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

2022年10月13日

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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

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