### 统计代写|贝叶斯分析代写Bayesian Analysis代考|Predictive Inference

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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 Population

Suppose the binomial case is again considered, where the posterior density of the binomial parameter $\theta$ is
\begin{aligned} \xi(\theta \mid X)=& {[\Gamma(\alpha+\beta) \Gamma(n+1) / \Gamma(\alpha) \Gamma(\beta) \Gamma(x+1) \Gamma(n-x+1)] } \ & \theta^{\alpha+x-1}(1-\theta)^{\beta+n-x-1} \end{aligned}
a beta with parameters $\alpha+x$ and $n-x+\beta$, and $X$ is the sum of the set of $n$ observations. The population mass function of a future observation $Z$ is

$f(z / \theta)=\theta^{z}(1-\theta)^{1-z}$, and the predictive mass function of $Z$, called the beta-binomial, is
\begin{aligned} &\mathrm{g}(\mathrm{z} \mid \mathrm{X})=\Gamma(\alpha+\beta) \Gamma(n+1) \Gamma\left(\alpha+\sum_{i=1}^{i=n} x_{i}+z\right) \Gamma(1+n+\beta-x-z) \div \ &\Gamma(\alpha) \Gamma(\beta) \Gamma(n-x+1) \Gamma(x+1) \Gamma(n+1+\alpha+\beta) \end{aligned}
where $z=0.1$. Note this function does not depend on the unknown parameter, and that the $n$ past observations are known, and that if $\alpha=\beta=1$, one is assuming a uniform prior density for $\theta$.

As an example for the predictive distribution of the binomial distribution, let
$$\mathrm{N}=\left(\begin{array}{l} 1,7,4,6,2 \ 5,3,4,6,5 \ 3,6,2,3,1 \ 5,7,2,5,4 \ 6,1,3,3,4 \end{array}\right)$$
be the transition counts for a five-state Markov chain and
$$\Phi=\left(\begin{array}{l} \phi_{11}, \phi_{12}, \phi_{13}, \phi_{14}, \phi_{15} \ \phi_{21}, \phi_{22}, \phi_{23}, \phi_{24}, \phi_{25} \ \phi_{31}, \phi_{32}, \phi_{33}, \phi_{34}, \phi_{35} \ \phi_{41}, \phi_{42}, \phi_{43}, \phi_{44}, \phi_{45} \ \phi_{51}, \phi_{52}, \phi_{53}, \phi_{54}, \phi_{55} \end{array}\right)$$
as the one-step transition matrix.
Our focus is on forecasting the number of transitions $Z_{11}$ from 1 to 1 , that is the number of times the chain remains in state 1 , assuming a total of $m$ replications for the first row of the chain, that is $Z_{11}=0,1,2, \ldots, m$. Using (2.77), one may show the predictive mass function of $Z_{11}$ is equation (2.78)
$$\begin{gathered} g\left(z \mid n_{11}=1\right)=\left(\begin{array}{c} m \ z \end{array}\right) \Gamma(\alpha+\beta) \Gamma(n+1) \Gamma\left(\alpha+z+n_{11}\right) \Gamma\left(\beta+n-z-n_{11}\right) / \ \Gamma(\alpha) \Gamma(\beta) \Gamma\left(n_{11}+1\right) \Gamma\left(n-n_{11}+1\right) \Gamma(\alpha+\beta+n) \end{gathered}$$
The relevant quantities of $(2.78)$ are $n=20, n_{11}=1$. Also remember that $\alpha$ and $\beta$ are the parameters of the prior distribution of $\phi_{11}$, the probability of remaining in state 1 , and that predictive inferences are conditional on $n=20$, the total transition counts for the first row of the one-step transition matrix of the chain.

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

It is imperative to check model adequacy in order to choose an appropriate model and to conduct a valid study. The approach taken here is based on many sources, including Gelman et al., ${ }^{2}$ Carlin and Louis, ${ }^{18}$ and Congdon ${ }^{16} \mathrm{Our}$ main focus will be on the likelihood function of the posterior distribution, and not the prior distribution, and to this end, graphical representations such as histograms, boxplots, and various probability plots of the original observations will be compared with those of the observations generated from the predictive distribution. In addition to graphical methods, Bayesian versions of overall goodness-offit-type operations are taken to check model validity. Methods presented at this juncture are just a small subset of those presented in more advanced works, including Gelman et al., Carlin and Louis, and Congdon.

