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

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|LASSO for Location Parameter

In this section, we define the LASSO estimator of $\theta$ introduced by Tibshirani (1996) in connection with the regression model.
Theorem 2.1 The LASSO estimator of $\theta$ is defined by
\begin{aligned} \hat{\theta}{n}^{\mathrm{LASSO}}(\lambda) &=\operatorname{argmin}{\theta}\left{\left|Y-\theta 1_{n}\right|^{2}+2 \lambda \sqrt{n} \sigma|\theta|\right} \ &=\operatorname{sgn}\left(\tilde{\theta}{n}\right)\left(\left|\tilde{\theta}{n}\right|-\lambda \frac{\sigma}{\sqrt{n}}\right)^{+} \end{aligned}

Proof: The derivative of objective function inside $\left{\left|Y-\theta \mathbf{1}{n}\right|^{2}+2 \lambda \sqrt{n} \sigma|\theta|\right}$ is given by $$-2 \mathbf{1}{n}^{\top}\left(Y-\hat{\theta}{n}^{\mathrm{LASSO}} \mathbf{1}{n}\right)+2 \lambda \sqrt{n} \sigma \operatorname{sgn}\left(\hat{\theta}{n}^{\mathrm{LASSO}}\right)=0$$ or $$\hat{\theta}{n}^{\mathrm{LASSO}}(\lambda)-\ddot{\theta}{n}+\lambda \frac{\sigma}{\sqrt{n}} \operatorname{sgn}\left(\hat{\theta}{n}^{\mathrm{LASSO}}\right)=0, \quad \ddot{\theta}{n}=\bar{Y}$$ or $$\frac{\sqrt{n} \hat{\theta}{n}^{\mathrm{LASSO}}(\lambda)}{\sigma}-\frac{\sqrt{n} \tilde{\theta}{n}}{\sigma}+\lambda \operatorname{sgn}\left(\hat{\theta}{n}^{\mathrm{LASSO}}(\lambda)\right)=0 .$$
(i) If $\operatorname{sgn}\left(\hat{\theta}{n}^{\mathrm{LASSO}}(\lambda)\right)=1$, then Eq. (2.24) reduces to $$\frac{\sqrt{n}}{\sigma} \hat{\theta}{n}^{\mathrm{LASSO}}(\lambda)-\frac{\sqrt{n} \tilde{\theta}{n}}{\sigma}+\lambda=0 .$$ Hence, using (2.21) $$0<\hat{\theta}{n}^{\mathrm{LASSO}}(\lambda)=\frac{\sigma}{\sqrt{n}}\left(Z_{n}-\lambda\right)=\frac{\sigma}{\sqrt{n}}\left(\left|Z_{n}\right|-\lambda\right),$$
with $Z_{n}>0$ and $\left|Z_{n}\right|>\lambda$.
(ii) If $\operatorname{sgn}\left(\hat{\theta}{n}^{\mathrm{LASSO}}(k)\right)=-1$, then we have $$0>\hat{\theta}{n}^{\mathrm{LASSO}}(\lambda)=\frac{\sigma}{\sqrt{n}}\left(Z_{n}+\lambda\right)=-\frac{\sigma}{\sqrt{n}}\left(\left|Z_{n}\right|-\lambda\right) .$$
(iii) If $\operatorname{sgn}\left(\hat{\theta}{n}^{\text {LAsso }}(\lambda)\right)=0$, we have $-Z{n}+\lambda \gamma=0, \gamma \in(-1,1)$. Hence, we obtain $Z_{n}=\lambda \gamma$, which implies $\left|Z_{n}\right|<\lambda$.
Combining (i)–(iii), we obtain
$$\hat{\theta}{n}^{\operatorname{LASsO}}(\lambda)=\frac{\sigma}{\sqrt{n}} \operatorname{sgn}\left(Z{n}\right)\left(\left|Z_{n}\right|-\lambda\right)^{+} .$$
Donoho and Johnstone (1994) defined this estimator as the “soft threshold estimator” (STE).

