Statistics 210A - 统计代写答疑辅导

标签： Statistics 210A

统计代写|统计推断作业代写statistical inference代考|Theories of Estimation

Posted on 2022年4月25日2022年4月25日 by statistics-lab

如果你也在怎样代写统计推断statistical inference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。

统计推断是利用数据分析来推断概率基础分布的属性的过程。推断性统计分析推断出人口的属性，例如通过测试假设和得出估计值。

statistics-lab™ 为您的留学生涯保驾护航在代写统计推断statistical inference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。我们的专家在代写统计推断statistical inference代写方面经验极为丰富，各种代写统计推断statistical inference相关的作业也就用不着说。

我们提供的统计推断statistical inference及其相关学科的代写，服务范围广, 其中包括但不限于:

Statistical Inference 统计推断
Statistical Computing 统计计算
Advanced Probability Theory 高等楖率论
Advanced Mathematical Statistics 高等数理统计学
(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

A Gentle Introduction to Bayesian Belief Networks — 统计代写|统计推断作业代写statistical inference代考|Theories of Estimation

统计代写|统计推断作业代写statistical inference代考|ELEMENTS OF POINT ESTIMATION

Essentially there are three stages of sophistication with regard to estimation of a parameter:

At the lowest level-a simple point estimate;
At a higher level-a point estimate along with some indication of the error of that estimate;
At the highest level-one conceives of estimating in terms of a “distribution” or probability of some sort of the potential values that can occur.

This entails the specification of some set of values presumably more restrictive than the entire set of values that the parameter can take on or relative plausibilities of those values or an interval or region.

Consider the I.Q. of University of Minnesota freshmen by taking a random sample of them. We could be satisfied with the sample average as reflective of that population. More insight, however, may be gained by considering the variability

of scores by estimating a variance. Finally one might decide that a highly likely interval for the entire average of freshmen would be more informative.

Sometimes a point estimate is about all you can do. Representing distances on a map, for example. At present there is really no way of reliably illustrating a standard error on a map-so a point estimate will suffice.

An “estimate” is a more or less reasonable guess at the true value of a magnitude or parameter or even a potential observation, and we are not necessarily interested in the consequences of estimation. We may only be concerned in what we should believe a true value to be rather than what action or what the consequences are of this belief. At this point we separate estimation theory from decision theory, though in many instances this is not the case.

统计代写|统计推断作业代写statistical inference代考|POINT ESTIMATION

An estimate might be considered “good” if it is in fact close to the true value on average or in the long run (pre-trial).
An estimate might be considered “good” if the data give good reason to believe the estimate will be close to the true value (post trial).
A system of estimation will be called an estimator
Choose estimators which on average or very often yield estimates which are close to the true value.
Choose an estimator for which the data give good reason to believe it will be close to the true value that is, a well-supported estimate (one that is suitable after the trials are made.)

With regard to the first type of estimators we do not reject one (theoretically) if it gives a poor result (differs greatly from the true value) in a particular case (though you would be foolish not to). We would only reject an estimation procedure if it gives bad results on average or in the long run. The merit of an estimator is judged, in general, by the distribution of estimates it gives rise to-the properties of its sampling distribution. One property sometimes stressed is unbiasedness. If

$T(D)$ is the estimator of $\theta$ then unbiasedness requires
$$
E[T(D)]=\theta .
$$
For example an unbiased estimator of a population variance $\sigma^{2}$ is
$$
(n-1)^{-1} \sum\left(x_{i}-\bar{x}\right)^{2}=s^{2}
$$
since $E\left(s^{2}\right)=\sigma^{2}$.
Suppose $Y_{1}, Y_{2}, \ldots$ are i.i.d. Bernoulli random variables $P\left(Y_{i}=1\right)=\theta$ and we sample until the first “one” comes up so that probability that the first one appears after $X=x$ zeroes is
$$
P(X=x \mid \theta)=\theta(1-\theta)^{x} \quad x=0,1, \ldots \quad 0<\theta<1 . $$ Seeking an unbiased estimator we have $$ \theta=E(T(Y))=\sum_{x=0}^{\infty} t(x) \theta(1-\theta)^{x}=t(0) \theta+t(1) \theta(1-\theta)+\cdots . $$ Equating the terms yields the unique solution $t(0)=1, t(x)=0$ for $x \geq 1$. This is flawed because this unique estimator always lies outside of the range of $\theta$. So unbiasedness alone can be a very poor guide. Prior to unbiasedness we should have consistency (which is an asymptotic type of unbiasedness, but considerably more). Another desideratum that many prefer is invariance of the estimation procedure. But if $E(X)=\theta$, then for $g(X)$ a smooth function of $X, E(g(X)) \neq g(\theta)$ unless $g(\cdot)$ is linear in $X$. Definitions of classical and Fisher consistency follow: Consistency: An estimator $T_{n}$ computed from a sample of size $n$ is said to be a consistent estimator of $\theta$ if for any arbitrary $\epsilon>0$ and $\delta>0$ there is some value, $N$, such that
$$
P\left[\left|T_{n}-\theta\right|<\epsilon\right]>1-\delta \quad \text { for all } n>N,
$$

统计代写|统计推断作业代写statistical inference代考|Fisher’s Definition of Consistency for i.i.d. Random Variables

“A function of the observed frequencies which takes on the exact parametric value when for those frequencies their expectations are substituted.”

For a discrete random variable with $P\left(X_{j}=x_{j} \mid \theta\right)=p_{j}(\theta)$ let $T_{n}$ be a function of the observed frequencies $n_{j}$ whose expectations are $E\left(n_{j}\right)=n p_{j}(\theta)$. Then the linear function of the frequencies $T_{n}=\frac{1}{n} \sum_{j} c_{j} n_{j}$ will assume the value
$$
\tau(\theta)=\Sigma c_{j} p_{j}(\theta)
$$
when $n p_{j}(\theta)$ is substituted for $n_{j}$ and thus $n^{-1} T_{n}$ is a consistent estimator of $\tau(\theta)$.
Another way of looking at this is:
Let $F_{n}(x)=\frac{1}{n} \times #$ of observations $\leq x$
$$
=\frac{i}{n} \text { for } x_{(i-1)}<x \leq x_{(i)}
$$
where $x_{(j)}$ is the $j$ th smallest observation. If $T_{n}=g\left(F_{n}(x)\right)$ and $g(F(x \mid \theta))=\tau(\theta)$ then $T_{n}$ is Fisher consistent for $\tau(\theta)$. Note if
$$
T_{n}=\int x d F_{n}(x)=\bar{x}{n} $$ and if $$ g(F)=\int x d F(x)=\mu $$ then $\bar{x}{n}$ is Fisher consistent for $\mu$.
On the other hand, if $T_{n}=\bar{x}{n}+\frac{1}{n}$, then this is not Fisher consistent but is consistent in the ordinary sense. Fisher Consistency is only defined for i.i.d. $X{1}, \ldots, X_{n}$.
However, as noted by Barnard (1974), “Fisher consistency can only with difficulty be invoked to justify specific procedures with finite samples” and also “fails because not all reasonable estimates are functions of relative frequencies.” He also presents an estimating procedure that does meet his requirements that the estimate lies within the parameter space and is invariant based on pivotal functions.

A Bayesian Network approach to diagnosing the root cause of failure from trouble tickets — 统计代写|统计推断作业代写statistical inference代考|Theories of Estimation

统计推断代考

统计代写|统计推断作业代写statistical inference代考|ELEMENTS OF POINT ESTIMATION

关于参数估计，本质上存在三个复杂阶段：

在最低层——简单的点估计；
在更高的层次上——一个点估计以及该估计错误的一些指示；
在最高级别，第一级设想根据可能发生的某种潜在值的“分布”或概率进行估计。

这需要指定一组值，可能比参数可以采用的整组值或这些值或区间或区域的相对合理性更具限制性。

通过随机抽取明尼苏达大学新生的样本来考虑他们的智商。我们可以对反映该总体的样本平均值感到满意。然而，通过考虑可变性可以获得更多的洞察力

通过估计方差来获得分数。最后，人们可能会决定，整个新生平均水平的一个极有可能的区间会提供更多信息。

有时，点估计就是你能做的一切。例如，在地图上表示距离。目前确实没有办法可靠地说明地图上的标准误差——所以点估计就足够了。

“估计”是对幅度或参数甚至潜在观察的真实值或多或少合理的猜测，我们不一定对估计的后果感兴趣。我们可能只关心我们应该相信一个真正的价值是什么，而不是这种信念的行动或后果是什么。在这一点上，我们将估计理论与决策理论分开，尽管在许多情况下并非如此。

统计代写|统计推断作业代写statistical inference代考|POINT ESTIMATION

如果实际上平均或长期（预审）接近真实值，则估计值可能被认为是“好”的。
如果数据有充分的理由相信估计值将接近真实值（试验后），则估计值可能被认为是“好的”。
估计系统将被称为估计器
选择平均或经常产生接近真实值的估计值的估计器。
选择一个数据有充分理由相信它会接近真实值的估计量，即一个有充分支持的估计值（在进行试验后适合的估计值。）

对于第一种类型的估计器，如果它在特定情况下给出的结果很差（与真实值相差很大），我们不会（理论上）拒绝它（尽管你不这样做是愚蠢的）。我们只会拒绝一个估计程序，如果它在平均或从长远来看给出了不好的结果。一般来说，估计量的优劣是通过它产生的估计分布来判断的——它的抽样分布的特性。有时强调的一个属性是不偏不倚。如果

吨(D)是的估计量θ那么公正需要
和[吨(D)]=θ.
例如总体方差的无偏估计σ2是
(n−1)−1∑(X一世−X¯)2=s2
自从和(s2)=σ2.
认为是1,是2,…是独立同分布的伯努利随机变量磷(是一世=1)=θ我们采样直到第一个“一个”出现，这样第一个出现在之后的概率X=X零是
磷(X=X∣θ)=θ(1−θ)XX=0,1,…0<θ<1.寻求一个无偏估计我们有θ=和(吨(是))=∑X=0∞吨(X)θ(1−θ)X=吨(0)θ+吨(1)θ(1−θ)+⋯.相等的项产生唯一的解决方案吨(0)=1,吨(X)=0为了X≥1. 这是有缺陷的，因为这个唯一的估计量总是在θ. 因此，仅凭公正可能是一个非常糟糕的指南。在无偏性之前，我们应该具有一致性（这是一种渐近类型的无偏性，但要多得多）。许多人更喜欢的另一个要求是估计过程的不变性。但如果和(X)=θ，那么对于G(X)的平滑函数X,和(G(X))≠G(θ)除非G(⋅)是线性的X. 经典一致性和 Fisher 一致性的定义如下：一致性：估计量吨n根据大小样本计算n据说是一致的估计量θ如果对于任何任意ε>0和d>0有一定的价值，ñ, 这样
磷[|吨n−θ|<ε]>1−d 对全部 n>ñ,

统计代写|统计推断作业代写statistical inference代考|Fisher’s Definition of Consistency for i.i.d. Random Variables

“观察到的频率的函数，当这些频率的期望被替换时，它具有精确的参数值。”

对于离散随机变量磷(Xj=Xj∣θ)=pj(θ)让吨n是观测频率的函数nj谁的期望是和(nj)=npj(θ). 然后是频率的线性函数吨n=1n∑jCjnj将假定值
τ(θ)=ΣCjpj(θ)
什么时候npj(θ)被取代nj因此n−1吨n是一致的估计量τ(θ).
另一种看待这个问题的方式是：
让F_{n}(x)=\frac{1}{n} \times #F_{n}(x)=\frac{1}{n} \times #观察≤X
=一世n 为了 X(一世−1)<X≤X(一世)
在哪里X(j)是个j最小的观察。如果吨n=G(Fn(X))和G(F(X∣θ))=τ(θ)然后吨n费雪是否一致τ(θ). 注意如果
吨n=∫XdFn(X)=X¯n而如果G(F)=∫XdF(X)=μ然后X¯n费雪是否一致μ.
另一方面，如果吨n=X¯n+1n，那么这不是Fisher一致，而是通常意义上的一致。Fisher 一致性仅针对 iid 定义X1,…,Xn.
然而，正如 Barnard (1974) 所指出的，“Fisher 一致性很难被用来证明具有有限样本的特定程序”并且“失败，因为并非所有合理的估计都是相对频率的函数”。他还提出了一个估计程序，该程序确实满足了他的要求，即估计值位于参数空间内并且基于关键函数是不变的。

统计代写|统计推断作业代写statistical inference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。

随机过程代考

在概率论概念中，随机过程是随机变量的集合。若一随机系统的样本点是随机函数，则称此函数为样本函数，这一随机系统全部样本函数的集合是一个随机过程。实际应用中，样本函数的一般定义在时间域或者空间域。 随机过程的实例如股票和汇率的波动、语音信号、视频信号、体温的变化，随机运动如布朗运动、随机徘徊等等。

贝叶斯方法代考

贝叶斯统计概念及数据分析表示使用概率陈述回答有关未知参数的研究问题以及统计范式。后验分布包括关于参数的先验分布，和基于观测数据提供关于参数的信息似然模型。根据选择的先验分布和似然模型，后验分布可以解析或近似，例如，马尔科夫链蒙特卡罗 (MCMC) 方法之一。贝叶斯统计概念及数据分析使用后验分布来形成模型参数的各种摘要，包括点估计，如后验平均值、中位数、百分位数和称为可信区间的区间估计。此外，所有关于模型参数的统计检验都可以表示为基于估计后验分布的概率报表。

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。

statistics-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。

机器学习代写

随着AI的大潮到来，Machine Learning逐渐成为一个新的学习热点。同时与传统CS相比，Machine Learning在其他领域也有着广泛的应用，因此这门学科成为不仅折磨CS专业同学的“小恶魔”，也是折磨生物、化学、统计等其他学科留学生的“大魔王”。学习Machine learning的一大绊脚石在于使用语言众多，跨学科范围广，所以学习起来尤其困难。但是不管你在学习Machine Learning时遇到任何难题，StudyGate专业导师团队都能为你轻松解决。

多元统计分析代考

基础数据: $N$ 个样本， $P$ 个变量数的单样本，组成的横列的数据表
变量定性: 分类和顺序；变量定量：数值
数学公式的角度分为: 因变量与自变量

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|统计推断作业代写statistical inference代考|Elements of Bayesianism

Posted on 2022年4月25日2022年4月25日 by statistics-lab

如果你也在怎样代写统计推断statistical inference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。

统计推断是利用数据分析来推断概率基础分布的属性的过程。推断性统计分析推断出人口的属性，例如通过测试假设和得出估计值。

我们提供的统计推断statistical inference及其相关学科的代写，服务范围广, 其中包括但不限于:

Statistical Inference 统计推断
Statistical Computing 统计计算
Advanced Probability Theory 高等楖率论
Advanced Mathematical Statistics 高等数理统计学
(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

Bayesian Belief Network in Artificial Intelligence - Javatpoint — 统计代写|统计推断作业代写statistical inference代考|Elements of Bayesianism

统计代写|统计推断作业代写statistical inference代考|Elements of Bayesianism

Bayesian inference involves placing a probability distribution on all unknown quantities in a statistical problem. So in addition to a probability model for the data, probability is also specified for any unknown parameters associated with it. If future observables are to be predicted, probability is posed for these as well. The Bayesian inference is thus the conditional distribution of unknown parameters (and/or future observables) given the data. This can be quite simple if everything being modeled is discrete, or quite complex if the class of models considered for the data is broad.
This chapter presents the fundamental elements of Bayesian testing for simple versus simple, composite versus composite and for point null versus composite alternative, hypotheses. Several applications are given. The use of Jeffreys “noninformative” priors for making general Bayesian inferences in binomial and negative binomial sampling, and methods for hypergeometric and negative hypergeometric sampling, are also discussed. This discussion leads to a presentation and proof of de Finetti’s theorem for Bernoulli trials. The chapter then gives a presentation of another de Finetti result that Bayesian assignment of probabilities is both necessary and sufficient for “coherence,” in a particular setting. The chapter concludes with a discussion and illustration of model selection.

