ k = n ( λ t ) k k ! e λ t = λ k = n ( λ t ) k 1 ( k 1 ) ! e λ t λ k = n ( λ t ) k k ! e λ t = λ n Γ ( n ) t n 1 e λ t \begin{aligned} p_n(t)=&\sum_{k=n}^\infty k\frac{\lambda^k t^{k-1}}{k!}e^{-\lambda t}-\lambda\sum_{k=n}^\infty \frac{(\lambda t)^k}{k!}e^{-\lambda t}\\ =&\lambda \sum_{k=n}^\infty \frac{(\lambda t)^{k-1}}{(k-1)!}e^{-\lambda t}-\lambda \sum_{k=n}^\infty \frac{(\lambda t)^{k}}{k!}e^{-\lambda t}\\ =&\frac{\lambda^n}{\Gamma(n)}t^{n-1}e^{-\lambda t} \end{aligned} 可见 S n S_n 服从伽马分布,参数为 n , λ n,\lambda ,其均值为 n λ \frac{n}{\lambda} ,接下来咱们来求 ( S 1 , S 2 , , S n ) (S_1,S_2,\cdots,S_n) 的联合分布:咱们假设其分布函数为 F ( s 1 , s 2 , , s n ) = P { S 1 s 1 , S 2 s 2 , , S n s n } F(s_1,s_2,\cdots,s_n)=P\{S_1\le s_1,S_2\le s_2,\cdots,S_n\le s_n\} 若是 i < j i<j ,但 s i > s j s_i>s_j ,那么 P { S 1 s 1 , , S i > s i , , S j s j , , S n s n } = P { S 1 s 1 , , S i 1 s i 1 , S i + 1 s i + 1 , , S n s n } F ( x 1 , x 2 , , x n ) = 0 \begin{aligned} &P\{S_1\le s_1,\cdots,S_i>s_i,\cdots,S_j\le s_j,\cdots,S_n\le s_n\}\\ =&P\{S_1\le s_1,\cdots,S_{i-1}\le s_{i-1},S_{i+1}\le s_{i+1},\cdots ,S_n\le s_n\}-\\ &F(x_1,x_2,\cdot,x_n)=0 \end{aligned} 两边对 x 1 , , x n x_1,\cdots,x_n 依次求偏导,则 p ( x 1 , , x n ) = 0 p(x_1,\cdots,x_n)=0 所以,咱们假设 s 1 s 2 s 3 s n s_1\le s_2\le s_3\cdots\le s_n ,咱们求分布函数 F ( s 1 , , s n ) F(s_1,\cdots,s_n) ,以 n = 2 n=2 为例, s 1 < s 2 s_1< s_2 ,则 F ( s 1 , s 2 ) = P { S 1 s 1 , S 2 s 2 } = P { S 2 s 2 } P { S 1 > s 1 , S 2 s 2 } \begin{aligned} F(s_1,s_2)=&P\{S_1\le s_1,S_2\le s_2\}\\ =&P\{S_2\le s_2\}-P\{S_1>s_1,S_2\le s_2\} \end{aligned} 而事件 { S 1 > s 1 , S 2 s 2 } \{S_1>s_1,S_2\le s_2\} 至关于在 ( 0 , s 1 ] (0,s_1] 段事件发生了0次, ( s 1 , s 2 ] (s_1,s_2] 段事件至少发生了两次,有 P { S 1 > s 1 , S 2 s 2 } = P { N ( s 1 ) = 0 , N ( s 2 ) N ( s 1 ) 2 } = P { N ( s 1 ) = 0 } P { N ( s 2 s 1 ) 2 } = e λ s 1 ( k = 2 λ k ( s 2 s 1 ) k k ! ) e λ ( s 2 s 1 ) = ( k = 2 λ k ( s 2 s 1 ) k k ! ) e λ s 2 \begin{aligned} P\{S_1>s_1,S_2\le s_2\}=&P\{N(s_1)=0,N(s_2)-N(s_1)\ge 2\}\\ =&P\{N(s_1)=0\}P\{N(s_2-s_1)\ge 2\}\\ =&e^{-\lambda s_1}(\sum_{k=2}^\infty\frac{\lambda^k(s_2-s_1)^k}{k!