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数学代写|随机过程代写Stochastic Porcess代考|Non-ageing property of geometric distribution

For a geometric r.v. $X$, we have
\begin{aligned} \operatorname{Pr}{X & =s+r \mid X \geq s}=\frac{q^{s+r} p}{q^s} \ & =q^r p=\operatorname{Pr}{X=r} . \end{aligned}
This property, called non-ageing (or memoryless) property, characterizes geometric distribution among all distributions of discrete non-negative integral r.v.’s.
Note: If $Y_1, Y_2, \ldots$, is a sequence of independent Bernoulli r.v.s, then
$$X_i=\min \left{i, Y_i=1\right}-1 \text { is a geometric r.v. }$$
Example 1(e). Logarithmic Series Distribution:
The r.v. $X$ has logarithmic series distribution if
\begin{aligned} & p_k=\operatorname{Pr}{X=k}=\frac{\alpha q^k}{k}, k=1,2,3, \ldots \ & \alpha=-1 /(\log p) \ & 0<q=1-p<1 . \end{aligned}
The p.g.f. of $X$ is
\begin{aligned} P(s) & =\sum_{k=1}^{\infty} p_k s^k=\sum_{k=1}^{\infty} \frac{\alpha q^k}{k} s^k \ & =-\alpha \log (1-s q)=\frac{\log (1-s q)}{\log (1-q)} . \end{aligned}

数学代写|随机过程代写Stochastic Porcess代考|Determination of $\left{p_k\right}$ from a given $P(s)$

From the above examples, we see how a single generating function $P(s)$ may be used to represent a whole set of probabilities
$$p_k=\operatorname{Pr}{X=k}, \quad k=0,1,2, \ldots$$
In these examples we were concerned with the problem of finding $P(s)$ for a given set of $p_k$ ‘s. In many cases the reverse problem arises: to determine $p_k$ from a given p.g.f. $P(s)$. Many situations arise, where it is easier to find the p.g.f. $P(s)$ of a variable rather than the probability distribution $\left{p_k\right}$ of the variable. One proceeds to find first the p.g.f. $P(s)$ and then to find the probability $p_k$ from the function $P(s)$. Even without finding the $p_k$ ‘s one can find the moments of the distribution from $P(s)$.
Again, $p_k$ can be (uniquely) determined from $P(s)$ as follows:
$p_k$ can be found from $P(s)$ by applying (1.2), i.e.
$$p_k=\frac{1}{k !}\left[\frac{d^k P(s)}{d s^k}\right]_{s=0} ;$$
$p_k$ is also given by the coefficient of $s^k$ in the expansion of $P(s)$ as a power series in $s$.
When $P(s)$ is of the form $P(s)=U(s) / V(s)$, it may be convenient to expand $P(s)$ in a power series in $s$ first by decomposing $P(s)$ into partial factions. Suppose that $s_1, \ldots, s_r$ are the distinct roots of $V(s)$, i.e. $V(s)=\left(s-s_1\right) \ldots\left(s-s_r\right)$ apart from a constant factor $c$, which, for simplicity, we take to be equal to $1)$, then $P(s)$ can be decomposed into partial fractions as
$$P(s)=\frac{a_1}{s_1-s}+\cdots+\frac{a_r}{s_r-s},$$
where $a_i$ ‘s can be determined. It may be verified that
$$a_i=-U\left(s_i\right) / V^{\prime}\left(s_i\right) .$$

数学代写|随机过程代写Stochastic Porcess代考|Non-ageing property of geometric distribution

\begin{aligned} \operatorname{Pr}{X & =s+r \mid X \geq s}=\frac{q^{s+r} p}{q^s} \ & =q^r p=\operatorname{Pr}{X=r} . \end{aligned}

$$X_i=\min \left{i, Y_i=1\right}-1 \text { is a geometric r.v. }$$

r.v. $X$具有对数级数分布
\begin{aligned} & p_k=\operatorname{Pr}{X=k}=\frac{\alpha q^k}{k}, k=1,2,3, \ldots \ & \alpha=-1 /(\log p) \ & 0<q=1-p<1 . \end{aligned}
$X$的p.g.f.是
\begin{aligned} P(s) & =\sum_{k=1}^{\infty} p_k s^k=\sum_{k=1}^{\infty} \frac{\alpha q^k}{k} s^k \ & =-\alpha \log (1-s q)=\frac{\log (1-s q)}{\log (1-q)} . \end{aligned}

