### 统计代写|广义线性模型代写generalized linear model代考|STATS3001

statistics-lab™ 为您的留学生涯保驾护航 在代写广义线性模型generalized linear model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。我们的专家在代写广义线性模型generalized linear model代写方面经验极为丰富，各种代写广义线性模型generalized linear model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|广义线性模型代写generalized linear model代考|Choice of Link Function

We must choose a link function to specify a binomial regression model. It is usually not possible to make this choice based on the data alone. For regions of moderate $p$, that is not close to zero or one, the link functions we have proposed are quite similar and so a very large amount of data would be necessary to distinguish between them. Larger differences are apparent in the tails, but for very small $p$, one needs a very large amount of data to obtain just a few successes, making it expensive to distinguish between link functions in this region. So usually, the choice of link function is made based on assumptions derived from physical knowledge or simple convenience. We now look at some of the advantages and disadvantages of the three proposed link functions and what motivates the choice.

Bliss (1935) analyzed some data on the numbers of insects dying at different levels of insecticide concentration. We fit all three link functions:

The lines in the left panel of Figure $2.3$ do not seem very different, but look at the relative differences:
$>$ matplot $(x$, cbind $(p p / p l,(1-p p) /(1-$
pl)), type=”1″, $x l a b=$ Dose”, $1 \mathrm{ab}=$ “Ratio”)
$>$ matplot (x, cbind (pc/pl, (1-pc) /(1-
pl)), type=”1″, 1 lab=”Dose”,ylab=”Ratio”)
as they appear in the second and third panels of Figure 2.3. We see that the probit and logit differ substantially in the tails. The same phenomenon is observed for the complementary log-log. This is problematic since the former plot indicates it would be difficult to distinguish between the two using the data we have. This is an issue in trials of potential carcinogens and other substances that must be tested for possible harmful effects on humans. Some substances are highly poisonous in that their effects become immediately obvious at doses that might normally be experienced in the environment. It is not difficult to detect such substances. However, there are other substances whose harmful effects only become apparent at large dosages where the observed probabilities are sufficiently larger than zero to become estimable without immense sample sizes. In order to estimate the probability of a harmful effect at a low dose, it would be necessary to select an appropriate link function and yet the data for high dosages will be of little help in doing this. As Paracelsus (1493-1541) said, “All substances are poisons; there is none which is not a poison. The right dose differentiates a poison.”

A good example of this problem is asbestos. Information regarding the harmful effects of asbestos derives from historical studies of workers in industries exposed to very high levels of asbestos dust. However, we would like to know the risk to individuals exposed to low levels of asbestos dust such as those found in old buildings. It is virtually impossible to accurately determine this risk. We cannot accurately measure exposure or outcome. This is not to argue that nothing should be done, but that decisions should be made in recognition of the uncertainties.

## 统计代写|广义线性模型代写generalized linear model代考|Goodness of Fit

The deviance is one measure of how well the model fits the data, but there are alternatives. The Pearson’s $X^2$ statistic takes the general form:
$$X^2=\sum_{i=1}^n \frac{\left(O_i-E_i\right)^2}{E_i}$$
where $O_i$ is the observed counts and $E_i$ are the expected counts for case $i$. For a binomial response, we count the number of successes for which $\theta_i=y_i$ while $E_i=n_i \hat{p}i$ and failures for which $O_i=n_i-y_i$ and $E_i=n_i\left(1-\hat{p}_i\right){\text {which results in: }}$
$$X^2=\sum_{i=1}^n \frac{\left(y_i-n_i \hat{p}i\right)^2}{n_i \hat{p}_i\left(1-\hat{p}_i\right)}$$ If we define Pearson residuals as: $$r_i^P=\left(y_i-n_i \hat{p}_i\right) / \sqrt{\operatorname{var} \hat{y}_i}$$ which can be viewed as a type of standardized residual, then $X^2=\sum{i=1}^n\left(r_i^P\right)^2$.So the Pearson’s $X^2$ is analogous to the residual sum of squares used in normal linear models.
The Pearson $X^2$ will typically be close in size to the deviance and can be used in the same manner. Alternative versions of the hypothesis tests described above might use the $X^2$ in place of the deviance with the same approximate null distributions.

However, some care is necessary because the model is fit to minimize the deviance and not the Pearson’s $X^2$. This means that it is possible, although unlikely, that the $X^2$ could increase as a predictor is added to the model. $X^2$ can be computed like this:

The proportion of variance explained or $R^2$ is a popular measure of fit for normal linear models. We might consider applying the same concept to binomial regression models by using the proportion of deviance explained. However, a better statistic is due to Naglekerke (1991):
$$R^2=\frac{1-\left(\hat{L}0 / \hat{L}\right)^{2 / n}}{1-\hat{L}_o^{2 / n}}=\frac{1-\exp \left(\left(D-D{\text {null }}\right) / n\right)}{1-\exp \left(-D_{\text {null }} / n\right)}$$

# 广义线性模型代考

## 统计代写|广义线性模型代写generalized linear model代考|Choice of Link Function

Bliss (1935) 分析了一些关于在不同杀虫剂浓度水平下死亡的昆蝼量的数据。我们适合所有 三个链㢺功能:

$>$ 绘图 $(x$, 绑定 $(p p / p l,(1-p p) /(1-$
pl)), type=”1″, $x l a b=$ 剂量”, $1 \mathrm{ab}=$ “比率”)
$>$ matplot $(x$, cbind $(\mathrm{pc} / \mathrm{pl},(1-\mathrm{pc}) /(1-$
pl)), type=”1″, 1 lab=”Dose”,ylab=”Ratio”)

## 统计代写|广义线性模型代写generalized linear model代考|Goodness of Fit

$$X^2=\sum_{i=1}^n \frac{\left(O_i-E_i\right)^2}{E_i}$$

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## MATLAB代写

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