### 统计代写|贝叶斯分析代写Bayesian Analysis代考|Posterior Information

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• Statistical Inference 统计推断
• Statistical Computing 统计计算
• (Generalized) Linear Models 广义线性模型
• Statistical Machine Learning 统计机器学习
• Longitudinal Data Analysis 纵向数据分析
• Foundations of Data Science 数据科学基础

## 统计代写|贝叶斯分析代写Bayesian Analysis代考|The Binomial Distribution

The preceding section explains how prior information is expressed in an informative or in a noninformative way. Several examples are given and will be revisited as illustrations for the determination of the posterior distribution of the parameters. Suppose a uniform prior distribution for the transition probability (of the five-state Markov chain) $p_{i j}$ is used. What is the posterior distribution of $p_{i j}$ ?
By Bayes theorem,
$$f\left(p_{i j} \mid N\right) \propto\left(\begin{array}{c} n \ n_{i j} \end{array}\right) p_{i j}^{n_{i j}}\left(1-p_{i j}\right)^{n-n_{i j}}$$
where $n_{i j}$ is the observed transitions from state $i$ to state $j$ and $n$ is the total cell counts for the five-by-five cell frequency matrix $N$. Of course, this is recognized as a beta $\left(n_{i j}+1, n-n_{i j}+1\right)$ distribution, and the posterior mean is $\left(n_{i j}+1 / n+2\right)$. However, if the Lhoste ${ }^{5}$ prior density (2.4) is used, the posterior distribution of $p_{i j}$ is beta $\left(n_{i j}, n-n_{i j}\right)$ with mean $n_{i j} / n$, which is the usual estimator of $p_{i j}$.

## 统计代写|贝叶斯分析代写Bayesian Analysis代考|The Normal Distribution

Consider a random sample $\mathrm{X}=\left(x_{1}, x_{2}, \ldots, x_{n}\right)$ of size $n$ from a normal $(\mu, 1 / \tau)$ population, where $\tau=1 / \sigma^{2}$ is the inverse of the variance, and suppose the prior information is vague and the Jeffreys-Lhoste prior $\xi(\mu, \tau) \propto 1 / \tau$ is appropriate, then the posterior density of the parameters is
$$\xi(\mu, \tau \mid \text { data }) \propto \tau^{n / 2-1} \exp -(\tau / 2)\left[n(\mu-\bar{x})^{2}+\sum_{i=1}^{i=n}\left(x_{i}-\bar{x}\right)^{2}\right]$$

Using the properties of the gamma density, $\tau$ is eliminated by integrating the joint density with respect to $\tau$ to give
$$\begin{gathered} \xi(\mu \mid \text { data }) \propto \ \left{\Gamma(n / 2) n^{1 / 2} /(n-1)^{1 / 2} S \pi^{1 / 2} \Gamma((n-10 / 2)} /\left[1+n(\mu-\bar{x})^{2} /(n-1) S^{2}\right]^{(n-1+1) / 2}\right. \end{gathered}$$
which is recognized as a $t$-distribution with $n-1$ degrees of freedom, location $\bar{x}$, and precision $n / S^{2}$. Transforming to $(\mu-\bar{x}) \sqrt{n} / S$, the resulting variable has a Student’s $t$-distribution with $n-1$ degrees of freedom. Note the mean of $\mu$ is the sample mean, while the variance is $[(n-1) / n(n-3)], n>3$.
Eliminating $\mu$ from (12) results in the marginal distribution of $\tau$ as
$$\xi\left(\tau \mid s^{2}\right) \propto \tau^{[(n-1) / 2 \mid-1} \exp -\tau(n-1) S^{2} / 2, \tau>0,$$
which is a gamma density with parameters $(n-1) / 2$ and $(n-1) s^{2} / 2$. This implies the posterior mean is $1 / s^{2}$ and the posterior variance is $2 /(n-1) s^{4}$. For example, consider the $\mathrm{AR}(1)$ (2.15) series where $\theta=.6$ and $\sigma^{2}=1$, then suppose $\mathrm{R}$ is used to generate a realization of $n=50$ from the series.