数学代写|随机过程代写Stochastic Porcess代考|Determination of $\left{p_k\right}$ from a given $P(s)$

$$p_k=\operatorname{Pr}{X=k}, \quad k=0,1,2, \ldots$$

$$p_k=\frac{1}{k !}\left[\frac{d^k P(s)}{d s^k}\right]_{s=0} ;$$
$p_k$也由$P(s)$展开为$s$的幂级数时的$s^k$的系数给出。

$$P(s)=\frac{a_1}{s_1-s}+\cdots+\frac{a_r}{s_r-s},$$

$$a_i=-U\left(s_i\right) / V^{\prime}\left(s_i\right) .$$

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。

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数学代写|随机过程代写Stochastic Porcess代考|Processes with continuous time

Let $\xi(t), t \geqslant 0$, be a homogeneous Markov branching process with continuous time (parameter). Let $\mathscr{X}$ denote as above the phase space of the process $\xi(t)$ which is an $m$-dimensional lattice of vectors with non-negative integer-valued components. The transition probabilities of the process $\xi(t)$ will be denoted by $p_t(x, y)$ and, as before, in place of $p_t\left(e_i, y\right), p_t\left(x, e_j\right)$ and $p_t\left(e_i, e_j\right)$ we shall write $p_t(i, y), p_t(x, j), p_t(i, j)$ respectively. We shall assume that the transition probabilities satisfy the condition
$$\lim {t \downarrow 0} p_t(x, y)=\delta(x, y) .$$ First we shall discuss Kolmogorov’s differential equations for branching processes. In accordance with the general theory of homogeneous Markov processes the limits $$\lim {t \downarrow 0} \frac{p_t(x, y)-\delta(x, y)}{t}=q(x, y)$$
exist. We shall consider only regular branching processes, i. e. we shall assume that the conditions
$$-q(x, y)<\infty, \quad \sum_{y \in \mathcal{X}} q(x, y)=0$$
are satisfied.

数学代写|随机过程代写Stochastic Porcess代考|Moments (continuous time)

Assume that
$$\sum_{x \in \mathscr{X}} q(i, x) x^k=\alpha_i^k \neq \infty \quad(k, i=1, \ldots, m) .$$
Since the functions $Q(i, w)$ are analytic in the domain $|w|<1$, one can differentiate equations (32) in this domain. We then obtain:
$$\frac{d a_j^k(t, w)}{d t}=\sum_{r=1}^m Q_j^r\left(g_t(w)\right) a_r^k(t, w), \quad a_j^k(0, w)=\delta_j^k,$$
where
$$Q_j^k(t, w)=\frac{\partial Q(j, w)}{\partial w_k}=\sum_{x \in \mathscr{X}} q(j, x) x^k w^{x-e_k}$$
and
$$a_j^k(t, w)=\frac{\partial g_t(j, w)}{\partial w_k} .$$
Assume that the components of the vector $w$ are positive and $w_k \uparrow 1$. Then, by Lebesgue’s theorem,
$$\lim {w \uparrow 1} a_j^k(t, w)=\lim {w \uparrow 1} \mathrm{E}j \xi^k(t) w^{\xi(t)}=\mathrm{E}_j \xi^k(t)=a_j^k(t),$$ and, by Dini’s theorem, $g_t(w) \rightarrow 1$ uniformly in $t$. Approaching the limit as $w \uparrow 1$ in $$a_j^k(t, w)=\delta_j^k+\int_0^t \sum{r=1}^m Q_j^r\left(g_s(w)\right) a_r^k(s, w) d s,$$
we obtain
$$a_j^k(t)=\delta_j^k+\int_0^t \sum_{r=1}^m \alpha_j^r a_r^k(s) d s .$$

数学代写|随机过程代写Stochastic Porcess代考|Processes with continuous time

$$\lim {t \downarrow 0} p_t(x, y)=\delta(x, y) .$$首先，我们将讨论柯尔莫哥洛夫分支过程的微分方程。根据齐次马尔可夫过程的一般理论，极限$$\lim {t \downarrow 0} \frac{p_t(x, y)-\delta(x, y)}{t}=q(x, y)$$

$$-q(x, y)<\infty, \quad \sum_{y \in \mathcal{X}} q(x, y)=0$$

数学代写|随机过程代写Stochastic Porcess代考|Moments (continuous time)

$$\sum_{x \in \mathscr{X}} q(i, x) x^k=\alpha_i^k \neq \infty \quad(k, i=1, \ldots, m) .$$

$$\frac{d a_j^k(t, w)}{d t}=\sum_{r=1}^m Q_j^r\left(g_t(w)\right) a_r^k(t, w), \quad a_j^k(0, w)=\delta_j^k,$$

$$Q_j^k(t, w)=\frac{\partial Q(j, w)}{\partial w_k}=\sum_{x \in \mathscr{X}} q(j, x) x^k w^{x-e_k}$$

