assignmentutor™您的专属作业导师

assignmentutor-lab™ 为您的留学生涯保驾护航 在代写贝叶斯统计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。我们的专家在代写贝叶斯统计代写方面经验极为丰富，各种代写贝叶斯统计相关的作业也就用不着说。

• 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|贝叶斯统计代写Bayesian statistics代考|UNIFORMITY AND DISCOVERY

The first line of defense against the charge that the induction rule fails to take cognizance of the mediating role of background theory is to insist that this knowledge must itself have been gleaned inductively via the rule. We saw that this contention is irrelevant, for even if it could be sustained, ‘rich’ contexts would have to be dealt with. But in this section we propose to scrutinize more fully the heuristic status of the rule, the claim that it alone enables us to uncover lawful regularity.

Given a table of data – mapping, e.g., the number of primes less than $x$ against $x$, or the sum of the first $n$ cubes against $n$ – the induction rule not only fails to determine which of many discernible patterns to extrapolate, it does not even serve to detect pattern. It is absurd to depict the discovery of, say, the Prime Number Theorem or the formula for the sum of the first $n$ cubes, as the result of extrapolating observed frequencies or concatenations of events. The whole difficulty in such problems is to find a simple formula that fits the instances already examined. Once the relationship has been found, its extrapolation is virtually automatic. The point is that the discovery of a law and its extrapolation are two separate problems. At best, the induction rule is pertinent to the second of these.

The deterministic or indeterministic character of the process under observation has no bearing at all on this point. Recursive sequences, like arithmetic progressions, are, I would suppose, deterministic paradigms. Yet, given the first $n$ terms of such a sequence, e.g. $4,12,29,57,98, \ldots$, the problem of finding the recursive equations (or a formula for the general term) is no less acute for knowing such equations exist. It is not even assured that the correct equations will become more apparent as the number of terms is increased. A similar point applies to Reichenbach’s discussion of the straight rule: knowing that sample proportions converge in the limit to the population proportion is no guarantee that the former will provide ever better estimates of the latter. We can only infer that this is true beyond some point. But that is no help unless we know which point. Similarly, if we knew that invalidity of first order formulae always shows up in a finite domain, that would not issue in a decision procedure unless we could set upper bounds! Nor would knowledge that at most finitely many causes govern an effect help us to isolate the ‘active’ cause. In any case, Reichenbach’s argument for the straight rule founders on the fact that infinitely many other estimation rules (indeed, all those of Carnap’s $\lambda$-continuum) are asymptotically indistinguishable from it.

## 统计代写|贝叶斯统计代写Bayesian statistics代考|CONDITIONALIZATION

There is no sound principle of induction as traditionally understood. Induction, or learning from experience, is just the process of revising probability assessments in the light of additional information. This is done by Bayes’ rule:
(3.1) $\quad P(H / x)=P(x / H) P(H) / P(x)$,
where $H$ is an hypothesis and $x$ an experimental outcome or datum. The problem of justifying induction is thus, in the first place, the problem of justifying conditionalization or Bayes’ rule. Many such justifications have been given ${ }^7$, but $I$ want to sketch a particularly intuitive and compelling one here.

EXAMPLE 1. We are given an urn containing three balls, each of them either black or white. Label the four possible compositions $0,1,2,3$ indicating the number of black balls, and let $H_i$ be the hypothesis that composition $i$ obtains, $i=0,1,2,3$. Initially, let us suppose, the hypotheses are equiprobable. The contents of the urn are examined by a stooge who reports only that there is at least one black ball, thereby excluding $H_0$. What new probabilities should be assigned our hypotheses in the light of this information $E$ ?

Well, clearly, $P\left(H_0 / E\right)=0$, since $E$ excludes $H_0$. While $E$ excludes $H_0$, it does no more than this, and so has no bearing on the relative probabilities of the three non-excluded hypotheses. Since they were equiprobable prior to receipt of $E$, they should remain equiprobable. However, probabilities must sum to one, and so $P\left(H_i / E\right)=1 / 3, i=1,2,3$.

