分类：随机过程代写

澳洲代写随机过程Stochastic process代考2023

Posted on 2023年10月10日2023年10月10日 by statistics-lab

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随机过程Stochastic process代考

在数学中，更确切地说，在概率论中，随机过程（或随机过程）是动态系统概念的概率版本。随机过程是某个参数（通常是时间）的实函数的有序集合，具有某些统计特性。一般来说，可以将这一过程确定为实数随机变量 $X(t)$ 的单参数族，代表从初始状态到一定时间 $t$ 之后的状态的变换。更精确地说，这是基于实数极限上的随机变量（如 $\mathbb{R}^n$，或函数空间，或实数的连续）。当时间参数也被考虑在内时，随机过程就是随机变量概念的扩展。

随机过程Stochastic process包含几个不同的主题，列举如下：

泛函分析Functional analysis代写代考

函数分析（也写作函数分析，Analyse fonctionnelle，法语。也称拓扑分析）。是数学（尤其是分析）的一个分支，起源于对傅立叶变换、微分方程和积分方程的研究。它研究函数空间的结构，为由某类函数组成的向量空间定义了某种拓扑结构，并研究由其公理化得到的线性拓扑空间。主要兴趣在于通过各种函数空间上由积分和微分定义的线性算子的行为，对积分和微分方程进行线性代数处理，这通常被视为无穷维向量空间上的线性代数。它也可被视为无穷维空间上的微分和积分微积分，因为它涉及无穷维空间上的导数（如弗雷谢特导数）。

傅立叶分析Fourier analysis代写代考

在数学分析中，傅里叶分析，也称为调和分析，是一个研究分支，始于让·巴蒂斯特·约瑟夫·傅里叶的研究，他在十九世纪初成功地用数学方法证明了如何将任何周期函数分解为无限“适当”正弦函数或分量（正弦和余弦）的总和，称为谐波。从这一观察中诞生了将复杂函数分解为一系列函数（称为傅里叶级数）的想法，使它们的分析更简单、更有利。傅里叶变换的概念以及相关的频域概念也源自傅里叶级数的数学概念。

其他相关科目课程代写：

数学模型Mathematical model
线性代数Linear algebra
概率学Probability

随机过程Stochastic process定义

随机或随机过程可以定义为由某个数学集合索引的随机变量集合，这意味着随机过程中的每个随机变量都与集合中的某个元素唯一相关。从历史上看，索引集是实线的某个子集，如自然数，这就赋予了索引集时间的解释[1]。例如，状态空间可以是整数、实线或……。n 维欧几里得空间]。增量是随机过程在两个指数值之间的变化量，通常被解释为两个时间点。由于随机性，随机过程可以有很多结果，随机过程的单一结果被称为样本函数或实现。

许多领域使用观测值作为时间的函数（或者，更罕见的是，空间变量）。在最简单的情况下，这些观察结果会产生一条明确的曲线。事实上，从地球科学到人文学科，观测或多或少都会出现不稳定的情况。因此，对这些观察结果的解释存在一定的不确定性，这可能反映在使用概率来表示它们上。

随机过程概括了概率中使用的随机变量的概念。它被定义为与所有值 t ∈ T（通常是时间）相关的一系列随机变量 X(t)。

从统计的角度来看，我们将所有可用的观测值x(t)视为过程的实现，这会带来一定的困难。第一个问题涉及以下事实：构建过程的持续时间通常是无限的，而实现则涵盖有限的持续时间。因此，不可能完美地再现现实。第二个更严重的困难是，与随机变量问题不同，有关过程的可用信息通常被简化为单个实现。

A stochastic or stochastic process can be defined as a collection of random variables indexed by some mathematical set, which means that each random variable in the stochastic process is uniquely related to some element in the set. Historically, the index set was some subset of the real lines, such as the natural numbers, which gave the index set a temporal interpretation. For example, the state space can be integers, solid lines, or… n-dimensional Euclidean space]. An increment is the amount of change of a random process between two exponential values, usually interpreted as two points in time. Due to randomness, a random process can have many outcomes, and a single outcome of a random process is called a sample function or realization.

Many fields use observations as functions of time (or, more rarely, spatial variables). In the simplest case, these observations yield a well-defined curve. In fact, from the earth sciences to the humanities, observations are more or less unstable. Therefore, there is a certain uncertainty in the interpretation of these observations, which may be reflected in the use of probabilities to express them.

Stochastic processes generalize the concept of random variables used in probability. It is defined as a sequence of random variables X(t) related to all values t ∈ T (usually time).

From a statistical perspective, we treat all available observations x(t) as realizations of the process, which creates certain difficulties. The first problem concerns the fact that the duration of the build process is usually infinite, whereas the implementation covers a finite duration. Therefore, it is impossible to perfectly reproduce reality. A second, more serious difficulty is that, unlike random variable problems, the available information about the process is often reduced to a single realization.

统计代写请认准statistics-lab™. statistics-lab™为您的留学生涯保驾护航。

随机过程Stochastic process的重难点

什么是伯努利过程Bernoulli process？

每个单一随机变量 $X_{mathrm{i}}$ 只能提供两种结果：成功 (1) 或失败 $(0)$，概率分别为 $p$ 和 $q=1$ – $p$ ：

$p$ e $q=1$ – $p$ :
$$
\begin{aligned}
& P\left(X_i=1\right)=p ; \
& P\left(X_i=0\right)=q=1-p .
\end{aligned}
$$

遵循二项式定律 $B(n,p)$，概率为

$$
P\left(S_n=k\right)=\left(\begin{array}{l}
n \
k
\end{array}\right) p^k q^{n-k}
$$

等于$k$成功和$n-k$失败的序列数乘以其中任何一个出现的概率。
获得一次成功所需的投掷次数由随机变量 $N$ 给出，该随机变量遵循几何比率定律 $q$。

$$
P(N=n)=P\left(S_{n-1}=0\right) \cdot P\left(X_n=1\right)=q^n \frac{p}{q} .
$$

更一般地说，获得 $k$ 成功所需的投掷次数由随机变量 $N_k$ 给出，其规律为
$P\left(N_k=n\right)=P\left(S_{n-1}=k-1\right) \cdot P\left(X_n=1\right)=\left(\begin{array}{c}n-1 \ k-1\end{array}\right) p^k q^{n-k}$
特别是，失败的次数由帕斯卡定律（或负二项式）的随机变量 $P_k=N_k-n$ $P(p, k)$ 给出

$$
P\left(P_k=r\right)=P\left(N_k=r+k\right)=\left(\begin{array}{c}
r+k-1 \
k-1
\end{array}\right) p^k q^r=(-1)^k\left(\begin{array}{c}
-r \
k
\end{array}\right) p^k q^r
$$
Applicazioni
[ modifica | modifica wikitesto ]

什么是随机漫步Random walk？

在一维随机游走中，研究了被约束在两个允许方向上沿直线移动的点粒子的运动。每次移动时，它都会（随机）向右移动一步（以固定概率 $p$）或以概率 $1-p$ 向左移动，并且每一步长度相等且彼此独立。我们的目标是计算在 $\mathrm{N}$ 运动后粒子将返回（如果它返回的话！）到起点的概率。我们引入以下随机变量 $X(N)$，它提供 $N$ 运动后向左迈出的步数；特别是，它模拟了经过适当操纵的硬币投掷 $N$ 后出现的正面的数量。显然这是一个二项式分布的离散随机变量。我们还注意到，事件“返回原点”相当于在 $2 N$ 总步数中向左精确走了 $N$ 步；因此，所寻求的概率相当于 $P{X=N}$，其中二项式 $X$ 的参数为 $n=2 N, k=N, p$ 因此

什么是因子分析Factor analysis？

该模型试图用一组 $k$ 的公共因子 $\left(f_{i, j}\right)$ 来解释 $n$ 个体中每个个体的 $p$ 观察值，其中每个单位的因子数量少于每个单位的观察值 $(k<p)$。每个个体都有自己的 $k$ 共同因子，这些因子通过因子载荷矩阵 $(left(L)(in\mathbb{R}^{p\times k}\right)$与观测值相关，对于单个观测值，根据

$$
P{X=N}=\frac{(2 N) !}{N !(2 N-N) !} p^N(1-p)^N=\left(\begin{array}{c}
2 N \
N
\end{array}\right)\left(p-p^2\right)^N .
$$

例如，如果粒子在每一步 $(p=1 / 2)$ 向左或向右移动的机会均等，则在 $2 N$ 步骤后返回原点的概率为

$$
P{X=N}=\left(\begin{array}{c}
2 N \
N
\end{array}\right)\left(\frac{1}{2}\right)^{2 N} \approx \frac{1}{\sqrt{N \pi}},
$$我们对足够大的 $N$ 应用斯特林近似，

$$
N ! \sim \sqrt{2 \pi N}\left(\frac{N}{e}\right)^N .
$$现在记住随机变量的期望值由下式给出

$$
E[X]=\sum_{n=0}^{\infty} n P(n)
$$

随机过程Stochastic process的相关课后作业范例

这是一篇关于多元统计分析Multivariate Statistical Analysis的作业

问题 1.

Show that in successive tosses of a fair die indefinitely, the probability of obtaining no 6 is 0 .

Solution: For $n \geq 1$, let $E_n$ be the event of at least one 6 in the first $n$ tosses of the die. Clearly,
$$
E_1 \subseteq E_2 \subseteq \cdots \subseteq E_n \subseteq E_{n+1} \subseteq \cdots .
$$
Therefore, $E_n$ ‘s form an increasing sequence of events. Note that $\lim {n \rightarrow \infty} E_n=\bigcup{i=1}^{\infty} E_i$ is the event that in successive tosses of the die indefinitely, eventually a 6 will occur. By the Continuity of the Probability Function (Theorem 1.8), we have
$$
P\left(\lim {n \rightarrow \infty} E_n\right)=\lim {n \rightarrow \infty} P\left(E_n\right)=\lim {n \rightarrow \infty}\left[1-\left(\frac{5}{6}\right)^n\right]=1-\lim {n \rightarrow \infty}\left(\frac{5}{6}\right)^n=1-0=1 .
$$
This shows that, with probability 1 , eventually a 6 will occur. Therefore, the probability of no 6 ever is 0 .

通过对随机过程Stochastic process各方面的介绍，想必您对这门课有了初步的认识。如果你仍然不确定或对这方面感到困难，你仍然可以依靠我们的代写和辅导服务。我们拥有各个领域、具有丰富经验的专家。他们将保证你的 essay、assignment或者作业都完全符合要求、100%原创、无抄袭、并一定能获得高分。需要如何学术帮助的话，随时联系我们的客服。

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统计代写|随机过程代写stochastic process代考|IE2084

Posted on 2023年8月29日2023年8月29日 by statistics-lab

如果你也在怎样代写随机过程Stochastic Porcesses 这个学科遇到相关的难题，请随时右上角联系我们的24/7代写客服。随机过程Stochastic Porcesses是指一个系统在特定的时间有观测值，结果，即每次的观测值是一个随机变量。

随机过程Stochastic Porcesses在是由一些数学集合索引的随机变量的集合。每个概率和随机过程都与集合中的一个元素唯一地相关联。索引集是用来索引随机变量的集合。索引集传统上是实数的子集，例如自然数，这为索引集提供了时间解释。

statistics-lab™ 为您的留学生涯保驾护航在代写随机过程stochastic process方面已经树立了自己的口碑, 保证靠谱, 高质且原创的统计Statistics代写服务。我们的专家在代写随机过程stochastic process代写方面经验极为丰富，各种代写随机过程stochastic process相关的作业也就用不着说。

