### 统计代写|似然估计作业代写Probability and Estimation代考|Location and Simple Linear Models

statistics-lab™ 为您的留学生涯保驾护航 在代写似然估计Probability and Estimation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。我们的专家在代写似然估计Probability and Estimation代写方面经验极为丰富，各种代写似然估计Probability and Estimation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|似然估计作业代写Probability and Estimation代考|Location Model

In this section, we introduce two basic penalty estimators, namely, the ridge regression estimator (RRE) and the least absolute shrinkage and selection operator (LASSO) estimator for the location parameter of a distribution. The penalty estimators have become viral in statistical literature. The subject evolved as the solution to ill-posed problems raised by Tikhonov (1963) in mathematics. In 1970 , Hoerl and Kennard applied the Tikhonov method of solution to obtain the $R R E$ for linear models. Further, we compare the estimators with the LSE in terms of $\mathrm{L}_{2}$-risk function.

Consider the simple location model,
$$Y=\theta \mathbf{1}{n}+\boldsymbol{\epsilon}, \quad n>1,$$ where $Y=\left(Y{1}, \ldots, Y_{n}\right)^{\top}, \mathbf{1}{n}=(1, \ldots, 1)^{\top}=n$-tuple of 1’s, and $\boldsymbol{\epsilon}=\left(\varepsilon{1}, \ldots, \epsilon_{n}\right)^{\top}$ $n$-vector of i.i.d. random errors such that $\mathbb{E}(\boldsymbol{\epsilon})=\mathbf{0}$ and $\mathbb{E}\left(\boldsymbol{\epsilon} \boldsymbol{\epsilon}^{\top}\right)=\sigma^{2} \boldsymbol{I}{n}, \boldsymbol{I}{n}$ is the identity matrix of rank $n(n \geq 2), \theta$ is the location parameter, and, in this case, $\sigma^{2}$ may be unknown.
The LSE of $\theta$ is obtained by
$$\min {\theta}\left{\left(Y-\theta \mathbf{1}{n}\right)^{\top}\left(\boldsymbol{Y}-\theta \mathbf{1}{n}\right)\right}=\tilde{\theta}{n}=\frac{1}{n} \mathbf{1}{n}^{\top} \boldsymbol{Y}=\bar{y} .$$ Alternatively, it is possible to minimize the log-likelihood function when the errors are normally distributed: $$l(\theta)=-n \log \sigma+\frac{n}{2} \log (2 \pi)-\frac{1}{2}\left(Y-\theta \mathbf{1}{n}\right)^{\top}\left(\boldsymbol{Y}-\theta \mathbf{1}{n}\right),$$ giving the same solution (2.4) as in the case of LSE. It is known that the $\ddot{\theta}{n}$ is unbiased, i.e. $\mathbb{E}\left(\tilde{\theta}{n}\right)=\theta$ and the variance of $\tilde{\theta}{n}$ is given by
$$\operatorname{Var}\left(\tilde{\theta}{n}\right)=\frac{\sigma^{2}}{n} .$$ The unbiased estimator of $\sigma^{2}$ is given by $$s{n}^{2}=(n-1)^{-1}\left(Y-\tilde{\theta}{n} \mathbf{1}{n}\right)^{\top}\left(Y-\tilde{\theta}{n} \mathbf{1}{n}\right) .$$
The mean squared error (MSE) of $\theta_{n}^{}$, any estimator of $\theta$, is defined as $$\operatorname{MSE}\left(\theta_{n}^{}\right)=\mathbb{E}\left(\theta_{n}^{*}-\theta\right)^{2} .$$
Test for $\theta=0$ when $\sigma^{2}$ is known:
For the test of null-hypothesis $\mathcal{H}{\mathrm{o}}: \theta=0$ vs. $\mathcal{H}{\mathrm{A}}: \theta \neq 0$, we use the test statistic
$$Z_{n}=\frac{\sqrt{n} \tilde{\theta}_{n}}{\sigma}$$

Under the assumption of normality of the errors, $Z_{n} \sim \mathcal{N}(\Delta, 1)$, where $\Delta=\frac{\sqrt{n} \theta}{\sigma}$. Hence, we reject $\mathcal{H}{\mathrm{o}}$ whenever $\left|Z{n}\right|$ exceeds the threshold value from the null distribution. An interesting threshold value is $\sqrt{2 \log 2}$.

For large samples, when the distribution of errors has zero mean and finite variance $\sigma^{2}$, under a sequence of local alternatives,
$$\mathcal{K}{(n)}: \theta{(n)}=n^{-\frac{1}{2}} \delta, \quad \delta \neq 0,$$
and assuming $\mathbb{E}\left(\epsilon_{j}\right)=0$ and $\mathbb{E}\left(\epsilon_{j}^{2}\right)=\sigma^{2}(<\infty), j=1, \ldots, n$, the asymptotic distribution of $\sqrt{n} \tilde{\theta}{n} / s{n}$ is $\mathcal{N}(\Delta, 1)$. Then the test procedure remains the same as before.

