# 统计代写|贝叶斯统计代写Bayesian statistics代考|FNR6560

## 统计代写|贝叶斯统计代写Bayesian statistics代考|INTERVAL HYPOTHESES

The previous considerations show that, of two normal hypotheses, that with the smaller variance is both simpler and better supported by ‘equally agreeing’ outcomes. The same must then be true if we are comparing, say, two interval hypotheses about a binomial parameter $p$, where agreement is measured in standard deviation units. Here I am thinking of connected (or ‘one-sided’) interval hypotheses, but let us see what happens when a central connected interval hypothesis is compared to the two-sided alternative.

As a special case, consider the test of a simple hypothesis, say $p=0.5$, against the two-sided (composite) alternative $p \neq 0.5$. The former, a point hypothesis, is obviously ‘simpler’ in my sense as well (a sort of pre-established harmony of usage). Assume for convenience that the prior distribution of $p$ is uniform, or, at any rate, uniform over values of $p \neq 0.5$. The aver age likelihood of the composite hypothesis, $p \neq 0.5$, is then
$$\int_0^1 p^x(1-p)^{n-x} \mathrm{~d} p=x !(n-x) ! /(n+1) !$$
omitting the binomial coefficient, and hence the ratio of the average likelihoods (the Bayes factor) in favor of $p \neq 0.5$ is $2^n x !(n-x) ! /(n+1) !$, which exceeds unity iff
(6.1) $\quad\left(\begin{array}{l}n \ x\end{array}\right)<2^n /(n+1)$,
i.e., iff the observed binomial coefficient, $\left(\begin{array}{l}n \ x\end{array}\right)$, is smaller than the average of the binomial coefficients $\left(\begin{array}{c}n \ i\end{array}\right), i=0,1, \ldots, n$. Consequently, the simpler hypothesis, $p=0.5$, is better supported, except at outcomes which lie far out in the tails of the binomial distribution based on $p=0.5$. This test has the desirable property that its power decreases systematically with sample size.

Consider next a test of an interval hypothesis, say $0.49 \leqslant p \leqslant 0.51$, against the two-sided composite alternative, $p<0.49$ or $p>0.51$. The ‘borderline’ outcome $x=$ nearest integer to $0.49 n$ agrees about equally well with both hypotheses, and so we must ask: which hypothesis is favored by this outcome?

## 统计代写|贝叶斯统计代写Bayesian statistics代考|OTHER APPROACHES

I was led to my own conception of simplicity as sample coverage by attempting to find what the logic of significance tests (cf. Chapter 9) has in common with the paucity-of-parameters criterion and the intuition that the simplest ‘strong generalization’ compatible with a sample is that which designates all and only those kinds present in the sample. My conception also has an obvious affinity with that of Sir Karl Popper, an affinity we explore in Section 4 of the next chapter. There have been a great many attempts to solve the problem of simplicity as here understood; some of the early work is critically discussed in Ackermann (1961).

Although, for a time, simplicity was given up for dead (it was relegated to the limbo of the ‘merely aesthetic’), there has been, in recent years, a strong revival of interest in the problem. It would be impossible to do justice to this ferment of ideas in a reasonably short space, but mention should at least be made of the works of Good, Goodman, Friedman, Sober and Kemeny cited in the references to this chapter. Too, there is work in progress on the fascinating concept of Kolmogoroff complexity (cf. Fine (1973), Chapter 5).
The focus of much of this work is a notion of descriptive simplicity. The ‘simplest theory’ in the descriptive sense is, roughly speaking, that which encodes the data in the most efficient way. Some fleeting contact will be made with this notion in Chapter 8 ; the question of its connection, if any, with the notion of inductive simplicity at issue in this chapter is, at any rate, deserving of attention. Given its intimate connection with evidential support, the concept of simplicity developed here is at least a plausible candidate for the title ‘inductive simplicity’. Evidence that scientists have tended to use ‘simplicity’ in something very close to my sense will be marshalled with reference to an important historical example in Chapter 7.

# 贝叶斯统计代考

## 统计代写|贝叶斯统计代写Bayesian statistics代考|INTERVAL HYPOTHESES

$$\int_0^1 p^x(1-p)^{n-x} \mathrm{~d} p=x !(n-x) ! /(n+1) !$$

(6.1) $(n x)<2^n /(n+1)$, 即，如果观察到的二项式系数， $(n x)$ ，小于二项式系数的平均值 $(n i), i=0,1, \ldots, n$. 因此，更简单 的假设， $p=0.5$ ，得到更好的支持，除了在基于二项分布尾部的结果 $p=0.5$. 该检验具有理想的特性， 即它的功效会随着样本量的增加而系统地降低。 接下来考虑一个区间假设的检验，比如说 $0.49 \leqslant p \leqslant 0.51$ ，针对两侧复合替代方案 $p<0.49$ 或者 $p>0.51$. “边缘”结果 $x=$ 最接近的整数 $0.49 n$ 与这两个假设都一致，因此我们必须问: 这个结果支持哪 个假设?

## 统计代写|贝叶斯统计代写Bayesian statistics代考|OTHER APPROACHES

myassignments-help数学代考价格说明

1、客户需提供物理代考的网址，相关账户，以及课程名称，Textbook等相关资料~客服会根据作业数量和持续时间给您定价~使收费透明，让您清楚的知道您的钱花在什么地方。

2、数学代写一般每篇报价约为600—1000rmb，费用根据持续时间、周作业量、成绩要求有所浮动(持续时间越长约便宜、周作业量越多约贵、成绩要求越高越贵)，报价后价格觉得合适，可以先付一周的款，我们帮你试做，满意后再继续，遇到Fail全额退款。

3、myassignments-help公司所有MATH作业代写服务支持付半款，全款，周付款，周付款一方面方便大家查阅自己的分数，一方面也方便大家资金周转，注意:每周固定周一时先预付下周的定金，不付定金不予继续做。物理代写一次性付清打9.5折。

Math作业代写、数学代写常见问题

myassignments-help擅长领域包含但不是全部: