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

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

## 统计代写|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
$$E(T(D)-\alpha \mid t(D)=t, \quad \theta \in w)=0 \quad \text { or } \quad E(T(D) \mid t(D)=t, \quad \theta \in w)=\alpha \text { or NS. }$$

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

1−b吨(θ)≤一种 为了 θ∈θH 1−b吨(θ)≥一种 为了 θ∈θķ

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

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

## 广义线性模型代考

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## MATLAB代写

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