统计代写|统计推断作业代写statistical inference代考|TESTING A COMPOSITE VS. A COMPOSITE

Suppose we can assume that the parameter or set of parameters $\theta \in \Theta$ is assigned a prior distribution (subjective or objective) that is, we may be sampling $\theta$ from a hypothetical population specifying a probability function $g(\theta)$, or our beliefs about $\theta$ can be summarized by a $g(\theta)$ or we may assume a $g(\theta)$ that purports to reflect our prior ignorance. We are interested in deciding whether
$$
H_{0}: \theta \in \Theta_{0} \quad \text { or } \quad H_{1}: \theta \in \Theta_{1} \quad \text { for } \quad \Theta_{0} \cap \Theta_{1}=\emptyset
$$

Now the posterior probability function is
$$
p(\theta \mid D) \propto L(\theta \mid D) g(\theta) \quad \text { or } \quad p(\theta \mid D)=\frac{L(\theta \mid D) g(\theta)}{\int_{\Theta} L(\theta \mid D) d G(\theta)}
$$
using the Lebesgue-Stieltjes integral representation in the denominator above.
Usually $\Theta_{0} \cup \Theta_{1}=\Theta$ but this is not necessary. We calculate
$$
\begin{aligned}
&P\left(\theta \in \Theta_{0} \mid D\right)=\int_{\Theta_{0}} d P(\theta \mid D) \
&P\left(\theta \in \Theta_{1} \mid D\right)=\int_{\Theta_{1}} d P(\theta \mid D)
\end{aligned}
$$
and calculate the posterior odds
$$
\frac{P\left(\theta \in \Theta_{0} \mid D\right)}{P\left(\theta \in \Theta_{1} \mid D\right)}
$$
and if this is greater than some predetermined value $k$ choose $H_{0}$, and if less choose $H_{1}$, and if equal to $k$ be indifferent.

统计代写|统计推断作业代写statistical inference代考|SOME REMARKS ON PRIORS FOR THE BINOMIAL

Personal Priors. If one can elicit a personal subjective prior for $\theta$ then his posterior for $\theta$ is personal as well and depending on the reasoning that went into it may or may not convince anyone else about the posterior on $\theta$. A convenient prior that is often used when subjective opinion can be molded into this prior is the beta prior
$$
g(\theta \mid a, b) \propto \theta^{a-1}(1-\theta)^{b-1}
$$
when this is combined with the likelihood to yield
$$
g(\theta \mid a, b, r) \propto \theta^{a+r-1}(1-\theta)^{b+n-r-1}
$$
So-Called Ignorance or Informationless or Reference Priors. It appears that in absence of information regarding $\theta$, it was interpreted by Laplace that Bayes used a uniform prior in his “Scholium”. An objection raised by Fisher to this is essentially on the grounds of a lack of invariance. He argued that setting a parameter $\theta$ to be uniform resulted in, say $\tau=\theta^{3}$ (or $\tau=\tau(\theta))$ and then why not set $\tau$ to be uniform so that $g(\tau)=1,0<\tau<1$ then implies that $g(\theta)=3 \theta^{2}$ instead $g(\theta)=1$. Hence one will get different answers depending on what function of the parameter is assumed uniform.
Jeffreys countered this lack of invariance with the following:
The Fisher Information quantity of a probability function $f(x \mid \theta)$ is
$$
I(\theta)=E\left(\frac{d \log f}{d \theta}\right)^{2}
$$
assuming it exists. Then set
$$
g(\theta)=I^{\frac{1}{2}}(\theta)
$$
Now suppose $\tau=\tau(\theta)$. Then
$$
\begin{aligned}
I(\tau) &=E\left(\frac{d \log f}{d \tau}\right)^{2}=E\left(\frac{d \log f}{d \theta} \times \frac{d \theta}{d \tau}\right)^{2} \
&=E\left(\frac{d \log f}{d \theta}\right)^{2} \times\left(\frac{d \theta}{d \tau}\right)^{2}
\end{aligned}
$$

Bayesian Statistics Explained in Simple English For Beginners — 统计代写|统计推断作业代写statistical inference代考|Elements of Bayesianism

统计推断代考

统计代写|统计推断作业代写statistical inference代考|Elements of Bayesianism

贝叶斯推理涉及在统计问题中对所有未知量进行概率分布。因此，除了数据的概率模型之外，还为与之关联的任何未知参数指定了概率。如果要预测未来的可观察到的，那么也为这些提出了概率。因此，贝叶斯推理是给定数据的未知参数（和/或未来可观察值）的条件分布。如果要建模的所有内容都是离散的，这可能非常简单，如果为数据考虑的模型类别很广泛，这可能会非常复杂。
本章介绍了简单与简单、复合与复合以及零点与复合替代假设的贝叶斯检验的基本要素。给出了几个应用程序。还讨论了使用 Jeffreys “非信息性”先验在二项式和负二项式采样中进行一般贝叶斯推断，以及超几何和负超几何采样的方法。这个讨论导致了对伯努利试验的 de Finetti 定理的介绍和证明。然后，本章介绍了另一个 de Finetti 结果，即在特定环境中，概率的贝叶斯分配对于“连贯性”是必要和充分的。本章最后对模型选择进行了讨论和说明。

统计代写|统计推断作业代写statistical inference代考|TESTING A COMPOSITE VS. A COMPOSITE

假设我们可以假设参数或参数集θ∈θ被分配了一个先验分布（主观或客观），也就是说，我们可能正在抽样θ来自指定概率函数的假设总体G(θ)，或者我们对θ可以概括为G(θ)或者我们可以假设G(θ)这旨在反映我们之前的无知。我们有兴趣决定是否
H0:θ∈θ0 或者 H1:θ∈θ1 为了 θ0∩θ1=∅

现在后验概率函数是
p(θ∣D)∝大号(θ∣D)G(θ) 或者 p(θ∣D)=大号(θ∣D)G(θ)∫θ大号(θ∣D)dG(θ)
在上面的分母中使用 Lebesgue-Stieltjes 积分表示。
通常θ0∪θ1=θ但这不是必需的。我们计算
磷(θ∈θ0∣D)=∫θ0d磷(θ∣D) 磷(θ∈θ1∣D)=∫θ1d磷(θ∣D)
并计算后验赔率
磷(θ∈θ0∣D)磷(θ∈θ1∣D)
如果这大于某个预定值ķ选择H0, 如果少选H1, 如果等于ķ无所谓。

统计代写|统计推断作业代写statistical inference代考|SOME REMARKS ON PRIORS FOR THE BINOMIAL

个人先验。如果可以引出个人主观先验θ然后他的后部为θ也是个人的，并且取决于进入它的推理可能会或可能不会说服其他人关于后验θ. 当主观意见可以被塑造成这个先验时，经常使用的一个方便的先验是 beta 先验
G(θ∣一种,b)∝θ一种−1(1−θ)b−1
当这与产生的可能性相结合时
G(θ∣一种,b,r)∝θ一种+r−1(1−θ)b+n−r−1
所谓的无知或无信息或参考先验。似乎在缺乏有关信息的情况下θ，拉普拉斯解释说，贝叶斯在他的“Scholium”中使用了统一先验。费舍尔对此提出的反对意见主要是基于缺乏不变性。他认为设置参数θ统一导致，说τ=θ3（或者τ=τ(θ))然后为什么不设置τ是统一的，这样G(τ)=1,0<τ<1那么意味着G(θ)=3θ2反而G(θ)=1. 因此，根据假设参数的函数是一致的，人们将得到不同的答案。
Jeffreys 用以下方法反驳了这种缺乏不变性：
概率函数的 Fisher 信息量F(X∣θ)是
一世(θ)=和(d日志⁡Fdθ)2
假设它存在。然后设置
G(θ)=一世12(θ)
现在假设τ=τ(θ). 然后
一世(τ)=和(d日志⁡Fdτ)2=和(d日志⁡Fdθ×dθdτ)2 =和(d日志⁡Fdθ)2×(dθdτ)2

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

随机过程代考

贝叶斯方法代考

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。

机器学习代写

多元统计分析代考

基础数据: $N$ 个样本， $P$ 个变量数的单样本，组成的横列的数据表
变量定性: 分类和顺序；变量定量：数值
数学公式的角度分为: 因变量与自变量

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|统计推断作业代写statistical inference代考|REMARKS ON N-P THEORY

Posted on 2022年4月25日2022年4月25日 by statistics-lab

如果你也在怎样代写统计推断statistical inference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。

统计推断是利用数据分析来推断概率基础分布的属性的过程。推断性统计分析推断出人口的属性，例如通过测试假设和得出估计值。

我们提供的统计推断statistical inference及其相关学科的代写，服务范围广, 其中包括但不限于:

Statistical Inference 统计推断
Statistical Computing 统计计算
Advanced Probability Theory 高等楖率论
Advanced Mathematical Statistics 高等数理统计学
(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

Sequential Nonparametric Testing with the Law of the Iterated Logarithm — 统计代写|统计推断作业代写statistical inference代考|REMARKS ON N-P THEORY

统计代写|统计推断作业代写statistical inference代考|REMARKS ON N-P THEORY

A synopsis of criticisms of N-P theory due to Hacking (1965) follows. The N-P hypothesis testing theory is then one of fixing a small size and searching for large power. The rationale behind it is to search for rules for governing our behavior with regard to hypotheses (without hoping to know whether any one of them is true or false) which will ensure that in the long run we shall not be wrong too often. To assert whether $H$ be rejected or not, calculate $D$ (the observables) and if $D \in s$ reject $H$, if $D \notin s$ accept $H$. Such a rule tells us nothing as to whether in a particular case $H$ is true when $D \notin s$ or false when $D \notin s$. If we behave in such a way we shall reject when it is true not more than $100 \alpha \%$ of the time and in addition we may have evidence that we shall reject $H$ sufficiently often when it is false.

Presumably if we behave in such a way and keep $\alpha$ fixed we shall reject hypotheses tested through our lifetimes that are true not more than $100 \alpha \%$ of the time (i.e., there is a very high likelihood that this will happen in the long run), and really one is no more or less certain about any of these hypotheses. But if we had to adopt a testing policy now and were bound to follow it for the rest of our lives so that for every false hypothesis we rejected we would have bestowed upon us $h$ heavenly units and likewise for every true hypothesis we accept, while we lose the same for each true one we reject and every false one we accept then this is the best life long policy-but no one has ever been in this situation. This is of course a pre-trial, not a post trial evaluation.

Now it may be that before a trial a decision must be made (because accuracy is difficult or tedious or it may be economical to do so) only to note whether a trial made is in $s$ or not. This may be wholly rational and in this case N-P theory is an economical post trial evaluation. Of course in some cases, say, a simple $H_{0}$ vs. a simple $H_{1}$, where a MP test exists, the extra knowledge will not change our evaluations of $H_{0}$. In these cases where a rational decision to discard or ignore data the N-P theory is a special case of likelihood.
In any event the N-P theory can be viewed from a likelihood perspective:
$$
\begin{aligned}
P(D \in s \mid H) &=L(H \mid D \in s) \leq \alpha(H) \quad \text { say small relative to } \
P(D \in s \mid K) &=L(K \mid D \in s)=1-\beta(K) \text { so } \
Q &=\frac{L(H \mid D \in s)}{L(K \mid D \in s)} \leq \frac{\alpha(H)}{1-\beta(K)} \quad \text { small. }
\end{aligned}
$$

统计代写|统计推断作业代写statistical inference代考|FURTHER REMARKS ON N-P THEORY

We now reconsider an example presented earlier.
Example $3.2$ (continued)
Suppose we make independent trials of Binary variables with probability $p$ then if we hold $n=$ number trials fixed and got $r$ successes then
$$
P(R=r \mid n)=\left(\begin{array}{l}
n \
r
\end{array}\right) p^{r}(1-p)^{n-r}
$$

For $n=5 H_{0}: p=\frac{1}{2} H_{1}: p=p_{1}<\frac{1}{2}$ and $\alpha=\frac{1}{16}$ a UMP N-P test is
$$
\text { so } \alpha=\left(\frac{5}{0}\right)\left(\frac{1}{2}\right)^{5}+\frac{1}{5}\left(\frac{5}{1}\right)\left(\frac{1}{2}\right)^{5}=\frac{1}{16} \text {. }
$$
Now suppose the trial was conducted until we got $r$ heads which took $n$ trials (random). Then
$$
P(n \mid r)=\left(\begin{array}{c}
n-1 \
r-1
\end{array}\right) p^{r}(1-p)^{n-r} \quad n=r, r+1, \ldots
$$
Note $L_{B}(p \mid r)=L_{N B}(p \mid n)=p^{r}(1-p)^{n-r}$, that is, the binomial and negative binomial likelihoods are the same.
Now suppose for $\alpha=1 / 16$ and $r=1$, so
$$
P\left(N=n \mid r=1, p=\frac{1}{2}\right)=\left(\begin{array}{l}
1 \
2
\end{array}\right)^{n}
$$
such that
$$
P(N \leq 4)=\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{10}=\frac{15}{16}
$$
or $P(N \geq 5)=1 / 16$, hence
$$
T(n)=\left{\begin{array}{lll}
1 & \text { if } & n \geq 5 \
0 & \text { if } \quad n \leq 4
\end{array}\right.
$$
Now if the data in both experiments were $r=1, n=5$ then for the Binomial trials we would reject with probability $\frac{1}{5}$ and with the negative binomial trials we would reject with probability 1 . So as previously noted the likelihood principle and likelihood tests are contradicted although in both cases
$$
L=p(1-p)^{4}
$$

统计代写|统计推断作业代写statistical inference代考|LAW OF THE ITERATED LOGARITHM

Let $X_{1}, X_{2}, \ldots$ be i.i.d. random variables with $E\left(X_{i}\right)=\mu \operatorname{Var}\left(X_{i}\right)=\sigma^{2}$ and $E|X|^{2+\delta}<\infty$ for some $\delta>0$. Then with probability 1 the inequality
$$
\sum_{i=1}^{n} X_{i}2$. Similarly
$$
\Sigma X_{i}>n \mu+\left(n \sigma^{2} \lambda \log \log n\right)^{\frac{1}{2}} \text { or } \frac{\Sigma X_{i}-n \mu}{\sigma \sqrt{n}}>(\lambda \log \log n)^{\frac{1}{2}}
$$
is satisfied for infinitely many $n$ if $\lambda<2$, but only for finitely many $n$ if $\lambda>2$. Further (almost surely)
$$
\limsup {n \rightarrow \infty} \frac{\left(\Sigma X{i}-n \mu\right) / \sigma \sqrt{n}}{(2 \log \log n)^{\leftarrow}}=1
$$
We apply the LIL to $n$ i.i.d. binary variables where $P\left(X_{i}=0\right)=P\left(X_{i}=1\right)=\frac{1}{2}$ where $E\left(X_{i}\right)=\frac{1}{2}, \operatorname{var}\left(X_{i}\right)=\frac{1}{4}$. Then for $\lambda=1$, consider the event
$$
r<\frac{n}{2}-\left(\frac{n}{4} \log \log n\right)^{\frac{1}{2}} . $$ Sooner or later if we continue sampling then for sufficiently large $n$ $$ (\log \log n)^{\frac{1}{2}}>z_{\alpha}
$$
since $z_{\alpha}$ is a constant. Then
$$
-\left(\frac{n}{4} \log \log n\right)^{\frac{1}{2}}<-\frac{\sqrt{n}}{2} z_{\alpha}
$$
and
$$
\frac{n}{2}-\left(\frac{n}{4} \log \log n\right)^{\frac{1}{2}}<\frac{n}{2}-\frac{\sqrt{n}}{2} z_{\alpha}
$$
Therefore, with probability 1 the inequality
$$
r<\frac{n}{2}-\frac{\sqrt{n}}{2} z_{\alpha}
$$