## 统计代写|似然估计作业代写Probability and Estimation代考|Stein-Type Estimation of Location Parameter

The PT heavily depends on the critical value of the test that $\theta$ may be zero. Thus, due to down effect of discreteness of the PTE, we define the Stein-type estimator of $\theta$ as given here assuming $\sigma$ is known
\begin{aligned} \hat{\theta}{n}^{5} &=\tilde{\theta}{n}\left(1-\lambda\left|Z_{n}\right|^{-1}\right), \quad Z_{n}=\frac{\sqrt{n} \tilde{\theta}{n}}{\sigma} \ &=\tilde{\theta}{n}\left(1-\frac{\lambda \sigma}{\sqrt{n}\left|\tilde{\theta}{n}\right|}\right) \ &=\frac{\sigma}{\sqrt{n}}\left(Z{n}-\lambda \operatorname{sgn}\left(Z_{n}\right)\right), \quad Z_{n} \sim \mathcal{N}(\Delta, 1), \quad \Delta=\frac{\sqrt{n} \theta}{\sigma} . \end{aligned}
The bias of $\hat{\theta}{n}^{\mathrm{S}}$ is $-\frac{\lambda \sigma}{n}[2 \Phi(\Delta)-1]$, and the MSE of $\hat{\theta}{n}^{\mathrm{s}}$ is given by
\begin{aligned} \operatorname{MSE}\left(\hat{\theta}{n}^{5}\right) &=\mathbb{E}\left(\hat{\theta}{n}^{\mathrm{S}}-\theta\right)^{2}=\mathbb{E}\left[\tilde{\theta}{n}-\theta\right]^{2}+\frac{\lambda^{2} \sigma^{2}}{n}-2 \frac{\lambda \sigma}{\sqrt{n}} \mathbb{E}\left[Z{n} \operatorname{sgn}\left(Z_{n}\right)\right] \ &=\frac{\sigma^{2}}{n}\left[1+\lambda^{2}-2 \lambda \sqrt{\frac{2}{\pi}} \exp \left{-\Delta^{2} / 2\right}\right] \end{aligned}
The value of $\lambda$ that minimizes $\operatorname{MSE}\left(\hat{\theta}{n}^{5}, \theta\right)$ is $\lambda{0}=\sqrt{\frac{2}{\pi}} \exp \left{-\Delta^{2} / 2\right}$, which is a decreasing function of $\Delta^{2}$ with a maximum at $\Delta=0$ and maximum value $\sqrt{\frac{2}{\pi}}$. Hence, the optimum value of MSE is
$$\operatorname{MSE}{\text {opt }}\left(\hat{\theta}{n}^{\mathrm{S}}\right)=\frac{\sigma^{2}}{n}\left[1-\frac{2}{\pi}\left(2 \exp \left{-\Delta^{2} / 2\right}-1\right)\right] .$$
The REff compared to LSE, $\tilde{\theta}{n}$ is $$\operatorname{REff}\left(\hat{\theta}{n}^{\mathrm{S}}: \tilde{\theta}{n}\right)=\left[1-\frac{2}{\pi}\left(2 \exp \left{-\Delta^{2} / 2\right}-1\right)\right]^{-1} .$$ In general, the $\operatorname{REff}\left(\hat{\theta}{n}^{\mathrm{s}}: \tilde{\theta}{n}\right)$ decreases from $\left(1-\frac{2}{\pi}\right)^{-1}$ at $\Delta^{2}=0$, then it crosses the 1 -line at $\Delta^{2}=2 \log 2$, and for $0<\Delta^{2}=2 \log 2$, $\hat{\theta}{n}^{\mathrm{s}}$ performs better than $\tilde{\theta}_{n^{*}}$

## 统计代写|似然估计作业代写Probability and Estimation代考|Comparison of LSE, PTE, Ridge, SE, and LASSO

We know the following MSE from previous sections:
\begin{aligned} \operatorname{MSE}\left(\tilde{\theta}{n}\right)=& \frac{\sigma^{2}}{n} \ \operatorname{MSE}\left(\hat{\theta}{n}^{\mathrm{PT}}(\lambda)\right)=& \frac{\sigma^{2}}{n}{[\Phi(\lambda-\Delta)+\Phi(\lambda+\Delta)]\ &+(\lambda-\Delta) \phi(\lambda-\Delta)+(\lambda+\Delta) \phi(\lambda+\Delta) \ &\left.+\Delta^{2}[\Phi(\lambda-\Delta)-\Phi(-\lambda-\Delta)]\right} \end{aligned}

