Posted on Categories:Stochastic Porcesses, 数学代写, 随机过程

# 数学代写|随机过程Stochastic Porcess代考|Stat150

avatest™

## avatest™帮您通过考试

avatest™的各个学科专家已帮了学生顺利通过达上千场考试。我们保证您快速准时完成各时长和类型的考试，包括in class、take home、online、proctor。写手整理各样的资源来或按照您学校的资料教您，创造模拟试题，提供所有的问题例子，以保证您在真实考试中取得的通过率是85%以上。如果您有即将到来的每周、季考、期中或期末考试，我们都能帮助您！

•最快12小时交付

•200+ 英语母语导师

•70分以下全额退款

## 数学代写|随机过程代写Stochastic Porcess代考|Finite-Dimensional Homogeneous Processes with Independent Increments

In this section we shall discuss homogeneous processes with independent increments with values in $\mathscr{R}^m$. The characteristic function of such a process is of the form
\begin{aligned} \mathrm{E} e^{i(z, \xi(t))} & =\exp {t K(z)} \ & =\exp \left{t\left[i(a, z)-\frac{1}{2}(B z, z)+\int\left(e^{i(z, x)}-1-\frac{i(z, x)}{1+|x|^2}\right) \Pi(d x)\right]\right}, \end{aligned}
where $z \in \mathscr{R}^m$ and $(z, y)$ is the scalar product in $\mathscr{R}^m$. In formula (1) $a \in \mathscr{R}^m, B$ is a non-negative symmetric linear operator in $\mathscr{R}^m$, and the measure $\Pi$ is defined on Borel sets and is such that
$$\int \frac{|x|^2}{1+|x|^2} \Pi(d x)<\infty .$$
Analogously to the one-dimensional case the function $K(z)$ appearing in (1) is called the cumulant of the process; it completely determines the marginal distributions of the processes. The processes under consideration are assumed to be separable and thus have no discontinuities of the second kind. Sample functions of the processes are assumed to be continuous from the right.

One can associate uniquely a homogeneous Markov process $\left{\mathscr{F}, \mathscr{N}, \mathrm{P}_x\right}$ with a homogeneous process with independent increments $\xi(t)$. This Markov process is of the form: the set of functions of the type $x_t=\xi(s+t)-\xi(s)+x$, $s \geqslant 0, x \in \mathscr{R}^m$, where $\xi(\cdot)$ are various sample functions of the process $\xi(t)$, is chosen as the set $\mathscr{F}$; the set $\mathscr{N}$ is defined in the usual manner as the minimal $\sigma$-algebra containing all the cylinders in $\mathscr{F}$. For each cylinder $A$ we have
$$\mathrm{P}x{A}=\mathrm{P}{x+\xi(\cdot) \in A}$$ (the probability on the r.h.s. is defined on the same probability space on which the process $\xi(t)$ is defined). The process is homogeneous Markov in view of the relation \begin{aligned} \mathrm{P}{x+\xi(t+s) \in A \mid \xi(u), u \leqslant s} & =\mathrm{P}{x+\xi(s)+\xi(t+s)-\xi(s) \in A \mid \xi(u), u \leqslant s} \ & =\mathrm{P}{y+\xi(t+s)-\xi(s) \in A}{y=x+\xi(s)} \ & =\mathrm{P}{y+\xi(t) \in A}_{y=x+\xi(s)}=\mathrm{P}_{x(s)}{x(t) \in A} . \end{aligned}

## 数学代写|随机过程代写Stochastic Porcess代考|Resolvent, characteristic and generating operators

Resolvent, characteristic and generating operators. Consider a resolvent of a Markov process associated with a process with independent increments (hereafter we shall refer to it as the resolvent of the process $\xi(t))$. Formula (3) implies
$$\mathbf{R}\lambda f(x)=\int_0^{\infty} e^{-\lambda t} \mathrm{E} f(x+\xi(t)) d t .$$ Let $F(t, A)=\mathrm{P}{\xi(t) \in A}$. where $$\mathbf{R}\lambda f(x)=\frac{1}{\lambda} \int f(x+y) F_\lambda(d y),$$
$$F_\lambda(A)=\lambda \int_0^{\infty} e^{-\lambda t} F(t, A) d t .$$
The function $F_\lambda(A)$ can be conveniently defined by means of the Fourier transform
$$\Phi_\lambda(z)=\int e^{i(z, y)} F_\lambda(d y) .$$
Utilizing (5) we obtain
$$\Phi_\lambda(z)=\lambda \int_0^{\infty} e^{-\lambda t} e^{t K(z)} d t=\frac{\lambda}{\lambda-K(z)} .$$