Large-Scale Simulation and Proof for Khinchin's Law of the Iterated Logarithm — 统计代写|统计推断作业代写statistical inference代考|REMARKS ON N-P THEORY

统计推断代考

统计代写|统计推断作业代写statistical inference代考|REMARKS ON N-P THEORY

Hacking (1965) 对 NP 理论的批评概要如下。NP假设检验理论是固定小尺寸并寻找大功率的理论之一。其背后的基本原理是寻找规则来管理我们关于假设的行为（不希望知道其中任何一个是真还是假），这将确保从长远来看我们不会经常犯错。断言是否H被拒绝与否，计算D（可观察的）如果D∈s拒绝H，如果D∉s接受H. 这样的规则并没有告诉我们在特定情况下是否H是真的D∉s或 false 时D∉s. 如果我们以这样的方式行事，我们将在其为真时拒绝不超过100一种%的时间，此外，我们可能有证据表明我们将拒绝H当它是错误的时，足够频繁。

大概如果我们以这样的方式行事并保持一种固定我们将拒绝通过我们的一生检验的假设，这些假设不超过100一种%时间（即，从长远来看，这种情况很有可能发生），实际上，人们对这些假设中的任何一个都没有或多或少的把握。但是，如果我们现在必须采用一项测试政策，并且注定要在我们的余生中遵循它，那么对于我们拒绝的每一个错误假设，我们都会赋予我们H天上的单位，同样对于我们接受的每一个真实假设，虽然我们对每一个我们拒绝的真实假设和每一个我们接受的虚假假设都失去了同样的假设，但这是最好的终身政策——但没有人遇到过这种情况。这当然是预审，而不是审后评估。

现在可能是在审判之前必须做出决定（因为准确性是困难或乏味的，或者这样做可能是经济的）只是要注意所进行的审判是否在s或不。这可能是完全合理的，在这种情况下，NP 理论是一种经济的试验后评估。当然在某些情况下，比如说，一个简单的H0与简单的H1，在存在 MP 测试的情况下，额外的知识不会改变我们对H0. 在这些合理决定丢弃或忽略数据的情况下，NP 理论是可能性的一个特例。
无论如何，可以从可能性的角度来看待 NP 理论：
磷(D∈s∣H)=大号(H∣D∈s)≤一种(H) 说相对于小磷(D∈s∣ķ)=大号(ķ∣D∈s)=1−b(ķ) 所以问=大号(H∣D∈s)大号(ķ∣D∈s)≤一种(H)1−b(ķ) 小的。

统计代写|统计推断作业代写statistical inference代考|FURTHER REMARKS ON N-P THEORY

我们现在重新考虑之前提出的一个例子。
例子3.2（续）
假设我们以概率对二元变量进行独立试验p那么如果我们持有n=数字试验固定并得到r然后成功
磷(R=r∣n)=(n r)pr(1−p)n−r

为了n=5H0:p=12H1:p=p1<12和一种=116UMP NP 测试是
所以一种=(50)(12)5+15(51)(12)5=116.
现在假设试验一直进行到我们得到r头颅n试验（随机）。然后
磷(n∣r)=(n−1 r−1)pr(1−p)n−rn=r,r+1,…
笔记大号乙(p∣r)=大号ñ乙(p∣n)=pr(1−p)n−r，即二项式和负二项式似然相同。
现在假设一种=1/16和r=1，所以
磷(ñ=n∣r=1,p=12)=(1 2)n
这样
磷(ñ≤4)=12+14+18+110=1516
或者磷(ñ≥5)=1/16，因此
$$
T(n)=\left{1 如果 n≥5 0 如果 n≤4\对。
ñ这在一世F吨H和d一种吨一种一世nb这吨H和Xp和r一世米和n吨s在和r和$r=1,n=5$吨H和nF这r吨H和乙一世n这米一世一种l吨r一世一种ls在和在这在ldr和j和C吨在一世吨Hpr这b一种b一世l一世吨是$15$一种nd在一世吨H吨H和n和G一种吨一世在和b一世n这米一世一种l吨r一世一种ls在和在这在ldr和j和C吨在一世吨Hpr这b一种b一世l一世吨是1.小号这一种spr和在一世这在sl是n这吨和d吨H和l一世ķ和l一世H这这dpr一世nC一世pl和一种ndl一世ķ和l一世H这这d吨和s吨s一种r和C这n吨r一种d一世C吨和d一种l吨H这在GH一世nb这吨HC一种s和s
L=p(1-p)^{4}
$$

统计代写|统计推断作业代写statistical inference代考|LAW OF THE ITERATED LOGARITHM

让X1,X2,…是独立同分布的随机变量和(X一世)=μ曾是⁡(X一世)=σ2和和|X|2+d<∞对于一些d>0. 然后以概率 1 不等式
∑一世=1nX一世2$.小号一世米一世l一种rl是
\Sigma X_{i}>n \mu+\left(n \sigma^{2} \lambda \log \log n\right)^{\frac{1}{2}} \text { 或 } \frac{\西格玛 X_{i}-n \mu}{\sigma \sqrt{n}}>(\lambda \log \log n)^{\frac{1}{2}}
一世ss一种吨一世sF一世和dF这r一世nF一世n一世吨和l是米一种n是$n$一世F$λ<2$,b在吨这nl是F这rF一世n一世吨和l是米一种n是$n$一世F$λ>2$.F在r吨H和r(一种l米这s吨s在r和l是)
\limsup {n \rightarrow \infty} \frac{\left(\Sigma X{i}-n \mu\right) / \sigma \sqrt{n}}{(2 \log \log n)^{\leftarrow }}=1
在和一种ppl是吨H和大号一世大号吨这$n$一世.一世.d.b一世n一种r是在一种r一世一种bl和s在H和r和$磷(X一世=0)=磷(X一世=1)=12$在H和r和$和(X一世)=12,曾是⁡(X一世)=14$.吨H和nF这r$λ=1$,C这ns一世d和r吨H和和在和n吨
r<\frac{n}{2}-\left(\frac{n}{4} \log \log n\right)^{\frac{1}{2}} 。小号这这n和r这rl一种吨和r一世F在和C这n吨一世n在和s一种米pl一世nG吨H和nF这rs在FF一世C一世和n吨l是l一种rG和$n$(\log \log n)^{\frac{1}{2}}>z_{\alpha}
s一世nC和$和一种$一世s一种C这ns吨一种n吨.吨H和n
-\left(\frac{n}{4} \log \log n\right)^{\frac{1}{2}}<-\frac{\sqrt{n}}{2} z_{\alpha}
一种nd
\frac{n}{2}-\left(\frac{n}{4} \log \log n\right)^{\frac{1}{2}}<\frac{n}{2}-\ frac{\sqrt{n}}{2} z_{\alpha}
吨H和r和F这r和,在一世吨Hpr这b一种b一世l一世吨是1吨H和一世n和q在一种l一世吨是
r<\frac{n}{2}-\frac{\sqrt{n}}{2} z_{\alpha}
$$

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

随机过程代考

贝叶斯方法代考

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。

机器学习代写

多元统计分析代考

基础数据: $N$ 个样本， $P$ 个变量数的单样本，组成的横列的数据表
变量定性: 分类和顺序；变量定量：数值
数学公式的角度分为: 因变量与自变量

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|统计推断作业代写statistical inference代考| INVARIANT TESTS

Posted on 2022年4月25日2022年4月25日 by statistics-lab

如果你也在怎样代写统计推断statistical inference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。

统计推断是利用数据分析来推断概率基础分布的属性的过程。推断性统计分析推断出人口的属性，例如通过测试假设和得出估计值。

我们提供的统计推断statistical inference及其相关学科的代写，服务范围广, 其中包括但不限于:

Statistical Inference 统计推断
Statistical Computing 统计计算
Advanced Probability Theory 高等楖率论
Advanced Mathematical Statistics 高等数理统计学
(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

Know-How about Pre-Construction Soil Testing — 统计代写|统计推断作业代写statistical inference代考| INVARIANT TESTS

统计代写|统计推断作业代写statistical inference代考|INVARIANT TESTS

More generally suppose we have $k$ normal populations with assumed $N\left(\mu_{i}, \sigma^{2}\right) i=$ $1, \ldots, k$ and random samples of size $n_{i}$ from each of the populations. Recall for testing $H_{0}: \mu_{1}=\mu_{2}=\cdots=\mu_{k}$ vs. $H_{1}$ : not all the $\mu_{i}$ ‘s are equal and $\sigma^{2}$ unspecified, that the F-test used to test this was
$$
\begin{aligned}
F_{k-1, n-k} &=\frac{\sum n_{i}\left(\bar{X}{i}-\bar{X}\right)^{2} /(k-1)}{s^{2}}, \quad n{i} \bar{x}{i}=\sum{j=1}^{n_{i}} X_{i j}, \quad n \bar{x}=\sum_{1}^{k} n_{i} \bar{X}{i} \ n &=\sum{1}^{k} n_{i}
\end{aligned}
$$
and
$$
(n-k) s^{2}=\sum_{i=1}^{k} \sum_{j=1}^{n_{j}}\left(X_{i j}-\bar{X}{i}\right)^{2} $$ If $Y{i j}=a X_{i j}+b$ for $a \neq 0$, then
$$
\frac{\sum n_{i}\left(\bar{Y}{i}-\bar{Y}\right)^{2} /(k-1)}{s{y}^{2}}=F_{k-1, n-k}
$$
as previously defined. The statistic is invariant under linear transformations. Note also that $H_{0}: \mu_{1}=\cdots=\mu_{k}$ is equivalent to $H_{0}^{\prime}: a \mu_{i}+b=\cdots=a \mu_{k}+b$ and $H_{1}$ : $\Longrightarrow$ to $H_{1}^{\prime}$. So it is an invariant test and it turns out to be UMP among invariant tests so we say it is UMPI. However this test is not UMPU.

In general, for the problem of
$$
H: \theta \in \Theta_{H} \quad \text { vs. } \quad K: \theta \in \Theta_{K}
$$
for any transformation $g$ on $D$ which leaves the problem invariant, it is natural to restrict our attention to all tests $T(D)$ such that $T(g D)=T(D)$ for all $D \in S$. A transformation $g$ is one which essentially changes the coordinates and a test is invariant if it is independent of the particular coordinate system in which the data are expressed.
We define invariance more precisely:
Definition: If for each $D \in S$ the function $t(D)=t(g D)$ for all $g \in G, G$ a group of transformations then $t$ is invariant with respect to $G$.
Recall that a set of elements $G$ is a group if under some operation it is
(a) closed: for all $g_{1}$ and $g_{2} \in G, g_{1} g_{2} \in G$;
(b) associative: $\left(g_{1} g_{2}\right) g_{3}=g_{1}\left(g_{2} g_{3}\right)$ for all $g_{1}, g_{2}, g_{3} \in G$;
(c) has an identity element: $g_{I} g=g g_{I}=g$ where $g_{I} \in G$;
(d) has an inverse: that is, if $g \in G$ then $g^{-1} \in G$ where $g g^{-1}=g^{-1} g=g_{I}$.

统计代写|统计推断作业代写statistical inference代考|LOCALLY BEST TESTS

When there are no UMP tests we may sometimes restrict the alternative parameter values to cases of presumably critical interest and look for high power against these alternatives. In particular if interest is focused on alternatives close to $H_{0}$ : say $\theta=\theta_{0}$, we could define $\delta(\theta)$ as a measure of the discrepancy of a close alternative from $H_{0}$.
Definition: A level $\alpha$ test $T$ is defined as locally most powerful (LMP) if for every other test $T^{}$ there exists a $\Delta$ such that $$ 1-\beta_{T}(\theta) \geq 1-\beta_{T^{}}(\theta) \text { for all } \theta \text { such that } 0<\delta(\theta)<\Delta .
$$

Theorem $5.8$ If among all unbiased level $\alpha$ tests $T^{*}, T$ is LMP then we say $T$ is LMPU.

For real $\theta$ for any $T^{}$ such that $1-\beta_{T^{}}(\theta)$ is continuously differentiable at $\theta=\theta_{0}$ with $H_{1}: \theta>\theta_{0}$ or $H_{1}: \theta<\theta_{0}$ (i.e., one sided tests) a LMP test exists and is defined such that for all level $\alpha$ tests $T^{}$ there is a unique $T$ such that $$ \arg \max {T^{}}\left[\frac{d\left(1-\beta{T^{*}}(\theta)\right)}{d \theta}\right]{\theta=\theta{0}}=T .
$$
For $\theta$ real valued and $1-\beta_{T^{}}(\theta)$ twice continuously differentiable at $\theta=\theta_{0}$ for all $T^{}$, then a LMPU level $\alpha$ test $T$ exists of $H_{0}: \theta=\theta_{0}$ vs. $H_{1}: \theta \neq \theta_{0}$ and is given by
$$
\arg \max {T^{}}\left[\frac{d^{2}\left(1-\beta{T^{2}}(\theta)\right)}{d \theta^{2}}\right]{\theta=\theta{1}}=T \Longrightarrow 1-\beta_{T}(\theta) \geq 1-\beta_{T^{}}(\theta)
$$
for $0<\delta(\theta)<\Delta$. All locally unbiased tests result in $$ \left.\frac{d\left[1-\beta_{T^{}}(\theta)\right]}{d \theta}\right|{\theta=\theta{1}}=0 $$ given the condition of being continuously differentiable. Proof of (I): For a test $T^{}$
$$
\begin{aligned}
\gamma_{T^{}}(\theta) & \equiv 1-\beta_{T^{}}(\theta)=\int T^{} f(D \mid \theta) d \mu, \ \gamma_{T^{}}^{\prime}(\theta) &=\int T^{} \frac{\partial f_{\theta}}{\partial \theta} d \mu . \end{aligned} $$ Suppose $\gamma_{T}^{\prime}\left(\theta_{0}\right) \geq \gamma_{T^{2}}\left(\theta_{0}\right)$. Then note that $$ \begin{aligned} &\gamma_{T}\left(\theta_{0}\right)=1-\beta_{T}\left(\theta_{0}\right)=1-\beta_{T}\left(\theta_{0}\right)=\gamma_{T^{}}\left(\theta_{0}\right) \
&\gamma_{T}^{\prime}\left(\theta_{0}\right)=\lim {\Delta \theta \rightarrow 0} \frac{\gamma{T}\left(\theta_{0}+\Delta \theta\right)-\gamma_{T}\left(\theta_{0}\right)}{\Delta \theta} \geq \lim {\Delta \theta \rightarrow 0} \frac{\gamma{T^{}}\left(\theta_{0}+\Delta \theta\right)-\gamma_{T^{}}\left(\theta_{0}\right)}{\Delta \theta}=\gamma_{T^{\prime}}^{\prime}\left(\theta_{0}\right) \
&\gamma_{T}^{\prime}\left(\theta_{0}\right)-\gamma_{T^{}}^{\prime}\left(\theta_{0}\right)=\lim {\Delta \theta \rightarrow 0} \frac{\gamma{T}\left(\theta_{0}+\Delta \theta\right)-\gamma_{T^{}}\left(\theta_{0}+\Delta \theta\right)}{\Delta \theta} \geq 0
\end{aligned}
$$

统计代写|统计推断作业代写statistical inference代考|TEST CONSTRUCTION

So far N-P theory has not really given a principle for constructing a test. It has indicated how we should compare tests that is, for a given size the test with the larger power is superior, or for sample space $S$, we want $T(D)$ to be such that for all tests $T^{}(D)$ $$ E\left(T^{} \mid H\right) \leq \alpha
$$
choose $T(D)$ such that
$$
E\left(T^{*} \mid K\right) \leq E(T \mid K)
$$
and as this doesn’t always happen we go on to other criteria, unbiasedness, invariance, and so forth.