\begin{aligned} \operatorname{MSE}\left(\hat{\theta}{n}^{\mathrm{RR}}(k)\right) &=\frac{\sigma^{2}}{n(1+k)^{2}}\left(1+k^{2} \Delta^{2}\right) \ &=\frac{\sigma^{2}}{n} \frac{\Delta^{2}}{1+\Delta^{2}}, \quad \text { for } k=\Delta^{-2} \ \operatorname{MSE}\left(\hat{\theta}{n}^{\mathrm{S}}\right) &=\frac{\sigma^{2}}{n}\left[1-\frac{2}{\pi}\left(2 \exp \left{-\Delta^{2} / 2\right}-1\right)\right] \ \operatorname{MSE}\left(\hat{\theta}{n}^{\mathrm{LASSO}}(\lambda)\right) &=\frac{\sigma^{2}}{n} \rho{\mathrm{ST}}(\lambda, \Delta), \quad \lambda=\sqrt{2 \log 2} . \end{aligned}
Hence, the REff expressions are given by
\begin{aligned} \operatorname{REff}\left(\hat{\theta}{n}^{\mathrm{pT}}(\alpha): \tilde{\theta}{n}\right)=&{[\tilde{\Phi}(\lambda-\Delta)+\tilde{\Phi}(\lambda+\Delta)]\ &+(\lambda-\Delta) \phi(\lambda-\Delta)+(\lambda+\Delta) \phi(\lambda+\Delta) \ &\left.+\Delta^{2}[\Phi(\lambda-\Delta)-\Phi(-\lambda-\Delta)]\right}^{-1} \ \operatorname{REff}\left(\hat{\theta}{n}^{\mathrm{RR}}\left(\Delta^{2}\right): \tilde{\theta}{n}\right)=& 1+\frac{1}{\Delta^{2}} \ \operatorname{REff}\left(\hat{\theta}{n}^{\mathrm{S}}, \tilde{\theta}{n}\right)=& {\left[1-\frac{2}{\pi}\left(2 \exp \left{-\Delta^{2} / 2\right}-1\right)\right]^{-1} } \ \operatorname{REff}\left(\hat{\theta}{n}^{\mathrm{LASSO}}(\lambda): \tilde{\theta}{n}\right)=& \rho_{\mathrm{ST}}^{-1}(\lambda, \Delta) \quad \lambda=\sqrt{2 \log 2} . \end{aligned}
It is seen from Table $2.1$ that the RRE dominates all other estimators uniformly and LASSO dominates UE, PTE, and $\hat{\theta}{n}^{\mathrm{PT}}>\hat{\theta}{n}^{\mathrm{s}}$ in an interval near 0 .

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

\begin{aligned} \hat{\theta}{n}^{\mathrm{LASSO}}(\lambda) &=\operatorname{argmin}{\theta}\left{\left|Y-\theta 1_{n }\right|^{2}+2 \lambda \sqrt{n} \sigma|\theta|\right} \ &=\operatorname{sgn}\left(\tilde{\theta}{n}\right)\左(\left|\tilde{\theta}{n}\right|-\lambda \frac{\sigma}{\sqrt{n}}\right)^{+} \end{对齐}\begin{aligned} \hat{\theta}{n}^{\mathrm{LASSO}}(\lambda) &=\operatorname{argmin}{\theta}\left{\left|Y-\theta 1_{n }\right|^{2}+2 \lambda \sqrt{n} \sigma|\theta|\right} \ &=\operatorname{sgn}\left(\tilde{\theta}{n}\right)\左(\left|\tilde{\theta}{n}\right|-\lambda \frac{\sigma}{\sqrt{n}}\right)^{+} \end{对齐}

(一) 如果sgn⁡(θ^n大号一种小号小号这(λ))=1，然后等式。（2.24）减少到nσθ^n大号一种小号小号这(λ)−nθ~nσ+λ=0.因此，使用 (2.21)0<θ^n大号一种小号小号这(λ)=σn(从n−λ)=σn(|从n|−λ),

(ii) 如果sgn⁡(θ^n大号一种小号小号这(ķ))=−1，那么我们有0>θ^n大号一种小号小号这(λ)=σn(从n+λ)=−σn(|从n|−λ).
(iii) 如果sgn⁡(θ^n套索 (λ))=0， 我们有−从n+λC=0,C∈(−1,1). 因此，我们得到从n=λC，这意味着|从n|<λ.

θ^n套索(λ)=σnsgn⁡(从n)(|从n|−λ)+.
Donoho 和 Johnstone (1994) 将此估计器定义为“软阈值估计器”(STE)。

## 统计代写|似然估计作业代写Probability and Estimation代考|Stein-Type Estimation of Location Parameter

PT 在很大程度上取决于测试的临界值，即θ可能为零。因此，由于 PTE 离散性的下降效应，我们定义了θ如此处给出的假设σ已知
θ^n5=θ~n(1−λ|从n|−1),从n=nθ~nσ =θ~n(1−λσn|θ~n|) =σn(从n−λsgn⁡(从n)),从n∼ñ(Δ,1),Δ=nθσ.

