### 统计代写|似然估计作业代写Probability and Estimation代考|Test of Significance

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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|似然估计作业代写Probability and Estimation代考|Test of Significance

For the test of $\mathcal{H}{\mathrm{o}}: \boldsymbol{\theta}{2}=\mathbf{0}$ vs. $\mathcal{H}{\mathrm{A}}: \boldsymbol{\theta}{2} \neq \mathbf{0}$, we consider the statistic $\mathcal{L}{n}$ given by \begin{aligned} \mathcal{L}{n} &=\frac{1}{\sigma^{2}} \tilde{\boldsymbol{\theta}}{2 n}^{\top} \boldsymbol{N}{2} \tilde{\boldsymbol{\theta}}{2 n}, \quad \text { if } \sigma^{2} \text { is known } \ &=\frac{1}{p{2} s_{n}^{2}} \tilde{\boldsymbol{\theta}}{2 n}^{\top} \boldsymbol{N}{2} \tilde{\boldsymbol{\theta}}{2 n}, \quad \text { if } \sigma^{2} \text { is unknown. } \end{aligned} Under a null-hypothesis $\mathcal{H}{0}$, the null distribution of $\mathcal{L}{n}$ is the central $\chi^{2}$ distribution with $p{2}$ DF. when $\sigma^{2}$ is known and the central $F$-distribution with $\left(p_{2}, m\right)$ DE in the case of $\sigma^{2}$ being unknown, respectively. Under the alternative hypothesis, $\mathcal{H}{A}$, the test statistics $\mathcal{L}{n}$ follows the noncentral version of the mentioned densities. In both cases, the noncentrality parameter is $\Delta^{2}=\theta_{2}^{\top} N_{2} \theta_{2} / \sigma^{2}$. In this paper, we always assume that $\sigma^{2}$ is known, then $\mathcal{L}{n}$ follows a chi-square distribution with $p{2} \mathrm{DF}$.
Further, we note that
$$\tilde{\theta}{j n} \sim \mathcal{N}\left(\theta{j}, \sigma^{2} n_{j}^{-1}\right), \quad j=1, \ldots, p$$
so that $\left.\mathcal{Z}{j}=\sqrt{n{j}} \tilde{\theta}{j n} / \sigma \sim \mathcal{N}^{(} \Delta{j}, 1\right)$, where $\Delta_{j}=\sqrt{n_{j}} \theta_{j} / \sigma$. Thus, one may use $\mathcal{Z}{j}$ to test the null-hypothesis $\mathcal{H}{\mathrm{o}}^{(j)}: \theta_{j}=0$ vs. $\mathcal{H}{\mathrm{A}}^{(j)}: \theta{j} \neq 0, j=p_{1}+1, \ldots, p$.

In this chapter, we are interested in studying three penalty estimators, namely,
(i) the subset rule called “hard threshold estimator” (HTE),
(ii) LASSO or the “soft threshold estimator” (STE),
(iii) the “ridge regression estimator” (RRE),
(iv) the classical PTE and shrinkage estimators such as “Stein estimator” (SE) and “positive-rule Stein-type estimator” (PRSE).

## 统计代写|似然估计作业代写Probability and Estimation代考|Penalty Estimators

In this section, we discuss the penalty estimators. Define the HTE as
\begin{aligned} \hat{\boldsymbol{\theta}}{n}^{\mathrm{HT}}(\kappa) &=\left(\tilde{\theta}{j n} I\left(\left|\tilde{\theta}{j n}\right|>\kappa \sigma n{j}^{-\frac{1}{2}}\right) \mid j=1, \ldots, p\right)^{\top} \ &=\left(\sigma n_{j}^{-\frac{1}{2}} \mathcal{Z}{j} I\left(\left|\mathcal{Z}{j}\right|>\kappa\right) \mid j=1, \ldots, p\right)^{\top} \end{aligned}
where $\kappa$ is a positive threshold parameter.
This estimator is discrete in nature and may be extremely variable and unstable due to the fact that small changes in the data can result in very different models and can reduce the prediction accuracy. As such we obtain the continuous version of $\hat{\theta}{n}^{\mathrm{HT}}(\kappa)$, and the LASSO is defined by $$\hat{\boldsymbol{\theta}}{n}^{\mathrm{L}}(\lambda)=\operatorname{argmin}{\theta}(\boldsymbol{Y}-\boldsymbol{B} \boldsymbol{\theta})^{\top}(\boldsymbol{Y}-\boldsymbol{B} \theta)+2 \lambda \sigma \sum{j=1}^{p} \sqrt{n_{j}} \kappa\left|\theta_{j}\right|,$$
where $|\theta|=\left(\left|\theta_{1}\right|, \ldots,\left|\theta_{p}\right|\right)^{\mathrm{T}}$, yielding the equation
$$\boldsymbol{B}^{\top} \boldsymbol{B} \boldsymbol{\theta}-\boldsymbol{B}^{\top} \boldsymbol{Y}+\lambda \sigma \boldsymbol{N}^{\frac{1}{2}} \operatorname{sgn}(\boldsymbol{\theta})=\mathbf{0}$$
or
$$\hat{\boldsymbol{\theta}}{n}^{\mathrm{L}}(\lambda)-\tilde{\boldsymbol{\theta}}{n}+\frac{1}{2} \lambda \sigma \boldsymbol{N}^{-\frac{1}{2}} \operatorname{sgn}\left(\hat{\boldsymbol{\theta}}{n}^{\mathrm{L}}(\lambda)\right)=\mathbf{0} .$$ Now, the $j$ th component of (3.5) is given by $$\hat{\theta}{j n}^{\mathrm{L}}(\lambda)-\ddot{\theta}{j n}+\lambda \sigma n{j}^{-\frac{1}{2}} \operatorname{sgn}\left(\hat{\theta}{j m}^{\mathrm{L}}(\lambda)\right)=0 .$$ Then, we consider three cases: (i) $\operatorname{sgn}\left(\hat{\theta}{j n}^{L}(\lambda)\right)=+1$, then, (3.6) reduces to
$$0<\frac{\hat{\theta}{j n}^{\mathrm{L}}(\lambda)}{\sigma n{j}^{-\frac{1}{2}}}-\frac{\tilde{\theta}{j n}}{\sigma n{j}^{-\frac{1}{2}}}+\lambda=0 .$$ Hence, $$0<\hat{\theta}{j n}^{L}(\lambda)=\sigma n{j}^{-\frac{1}{2}}\left(Z_{j}-\lambda\right)=\sigma n_{j}^{-\frac{1}{2}}\left(\left|Z_{j}\right|-\lambda\right),$$ with, clearly, $\mathcal{Z}{j}>0$ and $\left|\mathcal{Z}{j}\right|>\lambda$.