There are several test construction methods that are not dependent on the N-P approach but are often evaluated by the properties inherent in that approach. The most popular one is the Likelihood Ratio Test (LRT) criterion.

Specifically the Likelihood Ratio Test (LRT) criterion statistic for a set of parameters $\theta$ to test $H_{\theta}$ vs. $K_{\theta}$ is defined as
$$
\frac{\sup {\theta \in H{\theta}} L(\theta \mid D)}{\sup {\theta \in\left(H{\theta} \cup K_{\theta}\right)} L(\theta \mid D)}=\lambda(D) \leq 1
$$
with critical region defined as $\lambda<k_{\alpha}$ reject $H_{\theta}$. This yields
$$
P\left{\lambda(D)<k_{\alpha}\right} \leq \alpha
$$

PDF) Usability Dimensions and Behavioral Intention to Use Markdown-to-Moodle in Test Construction | Julius Garcia - Academia.edu — 统计代写|统计推断作业代写statistical inference代考| INVARIANT TESTS

统计推断代考

统计代写|统计推断作业代写statistical inference代考|INVARIANT TESTS

更一般地假设我们有ķ假设正常人群ñ(μ一世,σ2)一世= 1,…,ķ和大小的随机样本n一世来自每个人群。召回测试H0:μ1=μ2=⋯=μķ对比H1: 不是全部μ一世是相等的并且σ2未指定，用于测试的 F 检验是
Fķ−1,n−ķ=∑n一世(X¯一世−X¯)2/(ķ−1)s2,n一世X¯一世=∑j=1n一世X一世j,nX¯=∑1ķn一世X¯一世 n=∑1ķn一世
和
(n−ķ)s2=∑一世=1ķ∑j=1nj(X一世j−X¯一世)2如果是一世j=一种X一世j+b为了一种≠0，然后
∑n一世(是¯一世−是¯)2/(ķ−1)s是2=Fķ−1,n−ķ
如前所述。统计量在线性变换下是不变的。另请注意H0:μ1=⋯=μķ相当于H0′:一种μ一世+b=⋯=一种μķ+b和H1 : ⟹到H1′. 所以它是一个不变的测试，结果证明它是不变测试中的UMP，所以我们说它是UMPI。但是，此测试不是 UMPU。

一般来说，对于问题
H:θ∈θH 对比 ķ:θ∈θķ
对于任何转换G在D这使问题保持不变，很自然地将我们的注意力限制在所有测试上吨(D)这样吨(GD)=吨(D)对全部D∈小号. 转变G是一个本质上改变坐标的测试，如果它独立于表达数据的特定坐标系，则测试是不变的。
我们更精确地定义不变性：
定义：如果对于每个D∈小号功能吨(D)=吨(GD)对全部G∈G,G一组转换然后吨是不变的G.
回想一下一组元素G是一个组，如果在某些操作下它是
(a) 封闭的：对于所有G1和G2∈G,G1G2∈G;
(b) 联想：(G1G2)G3=G1(G2G3)对全部G1,G2,G3∈G;
(c) 具有标识元素：G一世G=GG一世=G在哪里G一世∈G;
(d) 有一个逆：即，如果G∈G然后G−1∈G在哪里GG−1=G−1G=G一世.

统计代写|统计推断作业代写statistical inference代考|LOCALLY BEST TESTS

当没有 UMP 测试时，我们有时可能会将替代参数值限制在可能具有关键兴趣的情况下，并针对这些替代方案寻找高功率。特别是如果兴趣集中在接近于H0：说θ=θ0，我们可以定义d(θ)衡量一个接近的替代方案的差异H0.
定义：A级一种测试吨如果对于其他所有测试，则定义为本地最强大 (LMP)吨存在一个Δ这样1−b吨(θ)≥1−b吨(θ) 对全部 θ 这样 0<d(θ)<Δ.

定理5.8如果在所有无偏水平中一种测试吨∗,吨是 LMP 那么我们说吨是LMPU。

真的θ对于任何吨这样1−b吨(θ)是连续可微的θ=θ0和H1:θ>θ0或者H1:θ<θ0（即，单面测试）存在 LMP 测试并定义为，对于所有级别一种测试吨有一个独特的吨这样参数⁡最大限度吨[d(1−b吨∗(θ))dθ]θ=θ0=吨.
为了θ真正有价值和1−b吨(θ)两次连续可微θ=θ0对全部吨，然后是LMPU级别一种测试吨存在的H0:θ=θ0对比H1:θ≠θ0并且由
参数⁡最大限度吨[d2(1−b吨2(θ))dθ2]θ=θ1=吨⟹1−b吨(θ)≥1−b吨(θ)
为了0<d(θ)<Δ. 所有局部无偏测试结果d[1−b吨(θ)]dθ|θ=θ1=0给定连续可微的条件。(I) 的证明：用于测试吨
C吨(θ)≡1−b吨(θ)=∫吨F(D∣θ)dμ, C吨′(θ)=∫吨∂Fθ∂θdμ.认为C吨′(θ0)≥C吨2(θ0). 然后注意C吨(θ0)=1−b吨(θ0)=1−b吨(θ0)=C吨(θ0) C吨′(θ0)=林Δθ→0C吨(θ0+Δθ)−C吨(θ0)Δθ≥林Δθ→0C吨(θ0+Δθ)−C吨(θ0)Δθ=C吨′′(θ0) C吨′(θ0)−C吨′(θ0)=林Δθ→0C吨(θ0+Δθ)−C吨(θ0+Δθ)Δθ≥0

统计代写|统计推断作业代写statistical inference代考|TEST CONSTRUCTION

到目前为止，NP 理论还没有真正给出构建测试的原则。它指出了我们应该如何比较测试，即对于给定的大小，具有较大功效的测试是优越的，或者对于样本空间小号，我们想要吨(D)是这样的，对于所有的测试吨(D)和(吨∣H)≤一种
选择吨(D)这样
和(吨∗∣ķ)≤和(吨∣ķ)
由于这并不总是发生，我们继续使用其他标准，无偏性，不变性等等。

有几种测试构造方法不依赖于 NP 方法，但通常通过该方法固有的属性进行评估。最流行的是似然比检验 (LRT) 标准。

特别是一组参数的似然比检验 (LRT) 标准统计量θ去测试Hθ对比ķθ定义为
支持θ∈Hθ大号(θ∣D)支持θ∈(Hθ∪ķθ)大号(θ∣D)=λ(D)≤1
临界区定义为λ<ķ一种拒绝Hθ. 这产生
P\left{\lambda(D)<k_{\alpha}\right} \leq \alpha

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

随机过程代考

贝叶斯方法代考

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。

机器学习代写

多元统计分析代考

基础数据: $N$ 个样本， $P$ 个变量数的单样本，组成的横列的数据表
变量定性: 分类和顺序；变量定量：数值
数学公式的角度分为: 因变量与自变量

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|统计推断作业代写statistical inference代考|Unbiased and Invariant Tests

Posted on 2022年4月25日2022年4月25日 by statistics-lab

如果你也在怎样代写统计推断statistical inference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。

统计推断是利用数据分析来推断概率基础分布的属性的过程。推断性统计分析推断出人口的属性，例如通过测试假设和得出估计值。

我们提供的统计推断statistical inference及其相关学科的代写，服务范围广, 其中包括但不限于:

Statistical Inference 统计推断
Statistical Computing 统计计算
Advanced Probability Theory 高等楖率论
Advanced Mathematical Statistics 高等数理统计学
(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

Fundamentals of prediction — 统计代写|统计推断作业代写statistical inference代考|Unbiased and Invariant Tests

统计代写|统计推断作业代写statistical inference代考|UNBIASED TESTS

Whenever a UMP test exists at level $\alpha$, we have shown that $\alpha \leq 1-\beta_{K}$, that is, the power is at least as large as the size. If this were not so then there would be a test $T^{} \equiv \alpha$ which did not use the data and had greater power. Further, such a test is sometimes termed as “worse than useless” since it has smaller power than the useless test $T^{} \equiv \alpha$.

Now UMP tests do not always exist except in fairly restricted situations. Therefore N-P theory proposes that in the absence of a UMP test, a test should at least be unbiased, that is, $1-\beta_{K} \geq \alpha$. Further, if among all unbiased tests at level $\alpha$, there exists one that is UMP then this is to be favored and termed UMPU. So for a class of problems for which UMP tests do not exist there may exist UMPU tests.

统计代写|统计推断作业代写statistical inference代考|ADMISSIBILITY AND TESTS SIMILAR ON THE BOUNDARY

A test $T$ at level $\alpha$ is called admissible if there exists no other test that has power no less than $T$ for all alternatives while actually exceeding it for some alternatives, that is, no other test has power that dominates the power of test $T$. Hence if $T$ is UMPU then it is admissible, that is, if a test $T^{*}$ dominates $T$ then it also must be unbiased and hence $T$ is not UMPU which contradicts the assumption of $T$ being UMPU. Hence $T$ is admissible.
Recall a test $T$ which satisfies
$$
\begin{array}{ll}
1-\beta_{T}(\theta) \leq \alpha & \text { for } \theta \in \Theta_{H} \
1-\beta_{T}(\theta) \geq \alpha & \text { for } \theta \in \Theta_{K}
\end{array}
$$
is an unbiased test. If $T$ is unbiased and $1-\beta_{T}(\theta)$ is a continuous function of $\theta$ there is a boundary of values $\theta \in w$ for which $1-\beta_{T}(\theta)=\alpha$. Such a level $\alpha$ test with boundary $w$ is said to be similar on the boundary.

Theorem $5.1$ If $f(D \mid \theta)$ is such that for every test $T$ at level $\alpha$ the power function $1-\beta_{T}(\theta)$ is continuous and a test $T^{\prime}$ is UMP among all level $\alpha$ tests similar on the boundary then $T^{\prime}$ is UMPU.

Proof: The class of tests similar on the boundary contains the class of unbiased tests, that is, if $1-\beta_{T}(\theta)$ is continuous then every unbiased test is similar on the boundary. Further if $T^{\prime}$ is UMP among similar tests it is at least as powerful as any unbiased test and hence as $T^{*} \equiv \alpha$ so that it is unbiased and hence UMPU.
Finding a UMP similar test is often easier than finding a UMPU test directly so that this may provide a way of finding UMPU tests.

统计代写|统计推断作业代写statistical inference代考|NEYMAN STRUCTURE AND COMPLETENESS

Now if there is a sufficient statistic $t(D)$ for $\theta$ then any test $T$ such that
$$
E_{D \mid t}(T(D) \mid t(D) ; \quad \theta \in w)=\alpha
$$
is called a test of Neyman structure (NS) with respect to $t(D)$. This is a level $\alpha$ test since
$$
E_{\theta}(T(D) \mid \theta \in w)=E_{t} E_{D \mid t}(T(D) \mid t(D) ; \theta \in w)=E_{t}(\alpha)=\alpha .
$$
Now it is often easy to obtain a MP test among tests having Neyman structure termed UMPNS. If every similar test has Neyman structure, this test would be MP among similar tests or UMPS. A sufficient condition for a similar test to have Neyman structure is bounded completeness of the family $f(t \mid \theta)$ of sufficient statistics.
A family $\mathcal{F}$ of probability functions is complete if for any $g(x)$ satisfying
$$
E(g(x))=0 \quad \text { for all } f \in \mathcal{F}
$$
then $g(x)=0$.
A slightly weaker condition is “boundedly complete” in which the above holds for all bounded functions $g$.

Theorem $5.3$ Suppose for $f(D \mid \theta), t(D)$ is a sufficient statistic. Then a necessary and sufficient condition for all similar tests to have NS with respect to $t(D)$ is that $f(t)$ be boundedly complete.
Proof of bounded completeness implies NS given similar tests:
Assume $f(t)$ is boundedly complete and let $T(D)$ be similar. Now $E(T(D)-\alpha)=0$ for $f(D \mid \theta)$ and if $E(T(D)-\alpha \mid t(D)=t, \theta \in w)=g(t)$ such that
$$
\left.0=E_{D}(T(D)-\alpha)=E_{t} E_{D \mid t}(T(D)-\alpha) \mid t(D)=t, \theta \in w\right)=E_{t} g(t)=0
$$
Since $T(D)-\alpha$ is bounded then so is $g(t)$ and from the bounded completeness of $f(t), E_{t}(g(t))=0$ implies $g(t)=0$ such that
$$
E(T(D)-\alpha \mid t(D)=t, \quad \theta \in w)=0 \quad \text { or } \quad E(T(D) \mid t(D)=t, \quad \theta \in w)=\alpha \text { or NS. }
$$

Neyman-Pearson Lemma: Definition - Statistics How To — 统计代写|统计推断作业代写statistical inference代考|Unbiased and Invariant Tests

统计推断代考

统计代写|统计推断作业代写statistical inference代考|UNBIASED TESTS

每当 UMP 测试存在于级别一种，我们已经证明一种≤1−bķ，即功率至少与尺寸一样大。如果不是这样，那么会有一个测试 $T^{ } \equiv \alpha在H一世CHd一世dn这吨在s和吨H和d一种吨一种一种ndH一种dGr和一种吨和rp这在和r.F在r吨H和r,s在CH一种吨和s吨一世ss这米和吨一世米和s吨和r米和d一种s“在这rs和吨H一种n在s和l和ss”s一世nC和一世吨H一种ss米一种ll和rp这在和r吨H一种n吨H和在s和l和ss吨和s吨T^{ } \equiv \alpha$。

现在 UMP 测试并不总是存在，除非在相当有限的情况下。因此 NP 理论提出，在没有 UMP 检验的情况下，检验至少应该是无偏的，即1−bķ≥一种. 此外，如果在水平的所有无偏测试中一种，存在一个是UMP，那么这是受到青睐并被称为UMPU。因此，对于不存在UMP 测试的一类问题，可能存在UMPU 测试。

统计代写|统计推断作业代写statistical inference代考|ADMISSIBILITY AND TESTS SIMILAR ON THE BOUNDARY

一个测试吨在水平一种如果不存在其他测试的效力不小于吨对于所有替代方案，而对于某些替代方案实际上超过了它，也就是说，没有其他测试具有支配测试力量的力量吨. 因此，如果吨是 UMPU 那么它是可接受的，也就是说，如果一个测试吨∗占主导地位吨那么它也必须是公正的，因此吨不是与以下假设相矛盾的 UMPU吨是UMPU。因此吨是可以接受的。
召回测试吨满足
1−b吨(θ)≤一种为了 θ∈θH 1−b吨(θ)≥一种为了 θ∈θķ
是一个无偏的测试。如果吨是公正的并且1−b吨(θ)是一个连续函数θ有价值观的界限θ∈在为此1−b吨(θ)=一种. 这样的水平一种带边界测试在据说在边界上是相似的。

定理5.1如果F(D∣θ)是这样的，对于每一个测试吨在水平一种幂函数1−b吨(θ)是连续的并且是一个测试吨′是所有级别中的UMP一种然后在边界上进行类似的测试吨′是UMPU。

证明：边界上相似的测试类包含无偏测试类，即如果1−b吨(θ)是连续的，那么每个无偏测试在边界上都是相似的。进一步如果吨′是类似测试中的 UMP 它至少与任何无偏测试一样强大，因此吨∗≡一种所以它是公正的，因此是UMPU。
查找 UMP 类似测试通常比直接查找 UMPU 测试更容易，因此这可以提供一种查找 UMPU 测试的方法。

统计代写|统计推断作业代写statistical inference代考|NEYMAN STRUCTURE AND COMPLETENESS

现在如果有足够的统计吨(D)为了θ然后任何测试吨这样
和D∣吨(吨(D)∣吨(D);θ∈在)=一种
称为 Neyman 结构检验 (NS)吨(D). 这是一个级别一种测试以来
和θ(吨(D)∣θ∈在)=和吨和D∣吨(吨(D)∣吨(D);θ∈在)=和吨(一种)=一种.
现在，在具有称为 UMPNS 的内曼结构的测试中，通常很容易获得 MP 测试。如果每个相似测试都具有 Neyman 结构，则该测试将是相似测试中的 MP 或 UMPS。类似检验具有 Neyman 结构的充分条件是家庭的有限完备性F(吨∣θ)的充分统计。
一个家庭F的概率函数是完整的，如果对于任何G(X)令人满意的
和(G(X))=0 对全部 F∈F
然后G(X)=0.
稍微弱一点的条件是“有界完全”，其中上述条件适用于所有有界函数G.