Of course, the prior distribution is an important component of the analysis, and if one is not sure of the ‘true’ prior, one should perform a sensitivity analysis to determine the robustness of posterior inferences to various alternative choices of prior information. See Gelman et al. or Carlin and Louis for details of performing a sensitivity study for prior information. Our approach is to either use informative or vague prior distributions, where the former is done when prior relevant experimental evidence determines the prior, or the latter is taken if there is none or very little germane experimental studies. In scientific studies, the most likely scenario is that there are relevant experimental studies providing informative prior information.

## 统计代写|贝叶斯分析代写Bayesian Analysis代考|Sampling from an Exponential, but Assuming a Normal Population

Consider a random sample of size 30 from an exponential distribution with mean 3 . An exponential distribution is often used to model the survival times of a screening test. The 30 exponential values are generated with $B C 2.4$.
BC $2.4$
model;
The sample mean and standard deviation are $4.13$ and $3.739$, respectively.
Assume the sample is from a normal population with unknown mean and variance, with an improper prior density
$\xi(\mu, \tau)=1 / \tau, \mu \in R$ and $\sigma>0$, then the posterior predictive density is a univariatet with $n-1=29$ degrees of freedom, mean $\bar{x}=3.744$, standard deviation $3.872$, and precision $p=.0645$. This is verified from the original observations $x$ and the formula for the precision. From the predictive distribution, 30 observations are generated with BC $2.5$.
BC 2.5.
{
For $(t$ in $1: 30){$
Z[t] $\mathrm{t}(3.744, .0645,29)}$
$\mathrm{z}=(2.76213,3.46370,2.88747,3.13581,4.50398,5.09963,4.39670,3.24032,3.58791$,
$5.60893,3.76411,3.15034,4.15961,2.83306,3.64620,3.48478,2.24699,2.44810$,
$3.39590,3.56703,4.04226,4.00720,4.33006,3.44320,5.03451,2.07679,2.30578$,
$5.99297,3.88463,2.52737)$
which gives a mean of $\bar{z}=3.634$ and standard deviation $S=.975$.
The histograms obviously are different, where for the original observations, a right skewness is depicted; however, this is lacking for the histogram of the predicted observations, which is for a $t$-distribution. Although the example seems trivial, it would not be for the first time that exponential observations were analyzed as if they were generated from a normal population! Of course,

we have seen the relevance of the exponential distribution to Markov jump processes such as the Poisson process of Section 3.4.3, where the inter-arrival times are independent and identically distributed with a mean that is the reciprocal of the mean rate of event happenings.

It would be interesting to generate more replicate samples from the predictive distribution in order to see if these conclusions hold firm.

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

X(θ∣X)=[Γ(一个+b)Γ(n+1)/Γ(一个)Γ(b)Γ(X+1)Γ(n−X+1)] θ一个+X−1(1−θ)b+n−X−1

F(和/θ)=θ和(1−θ)1−和，以及预测质量函数从，称为 beta-二项式，是

G(和∣X)=Γ(一个+b)Γ(n+1)Γ(一个+∑一世=1一世=nX一世+和)Γ(1+n+b−X−和)÷ Γ(一个)Γ(b)Γ(n−X+1)Γ(X+1)Γ(n+1+一个+b)

ñ=(1,7,4,6,2 5,3,4,6,5 3,6,2,3,1 5,7,2,5,4 6,1,3,3,4)

G(和∣n11=1)=(米 和)Γ(一个+b)Γ(n+1)Γ(一个+和+n11)Γ(b+n−和−n11)/ Γ(一个)Γ(b)Γ(n11+1)Γ(n−n11+1)Γ(一个+b+n)

## 统计代写|贝叶斯分析代写Bayesian Analysis代考|Sampling from an Exponential, but Assuming a Normal Population

X(μ,τ)=1/τ,μ∈R和σ>0, 那么后验预测密度是一个单变量n−1=29自由度，平均值X¯=3.744, 标准差3.872, 和精度p=.0645. 这从原始观察中得到验证X以及精度公式。根据预测分布，使用 BC 生成 30 个观测值2.5.

{

5.60893,3.76411,3.15034,4.15961,2.83306,3.64620,3.48478,2.24699,2.44810,
3.39590,3.56703,4.04226,4.00720,4.33006,3.44320,5.03451,2.07679,2.30578,
5.99297,3.88463,2.52737)

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