泊松过程¶
\(\renewcommand{\geq}{\geqslant}\renewcommand{\leq}{\leqslant}\) 满足如下条件的随机过程\(\{N(t), t\geq 0\}\)为时齐泊松过程。
- \(\{N(t), t\geq 0\}\)为计数过程，且\(N(0) = 0\)；
- \(\{N(t), t\geq 0\}\)为独立增量过程；
- \(\{N(t), t\geq 0\}\)具备增量平稳性，即\(N(t) - N(s)\)的分布仅与\(t - s\)有关
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对于充分小的\(\Delta t > 0\)，有
\[ \left\{ \begin{aligned} P(N(t + \Delta t) - N(t) = 1) &= \lambda \Delta t + o(\Delta t) \\ P(N(t + \Delta t) - N(t) \geq 2) &= o(\Delta t) \end{aligned} \right. \]
增量的概率分布¶
泊松过程的增量\(N(t + s) - N(s)\)服从参数为\(\lambda t\)的泊松分布，即
证明
根据泊松过程的增量平稳性，有
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\(n = 0\)时，有\(\{N(h + t) = 0, h > 0\} = \{N(h+ t) - N(t) = 0, N(t) = 0, h > 0\}\)，根据增量独立性，有
\[ \begin{aligned} P(N(h + t) = 0) &= P(N(h + t) - N(t) = 0) P(N(t) = 0) \\ &= P(N(h) = 0) P(N(t) = 0) \end{aligned} \]根据泊松过程的性质\(P(N(t + \Delta t) - N(t) = 1) = \lambda \Delta t + o(\Delta t)\)，有：
\[ \begin{aligned} \lim_{h\rightarrow 0} \frac{P(t + h) - P(t)}{h} &= \lim_{h\rightarrow 0} \frac{P(t)(1 - \lambda h - o(h)) - P(t)}{h} \\ &= -\lambda P(t) - \lim_{h\rightarrow 0} \frac{o(h)}{h} \\ &= -\lambda P(t) \end{aligned} \]解微分方程\(P_0'(t) = -\lambda P_0(t)\)，解得\(P_0(t) = e^{-\lambda t + C}\)，又因为\(N(0) = 0, P_0(N(0) = 0) = 1\)，得\(C = 0\)，从而有
\[ P_0(t) = e^{-\lambda t} \] -
\(n > 0\)时，\(\{N(t + h) = n\} = \bigcup_{k = 0} ^ n \{N(t) = k, N(t + h) - N(t) = (n - k)\}\)，而\(n - k \geq 2\)时，有\(P(N(t + h) - N(t) = n - k) = o(h)\)，由增量独立性
\[ \begin{aligned} P(N(t + h) = n) &= \sum_{k = 0} ^ n P(N(t) = k) P(N(t + h) - N(t) = n - k) \\ &= P(N(t) = n)(1 - \lambda t - o(h)) + P(N(t) = n - 1)(\lambda t + o(h)) \\ &+ \sum_{k = 2}^n P(N(t) = n - k)o(h) \end{aligned} \]由此
\[ \begin{aligned} &\lim_{h\rightarrow 0} \frac{P(N(t + h) = n) - P(N(t) = n)}{h} \\ =& \lim_{h\rightarrow 0}\frac{P(N(t) = n)(\sum_{k = 2}^n P(N(t) = n - k)o(h) - P(N(t) = n)}{h} \\ +& \lim_{h\rightarrow 0}\frac{\sum_{k = 2}^n P(N(t) = n - k)o(h) - P(N(t) = n)}{h} \\ =& \lim_{h\rightarrow 0}\frac{P(N(t) = n)(- \lambda t) + P(N(t) = n - 1)(\lambda t)}{h} \\ \end{aligned} \]得到
\[ P_n'(t) = -\lambda P_n(t) + \lambda P_{n-1}(t) \]考虑约束条件\(P_n(0) = 0\)，解得
\[ \frac{\mathrm d}{\mathrm dt}[e^{\lambda t} P_n(t)] = \lambda e^{\lambda t}P_{n - 1}(t) \]- \(n = 0\)时，有\(P_n(t) = e^{-\lambda t} = \frac{(\lambda t)^0}{0!}e^{-\lambda t}\)
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设对于\(n < k\)时命题成立，只需证明\(n = k\)时命题成立即可，有
\[ \begin{aligned} P_k(t) &= e^{-\lambda t}\int \frac{\mathrm d}{\mathrm dt}[e^{\lambda t} P_k(t)] \mathrm dt \\ &= e^{-\lambda t}\int \lambda e^{\lambda t}P_{k - 1}(t) \mathrm dt \\ &= e^{-\lambda t}\int \lambda \frac{(\lambda t)^{k-1}}{(k-1)!} \mathrm dt \\ &= \frac{(\lambda t)^k}{k!}e^{-\lambda t} \end{aligned} \]
相邻事件时间间隔的概率分布¶
设\(\{N(t), t \geq 0\}\)为计数过程，设\(S_n\)表示第\(n\)个时间发生的时刻，即\(S_n = \inf\{t, t > S_{n - 1}, N(t) = n\}\)，特别地，令\(S_0 = 0\)。\(S_n, N(t)\)的事件满足
由于\(\{S_n \leq t\} \equiv \{N(t) \geq n\}\)，可以计算\(S_n\)的分布函数
特别地，当\(n = 1\)时，有\(P(X_1\leq t) = P(S_1\leq t) = (1 - e^{-\lambda t}) I_{\{t\geq 0\}}\)