$$a_j^k(t, w)=\frac{\partial g_t(j, w)}{\partial w_k} .$$

$$\lim {w \uparrow 1} a_j^k(t, w)=\lim {w \uparrow 1} \mathrm{E}j \xi^k(t) w^{\xi(t)}=\mathrm{E}j \xi^k(t)=a_j^k(t),$$，根据迪尼定理，$g_t(w) \rightarrow 1$均匀分布于$t$。接近$$a_j^k(t, w)=\delta_j^k+\int_0^t \sum{r=1}^m Q_j^r\left(g_s(w)\right) a_r^k(s, w) d s,$$中的$w \uparrow 1$的极限 我们得到 $$a_j^k(t)=\delta_j^k+\int_0^t \sum{r=1}^m \alpha_j^r a_r^k(s) d s .$$

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。

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

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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。

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数学代写|随机过程代写Stochastic Porcess代考|INDEPENDENCE

Let $A$ and $B$ be two events of a sample space $S$, and assume that $P(A)>0$ and $P(B)>0$. We have seen that, in general, the conditional probability of $A$ given $B$ is not equal to the probability of $A$. However, if it is, that is, if $P(A \mid B)=P(A)$, we say that $A$ is independent of $B$. This means that if $A$ is independent of $B$, knowledge regarding the occurrence of $B$ does not change the chance of the occurrence of $A$. The relation $P(A \mid B)=P(A)$ is equivalent to the relations $P(A B) / P(B)=P(A), P(A B)=P(A) P(B), P(B A) / P(A)=P(B)$, and $P(B \mid A)=P(B)$. The equivalence of the first and last of these relations implies that if $A$ is independent of $B$, then $B$ is independent of $A$. In other words, if knowledge regarding the occurrence of $B$ does not change the chance of occurrence of $A$, then knowledge regarding the occurrence of $A$ does not change the chance of occurrence of $B$. Hence independence is a symmetric relation on the set of all events of a sample space. As a result of this property, instead of making the definitions ” $A$ is independent of $B$ ” and ” $B$ is independent of $A$,” we simply define the concept of the “independence of $A$ and $B$.” To do so, we take $P(A B)=$ $P(A) P(B)$ as the definition. We do this because a symmetrical definition relating $A$ and $B$ does not readily follow from either of the other relations given [i.e., $P(A \mid B)=P(A)$ or $P(B \mid A)=P(B)]$. Moreover, these relations require either that $P(B)>0$ or $P(A)>0$, whereas our definition does not.
Definition 3.3 Two events $A$ and $B$ are called independent if
$$P(A B)=P(A) P(B) .$$
If two events are not independent, they are called dependent. If $A$ and $B$ are independent, we say that ${A, B}$ is an independent set of events.

Note that in this definition we did not require $P(A)$ or $P(B)$ to be strictly positive. Hence by this definition any event $A$ with $P(A)=0$ or 1 is independent of every event $B$ (see Exercise 16).

数学代写|随机过程代写Stochastic Porcess代考|RANDOM VARIABLES

In real-world problems we are often faced with one or more quantities that do not have fixed values. The values of such quantities depend on random actions, and they usually change from one experiment to another. For example, the number of babies born in a certain hospital each day is not a fixed quantity. It is a complicated function of many random factors that vary from one day to another. So are the following quantities: the arrival time of a bus at a station, the sum of the outcomes of two dice when thrown, the amount of rainfall in Seattle during a given year, the number of earthquakes that occur in California per month, and the weight of grains of wheat grown on a certain plot of land (it varies from one grain to another). In probability, quantities introduced in these diverse examples are called random variables. The numerical values of random variables are unknown. They depend on random elements occurring at the time of the experiment and over which we have no control. For example, if in rolling two fair dice, $X$ is the sum, then $X$ can only assume the values $2,3,4, \ldots, 12$ with the following probabilities:
\begin{aligned} & P(X=2)=P({(1,1)})=1 / 36, \ & P(X=3)=P({(1,2),(2,1)})=2 / 36, \ & P(X=4)=P({(1,3),(2,2),(3,1)})=3 / 36, \end{aligned}
and, similarly,
\begin{tabular}{c|cccccccc}
Sum, $i$ & 5 & 6 & 7 & 8 & 9 & 10 & 11 & 12 \
\hline$P(X=i)$ & $4 / 36$ & $5 / 36$ & $6 / 36$ & $5 / 36$ & $4 / 36$ & $3 / 36$ & $2 / 36$ & $1 / 36$
\end{tabular}
Clearly, ${2,3,4, \ldots, 12}$ is the set of possible values of $X$. Since $X \in{2,3,4, \ldots, 12}$, we should have $\sum_{i=2}^{12} P(X=i)=1$, which is readily verified. The numerical value of a random variable depends on the outcome of the experiment. In this example, for instance, if the outcome is $(2,3)$, then $X$ is 5 , and if it is $(5,6)$, then $X$ is $11 . X$ is not defined for points that do not belong to $S$, the sample space of the experiment. Thus $X$ is a real-valued function on $S$. However, not all real-valued functions on $S$ are considered to be random variables. For theoretical reasons, it is necessary that the inverse image of an interval in $\mathbf{R}$ be an event of $S$, which motivates the following definition.