Essentially the same reasoning shows that, where initial probabilities are unequal, an outcome $E$ which logically excludes $H_0$ but no more, should make the probability of $H_0$ zero and leave the relative probabilities of the non-excluded hypotheses unchanged. Thus, there is a constant of proportionality, $c$, such that $P\left(H_i / E\right)=c P\left(H_i\right), i=1,2,3$, and determined from the requirement that the new probabilities must continue to sum to one (i.e., $c$ is a normalization constant). Solving $c P\left(H_1\right)+c P\left(H_2\right)+c P\left(H_3\right)=1$ for $c$ gives $c=1 /\left(P\left(H_1\right)+P\left(H_2\right)+P\left(H_3\right)\right)=1 /\left(1-P\left(H_0\right)\right)$. To say $E$ logically excludes $H_0$ and no more, is to identify $E$ with the denial of $H_0$, in which case our equation for $c$ becomes $c=1 / P(E)$. Then the equation $P\left(H_i / E\right)=c P\left(H_i\right)$ reduces to $P\left(H_i / E\right)=P\left(H_i\right) / P(E)$, the special case $P\left(E / H_i\right)=1$ of Bayes’ rule. The probability $P\left(E / H_i\right)$ is conditional on an hypothesis; it is not ‘conditional’ in the sense of ‘revised’. And these probabilities conditional on an hypothesis are easily computed for our problem. Given that $H_i$ is true, for $i=1$, 2, or 3 , the probability is one that the contents of the urn include at least one black ball. I.e., $P\left(E / H_i\right)$ is indeed equal to one.

# 贝叶斯统计代写

## 统计代写|贝叶斯统计代写Bayesian statistics代考|CONDITIONALIZATION

(3.1)磷(H/X)=磷(X/H)磷(H)/磷(X),

## 有限元方法代写

assignmentutor™作为专业的留学生服务机构，多年来已为美国、英国、加拿大、澳洲等留学热门地的学生提供专业的学术服务，包括但不限于Essay代写，Assignment代写，Dissertation代写，Report代写，小组作业代写，Proposal代写，Paper代写，Presentation代写，计算机作业代写，论文修改和润色，网课代做，exam代考等等。写作范围涵盖高中，本科，研究生等海外留学全阶段，辐射金融，经济学，会计学，审计学，管理学等全球99%专业科目。写作团队既有专业英语母语作者，也有海外名校硕博留学生，每位写作老师都拥有过硬的语言能力，专业的学科背景和学术写作经验。我们承诺100%原创，100%专业，100%准时，100%满意。

## MATLAB代写

MATLAB 是一种用于技术计算的高性能语言。它将计算、可视化和编程集成在一个易于使用的环境中，其中问题和解决方案以熟悉的数学符号表示。典型用途包括：数学和计算算法开发建模、仿真和原型制作数据分析、探索和可视化科学和工程图形应用程序开发，包括图形用户界面构建MATLAB 是一个交互式系统，其基本数据元素是一个不需要维度的数组。这使您可以解决许多技术计算问题，尤其是那些具有矩阵和向量公式的问题，而只需用 C 或 Fortran 等标量非交互式语言编写程序所需的时间的一小部分。MATLAB 名称代表矩阵实验室。MATLAB 最初的编写目的是提供对由 LINPACK 和 EISPACK 项目开发的矩阵软件的轻松访问，这两个项目共同代表了矩阵计算软件的最新技术。MATLAB 经过多年的发展，得到了许多用户的投入。在大学环境中，它是数学、工程和科学入门和高级课程的标准教学工具。在工业领域，MATLAB 是高效研究、开发和分析的首选工具。MATLAB 具有一系列称为工具箱的特定于应用程序的解决方案。对于大多数 MATLAB 用户来说非常重要，工具箱允许您学习应用专业技术。工具箱是 MATLAB 函数（M 文件）的综合集合，可扩展 MATLAB 环境以解决特定类别的问题。可用工具箱的领域包括信号处理、控制系统、神经网络、模糊逻辑、小波、仿真等。

assignmentutor™您的专属作业导师
assignmentutor™您的专属作业导师