## 统计代写|似然估计作业代写Probability and Estimation代考|Preliminary Test and Stein-Type Estimators

We recall that the unrestricted estimator of $\theta=\left(\theta_{1}^{\top}, \theta_{2}^{\top}\right)^{\top}$ is given by $\left(\tilde{\theta}{1 n}^{\top}, \tilde{\theta}{2 n}^{\top}\right)^{\top}$ with marginal distribution $\tilde{\theta}{1 n} \sim \mathcal{N}{p_{1}}\left(\theta_{1}, \sigma^{2} N_{1}^{-1}\right)$ and $\tilde{\theta}{2 n} \sim \mathcal{N}{p_{2}}\left(\theta_{2}, \sigma^{2} N_{2}^{-1}\right)$, respectively. The restricted estimator of $\left(\theta_{1}^{\top}, \mathbf{0}^{\top}\right)^{\top}$ is $\left(\tilde{\theta}{1 n}^{\top}, \mathbf{0}^{\top}\right)^{\top}$. Similarly, the PTE of $\theta$ is given by $$\hat{\theta}{n}^{\mathrm{PT}}(\alpha)=\left(\begin{array}{c} \tilde{\theta}{1 n} \ \tilde{\theta}{2 n} I\left(\mathcal{L}{n}>c{\alpha}\right) \end{array}\right),$$
where $I(A)$ is the indicator function of the set $A, \mathcal{L}{n}$ is the test statistic given in Section $2.2$, and $c{\alpha}$ is the $\alpha$-level critical value.
Similarly, the Stein estimator (SE) is given by
$$\hat{\boldsymbol{\theta}}{n}^{\mathrm{S}}=\left(\begin{array}{c} \tilde{\boldsymbol{\theta}}{1 n} \ \tilde{\boldsymbol{\theta}}{2 n}\left(1-\left(p{2}-2\right) \mathcal{L}{n}^{-1}\right) \end{array}\right), \quad p{2} \geq 3$$

and the positive-rule Stein-type estimator (PRSE) is given by
$$\hat{\boldsymbol{\theta}}{n}^{\mathrm{S+}}=\left(\begin{array}{c} \tilde{\boldsymbol{\theta}}{1 n} \ \hat{\theta}{2 n}^{\mathrm{S}} I\left(\mathcal{L}{n}>p_{2}-2\right) \end{array}\right)$$

## 统计代写|似然估计作业代写Probability and Estimation代考|Test of Significance

θ~jn∼ñ(θj,σ2nj−1),j=1,…,p

（i）称为“硬阈值估计器”（HTE）的子集规则，
（ii）LASSO 或“软阈值估计器”（STE），
（iii） “岭回归估计器”（RRE），
（iv）经典的 PTE 和收缩估计器，例如“Stein 估计器”（SE）和“正规则 Stein 型估计器”（PRSE）。

## 统计代写|似然估计作业代写Probability and Estimation代考|Penalty Estimators

θ^nH吨(ķ)=(θ~jn一世(|θ~jn|>ķσnj−12)∣j=1,…,p)⊤ =(σnj−12从j一世(|从j|>ķ)∣j=1,…,p)⊤

θ^n大号(λ)−θ~n+12λσñ−12sgn⁡(θ^n大号(λ))=0.现在j(3.5) 的第 th 个分量由下式给出θ^jn大号(λ)−θ¨jn+λσnj−12sgn⁡(θ^j米大号(λ))=0.然后，我们考虑三种情况：（i）sgn⁡(θ^jn大号(λ))=+1，那么，(3.6) 式简化为
0<θ^jn大号(λ)σnj−12−θ~jnσnj−12+λ=0.因此，0<θ^jn大号(λ)=σnj−12(从j−λ)=σnj−12(|从j|−λ),显然，从j>0和|从j|>λ.

## 统计代写|似然估计作业代写Probability and Estimation代考|Preliminary Test and Stein-Type Estimators

θ^n小号=(θ~1n θ~2n(1−(p2−2)大号n−1)),p2≥3

θ^n小号+=(θ~1n θ^2n小号一世(大号n>p2−2))

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