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。

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数学代写|随机过程代写Stochastic Porcess代考|DISCRETE RANDOM VARIABLES

In Section 4.1 we observed that the set of possible values of a random variable might be finite, infinite but countable, or uncountable. For example, let $X, Y$, and $Z$ be three random variables representing the respective number of tails in flipping a coin twice, the number of flips until the first heads, and the amount of next year’s rainfall. Then the sets of possible values for $X, Y$, and $Z$ are the finite set ${0,1,2}$, the countable set ${1,2,3,4, \ldots}$, and the uncountable set ${x: x \geq 0}$, respectively. Whenever the set of possible values that a random variable $X$ can assume is at most countable, $X$ is called discrete. Therefore, $X$ is discrete if either the set of its possible values is finite or it is countably infinite. To each discrete random variable, a real-valued function $p: \mathbf{R} \rightarrow \mathbf{R}$, defined by $p(x)=P(X=x)$, is assigned and is called the probability mass function of $X$. (It is also called the probability function of $X$ or the discrete probability function of $X$.) Since the set of values of $X$ is countable, $p(x)$ is positive at most for a countable set. It is zero elsewhere; that is, if possible values of $X$ are $x_1, x_2, x_3, \ldots$, then $p\left(x_i\right) \geq 0(i=1,2,3, \ldots)$ and $p(x)=0$ if $x \notin\left{x_1, x_2, x_3, \ldots\right}$. Now, clearly, the occurrence of the event $\left{x_1, x_2, x_3, \ldots\right}$ is certain. Therefore, we have that $\sum_{i=1}^{\infty} P\left(X=x_i\right)=1$ or, equivalently, $\sum_{i=1}^{\infty} p\left(x_i\right)=1$.

Definition 4.3 The probability mass function $p$ of a random variable $X$ whose set of possible values is $\left{x_1, x_2, x_3, \ldots\right}$ is a function from $\mathbf{R}$ to $\mathbf{R}$ that satisfies the following properties.
(a) $p(x)=0$ if $x \notin\left{x_1, x_2, x_3, \ldots\right}$.
(b) $p\left(x_i\right)=P\left(X=x_i\right)$ and hence $p\left(x_i\right) \geq 0(i=1,2,3, \ldots)$.
(c) $\sum_{i=1}^{\infty} p\left(x_i\right)=1$.

数学代写|随机过程代写Stochastic Porcess代考|EXPECTATIONS OF DISCRETE RANDOM VARIABLES

To clarify the concept of expectation, consider a casino game in which the probability of losing $\$ 1$per game is 0.6 , and the probabilities of winning$\$1, \$ 2$, and$\$3$ per game are $0.3,0.08$, and 0.02 , respectively. The gain or loss of a gambler who plays this game only a few times depends on his luck more than anything else. For example, in one play of the game, a lucky gambler might win $\$ 3$, but he has a$60 \%$chance of losing$\$1$. However, if a gambler decides to play the game a large number of times, his loss or gain depends more on the number of plays than on his luck. A calculating player argues that if he plays the game $n$ times, for a large $n$, then in approximately $(0.6) n$ games he will lose $\$ 1$per game, and in approximately$(0.3) n,(0.08) n$, and$(0.02) n$games he will win$\$1, \$ 2$, and$\$3$, respectively. Therefore, his total gain is
$$(0.6) n \cdot(-1)+(0.3) n \cdot 1+(0.08) n \cdot 2+(0.02) n \cdot 3=(-0.08) n .$$
This gives an average of $\$-0.08$, or about 8 cents of loss per game. The more the gambler plays, the less luck interferes and the closer his loss comes to$\$0.08$ per game. If $X$ is the random variable denoting the gain in one play, then the number -0.08 is called the expected value of $X$. We write $E(X)=-0.08 . E(X)$ is the average value of $X$. That is, if we play the game $n$ times and find the average of the values of $X$, then as $n \rightarrow \infty, E(X)$ is obtained. Since, for this game, $E(X)<0$, we have that, on the average, the more we play, the more we lose. If for some game $E(X)=0$, then in the long run the player neither loses nor wins. Such games are called fair. In this example, $X$ is a discrete random variable with the set of possible values ${-1,1,2,3}$. The probability mass function of $X, p(x)$, is given by
\begin{tabular}{c|cccc}
$i$ & -1 & 1 & 2 & 3 \
\hline$p(i)=P(X=i)$ & 0.6 & 0.3 & 0.08 & 0.02
\end{tabular}
and $p(x)=0$ if $x \notin{-1,1,2,3}$. Dividing both sides of (4.1) by $n$, we obtain
$$(0.6) \cdot(-1)+(0.3) \cdot 1+(0.08) \cdot 2+(0.02) \cdot 3=-0.08 \text {. }$$