\begin{aligned} \operatorname{MSE}\left(\hat{\theta}{n}^{5}\right) &=\mathbb{E}\left(\hat{\theta}{n}^{ \mathrm{S}}-\theta\right)^{2}=\mathbb{E}\left[\tilde{\theta}{n}-\theta\right]^{2}+\frac{\lambda ^{2} \sigma^{2}}{n}-2 \frac{\lambda \sigma}{\sqrt{n}} \mathbb{E}\left[Z{n} \operatorname{sgn}\left (Z_{n}\right)\right] \ &=\frac{\sigma^{2}}{n}\left[1+\lambda^{2}-2 \lambda \sqrt{\frac{2} {\pi}} \exp \left{-\Delta^{2} / 2\right}\right] \end{aligned}\begin{aligned} \operatorname{MSE}\left(\hat{\theta}{n}^{5}\right) &=\mathbb{E}\left(\hat{\theta}{n}^{ \mathrm{S}}-\theta\right)^{2}=\mathbb{E}\left[\tilde{\theta}{n}-\theta\right]^{2}+\frac{\lambda ^{2} \sigma^{2}}{n}-2 \frac{\lambda \sigma}{\sqrt{n}} \mathbb{E}\left[Z{n} \operatorname{sgn}\left (Z_{n}\right)\right] \ &=\frac{\sigma^{2}}{n}\left[1+\lambda^{2}-2 \lambda \sqrt{\frac{2} {\pi}} \exp \left{-\Delta^{2} / 2\right}\right] \end{aligned}

\operatorname{MSE}{\text {opt }}\left(\hat{\theta}{n}^{\mathrm{S}}\right)=\frac{\sigma^{2}}{n}\左[1-\frac{2}{\pi}\left(2 \exp \left{-\Delta^{2} / 2\right}-1\right)\right] 。\operatorname{MSE}{\text {opt }}\left(\hat{\theta}{n}^{\mathrm{S}}\right)=\frac{\sigma^{2}}{n}\左[1-\frac{2}{\pi}\left(2 \exp \left{-\Delta^{2} / 2\right}-1\right)\right] 。
REff 与 LSE 相比，θ~n是\operatorname{REff}\left(\hat{\theta}{n}^{\mathrm{S}}: \tilde{\theta}{n}\right)=\left[1-\frac{2}{ \pi}\left(2 \exp \left{-\Delta^{2} / 2\right}-1\right)\right]^{-1} 。\operatorname{REff}\left(\hat{\theta}{n}^{\mathrm{S}}: \tilde{\theta}{n}\right)=\left[1-\frac{2}{ \pi}\left(2 \exp \left{-\Delta^{2} / 2\right}-1\right)\right]^{-1} 。一般来说，REff⁡(θ^ns:θ~n)从减少(1−2圆周率)−1在Δ2=0，然后它穿过 1 线在Δ2=2日志⁡2，并且对于0<Δ2=2日志⁡2, θ^ns表现优于θ~n∗

