# 统计代写|线性回归代写linear regression代考|MATH839

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## 统计代写|线性回归代写linear regression代考|Complements

The Least Squares Central Limit Theorem $2.8$ is often a good approximation if $n \geq 10 p$ and the error distribution has “light tails,” i.e. the probability of an outlier is nearly 0 and the tails go to zero at an exponential rate or faster. For error distributions with heavier tails, much larger samples are needed, and the assumption that the variance $\sigma^2$ exists is crucial, e.g. Cauchy errors are not allowed. Norman and Streiner (1986, p. 63) recommend $n \geq 5 p$.
The classical MLR prediction interval does not work well and should be replaced by the Olive (2007) asymptotically optimal PI (2.20). Lei and Wasserman (2014) provide an alternative: use the Lei et al. (2013) PI $\left[\tilde{r}_L, \tilde{r}_L\right]$ on the residuals, then the PI for $Y_f$ is
$$\left[\hat{Y}_f+\tilde{r}_L, \hat{Y}_f+\tilde{r}_U\right] .$$
Bootstrap PIs need more theory and instead of using $B=1000$ samples, use $B=\max (1000, n)$. See Olive (2014, pp. 279-285).

For the additive error regression model $Y=m(\boldsymbol{x})+e$, the response plot of $\hat{Y}=\hat{m}(\boldsymbol{x})$ vs. $Y$, with the identity line added as a visual aid, is used like the MLR response plot. We want $n \geq 10 d f$ where $d f$ is the degrees of freedom from fitting $\hat{m}$. Olive (2013a) provides PIs for this model, including the location model. These PIs are large sample PIs provided that the sample quantiles of the residuals are consistent estimators of the population quantiles of the errors. The response plot and PIs could also be used for methods described in James et al. (2013) such as ridge regression, lasso, principal components regression, and partial least squares. See Pelawa Watagoda and Olive (2017) if $n$ is not large compared to $p$.

In addition to large sample theory, we want the PIs to work well on a single data set as future observations are gathered, but only have the training data $\left(\boldsymbol{x}_1, Y_1\right), \ldots,\left(\boldsymbol{x}_n, Y_n\right)$. Much like $k$-fold cross validation for discriminant analysis, randomly divide the data set into $k=5$ groups of approximately equal size. Compute the model from 4 groups and use the 5th group as a validation set: compute the PI for $\boldsymbol{x}_f=\boldsymbol{x}_j$ for each $j$ in the 5 th group. Repeat so each of the 5 groups is used as a validation set. Compute the proportion of times $Y_i$ was in its PI for $i=1, \ldots, n$ as well as the average length of the $n$ PIs. We want the proportion near the nominal proportion and short average length if two or more models or PIs are being considered.
Following Chapter 11, under the regularity conditions, much of the inference that is valid for the normal MLR model is approximately valid for the unimodal MLR model when the sample size is large. For example, confidence intervals for $\beta_i$ are asymptotically correct, as are $t$ tests for $\beta_i=0$ (see $\mathrm{Li}$ and Duan (1989, p. 1035)), the MSE is an estimator of $\sigma^2$ by Theorems $2.6$ and 2.7, and variable selection procedures perform well (see Chapter 3 and Olive and Hawkins 2005).

## 统计代写|线性回归代写linear regression代考|Lack of Fit Tests

Then $M S P E=S S P E /(n-c)$ is an unbiased estimator of $\sigma^2$ when model (2.29) holds, regardless of the form of $m$. The PE in SSPE stands for “pure error.”

Now SSLF $=S S E-S S P E=\sum_{j=1}^c n_j\left(\bar{Y}_j-\hat{Y}_j\right)^2$. Notice that $\bar{Y}_j$ is an unbiased estimator of $m\left(\boldsymbol{x}_j\right)$ while $\hat{Y}_j$ is an estimator of $m$ if the MLR model is appropriate: $m\left(\boldsymbol{x}_j\right)=\boldsymbol{x}_j^T \boldsymbol{\beta}$. Hence SSLF and MSLF can be very large if the MLR model is not appropriate.

The 4 step lack of fit test is i) Ho: no evidence of MLR lack of fit, $H_A$ : there is lack of fit for the MLR model.
ii) $F_{L F}=M S L F / M S P E$.
iii) The pval $=P\left(F_{c-p, n-c}>F_{L F}\right)$.
iv) Reject Ho if pval $\leq \delta$ and state the $H_A$ claim that there is lack of fit. Otherwise, fail to reject Ho and state that there is not enough evidence to conclude that there is MLR lack of fit.

Although the lack of fit test seems clever, examining the response plot and residual plot is a much more effective method for examining whether or not the MLR model fits the data well provided that $n \geq 10 p$. A graphical version of the lack of fit test would compute the $\bar{Y}_j$ and see whether they scatter about the identity line in the response plot. When there are no replicates, the range of $\hat{Y}$ could be divided into several narrow nonoverlapping intervals called slices. Then the mean $\bar{Y}_j$ of each slice could be computed and a step function with step height $\bar{Y}_j$ at the $j$ th slice could be plotted. If the step function follows the identity line, then there is no evidence of lack of fit. However, it is easier to check whether the $Y_i$ are scattered about the identity line. Examining the residual plot is useful because it magnifies deviations from the identity line that may be difficult to see until the linear trend is removed. The lack of fit test may be sensitive to the assumption that the errors are iid $N\left(0, \sigma^2\right)$.

# 线性回归代写

## 统计代写|线性回归代写线性回归代考|补充

$$\left[\hat{Y}_f+\tilde{r}_L, \hat{Y}_f+\tilde{r}_U\right] .$$
Bootstrap PI需要更多的理论，而不是使用$B=1000$样本，使用$B=\max (1000, n)$。参见Olive (2014, pp. 279-285)

## 统计代写|线性回归代写线性回归代考|缺乏适合测试

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4步缺乏适合度检验是i) Ho:没有证据表明MLR缺乏适合度， $H_A$ : MLR模型缺乏拟合。
ii) $F_{L F}=M S L F / M S P E$.
iii) pval $=P\left(F_{c-p, n-c}>F_{L F}\right)$.
iv)如果没有拒绝Ho $\leq \delta$ and state the $H_A$ 声称缺乏契合度。否则，不能拒绝Ho，并声明没有足够的证据得出MLR缺乏拟合的结论。 虽然缺乏拟合检验似乎很聪明，但检验响应图和残差图是检验MLR模型是否很好地拟合数据的一种更有效的方法，条件是$n \geq 10 p$。缺乏拟合检验的图形版本将计算$\bar{Y}_j$并查看它们是否分散在响应图中的标识线上。当没有重复时，$\hat{Y}$的范围可以被划分为几个狭窄的不重叠的区间，称为片。然后计算出每个切片的平均值$\bar{Y}_j$，并在第$j$切片处绘制出阶梯高度为$\bar{Y}_j$的阶梯函数。如果阶跃函数遵循恒等线，则没有缺乏拟合的证据。然而，更容易检查$Y_i$是否分散在标识线上。检查残差图是有用的，因为它放大了偏离恒等线的偏差，这些偏差在去除线性趋势之前可能很难看到。缺乏拟合检验可能对错误为iid $N\left(0, \sigma^2\right)$ .

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## 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。

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