定理5.3假设为F(D∣θ),吨(D)是一个充分的统计量。然后是所有类似测试都具有 NS 的充要条件吨(D)就是它F(吨)有界地完整。
有界完整性的证明意味着 NS 给定类似的测试：
假设F(吨)是有限完备的，让吨(D)相似。现在和(吨(D)−一种)=0为了F(D∣θ)而如果和(吨(D)−一种∣吨(D)=吨,θ∈在)=G(吨)这样
0=和D(吨(D)−一种)=和吨和D∣吨(吨(D)−一种)∣吨(D)=吨,θ∈在)=和吨G(吨)=0
自从吨(D)−一种是有界的，那么也是G(吨)并且从有界完整性F(吨),和吨(G(吨))=0暗示G(吨)=0这样
和(吨(D)−一种∣吨(D)=吨,θ∈在)=0 或者和(吨(D)∣吨(D)=吨,θ∈在)=一种或NS。

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

随机过程代考

贝叶斯方法代考

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。

机器学习代写

多元统计分析代考

基础数据: $N$ 个样本， $P$ 个变量数的单样本，组成的横列的数据表
变量定性: 分类和顺序；变量定量：数值
数学公式的角度分为: 因变量与自变量

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|统计推断作业代写statistical inference代考|MONOTONE LIKELIHOOD RATIO PROPERTY

Posted on 2022年4月25日2022年4月25日 by statistics-lab

如果你也在怎样代写统计推断statistical inference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。

统计推断是利用数据分析来推断概率基础分布的属性的过程。推断性统计分析推断出人口的属性，例如通过测试假设和得出估计值。

我们提供的统计推断statistical inference及其相关学科的代写，服务范围广, 其中包括但不限于:

Statistical Inference 统计推断
Statistical Computing 统计计算
Advanced Probability Theory 高等楖率论
Advanced Mathematical Statistics 高等数理统计学
(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|统计推断作业代写statistical inference代考|MONOTONE LIKELIHOOD RATIO PROPERTY

For scalar $\theta$ we suppose $f(\theta \mid D)=L(\theta \mid D)=L(\theta \mid t(D))$ and for all $\theta^{\prime}<\theta^{\prime \prime}$,
$$
Q\left(t(D) \mid \theta^{\prime}, \theta^{\prime \prime}\right)=\frac{L\left(\theta^{\prime} \mid t(D)\right)}{L\left(\theta^{\prime \prime} \mid t(D)\right)}
$$
is a non-decreasing function of $t(D)$ or as $t(D) \uparrow Q \uparrow$, that is, the support for $\theta^{\prime}$ vs. $\theta^{\prime}$ does not decrease as $t$ increases and does not increase as $t$ decreases. We call this a monotone likelihood ratio (MLR) property of $L(\theta \mid t(D))$.

Theorem 4.1 Let $f(D \mid \theta)$ have the MLR property. To test $H: \theta \leq \theta_{0}$ vs. $K: \theta>\theta_{0}$ the following test $T(D)$ is UMP at level $\alpha$.
$$
T(D)=\left{\begin{array}{llll}
1 & \text { if } \quad t(D)k & \text { e.g., accept } H,
\end{array}\right.
$$
where $k$ and $p$ are determined by
$$
\alpha=E_{\theta_{0}}(T(D))=P\left(t(D)\theta_{0}$ so there exists a $k$ and a $p$ such that the test satisfies (a) and (b), that is,
$Q=\frac{f\left(D \mid \theta_{0}\right)}{f\left(D \mid \theta_{1}\right)}\theta^{\prime} \text { at level } \alpha^{\prime}=1-\beta\left(\theta^{\prime}\right) \text {. }
$$

统计代写|统计推断作业代写statistical inference代考|DECISION THEORY

Suppose we now consider the testing problem from a decision theoretic point of view, that is, testing $H: \theta \leq \theta_{0}$ vs. $K: \theta>\theta_{0}$ with $d_{0}$ the decision to accept $H$ and $d_{1}$ the decision to accept $K$. Assume a loss function $l\left(\theta, d_{i}\right)$. Now it appears to be sensible to assume
$$
\begin{array}{lll}
l\left(\theta, d_{0}\right)=0 & \text { if } \quad \theta \leq \theta_{0} \text { and } \uparrow \text { for } \theta>\theta_{0}, \
l\left(\theta, d_{1}\right)=0 & \text { if } \quad \theta>\theta_{0} \text {, and } \uparrow \text { for } \theta \leq \theta_{0} .
\end{array}
$$
Hence
(1) $l\left(\theta, d_{1}\right)-l(\theta, d 0) \quad \begin{cases}>0 & \text { for } \theta \leq \theta_{0} \ <0 & \text { for } \theta>\theta_{0} .\end{cases}$
Now recall that the risk function of a test $T$ for a given $\theta$ is the average loss
$$
\begin{aligned}
&l(\theta, T)=T(D) l\left(\theta, d_{1}\right)+(1-T(D)) l\left(\theta, d_{0}\right) \
&R(\theta, T)=E_{D} l(\theta, T)=E_{D}\left[T(D) l\left(\theta, d_{1}\right)+(1-T) l\left(\theta, d_{0}\right)\right]
\end{aligned}
$$
Further suppose for tests $T(D)$ and $T^{}(D)$ that (2) $R\left(\theta, T^{}\right) \leq R(\theta, T)$ for all $\theta$,
(3) $R\left(\theta, T^{}\right)}$ dominates $T$ and we say $T$ is inadmissible. On the other hand if no such $T^{}$ exists then $T$ is admissible. A class $C$ of test (decision) procedures is complete if for any test $T$ not in $C$ there exists a test $T^{}$ in $C$ dominating it. (A complete class is minimal if it does not contain a complete subclass). A class $C$ is essentially complete if at least (2) holds. (A complete class is also essentially complete, a class is minimal essentially complete if it does not contain an essentially complete subclass).

统计代写|统计推断作业代写statistical inference代考|Unbiased and Invariant Tests

Since UMP tests don’t always exist, statisticians have proceeded to find optimal tests in more restricted classes. One such restriction is unbiasedness. Another is invariance. This chapter develops the theory of uniformly most powerful unbiased (UMPU) and invariant (UMPI) tests. When it is not possible to optimize in these ways, it is still possible to make progress, at least on the mathematical front. Locally most powerful (LMP) tests are those that have greatest power in a neighborhood of the null, and locally most powerful unbiased (LMPU) tests are most powerful in a neighborhood of the null, among unbiased tests. These concepts and main results related to them are presented here. The theory is illustrated with examples and moreover, examples are given that illustrate potential flaws. The concept of a “worse than useless” test is illustrated using a commonly accepted procedure. The sequential probability ratio test is also presented.

Decision Theory and large deviations for dynamical hypotheses tests: The Neyman-Pearson Lemma, Min-Max and Bayesian tests — 统计代写|统计推断作业代写statistical inference代考|MONOTONE LIKELIHOOD RATIO PROPERTY

统计推断代考

统计代写|统计推断作业代写statistical inference代考|MONOTONE LIKELIHOOD RATIO PROPERTY

对于标量θ我们假设F(θ∣D)=大号(θ∣D)=大号(θ∣吨(D))并为所有人θ′<θ′′,
问(吨(D)∣θ′,θ′′)=大号(θ′∣吨(D))大号(θ′′∣吨(D))
是一个非减函数吨(D)或作为吨(D)↑问↑，也就是支持θ′对比θ′不会减少吨增加和不增加吨减少。我们称其为单调似然比 (MLR) 属性大号(θ∣吨(D)).

定理 4.1 让F(D∣θ)具有 MLR 属性。去测试H:θ≤θ0对比ķ:θ>θ0以下测试吨(D)是水平的UMP一种.
$$
T(D)=\左{1 如果吨(D)ķ 例如，接受 H,\对。
在H和r和$ķ$一种nd$p$一种r和d和吨和r米一世n和db是
\alpha=E_{\theta_{0}}(T(D))=P\left(t(D)\theta_{0}s这吨H和r和和X一世s吨s一种ķ一种nd一种ps在CH吨H一种吨吨H和吨和s吨s一种吨一世sF一世和s(一种)一种nd(b),吨H一种吨一世s,Q=\frac{f\left(D \mid \theta_{0}\right)}{f\left(D \mid \theta_{1}\right)}\theta^{\prime} \text { 在级别} \alpha^{\prime}=1-\beta\left(\theta^{\prime}\right) \text {. }
$$

统计代写|统计推断作业代写statistical inference代考|DECISION THEORY

假设我们现在从决策论的角度考虑测试问题，即测试H:θ≤θ0对比ķ:θ>θ0和d0接受的决定H和d1接受的决定ķ. 假设一个损失函数l(θ,d一世). 现在假设是明智的
l(θ,d0)=0 如果 θ≤θ0 和 ↑ 为了 θ>θ0, l(θ,d1)=0 如果 θ>θ0，和 ↑ 为了 θ≤θ0.
因此
(1)l(θ,d1)−l(θ,d0){>0 为了 θ≤θ0 <0 为了 θ>θ0.
现在回想一下测试的风险函数吨对于给定的θ是平均损失
l(θ,吨)=吨(D)l(θ,d1)+(1−吨(D))l(θ,d0) R(θ,吨)=和Dl(θ,吨)=和D[吨(D)l(θ,d1)+(1−吨)l(θ,d0)]
进一步假设测试吨(D)和吨(D)(2)R(θ,吨)≤R(θ,吨)对全部θ,
(3) R\left(\theta, T^{}\right)}R\left(\theta, T^{}\right)}占主导地位吨我们说吨是不可接受的。另一方面，如果没有这样吨那么存在吨是可以接受的。一类C的测试（决定）程序是完整的，如果对于任何测试吨不在C存在一个测试吨在C支配它。（如果一个完整的类不包含一个完整的子类，那么它就是最小的）。一类C如果至少 (2) 成立，则基本上是完整的。（一个完整的类也是本质上完整的，如果一个类不包含本质上完整的子类，那么它就是最小的本质上完整的类）。

统计代写|统计推断作业代写statistical inference代考|Unbiased and Invariant Tests

由于 UMP 检验并不总是存在，统计学家已经着手在更受限制的类别中寻找最佳检验。一种这样的限制是不偏不倚。另一个是不变性。本章发展了一致最强大无偏（UMPU）和不变（UMPI）检验的理论。当无法以这些方式进行优化时，仍然有可能取得进展，至少在数学方面是这样。局部最强大 (LMP) 检验是在零附近具有最大功效的检验，在无偏检验中，局部最强大无偏 (LMPU) 检验在零附近最强大。这里介绍了这些概念和与它们相关的主要结果。该理论通过示例进行说明，此外，还给出了说明潜在缺陷的示例。使用普遍接受的程序来说明“比无用更糟糕”的测试概念。还提出了顺序概率比检验。

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

随机过程代考

贝叶斯方法代考

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。

机器学习代写

多元统计分析代考

基础数据: $N$ 个样本， $P$ 个变量数的单样本，组成的横列的数据表
变量定性: 分类和顺序；变量定量：数值
数学公式的角度分为: 因变量与自变量

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|统计推断作业代写statistical inference代考|Testing Hypotheses

Posted on 2022年4月25日2022年4月25日 by statistics-lab

如果你也在怎样代写统计推断statistical inference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。

统计推断是利用数据分析来推断概率基础分布的属性的过程。推断性统计分析推断出人口的属性，例如通过测试假设和得出估计值。

我们提供的统计推断statistical inference及其相关学科的代写，服务范围广, 其中包括但不限于:

Statistical Inference 统计推断
Statistical Computing 统计计算
Advanced Probability Theory 高等楖率论
Advanced Mathematical Statistics 高等数理统计学
(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

Comparing Variations of the Neyman-Pearson Lemma — 统计代写|统计推断作业代写statistical inference代考|Testing Hypotheses

统计代写|统计推断作业代写statistical inference代考|HYPOTHESIS TESTING VIA THE REPEATED

Neyman-Pearson (N-P) (1933) Theory of Hypotheses Testing is based on the repeated sampling principle and is basically a two decision procedure.

We will collect data $D$ and assume we have two rival hypotheses about the populations from which the data were generated: $H_{0}$ (the null hypothesis) and $H_{1}$ (the alternative hypothesis). We assume a sample space $S$ of all possible outcomes of the data $D$. A rule is then formulated for the rejection of $H_{0}$ or $H_{1}$ in the following manner. Choose a subset $s$ (critical region) of $S$ and
if $D \in s$ reject $H_{0}$ and accept $H_{1}$
if $D \in S-s$ reject $H_{1}$ and accept $H_{0}$
according to
$P\left(D \in s \mid H_{0}\right)=\epsilon$ (size) $\leq \alpha$ (level) associated with the test (Type 1 error)
$P\left(D \in s \mid H_{1}\right)=1-\beta$ (power of the test)
$P\left(D \in S-s \mid H_{1}\right)=\beta$ (Type 2 error)

The two basic concepts are size and power and N-P theory dictates that we choose a test (critical region) which results in small size and large power. At this juncture we assume size equals level and later show how size and level can be equated. Now
$$
\alpha=P\left(\text { accepting } H_{1} \mid H_{0}\right) \text { or } 1-\alpha=P\left(\text { accepting } H_{0} \mid H_{0}\right)
$$
and
$$
1-\beta=P\left(\text { accepting } H_{1} \mid H_{1}\right) \text { or } \quad \beta=P\left(\text { accepting } H_{0} \mid H_{1}\right) .
$$
Since there is no way of jointly minimizing size and maximizing power, N-P suggest choosing small size $\alpha$ and then maximize power $1-\beta$ (or minimize $\beta$ ).

统计代写|统计推断作业代写statistical inference代考|REMARKS ON SIZE

Before you have the data $D$ it would be reasonable to require that you should have small size (frequency of rejecting $H_{0}$ when $H_{0}$ is true). But this may mislead you once you have the data, for example, suppose you want size $=.05$ where the probabilities for the data $D=\left(D_{1}, D_{2}, D_{3}\right)$ are given in Table $4.1$.