统计代写|随机过程代写stochastic process代考|Probability distribution

It may be seen that the probability distribution of $X_r, X_{r+1}, \ldots, X_{r+n}$ can be computed in terms of the transition probabilities $p_{j k}$ and the initial distribution of $X_r$. Suppose, for simplicity, that $r=0$, then
$$
\begin{aligned}
\operatorname{Pr}\left{X_0\right. & \left.=a, X_1=b, \ldots, X_{n-2}=i, X_{n-1}=j, X_n=k\right} \
& =\operatorname{Pr}\left{X_n=k \mid X_{n-1}=j, \cdots, X_0=a\right} \operatorname{Pr}\left{X_{n-1}=j, \cdots, X_0=a\right} \
& =\operatorname{Pr}\left{X_n=k \mid X_{n-1}=j\right} \operatorname{Pr}\left{X_{n-1}=j \mid X_{n-2}=i\right} \operatorname{Pr}\left{X_{n-2}=i, \cdots, X_0=a\right} \
& =\operatorname{Pr}\left{X_n=k \mid X_{n-1}=j\right} \operatorname{Pr}\left{X_{n-1}=j \mid X_{n-2}=i\right} \cdots \operatorname{Pr}\left{X_1=b \mid X_0=a\right} \operatorname{Pr}\left{X_0=a\right} \
& =\left{\operatorname{Pr}\left(X_0=a\right)\right} p_{a b} \cdots p_{i j} p_{j k} .
\end{aligned}
$$
Thus,
$$
\begin{aligned}
\operatorname{Pr}\left{X_r\right. & \left.=a, X_{r+1}=b, \ldots, X_{r+n-2}=i, X_{r+n-1}=j, X_{r+n}=k\right} \
& =\left{\operatorname{Pr}\left(X_r=a\right)\right} p_{a b} \ldots p_{i j} p_{j k} .
\end{aligned}
$$
Example 1(g). Let $\left{X_n, n \geq 0\right}$ be a Markov chain with three states $0,1,2$ and with transition matrix
$$
\left(\begin{array}{ccc}
3 / 4 & 1 / 4 & 0 \
1 / 4 & 1 / 2 & 1 / 4 \
0 & 3 / 4 & 1 / 4
\end{array}\right)
$$
and the initial distribution $\operatorname{Pr}\left{X_0=i\right}=\frac{1}{3}, i=0,1,2$.
We have
$$
\begin{aligned}
& \operatorname{Pr}\left{X_1=1 \mid X_0=2\right}=\frac{3}{4} \
& \operatorname{Pr}\left{X_2=2 \mid X_1=1\right}=\frac{1}{4} \
& \operatorname{Pr}\left{X_2=2, X_1=1 \mid X_0=2\right} \
& =\operatorname{Pr}\left{X_2=2 \mid X_1=1\right} \operatorname{Pr}\left{X_1=1 \mid X_0=2\right}=\frac{1}{4} \cdot \frac{3}{4}=\frac{3}{16}
\end{aligned}
$$

\begin{aligned}
\operatorname{Pr}\left{X_2\right. & \left.=2, X_1=1, X_0=2\right} \
& =\operatorname{Pr}\left{X_2=2, X_1=1 \mid X_0=2\right} \operatorname{Pr}\left{X_0=2\right}=\frac{3}{16} \cdot \frac{1}{3}=\frac{1}{16} \
\operatorname{Pr}\left{X_3\right. & \left.=1, X_2=2, X_1=1, X_0=2\right} \
& =\operatorname{Pr}\left{X_3=1 \mid X_2=2, X_1=1, X_0=2\right} \times \operatorname{Pr}\left{X_2=2, X_1=1, X_0=2\right} \
& =\operatorname{Pr}\left{X_3=1 \mid X_2=2\right}\left(\frac{1}{16}\right)=\frac{3}{4} \cdot \frac{1}{16}=\frac{3}{64} .
\end{aligned}

统计代写|随机过程代写stochastic process代考|Strong Markov property

Stopping time for a sequence of r.v.’s $\left{X_n\right}$ is a random variable (see Sec. 6.4.1).
Let $N$ be a stopping time for a Markov chain $\left{X_n, n>0\right}$ and let $A$ and $B$ to two events (relating to $X_n$ and happening) prior and posterior respectively to $N$. Then
$$
\operatorname{Pr}\left{B \mid X_N=i, A\right}=\operatorname{Pr}\left{B \mid X_N=i\right} .
$$
This is called the strong Markov property. It shows that if $N$ is a stopping time for a Markov chain $\left{X_n, n>0\right}$, then the evolution of the chain starts afresh from the state reached at time $N$.

Strong Markov property is implied by the Markov property; both the properties are equivalent when $N$ is constant (a degenerate r.v.).
Every discrete time Markov chain $\left{X_n, n \geq 0\right}$ possesses the strong Markov property.

Definition. A Markov chain $\left{X_n\right}$ is said to be of order $s(s=1,2,3, \ldots)$, if, for all $n$,
$$
\begin{aligned}
\operatorname{Pr}\left{X_n\right. & \left.=k \mid X_{n-1}=j, X_{n-2}=j_1, \ldots, X_{n-s}=j_{s-1}, \ldots\right} \
& =\operatorname{Pr}\left{X_n=k \mid X_{n-1}=j, \ldots, X_{n-s}=j_{s-1}\right} .
\end{aligned}
$$
whenever the 1.h.s. is defined.
A Markov chain $\left{X_n\right}$ is said to be of order one (or simply a Markov chain) if
$$
\begin{aligned}
\operatorname{Pr}\left{X_n\right. & \left.=k \mid X_{n-1}=j, X_{n-2}=j_1, \ldots\right}=\operatorname{Pr}\left{X_n=k \mid X_{n-1}=j\right} \
& =p_{j k} .
\end{aligned}
$$
whenever $\operatorname{Pr}\left{X_{n-1}=j, X_{n-2}=j_1, \ldots\right}>0$.
Unless explicitly stated otherwise, we shall mean by Markov chain, a chain of order one, to which we shall mostly confine ourselves here. A chain is said to be of order zero if $p_{j k}=p_k$ for all $j$. This implies independence of $X_n$ and $X_{n-1}$. For example, for the Bernoulli coin tossing experiment, the t.p.m. is $\left(\begin{array}{ll}q & p \ q & p\end{array}\right)=\left(\begin{array}{l}1 \ 1\end{array}\right)(q, p)=\mathbf{e}(q, p)$.
Denote the state that a day is rainy by 1 and that a day is not rainy by 0 .
Let (1.5) hold for $s=2$ and let
$p_{i j k}=\operatorname{Pr}{$ actual day is in state $k \mid$ the preceding day was in state $j$, the day before the preceding was in state $i}, i, j, k=0,1$.
We then have a Markov chain of order two. Note that the matrix $\left(p_{i j k}\right)$ is not a stochastic matrix. It is a $(4 \times 2)$ matrix and not a square matrix.
Let (1.5) hold for $s=1$ and let
$p_{j k}=\operatorname{Pr}{$ actual day is in state $k \mid$ preceding day was in state $j}$. We then have a Markov chain (i.e. a chain of order one) with t.p.m. $\left(p_{j k}\right), j, k=0,1$.

随机过程代考

统计代写|随机过程代写stochastic process代考|Probability distribution

可以看出，$X_r, X_{r+1}, \ldots, X_{r+n}$的概率分布可以用过渡概率$p_{j k}$和$X_r$的初始分布来计算。为简单起见，假设是$r=0$
$$
\begin{aligned}
\operatorname{Pr}\left{X_0\right. & \left.=a, X_1=b, \ldots, X_{n-2}=i, X_{n-1}=j, X_n=k\right} \
& =\operatorname{Pr}\left{X_n=k \mid X_{n-1}=j, \cdots, X_0=a\right} \operatorname{Pr}\left{X_{n-1}=j, \cdots, X_0=a\right} \
& =\operatorname{Pr}\left{X_n=k \mid X_{n-1}=j\right} \operatorname{Pr}\left{X_{n-1}=j \mid X_{n-2}=i\right} \operatorname{Pr}\left{X_{n-2}=i, \cdots, X_0=a\right} \
& =\operatorname{Pr}\left{X_n=k \mid X_{n-1}=j\right} \operatorname{Pr}\left{X_{n-1}=j \mid X_{n-2}=i\right} \cdots \operatorname{Pr}\left{X_1=b \mid X_0=a\right} \operatorname{Pr}\left{X_0=a\right} \
& =\left{\operatorname{Pr}\left(X_0=a\right)\right} p_{a b} \cdots p_{i j} p_{j k} .
\end{aligned}
$$
因此，
$$
\begin{aligned}
\operatorname{Pr}\left{X_r\right. & \left.=a, X_{r+1}=b, \ldots, X_{r+n-2}=i, X_{r+n-1}=j, X_{r+n}=k\right} \
& =\left{\operatorname{Pr}\left(X_r=a\right)\right} p_{a b} \ldots p_{i j} p_{j k} .
\end{aligned}
$$
例1(g)。设$\left{X_n, n \geq 0\right}$为具有三态$0,1,2$和转移矩阵的马尔可夫链
$$
\left(\begin{array}{ccc}
3 / 4 & 1 / 4 & 0 \
1 / 4 & 1 / 2 & 1 / 4 \
0 & 3 / 4 & 1 / 4
\end{array}\right)
$$
初始分布$\operatorname{Pr}\left{X_0=i\right}=\frac{1}{3}, i=0,1,2$。
我们有
$$
\begin{aligned}
& \operatorname{Pr}\left{X_1=1 \mid X_0=2\right}=\frac{3}{4} \
& \operatorname{Pr}\left{X_2=2 \mid X_1=1\right}=\frac{1}{4} \
& \operatorname{Pr}\left{X_2=2, X_1=1 \mid X_0=2\right} \
& =\operatorname{Pr}\left{X_2=2 \mid X_1=1\right} \operatorname{Pr}\left{X_1=1 \mid X_0=2\right}=\frac{1}{4} \cdot \frac{3}{4}=\frac{3}{16}
\end{aligned}
$$

\begin{aligned}
\operatorname{Pr}\left{X_2\right. & \left.=2, X_1=1, X_0=2\right} \＆ =\operatorname{Pr}\left{X_2=2, X_1=1 \mid X_0=2\right} \operatorname{Pr}\left{X_0=2\right}＝\frac{3}{16} \cdot \frac{1}{3}＝\frac{1}{16} \operatorname{Pr}\left{X_3\right. & \left.=1, X_2=2, X_1=1, X_0=2\right} \＆ =\operatorname{Pr}\left{X_3=1 \mid X_2=2, X_1=1, X_0=2\right} \times \operatorname{Pr}\left{X_2=2, X_1=1, X_0=2\right} \＆ =\operatorname{Pr}\left{X_3=1 \mid X_2=2\right}\left（\frac{1}{16}\right）=\frac{3}{4} \cdot \frac{1}{16}＝\frac{3}{64} ．
\end{aligned}

统计代写|随机过程代写stochastic process代考|Strong Markov property

rv序列的停止时间。的$\left{X_n\right}$是一个随机变量(见第6.4.1节)。
设$N$为马尔可夫链$\left{X_n, n>0\right}$的停止时间，并设$A$和$B$分别表示两个事件(与$X_n$和正在发生的事件有关)在$N$之前和之后。然后
$$
\operatorname{Pr}\left{B \mid X_N=i, A\right}=\operatorname{Pr}\left{B \mid X_N=i\right} .
$$
这被称为强马尔可夫性质。它表明，如果$N$是马尔可夫链$\left{X_n, n>0\right}$的停止时间，则该链的进化从时间$N$所达到的状态重新开始。

强马尔可夫性质由马尔可夫性质隐含;当$N$为常数时，这两个属性是等价的(简并的r.v.)。
每一个离散时间马尔可夫链$\left{X_n, n \geq 0\right}$都具有强马尔可夫性。

定义。一个马尔可夫链$\left{X_n\right}$的阶为$s(s=1,2,3, \ldots)$，如果对于所有$n$，
$$
\begin{aligned}
\operatorname{Pr}\left{X_n\right. & \left.=k \mid X_{n-1}=j, X_{n-2}=j_1, \ldots, X_{n-s}=j_{s-1}, \ldots\right} \
& =\operatorname{Pr}\left{X_n=k \mid X_{n-1}=j, \ldots, X_{n-s}=j_{s-1}\right} .
\end{aligned}
$$
每当1。h。s。已定义。
一个马尔可夫链$\left{X_n\right}$被认为是一阶(或简单的马尔可夫链)，如果
$$
\begin{aligned}
\operatorname{Pr}\left{X_n\right. & \left.=k \mid X_{n-1}=j, X_{n-2}=j_1, \ldots\right}=\operatorname{Pr}\left{X_n=k \mid X_{n-1}=j\right} \
& =p_{j k} .
\end{aligned}
$$
每当$\operatorname{Pr}\left{X_{n-1}=j, X_{n-2}=j_1, \ldots\right}>0$。
除非另有明确说明，我们所说的马尔可夫链是指一阶的链，我们在这里主要限于此。如果$p_{j k}=p_k$对于所有的$j$链都是0阶。这意味着$X_n$和$X_{n-1}$的独立性。例如，在伯努利抛硬币实验中，t.p.m.是$\left(\begin{array}{ll}q & p \ q & p\end{array}\right)=\left(\begin{array}{l}1 \ 1\end{array}\right)(q, p)=\mathbf{e}(q, p)$。
表示某天下雨的状态为1，某天不下雨的状态为0。
Let (1.5) hold for $s=2$, Let
$p_{i j k}=\operatorname{Pr}{$实际日期为状态$k \mid$前一天为状态$j$，前一天为状态$i}, i, j, k=0,1$。
然后我们有一个二阶马尔可夫链。注意，矩阵$\left(p_{i j k}\right)$不是一个随机矩阵。它是一个$(4 \times 2)$矩阵而不是一个方阵。
Let (1.5) hold for $s=1$, Let
$p_{j k}=\operatorname{Pr}{$实际日期为$k \mid$前一天为$j}$。然后我们有一个带有t.p.m. $\left(p_{j k}\right), j, k=0,1$的马尔可夫链(即一阶链)。