## 统计代写|贝叶斯分析代写Bayesian Analysis代考|The Poisson Distribution

The Poisson distribution often occurs as a population for a discrete random variable with mass function
$$\mathrm{f}(\mathrm{X} \mid \theta)=e^{-\theta} \theta^{x} / x !$$

where the gamma density
$$\xi(\theta)=\left[\beta^{\alpha} / \Gamma(\alpha)\right] \theta^{\alpha-1} e^{-\theta \beta},$$
is a conjugate distribution that expresses informative prior information. For example, in a previous experiment with $m$ observations, the prior density would be gamma with the appropriate values of alpha and beta. Based on a random sample of size $n$, the posterior density is
$$\xi(\theta \mid \text { data }) \propto \sum_{\theta=1}^{i n n} x_{i}+\alpha-1, e^{-\theta(n+\beta)},$$
which is identified as a gamma density with parameters $\alpha^{\prime}=\sum_{i=1}^{i=n} x_{i}+\alpha$ and $\beta^{\prime}=n+\beta$. Remember the posterior mean is $\alpha^{\prime} / \beta^{\prime}$, median $\left(\alpha^{\prime}-1\right) / \beta^{\prime}$, and variance $\alpha^{\prime} /\left(\beta^{r}\right)^{2}$.
One of the most important time series is the Poisson process.
The Poisson process $N(t)$ with parameter $\lambda>0$ is defined as follows:

1. $N(t)$ is the number of events occurring over time 0 to $t$ with $N(0)=0$ and the process has independent increments.
2. For all $t>0,0\langle P[N(t)\rangle 0]<1$, that is to say for all intervals, no matter how small, there is a positive probability that an event will occur, but it is not certain an event will occur.
3. For all $t \geq 0$,
$$\lim {P[N(t+h)-N(t) \geq 2] / P[N(t+h)-N(t)=1]},$$
where the limit is as $h$ approaches 0 . This implies that events cannot occur simultaneously.
4. The process has stationary independent increments; thus for all points $t>s \geq 0$ and $h>0$, the two random variables $N(t+h)-N(s+h)$ and $N(t)-N(s)$ are identically distributed and are independent.

## 统计代写|贝叶斯分析代写Bayesian Analysis代考|The Binomial Distribution

F(p一世j∣ñ)∝(n n一世j)p一世jn一世j(1−p一世j)n−n一世j

## 统计代写|贝叶斯分析代写Bayesian Analysis代考|The Normal Distribution

X(μ,τ∣ 数据 )∝τn/2−1经验−(τ/2)[n(μ−X¯)2+∑一世=1一世=n(X一世−X¯)2]

\begin{聚集}\xi(\mu\mid\text{data})\propto\\left{\Gamma(n/2)n^{1/2}/(n-1)^{1/2} S \pi^{1/2} \Gamma((n-10/2)} /\left[1+n(\mu-\bar{x})^{2} /(n-1) S^{ 2}\right]^{(n-1+1)/2}\right.\end{聚集}\begin{聚集}\xi(\mu\mid\text{data})\propto\\left{\Gamma(n/2)n^{1/2}/(n-1)^{1/2} S \pi^{1/2} \Gamma((n-10/2)} /\left[1+n(\mu-\bar{x})^{2} /(n-1) S^{ 2}\right]^{(n-1+1)/2}\right.\end{聚集}

X(τ∣s2)∝τ[(n−1)/2∣−1经验−τ(n−1)小号2/2,τ>0,

## 统计代写|贝叶斯分析代写Bayesian Analysis代考|The Poisson Distribution

F(X∣θ)=和−θθX/X!

X(θ)=[b一个/Γ(一个)]θ一个−1和−θb,

X(θ∣ 数据 )∝∑θ=1一世nnX一世+一个−1,和−θ(n+b),

1. ñ(吨)是在时间 0 到吨和ñ(0)=0并且该过程具有独立的增量。
2. 对所有人吨>0,0⟨磷[ñ(吨)⟩0]<1，也就是说对于所有的区间，无论多小，都有一个事件发生的正概率，但并不确定某个事件是否会发生。
3. 对所有人吨≥0,
林磷[ñ(吨+H)−ñ(吨)≥2]/磷[ñ(吨+H)−ñ(吨)=1],
限制在哪里H接近 0 。这意味着事件不能同时发生。
4. 该过程具有平稳的独立增量；因此对于所有点吨>s≥0和H>0, 两个随机变量ñ(吨+H)−ñ(s+H)和ñ(吨)−ñ(s)同分布且独立。

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## MATLAB代写

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