Presumably if you want size $=.05$ you reject $H_{0}$ if event $D_{1}$ occurs and accept if $D_{2}$ or $D_{3}$ occur. However if $D_{1}$ occurs you are surely wrong to reject $H_{0}$ since $P\left(D_{1} \mid H_{1}\right)=0$. So you need more than size. Note that before making the test, all tests of the same size provide us with the same chance of rejecting $H_{0}$, but after the data are in hand not all tests of the same size are equally good. In the N-P set up we are forced to choose a test before we know what the sample value actually is, even when our interest is in evaluating hypotheses with regard to the sample data we have. Therefore if two tests $T_{1}$ and $T_{2}$ have the same size one might be led to choose the test with greater power. That this is not necessarily the best course is demonstrated in the following Example $4.1$ and its variations, Hacking (1965).

统计代写|统计推断作业代写statistical inference代考|NEYMAN-PEARSON FUNDAMENTAL LEMMA

Lemma 4.1 Let $F_{0}(D)$ and $F_{1}(D)$ be distribution functions possessing generalized densities $f_{0}(D)$ and $f_{1}(D)$ with respect to $\mu$. Let $H_{0}: f_{0}(D)$ vs. $H_{1}: f_{1}(D)$ and $T(D)=P\left[\right.$ rejecting $H_{0}$ ]. Then

Existence: For testing $H_{0}$ vs. $H_{1}$ there exists a test $T(D)$ and a constant $k$ such that for any given $\alpha, 0 \leq \alpha \leq 1$
(a) $E_{H_{0}}(T(D))=\alpha=\int_{S} T(D) f_{0}(D) d \mu$
(b) $T(D)=\left{\begin{array}{lll}1 & \text { if } & f_{0}(D)k f_{1}(D)^{*}\end{array}\right.$.
Sufficiency: If a test satisfies (a) and (b) then it is the most powerful (MP) for testing $H_{0}$ vs. $H_{1}$ at level $\alpha$.

Necessity: If $T(D)$ is most powerful $(M P)$ at level $\alpha$ for $H_{0}$ vs. $H_{1}$ then for some $k$ it satisfies (a) and (b) unless there exists a test of size less than $\alpha$ and power $1 .$

Proof: For $\alpha=0$ or $\alpha=1$ let $k=0$ or $\infty$ respectively, hence we restrict $\alpha$ such that $0<\alpha<1$.

Existence: Let
$$
\alpha(k)=P\left(f_{0} \leq k f_{1} \mid H_{0}\right)=P\left(f_{0}k f_{1}
\end{array}\right.
$$
If $\alpha(k)=\alpha(k-0)$ there is no need for the middle term since
$$
P\left(f_{0}=k f_{1}\right)=0 .
$$
Hence we can produce a test with properties (a) and (b).
Sufficiency: If a test satisfies (a) and (b) then it is most powerful for testing $H_{0}$ against $H_{1}$ at level $\alpha$.
Proof: Let $T^{}$ (D) be any other test such that $$ E_{H_{0}}\left(T^{}(D)\right) \leq \alpha .
$$
Now consider
$$
E_{H_{1}}\left(T^{}(D)\right)=\int_{S} T^{} f_{1} d \mu=1-\beta^{*}
$$

Neyman-Pearson Lemma and Receiver Operating Characteristic Curve - Rhea — 统计代写|统计推断作业代写statistical inference代考|Testing Hypotheses

统计推断代考

统计代写|统计推断作业代写statistical inference代考|HYPOTHESIS TESTING VIA THE REPEATED

Neyman-Pearson (NP) (1933) 假设检验理论基于重复抽样原则，基本上是一个两决定程序。

我们将收集数据D并假设我们对生成数据的人群有两个相互竞争的假设：H0（零假设）和H1（备择假设）。我们假设一个样本空间小号数据的所有可能结果D. 然后制定规则来拒绝H0或者H1以下列方式。选择一个子集s（临界区）的小号
如果_D∈s拒绝H0并接受H1
如果D∈小号−s拒绝H1并接受H0
根据
磷(D∈s∣H0)=ε（尺寸）≤一种（级别）与测试相关联（类型 1 错误）
磷(D∈s∣H1)=1−b（测试的力量）
磷(D∈小号−s∣H1)=b（类型 2 错误）

两个基本概念是大小和功率，NP 理论要求我们选择一个测试（临界区域），它会导致小尺寸和大功率。在这个关头，我们假设大小等于级别，稍后将说明如何使大小和级别相等。现在
一种=磷( 接受 H1∣H0) 或者 1−一种=磷( 接受 H0∣H0)
和
1−b=磷( 接受 H1∣H1) 或者 b=磷( 接受 H0∣H1).
由于没有办法联合最小化尺寸和最大化功率，NP建议选择小尺寸一种然后最大化功率1−b（或最小化b ).

统计代写|统计推断作业代写statistical inference代考|REMARKS ON SIZE

在你有数据之前D要求你应该有小尺寸是合理的（拒绝的频率H0什么时候H0是真的）。但是一旦你有了数据，这可能会误导你，例如，假设你想要大小=.05其中数据的概率D=(D1,D2,D3)在表中给出4.1.

大概如果你想要大小=.05你拒绝H0如果事件D1发生并接受如果D2或者D3发生。然而，如果D1发生你拒绝肯定是错误的H0自从磷(D1∣H1)=0. 所以你需要的不仅仅是尺寸。请注意，在进行测试之前，所有相同大小的测试都为我们提供了相同的拒绝机会H0，但在获得数据后，并非所有相同大小的测试都同样好。在 NP 设置中，我们被迫在知道样本值实际是什么之前选择测试，即使我们的兴趣是评估与我们拥有的样本数据有关的假设。因此，如果两个测试吨1和吨2具有相同大小的人可能会被引导选择具有更大功率的测试。这不一定是最好的课程在下面的例子中得到证明4.1及其变体，Hacking (1965)。

统计代写|统计推断作业代写statistical inference代考|NEYMAN-PEARSON FUNDAMENTAL LEMMA

引理 4.1 让F0(D)和F1(D)是具有广义密度的分布函数F0(D)和F1(D)关于μ. 让H0:F0(D)对比H1:F1(D)和吨(D)=磷[拒绝H0]。然后

存在：用于测试H0对比H1存在一个测试吨(D)和一个常数ķ这样对于任何给定的一种,0≤一种≤1
（一种）和H0(吨(D))=一种=∫小号吨(D)F0(D)dμ
(b) $T(D)=\左{1 如果 F0(D)ķF1(D)∗\对。$。
充分性：如果一个测试满足（a）和（b），那么它是最强大的（MP）测试H0对比H1在水平一种.
必要性：如果吨(D)是最强大的(米磷)在水平一种为了H0对比H1然后对于一些ķ它满足 (a) 和 (b) 除非存在规模小于一种和权力1.

证明：对于一种=0或者一种=1让ķ=0或者∞分别，因此我们限制一种这样0<一种<1.

存在：让
\alpha(k)=P\left(f_{0} \leq k f_{1} \mid H_{0}\right)=P\left(f_{0}k f_{1} \end{array}\对。\alpha(k)=P\left(f_{0} \leq k f_{1} \mid H_{0}\right)=P\left(f_{0}k f_{1} \end{array}\对。
如果一种(ķ)=一种(ķ−0)不需要中期，因为
磷(F0=ķF1)=0.
因此，我们可以生成具有属性 (a) 和 (b) 的测试。
充分性：如果一个测试满足（a）和（b），那么它对测试最有效H0反对H1在水平一种.
证明：让吨(D) 是任何其他测试，使得和H0(吨(D))≤一种.
现在考虑
和H1(吨(D))=∫小号吨F1dμ=1−b∗

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

随机过程代考

贝叶斯方法代考

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。

机器学习代写

多元统计分析代考

基础数据: $N$ 个样本， $P$ 个变量数的单样本，组成的横列的数据表
变量定性: 分类和顺序；变量定量：数值
数学公式的角度分为: 因变量与自变量

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|统计推断作业代写statistical inference代考|SAMPLING ISSUES

Posted on 2022年4月25日2022年4月25日 by statistics-lab

如果你也在怎样代写统计推断statistical inference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。

统计推断是利用数据分析来推断概率基础分布的属性的过程。推断性统计分析推断出人口的属性，例如通过测试假设和得出估计值。

我们提供的统计推断statistical inference及其相关学科的代写，服务范围广, 其中包括但不限于:

Statistical Inference 统计推断
Statistical Computing 统计计算
Advanced Probability Theory 高等楖率论
Advanced Mathematical Statistics 高等数理统计学
(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

Maximum-Likelihood Methods, Score Tests, and Constructed Variables - SAGE Research Methods — 统计代写|统计推断作业代写statistical inference代考|SAMPLING ISSUES

统计代写|统计推断作业代写statistical inference代考|The “Mixed” Sampling Case

Another negative multinomial sampling approach stops when $r_{1}$, say, attains a given value. Here
$$
\begin{aligned}
f_{r_{1}}=& f\left(r_{2}, n_{1}, n \mid r_{1}\right)=f\left(r_{2}, n \mid n_{1}, r_{1}\right) f\left(n_{1} \mid r_{1}\right) \
=& f\left(n_{1} \mid r_{1}\right) f\left(r_{2} \mid n, n_{1}, r_{1}\right) f\left(n \mid n_{1}, r_{1}\right) \
=&\left(\begin{array}{c}
n_{1}-1 \
r_{1}-1
\end{array}\right) p_{1}^{n_{1}}\left(1-p_{1}\right)^{n_{1}-r_{1}}\left(\begin{array}{c}
n_{1} \
r_{2}
\end{array}\right) \
& \times p_{2}^{r_{2}}\left(1-p_{2}\right)^{n_{2}-r_{2}}\left(\begin{array}{c}
n-1 \
n_{1}-1
\end{array}\right) \gamma^{n_{1}}(1-\gamma)^{n-n_{1}}
\end{aligned}
$$
Again, the likelihood for $p_{1}$ and $p_{2}$ is preserved for inference that respects ELP but here we now encounter a difficulty for conditional frequentist inference regarding $p_{1}$ and $p_{2}$. What does one condition on to obtain an exact significance test on $p_{1}=p_{2}$ ? Of course, this problem would also occur when we start with one sample that is binomial, say, a control, and the other negative binomial, for say a new treatment where one would like to stop the latter trial after a given number of failures. Note that this problem would persist for $f_{r_{2}}, f_{n_{1}-r_{1}}$ and $f_{n_{2}-r_{2}}$. Hence in these four cases there is no exact conditional Fisher type test for $p_{1}=p_{2}$.

Next we examine these issues for the $2 \times 2$ table. Here we list the various ways one can sample in constructing a $2 \times 2$ table such that one of the nine values is fixed, that is, when that value appears sampling ceases. For 7 out of the 9 cases the entire likelihood is the same, where
$$
L=\gamma^{n_{1}}(1-\gamma)^{n_{2}} \prod_{i-1}^{2} p_{i}^{r_{1}}\left(1-p_{i}\right)^{n_{i}-r_{i}}=L(\gamma) L\left(p_{1}, p_{2}\right)
$$
with sampling probabilities
$$
\begin{aligned}
f_{n} &=\left(\begin{array}{l}
n_{1} \
r_{1}
\end{array}\right)\left(\begin{array}{l}
n_{2} \
r_{2}
\end{array}\right)\left(\begin{array}{c}
n \
n_{1}
\end{array}\right) L \
f_{n_{1}} &=\left(\begin{array}{l}
n_{1} \
r_{1}
\end{array}\right)\left(\begin{array}{l}
n_{2} \
r_{2}
\end{array}\right)\left(\begin{array}{c}
n-1 \
n_{1}-1
\end{array}\right) L=f\left(r_{1}, r_{2}, n \mid n_{1}\right) \
&=f\left(r_{1}, r_{2} \mid n, n_{1}\right) f\left(n \mid n_{1}\right) \
f_{n_{2}} &=\left(\begin{array}{l}
n_{1} \
r_{1}
\end{array}\right)\left(\begin{array}{c}
n_{2} \
r_{2}
\end{array}\right)\left(\begin{array}{c}
n-1 \
n_{2}-1
\end{array}\right) L=f\left(r_{1}, r_{2} \mid n_{2}, n\right) f\left(n \mid n_{2}\right)
\end{aligned}
$$

统计代写|统计推断作业代写statistical inference代考|OTHER PRINCIPLES

Restricted Conditionality Principle RCP: Same preliminaries as LP. $\mathcal{E}=(S, \mu, \theta, f)$ is a mixture of experiments $\mathcal{E}{i}=\left(S{i}, \mu_{i}, \theta, f_{i}\right)$ with mixture probabilities $q_{i}$ independent of $\theta$. First we randomly select $\mathcal{E}{1}$ or $\mathcal{E}{2}$ with probabilities $q_{1}$ and $q_{2}=1-q_{1}$, and then perform the chosen experiment $\mathcal{E}{i}$. Then we recognize the sample $D=\left(i, D{i}\right)$ and $f(D \mid \theta)=q_{i} f_{i}\left(D_{i} \mid \theta\right), i=1,2$. Then RCP asserts $\operatorname{Inf}(\mathcal{E}, D)=\operatorname{Inf}\left(\mathcal{E}{i}, D{i}\right)$.

Definition of Ancillary Statistic: A statistic $C=C(D)$ is ancillary with respect to $\theta$ if $f_{C}(c \mid \theta)$ is independent of $\theta$, so that an ancillary is non-informative about $\theta$. $C(D)$ maps $S \rightarrow S_{c}$ where each $c \in S_{c}$ defines $S_{c}=(D \mid C(D)=c)$. Define the conditional experiment $\mathcal{E}{D \mid C}=\left(S{c}, \mu, \theta, f_{D \mid C}(D \mid c)\right)$, and the marginal experiment $\mathcal{E}{C}=\left(S{c}, \mu, \theta, f_{c}(c)\right)$, where $\mathcal{E}{C}=$ sample from $S{c}$ or sample from $S$ and observe $c$ and $\mathcal{E}{D \mid C}=$ conditional on $C=c$, sample from $S{c}$.
Unrestricted Conditionality Principle $(U C P)$ : When $C$ is an ancillary $\operatorname{Inf}(\mathcal{E}, D)=\operatorname{Inf}\left(\mathcal{E}{D \mid C}, D\right)$ concerning $\theta$. It is as if we performed $\mathcal{E}{C}$ and then performed $\mathcal{E}_{D \mid C}$.

