## 澳洲代写 随机过程Stochastic process代考2023

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## 随机过程Stochastic process代考

#### 傅立叶分析Fourier analysis代写代考

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

## 随机过程Stochastic process定义

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.

## 随机过程Stochastic process的重难点

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

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

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

$P\left(N_k=n\right)=P\left(S_{n-1}=k-1\right) \cdot P\left(X_n=1\right)=\left(\begin{array}{c}n-1 \ k-1\end{array}\right) p^k q^{n-k}$

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

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

$$P{X=N}=\left(\begin{array}{c} 2 N \ N \end{array}\right)\left(\frac{1}{2}\right)^{2 N} \approx \frac{1}{\sqrt{N \pi}},$$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

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 .

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

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## 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

\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}

\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}

$$\left(\begin{array}{ccc} 3 / 4 & 1 / 4 & 0 \ 1 / 4 & 1 / 2 & 1 / 4 \ 0 & 3 / 4 & 1 / 4 \end{array}\right)$$

\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)。

$$\operatorname{Pr}\left{B \mid X_N=i, A\right}=\operatorname{Pr}\left{B \mid X_N=i\right} .$$

\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}

\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}

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$i}, i, j, k=0,1$。

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)。

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## MATLAB代写

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

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## 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

$$\operatorname{Pr}\left(X_n=1\right)=p, \operatorname{Pr}\left{X_n=0\right}=q .$$

\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}

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

$$p_{j k} \geq 0, \quad \sum_k p_{j k}=1 \text { for all } j .$$

$$P=\left(\begin{array}{cccc} p_{11} & p_{12} & p_{13} & \cdots \ p_{21} & p_{22} & p_{23} & \cdots \ \cdots & \cdots & \ldots & \ldots \ \cdots & \ldots & \ldots & \ldots \end{array}\right)$$

$$\begin{array}{ll} \operatorname{Pr}\left{X_n=r+1 \mid X_{n-1}=r\right}=p & \ \operatorname{Pr}\left{X_n=r-1 \mid X_{n-1}=r\right}=q & 0<r<4 \end{array}$$

\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}

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

## 有限元方法代写

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。

## 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。

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

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|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

$$v_\lambda(G) f(x)=\mathrm{E} e^{-\lambda \tau(x)} f(x+\xi(\tau(x))) .$$

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

## 有限元方法代写

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。

## 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。