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

In this section, we consider a shrinkage estimator of the location parameter $\theta$ of the form
$$\hat{\theta}{n}^{\text {Shrinkage }}(c)=c \tilde{\theta}{n}, \quad 0 \leq c \leq 1,$$
where $\tilde{\theta}{n} \sim \mathcal{N}\left(\theta, \sigma^{2} / n\right)$. The bias and the $\operatorname{MSE}$ of $\hat{\theta}{n}^{\text {Shrinkage }}(c)$ are given by
\begin{aligned} \mathrm{b}\left(\hat{\theta}{n}^{\text {Shrinkage }}(c)\right) &=\mathbb{E}\left[c \tilde{\theta}{n}\right]-\theta=c \theta-\theta=-(1-c) \theta \ \operatorname{MSE}\left(\hat{\theta}{n}^{\text {Shrinkage }}(c)\right) &=c^{2} \frac{\sigma^{2}}{n}+(1-c)^{2} \theta^{2} \end{aligned} Minimizing $\operatorname{MSE}\left(\hat{\theta}{n}^{\text {Shrinkage }}(c)\right)$ w.r.t. $c$, we obtain
$$c^{}=\frac{\Delta^{2}}{1+\Delta^{2}}, \quad \Delta^{2}=\frac{n \theta^{2}}{\sigma^{2}}$$ So that $$\operatorname{MSE}\left(\hat{\theta}{n}^{\text {Shrinkage }}\left(c^{}\right)\right)=\frac{\sigma^{2}}{n} \frac{\Delta^{2}}{1+\Delta^{2}} .$$
Thus, $\operatorname{MSE}\left(\hat{\theta}{n}^{\text {Shrinkage }}\left(c^{}\right)\right)$ is an increasing function of $\Delta^{2} \geq 0$ and the relative efficiency (REff) of $\hat{\theta}{n}^{\text {Shrinkage }}\left(c^{}\right)$ compared to $\tilde{\theta}{n}$ is
$$\operatorname{REff}\left(\hat{\theta}{n}^{\text {Shrinkage }}\left(c^{}\right) ; \tilde{\theta}{n}\right)=1+\frac{1}{\Delta^{2}}, \quad \Delta^{2} \geq 0 .$$
Further, the MSE difference is
$$\frac{\sigma^{2}}{n}-\frac{\sigma^{2}}{n} \frac{\Delta^{2}}{1+\Delta^{2}}=\frac{\sigma^{2}}{n} \frac{1}{1+\Delta^{2}} \geq 0, \quad \forall \Delta^{2} \geq 0 .$$
Hence, $\hat{\theta}{n}^{\text {Shrinkage }}\left(c^{}\right)$ outperforms the $\tilde{\theta}{n}$ uniformly.

## 统计代写|似然估计作业代写Probability and Estimation代考|Ridge Regression–Type Estimation of Location Parameter

Consider the problem of estimating $\theta$ when one suspects that $\theta$ may be 0 . Then following Hoerl and Kennard (1970), if we define
$$\tilde{\theta}{n}^{\mathrm{R}}(k)=\operatorname{argmin}{\theta}\left{\left|Y-\theta \mathbf{1}{n}\right|^{2}+n k \theta^{2}\right}, \quad k \geq 0 .$$ Then, we obtain the ridge regression-type estimate of $\theta$ as $$n \theta+n k \theta=n \tilde{\theta}{n}$$
or
$$\tilde{\theta}{n}^{\mathrm{R}}(k)=\frac{\tilde{\theta}{n}}{1+k}$$
Note that it is the same as taking $c=1 /(1+k)$ in (2.8).
Hence, the bias and MSE of $\theta_{n}^{\mathrm{R}}(k)$ are given by
$$b\left(\tilde{\theta}{n}^{\mathrm{R}}(k)\right)=-\frac{k}{1+k} \theta$$ and \begin{aligned} \operatorname{MSE}\left(\tilde{\theta}{n}^{\mathrm{R}}(k)\right) &=\frac{\sigma^{2}}{n(1+k)^{2}}+\frac{k^{2} \theta^{2}}{(1+k)^{2}} \ &=\frac{\sigma^{2}}{n(1+k)^{2}}\left(1+k^{2} \Delta^{2}\right) \end{aligned}
It may be seen that the optimum value of $k$ is $\Delta^{-2}$ and MSE at (2.18) equals
$$\frac{\sigma^{2}}{n}\left(\frac{\Delta^{2}}{1+\Delta^{2}}\right)$$
Further, the MSE difference equals
$$\frac{\sigma^{2}}{n}-\frac{\sigma^{2}}{n} \frac{\Delta^{2}}{1+\Delta^{2}}=\frac{\sigma^{2}}{n} \frac{1}{1+\Delta^{2}} \geq 0,$$
which shows $\tilde{\theta}{n}^{\mathrm{R}}\left(\Delta^{-2}\right)$ uniformly dominates $\tilde{\theta}{n}$.
The REff of $\theta_{n}^{\mathrm{R}}\left(\Delta^{-2}\right)$ is given by
$$\operatorname{REff}\left(\tilde{\theta}{n}^{\mathrm{R}}\left(\Delta^{-2}\right): \tilde{\theta}{n}\right)=1+\frac{1}{\Delta^{2}} .$$