统计代写|统计推断作业代写statistical inference代考|ULP

Note this is just a special case of ULP.
We show that ULP $\Leftrightarrow$ (RCP, MEP). First assume ULP, so that $\operatorname{Inf}(\mathcal{E}, D)=$ $\operatorname{Inf}\left(\mathcal{E}^{\prime}, D^{\prime}\right)$. Now suppose $f(D \mid \theta)=f\left(D^{\prime} \mid \theta\right)$, then apply ULP so $\operatorname{Inf}\left(\mathcal{E}, D^{\prime}\right)=$ $\operatorname{Inf}\left(\mathcal{E}, D^{\prime}\right)$, which is MEP. Further suppose, where $C$ is an ancillary, and hence that
$$
f(D \mid \theta)=f(D \mid c, \theta) h(c), \quad f\left(D^{\prime} \mid \theta\right)=f\left(D^{\prime} \mid c, \theta\right) h(c) .
$$
Hence ULP implies that
$$
\operatorname{Inf}(\mathcal{E}, D)=\operatorname{Inf}(\mathcal{E}, D \mid C)
$$
or UCP, and UCP implies RCP.
Conversely, assume (RCP, MEP) and that $\left(\mathcal{E}{1}, D{1}\right)$ and $\left(\mathcal{E}{2}, D{2}\right)$ generate equivalent likelihoods
$$
f_{1}\left(D_{1} \mid \theta\right)=h\left(D_{1}, D_{2}\right) f_{2}\left(D_{2} \mid \theta\right)
$$
We will use (RCP, MEP) to show $\operatorname{Inf}\left(\mathcal{E}{1}, D{1}\right)=\operatorname{Inf}\left(\mathcal{E}{2}, D{2}\right)$. Now let $\mathcal{E}$ be the mixture experiment with probabilities $(1+h)^{-1}$ and $h(1+h)^{-1}$. Hence the sample points in $\mathcal{E}$ are $\left(1, D_{1}\right)$ and $\left(2, D_{2}\right)$ and
$$
\begin{aligned}
&f\left(1, D_{1} \mid \theta\right)=(1+h)^{-1} f_{1}\left(D_{1} \mid \theta\right)=h(1+h)^{-1} f_{2}\left(D_{2} \mid \theta\right) \
&f\left(2, D_{2} \mid \theta\right)=h(1+h)^{-1} f_{2}\left(D_{2} \mid \theta\right)
\end{aligned}
$$
Therefore $f\left(1, D_{1} \mid \theta\right)=f\left(2, D_{2} \mid \theta\right)$ so from MEP
$$
\operatorname{Inf}\left(\mathcal{E},\left(1, D_{1}\right)\right)=\operatorname{Inf}\left(\mathcal{E},\left(2, D_{2}\right)\right.
$$
Now apply RCP to both sides above so that
$$
\begin{aligned}
&\operatorname{Inf}\left(\mathcal{E}{1}, D{1}\right)=\operatorname{Inf}\left(\mathcal{E}, \theta,\left(1, D_{1}\right)\right)=\operatorname{Inf}\left(\mathcal{E}, \theta,\left(2, D_{2}\right)\right)=\operatorname{Inf}\left(\mathcal{E}{2}, D{2}\right) \
&\text { therefore }(\operatorname{RCP}, \mathrm{MEP}) \Longrightarrow \text { ULP }
\end{aligned}
$$

Intersections of likelihood ratio test significance. Overlaps of... | Download Scientific Diagram — 统计代写|统计推断作业代写statistical inference代考|SAMPLING ISSUES

统计推断代考

统计代写|统计推断作业代写statistical inference代考|The “Mixed” Sampling Case

另一种负多项式抽样方法在以下情况下停止r1，比如说，达到一个给定的值。这里
Fr1=F(r2,n1,n∣r1)=F(r2,n∣n1,r1)F(n1∣r1) =F(n1∣r1)F(r2∣n,n1,r1)F(n∣n1,r1) =(n1−1 r1−1)p1n1(1−p1)n1−r1(n1 r2) ×p2r2(1−p2)n2−r2(n−1 n1−1)Cn1(1−C)n−n1
再次，可能性为p1和p2被保留用于尊重 ELP 的推理，但在这里我们现在遇到了关于条件频率论推理的困难p1和p2. 获得精确显着性检验的条件是什么p1=p2? 当然，当我们从一个二项式样本开始时也会出现这个问题，比如说，一个控制，另一个负二项式，比如说一种新的治疗方法，在给定次数的失败后想要停止后一种试验。请注意，此问题将持续存在Fr2,Fn1−r1和Fn2−r2. 因此，在这四种情况下，没有精确的条件 Fisher 类型检验p1=p2.

接下来我们来分析一下这些问题2×2桌子。在这里，我们列出了构建一个可以采样的各种方法2×2表使得九个值之一是固定的，也就是说，当该值出现时，采样停止。对于 9 个案例中的 7 个，整个可能性是相同的，其中
大号=Cn1(1−C)n2∏一世−12p一世r1(1−p一世)n一世−r一世=大号(C)大号(p1,p2)
有抽样概率
Fn=(n1 r1)(n2 r2)(n n1)大号 Fn1=(n1 r1)(n2 r2)(n−1 n1−1)大号=F(r1,r2,n∣n1) =F(r1,r2∣n,n1)F(n∣n1) Fn2=(n1 r1)(n2 r2)(n−1 n2−1)大号=F(r1,r2∣n2,n)F(n∣n2)

统计代写|统计推断作业代写statistical inference代考|OTHER PRINCIPLES

限制条件原则 RCP：与 LP 相同的预备条件。和=(小号,μ,θ,F)是实验的混合体和一世=(小号一世,μ一世,θ,F一世)具有混合概率q一世独立于θ. 首先我们随机选择和1或者和2有概率q1和q2=1−q1，然后执行选择的实验和一世. 然后我们识别样本D=(一世,D一世)和F(D∣θ)=q一世F一世(D一世∣θ),一世=1,2. 然后 RCP 断言信息⁡(和,D)=信息⁡(和一世,D一世).

辅助统计的定义：一个统计C=C(D)是辅助的θ如果FC(C∣θ)独立于θ，因此辅助项是非信息性的θ. C(D)地图小号→小号C其中每个C∈小号C定义小号C=(D∣C(D)=C). 定义条件实验和D∣C=(小号C,μ,θ,FD∣C(D∣C)), 和边际实验和C=(小号C,μ,θ,FC(C))，在哪里和C=样本来自小号C或样本来自小号并观察C和和D∣C=有条件的C=C, 样本来自小号C.
无限制条件原则(在C磷)：什么时候C是辅助信息⁡(和,D)=信息⁡(和D∣C,D)关于θ. 就好像我们表演了和C然后执行和D∣C.

统计代写|统计推断作业代写statistical inference代考|ULP

请注意，这只是 ULP 的一个特例。
我们证明 ULP⇔（RCP，环保部）。首先假设 ULP，所以信息⁡(和,D)= 信息⁡(和′,D′). 现在假设F(D∣θ)=F(D′∣θ)，然后应用 ULP 所以信息⁡(和,D′)= 信息⁡(和,D′)，即 MEP。进一步假设，其中C是辅助的，因此
F(D∣θ)=F(D∣C,θ)H(C),F(D′∣θ)=F(D′∣C,θ)H(C).
因此 ULP 意味着
信息⁡(和,D)=信息⁡(和,D∣C)
或 UCP，而 UCP 意味着 RCP。
相反，假设 (RCP, MEP) 并且(和1,D1)和(和2,D2)产生等价的可能性
F1(D1∣θ)=H(D1,D2)F2(D2∣θ)
我们将使用 (RCP, MEP) 来展示信息⁡(和1,D1)=信息⁡(和2,D2). 现在让和是概率的混合实验(1+H)−1和H(1+H)−1. 因此样本点在和是(1,D1)和(2,D2)和
F(1,D1∣θ)=(1+H)−1F1(D1∣θ)=H(1+H)−1F2(D2∣θ) F(2,D2∣θ)=H(1+H)−1F2(D2∣θ)
所以F(1,D1∣θ)=F(2,D2∣θ)所以从 MEP
信息⁡(和,(1,D1))=信息⁡(和,(2,D2)
现在将 RCP 应用于上面的两侧，以便
信息⁡(和1,D1)=信息⁡(和,θ,(1,D1))=信息⁡(和,θ,(2,D2))=信息⁡(和2,D2) 所以 (RCP,米和磷)⟹ ULP

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

随机过程代考

贝叶斯方法代考

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。

机器学习代写

多元统计分析代考

基础数据: $N$ 个样本， $P$ 个变量数的单样本，组成的横列的数据表
变量定性: 分类和顺序；变量定量：数值
数学公式的角度分为: 因变量与自变量

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|统计推断作业代写statistical inference代考|FURTHER REMARKS ON TESTS OF SIGNIFICANCE

Posted on 2022年4月25日2022年4月25日 by statistics-lab

如果你也在怎样代写统计推断statistical inference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。

统计推断是利用数据分析来推断概率基础分布的属性的过程。推断性统计分析推断出人口的属性，例如通过测试假设和得出估计值。

我们提供的统计推断statistical inference及其相关学科的代写，服务范围广, 其中包括但不限于:

Statistical Inference 统计推断
Statistical Computing 统计计算
Advanced Probability Theory 高等楖率论
Advanced Mathematical Statistics 高等数理统计学
(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

PDF] Bayesian Hypothesis Testing: An Alternative to Null Hypothesis Significance Testing (NHST) in Psychology and Social Sciences | Semantic Scholar — 统计代写|统计推断作业代写statistical inference代考|FURTHER REMARKS ON TESTS OF SIGNIFICANCE

统计代写|统计推断作业代写statistical inference代考|FURTHER REMARKS ON TESTS OF SIGNIFICANCE

The claim for significance tests are for those cases where alternative hypotheses are not sensible. Note that Goodness-of-Fit tests fall into this category, that is, Do the data fit a normal distribution? Here $H_{0}$ is merely a family of distributions rather than a specification of parameter values. Note also that a test of significance considers not only the event that occurred but essentially puts equal weight on more discrepent events that did not occur as opposed to a test which only considers what did occur.
A poignant criticism of Fisherian significance testing is made by Jeffreys (1961). He said

What the use of the P implies, therefore, is that a hypothesis that may be true may be rejected because it has not predicted observable results that have not occurred.

Fisher (1956b) gave as an example of a pure test of significance the following by commenting on the work of astronomer J. Michell. Michell supposed that there were in all 1500 stars of the required magnitude and sought to calculate the probability, on the hypothesis that they are individually distributed at random, that any one of them should have five neighbors within a distance of $a$ minutes of arc from it. Fisher found the details of Michell’s calculation obscure, and suggested the following argument.
“The fraction of the celestial sphere within a circle of radius $a$ minutes is, to a satisfactory approximation,
$$
p=\left(\frac{a}{6875.5}\right)^{2}
$$
in which the denominator of the fraction within brackets is the number of minutes in two radians. So, if $a$ is 49 , the number of minutes from Maia to its fifth nearest neighbor, Atlas, we have
$$
p=\frac{1}{(140.316)^{2}}=\frac{1}{19689}
$$
Out of 1499 stars other than Maia of the requisite magnitude the expected number within this distance is therefore
$$
m=\frac{1499}{19689}=\frac{1}{13.1345}=.07613
$$
The frequency with which five stars should fall within the prescribed area is then given approximately by the term of the Poisson series
$$
e^{-m} \frac{m^{5}}{5 !}
$$
or, about 1 in $50,000,000$, the probabilities of having 6 or more close neighbors adding very little to this frequency. Since 1500 stars each have this probability of being the center of such a close cluster of 6 , although these probabilities are not strictly independent,

统计代写|统计推断作业代写statistical inference代考|FORMS OF THE LIKELIHOOD PRINCIPLE

The model for experiment $\mathcal{E}$ consists of a sample space $S$ and a parameter space $\Theta$, a measure $\mu$, and a family of probability functions $f: S \times \Theta \rightarrow R^{+}$such that for all $\theta \in \Theta$
$$
\int_{S} f d \mu=1
$$

Unrestricted LP (ULP)
For two such experiments modeled as $\mathcal{E}=(S, \mu, \Theta, f)$ and $\mathcal{E}^{\prime}=\left(S^{\prime}, \mu^{\prime}, \Theta, f^{\prime}\right)$, and for realization $D \in S$ and $D^{\prime} \in S^{\prime}$, if
$$
f(D \mid \theta)=g\left(D, D^{\prime}\right) f^{\prime}\left(D^{\prime} \mid \theta\right) \quad \text { for } g>0
$$
for all $\theta$ and the choice of $\mathcal{E}$ or $\mathcal{E}^{\prime}$ is uniformative with regard to $\theta$, then the evidence or inferential content or information concerning $\theta$, all denoted by Inf is such that
$$
\operatorname{Inf}(\mathcal{E}, D)=\operatorname{Inf}\left(\mathcal{E}^{\prime}, D^{\prime}\right)
$$
Note this implies that all of the statistical evidence provided by the data is conveyed by the likelihood function. There is an often useful extension, namely, when $\theta=\left(\theta_{1}, \theta_{2}\right), \delta_{1}=h\left(\theta_{1}, \theta_{2}\right), \delta_{2}=k\left(\theta_{1}, \theta_{2}\right)$, and
$$
f(D \mid \theta)=g\left(D, D^{\prime}, \delta_{1}\right) f^{\prime}\left(D^{\prime} \mid \delta_{2}\right)
$$
then $\operatorname{Inf}(\mathcal{E}, D)=\operatorname{Inf}\left(\mathcal{E}^{\prime}, D^{\prime}\right)$ concerning $\delta_{2}$.
Weakly restricted LP (RLP)
LP is applicable whenever a) $(S, \mu, \Theta, f)=\left(S^{\prime}, \mu^{\prime}, \Theta, f^{\prime}\right)$ and b) $(S, \mu, \Theta, f) \neq$ $\left(S^{\prime}, \mu^{\prime}, \Theta, f^{\prime}\right)$ when there are no structural features of $(S, \mu, \Theta, f)$ which have inferential relevance and which are not present in $\left(S^{\prime}, \mu^{\prime}, \Theta, f^{\prime}\right)$.
Strongly restricted LP (SLP)
Applicable only when $(S, \mu, \Theta, f)=\left(S^{\prime}, \mu^{\prime}, \Theta, f^{\prime}\right)$.
Extended $L P$ (ELP)
When $\theta=(p, \gamma)$ and $f\left(D, D^{\prime} \mid p, \gamma\right)=g\left(D, D^{\prime} \mid p\right) f^{\prime}\left(D^{\prime} \mid \gamma\right)$ it is plausible to extend LP to
$$
\operatorname{Inf}(\mathcal{E}, D)=\operatorname{Inf}\left(\mathcal{E}^{\prime}, D^{\prime}\right)
$$
concerning $p$ assuming $p$ and $\gamma$ are unrelated.

统计代写|统计推断作业代写statistical inference代考|LIKELIHOOD AND SIGNIFICANCE TESTING

We now compare the use of likelihood analysis with a significance test. Suppose we are only told that in a series of independent and identically distributed binary trials there were $r$ successes and $n-r$ failures, and the sampling was conducted in one of three ways:

The number of trials was fixed at $n$.
Sampling was stopped at the $r$ th success.
Sampling was stopped when $n-r$ failures were obtained.
Now while the three sampling probabilities differ they all have the same likelihood:
$$
L=p^{r}(1-p)^{n-r}
$$
The probabilities of $r$ successes and $n-r$ failures under these sampling methods are:
$$
\begin{aligned}
f_{n} &=\left(\begin{array}{l}
n \
r
\end{array}\right) L \quad r=0,1,2, \ldots, n \
f_{r} &=\left(\begin{array}{c}
n-1 \
r-1
\end{array}\right) L \quad n=r, r+1, \ldots \
f_{n-r} &=\left(\begin{array}{c}
n-1 \
n-r-1
\end{array}\right) L \quad n=r+1, r+2, \ldots
\end{aligned}
$$
where $f_{a}$ denotes the probability where subscript $a$ is fixed for sampling.