})e^{-\lambda(s_2-s_1)}\\ =&(\sum_{k=2}^\infty\frac{\lambda^k(s_2-s_1)^k}{k!})e^{-\lambda s_2} \end{aligned} 则密度函数 p ( s 1 , s 2 ) = 2 s 1 s 2 F ( s 1 , s 2 ) = λ 2 e λ s 2 p(s_1,s_2)=\frac{\partial^2}{\partial s_1\partial s_2}F(s_1,s_2)=\lambda^2e^{-\lambda s_2} 所以 ( S 1 , S 2 ) (S_1,S_2) 的密度函数为 p ( s 1 , s 2 ) = λ 2 e λ s 2 I s 1 < s 2 p(s_1,s_2)=\lambda^2e^{-\lambda s_2}I_{s_1<s_2} ,再求 n = 3 n=3 的情形,设 s 1 < s 2 < s 3 s_1< s_2< s_3 ,那么 F ( s 1 , s 2 , s 3 ) = P { S 1 s 1 , S 2 s 2 , S 3 s 3 } = P { S 2 s 2 , S 3 s 3 } P { S 1 > s 1 , S 2 s 2 , S 3 s 3 } = P { S 2 s 2 , S 3 s 3 } P { S 1 > s 1 , S 2 s 2 } + P { S 1 > s 1 , S 2 s 2 , S 3 > s 3 } \begin{aligned} F(s_1,s_2,s_3)=&P\{S_1\le s_1,S_2\le s_2,S_3\le s_3\}\\ =&P\{S_2\le s_2,S_3\le s_3\}-P\{S_1>s_1,S_2\le s_2,S_3\le s_3\}\\ =&P\{S_2\le s_2,S_3\le s_3\}-P\{S_1>s_1,S_2\le s_2\}+\\ &P\{S_1>s_1,S_2\le s_2,S_3>s_3\} \end{aligned} 从下图能够看出 { S 1 > s 1 , S 2 s 2 , S 3 > s 3 } = { N ( s 1 ) = 0 , N ( s 2 ) N ( s 1 ) = 2 , N ( s 3 ) N ( s 2 ) = 0 } \{S_1>s_1,S_2\le s_2,S_3>s_3\}=\{N(s_1)=0,N(s_2)-N(s_1)=2,N(s_3)-N(s_2)=0\} 在这里插入图片描述
P { S 1 > s 1 , S 2 s 2 , S 3 > s 3 } = P { N ( s 1 ) = 0 } P { N ( s 2 ) N ( s 1 ) = 2 } P { N ( s 3 ) N ( s 2 ) = 0 } = e λ s 1 λ 2 ( s 2 s 1 ) 2 2 e λ ( s 2 s 1 ) e λ ( s 3 s 2 ) = λ 2 ( s 2 s 1 ) 2 2 e λ s 3 \begin{aligned} &P\{S_1>s_1,S_2\le s_2,S_3>s_3\}\\ =&P\{N(s_1)=0\}P\{N(s_2)-N(s_1)=2\}P\{N(s_3)-N(s_2)=0\}\\ =&e^{-\lambda s_1}\frac{\lambda^2(s_2-s_1)^2}{2}e^{-\lambda(s_2-s_1)}e^{-\lambda(s_3-s_2)}=\frac{\lambda^2(s_2-s_1)^2}{2}e^{-\lambda s_3} \end{aligned} 一样地方法能够获得 ( S 1 , S 2 , S 3 ) (S_1,S_2,S_3) 的密度函数为 p ( s 1 , s 2 , s 3 ) = λ 3 e λ s 3 I s 1 < s 2 < s 3 p(s_1,s_2,s_3)=\lambda^3e^{-\lambda s_3}I_{s_1<s_2<s_3} n = 2 n=2 n = 3 n=3 情形已经给出求 ( S 1 , , S n ) (S_1,\cdots,S_n) 的通常方法,当 n = 2 k + 1 n=2k+1 s 1 < s 2 < < s 2 k s_1<s_2<\cdots<s_{2k} 时,首先,比较容易求出的是如下几率 G k ( s 1 , s 2 , , s 2 k ) = P { S 1 > s 1 , S 2 s 2 , S 3 > s 3 , S 4 s 4 , , S 2 k 1 > s 2 k 1 , S 2 k s 2 k } = i = 1 n P { N ( s 2 i 1 ) N ( s 2 i 2 ) = 0 } . i = 1 n 1 P { N ( s 2 i ) N ( s 2 i 1 ) = 2 } × P { N ( s 2 k ) N ( s 2 k 1 ) 2 } = i = 1 n 1 λ 2 ( s 2 i s 2 i 1 ) 2 2 ( j = 2 λ j ( s 2 k s 2 k 1 ) j j ! ) e λ s 2 k \begin{aligned} &G_k(s_1,s_2,\cdots,s_{2k})\\ =&P\{S_1>s_1,S_2\le s_2,S_3>s_3,S_4\le s_4,\cdots,S_{2k-1}>s_{2k-1},S_{2k}\le s_{2k}\}\\ =&\prod_{i=1}^nP\{N(s_{2i-1})-N(s_{2i-2})=0\}.\prod_{i=1}^{n-1}P\{N(s_{2i})-N(s_{2i-1})=2\}\\ \times&P\{N(s_{2k})-N(s_{2k-1})\ge 2\}\\ =&\prod_{i=1}^{n-1}\frac{\lambda^2(s_{2i}-s_{2i-1})^2}{2}(\sum_{j=2}^\infty\frac{\lambda^j(s_{2k}-s_{2k-1})^j}{j!})e^{-\lambda s_{2k}} \end{aligned} 其中规定 s 0 = 0 s_0=0 ,先对 s 1 , , s 2 k 2 s_1,\cdots,s_{2k-2} 依次求偏导 2 k 2 s 1 s 2 s 2 k 2 G k = λ 2 k 2 ( j = 2 λ j ( s 2 k s 2 k 1 ) j j ! ) e λ s 2 k \frac{\partial^{2k-2}}{\partial s_1\partial s_2\cdots\partial s_{2k-2}}G_k=\lambda^{2k-2}(\sum_{j=2}^\infty\frac{\lambda^j(s_{2k}-s_{2k-1})^j}{j!})e^{-\lambda s_{2k}} 这样对其他两个变元求偏导就轻松不少,具体过程再也不详述,最终获得 2 k s 1 s 2 s 2 k G k = ( 1 ) k λ 2 k e λ s 2 k \frac{\partial^{2k}}{{\partial s_1\partial s_2\cdots\partial s_{2k}}}G_k=(-1)^k\lambda^{2k}e^{-\lambda s_{2k}} 如今,咱们令 G 1 ( s 1 , s 2 , , s n ) = P { S 1 > s 1 , S 2 s 2 , S 3 s 3 , , S 2 k s 2 k } G 2 ( s 1 , s 2 , , s n ) = P { S 1 > s 1 , S 2 s 2 , S 3 > s 3 , S 4 s 4 , S 5 s 5 , , S 2 k s 2 k } G i ( s 1 , s 2 , , s n ) = P { S 1 > s 1 , S 2 s 2 , , S 2 i 1 > s 2 i 1 , S j s j , n j > 2 i 1 } G k ( s 1 , s 2 , , s n ) = P { S 1 > s 1 , S 2 s 2 , S 3 > s 3 , S 4 s 4 , , S 2 k 1 > s 2 k 1 , S 2 k s 2 k } \begin{aligned} G_1(s_1,s_2,\cdots,s_n)=&P\{S_1>s_1,S_2\le s_2,S_3\le s_3,\cdots,S_{2k}\le s_{2k}\}\\ G_2(s_1,s_2,\cdots,s_n)=&P\{S_1>s_1,S_2\le s_2,S_3>s_3,S_4\le s_4,S_5\le s_5,\cdots,S_{2k}\le s_{2k}\}\\ \cdots\\ G_i(s_1,s_2,\cdots,s_n)=&P\{S_1>s_1,S_2\le s_2,\cdots,S_{2i-1}>s_{2i-1},S_j\le s_j,n\ge j>2i-1\}\\ \cdots\\ G_k(s_1,s_2,\cdots,s_n)=&P\{S_1>s_1,S_2\le