计数过程\(\{N(t), t\geq 0\}\)是泊松过程的充要条件是\(\{X_n = S_{n} - S_{n-1}, n\geq 1\}\)独立同分布，服从的分布为参数\(\lambda\)的指数分布。
证明
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\(\Leftarrow\)：必要性证明，先求\(S_1, \cdots, S_n\)的联合概率密度函数，设\(t_1 < t_2 < \cdots < t_n\)，设\(h\)为充分小的正数，满足
\[ t_1 + \frac h2 < t_2 - \frac h2 < t_2 + \frac h2 < \cdots < t_n - \frac h2 \]考虑事件\(A\)
\[ \begin{aligned} A& \\ &\equiv \left\{t_1 - \frac h2 < S_1 \leq t_1 + \frac h2, t_2 - \frac h2 < S_2 \leq t_2 + \frac h2, \cdots, t_n - \frac h2 < S_n \leq t_n + \frac h2\right\} \\ \equiv & \left\{N\left(t_1 - \frac h2\right) = 0, N\left(t_{i+1} - \frac h2\right) - N\left(t_i + \frac h2\right) = 0, N\left(t_{i} + \frac h2\right) - N\left(t_{i} - \frac h2\right) = 1\right\} \\ & 0\leq i\leq n - 1 \end{aligned} \]而根据泊松过程的性质
\[ \begin{aligned} P\left(N\left(t_n + \frac h2\right) - N\left(t_n - \frac h2\right)= 1\right) &= \lambda h + o(h) \\ P\left(N\left(t_n + \frac h2\right) - N\left(t_n - \frac h2\right)\geq 2\right) &= o(h) \\ \end{aligned} \]有
\[ P(A) = (\lambda h)^ne^{-\lambda (t_n + h/2)} + o\left(h^n\right) = \lambda ^nh^ne^{-\lambda t_n} + o\left(h^n\right) \]因此，\(\left(S_1, \cdots, S_n\right)\)的概率密度为
\[ f(t_1, \cdots, t_n) = \left\{ \begin{aligned} & \lambda^n e^{-\lambda t_n} & 0 < t_1 < \cdots < t_n \\ & 0 & \mathrm{otherwise} \end{aligned} \right. \]而\(X_n = S_n - S_{n-1}\)，设\(x_n = t_n - t_{n - 1}\)，则\(t_n = \sum_{i=1}^n x_n\)，从而变换的Jacob矩阵为
\[ J = \frac{\partial (t_1, \cdots, t_n)}{\partial (x_1, \cdots, x_n)} = \begin{vmatrix} 1 & 0 & \cdots & 0 \\ 1 & 1 & \cdots & 0 \\ \vdots & \vdots & \ddots & \vdots \\ 1 & 1 & \cdots & 1 \end{vmatrix} \]因此\(X_1, \cdots, X_n\)的概率密度为
\[ g(x_1, \cdots, x_n) = \lambda^n e^{-\lambda (x_1 + \cdots + x_n)} = \prod_{i=1}^n \left(\lambda e^{-\lambda x_i}\right), x_i \geq 0 \]因此\(X_1, \cdots, X_n\)独立同指数分布。
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\(\Rightarrow\)：充分性证明，设\(\{X_k, k\geq 1\}\)独立同指数分布，设\(S_0 = 0, S_n = S_0 + \sum_{i = 1}^n X_i\)。定义
\[ N(t) = \sum_{n=1}^\infty I_{\{S_n\leq t\}} \]- 由于\(I_{\{S_n\leq t\}}\in \mathbb N\)，则\(N(t)\in \mathbb N\)。
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对于\(t > s\)，有
\[ N(t) - N(s) = \sum_{n = 1}^\infty I_{\{S_n\leq t\}}- I_{\{S_n\leq s\}} \]而\(S_n - S_{n-1} = X_n > 0\)，因此\(S_n\)为单调增序列，从而\(\{S_n\leq s\}\subset \{S_n\leq t\}\)，即\(I_{\{S_n\leq t\}}- I_{\{S_n\leq s\}} \geq 0\)，因此\(N(t) - N(s)\)
可得\(N(t)\)为计数过程。以下计算\(S_n\)的分布。已知\(X_1, \cdots, X_n\)独立同指数分布，设\(t_n = \sum_{i=1}^n x_i\)，则\(x_n = t_n - t_{n-1}\)，从而Jacob矩阵为
\[ J = \frac{\partial (x_1, \cdots, x_n)}{\partial (t_1, \cdots, t_n)} = \begin{vmatrix} 1 & 0 & 0 & \cdots & 0 \\ -1 & 1 & 0 & \cdots & 0 \\ 0 & -1 & 1 & \cdots & 0 \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & 0 & \cdots & 1 \end{vmatrix} \]而\(X_1, \cdots, X_n\)的联合概率密度为
\[ g(x_1, \cdots, x_n) = \lambda^n e^{-\lambda (x_1 + \cdots + x_n)} = \prod_{i=1}^n \left(\lambda e^{-\lambda x_i}\right), x_i \geq 0 \]因此\(S_1, \cdots, S_n\)的联合概率密度为
\[ f(t_1, \cdots, t_n) = \left\{ \begin{aligned} & \lambda^n e^{-\lambda t_n} & 0 < t_1 < \cdots < t_n \\ & 0 & \mathrm{otherwise} \end{aligned} \right. \]