latex常见用法如下:https://blog.csdn.net/ViatorSun/article/details/82826664
高等教育出版社 **浙江大学《概率论与数理统计》**一 书关于泊松过程的推导如下:
理解了上面的思路才能更好的理解泊松过程的数学模型和本质。
上面的思路是:
- 将时间过程分解为两部分,第一部分是事件不可微分的时间间隔,第二部分是事件发生次数可以微分的时间间隔,这两个部分是概率相乘的关系(而不是相加)。
- 先算出k=0时的泊松过程,然后计算k>=1时的泊松过程方程,依赖的数学知识是导数和求解微分方程,整个过程严密有条理。
- 复现推导过程。
(1)当k = 0时:
P 0 ( t 0 , t + Δ t ) = P 0 ( t 0 , t ) P 0 ( t , t + Δ t ) = P 0 ( t 0 , t ) ( 1 − λ Δ t − o ( Δ t ) ) 即 : lim Δ t → 0 P 0 ( t 0 , t + Δ t ) − P 0 ( t 0 , t ) Δ t = − λ P 0 ( t 0 , t ) − o ( Δ t ) P 0 ( t 0 , t ) Δ t 即: P 0 ( t 0 , t ) = e − λ ( t − t 0 ) P_0(t_0,t+\Delta t) = P_0(t_0,t) P_0(t,t+ \Delta t) = P_0(t_0,t)(1 - \lambda \Delta t - o( \Delta t) ) \\ \\即:\\ \lim_{\Delta t \to 0} \frac {P_0(t_0,t+\Delta t) - P_0(t_0,t) }{\Delta t} = -\lambda P_0(t_0,t) - \frac {o(\Delta t) P_0(t_0,t)}{\Delta t } \\即:P_0(t_0,t) = e^{-\lambda (t - t_0)} P0(t0,t+Δt)=P0(t0,t)P0(t,t+Δt)=P0(t0,t)(1−λΔt−o(Δt))即:Δt→0limΔtP0(t0,t+Δt)−P0(t0,t)=−λP0(t0,t)−Δto(Δt)P0(t0,t)即:P0(t0,t)=e−λ(t−t0)
(2)当k大于等于1时:
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注意:此处需要求解一阶非齐次微分方程
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P_k(t_0,t+\Delta t) = P_{k-j}(t_0,t) P_{j}(t,t+ \Delta t) = \\\\ P_k(t_0,t)P_{0}(t,t+ \Delta t) + P_{k-1}(t_0,t)P_{1}(t,t+ \Delta t) + o(\Delta t) =\\ P_k(t_0,t)(1 - \lambda \Delta t - o( \Delta t) ) + P_{k-1}(t_0,t)(\lambda \Delta t ) + o(\Delta t) \\ 即:\\ \lim_{\Delta t \to 0} \frac {P_k(t_0,t+\Delta t) - P_k(t_0,t) }{\Delta t} = -\lambda P_k(t_0,t) + \lambda P_{k-1}(t_0,t)+ o(\Delta t) \\ 即:P_k(t_0,t) = e^{-\lambda (t - t_0)} \\ \color {red}注意:此处需要求解一阶非齐次微分方程:\\ 假设u' = \lambda u,即du/dt = \lambda dt ,可得:u = C_1 e^{\lambda t} 带入上述一元非齐次微分方程可得: \\ dP_1/dt = -\lambda P_1 + \lambda e^{-\lambda t}\\ dP_1 u + \lambda P_1 u = u \lambda e^{-\lambda t} dt\\ (P_1 u )' = u \lambda e^{-\lambda t} dt \\ P_1 u = \int u \lambda e^{-\lambda t} dt = \int \lambda e^{-\lambda t} e^{\lambda t} dt = \int \lambda dt \\ P_1 = e^{-\lambda t} \int \lambda dt = \lambda t e^{-\lambda t} \\
Pk(t0,t+Δt)=Pk−j(t0,t)Pj(t,t+Δt)=Pk(t0,t)P0(t,t+Δt)+Pk−1(t0,t)P1(t,t+Δt)+o(Δt)=Pk(t0,t)(1−λΔt−o(Δt))+Pk−1(t0,t)(λΔt)+o(Δt)即:Δt→0limΔtPk(t0,t+Δt)−Pk(t0,t)=−λPk(t0,t)+λPk−1(t0,t)+o(Δt)即:Pk(t0,t)=e−λ(t−t0)注意:此处需要求解一阶非齐次微分方程:假设u′=λu,即du/dt=λdt,可得:u=C1eλt带入上述一元非齐次微分方程可得:dP1/dt=−λP1+λe−λtdP1u+λP1u=uλe−λtdt(P1u)′=uλe−λtdtP1u=∫uλe−λtdt=∫λe−λteλtdt=∫λdtP1=e−λt∫λdt=λte−λt
(3)当k大于等于2时:
P k = ( λ ( t − t 0 ) ) k ! k e − λ ( t − t 0 ) P_k = \frac{ (\lambda (t-t_0))^k }{!k} e^{-\lambda (t-t_0)} Pk=!k(λ(t−t0))ke−λ(t−t0)
(4)当
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P_k = \frac{ (\int \lambda (t-t_0) dt)^k }{!k} e^{-\int \lambda (t-t_0) dt}
Pk=!k(∫λ(t−t0)dt)ke−∫λ(t−t0)dt