数学代写|随机过程统计代写Stochastic process statistics代考请认准statistics-lab™

统计代写请认准statistics-lab™. statistics-lab™为您的留学生涯保驾护航。

金融工程代写

金融工程是使用数学技术来解决金融问题。金融工程使用计算机科学、统计学、经济学和应用数学领域的工具和知识来解决当前的金融问题，以及设计新的和创新的金融产品。

非参数统计代写

非参数统计指的是一种统计方法，其中不假设数据来自于由少数参数决定的规定模型；这种模型的例子包括正态分布模型和线性回归模型。

广义线性模型代考

广义线性模型（GLM）归属统计学领域，是一种应用灵活的线性回归模型。该模型允许因变量的偏差分布有除了正态分布之外的其它分布。

术语广义线性模型（GLM）通常是指给定连续和/或分类预测因素的连续响应变量的常规线性回归模型。它包括多元线性回归，以及方差分析和方差分析（仅含固定效应）。

有限元方法代写

有限元方法（FEM）是一种流行的方法，用于数值解决工程和数学建模中出现的微分方程。典型的问题领域包括结构分析、传热、流体流动、质量运输和电磁势等传统领域。

有限元是一种通用的数值方法，用于解决两个或三个空间变量的偏微分方程（即一些边界值问题）。为了解决一个问题，有限元将一个大系统细分为更小、更简单的部分，称为有限元。这是通过在空间维度上的特定空间离散化来实现的，它是通过构建对象的网格来实现的：用于求解的数值域，它有有限数量的点。边界值问题的有限元方法表述最终导致一个代数方程组。该方法在域上对未知函数进行逼近。[1] 然后将模拟这些有限元的简单方程组合成一个更大的方程系统，以模拟整个问题。然后，有限元通过变化微积分使相关的误差函数最小化来逼近一个解决方案。

tatistics-lab作为专业的留学生服务机构，多年来已为美国、英国、加拿大、澳洲等留学热门地的学生提供专业的学术服务，包括但不限于Essay代写，Assignment代写，Dissertation代写，Report代写，小组作业代写，Proposal代写，Paper代写，Presentation代写，计算机作业代写，论文修改和润色，网课代做，exam代考等等。写作范围涵盖高中，本科，研究生等海外留学全阶段，辐射金融，经济学，会计学，审计学，管理学等全球99%专业科目。写作团队既有专业英语母语作者，也有海外名校硕博留学生，每位写作老师都拥有过硬的语言能力，专业的学科背景和学术写作经验。我们承诺100%原创，100%专业，100%准时，100%满意。

随机分析代写

随机微积分是数学的一个分支，对随机过程进行操作。它允许为随机过程的积分定义一个关于随机过程的一致的积分理论。这个领域是由日本数学家伊藤清在第二次世界大战期间创建并开始的。

时间序列分析代写

随机过程，是依赖于参数的一组随机变量的全体，参数通常是时间。随机变量是随机现象的数量表现，其时间序列是一组按照时间发生先后顺序进行排列的数据点序列。通常一组时间序列的时间间隔为一恒定值（如1秒，5分钟，12小时，7天，1年），因此时间序列可以作为离散时间数据进行分析处理。研究时间序列数据的意义在于现实中，往往需要研究某个事物其随时间发展变化的规律。这就需要通过研究该事物过去发展的历史记录，以得到其自身发展的规律。

回归分析代写

多元回归分析渐进（Multiple Regression Analysis Asymptotics）属于计量经济学领域，主要是一种数学上的统计分析方法，可以分析复杂情况下各影响因素的数学关系，在自然科学、社会和经济学等多个领域内应用广泛。

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，工具箱允许您学习和应用专业技术。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。

R语言代写	问卷设计与分析代写
PYTHON代写	回归分析与线性模型代写
MATLAB代写	方差分析与试验设计代写
STATA代写	机器学习/统计学习代写
SPSS代写	计量经济学代写
EVIEWS代写	时间序列分析代写
EXCEL代写	深度学习代写
SQL代写	各种数据建模与可视化代写

统计代写|随机过程代写stochastic process代考|STAT433

Posted on 2023年8月29日2023年8月29日 by statistics-lab

统计代写|随机过程代写stochastic process代考|DEFINITION AND EXAMPLES

Consider a simple coin tossing experiment repeated for a number of times. The possible outcomes at each trial are two: head with probability, say, $p$ and tail with probability $q, p+q=1$. Let us denote head by 1 and tail by 0 and the random variable denoting the result of the $n$th toss by $X_n$. Then for $n=1,2$, $3, \ldots$,
$$
\operatorname{Pr}\left(X_n=1\right)=p, \operatorname{Pr}\left{X_n=0\right}=q .
$$
Thus we have a sequence of random variables $X_1, X_2, \ldots$. The trials are independent and the result of the $n$th trial does not depend in any way on the previous trials numbered $1,2, \ldots,(n-1)$. The random variables are independent.

Consider now the random variable given by the partial sum $S_n=X_1+\cdots+X_n$. The sum $S_n$ gives the accumulated number of heads in the first $n$ trials and its possible values are $0,1, \ldots, n$.

We have $S_{n+1}=S_n+X_{n+1}$. Given that $S_n=j(j=0,1, \ldots, n)$, the r.v. $S_{n+1}$ can assume only two possible values: $S_{n+1}=j$ with probability $q$ and $S_{n+1}=j+1$ with probability $p$; these probabilities are not at all affected by the values of the variables $S_1, \ldots, S_{n-1}$. Thus
$$
\begin{aligned}
& \operatorname{Pr}\left{S_{n+1}=j+1 \mid S_n=j\right}=p \
& \operatorname{Pr}\left{S_{n+1}=j \mid S_n=j\right}=q .
\end{aligned}
$$
We have here an example of a Markov* chain, a case of simple dependence that the outcome of $(n+1)$ st trial depends directly on that of $n$th trial and only on it. The conditional probability of $S_{n+1}$ given $S_n$ depends on the value of $S_n$ and the manner in which the value of $S_n$ was reached is of no consequence.

统计代写|随机过程代写stochastic process代考|Transition Matrix (or Matrix of Transition Probabilities)

The transition probabilities $p_{j k}$ satisfy
$$
p_{j k} \geq 0, \quad \sum_k p_{j k}=1 \text { for all } j .
$$
These probabilities may be written in the matrix form
$$
P=\left(\begin{array}{cccc}
p_{11} & p_{12} & p_{13} & \cdots \
p_{21} & p_{22} & p_{23} & \cdots \
\cdots & \cdots & \ldots & \ldots \
\cdots & \ldots & \ldots & \ldots
\end{array}\right)
$$
This is called the transition probability matrix or matrix of transition probabilities (t.p.m.) of the Markov chain. $P$ is a stochastic matrix i.e. a square matrix with non-negative elements and unit row sums.

Example 1(b). A particle performs a random walk with absorbing barriers, say, as 0 and 4 . Whenever it is at any position $r(0<r<4)$, it moves to $r+1$ with probability $P$ or to $(r-1)$ with probability $q$, $p+q=1$. But as soon as it reaches 0 or 4 it remains there itself. Let $X_n$ be the position of the particle after $n$ moves. The different states of $X_n$ are the different positions of the particle. $\left{X_n\right}$ is a Markov chain whose unit-step transition probabilities are given by
$$
\begin{array}{ll}
\operatorname{Pr}\left{X_n=r+1 \mid X_{n-1}=r\right}=p & \
\operatorname{Pr}\left{X_n=r-1 \mid X_{n-1}=r\right}=q & 0<r<4
\end{array}
$$
and
$$
\begin{aligned}
& \operatorname{Pr}\left{X_n=0 \mid X_{n-1}=0\right}=1, \
& \operatorname{Pr}\left{X_n=4 \mid X_{n-1}=4\right}=1 .
\end{aligned}
$$

The transition matrix is given by
$$
\left.\begin{array}{ccccccc}
\text { States of } X_{n-1} & \multicolumn{7}{c}{\text { States of } X_n} \
& & 0 & 1 & 2 & 3 & 4 \
& 1 & & 0 & 0 & 0 & 0 \
& 2 & 0 & p & 0 & 0 \
& 3 \
& 4 & q & 0 & p & 0 \
0 & 0 & q & 0 & p \
0 & 0 & 0 & 0 & 1
\end{array}\right)
$$

随机过程代考

统计代写|随机过程代写stochastic process代考|DEFINITION AND EXAMPLES

考虑一个重复多次的简单抛硬币实验。每次试验都有两种可能的结果:有概率的正面，比如， $p$ 尾部是概率 $q, p+q=1$．让我们用1表示头，用0表示尾随机变量表示的结果 $n$扔过去 $X_n$．然后是 $n=1,2$， $3, \ldots$，
$$
\operatorname{Pr}\left(X_n=1\right)=p, \operatorname{Pr}\left{X_n=0\right}=q .
$$
这样我们就得到了一个随机变量序列 $X_1, X_2, \ldots$．试验是独立的，结果是 $n$该试验不以任何方式依赖于先前试验的编号 $1,2, \ldots,(n-1)$．随机变量是独立的。

现在考虑部分和$S_n=X_1+\cdots+X_n$给出的随机变量。和$S_n$给出了在第一次$n$次试验中累积的正面数，它的可能值是$0,1, \ldots, n$。

我们有$S_{n+1}=S_n+X_{n+1}$。给定$S_n=j(j=0,1, \ldots, n)$, rv $S_{n+1}$只能假设两个可能的值:$S_{n+1}=j$的概率为$q$, $S_{n+1}=j+1$的概率为$p$;这些概率完全不受$S_1, \ldots, S_{n-1}$变量值的影响。因此
$$
\begin{aligned}
& \operatorname{Pr}\left{S_{n+1}=j+1 \mid S_n=j\right}=p \
& \operatorname{Pr}\left{S_{n+1}=j \mid S_n=j\right}=q .
\end{aligned}
$$
我们这里有一个马尔可夫链的例子，这是一个简单依赖的例子，即$(n+1)$第一次试验的结果直接依赖于$n$第二次试验的结果，而且只依赖于它。给定$S_n$的$S_{n+1}$的条件概率取决于$S_n$的值，而达到$S_n$值的方式无关紧要。

统计代写|随机过程代写stochastic process代考|Transition Matrix (or Matrix of Transition Probabilities)

跃迁概率$p_{j k}$满足
$$
p_{j k} \geq 0, \quad \sum_k p_{j k}=1 \text { for all } j .
$$
这些概率可以写成矩阵形式
$$
P=\left(\begin{array}{cccc}
p_{11} & p_{12} & p_{13} & \cdots \
p_{21} & p_{22} & p_{23} & \cdots \
\cdots & \cdots & \ldots & \ldots \
\cdots & \ldots & \ldots & \ldots
\end{array}\right)
$$
这被称为马尔可夫链的转移概率矩阵或转移概率矩阵。$P$是一个随机矩阵，即具有非负元素和单位行和的方阵。

例1(b)。粒子进行随机游走，有吸收障碍，比如0和4。当它在任何位置$r(0<r<4)$时，它以$P$的概率移动到$r+1$，或以$q$, $p+q=1$的概率移动到$(r-1)$。但一旦它达到0或4，它就会保持在那里。设$X_n$为$n$移动后粒子的位置。$X_n$的不同状态是粒子的不同位置。$\left{X_n\right}$是一个马尔可夫链，其单位阶跃转移概率由式给出
$$
\begin{array}{ll}
\operatorname{Pr}\left{X_n=r+1 \mid X_{n-1}=r\right}=p & \
\operatorname{Pr}\left{X_n=r-1 \mid X_{n-1}=r\right}=q & 0<r<4
\end{array}
$$
和
$$
\begin{aligned}
& \operatorname{Pr}\left{X_n=0 \mid X_{n-1}=0\right}=1, \
& \operatorname{Pr}\left{X_n=4 \mid X_{n-1}=4\right}=1 .
\end{aligned}
$$

转移矩阵由
$$
\left.\begin{array}{ccccccc}
\text { States of } X_{n-1} & \multicolumn{7}{c}{\text { States of } X_n} \
& & 0 & 1 & 2 & 3 & 4 \
& 1 & & 0 & 0 & 0 & 0 \
& 2 & 0 & p & 0 & 0 \
& 3 \
& 4 & q & 0 & p & 0 \
0 & 0 & q & 0 & p \
0 & 0 & 0 & 0 & 1
\end{array}\right)
$$

统计代写请认准statistics-lab™. statistics-lab™为您的留学生涯保驾护航。

金融工程代写

非参数统计代写

非参数统计指的是一种统计方法，其中不假设数据来自于由少数参数决定的规定模型；这种模型的例子包括正态分布模型和线性回归模型。

广义线性模型代考

广义线性模型（GLM）归属统计学领域，是一种应用灵活的线性回归模型。该模型允许因变量的偏差分布有除了正态分布之外的其它分布。

有限元方法代写

随机分析代写

时间序列分析代写

回归分析代写

MATLAB代写

R语言代写	问卷设计与分析代写
PYTHON代写	回归分析与线性模型代写
MATLAB代写	方差分析与试验设计代写
STATA代写	机器学习/统计学习代写
SPSS代写	计量经济学代写
EVIEWS代写	时间序列分析代写
EXCEL代写	深度学习代写
SQL代写	各种数据建模与可视化代写

统计代写|随机过程代写stochastic process代考|BISSP2023

Posted on 2023年8月29日2023年8月29日 by statistics-lab

统计代写|随机过程代写stochastic process代考|Specification of Stochastic Processes

The set of possible values of a single random variable $X_n$ of a stochastic process $\left{X_n, n \geq 1\right}$ is known as its state space. The state space is discrete if it contains a finite or a denumerable infinity of points; otherwise, it is continuous. For example, if $X_n$ is the total number of sixes appearing in the first $n$ throws of a die, the set of possible values of $X_n$ is the finite set of non-negative integers $0,1, \ldots, n$. Here, the state space of $X_n$ is discrete. We can write $X_n=Y_1+\cdots+Y_n$, where $Y_i$ is a discrete r.v. denoting the outcome of the $i$ th throw and $Y_i=1$ or 0 according as the $i$ th throw shows six or not. Secondly, consider $X_n=Z_1$ $+\ldots+Z_n$, where $Z_i$ is a continuous r.v. assuming values in $[0, \infty]$. Here, the set of possible values of $X_n$ is the interval $[0, \infty]$, and so the state space of $X_n$ is continuous.

