Abstract

Let {Xj,j≥0}$\{{\mathbf{X}}_j,j\ge 0\}$$\{{\mathbf{X}}_j,j\ge 0\}$ denote a Markov process on [−N−1,N+1]∪{c}$[-N-1,N+1]\cup \left\{\mathbf{c}\right\}$$[-N-1,N+1]\cup \left\{\mathbf{c}\right\}$. Suppose P(Xj+1=m+1|Xj=m)=ph,P(Xj+1=m−1|Xj=m)=(1−p)h$\mathbb{P}({\mathbf{X}}_{j+1}=m+1|{\mathbf{X}}_j=m)=ph,\mathbb{P}({\mathbf{X}}_{j+1}=m-1|{\mathbf{X}}_j=m)=(1-p)h$$\mathbb{P}({\mathbf{X}}_{j+1}=m+1|{\mathbf{X}}_j=m)=ph,\mathbb{P}({\mathbf{X}}_{j+1}=m-1|{\mathbf{X}}_j=m)=(1-p)h$, all j≥1$j\ge 1$$j\ge 1$ and |m|≤N$\left|m\right|\le N$, where $p=\frac{1}{2}+\frac{b}{N}$ and $h=1-c_N$ for $c_N=\frac{1}{2}{a}^2/N^2$. Define $\mathbb{P}({\mathbf{X}}_{j+1}=\mathbf{c}|{\mathbf{X}}_j=m)=c_N,j\ge 0,|m|\le N$. $\left\{{\mathbf{X}}_j\right\}$ terminates at the first j such that ${\mathbf{X}}_j\in \{-N-1,N+1,\mathbf{c}\}$. Let $\mathcal{L}=\max \{j\ge 0:{\mathbf{X}}_j=0\}$. On ${\Omega }^{°}=\left\{{\mathbf{X}}_j\text{ terminates at }\mathbf{c}\right\}$, denote by ${\mathcal{R}}^{°}$ and ${\mathcal{L}}^{°}$, respectively, as the numbers of runs and steps from $\mathcal{L}$ until termination. Denote ${\Delta }^{°}={\mathcal{L}}^{°}-2{\mathcal{R}}^{°}$. Then $\underset{N\to \infty }{\lim }\mathbb{E}\left\{e^{\frac{it}{N}{\Delta }^{°}}\right|{\Omega }^{°}\}=C_{a,b}\frac{\sqrt{c^2+t^2}(\cosh \sqrt{c^2+t^2}\text{-}\cosh (2b))}{(a^2+t^2)\sinh \sqrt{c^2+t^2}}$, where $c^2=a^2+4b^2$.

摘要

让{Xj,j ≥ 0 }$\{{\mathbf{X}}_j,j\ge 0\}$$\{{\mathbf{X}}_j,j\ge 0\}$表示马尔可夫过程[ - N− 1 , N+ 1 ] ∪ { c }$[-氮-1,氮+1]\cup \left\{\mathbf{C}\right\}$$[-N-1,N+1]\cup \left\{\mathbf{c}\right\}$。认为P (Xj + 1= m + 1 |Xj= m ) = pH , _P (Xj + 1= m − 1 |Xj= m ) = ( 1 − p ) h$\mathbb{磷}（{\mathbf{X}}_{j+1}=米+1|{\mathbf{X}}_j=米）=pH,\mathbb{磷}（{\mathbf{X}}_{j+1}=米-1|{\mathbf{X}}_j=米）=（1-p）H$$\mathbb{P}({\mathbf{X}}_{j+1}=m+1|{\mathbf{X}}_j=m)=ph,\mathbb{P}({\mathbf{X}}_{j+1}=m-1|{\mathbf{X}}_j=m)=(1-p)h$, 阿勒斯j≥1 _ _$j\ge 1$$j\ge 1$其他| 米| ≤ N$\left|m\right|\le N$， 在哪里$p=\frac{1}{2}+\frac{乙}{氮}$其他$H=1-C_{氮}$为了$C_{氮}=\frac{1}{2}{A}^2/{氮}^2$。定义$\mathbb{磷}（{\mathbf{X}}_{j+1}=\mathbf{C}|{\mathbf{X}}_j=米）=C_{氮},j\ge 0,|米|\le 氮$。$\left\{{\mathbf{X}}_j\right\}$终止于第一个j使得${\mathbf{X}}_j\epsilon \{-氮-1,氮+1,\mathbf{C}\}$。让$\mathcal{L}=\mathrm{最大限度}\{j\ge 0:{\mathbf{X}}_j=0\}$。在${\Omega }^{°}=\left\{{\mathbf{X}}_j\text{ 终止于 }\mathbf{C}\right\}$，表示为${\mathcal{右}}^{°}$其他${\mathcal{L}}^{°}$分别为运行次数和步数$\mathcal{L}$直至终止。表示${\Delta }^{°}={\mathcal{L}}^{°}-2{\mathcal{右}}^{°}$。然后$\underset{氮\to \mathrm{无穷大}}{\mathrm{有限的}}\mathbb{乙}\left\{e^{\frac{我t}{氮}{\Delta }^{°}}\right|{\Omega }^{°}\}=C_{A,乙}\frac{\sqrt{C^2+t^2}（\mathrm{科什}\sqrt{C^2+t^2}\text{-}\mathrm{科什}（2乙））}{（A^2+t^2）\mathrm{新}\sqrt{C^2+t^2}}$， 在哪里$C^2=A^2+4{乙}^2$。