s_2,S_3>s_3,S_4\le s_4,\cdots,S_{2k-1}>s_{2k-1},S_{2k}\le s_{2k}\} \end{aligned} 因而 F ( s 1 , s 2 , , s n ) = P { S 2 s 2 , S 3 s 3 , , S 2 k s 2 k } G 1 ( s 1 , , s n ) G 1 ( s 1 , , s n ) = P { S 1 > s 1 , S 2 s 2 , S 4 s 4 , , S 2 k s 2 k } G 2 ( s 1 , , s n ) F(s_1,s_2,\cdots,s_n)=P\{S_2\le s_2,S_3\le s_3,\cdots,S_{2k}\le s_{2k}\}-G_1(s_1,\cdots,s_n)\\ G_1(s_1,\cdots,s_n)=P\{S_1>s_1,S_2\le s_2,S_4\le s_4,\cdots,S_{2k}\le s_{2k}\}-G_2(s_1,\cdots,s_n)\\ \cdots 这样看 F F G k G_k 的偏导数的关系应该是 2 k s 1 s 2 s 2 k F = ( 1 ) k 2 k s 1 s 2 s 2 k G k \frac{\partial^{2k}}{{\partial s_1\partial s_2\cdots\partial s_{2k}}}F=(-1)^k\frac{\partial^{2k}}{{\partial s_1\partial s_2\cdots\partial s_{2k}}}G_k 从而获得 ( S 1 , , S 2 k ) (S_1,\cdots,S_{2k}) 的联合密度应该是 p ( s 1 , , s n ) = λ 2 k e λ s 2 k I 0 < s 1 < < s 2 k p(s_1,\cdots,s_n)=\lambda^{2k}e^{-\lambda s_{2k}}I_{0<s_1<\cdots<s_{2k}} n n 为奇数的时候也进行相似的讨论,就获得 ( S 1 , , S n ) (S_1,\cdots,S_n) 的联合密度为 p ( s 1 , , s n ) = λ n e λ s n I 0 < s 1 < s 2 < < s n p(s_1,\cdots,s_n)=\lambda^ne^{-\lambda s_n}I_{0<s_1<s_2<\cdots<s_n}
总结 { S n } \{S_n\} 是强度为 λ \lambda 的泊松过程 { N ( t ) } \{N(t)\} 的呼叫流,则
(1) S n S_n 服从 Γ ( n , λ ) \Gamma(n,\lambda)
(2) ( S 1 , , S n ) (S_1,\cdots,S_n) 的联合密度为 p ( s 1 , , s n ) = λ n e λ s n I 0 < s 1 < s 2 < < s n p(s_1,\cdots,s_n)=\lambda^ne^{-\lambda s_n}I_{0<s_1<s_2<\cdots<s_n} blog

呼叫流的条件分布

下面给定 N ( t ) = n N(t)=n 的条件, ( S 1 , , S n ) (S_1,\cdots,S_n) 的分布:
按照条件几率的定义 P ( S 1 s 1 , , S n s n N ( t ) = n ) = P { S i s i , i = 1 , 2 , , n , N ( t ) = n } P { N ( t ) = n } P(S_1\le s_1,\cdots,S_n \le s_n|N(t)=n)=\frac{P\{S_i\le s_i,i=1,2,\cdots,n,N(t)=n\}}{P\{N(t)=n\}} 同非条件的联合分布的讨论同样,若是 ( s 1 , , s n ) (s_1,\cdots,s_n) 不知足 0 < s 1 < < s n < t 0<s_1<\cdots<s_n<t ,则对 s 1 , , s n s_1,\cdots,s_n 依次求偏导等于0,所以咱们假设 0 < s 1 < < s n < t 0<s_1<\cdots<s_n<t 。若是 n = 2 k n=2k ,比较容易求解的一个几率是 P { S 1 > s 1 , S 2 s 2 , S 3 > s 3 , S 4 s 4 , , S 2 k 1 > s 2 k 1 , S 2 k s 2 k , N ( t ) = n } P\{S_1>s_1,S_2\le s_2,S_3>s_3,S_4\le s_4,\cdots,S_{2k-1}>s_{2k-1},S_{2k}\le s_{2k},N(t)=n\} 咱们把这些时刻画在时间轴上