In the above two examples we assume that the parameter $n$ of $X_n$ is restricted to the non-negative integers $n=0,1,2, \ldots$ We consider the state of the system at distinct time points $n=0,1,2, \ldots$, only. Here the word time is used in a wide sense. We note that in the first case considered above the “time $n$ ” implies throw number $n$.

On the other hand, one can visualise a family of random variables $\left{X_v, t \in T\right}$ (or ${X(t), t \in T}$ ) such that the state of the system is characterized at every instant over a finite or infinite interval. The system is then defined for a continuous range of time and we say that we have a family of r.v. in continuous time. A stochastic process in continuous time may have either a discrete or a continuous state space. For example, suppose that $X(t)$ gives the number of incoming calls at a switchboard in an interval $(0, t)$. Here the state space of $X(t)$ is discrete though $X(t)$ is defined for a continuous range of time. We have a process in continuous time having a discrete state space. Suppose that $X(t)$ represents the maximum temperature at a particular place in $(0, t)$, then the set of possible values of $X(t)$ is continuous. Here we have a system in continuous time having a continuous state space.

So far we have assumed that the values assumed by the r.v. $X_n$ (or $\left.X(t)\right)$ are one-dimensional, but the process $\left{X_n\right}$ (or ${X(t)}$ ) may be multi-dimensional. Consider $X(t)=\left(X_1(t), X_2(t)\right.$ ), where $X_1$ represents the maximum and $X_2$ the minimum temperature at a place in an interval of time $(0, t)$. We have here a two-dimensional stochastic process in continuous time having continuous state space. One can similarly have multi-dimensional processes. One-dimensional processes can be classified, in general, into the following four types of processes:
(i) Discrete time, discrete state space
(ii) Discrete time, continuous state space
(iii) Continuous time, discrete state space
(iv) Continuous time, continuous state space.

统计代写|随机过程代写stochastic process代考|Processes with independent increments

If for all $t_1, \ldots, t_n, t_1s$, do not depend on the values of $X(u), u<s$, then the process is said to be a Markov process.
A definition of such a process is given below.
If, for, $t_1<t_2<\ldots<t_n<t$
$$
\begin{aligned}
\operatorname{Pr}\left{a \leq X(t) \leq b \mid X\left(t_1\right)\right. & \left.=x_1, \ldots, X\left(t_n\right)=x_n\right} \
& =\operatorname{Pr}\left{a \leq X(t) \leq b \mid X\left(t_n\right)=x_n\right}
\end{aligned}
$$
the process ${X(t), t \in T}$ is a Markov process.
A discrete parameter Markov process is known as a Markov chain.

随机过程代考

统计代写|随机过程代写stochastic process代考|Specification of Stochastic Processes

随机过程$\left{X_n, n \geq 1\right}$的单个随机变量$X_n$的可能值的集合称为其状态空间。如果状态空间包含有限或无限的点，则状态空间是离散的;否则为连续。例如，如果$X_n$是第一次$n$次掷骰子中出现的6的总数，那么$X_n$的可能值的集合就是非负整数$0,1, \ldots, n$的有限集合。这里，$X_n$的状态空间是离散的。我们可以写$X_n=Y_1+\cdots+Y_n$，其中$Y_i$是一个离散的rv，表示$i$次投掷的结果，根据$i$次投掷是否显示6，表示$Y_i=1$或0。其次，考虑$X_n=Z_1$$+\ldots+Z_n$，其中$Z_i$是一个连续的rv，假设$[0, \infty]$中的值。这里，$X_n$的可能值的集合是区间$[0, \infty]$，因此$X_n$的状态空间是连续的。

在上面的两个例子中，我们假设$X_n$的参数$n$被限制为非负整数$n=0,1,2, \ldots$，我们只考虑系统在不同时间点$n=0,1,2, \ldots$的状态。在这里，时间这个词是广义的。我们注意到，在上面考虑的第一种情况中，“时间$n$”意味着抛出数$n$。

另一方面，人们可以想象一组随机变量$\left{X_v, t \in T\right}$(或${X(t), t \in T}$)，这样系统的状态在有限或无限区间内的每个瞬间都是有特征的。然后在连续时间范围内定义系统我们说我们有一个连续时间的rv族。连续时间的随机过程可以具有离散状态空间，也可以具有连续状态空间。例如，假设$X(t)$给出了在一个间隔$(0, t)$内总机的入站呼叫数。这里$X(t)$的状态空间是离散的，尽管$X(t)$是在连续时间范围内定义的。我们有一个连续时间的过程有一个离散的状态空间。假设$X(t)$代表$(0, t)$某一特定地点的最高温度，那么$X(t)$的可能值集是连续的。这里我们有一个连续时间系统有一个连续状态空间。

到目前为止，我们已经假定rv $X_n$(或$\left.X(t)\right)$)所假定的值是一维的，但是过程$\left{X_n\right}$(或${X(t)}$)可能是多维的。考虑$X(t)=\left(X_1(t), X_2(t)\right.$)，其中$X_1$表示一段时间内某一地点的最高温度，$X_2$表示一段时间内某一地点的最低温度$(0, t)$。我们这里有一个连续时间的二维随机过程具有连续状态空间。一个人也可以有类似的多维过程。一维过程一般可分为以下四类过程:
(i)离散时间，离散状态空间
(ii)离散时间，连续状态空间
(iii)连续时间，离散状态空间
(iv)连续时间，连续状态空间。

统计代写|随机过程代写stochastic process代考|Processes with independent increments

如果对于所有的$t_1, \ldots, t_n, t_1s$，不依赖于$X(u), u<s$的值，那么这个过程就是一个马尔可夫过程。
下面给出了这种过程的定义。
如果，for, $t_1<t_2<\ldots<t_n<t$
$$
\begin{aligned}
\operatorname{Pr}\left{a \leq X(t) \leq b \mid X\left(t_1\right)\right. & \left.=x_1, \ldots, X\left(t_n\right)=x_n\right} \
& =\operatorname{Pr}\left{a \leq X(t) \leq b \mid X\left(t_n\right)=x_n\right}
\end{aligned}
$$
这个过程${X(t), t \in T}$是一个马尔可夫过程。
离散参数马尔可夫过程称为马尔可夫链。

统计代写请认准statistics-lab™. statistics-lab™为您的留学生涯保驾护航。

金融工程代写

非参数统计代写

非参数统计指的是一种统计方法，其中不假设数据来自于由少数参数决定的规定模型；这种模型的例子包括正态分布模型和线性回归模型。

广义线性模型代考

广义线性模型（GLM）归属统计学领域，是一种应用灵活的线性回归模型。该模型允许因变量的偏差分布有除了正态分布之外的其它分布。

有限元方法代写

随机分析代写

时间序列分析代写

回归分析代写

MATLAB代写

R语言代写	问卷设计与分析代写
PYTHON代写	回归分析与线性模型代写
MATLAB代写	方差分析与试验设计代写
STATA代写	机器学习/统计学习代写
SPSS代写	计量经济学代写
EVIEWS代写	时间序列分析代写
EXCEL代写	深度学习代写
SQL代写	各种数据建模与可视化代写

统计代写|随机过程代写stochastic process代考|ST747

Posted on 2023年8月15日2023年8月28日 by statistics-lab

统计代写|随机过程代写stochastic process代考|Geometric and Exponential Distributions

The geometric and exponential distributions are unique in the sense that these are the only distributions with memoryless or non-aging property. In fact, there is a close connection between these two distributions. Exponential distribution is the continuous analogue of geometric distribution and discretization of exponential distribution leads to geometric distributions.
We show below how one can get the exponential distribution from the geometric distribution.
Suppose that the occurrence (success) or non-occurrence (failure) of an event $E$ may happen only at the discrete time points $\delta, 2 \delta, 3 \delta$,…each subinterval being of length $\delta$, the probability of occurrence being $p=\lambda \delta$ and that of non-occurrence being $q=1-\lambda \delta$.
The number of failures $N$ preceding the first success has a geometric distribution.
$$
\operatorname{Pr}{N=k}=q^k p, k=0,1,2, \ldots
$$
Now the event that the first success occurs at trial number $(k+1)$ or beyond has the probability $\sum_{r=k}^{\infty} q^r p=q^k p /(1-q)=q^k$, and therefore the probability that the time $T$ required for the first success exceeds $k \delta$ is given by
$$
\operatorname{Pr}{T>k \delta}=\operatorname{Pr}{N \geq k}=\sum_{r=k}^{\infty} q^r p=q^k=(1-\lambda \delta)^k
$$

Now suppose that $\delta \rightarrow 0$ as $k \rightarrow \infty$, such that $k \delta=t$ remains fixed, then as $k \rightarrow \infty$.
$$
\operatorname{Pr}(T>t)=(1-\lambda t / k)^k=\left{(1-\lambda t / k)^{k / \lambda t}\right}^{\lambda t} \rightarrow\left{e^{-1}\right}^{\lambda t}=e^{-\lambda t} .
$$
Thus $T$ has an exponential distribution with mean
$$
\frac{1}{\lambda}=\frac{\delta}{p}
$$

统计代写|随机过程代写stochastic process代考|Sum of a Random Number of Continuous Random Variables Stochastic δm

In Sec. 1.1 .4 we considered the sum $S_N$ of a random number $N$ of random variables $X_i$. Theorem 1.3 is stated in terms of the p.g.f. of $X_i$ on the assumption that the $X_i$ ‘s are discrete variables (with non-negative integral values). Here we assume that the $X_i$ ‘s are mutually independent and identically distributed nonnegative random variables which may be continuous as well.