数学代写|随机过程代写Stochastic Porcess代考|DISCRETE RANDOM VARIABLES

4.3随机变量$X$的可能值集为$\left{x_1, x_2, x_3, \ldots\right}$，其概率质量函数$p$是一个从$\mathbf{R}$到$\mathbf{R}$的函数，满足以下性质。
(a) $p(x)=0$如果$x \notin\left{x_1, x_2, x_3, \ldots\right}$。
(b) $p\left(x_i\right)=P\left(X=x_i\right)$，因此$p\left(x_i\right) \geq 0(i=1,2,3, \ldots)$。
(c) $\sum_{i=1}^{\infty} p\left(x_i\right)=1$。

数学代写|随机过程代写Stochastic Porcess代考|EXPECTATIONS OF DISCRETE RANDOM VARIABLES

$$(0.6) n \cdot(-1)+(0.3) n \cdot 1+(0.08) n \cdot 2+(0.02) n \cdot 3=(-0.08) n .$$

\begin{tabular}{c|cccc}
$i$ & -1 & 1 & 2 & 3 \hline$p(i)=P(X=i)$ & 0.6 & 0.3 & 0.08 & 0.02
\end{tabular}

$$(0.6) \cdot(-1)+(0.3) \cdot 1+(0.08) \cdot 2+(0.02) \cdot 3=-0.08 \text {. }$$

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。

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数学代写|随机过程代写Stochastic Porcess代考|THE MULTIPLICATION RULE

The relation
$$P(A \mid B)=\frac{P(A B)}{P(B)}$$
is also useful for calculating $P(A B)$. If we multiply both sides of this relation by $P(B)$ [note that $P(B)>0]$, we get
$$P(A B)=P(B) P(A \mid B)$$
which means that the probability of the joint occurrence of $A$ and $B$ is the product of the probability of $B$ and the conditional probability of $A$ given that $B$ has occurred. If $P(A)>0$, then by letting $A=B$ and $B=A$ in (3.4), we obtain
$$P(B A)=P(A) P(B \mid A)$$
Since $P(B A)=P(A B)$, this relation gives
$$P(A B)=P(A) P(B \mid A)$$
Thus, to calculate $P(A B)$, depending on which of the quantities $P(A \mid B)$ and $P(B \mid A)$ is known, we may use (3.4) or (3.5), respectively. The following example clarifies the usefulness of these relations.

数学代写|随机过程代写Stochastic Porcess代考|LAW OF TOTAL PROBABILITY

Sometimes it is not possible to calculate directly the probability of the occurrence of an event $A$, but it is possible to find $P(A \mid B)$ and $P\left(A \mid B^c\right)$ for some event $B$. In such cases, the following theorem, which is conceptually rich and has widespread applications, is used. It states that $P(A)$ is the weighted average of the probability of $A$ given that $B$ has occurred and probability of $A$ given that it has not occurred.

Theorem 3.3 (Law of Total Probability) Let $B$ be an event with $P(B)>0$ and $P\left(B^c\right)>0$. Then for any event $A$,
$$P(A)=P(A \mid B) P(B)+P\left(A \mid B^c\right) P\left(B^c\right)$$
Proof: By Theorem 1.7,
$$P(A)=P(A B)+P\left(A B^c\right)$$
Now $P(B)>0$ and $P\left(B^c\right)>0$. These imply that $P(A B)=P(A \mid B) P(B)$ and $P\left(A B^c\right)=P\left(A \mid B^c\right) P\left(B^c\right)$. Putting these in (3.7), we have proved the theorem.

Example 3.13 An insurance company rents $35 \%$ of the cars for its customers from agency I and $65 \%$ from agency II. If $8 \%$ of the cars of agency I and $5 \%$ of the cars of agency II break down during the rental periods, what is the probability that a car rented by this insurance company breaks down?