## 统计代写|似然估计作业代写Probability and Estimation代考|Comparison of LSE, PTE, Ridge, SE, and LASSO

\begin{aligned} \operatorname{MSE}\left(\tilde{\theta}{n}\right)=& \frac{\sigma^{2}}{n} \ \operatorname{MSE}\left(\帽子{\theta}{n}^{\mathrm{PT}}(\lambda)\right)=& \frac{\sigma^{2}}{n}{[\Phi(\lambda-\Delta)+ \Phi(\lambda+\Delta)]\ &+(\lambda-\Delta) \phi(\lambda-\Delta)+(\lambda+\Delta) \phi(\lambda+\Delta) \ &\left.+\ Delta^{2}[\Phi(\lambda-\Delta)-\Phi(-\lambda-\Delta)]\right} \end{对齐}\begin{aligned} \operatorname{MSE}\left(\tilde{\theta}{n}\right)=& \frac{\sigma^{2}}{n} \ \operatorname{MSE}\left(\帽子{\theta}{n}^{\mathrm{PT}}(\lambda)\right)=& \frac{\sigma^{2}}{n}{[\Phi(\lambda-\Delta)+ \Phi(\lambda+\Delta)]\ &+(\lambda-\Delta) \phi(\lambda-\Delta)+(\lambda+\Delta) \phi(\lambda+\Delta) \ &\left.+\ Delta^{2}[\Phi(\lambda-\Delta)-\Phi(-\lambda-\Delta)]\right} \end{对齐}\begin{aligned} \operatorname{MSE}\left(\hat{\theta}{n}^{\mathrm{RR}}(k)\right) &=\frac{\sigma^{2}}{n (1+k)^{2}}\left(1+k^{2} \Delta^{2}\right) \ &=\frac{\sigma^{2}}{n} \frac{\Delta ^{2}}{1+\Delta^{2}}, \quad \text { for } k=\Delta^{-2} \ \operatorname{MSE}\left(\hat{\theta}{n} ^{\mathrm{S}}\right) &=\frac{\sigma^{2}}{n}\left[1-\frac{2}{\pi}\left(2 \exp \left{- \Delta^{2} / 2\right}-1\right)\right] \ \operatorname{MSE}\left(\hat{\theta}{n}^{\mathrm{LASSO}}(\lambda)\对) &=\frac{\sigma^{2}}{n} \rho{\mathrm{ST}}(\lambda, \Delta), \quad \lambda=\sqrt{2 \log 2} 。\end{对齐}\begin{aligned} \operatorname{MSE}\left(\hat{\theta}{n}^{\mathrm{RR}}(k)\right) &=\frac{\sigma^{2}}{n (1+k)^{2}}\left(1+k^{2} \Delta^{2}\right) \ &=\frac{\sigma^{2}}{n} \frac{\Delta ^{2}}{1+\Delta^{2}}, \quad \text { for } k=\Delta^{-2} \ \operatorname{MSE}\left(\hat{\theta}{n} ^{\mathrm{S}}\right) &=\frac{\sigma^{2}}{n}\left[1-\frac{2}{\pi}\left(2 \exp \left{- \Delta^{2} / 2\right}-1\right)\right] \ \operatorname{MSE}\left(\hat{\theta}{n}^{\mathrm{LASSO}}(\lambda)\对) &=\frac{\sigma^{2}}{n} \rho{\mathrm{ST}}(\lambda, \Delta), \quad \lambda=\sqrt{2 \log 2} 。\end{对齐}

\begin{aligned} \operatorname{REff}\left(\hat{\theta}{n}^{\mathrm{pT}}(\alpha): \tilde{\theta}{n}\right)=&{ [\波浪号{\Phi}(\lambda-\Delta)+\波浪号{\Phi}(\lambda+\Delta)]\ &+(\lambda-\Delta) \phi(\lambda-\Delta)+(\ lambda+\Delta) \phi(\lambda+\Delta) \ &\left.+\Delta^{2}[\Phi(\lambda-\Delta)-\Phi(-\lambda-\Delta)]\right}^ {-1} \ \operatorname{REff}\left(\hat{\theta}{n}^{\mathrm{RR}}\left(\Delta^{2}\right): \tilde{\theta}{ n}\right)=& 1+\frac{1}{\Delta^{2}} \ \算子名{REff}\left(\hat{\theta}{n}^{\mathrm{S}}, \波浪线{\theta}{n}\right)=& {\left[1-\frac{2}{\pi}\left(2 \exp \left{-\Delta^{2} / 2\right}- 1\right)\right]^{-1} } \ \operatorname{REff}\left(\hat{\theta}{n}^{\mathrm{LASSO}}(\lambda): \tilde{\theta} {n}\right)=& \rho_{\mathrm{ST}}^{-1}(\lambda, \Delta) \quad \lambda=\sqrt{2 \log 2} 。\end{对齐}\begin{aligned} \operatorname{REff}\left(\hat{\theta}{n}^{\mathrm{pT}}(\alpha): \tilde{\theta}{n}\right)=&{ [\波浪号{\Phi}(\lambda-\Delta)+\波浪号{\Phi}(\lambda+\Delta)]\ &+(\lambda-\Delta) \phi(\lambda-\Delta)+(\ lambda+\Delta) \phi(\lambda+\Delta) \ &\left.+\Delta^{2}[\Phi(\lambda-\Delta)-\Phi(-\lambda-\Delta)]\right}^ {-1} \ \operatorname{REff}\left(\hat{\theta}{n}^{\mathrm{RR}}\left(\Delta^{2}\right): \tilde{\theta}{ n}\right)=& 1+\frac{1}{\Delta^{2}} \ \算子名{REff}\left(\hat{\theta}{n}^{\mathrm{S}}, \波浪线{\theta}{n}\right)=& {\left[1-\frac{2}{\pi}\left(2 \exp \left{-\Delta^{2} / 2\right}- 1\right)\right]^{-1} } \ \operatorname{REff}\left(\hat{\theta}{n}^{\mathrm{LASSO}}(\lambda): \tilde{\theta} {n}\right)=& \rho_{\mathrm{ST}}^{-1}(\lambda, \Delta) \quad \lambda=\sqrt{2 \log 2} 。\end{对齐}

## 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。