## 统计代写|似然估计作业代写Probability and Estimation代考|Location Model

\min {\theta}\left{\left(Y-\theta \mathbf{1}{n}\right)^{\top}\left(\boldsymbol{Y}-\theta \mathbf{1}{n }\right)\right}=\tilde{\theta}{n}=\frac{1}{n} \mathbf{1}{n}^{\top} \boldsymbol{Y}=\bar{y} .\min {\theta}\left{\left(Y-\theta \mathbf{1}{n}\right)^{\top}\left(\boldsymbol{Y}-\theta \mathbf{1}{n }\right)\right}=\tilde{\theta}{n}=\frac{1}{n} \mathbf{1}{n}^{\top} \boldsymbol{Y}=\bar{y} .或者， 当误差呈正态分布时，可以最小化对数似然函数：l(θ)=−n日志⁡σ+n2日志⁡(2圆周率)−12(是−θ1n)⊤(是−θ1n),给出与 LSE 相同的解（2.4）。据了解，θ¨n是无偏的，即和(θ~n)=θ和方差θ~n是（谁）给的

ķ(n):θ(n)=n−12d,d≠0,

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

$$\hat{\theta} {n}^{\text {收缩}}(c)=c \tilde{\theta} {n}, \quad 0 \leq c \leq 1, 在H和r和θ~n∼ñ(θ,σ2/n).吨H和b一世一种s一种nd吨H和MSE⁡这Fθ^n收缩 (C)一种r和G一世在和nb是 b(θ^n收缩 (C))=和[Cθ~n]−θ=Cθ−θ=−(1−C)θ MSE⁡(θ^n收缩 (C))=C2σ2n+(1−C)2θ2米一世n一世米一世和一世nGMSE⁡(θ^n收缩 (C))在.r.吨.C,在和这b吨一种一世n c^{}=\frac{\Delta^{2}}{1+\Delta^{2}}, \quad \Delta^{2}=\frac{n \theta^{2}}{\sigma^ {2}}小号这吨H一种吨\operatorname{MSE}\left(\hat{\theta}{n}^{\text {收缩}}\left(c^{}\right)\right)=\frac{\sigma^{2}}{ n} \frac{\Delta^{2}}{1+\Delta^{2}} 。 吨H在s,MSE⁡(θ^n收缩 (C))一世s一种n一世nCr和一种s一世nGF在nC吨一世这n这FΔ2≥0一种nd吨H和r和l一种吨一世在和和FF一世C一世和nC是(R和FF)这Fθ^n收缩 (C)C这米p一种r和d吨这θ~n一世s \operatorname{REff}\left(\hat{\theta}{n}^{\text {收缩 }}\left(c^{}\right) ; \tilde{\theta}{n}\right)=1 +\frac{1}{\Delta^{2}}, \quad \Delta^{2} \geq 0 。 F在r吨H和r,吨H和米小号和d一世FF和r和nC和一世s \frac{\sigma^{2}}{n}-\frac{\sigma^{2}}{n} \frac{\Delta^{2}}{1+\Delta^{2}}=\frac {\sigma^{2}}{n} \frac{1}{1+\Delta^{2}} \geq 0, \quad \forall \Delta^{2} \geq 0 。$$

## 统计代写|似然估计作业代写Probability and Estimation代考|Ridge Regression–Type Estimation of Location Parameter

\tilde{\theta}{n}^{\mathrm{R}}(k)=\operatorname{argmin}{\theta}\left{\left|Y-\theta \mathbf{1}{n}\right |^{2}+n k \theta^{2}\right}, \quad k \geq 0 。\tilde{\theta}{n}^{\mathrm{R}}(k)=\operatorname{argmin}{\theta}\left{\left|Y-\theta \mathbf{1}{n}\right |^{2}+n k \theta^{2}\right}, \quad k \geq 0 。然后，我们得到岭回归型估计θ作为nθ+nķθ=nθ~n

θ~nR(ķ)=θ~n1+ķ

b(θ~nR(ķ))=−ķ1+ķθ和MSE⁡(θ~nR(ķ))=σ2n(1+ķ)2+ķ2θ2(1+ķ)2 =σ2n(1+ķ)2(1+ķ2Δ2)

σ2n(Δ21+Δ2)

σ2n−σ2nΔ21+Δ2=σ2n11+Δ2≥0,

REff⁡(θ~nR(Δ−2):θ~n)=1+1Δ2.

## 有限元方法代写

tatistics-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。

## 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。