Process:
\[\begin{aligned} \ce{x ->[\lambda] x + 1} \\ \ce{x ->[\beta x] x - 1} \end{aligned}\]Chemical master equation:
\[\begin{aligned} \frac{d}{dt}P(x=m) &= \lambda P(x=m-1) - \lambda P(x=m)\\ &\quad + \beta(m+1)P(x=m+1) - \beta mP(x=m) \end{aligned}\]Define the factorial moment
\[m^{\underline{k}} = \frac{m(m-1)\cdots(m-k+1)}{k!} = \binom{m}{k}\]Summing over the chemical master equation by $\sum_{m=0}^\infty m^{\underline{k}}$, we can get the factorial moment balance equation
\[\begin{aligned} \frac{d}{dt} \langle x \rangle &= \sum_{m=0}^\infty [\lambda(m-1)^{\underline{k}}P(x=m-1) - \lambda m^{\underline{k}}P(x=m)\\ &\quad + \beta(m+1)m^{\underline{k}}P(x=m+1) - \beta(m)m^{\underline{k}}P(x=m+1)]\\ &= \lambda[(m+1)^{\underline{k}}-m^{\underline{k}}]P(x=m)\\ &\quad - \beta [m(m-1)^{\underline{k}}-mm^{\underline{k}}]P(x=m)\\ &=\lambda k\langle x^{\underline{k-1}}\rangle - \beta k\langle x^{\underline{k}}\rangle \end{aligned}\]Stationary solution ($\frac{d}{dt}\langle x \rangle=0$):
\[\lambda\langle x^{\underline{k-1}}\rangle = \beta \langle x^{\underline{k}}\rangle\]For $k=1$ we get
\[\lambda=\beta\langle x\rangle\]Then we get
\[\langle x^{\underline{k}}\rangle = \frac{\langle x\rangle^k}{k!}\]Define the generating function
\[G(z)=\sum_{m=0}^\infty z^mP(x=m)\]We find
\[\frac{\partial^k}{\partial z^k}G(z)\vert_{z=1}=k!\langle x^{\underline{k}}\rangle=k!\langle x\rangle^k\]So a solution to the generating function is
\[G(z) = \exp(z\langle x\rangle)\exp(-\langle x\rangle)\]Then get back the distribution by
\[P(x=m) = \frac{1}{m!}\frac{\partial^m}{\partial z^m} G(z)\vert_{z=0}\]That says
\[P(x=m) = \frac{\langle x\rangle^m\exp(-\langle x\rangle)}{m!}\]文档信息
- 本文作者：L Shi
- 本文链接：https://shi200005-github-io.pages.dev/2026/08/05/GF-Poisson/
- 版权声明：自由转载-非商用-非衍生-保持署名（创意共享3.0许可证）