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# 统计代写|统计推断代考Statistical Inference代写|Continuous-time Markov chains

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## 统计代写|统计推断代考Statistical Inference代写|Continuous-time Markov chains

A stochastic process ${X(t): t \geq 0}$ with state space $S={1,2, \ldots, M}$ is a continuoustime Markov chain if, for all $n \in \mathbb{Z}^{+}, x_1, x_2, \ldots, x_n \in S$, and $0 \leq t_1<t_2<\ldots<t_n$, it holds that
$$\mathrm{P}\left(X\left(t_n\right)=x_n \mid X\left(t_{n-1}\right)=x_{n-1}, \ldots, X\left(t_1\right)=x_1\right)=\mathrm{P}\left(X\left(t_n\right)=x_n \mid X\left(t_{n-1}\right)=x_{n-1}\right) .$$
If the chain is homogeneous (defined in the usual way) the transition probability, for $i, j \in S$ and $t, h \geq 0$, is given by
$$\mathrm{P}(X(t+h)=j \mid X(t)=i)= \begin{cases}1-g_{i, i} h+o(h) & \text { for } j=i, \ g_{i, j} h+o(h) & \text { for } j \neq i,\end{cases}$$
where $g_{i, j} \geq 0$ is the $(i, j)^{\text {th }}$ entry of the generator matrix, $\boldsymbol{G}$, a matrix with rows that sum to 0 . The similarity with Definition 6.7.7 is not coincidental; the Poisson process is a special case of a continuous-time Markov chain. Similarly to a Poisson process, a continuous-time Markov chain remains in each state for an exponentially distributed amount of time before jumping to the next state.

## 统计代写|统计推断代考Statistical Inference代写|Further exercises

For the one-way ANOVA model in section 6.3 .3 , let $n_j$ be the number of observations in the sample that came from population $j$,
$$n_j=\sum_{i=1}^n x_{i, j},$$
and define
$$\bar{Y}j=\frac{1}{n_j} \sum{i=1}^n x_{i, j} Y_i,$$
the mean of these observations; we refer to this as the $j^{\text {th }}$ group mean. Show that
\begin{aligned} & \underbrace{\sum_{i=1}^n\left(Y_i-\bar{Y}\right)^2}{\text {Total sum }}=\underbrace{\sum{j=0}^{k-1} n_j\left(\bar{Y}j-\bar{Y}\right)^2}{\text {Between-group }}+\underbrace{\sum_{i=1}^n\left(Y_i-\sum_{j=0}^{k-1} x_{i, j} \bar{Y}j\right)^2}{\text {Within-group }} . \ & \text { of squares sum of squares sum of squares } \ & \end{aligned}

[The total sum of squares is a measure of the variation in the response variable. The expression above shows that we can attribute some of this variation to the differences between the groups, and the rest to the differences within each group, which we typically consider to be random error.]

There is another way to formulate the logistic regression model, which uses the properties of the logistic distribution. If $Y \sim \operatorname{Logistic}(\mu, \tau)$, then the $\operatorname{CDF}$ of $Y$ is
$$F_Y(y)=\left[1+\exp \left(-\frac{y-\mu}{\tau}\right)\right]^{-1} \text { for } y \in \mathbb{R},$$
where $\mu$ is the location parameter and $\tau>0$ is the scale parameter. Suppose that we have binary data $Y_1, \ldots, Y_n$, where the value of $Y_i$ is dependent on the (unobserved) latent variable $Z_i \sim \operatorname{Logistic}\left(\mathbf{x}_i \beta, 1\right)$. As before, $\mathbf{x}_i$ denotes a vector of explanatory variables, and $\beta$ are the model parameters. We observe $Y_i=1$ if $Z_i>0$, and $Y_i=0$ otherwise. Show that this model is equivalent to the logistic regression model. What if the distribution of $Z_i$ were Normal with mean $\mathbf{x}_i \boldsymbol{\beta}$ and variance 1? Explain why we need to fix the variance parameter to 1 .

Suppose we are interested in whether people at a particular city commute to work by public transport, private car, bike, or on foot. This is a categorical variable so we model it with the categorical distribution, which is the multinomial distribution with $n=1$ (but slightly different notation). We write $Y_i=1$ if the $i^{\text {th }}$ person takes public transport, $Y_i=2$ if they use a private car, etc. The quantities of interest are $p_{i j}=P\left(Y_i=j\right)$, for $i=1, \ldots, n$ and $j=1, \ldots, k$, where $k$ is the number of categories. The link function is
$$\log \left(\frac{p_{i j}}{p_{i 1}}\right)=\boldsymbol{x}_i \boldsymbol{\beta}_j \text { for } j=2, \ldots, k,$$
where $\mathbf{x}_i$ is a vector of explanatory variables, and $\boldsymbol{\beta}_2, \ldots, \boldsymbol{\beta}_k$ are vectors of parameters.

# 统计推断代写

## 统计代写|统计推断代考Statistical Inference代写|Continuous-time Markov chains

$$\mathrm{P}\left(X\left(t_n\right)=x_n \mid X\left(t_{n-1}\right)=x_{n-1}, \ldots, X\left(t_1\right)=x_1\right)=\mathrm{P}\left(X\left(t_n\right)=x_n \mid X\left(t_{n-1}\right)=x_{n-1}\right) .$$

$$\mathrm{P}(X(t+h)=j \mid X(t)=i)=\left{1-g_{i, i} h+o(h) \quad \text { for } j=i, g_{i, j} h+o(h) \quad \text { for } j \neq i,\right.$$

## 统计代写|统计推断代考Statistical Inference代写|Further exercises

$$n_j=\sum_{i=1}^n x_{i, j}$$

$$\bar{Y} j=\frac{1}{n_j} \sum i=1^n x_{i, j} Y_i$$

$$\underbrace{\sum_{i=1}^n\left(Y_i-\bar{Y}\right)^2} \text { Total sum }=\underbrace{\sum j=0^{k-1} n_j(\bar{Y} j-\bar{Y})^2} \text { Between-group }+\underbrace{\sum_{i=1}^n\left(Y_i-\sum_{j=0}^{k-1} x_{i, j} \bar{Y} j\right)^2} \text { Within-group . }$$
[总平方和是响应变量变化的度量。上面的表达式表明，我们可以将这种变化的一部分归因于组之间的差异，其余的归因于每个组 内的差异，我们通常将其视为随机误差。]

$$F_Y(y)=\left[1+\exp \left(-\frac{y-\mu}{\tau}\right)\right]^{-1} \text { for } y \in \mathbb{R}$$

$$\log \left(\frac{p_{i j}}{p_{i 1}}\right)=\boldsymbol{x}_i \boldsymbol{\beta}_j \text { for } j=2, \ldots, k$$

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## 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。