Replacing the p.g.f $P(s)$ of $X_i$ by its L.T. $F^*(s)$, we can get an analogous result. This may be stated as follows:

Theorem 1.11. If $X_i$ are identically and independentally distributed random variables having for their L.T.’s
$$
F^(s)=E\left{\exp \left(-s X_i\right)\right} \text { for all } i $$ and if $N$ (independent of $X_i^{\prime}$ ‘s) is a random variable having p.g.f. $$ G(s)=\sum_n \operatorname{Pr}{N=n} s^n=\sum_n g_n s^n, $$ then the L.T. $F_{S_N}^(s)$ of the sum $S_N=X_1+\ldots+X_N$, is given by
$$
F_{S_N}^(s)=G\left(F^(s)\right)
$$
The proof is identical with that of Theorem 1.3. It also follows that
$$
E\left{S_N\right}=E\left{X_i\right} E{N} .
$$
It is to be noted that $S_N$ is a continuous random variable.
Corresponding to $h_j=\operatorname{Pr}\left{S_N=j\right}$ we shall have density function $f(x)$ of $S_N$, so that
$$
\begin{aligned}
f(x) d x=\operatorname{Pr}\left{x \leq S_N<x+d x\right} & =\sum_{n=0}^{\infty} \operatorname{Pr}\left{N=n \text { and } x \leq S_n<x+d x\right} \
& =\sum_{n=0}^{\infty} \operatorname{Pr}{N=n} \operatorname{Pr}\left{x \leq S_n<x+d x\right} \
& =\sum_n g_n \operatorname{Pr}\left{x \leq S_n<x+d x\right} .
\end{aligned}
$$

随机过程代考

统计代写|随机过程代写stochastic process代考|Geometric and Exponential Distributions

几何和指数分布是独一无二的，因为它们是唯一具有无记忆或非老化特性的分布。事实上，这两种分布之间有着密切的联系。指数分布是几何分布的连续模拟，指数分布的离散化导致几何分布。
我们将在下面展示如何从几何分布中得到指数分布。
假设一个事件$E$的发生(成功)或不发生(失败)可能只发生在离散时间点$\delta, 2 \delta, 3 \delta$，…每个子区间的长度为$\delta$，发生的概率为$p=\lambda \delta$，不发生的概率为$q=1-\lambda \delta$。
第一次成功之前的失败次数$N$呈几何分布。
$$
\operatorname{Pr}{N=k}=q^k p, k=0,1,2, \ldots
$$
现在，第一次成功发生在试验号$(k+1)$或以上的事件的概率为$\sum_{r=k}^{\infty} q^r p=q^k p /(1-q)=q^k$，因此，第一次成功所需的时间$T$超过$k \delta$的概率为
$$
\operatorname{Pr}{T>k \delta}=\operatorname{Pr}{N \geq k}=\sum_{r=k}^{\infty} q^r p=q^k=(1-\lambda \delta)^k
$$

现在假设$\delta \rightarrow 0$为$k \rightarrow \infty$，这样$k \delta=t$保持不变，然后是$k \rightarrow \infty$。
$$
\operatorname{Pr}(T>t)=(1-\lambda t / k)^k=\left{(1-\lambda t / k)^{k / \lambda t}\right}^{\lambda t} \rightarrow\left{e^{-1}\right}^{\lambda t}=e^{-\lambda t} .
$$
因此$T$与均值呈指数分布
$$
\frac{1}{\lambda}=\frac{\delta}{p}
$$

统计代写|随机过程代写stochastic process代考|Sum of a Random Number of Continuous Random Variables Stochastic δm

在第1.1 .4节中，我们考虑了随机变量$X_i$的随机数$N$的和$S_N$。定理1.3是在假设$X_i$是离散变量(具有非负整数值)的情况下，用$X_i$的p.g.f.来表述的。这里我们假设$X_i$是相互独立的同分布的非负随机变量，也可以是连续的。

用其L.T. $F^*(s)$代替$X_i$的p.g.f $P(s)$，我们可以得到类似的结果。这可以表述如下:

定理1.11。如果$X_i$是相同且独立分布的随机变量，它们的lt
$$
F^(s)=E\left{\exp \left(-s X_i\right)\right} \text { for all } i $$，如果$N$(独立于$X_i^{\prime}$)是一个随机变量，具有p.g.f. $$ G(s)=\sum_n \operatorname{Pr}{N=n} s^n=\sum_n g_n s^n, $$，那么和$S_N=X_1+\ldots+X_N$的L.T. $F_{S_N}^(s)$由
$$
F_{S_N}^(s)=G\left(F^(s)\right)
$$
这个证明与定理1.3的证明是相同的。这也意味着
$$
E\left{S_N\right}=E\left{X_i\right} E{N} .
$$
需要注意的是，$S_N$是一个连续随机变量。
对应$h_j=\operatorname{Pr}\left{S_N=j\right}$我们将得到$S_N$的密度函数$f(x)$，因此
$$
\begin{aligned}
f(x) d x=\operatorname{Pr}\left{x \leq S_N<x+d x\right} & =\sum_{n=0}^{\infty} \operatorname{Pr}\left{N=n \text { and } x \leq S_n<x+d x\right} \
& =\sum_{n=0}^{\infty} \operatorname{Pr}{N=n} \operatorname{Pr}\left{x \leq S_n<x+d x\right} \
& =\sum_n g_n \operatorname{Pr}\left{x \leq S_n<x+d x\right} .
\end{aligned}
$$

统计代写请认准statistics-lab™. statistics-lab™为您的留学生涯保驾护航。

金融工程代写

非参数统计代写

非参数统计指的是一种统计方法，其中不假设数据来自于由少数参数决定的规定模型；这种模型的例子包括正态分布模型和线性回归模型。

广义线性模型代考

广义线性模型（GLM）归属统计学领域，是一种应用灵活的线性回归模型。该模型允许因变量的偏差分布有除了正态分布之外的其它分布。

有限元方法代写

随机分析代写

时间序列分析代写

回归分析代写

MATLAB代写

R语言代写	问卷设计与分析代写
PYTHON代写	回归分析与线性模型代写
MATLAB代写	方差分析与试验设计代写
STATA代写	机器学习/统计学习代写
SPSS代写	计量经济学代写
EVIEWS代写	时间序列分析代写
EXCEL代写	深度学习代写
SQL代写	各种数据建模与可视化代写

统计代写|随机过程代写stochastic process代考|ST747

Posted on 2023年8月15日2023年8月28日 by statistics-lab

统计代写|随机过程代写stochastic process代考|Pareto distribution

As formulated by Pareto ${ }^1$, the number of persons $N$ having income in excess of $x$ is given by
$$
N=A x^{-r},
$$
where $A$ and $r$ are parameters.
The Pareto d.f. may be written as,
$$
F(x)=\operatorname{Pr}(X \leq x)=1-\left(\frac{k}{x}\right)^a, k>0, a>0, x \geq k
$$
The p.d.f. can be put as
$$
f(x)=\frac{a k^a}{x^{a+1}}, \quad a>0, x \geq k>0
$$
The $r^{\text {th }}$ moment is
$$
E\left(X^r\right)=\frac{a k^r}{a-r}, 1 \leq r2
\end{gathered}
$$
If we take $E(X)=1$, then
$$
\operatorname{var}(X)=\frac{1}{a(a-2)}, a>2,
$$
which is large for $a$ close to 2 .

统计代写|随机过程代写stochastic process代考|Coxian distribution (Branching Erlang)

Exponential, Erlang, hyper- and hypo-exponential distributions all have their L.T.’s as rational functions of $s$. The poles are on the negative real axis of the complex $s$-plane. A family of distributions is obtained by a generalisation of an idea (due to Erlang) by Cox (1955). The family of distributions has as L.T. a rational function of $s$ with the degree of polynomial in the numerator less than or equal to that of the denominator. This family has been increasingly considered in the literature as a more general service time distribution (in queueing context) and repair time distribution (in reliability context). Extensive references to Coxian distributions appear also in computer science and telecommunication literature.

Consider a service facility, with $m$ phases of service channels (nodes); the system is a subnetwork of $m$ nodes. The service time distribution in the node (phase) $i, i=1,2, \ldots, m$ is exponential with rate $\mu_i$, service time in any node being independent of the service time in other nodes. The job (or the customer) needing service can be at one of the $m$ stages at a given time and no other job can be admitted for service until the job receiving service at one of the nodes (phases) has completed his service and departs from the system.

A job enters from the left and moves to the right. After receiving service at node $i$, the job may leave the system with probability $b_i$ or move for further service to the next node $(i+1)$ with probability $a_i$, $a_i+b_i=1, i=1, . . m$; we can include $i=0$ such that $a_0=0$ indicates that the job does not require any service from any of the nodes and departs from the system without receiving any service from any node, whereas $a_0=1$ indicates that it needs service at least from the first note. After receiving service at the last node $m$, if it reaches that node, the job departs from the system, so that $b_m=1$.

The distribution is denoted by $K_m$ (or $C_m$ as a tribute to Cox). The probability that a job receives (requires) services at nodes $1,2, \ldots, k(k \leq m)$ and departs from the system after service completion at the facility equals $A_k b_k$,
where $A_k=a_0 a_1 \ldots a_{k-1}$. We have
$$
b_0=\sum_{k=1}^m A_k b_k=1 .
$$
Let $\gamma$ be the total service time of a job. The L.T. of the r.v. $\gamma$ is given by
$$
\begin{aligned}
F^*(s) & =b_0+\sum_{k=1}^m A_k b_k\left{\prod_{i=1}^k \frac{\mu_i}{s+\mu_i}\right} \
& =b_0+A_1 b_1\left(\frac{\mu_1}{s+\mu_1}\right)+\sum_{k=2}^m\left{a_1 a_2 \ldots a_{k-1}\right} b_k\left[\prod_{i=1}^k \frac{\mu_i}{s+\mu_i}\right] .
\end{aligned}
$$
Th last term can also be written as
$$
\sum_{k=0}^m \frac{b_k \mu_k}{s+\mu_k}\left[\prod_{i=0}^{k-1}\left(1-b_i\right) \frac{\mu_i}{s+\mu_i}\right]
$$

随机过程代考

统计代写|随机过程代写stochastic process代考|Pareto distribution

根据帕累托${ }^1$的公式，收入超过$x$的人数$N$为
$$
N=A x^{-r},
$$
其中$A$和$r$为参数。
帕累托方程可以写成:
$$
F(x)=\operatorname{Pr}(X \leq x)=1-\left(\frac{k}{x}\right)^a, k>0, a>0, x \geq k
$$
p.d.f.可以写成
$$
f(x)=\frac{a k^a}{x^{a+1}}, \quad a>0, x \geq k>0
$$
$r^{\text {th }}$时刻是
$$
E\left(X^r\right)=\frac{a k^r}{a-r}, 1 \leq r2
\end{gathered}
$$
如果我们取$E(X)=1$，那么
$$
\operatorname{var}(X)=\frac{1}{a(a-2)}, a>2,
$$
这对于$a$来说是很大的，接近2。

统计代写|随机过程代写stochastic process代考|Coxian distribution (Branching Erlang)

指数分布、Erlang分布、超指数分布和次指数分布都有它们的ltt为$s$的有理函数。极点位于复$s$ -平面的负实轴上。一组分布是通过Cox(1955)对一个思想(由于Erlang)的推广得到的。该分布族有一个有理函数$s$，其分子的多项式次数小于或等于分母的多项式次数。这个家族在文献中越来越多地被认为是一个更通用的服务时间分布(在排队上下文中)和维修时间分布(在可靠性上下文中)。在计算机科学和电信文献中也出现了大量关于Coxian分布的参考文献。

考虑一个服务设施，它具有$m$阶段的服务通道(节点);系统为$m$节点的子网。节点(阶段)$i, i=1,2, \ldots, m$的服务时间分布与速率$\mu_i$呈指数关系，任意节点的服务时间与其他节点的服务时间无关。需要服务的作业(或客户)可以在给定时间处于$m$阶段之一，并且在其中一个节点(阶段)接受服务的作业完成其服务并离开系统之前，不能允许其他作业接受服务。

作业从左侧进入并移至右侧。在节点$i$接受服务后，作业有可能以$b_i$的概率离开系统，也有可能以$a_i$, $a_i+b_i=1, i=1, . . m$的概率移动到下一个节点$(i+1)$继续服务;我们可以包含$i=0$，这样$a_0=0$表示作业不需要来自任何节点的任何服务，并且在不从任何节点接收任何服务的情况下离开系统，而$a_0=1$表示它至少从第一个注释开始就需要服务。在最后一个节点$m$接收到服务后，如果到达该节点，作业将离开系统，因此$b_m=1$。

该分布用$K_m$表示(或$C_m$表示对Cox的敬意)。作业在节点$1,2, \ldots, k(k \leq m)$接收(需要)服务，并在设施完成服务后离开系统的概率等于$A_k b_k$;
在哪里$A_k=a_0 a_1 \ldots a_{k-1}$。我们有
$$
b_0=\sum_{k=1}^m A_k b_k=1 .
$$
设$\gamma$为一个工作的总服务时间。旅行车$\gamma$的L.T.由
$$
\begin{aligned}
F^*(s) & =b_0+\sum_{k=1}^m A_k b_k\left{\prod_{i=1}^k \frac{\mu_i}{s+\mu_i}\right} \
& =b_0+A_1 b_1\left(\frac{\mu_1}{s+\mu_1}\right)+\sum_{k=2}^m\left{a_1 a_2 \ldots a_{k-1}\right} b_k\left[\prod_{i=1}^k \frac{\mu_i}{s+\mu_i}\right] .
\end{aligned}
$$
最后一项也可以写成
$$
\sum_{k=0}^m \frac{b_k \mu_k}{s+\mu_k}\left[\prod_{i=0}^{k-1}\left(1-b_i\right) \frac{\mu_i}{s+\mu_i}\right]
$$

统计代写请认准statistics-lab™. statistics-lab™为您的留学生涯保驾护航。

金融工程代写

非参数统计代写

非参数统计指的是一种统计方法，其中不假设数据来自于由少数参数决定的规定模型；这种模型的例子包括正态分布模型和线性回归模型。

广义线性模型代考

广义线性模型（GLM）归属统计学领域，是一种应用灵活的线性回归模型。该模型允许因变量的偏差分布有除了正态分布之外的其它分布。

有限元方法代写

随机分析代写

时间序列分析代写

回归分析代写

MATLAB代写

R语言代写	问卷设计与分析代写
PYTHON代写	回归分析与线性模型代写
MATLAB代写	方差分析与试验设计代写
STATA代写	机器学习/统计学习代写
SPSS代写	计量经济学代写
EVIEWS代写	时间序列分析代写
EXCEL代写	深度学习代写
SQL代写	各种数据建模与可视化代写

统计代写|随机过程代写stochastic process代考|MATH11029

Posted on 2023年8月15日2023年8月28日 by statistics-lab

统计代写|随机过程代写stochastic process代考|Some Important Distributions

Suppose that $X$ is a random variable whose whole mass is concentrated in one single point, say, point a. This implies that the variable is ‘almost always’ equal to $a$, i.e. $\operatorname{Pr}(X=a)=1$ and $\operatorname{Pr}(X \neq a)=0$. Its distribution function is given by
$$
\begin{aligned}
F(x)=\operatorname{Pr}{X \leq x} & =0 & & \text { for } x<a . \
& =1 & & \text { for } x \geq a .
\end{aligned}
$$
We have $E(X)=a$ and $\operatorname{var}(X)=E\left(X^2\right)-(E(X))^2=a^2-a^2=0$.
The L.T. of the distribution is given by
$$
F^*(s)=E{\exp (-s X)}=\exp (-s a) \operatorname{Pr}(X=a)=\exp (-s a) .
$$
This is the L.T. of Dirac-delta function located at $a$. The distribution is known as degenerate distribution.