Solution: Let $A$ be the event that a car rented by this insurance company breaks down. Let I and II be the events that it is rented from agencies I and II, respectively. Then by the law of total probability,
\begin{aligned} P(A) & =P(A \mid \mathrm{I}) P(\mathrm{I})+P(A \mid \mathrm{II}) P(\mathrm{II}) \ & =(0.08)(0.35)+(0.05)(0.65)=0.0605 . \end{aligned}
Tree diagrams facilitate solutions to this kind of problem. Let $B$ and $B^c$ stand for breakdown and not breakdown during the rental period, respectively. Then, as Figure 3.2 shows, to find the probability that a car breaks down, all we need to do is to compute, by multiplication, the probability of each path that leads to a point $B$ and then add them up. So, as seen from the tree, the probability that the car breaks down is $(0.35)(0.08)+(0.65)(0.05)=0.06$, and the probability that it does not break down is $(0.35)(0.92)+(0.65)(0.95)=0.94$.

数学代写|随机过程代写Stochastic Porcess代考|THE MULTIPLICATION RULE

$$P(A \mid B)=\frac{P(A B)}{P(B)}$$

$$P(A B)=P(B) P(A \mid B)$$

$$P(B A)=P(A) P(B \mid A)$$

$$P(A B)=P(A) P(B \mid A)$$

数学代写|随机过程代写Stochastic Porcess代考|LAW OF TOTAL PROBABILITY

$$P(A)=P(A \mid B) P(B)+P\left(A \mid B^c\right) P\left(B^c\right)$$

$$P(A)=P(A B)+P\left(A B^c\right)$$

\begin{aligned} P(A) & =P(A \mid \mathrm{I}) P(\mathrm{I})+P(A \mid \mathrm{II}) P(\mathrm{II}) \ & =(0.08)(0.35)+(0.05)(0.65)=0.0605 . \end{aligned}

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。

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

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数学代写|随机过程代写Stochastic Porcess代考|CONTINUITY OF PROBABILITY FUNCTIONS

Let $\mathbf{R}$ denote (here and everywhere else throughout the book) the set of all real numbers. We know from calculus that a function $f: \mathbf{R} \rightarrow \mathbf{R}$ is called continuous at a point $c \in \mathbf{R}$ if $\lim {x \rightarrow c} f(x)=f(c)$. It is called continuous on $\mathbf{R}$ if it is continuous at all points $c \in \mathbf{R}$. We also know that this definition is equivalent to the sequential criterion $f: \mathbf{R} \rightarrow \mathbf{R}$ is continuous on $\mathbf{R}$ if and only if, for every convergent sequence $\left{x_n\right}{n=1}^{\infty}$ in $\mathbf{R}$,
$$\lim {n \rightarrow \infty} f\left(x_n\right)=f\left(\lim {n \rightarrow \infty} x_n\right)$$
This property, in some sense, is shared by the probability function. To explain this, we need to introduce some definitions. But first recall that probability is a set function from $\mathcal{P}(S)$, the set of all possible events of the sample space $S$, to $[0,1]$.
A sequence $\left{E_n, n \geq 1\right}$ of events of a sample space is called increasing if
$$E_1 \subseteq E_2 \subseteq E_3 \subseteq \cdots \subseteq E_n \subseteq E_{n+1} \cdots ;$$

it is called decreasing if
$$E_1 \supseteq E_2 \supseteq E_3 \supseteq \cdots \supseteq E_n \supseteq E_{n+1} \supseteq \cdots$$
For an increasing sequence of events $\left{E_n, n \geq 1\right}$, by $\lim {n \rightarrow \infty} E_n$ we mean the event that at least one $E_i, 1 \leq i<\infty$ occurs. Therefore, $$\lim {n \rightarrow \infty} E_n=\bigcup_{i=1}^{\infty} E_i .$$
Similarly, for a decreasing sequence of events $\left{E_n, n \geq 1\right}$, by $\lim {n \rightarrow \infty} E_n$ we mean the event that every $E_i$ occurs. Thus in this case $$\lim {n \rightarrow \infty} E_n=\bigcap_{i=1}^{\infty} E_i .$$
The following theorem expresses the property of probability function that is analogous to (1.4).

数学代写|随机过程代写Stochastic Porcess代考|PROBABILITIES 0 AND 1

Events with probabilities 1 and 0 should not be misinterpreted. If $E$ and $F$ are events with probabilities 1 and 0 , respectively, it is not correct to say that $E$ is the sample space $S$ and $F$ is the empty set $\emptyset$. In fact, there are experiments in which there exist infinitely many events each with probability 1 , and infinitely many events each with probability 0 . An example follows.
Suppose that an experiment consists of selecting a random point from the interval $(0,1)$. Since every point in $(0,1)$ has a decimal representation such as
$0.529387043219721 \cdots$,
the experiment is equivalent to picking an endless decimal from $(0,1)$ at random (note that if a decimal terminates, all of its digits from some point on are 0 ). In such an experiment we want to compute the probability of selecting the point $1 / 3$. In other words, we want to compute the probability of choosing $0.333333 \cdots$ in a random selection of an endless decimal. Let $A_n$ be the event that the selected decimal has 3 as its first $n$ digits; then
$$A_1 \supset A_2 \supset A_3 \supset A_4 \supset \cdots \supset A_n \supset A_{n+1} \supset \cdots$$

