### 统计代写|广义线性模型代写generalized linear model代考|MAST30025

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|Poisson Regression

If $Y$ is Poisson with mean $\mu>0$, then:
$$P(Y=y)=\frac{e^{-\mu} \mu^{y}}{y !}, \quad y=0,1,2, \ldots$$
Three examples of the Poisson density are depicted in Figure 5.1. In the left panel, we see a distribution that gives highest probability to $y=0$ and falls rapidly as $y$ increases. In the center panel, we see a skew distribution with longer tail on the right. Even for a not so large $\mu=5$, we see the distribution become more normally shaped. This becomes more pronounced as $\mu$ increases.
barplot (dpois $(0: 5,0.5)$, xlab=” $y$ “, ylab=”Probability”, names=0:5, main=”
$\hookrightarrow$ mean $\left.=0.5^{\prime \prime}\right)$
barplot (dpois $(0: 10,2), x 1 a b=” y ” y$ labm”Probability”, names= $0: 10$, main $=$ “
$\rightarrow$ mean $\left.=2^{\prime \prime}\right)$
barplot (dpois $(0: 15,5)$, xlab=” ” $”, y l a b=”$ Probability”, names $0: 15$, main ” “
$\rightarrow$ mean $\left.=5^{\prime \prime}\right)$
The expectation and variance of a Poisson are the same: $E Y=\operatorname{var} Y=\mu$. The Poisson distribution arises naturally in several ways:

1. If the count is some number out of some possible total, then the response would be more appropriately modeled as a binomial. However, for small success probabilities and large totals, the Poisson is a good approximation and can be used. For example, in modeling the incidence of rare forms of cancer, the number of people affected is a small proportion of the population in a given geographical area. Specifically, if $\mu=n p$ while $n \rightarrow \infty$, then $B(n, p)$ is well approximated by Pois $(\mu)$. Also, for small $p$, note that $\operatorname{logit}(p) \approx \log p$, so that the use of the Poisson with a log link is comparable to the binomial with a logit link. Where $n$ varies between cases, a rate model can be used as described in Section 5.3.

## 统计代写|广义线性模型代写generalized linear model代考|Dispersed Poisson Model

We can modify the standard Poisson model to allow for more variation in the response. But before we do that, we must check whether the large size of deviance might be related to some other cause.

In the Galápagos example, we check the residuals to see if the large deviance can be explained by an outlier:
halfnorm (residuals (modp))
The half-normal plot of the (absolute value of the) residuals shown in Figure $5.3$ shows no outliers. It could be that the structural form of the model needs some improvement, but some experimentation with different forms for the predictors will reveal that there is little scope for improvement. Furthermore, the proportion of deviance explained by this model, $1-717 / 3510=0.796$, is about the same as in the linear model above.