Let $X$ have Poisson distribution with mean $\lambda$. Its p.g.f. is $P(s)=\exp {\lambda(s-1)}$. From (3.4), its L.T. is given by
$$
F^*(s)=P{\exp (-s)}=\exp [\lambda{\exp (-s)-1}] .
$$

Let $X$ have the negative exponential distribution with parameter $\lambda$. Its density function is
$$
\begin{aligned}
f(x) & =\lambda \exp (-\lambda x), & \lambda>0, & 0 \leq x<\infty, \
& =0, & & x<0 .
\end{aligned}
$$
The distribution function $F(x)=\operatorname{Pr}[X \leq x]=1-e^{-\lambda x}$ and $\operatorname{Pr}[X \geq x]=e^{-\lambda x}$.
The L.T. $F^(s)$ of the r.v. $X$ is $$ F^(s)=\int_0^{\infty}{\exp (-s x)}{\lambda \exp (-\lambda x)} d x=\frac{\lambda}{(s+\lambda)} .
$$

统计代写|随机过程代写stochastic process代考|Properties of exponential distribution

(a) Non-ageing (or memoryless or Markov) property
For a r.v. $X$ having exponential distribution with parameter $\lambda$, we have, for all $x, y \geq 0$,
$$
\begin{aligned}
\operatorname{Pr}{X \geq x+y \mid X \geq x} & =\frac{e^{-\lambda(x+y)}}{e^{-\lambda x}} \
& =e^{-\lambda y}=\operatorname{Pr}{X \geq y} . \
\operatorname{Pr}{X \geq x+y} & =\operatorname{Pr}{X \geq x} \operatorname{Pr}{X \geq y}
\end{aligned}
$$
or,
This implies that if the duration $X$ of a certain activity has exponential distribution with parameter $\lambda$ and if the activity is observed after a length of time $x$ after its commencement, then the remaining duration of the activity is independent of $x$ and is also distributed as an exponential r.v. with the same parameter $\lambda$.

Again, if $X$ is a non-negative continuous r.v. with non-ageing property, then it can be shown that $X$ must be exponential.

This non-ageing or momoryless property characterizes exponential distribution among all distributions of continuous non-negative r.v’s.
(b) Minimum of two independent exponential distributions
Suppose that $X_1, X_2$ have independent exponential distributions with parameters $\lambda_1, \lambda_2$ respectively. Let
$$
Z=\min \left(X_1, X_2\right)
$$
We have
$$
\begin{aligned}
\operatorname{Pr}{Z \geq x} & =\operatorname{Pr}\left{X_1 \geq x\right} \operatorname{Pr}\left{X_2 \geq x\right} \
& =e^{-\lambda_1 x} e^{-\lambda_2 x} \
& =e^{-\left(\lambda_1+\lambda_2\right) x}
\end{aligned}
$$
so that $Z$ is exponential with parameter $\left(\lambda_1+\lambda_2\right)$.
This implies that if the durations $X_1$ and $X_2$ of two activities $A_1$ and $A_2$ have independent exponential distributions with parameters $\lambda_1, \lambda_2$ respectively and these activities are observed when neither has been completed, then the duration of the interval $Z$ up to the first completion of one of the activities has also exponential distribution with parameter $\lambda_1+\lambda_2$.
The probability that the activity $A_1$ will be completed earlier than the activity $A_2$ is given by
$$
\begin{aligned}
\operatorname{Pr}\left{X_1x\right} \
& =\int_0^{\infty} \lambda_1 e^{-\lambda_1 x} d x e^{-\lambda_2 x} \
& =\frac{\lambda_1}{\lambda_1+\lambda_2} .
\end{aligned}
$$

随机过程代考

统计代写|随机过程代写stochastic process代考|Some Important Distributions

假设$X$是一个随机变量，它的全部质量都集中在一个点上，比如a点。这意味着该变量“几乎总是”等于$a$，即$\operatorname{Pr}(X=a)=1$和$\operatorname{Pr}(X \neq a)=0$。它的分布函数为
$$
\begin{aligned}
F(x)=\operatorname{Pr}{X \leq x} & =0 & & \text { for } x<a . \
& =1 & & \text { for } x \geq a .
\end{aligned}
$$
我们有$E(X)=a$和$\operatorname{var}(X)=E\left(X^2\right)-(E(X))^2=a^2-a^2=0$。
分布的L.T.由
$$
F^*(s)=E{\exp (-s X)}=\exp (-s a) \operatorname{Pr}(X=a)=\exp (-s a) .
$$
这是狄拉克函数的L.T.位于$a$。这种分布称为简并分布。

设$X$有均值为$\lambda$的泊松分布。它的p.g.f.是$P(s)=\exp {\lambda(s-1)}$。由式(3.4)，其L.T.由
$$
F^*(s)=P{\exp (-s)}=\exp [\lambda{\exp (-s)-1}] .
$$

设$X$为参数$\lambda$的负指数分布。其密度函数为
$$
\begin{aligned}
f(x) & =\lambda \exp (-\lambda x), & \lambda>0, & 0 \leq x<\infty, \
& =0, & & x<0 .
\end{aligned}
$$
分布函数$F(x)=\operatorname{Pr}[X \leq x]=1-e^{-\lambda x}$和$\operatorname{Pr}[X \geq x]=e^{-\lambda x}$。
旅行车$X$的L.T. $F^(s)$是 $$ F^(s)=\int_0^{\infty}{\exp (-s x)}{\lambda \exp (-\lambda x)} d x=\frac{\lambda}{(s+\lambda)} .
$$

统计代写|随机过程代写stochastic process代考|Properties of exponential distribution

(a)非老化(或无记忆或马尔可夫)特性
对于参数为$\lambda$的r。v。$X$的指数分布，对于所有的$x, y \geq 0$，
$$
\begin{aligned}
\operatorname{Pr}{X \geq x+y \mid X \geq x} & =\frac{e^{-\lambda(x+y)}}{e^{-\lambda x}} \
& =e^{-\lambda y}=\operatorname{Pr}{X \geq y} . \
\operatorname{Pr}{X \geq x+y} & =\operatorname{Pr}{X \geq x} \operatorname{Pr}{X \geq y}
\end{aligned}
$$
或者，
这意味着，如果某一活动的持续时间$X$与参数$\lambda$呈指数分布，如果该活动在开始后一段时间$x$后观察到，则该活动的剩余持续时间与$x$无关，并且也以相同参数$\lambda$的指数rv分布。

同样，如果$X$是具有非老化性质的非负连续rv，则可以证明$X$一定是指数的。

这种不老化或无单调的性质表征了连续非负rv的所有分布中的指数分布。
(b)两个独立指数分布的最小值
设$X_1, X_2$具有独立的指数分布，参数分别为$\lambda_1, \lambda_2$。让
$$
Z=\min \left(X_1, X_2\right)
$$
我们有
$$
\begin{aligned}
\operatorname{Pr}{Z \geq x} & =\operatorname{Pr}\left{X_1 \geq x\right} \operatorname{Pr}\left{X_2 \geq x\right} \
& =e^{-\lambda_1 x} e^{-\lambda_2 x} \
& =e^{-\left(\lambda_1+\lambda_2\right) x}
\end{aligned}
$$
所以$Z$是带参数$\left(\lambda_1+\lambda_2\right)$的指数。
这意味着，如果两个活动$A_1$和$A_2$的持续时间$X_1$和$X_2$分别具有带参数$\lambda_1, \lambda_2$的独立指数分布，并且当两个活动都没有完成时观察到这些活动，那么到其中一个活动首次完成的间隔$Z$的持续时间也具有带参数$\lambda_1+\lambda_2$的指数分布。
活动$A_1$将比活动$A_2$更早完成的概率由
$$
\begin{aligned}
\operatorname{Pr}\left{X_1x\right} \
& =\int_0^{\infty} \lambda_1 e^{-\lambda_1 x} d x e^{-\lambda_2 x} \
& =\frac{\lambda_1}{\lambda_1+\lambda_2} .
\end{aligned}
$$

统计代写请认准statistics-lab™. statistics-lab™为您的留学生涯保驾护航。

金融工程代写

非参数统计代写

非参数统计指的是一种统计方法，其中不假设数据来自于由少数参数决定的规定模型；这种模型的例子包括正态分布模型和线性回归模型。

广义线性模型代考

广义线性模型（GLM）归属统计学领域，是一种应用灵活的线性回归模型。该模型允许因变量的偏差分布有除了正态分布之外的其它分布。

有限元方法代写

随机分析代写

时间序列分析代写

回归分析代写

MATLAB代写

R语言代写	问卷设计与分析代写
PYTHON代写	回归分析与线性模型代写
MATLAB代写	方差分析与试验设计代写
STATA代写	机器学习/统计学习代写
SPSS代写	计量经济学代写
EVIEWS代写	时间序列分析代写
EXCEL代写	深度学习代写
SQL代写	各种数据建模与可视化代写

统计代写|随机过程代写stochastic process代考|STATS217

Posted on 2023年7月27日2023年8月25日 by statistics-lab

统计代写|随机过程代写stochastic process代考|Generating Function of Bivariate Distribution

Bivariate Distribution: Suppose that $X, Y$ is a pair of integral-valued random variables with joint probability distribution given by
$$
\operatorname{Pr}{X=j, Y=k}=p_{j k}, \quad j, k=0,1,2, \ldots, \sum_{j, k} p_{j k}=1 .
$$
The marginal distributions are given by
$$
\begin{aligned}
& \operatorname{Pr}{X=j}=\sum_{k=0}^{\infty} p_{j k}=f_j, \quad j=0,1,2, \ldots \
& \operatorname{Pr}{Y=k}=\sum_{j=0}^{\infty} p_{j k}=g_k, \quad k=0,1,2, \ldots
\end{aligned}
$$
If $(j, k)$ is any point at which $f_j \neq 0$ (i.e. $f_j>0$ ), then the conditional distribution of $Y$ given $X=j$ is given by
$$
\operatorname{Pr}{Y=k \mid X=j}=\frac{\operatorname{Pr}{Y=k, X=j}}{\operatorname{Pr}(X=j)}=\frac{p_{j k}}{f_j}, k=0,1,2, \ldots
$$
The conditional expectation $E{y \mid X=j}$ is given by
$$
\begin{aligned}
& E{Y \mid X=j}=\sum_k k \operatorname{Pr}{Y=k \mid X=j} \
& =\sum_k k \frac{p_{j k}}{f_j}=\frac{\sum_k k p_{j k}}{f_j}, \quad j=0,1,2, \ldots
\end{aligned}
$$

统计代写|随机过程代写stochastic process代考|LAPLACE TRANSFORM

Laplace transform is a generalization of generating function. Laplace transforms serve as very powerful tools in many situations. They provide an effective means for the solution of many problems arising in our study. For example, the transforms are very effective for solving linear differential equations. The Laplace transformation reduces a linear differential equation to an algebraic equation. In the study of some probability distributions, the method could be used with great advantage, for it happens quite often that it is easier to find the Laplace transform of a probability distribution rather than the distribution itself.

Definition. Let $f(t)$ be a function of a positive real variable $t$. Then the Laplace transform (L.T.) of $f(t)$ is defined by
$$
\bar{f}(s)=\int_0^{\infty} \exp (-s t) f(t) d t
$$
for the range of values of $s$ for which the integral exists.

We shall write $\bar{f}(s)=L{f(t)}$ to denote the Laplace transform of $f(t)$.