在这里插入图片描述
这样,如何求解这个几率值就一目了然了,这个事件等价于在 ( 0 , s 1 ] (0,s_1] 段事件发生了0次,在 ( s 1 , s 2 ] (s_1,s_2] 段事件发生了2次, ( s 2 , s 3 ] (s_2,s_3] 段事件发生了0次, ( s 3 , s 4 ] (s_3,s_4] 段事件发生了2次, \cdots ,在 ( s 2 k 1 , s 2 k ] (s_{2k-1},s_{2k}] 段事件发生了两次,在 ( s 2 k , t ] (s_{2k},t] 事件发生了0次。所以 G ( s 1 , , s n ) = P { S 1 > s 1 , S 2 s 2 , S 3 > s 3 , S 4 s 4 , , S 2 k 1 > s 2 k 1 , S 2 k s 2 k , N ( t ) = n } = i = 1 k ( λ 2 ( s 2 i s 2 i 1 ) 2 2 e λ ( s 2 i s 2 i 1 ) ) i = 1 k e λ ( s 2 i 1 s 2 i 2 ) . e λ ( t s 2 k ) = i = 1 k [ λ 2 ( s 2 i s 2 i 1 ) 2 2 ] e λ t \begin{aligned} &G(s_1,\cdots,s_n)\\ =&P\{S_1>s_1,S_2\le s_2,S_3>s_3,S_4\le s_4,\cdots,S_{2k-1}>s_{2k-1},S_{2k}\le s_{2k},N(t)=n\}\\ =&\prod_{i=1}^{k}(\frac{\lambda^2(s_{2i}-s_{2i-1})^2}{2}e^{-\lambda(s_{2i}-s_{2i-1})}) \prod_{i=1}^ke^{-\lambda(s_{2i-1}-s_{2i-2})}.e^{-\lambda(t-s_{2k})}\\ =&\prod_{i=1}^k[\frac{\lambda^2(s_{2i}-s_{2i-1})^2}{2}]e^{-\lambda t} \end{aligned} 其对 s 1 , , s 2 k s_1,\cdots,s_{2k} 依次求偏导,获得 2 k s 1 s 2 k G = ( 1 ) k λ 2 k e λ t \frac{\partial^{2k}}{\partial s_1\cdots\partial s_{2k}}G=(-1)^k\lambda^{2k}e^{-\lambda t} 相似于非条件分布的讨论,令 F ( s 1 , , s n ) = P { S 1 s 1 , , S n s n , N ( t ) = n } F(s_1,\cdots,s_n)=P\{S_1\le s_1,\cdots,S_n\le s_n,N(t)=n\} ,就有 2 k s 1 s 2 k F = λ 2 k e λ t \frac{\partial^{2k}}{\partial s_1\cdots\partial s_{2k}}F=\lambda^{2k}e^{-\lambda t} 因而,在给定 N ( t ) = n N(t)=n 条件下,条件密度为 p ( s 1 , , s n ) = λ 2 k e λ t ( λ t ) 2 k ( 2 k ) ! e λ t = ( 2 k ) ! t 2 k = n ! t n I 0 < s 1 < < s n < t p(s_1,\cdots,s_n)=\frac{\lambda^{2k}e^{-\lambda t}}{\frac{(\lambda t)^{2k}}{(2k)!}e^{-\lambda t}}=\frac{(2k)!}{t^{2k}}=\frac{n!}{t^n}I_{0<s_1<\cdots<s_n<t} n n 为奇数时讨论也相似,获得的条件密度函数也是上式。实际上,这也是来自均匀分布 U ( 0 , t ) U(0,t) 的样本 U 1 , , U n U_1,\cdots,U_n 的顺序统计量 U ( 1 ) , , U ( n ) U_{(1)},\cdots,U_{(n)} 的联合分布。也就是说,在给定 N ( t ) = n N(t)=n 的条件下,这 n n 个事件发生的时间点在 ( 0 , t ] (0,t] 内均匀分布,再进行一个排列就获得 S 1 , , S n S_1,\cdots,S_n

结论 S 1 , S 2 , , S n N (