Analogously to the one-dimensional case, it follows that $\Phi_\lambda(z)$, for $\lambda>0$, is the characteristic function of an infinitely divisible distribution since
$$\Phi_\lambda(z)=\exp \left{\int_0^{\infty} e^{-\lambda t} \frac{e^{t \mathbf{K}(z)}-1}{t} d t\right},$$
so that
$$\Phi_\lambda(z)=\lim {\varepsilon \rightarrow 0} \exp \left{\int{\varepsilon}^{\infty} e^{-\lambda t} \frac{e^{t K(z)}-1}{t} d t\right}=\lim {\varepsilon \rightarrow 0} \exp \left{\int{\varepsilon}^{\infty} \frac{e^{-\lambda t} e^{t K(z)}}{t} d t-\int_{\varepsilon}^{\infty} \frac{e^{-\lambda t}}{t} d t\right} ;$$
the function $\int_{\varepsilon}^{\infty}\left(e^{-\lambda t} e^{t K(z)} / t\right) d t$ is positive definite since $e^{t K(z)}$ is such a function. The compound function $\exp {\Phi(z)-\Phi(a)}$, where $\Phi(z)$ is positive definite, is infinitely divisible and finally the limit of infinitely divisible functions is also infinitely divisible. The infinite divisability of $\Phi_\lambda(z)$ implies the existence of $a_\lambda$, $B_\lambda$ and $\Pi_\lambda$ such that
$$\Phi_\lambda(z)=\exp \left{K_\lambda(z)\right},$$
where
$$K_\lambda(z)=i\left(a_\lambda z\right)-\frac{1}{2}\left(B_\lambda z, z\right)+\int\left(e^{i(z, x)}-1-\frac{i(z, x)}{1+|x|^2}\right) \Pi_\lambda(d x) .$$

## 数学代写|随机过程代写Stochastic Porcess代考|Finite-Dimensional Homogeneous Processes with Independent Increments

\begin{aligned} \mathrm{E} e^{i(z, \xi(t))} & =\exp {t K(z)} \ & =\exp \left{t\left[i(a, z)-\frac{1}{2}(B z, z)+\int\left(e^{i(z, x)}-1-\frac{i(z, x)}{1+|x|^2}\right) \Pi(d x)\right]\right}, \end{aligned}

$$\int \frac{|x|^2}{1+|x|^2} \Pi(d x)<\infty .$$

## 数学代写|随机过程代写Stochastic Porcess代考|Resolvent, characteristic and generating operators

$$\mathbf{R}\lambda f(x)=\int_0^{\infty} e^{-\lambda t} \mathrm{E} f(x+\xi(t)) d t .$$让$F(t, A)=\mathrm{P}{\xi(t) \in A}$。在哪里$$\mathbf{R}\lambda f(x)=\frac{1}{\lambda} \int f(x+y) F_\lambda(d y),$$
$$F_\lambda(A)=\lambda \int_0^{\infty} e^{-\lambda t} F(t, A) d t .$$

$$\Phi_\lambda(z)=\int e^{i(z, y)} F_\lambda(d y) .$$

$$\Phi_\lambda(z)=\lambda \int_0^{\infty} e^{-\lambda t} e^{t K(z)} d t=\frac{\lambda}{\lambda-K(z)} .$$

$$\Phi_\lambda(z)=\exp \left{\int_0^{\infty} e^{-\lambda t} \frac{e^{t \mathbf{K}(z)}-1}{t} d t\right},$$

$$\Phi_\lambda(z)=\lim {\varepsilon \rightarrow 0} \exp \left{\int{\varepsilon}^{\infty} e^{-\lambda t} \frac{e^{t K(z)}-1}{t} d t\right}=\lim {\varepsilon \rightarrow 0} \exp \left{\int{\varepsilon}^{\infty} \frac{e^{-\lambda t} e^{t K(z)}}{t} d t-\int_{\varepsilon}^{\infty} \frac{e^{-\lambda t}}{t} d t\right} ;$$

$$\Phi_\lambda(z)=\exp \left{K_\lambda(z)\right},$$

$$K_\lambda(z)=i\left(a_\lambda z\right)-\frac{1}{2}\left(B_\lambda z, z\right)+\int\left(e^{i(z, x)}-1-\frac{i(z, x)}{1+|x|^2}\right) \Pi_\lambda(d x) .$$

## MATLAB代写

MATLAB 是一种用于技术计算的高性能语言。它将计算、可视化和编程集成在一个易于使用的环境中，其中问题和解决方案以熟悉的数学符号表示。典型用途包括：数学和计算算法开发建模、仿真和原型制作数据分析、探索和可视化科学和工程图形应用程序开发，包括图形用户界面构建MATLAB 是一个交互式系统，其基本数据元素是一个不需要维度的数组。这使您可以解决许多技术计算问题，尤其是那些具有矩阵和向量公式的问题，而只需用 C 或 Fortran 等标量非交互式语言编写程序所需的时间的一小部分。MATLAB 名称代表矩阵实验室。MATLAB 最初的编写目的是提供对由 LINPACK 和 EISPACK 项目开发的矩阵软件的轻松访问，这两个项目共同代表了矩阵计算软件的最新技术。MATLAB 经过多年的发展，得到了许多用户的投入。在大学环境中，它是数学、工程和科学入门和高级课程的标准教学工具。在工业领域，MATLAB 是高效研究、开发和分析的首选工具。MATLAB 具有一系列称为工具箱的特定于应用程序的解决方案。对于大多数 MATLAB 用户来说非常重要，工具箱允许您学习应用专业技术。工具箱是 MATLAB 函数（M 文件）的综合集合，可扩展 MATLAB 环境以解决特定类别的问题。可用工具箱的领域包括信号处理、控制系统、神经网络、模糊逻辑、小波、仿真等。