统计推断代考

统计代写|统计推断作业代写statistical inference代考|FURTHER REMARKS ON TESTS OF SIGNIFICANCE

显着性检验的主张适用于替代假设不合理的情况。请注意，拟合优度检验属于这一类，即数据是否符合正态分布？这里H0只是一个分布族，而不是参数值的规范。还要注意，显着性检验不仅考虑已发生的事件，而且基本上对未发生的更多差异事件给予同等权重，而不是仅考虑已发生事件的检验。
Jeffreys (1961) 对费希尔显着性检验提出了尖锐的批评。他说

因此，P 的使用意味着一个可能为真的假设可能会被拒绝，因为它没有预测到尚未发生的可观察结果。

费舍尔 (1956b) 通过评论天文学家 J. Michell 的工作，给出了以下作为纯显着性检验的示例。米歇尔假设在所有 1500 颗恒星中都有所需的星等，并试图计算概率，假设它们是随机分布的，其中任何一颗应该在距离为一种从它的弧分钟。费舍尔发现米歇尔计算的细节晦涩难懂，并提出了以下论点。
“天球在半径圆内的分数一种分钟是，一个令人满意的近似值，
p=(一种6875.5)2
其中括号内分数的分母是以两个弧度为单位的分钟数。因此，如果一种是 49 ，从 Maia 到它的第五个最近邻居 Atlas 的分钟数，我们有
p=1(140.316)2=119689
因此，在必要星等的玛雅以外的 1499 颗恒星中，该距离内的预期数量为
米=149919689=113.1345=.07613
然后，五颗星应该落在规定区域内的频率大约由泊松级数的项给出
和−米米55!
或者，大约 1 英寸50,000,000，有 6 个或更多近邻的概率对这个频率的影响很小。由于 1500 颗恒星每颗都有这个概率成为如此紧密的 6 星团的中心，尽管这些概率不是严格独立的，

统计代写|统计推断作业代写statistical inference代考|FORMS OF THE LIKELIHOOD PRINCIPLE

实验模型和由一个样本空间组成小号和参数空间θ，一种方法μ, 和一系列概率函数F:小号×θ→R+这样对于所有人θ∈θ
∫小号Fdμ=1

Unrestricted LP (ULP)
对于两个这样的实验，建模为和=(小号,μ,θ,F)和和′=(小号′,μ′,θ,F′)，并为实现D∈小号和D′∈小号′，如果
F(D∣θ)=G(D,D′)F′(D′∣θ) 为了 G>0
对全部θ和选择和或者和′是统一的θ，然后是有关的证据或推论内容或信息θ, 所有由 Inf 表示的都是这样的
信息⁡(和,D)=信息⁡(和′,D′)
请注意，这意味着数据提供的所有统计证据都由似然函数传达。有一个经常有用的扩展，即当θ=(θ1,θ2),d1=H(θ1,θ2),d2=ķ(θ1,θ2)，和
F(D∣θ)=G(D,D′,d1)F′(D′∣d2)
然后信息⁡(和,D)=信息⁡(和′,D′)关于d2.
弱限制 LP (RLP)
LP 适用于 a)(小号,μ,θ,F)=(小号′,μ′,θ,F′)b)(小号,μ,θ,F)≠ (小号′,μ′,θ,F′)当没有结构特征时(小号,μ,θ,F)哪些具有推理相关性，哪些不存在于(小号′,μ′,θ,F′).
严格限制 LP (SLP)
仅适用于(小号,μ,θ,F)=(小号′,μ′,θ,F′).
扩展大号磷(ELP)
什么时候θ=(p,C)和F(D,D′∣p,C)=G(D,D′∣p)F′(D′∣C)将 LP 扩展到
信息⁡(和,D)=信息⁡(和′,D′)
关于p假设p和C是无关的。

统计代写|统计推断作业代写statistical inference代考|LIKELIHOOD AND SIGNIFICANCE TESTING

我们现在将似然分析的使用与显着性检验进行比较。假设我们只被告知在一系列独立且同分布的二元试验中r成功和n−r失败，并以以下三种方式之一进行抽样：

试验次数固定为n.
采样停止在r成功。
采样停止时n−r失败了。
现在，虽然三个采样概率不同，但它们都具有相同的可能性：
大号=pr(1−p)n−r
的概率r成功和n−r这些抽样方法的失败是：
Fn=(n r)大号r=0,1,2,…,n Fr=(n−1 r−1)大号n=r,r+1,… Fn−r=(n−1 n−r−1)大号n=r+1,r+2,…
在哪里F一种表示下标的概率一种固定用于采样。

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

随机过程代考

贝叶斯方法代考

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。

机器学习代写

多元统计分析代考

基础数据: $N$ 个样本， $P$ 个变量数的单样本，组成的横列的数据表
变量定性: 分类和顺序；变量定量：数值
数学公式的角度分为: 因变量与自变量

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|统计推断作业代写statistical inference代考|A Forerunner

Posted on 2022年4月25日2022年4月25日 by statistics-lab

如果你也在怎样代写统计推断statistical inference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。

统计推断是利用数据分析来推断概率基础分布的属性的过程。推断性统计分析推断出人口的属性，例如通过测试假设和得出估计值。

我们提供的统计推断statistical inference及其相关学科的代写，服务范围广, 其中包括但不限于:

Statistical Inference 统计推断
Statistical Computing 统计计算
Advanced Probability Theory 高等楖率论
Advanced Mathematical Statistics 高等数理统计学
(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|统计推断作业代写statistical inference代考|A Forerunner

统计代写|统计推断作业代写statistical inference代考|PROBABILISTIC INFERENCE

An early use of inferred probabilistic reasoning is described by Rabinovitch $(1970)$.

In the Book of Numbers, Chapter 18 , verse 5 , there is a biblical injunction which enjoins the father to redeem his wife’s first-born male child by payment of five pieces of silver to a priest (Laws of First Fruits). In the 12 th Century the theologian, physician and philosopher, Maimonides addressed himself to the following problem with a solution. Suppose one or more women have given birth to a number of children and the order of birth is unknown, nor is it known how many children each mother bore, nor which child belongs to which mother. What is the probability that a particular woman bore boys and girls in a specified sequence? (All permutations are assumed equally likely and the chances of male or female births is equal.)

Maimonides ruled as follows: Two wives of different husbands, one primiparous $(P)$ (a woman who has given birth to her first child) and one not $(\bar{P})$. Let $H$ be the event that the husband of $P$ pays the priest. If they gave birth to two males (and they were mixed up), $P(H)=1-$ if they bore a male $(M)$ and a female $(F)-P(H)=0$ (since the probability is only $1 / 2$ that the primipara gave birth to the boy). Now if they bore 2 males and a female, $P(H)=1$.
$$
\begin{array}{cccccc}
\frac{\text { Case 1 }}{\mathrm{M}, \mathrm{M}} & \frac{(P)}{\mathrm{M}} & \frac{(\bar{P})}{\mathrm{M}} & \frac{\text { Case } 2}{\mathrm{M}, \mathrm{F}} & \frac{(P)}{\mathrm{M}} & \frac{(\bar{P})}{\mathrm{F}} \
& & & \mathrm{F} & \mathrm{M} \
P(H)=1 & & P(H)=\frac{1}{2}
\end{array}
$$

$P A Y M E N T$
$\begin{array}{ccccc}\text { Case 3 } & \frac{(P)}{(\bar{P})} & Y e s & N o & \ \mathrm{M}, \mathrm{M}, \mathrm{F} & \mathrm{M}, \mathrm{M} & \mathrm{F} & \sqrt{ } & \ \mathrm{F}, \mathrm{M} & \mathrm{M} & & \sqrt{ } \ \mathrm{M}, \mathrm{F} & \mathrm{M} & \sqrt{ } & \ \mathrm{F} & \mathrm{M}, \mathrm{M} & & \sqrt{ } \ \mathrm{M} & \mathrm{F}, \mathrm{M} & \sqrt{\mathrm{M}} & \ \mathrm{M} & \mathrm{M}, \mathrm{F} & \sqrt{ } & & P(H)=\frac{2}{3}\end{array}$
Maimonides ruled that the husband of $P$ pays in Case 3 . This indicates that a probability of $2 / 3$ is sufficient for the priest to receive his 5 pieces of silver but $1 / 2$ is not. This leaves a gap in which the minimum probability is determined for payment.

What has been illustrated here is that the conception of equally likely events, independence of events, and the use of probability in making decisions were not unknown during the 12 th century, although it took many additional centuries to understand the use of sampling in determining probabilities.

统计代写|统计推断作业代写statistical inference代考|TESTING USING RELATIVE FREQUENCY

One of the earliest uses of relative frequency to test a Hypothesis was made by Arbuthnot $(1710)$, who questioned whether the births were equally likely to be male or female. He had available the births from London for 82 years. In every year male births exceeded females. Then he tested the hypothesis that there is an even chance whether a birth is male or female or the probability $p=\frac{1}{2}$. Given this hypothesis he calculated the chance of getting all 82 years of male exceedances $\left(\frac{1}{2}\right)^{82}$. Being that this is basically infinitesimal, the hypothesis was rejected. It is not clear how he would have argued if some other result had occurred since any particular result is small-the largest for equal numbers of male and female exceedances is less than $\frac{1}{10}$.

统计代写|统计推断作业代写statistical inference代考|PRINCIPLES GUIDING FREQUENTISM

Classical statistical inference is based on relative frequency considerations. A particular formal expression is given by Cox and Hinkley (1974) as follows:

Repeated Sampling Principle. Statistical procedures are to be assessed by their behavior in hypothetical repetition under the same conditions.
Two facets:

Measures of uncertainty are to be interpreted as hypothetical frequencies in long run repetitions.

Criteria of optimality are to be formulated in terms of sensitive behavior in hypothetical repetitions.
(Question: What is the appropriate space which generates these hypothetical repetitions? Is it the sample space $S$ or some other reference set?)

Restricted (Weak) Repeated Sampling Principle. Do not use a procedure which for some possible parameter values gives, in hypothetical repetitions, misleading conclusions most of the time (too vague and imprecise to be constructive). The argument for repeated sampling ensures a physical meaning for the quantities we calculate and that it ensures a close relation between the analysis we make and the underlying model which is regarded as representing the “true” state of affairs.

An early form of frequentist inferences were Tests of Significance. They were long in use before their logical grounds were given by Fisher (1956b) and further elaborated by Barnard (unpublished lectures).
Prior assumption: There is a null hypothesis with no discernible alternatives. Features of a significance test (Fisher-Barnard)

A significance test procedure requires a reference set $R$ (not necessarily the entire sample space) of possible results comparable with the observed result $X=x_{0}$ which also belongs to $R$.
A ranking of all possible results in $R$ in order of their significance or meaning or departure from a null hypothesis $H_{0}$. More specifically we adopt a criterion $T(X)$ such that if $x_{1}>x_{2}$ (where $x_{1}$ departs further in rank than $x_{2}$ both elements of the reference set $R)$ then $T\left(x_{1}\right)>T\left(x_{2}\right)$ [if there is doubt about the ranking then there will be corresponding doubt about how the results of the significance test should be interpreted].
$H_{0}$ specifies a probability distribution for $T(X)$. We then evaluate the observed result $x_{0}$ and the null hypothesis.

统计推断代考

统计代写|统计推断作业代写statistical inference代考|PROBABILISTIC INFERENCE

Rabinovitch 描述了推断概率推理的早期使用(1970).

在《民数记》第 18 章第 5 节中，有一条圣经命令要求父亲通过向祭司支付五块银子来赎回他妻子的长子（初熟果实的法则）。在 12 世纪，神学家、医生和哲学家迈蒙尼德自己解决了以下问题。假设一个或多个女人生了几个孩子，出生顺序不知道，也不知道每个母亲生了多少孩子，也不知道哪个孩子属于哪个母亲。一个特定的女人以特定的顺序生男孩和女孩的概率是多少？（假设所有排列的可能性相同，并且男性或女性出生的机会相同。）

迈蒙尼德统治如下：两个不同丈夫的妻子，一个初产(磷)（生了第一个孩子的女人）和一个没有(磷¯). 让H成为丈夫的事件磷付钱给牧师。如果他们生了两个雄性（而且他们混在一起了），磷(H)=1−如果他们生了一个男性(米)和一个女性(F)−磷(H)=0（因为概率只有1/2初产妇生下了这个男孩）。现在，如果他们生了两男一女，磷(H)=1.
情况1 米,米(磷)米(磷¯)米案子 2米,F(磷)米(磷¯)F F米磷(H)=1磷(H)=12

$付款\ begin {array {ccccc} \ text {案例3} & \ frac {(P) {{(\ bar {P})} & Y es & N o & \ \ mathrm {M}, \ mathrm {M , \ mathrm {F} & \ mathrm {M}, \ mathrm {M} & \ mathrm {F} & \ sqrt {} & \ \ mathrm F}, \ mathrm {M} & \ mathrm {M} & & \ sqrt} \ \ mathrm {M}, \ mathrm {F} & \ mathrm {M} & \ sqrt {} & \ \ mathrm {F} & \ mathrm {M}, \ mathrm {M} & & \ sqrt } \ \ mathrm {M} & \ mathrm {F}, \ mathrm {M} & \ sqrt {\ mathrm {M}} & \ \ mathrm {M} & \ mathrm {M}, \ mathrm {F} & \ sqrt} & & P (H) = \ frac {2} {3} \ end {数组米一种一世米这n一世d和sr在l和d吨H一种吨吨H和H在sb一种nd这F磷p一种是s一世nC一种s和3.吨H一世s一世nd一世C一种吨和s吨H一种吨一种pr这b一种b一世l一世吨是这F2 / 3一世ss在FF一世C一世和n吨F这r吨H和pr一世和s吨吨这r和C和一世在和H一世s5p一世和C和s这Fs一世l在和rb在吨1 / 2$ 不是。这留下了确定付款的最小概率的空白。

这里说明的是，在 12 世纪，同等可能性事件的概念、事件的独立性和概率在决策中的使用并不为人所知，尽管要理解抽样在确定概率中的使用又花了很多个世纪。 .

统计代写|统计推断作业代写statistical inference代考|TESTING USING RELATIVE FREQUENCY

Arbuthnot 最早使用相对频率来检验假设之一(1710)，谁质疑出生的男性或女性的可能性是否相同。他已经在伦敦接生了 82 年。每年男性的出生人数都超过女性。然后他检验了一个假设，即出生是男性还是女性的概率或概率是均等的。p=12. 鉴于这个假设，他计算了获得所有 82 年男性超额成绩的机会(12)82. 由于这基本上是无穷小的，因此该假设被拒绝。尚不清楚如果出现了其他结果，他会如何辩解，因为任何特定结果都很小——男性和女性数量相同的最大超标小于110.

统计代写|统计推断作业代写statistical inference代考|PRINCIPLES GUIDING FREQUENTISM

经典统计推断基于相对频率考虑。Cox 和 Hinkley (1974) 给出了一个特定的正式表达式，如下所示：

重复抽样原则。统计程序将通过它们在相同条件下假设重复的行为来评估。
两个方面：

不确定性的度量将被解释为长期重复中的假设频率。
最优标准将根据假设重复中的敏感行为来制定。
（问题：产生这些假设重复的适当空间是什么？是样本空间吗？小号或其他一些参考集？）

受限（弱）重复采样原则。不要使用对于某些可能的参数值在大多数情况下在假设的重复中给出误导性结论的程序（太模糊和不精确而无法建设性）。重复抽样的论点确保了我们计算的数量的物理意义，并确保了我们所做的分析与被视为代表“真实”事态的基础模型之间的密切关系。

频率论推论的早期形式是意义测试。在 Fisher (1956b) 给出逻辑依据并由 Barnard 进一步阐述（未发表的讲座）之前，它们已经使用了很长时间。
先验假设：存在一个零假设，没有可辨别的替代方案。显着性检验的特征（Fisher-Barnard）

显着性检验程序需要参考集R（不一定是整个样本空间）与观察结果相当的可能结果X=X0这也属于R.
所有可能结果的排名R按照它们的重要性或意义或偏离原假设的顺序H0. 更具体地说，我们采用一个标准吨(X)这样如果X1>X2（在哪里X1排名比离得更远X2参考集的两个元素R)然后吨(X1)>吨(X2)[如果对排名有疑问，那么对显着性检验的结果应该如何解释也会有相应的疑问]。
H0指定概率分布吨(X). 然后我们评估观察到的结果X0和原假设。

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

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