since the occurrence of $A_{n+1}$ guarantees the occurrence of $A_n$. Now $P\left(A_1\right)=1 / 10$ because there are 10 choices $0,1,2, \ldots, 9$ for the first digit, and we want only one of them, namely 3 , to occur. $P\left(A_2\right)=1 / 100$ since there are 100 choices $00,01, \ldots, 09,10,11, \ldots, 19,20$, $\ldots, 99$ for the first two digits, and we want only one of them, 33, to occur. $P\left(A_3\right)=1 / 1000$ because there are 1000 choices $000,001, \ldots, 999$ for the first three digits, and we want only one of them, 333, to occur. Continuing this argument, we have $P\left(A_n\right)=(1 / 10)^n$. Since $\bigcap_{n=1}^{\infty} A_n={1 / 3}$, by Theorem 1.8 ,
$$P\left(\frac{1}{3} \text { is selected }\right)=P\left(\bigcap_{n=1}^{\infty} A_n\right)=\lim {n \rightarrow \infty} P\left(A_n\right)=\lim {n \rightarrow \infty}\left(\frac{1}{10}\right)^n=0 .$$
Note that there is nothing special about the point $1 / 3$. For any other point $0 . \alpha_1 \alpha_2 \alpha_3 \alpha_4 \cdots$ from $(0,1)$, the same argument could be used to show that the probability of its occurrence is 0 (define $A_n$ to be the event that the first $n$ digits of the selected decimal are $\alpha_1, \alpha_2, \ldots, \alpha_n$, respectively, and repeat the same argument). We have shown that in random selection of points from $(0,1)$, the probability of the occurrence of any particular point is 0 . Now for $t \in(0,1)$, let $B_t=(0,1)-{t}$. Then $P({t})=0$ implies that
$$P\left(B_t\right)=P\left({t}^c\right)=1-P({t})=1$$

数学代写|随机过程代写Stochastic Porcess代考|CONTINUITY OF PROBABILITY FUNCTIONS

$$\lim {n \rightarrow \infty} f\left(x_n\right)=f\left(\lim {n \rightarrow \infty} x_n\right)$$

$$E_1 \subseteq E_2 \subseteq E_3 \subseteq \cdots \subseteq E_n \subseteq E_{n+1} \cdots ;$$

$$E_1 \supseteq E_2 \supseteq E_3 \supseteq \cdots \supseteq E_n \supseteq E_{n+1} \supseteq \cdots$$

数学代写|随机过程代写Stochastic Porcess代考|PROBABILITIES 0 AND 1

$0.529387043219721 \cdots$，

$$A_1 \supset A_2 \supset A_3 \supset A_4 \supset \cdots \supset A_n \supset A_{n+1} \supset \cdots$$

$$P\left(\frac{1}{3} \text { is selected }\right)=P\left(\bigcap_{n=1}^{\infty} A_n\right)=\lim {n \rightarrow \infty} P\left(A_n\right)=\lim {n \rightarrow \infty}\left(\frac{1}{10}\right)^n=0 .$$

$$P\left(B_t\right)=P\left({t}^c\right)=1-P({t})=1$$

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。

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数学代写|随机过程代写Stochastic Porcess代考|Definitions. General Properties

Let $\mathscr{X}$ be a linear space with $\sigma$-algebra $\mathfrak{B}$ possessing the following properties: a) for all $x \in \mathscr{X}$ and $A \in \mathfrak{B}$ the set $A_x={y: y-x \in A}$ belongs to $\mathfrak{B}$ (i. e. $\sigma$-algebra $\mathfrak{B}$ is such that all the shifts in $\mathscr{X}$ are measurable with respect to $\mathfrak{B}) ;$ b) for any $A \in \mathfrak{B}$ the set ${(x+y): x+y \in A}$ is $\mathfrak{B} \times \mathfrak{B}$-measurable in $\mathscr{X} \times \mathscr{X}$.

A process $\xi(t)$ defined on a set $T \subset \mathscr{R}$ and taking values in $\mathscr{X}$ is called a process with independent increments if, for all $t_0<t_1<\cdots<t_n$ belonging to $T$, the random variables $\xi\left(t_0\right), \xi\left(t_1\right)-\xi\left(t_0\right), \ldots, \xi\left(t_n\right)-\xi\left(t_{n-1}\right)$ are independent. Conditions imposed on $\mathfrak{B}$ assure that $\xi(t)-\xi\left(t_1\right)$ is a random variable.