随机过程代考

统计代写|随机过程代写stochastic process代考|Generating Function of Bivariate Distribution

二元分布:设$X, Y$是一对积分值随机变量，其联合概率分布为
$$
\operatorname{Pr}{X=j, Y=k}=p_{j k}, \quad j, k=0,1,2, \ldots, \sum_{j, k} p_{j k}=1 .
$$
边际分布由
$$
\begin{aligned}
& \operatorname{Pr}{X=j}=\sum_{k=0}^{\infty} p_{j k}=f_j, \quad j=0,1,2, \ldots \
& \operatorname{Pr}{Y=k}=\sum_{j=0}^{\infty} p_{j k}=g_k, \quad k=0,1,2, \ldots
\end{aligned}
$$
如果$(j, k)$是$f_j \neq 0$(即$f_j>0$)所在的任何点，则给定$X=j$的$Y$的条件分布由
$$
\operatorname{Pr}{Y=k \mid X=j}=\frac{\operatorname{Pr}{Y=k, X=j}}{\operatorname{Pr}(X=j)}=\frac{p_{j k}}{f_j}, k=0,1,2, \ldots
$$
条件期望$E{y \mid X=j}$由
$$
\begin{aligned}
& E{Y \mid X=j}=\sum_k k \operatorname{Pr}{Y=k \mid X=j} \
& =\sum_k k \frac{p_{j k}}{f_j}=\frac{\sum_k k p_{j k}}{f_j}, \quad j=0,1,2, \ldots
\end{aligned}
$$

统计代写|随机过程代写stochastic process代考|LAPLACE TRANSFORM

拉普拉斯变换是生成函数的一种推广。拉普拉斯变换在很多情况下都是非常强大的工具。它们为解决我们学习中出现的许多问题提供了有效的手段。例如，变换对于求解线性微分方程是非常有效的。拉普拉斯变换把一个线性微分方程化为一个代数方程。在一些概率分布的研究中，这种方法可以有很大的优势，因为通常情况下，找到概率分布的拉普拉斯变换比找到分布本身更容易。

定义。设$f(t)$为正实变量$t$的函数。那么$f(t)$的拉普拉斯变换(L.T.)定义为
$$
\bar{f}(s)=\int_0^{\infty} \exp (-s t) f(t) d t
$$
对于积分存在的$s$值范围。

我们用$\bar{f}(s)=L{f(t)}$表示$f(t)$的拉普拉斯变换。

统计代写请认准statistics-lab™. statistics-lab™为您的留学生涯保驾护航。

金融工程代写

非参数统计代写

非参数统计指的是一种统计方法，其中不假设数据来自于由少数参数决定的规定模型；这种模型的例子包括正态分布模型和线性回归模型。

广义线性模型代考

广义线性模型（GLM）归属统计学领域，是一种应用灵活的线性回归模型。该模型允许因变量的偏差分布有除了正态分布之外的其它分布。

有限元方法代写

随机分析代写

时间序列分析代写

回归分析代写

MATLAB代写

R语言代写	问卷设计与分析代写
PYTHON代写	回归分析与线性模型代写
MATLAB代写	方差分析与试验设计代写
STATA代写	机器学习/统计学习代写
SPSS代写	计量经济学代写
EVIEWS代写	时间序列分析代写
EXCEL代写	深度学习代写
SQL代写	各种数据建模与可视化代写

统计代写|随机过程代写stochastic process代考|MATH544

Posted on 2023年7月27日2023年8月25日 by statistics-lab

统计代写|随机过程代写stochastic process代考|Construction of a Markov process

To study a Markov branching process by means of the methods developed in the theory of Markov processes it is desirable to attempt to describe such a process using a special class of Markov processes (as was done in the case of semi-Markov processes). In this subsection we shall solve this problem. First we shall note that it is sufficient to consider processes with one type of particle since in the case of processes with a set $\mathscr{A}$ of various types (where $\mathscr{A}$ can be arbitrary) one can consider a one-type process on the set $\mathscr{X} \times \mathscr{A}$, with a phase point $(x, a), x \in \mathscr{X}, a \in \mathscr{A}$ which simultaneously defines the position $x$ of a particle in the phase space $\mathscr{X}$ and its type $a$. Clearly, the behavior of the process in $\mathscr{X} \times \mathscr{A}$ up to the transformation moment is completely determined by the collection of Markov process $\left{\mathscr{F}^{(a)}, \mathscr{N}^{(a)}, \mathrm{P}_x^{(a)}\right}$ since up to the moment of transformation the component $a$ of the process remains unchanged.
Thus let a certain phase space ${\mathscr{X}, B}$ be given. We shall consider a process with one type of particle described in the preceeding subsection. The state of such a process is completely determined by the number of particles and their positions in the phase space. Since the particles are identical a permutation of coordinates of particles does not alter the state of the process (the coordinate here is meant to be the location of a particle in the phase space).

We now construct the phase space of the process in the following manner.
Let the spaces $\tilde{\mathscr{X}}_n, n \geqslant 1$ be obtained from $\mathscr{X}^n$ by identifying all the points $\left(x_1, \ldots, x_n\right)$ which can be obtained by permutation of the coordinates. Denote by $\tilde{\mathfrak{B}}_n$ the image of the $\sigma$-algebra $\mathfrak{B}^n$ under this mapping of $\mathscr{X}^n$ into $\tilde{\mathscr{X}}_n$. We also introduce the space $\tilde{\mathscr{X}}_0$ consisting of a single point which is also denoted by $\tilde{\mathscr{X}}_0$. The process will be in state $\tilde{\mathscr{X}}_0$ if the phase space contains no particles. If the phase space contains $n$ particles with coordinates $x_1, \ldots, x_n$ then the process will be located at point $\left(x_1, \ldots, x_1\right) \in \tilde{\mathscr{X}}_n$. Thus the phase space of the process will consist of the union of all sets $\tilde{\mathscr{X}}_n, n=0,1, \ldots$ Denote this union by $\tilde{\mathscr{X}}$. Denote by $\tilde{\mathfrak{B}}$ the minimal $\sigma$-algebra of subsets of $\tilde{\mathscr{X}}$ containing the subset $\tilde{\mathscr{X}}_0$ and all the $\sigma$-algebras $\tilde{\mathfrak{B}}_n$. A Markov process in ${\tilde{\mathscr{X}}, \tilde{\mathfrak{B}}}$ can be associated with a Markov process with branching. If the process with branching is homogeneous then the process in ${\tilde{\mathscr{X}}, \tilde{\mathfrak{B}}}$ will also be homogeneous. It is natural to confine our attention to processes for which the exit time from each one of the sets $\tilde{\mathscr{X}}_n$ is positive given that the process is located in $\tilde{\mathscr{X}}_n$ at the initial moment. It follows from the other conditions imposed on the process that in order to achieve this it is sufficient that the exit time from $\tilde{\mathscr{X}}$, will be positive given that the process is situated at the initial time in $\tilde{\mathscr{X}}_1$.

统计代写|随机过程代写stochastic process代考|The characteristic operator of a process

The characteristic operator of a process. Let $\mathscr{X}$ be a topological space, $\mathfrak{B}$ be a $\sigma$-algebra generated by the continuous functions on $\mathscr{X}$. Then a topology is defined on each $\tilde{\mathscr{X}}m$ compatible with the topology on $\mathscr{X}$ and the $\sigma$-algebra $\tilde{\mathfrak{B}}_m$ is generated by functions which are continuous on $\mathscr{\mathscr { X }}_m$. The topology on $\tilde{\mathscr{X}}$ is defined by taking the neighborhoods of a point $\tilde{x} \in \tilde{\mathscr{X}}_m$ to be its neighborhoods in $\tilde{\mathscr{X}}$. Then each set $\tilde{\mathscr{X}}_m$ will be closed as well as open in $\tilde{\mathscr{X}}$. Let $\tilde{U}$ be a neighborhood of a point $\tilde{x}$ located entirely in the set $\tilde{\mathscr{X}}_m$ to which the point $\tilde{x}$ belongs. Assume that $\tau_U$ is the Markov moment of the first exit of the process out of the neighborhood $U$. Next let $\zeta$ be the moment of the first exit of the process from the set $\tilde{\mathscr{X}}_m$. $\left(\tilde{x} \in \tilde{\mathscr{X}}_m\right.$ is the initial position of the process.) Then for any bounded continuous function $f$ on $\hat{\mathscr{X}}$ we have $$ \mathrm{E}{\tilde{x}} f\left(\tilde{x}\left(\tau_U\right)\right)=\mathrm{E}{\tilde{x}} f\left(\tilde{x}\left(\tau_U\right)\right) \chi{\left{\tau_U<\zeta\right}}+\mathrm{E}{\tilde{x}} f\left(\tilde{x}\left(\tau_U\right)\right) \chi{\left{\tau_V \geqslant \zeta\right}},
$$
where $\tilde{x}(t)$ denotes the trajectory of a Markov branching process in the phase space ${\hat{\mathscr{X}}, \tilde{\mathfrak{B}}}$ and $\mathrm{E}{\tilde{x}}$ and $\mathrm{P}{\tilde{x}}$ denote here (and hereafter) respectively the mathematical expectation and probability associated with the process. We now observe that
$$
\mathrm{E}{\tilde{x}} f\left(\tilde{x}\left(\tau_U\right)\right) \chi{\left.\mid \tau_U=\zeta\right}}=\mathrm{E}{\tilde{x}} f(\tilde{x}(\zeta)) \chi{\left\langle\tau_U=\zeta\right}}=\mathrm{E}{\tilde{x}} f(\tilde{x}(\zeta))-\mathrm{E}{\tilde{x}} f(\tilde{x}(\zeta)) \chi_{\left\langle\tau_U<\zeta\right}} .
$$
Assume that the process $\tilde{x}(t)$ is strong Markov. Then
$$
\mathrm{E}{\tilde{x}} f(\tilde{x}(\zeta)) \chi{\left{\tau_U<\zeta\right}}=\mathrm{E}{\tilde{x}}\left[\mathrm{E}{\tilde{x}\left(\tau_U\right)} f(\tilde{x}(\zeta))\right] \chi_{\left{\tau_U<\zeta\right}} .
$$
We introduce the operator $\mathscr{T}: \mathscr{T} f(\tilde{x})=E_{\tilde{x}} f(\tilde{x}(\zeta))$. Then
$$
\mathrm{E}{\tilde{x}} f\left(\tilde{x}\left(\tau_U\right)\right)=\mathrm{E}{\tilde{x}} f\left(\tilde{x}\left(\tau_U\right)\right) \chi_{\left{\tau_U<\zeta\right}}+\mathscr{T} f(\tilde{x})-\mathrm{E}{\tilde{x}} \mathscr{T} f\left(\tilde{x}\left(\tau_U\right)\right) \chi{\left{\tau_U<\zeta\right}} .
$$

随机过程代考

统计代写|随机过程代写stochastic process代考|Construction of a Markov process

为了利用马尔可夫过程理论中发展的方法来研究马尔可夫分支过程，尝试使用一类特殊的马尔可夫过程来描述这样的过程是可取的(就像在半马尔可夫过程的情况下所做的那样)。在本节中，我们将解决这个问题。首先，我们应该注意到，考虑具有一种粒子类型的过程是足够的，因为在具有各种类型集合$\mathscr{A}$的过程的情况下($\mathscr{A}$可以是任意的)，可以考虑集合$\mathscr{X} \times \mathscr{A}$上的一种类型的过程，具有一个相点$(x, a), x \in \mathscr{X}, a \in \mathscr{A}$，该相点同时定义了粒子在相空间$\mathscr{X}$中的位置$x$及其类型$a$。显然，到转换时刻为止，$\mathscr{X} \times \mathscr{A}$中过程的行为完全由马尔可夫过程$\left{\mathscr{F}^{(a)}, \mathscr{N}^{(a)}, \mathrm{P}_x^{(a)}\right}$的集合决定，因为到转换时刻为止，过程的组件$a$保持不变。
因此，设一定的相空间${\mathscr{X}, B}$。我们将考虑具有上一小节中描述的一种粒子的过程。这种过程的状态完全由粒子的数量和它们在相空间中的位置决定。由于粒子是相同的，粒子坐标的排列不会改变过程的状态(这里的坐标是指粒子在相空间中的位置)。