For a Poisson distribution, the mean is equal to the variance. Let’s investigate this relationship for this model. It is difficult to estimate the variance for a given value of the mean, but $(y-\hat{\mu})^{2}$ does serve as a crude approximation. We plot this estimated variance against the mean, as seen in the second panel of Figure 5.3:
plot ( $\log ($ fitted (modp) ), $\log (($ gala\$Species-fitted (modp) ) 2$), \quad x l a b=\hookrightarrow$expression (hat (mu)), ylab=expression$\left.\left((y-h a t(m u))^{\wedge} 2\right)\right)$abline$(0,1)$We see that the variance is proportional to, but larger than, the mean. When the variance assumption of the Poisson regression model is broken but the link function and choice of predictors are correct, the estimates of$\beta$are consistent, but the standard er- rors will be wrong. We cannot determine which predictors are statistically significant in the above model using the output we have. The Poisson distribution has only one parameter and so is not very flexible for empirical fitting purposes. We can generalize by allowing ourselves a dispersion parameter. Over- or underdispersion can occur in various ways in Poisson models. For example, suppose the Poisson response$Y$has rate$\lambda$which is itself a random variable. The tendency to fail for a machine may vary from unit to unit even though they are the same model. We can model this by letting$\lambda$be gamma distributed with$E \lambda=\mu$and var$\lambda=\mu / \phi$. Now$Y$is negative binomial with mean$E Y=\mu$. The mean is the same as the Poisson, but the variance var$Y=\mu(1+\phi) / \phi$which is not equal to$\mu$. In this case, overdispersion would occur and could be modeled using a negative binomial model as demonstrated in Section 5.4. If we know the specific mechanism, as in the above example, we could model the response as a negative binomial or other more flexible distribution. If the mechanism is not known, we can introduce a dispersion parameter$\phi$such that var$Y=\phi E Y=\phi \mu$.$\phi=1$is the regular Poisson regression case, while$\phi>1$is overdispersion and$\phi<1$is underdispersion. The dispersion parameter may be estimated using: $$\hat{\phi}=\frac{X^{2}}{n-p}=\frac{\sum_{i}\left(y_{i}-\hat{\mu}{i}\right)^{2} / \hat{\mu}{i}}{n-p}$$ ## 统计代写|广义线性模型代写generalized linear model代考|Rate Models The number of events observed may depend on a size variable that determines the number of opportunities for the events to occur. For example, if we record the number of burglaries reported in different cities, the observed number will depend on the number of households in these cities. In other cases, the size variable may be time. For example, if we record the number of customers served by a sales worker, we must take account of the differing amounts of time worked. Sometimes, it is possible to analyze such data using a binomial response model. For the burglary example above, we might model the number of burglaries out of the number of households. However, if the proportion is small, the Poisson approxima- tion to the binomial is effective. Furthermore, in some examples, the total number of potential cases may not be known exactly. The modeling of rare diseases illustrates this issue as we may know the number of cases but not have precise population data. Sometimes, the binomial model simply cannot be used. In the burglary example, some households may be robbed more than once. In the customer service example, the size variable is not a count. An alternative approach is to model the ratio. However, there are often difficulties with normality and unequal variance when taking this approach, particularly if the counts are small. In Purott and Reeder (1976), some data is presented from an experiment conducted to determine the effect of gamma radiation on the numbers of chromosomal abnormalities (ca) observed. The number (cells), in hundreds of cells exposed in each run, differs. The dose amount (doseamt) and the rate (doserate) at which the dose is applied are the predictors of interest. We may format the data for observation like this: data (dicentric, package=”faraway”) round (xtabs (ca/cells doseamt+doserate, dicentric),2) ## 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|Poisson Regression 如果是是泊松的均值μ>0， 然后： 磷(是=是)=和−μμ是是!,是=0,1,2,… 图 5.1 描述了泊松密度的三个示例。在左侧面板中，我们看到一个分布，它给出的概率最高是=0并迅速下降是增加。在中心面板中，我们看到右侧有较长尾部的偏斜分布。即使对于一个不是那么大的μ=5，我们看到分布变得更正常。这变得更加明显μ增加。 条形图（dpois(0:5,0.5), xlab =”是“, ylab=”概率”, 名称=0:5, main=” 意思是=0.5′′) 条形图（dpois(0:10,2),X1一个b=”是”是实验室“概率”，名称=0:10， 主要的= “ →意思是=2′′) 条形图（dpois(0:15,5), xlab =” ””,是l一个b=”概率”，名称0:15， 主要的 ” ” →意思是=5′′) 泊松的期望和方差是相同的：和是=曾是⁡是=μ. 泊松分布以多种方式自然产生： 1. 如果计数是某个可能总数中的某个数字，则响应将更适合建模为二项式。但是，对于较小的成功概率和较大的总数，泊松是一个很好的近似值，可以使用。例如，在模拟罕见癌症的发病率时，受影响的人数只是特定地理区域内人口的一小部分。具体来说，如果μ=np尽管n→∞， 然后乙(n,p)由 Pois 很好地逼近(μ). 另外，对于小p， 注意罗吉特⁡(p)≈日志⁡p，因此使用对数链接的泊松与使用对数链接的二项式相当。在哪里n不同情况下的不同，可以使用第 5.3 节中描述的费率模型。 ## 统计代写|广义线性模型代写generalized linear model代考|Dispersed Poisson Model 我们可以修改标准泊松模型以允许响应的更多变化。但在我们这样做之前，我们必须检查较大的偏差是否与其他原因有关。 在加拉帕戈斯的例子中，我们检查残差，看看是否可以用异常值来解释大偏差： halfnorm (residuals (modp)) 残差（ 的绝对值）的半正态图如图所示5.3显示没有异常值。可能是模型的结构形式需要一些改进，但是对预测变量的不同形式进行一些实验会发现改进的余地很小。此外，该模型解释的偏差比例，1−717/3510=0.796, 与上述线性模型中的大致相同。 对于泊松分布，均值等于方差。让我们研究这个模型的这种关系。对于给定的均值，很难估计方差，但是(是−μ^)2确实可以作为粗略的近似值。 我们将这个估计的方差与平均值作图，如图 5.3 的第二个面板所示：日志⁡(安装（modp）），日志⁡((晚会$物种拟合 (modp) ) 2),Xl一个b=

rors 将是错误的。我们无法使用我们拥有的输出确定哪些预测变量在上述模型中具有统计显着性。

φ^=X2n−p=∑一世(是一世−μ^一世)2/μ^一世n−p

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

data (dicentric, package=”faraway”)
round (xtabs (ca/cells doseamt​​+doserate, dicentric),2)

## 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。