Marginal distributions of a process with independent increments are determined by the one-dimensional distributions and the distributions of the increments of the process. Let
$$\mu_t(A)=\mathrm{P}{\xi(t) \in A}, \quad \Phi_{t_1, t_2}(A)=\mathrm{P}\left{\xi\left(t_2\right)-\xi\left(t_1\right) \in A\right}$$
Then
$$\mathrm{P}\left{\xi\left(t_0\right) \in A_0, \xi\left(t_1\right) \in A_1, \ldots, \xi\left(t_n\right) \in A_n\right}=\int \cdots \int \mu_{t_0}\left(d x_0\right) \Phi_{t_0, t_1}\left(d x_1\right), \ldots, \Phi_{t_{n-1}, t_n}\left(d x_n\right),$$
where integration is carried out over the set
$$\left{\left(x_0, \ldots, x_n\right): x_0 \in A_0, x_0+x_1 \in A_1, \ldots, x_0+\cdots+x_n \in A_n\right}$$

数学代写|随机过程代写Stochastic Porcess代考|One-dimensional processes with independent increments

One-dimensional processes with independent increments. Note that for any $\mathfrak{B}$-measurable linear functional $l(x)$ the process $\eta(t)=l(\xi(t))$ will be a onedimensional process with independent increments. Therefore a study of onedimensional processes with independent increments may supply (and, as we shall see in the sequel, indeed does supply) non-trivial information about processes in more complex spaces. On the other hand, the space $\mathscr{R}^1$ is the simplest linear space so that processes in this space are also the simplest in a certain sense.
Thus we shall consider a process $\xi(t)$ taking on real values; the $\sigma$-algebra of all Borel sets on $\mathscr{R}^1$ will serve as the $\sigma$-algebra $\mathfrak{B}$. We shall assume that the process is defined on a set $T$.

Let $\varphi_t(\lambda)$ and $\varphi_{t_1, t_2}(\lambda)$ be characteristic functions of $\xi(t)$ and $\xi\left(t_2\right)-\xi\left(t_1\right)$ respectively. The functions $h_t(\lambda)=\left|\varphi_t(\lambda)\right|^2$ and $h_{t_1, t_2}(\lambda)=\left|\varphi_{t_1, t_2}(\lambda)\right|^2$ are non-negative and, moreover, $h_{t_1, t_2}(\lambda) \leqslant 1$. It follows from the relation
$$h_s(\lambda)=h_t(\lambda) h_{t, s}(\lambda), \quad t<s$$

that $h_s(\lambda)$ is a monotonically non-increasing bounded function of $s$. Consequently the limits $h_{t-0}(\lambda)$ and $h_{t+0}(\lambda)$ exist for $t$ belonging to the closure of $T$ (or only one limit exists if $t$ is a one-sided limit point; these limits are not defined for isolated points).

Consider in addition to $\xi(t)$ a process $\tilde{\xi}(t)$ with the same finite-dimensional distributions as those of $\xi(t)$ but independent of $\xi(t)$. To construct such a process, we take two copies of the same probability space and view the process $\xi(t)$ on the first space and the identical process $\tilde{\xi}(t)$ on the second as processes on the product of these probability spaces. Next set $\xi^(t)=\xi(t)-\tilde{\xi}(t)$. It is easy to see that $$\mathrm{E} e^{i \lambda \xi^(t)}=h_t(\lambda), \quad \mathrm{E} e^{i \lambda\left[\xi^\left(t_2\right)-\xi^\left(t_1\right)\right]}=h_{t_1, t_2}(\lambda)$$

数学代写|随机过程代写Stochastic Porcess代考|Definitions. General Properties

$$\mu_t(A)=\mathrm{P}{\xi(t) \in A}, \quad \Phi_{t_1, t_2}(A)=\mathrm{P}\left{\xi\left(t_2\right)-\xi\left(t_1\right) \in A\right}$$

$$\mathrm{P}\left{\xi\left(t_0\right) \in A_0, \xi\left(t_1\right) \in A_1, \ldots, \xi\left(t_n\right) \in A_n\right}=\int \cdots \int \mu_{t_0}\left(d x_0\right) \Phi_{t_0, t_1}\left(d x_1\right), \ldots, \Phi_{t_{n-1}, t_n}\left(d x_n\right),$$

$$\left{\left(x_0, \ldots, x_n\right): x_0 \in A_0, x_0+x_1 \in A_1, \ldots, x_0+\cdots+x_n \in A_n\right}$$

数学代写|随机过程代写Stochastic Porcess代考|One-dimensional processes with independent increments

$$h_s(\lambda)=h_t(\lambda) h_{t, s}(\lambda), \quad t<s$$

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。