我们现在用下面的方法构造这个过程的相空间。
将坐标置换得到的所有点$\left(x_1, \ldots, x_n\right)$，从$\mathscr{X}^n$得到空间$\tilde{\mathscr{X}}_n, n \geqslant 1$。用$\tilde{\mathfrak{B}}_n$表示$\sigma$ -代数$\mathfrak{B}^n$在$\mathscr{X}^n$到$\tilde{\mathscr{X}}_n$的映射下的图像。我们还引入了由单点组成的空间$\tilde{\mathscr{X}}_0$，该空间也用$\tilde{\mathscr{X}}_0$表示。如果相空间不包含粒子，则该过程将处于状态$\tilde{\mathscr{X}}_0$。如果相空间包含坐标为$x_1, \ldots, x_n$的$n$粒子，则过程将位于$\left(x_1, \ldots, x_1\right) \in \tilde{\mathscr{X}}_n$点。因此过程的相空间将由所有集合的并集组成$\tilde{\mathscr{X}}_n, n=0,1, \ldots$用$\tilde{\mathscr{X}}$表示这个并集。用$\tilde{\mathfrak{B}}$表示包含子集$\tilde{\mathscr{X}}_0$和所有$\sigma$ -代数$\tilde{\mathfrak{B}}_n$的$\tilde{\mathscr{X}}$子集的最小$\sigma$ -代数。${\tilde{\mathscr{X}}, \tilde{\mathfrak{B}}}$中的马尔可夫过程可以与具有分支的马尔可夫过程相关联。如果带有分支的流程是同构的，那么${\tilde{\mathscr{X}}, \tilde{\mathfrak{B}}}$中的流程也将是同构的。考虑到进程在初始时刻位于$\tilde{\mathscr{X}}_n$，很自然地将我们的注意力限制在每个集合$\tilde{\mathscr{X}}_n$的退出时间为正的进程上。从过程中施加的其他条件可以看出，为了实现这一目标，如果过程位于$\tilde{\mathscr{X}}_1$的初始时间，则从$\tilde{\mathscr{X}}$的退出时间将是正的，这就足够了。

统计代写|随机过程代写stochastic process代考|The characteristic operator of a process

过程的特征操作符。设$\mathscr{X}$为拓扑空间，$\mathfrak{B}$为由$\mathscr{X}$上的连续函数生成的$\sigma$ -代数。然后在每个$\tilde{\mathscr{X}}m$上定义一个与$\mathscr{X}$上的拓扑兼容的拓扑，并由$\mathscr{\mathscr { X }}m$上连续的函数生成$\sigma$ -代数$\tilde{\mathfrak{B}}_m$。通过取点$\tilde{x} \in \tilde{\mathscr{X}}_m$的邻域作为其在$\tilde{\mathscr{X}}$中的邻域来定义$\tilde{\mathscr{X}}$上的拓扑。然后，每个集$\tilde{\mathscr{X}}_m$将在$\tilde{\mathscr{X}}$中关闭和打开。设$\tilde{U}$为点$\tilde{x}$的邻域，它完全位于点$\tilde{x}$所属的集合$\tilde{\mathscr{X}}_m$中。假设$\tau_U$是该过程第一个出口出邻近区域$U$的马尔可夫矩。接下来，让$\zeta$作为进程第一次从集合$\tilde{\mathscr{X}}_m$退出的时刻。$\left(\tilde{x} \in \tilde{\mathscr{X}}_m\right.$是该过程的初始位置。)对于任意有界连续函数$f$在$\hat{\mathscr{X}}$上，我们有$$ \mathrm{E}{\tilde{x}} f\left(\tilde{x}\left(\tau_U\right)\right)=\mathrm{E}{\tilde{x}} f\left(\tilde{x}\left(\tau_U\right)\right) \chi{\left{\tau_U<\zeta\right}}+\mathrm{E}{\tilde{x}} f\left(\tilde{x}\left(\tau_U\right)\right) \chi{\left{\tau_V \geqslant \zeta\right}}, $$ 其中$\tilde{x}(t)$表示相空间中马尔可夫分支过程的轨迹${\hat{\mathscr{X}}, \tilde{\mathfrak{B}}}$, $\mathrm{E}{\tilde{x}}$和$\mathrm{P}{\tilde{x}}$分别表示与该过程相关的数学期望和概率。我们现在观察到 $$ \mathrm{E}{\tilde{x}} f\left(\tilde{x}\left(\tau_U\right)\right) \chi{\left.\mid \tau_U=\zeta\right}}=\mathrm{E}{\tilde{x}} f(\tilde{x}(\zeta)) \chi{\left\langle\tau_U=\zeta\right}}=\mathrm{E}{\tilde{x}} f(\tilde{x}(\zeta))-\mathrm{E}{\tilde{x}} f(\tilde{x}(\zeta)) \chi{\left\langle\tau_U<\zeta\right}} .
$$
假设过程$\tilde{x}(t)$是强马尔可夫的。然后
$$
\mathrm{E}{\tilde{x}} f(\tilde{x}(\zeta)) \chi{\left{\tau_U<\zeta\right}}=\mathrm{E}{\tilde{x}}\left[\mathrm{E}{\tilde{x}\left(\tau_U\right)} f(\tilde{x}(\zeta))\right] \chi_{\left{\tau_U<\zeta\right}} .
$$
我们引入算子$\mathscr{T}: \mathscr{T} f(\tilde{x})=E_{\tilde{x}} f(\tilde{x}(\zeta))$。然后
$$
\mathrm{E}{\tilde{x}} f\left(\tilde{x}\left(\tau_U\right)\right)=\mathrm{E}{\tilde{x}} f\left(\tilde{x}\left(\tau_U\right)\right) \chi_{\left{\tau_U<\zeta\right}}+\mathscr{T} f(\tilde{x})-\mathrm{E}{\tilde{x}} \mathscr{T} f\left(\tilde{x}\left(\tau_U\right)\right) \chi{\left{\tau_U<\zeta\right}} .
$$

统计代写请认准statistics-lab™. statistics-lab™为您的留学生涯保驾护航。

金融工程代写

非参数统计代写

非参数统计指的是一种统计方法，其中不假设数据来自于由少数参数决定的规定模型；这种模型的例子包括正态分布模型和线性回归模型。

广义线性模型代考

广义线性模型（GLM）归属统计学领域，是一种应用灵活的线性回归模型。该模型允许因变量的偏差分布有除了正态分布之外的其它分布。

有限元方法代写

随机分析代写

时间序列分析代写

回归分析代写

MATLAB代写

R语言代写	问卷设计与分析代写
PYTHON代写	回归分析与线性模型代写
MATLAB代写	方差分析与试验设计代写
STATA代写	机器学习/统计学习代写
SPSS代写	计量经济学代写
EVIEWS代写	时间序列分析代写
EXCEL代写	深度学习代写
SQL代写	各种数据建模与可视化代写

统计代写|随机过程代写stochastic process代考|Stat150

Posted on 2023年7月27日2023年8月25日 by statistics-lab

统计代写|随机过程代写stochastic process代考|Duration of stay in a region and the value at the exit time

Let $G$ be a region in $\mathscr{R}^m, \tau(x)$ be the first exit time of the process $\xi(t)+x(x \in G)$ from the region $G$, and $\xi(\tau(x))+x$ be the location of the process at the exit time. In this subsection the joint distribution of the variables $\tau(x)$ and $\xi(\tau(x))+x$ is studied.
First consider the process with the cumulant
$$
K(z)=\int\left(e^{i(z, y)}-1\right) \Pi(d y),
$$
where the measure $\Pi$ is finite. In this case $\xi(t)$ is a step process. If $\zeta$ is the time of the first jump of the process then $\zeta$ has an exponential distribution with parameter $\Pi\left(\mathscr{R}^m\right)$, and $\xi(\zeta)$ does not depend on $\zeta$ and
$$
\mathrm{P}{\xi(\zeta) \in B}=\frac{1}{\Pi\left(\mathscr{R}^m\right)} \Pi(B) .
$$
Therefore for any $t>0$ and any set $C \subset \mathscr{R}^m-G$ we have
$$
\begin{aligned}
\mathrm{P}{\tau(x)0$. Set
$$
v_\lambda(G) f(x)=\mathrm{E} e^{-\lambda \tau(x)} f(x+\xi(\tau(x))) .
$$

统计代写|随机过程代写stochastic process代考|Non-negative additive functionals

One can associate a homogeneous Markov process $x_t=\xi(t)+x$ with $\xi(t)$ in the manner indicated in the beginning of this section. Consider an additive functional on this process, i. e. a family of variables $\varphi_t$ defined for $t \geqslant 0$ and satisfying the following conditions:
a) $\varphi_t$ is measurable with respect to the $\sigma$-algebra $\mathscr{N}t$ generated by the variables $x_s$ for $s \leqslant t$; b) if $\theta_t$ is a shift operator associated with the Markov process $x_t$, then for all $h>0$ we have with probability $\mathrm{P}_x=1$ for any $x \in \mathscr{R}^m$ $$ \theta_h \varphi_t=\varphi{t+h}-\varphi_h .
$$
We shall discuss only non-negative continuous functionals; such functionals have the property that $\varphi_t$ is a continuous function of $t$ and is also non-decreasing. Such functionals for general homogeneous Markov processes were studied in detail in Section 6 of Chapter II. In particular the following assertions were proved there:
1) for any continuous non-negative additive functional $\varphi_t$ there exists a sequence of bounded Borel functions $f_n(x)$ such that
$$
\varphi_t=\lim \int_0^t f_n\left(x_s\right) d s
$$
in the measure $\mathbf{P}x$ for any $x \in \mathscr{R}^m$; 2) if $f(x)$ is a bounded non-negative Borel function and $\varphi_t$ is a continuous additive functional, then $$ \psi_t=\int_0^t f\left(x_s\right) d \varphi_s $$ is a homogeneous continuous non-negative additive functional and moreover the function $f(x)$ can be chosen in such a manner that $$ \sup {x \in \mathscr{\overparen { R }}^m} \mathrm{E}_x \psi_t<\infty,
$$
i. e. that $\psi$, will be a $W$-functional.

随机过程代考

统计代写|随机过程代写stochastic process代考|Duration of stay in a region and the value at the exit time

设$G$是$\mathscr{R}^m中的一个区域，$ tau(x)$是进程$\xi(t)+x(x \in G)$从区域$G$的第一次退出时间，$\xi(\tau(x))+x$是进程在退出时间的位置。本节研究了变量$\tau(x)$和$\xi(\tau(x))+x$的联合分布。
首先考虑累积量的过程
$ $
K(z)=\int\left(e^{i(z, y)}-1\right) \Pi(d y)，
$ $
其中度量$\Pi$是有限的。在这种情况下，$\xi(t)$是一个阶跃过程。如果$\zeta$是过程的第一次跳跃的时间，那么$\zeta$具有参数$\Pi\左(\mathscr{R}^m\右)$的指数分布，并且$\xi(\zeta)$不依赖于$\zeta$和
$ $

$$
\mathrm{P}{\xi(\zeta) \in B}=\frac{1}{\Pi\left(\mathscr{R}^m\right)} \Pi(B) .
$$
Therefore for any $t>0$ and any set $C \subset \mathscr{R}^m-G$ we have
$$
\begin{aligned}
\mathrm{P}{\tau(x)0$. Set
$$
v_\lambda(G) f(x)=\mathrm{E} e^{-\lambda \tau(x)} f(x+\xi(\tau(x))) .
$$

统计代写|随机过程代写stochastic process代考|Non-negative additive functionals

我们可以将齐次马尔可夫过程$x_t=\xi(t)+x$与$\xi(t)$以本节开头所示的方式联系起来。考虑这个过程上的一个加性泛函，即一组变量$\varphi_t$定义为$t \geqslant 0$，并满足以下条件:
a) $\varphi_t$对于$\sigma$-代数$\mathscr{N}t$是可测量的，该代数$\mathscr{N}t$是由变量$x_s$对于$s \leqslant t$产生的;b)如果$\theta_t$是与马尔可夫过程$x_t$相关联的移位算子，那么对于所有$h>0$，我们有概率$\ mathscr{R}^m$ $\ theta_h \varphi_t=\varphi{t+h}-\varphi_h中任意$x =1$。
$ $
我们只讨论非负连续泛函;这样的泛函具有$\varphi_t$是$t$的连续函数并且是非递减的性质。第2章第6节详细研究了一般齐次马尔可夫过程的泛函。其中特别证明了下列断言:
1)对于任意连续非负加性泛函$\varphi_t$存在一个有界Borel函数序列$f_n(x)$满足
$ $
\varphi_t=\lim \int_0^t f_n\left(x_s\right) d s
$ $
在测度$\mathbf{P}x$中对于任意$x \in \mathscr{R}^m$;2)如果$f(x)$是有界非负Borel函数，$\varphi_t$是连续可加泛函，则$$ \psi_t=\int_0^t f\左(x_s\右)d \varphi_s $$是齐次连续非负可加泛函，并且函数$f(x)$可以这样选择$$ \sup {x \in \mathscr{\ overparenn {R}}^m} \ maththrm {E} x \psi_t<\infty;
$ $
即$\psi$，将是$W$-函数。

统计代写请认准statistics-lab™. statistics-lab™为您的留学生涯保驾护航。

R语言代写	问卷设计与分析代写
PYTHON代写	回归分析与线性模型代写
MATLAB代写	方差分析与试验设计代写
STATA代写	机器学习/统计学习代写
SPSS代写	计量经济学代写
EVIEWS代写	时间序列分析代写
EXCEL代写	深度学习代写
SQL代写	各种数据建模与可视化代写