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统计代写|广义线性模型代写generalized linear model代考|STAT3022

Posted on 2022年6月10日2022年6月10日 by statistics-lab

如果你也在怎样代写广义线性模型generalized linear model这个学科遇到相关的难题，请随时右上角联系我们的24/7代写客服。

广义线性模型（GLiM，或GLM）是John Nelder和Robert Wedderburn在1972年制定的一种高级统计建模技术。它是一个包含许多其他模型的总称，它允许响应变量y具有除正态分布以外的误差分布。

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。

我们提供的广义线性模型generalized linear model及其相关学科的代写，服务范围广, 其中包括但不限于:

Statistical Inference 统计推断
Statistical Computing 统计计算
Advanced Probability Theory 高等概率论
Advanced Mathematical Statistics 高等数理统计学
(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|STAT3022

统计代写|广义线性模型代写generalized linear model代考|Correspondence Analysis

The analysis of the hair-eye color data in the previous section revealed how hair and eye color are dependent. But this does not tell us how they are dependent. To study this, we can use a kind of residual analysis for contingency tables called correspondence analysis.

Compute the Pearson residuals $r_{P}$ and write them in the matrix form $R_{i j}$, where $i=1, \ldots, r$ and $j=1, \ldots, c$, according to the structure of the data. Perform the singular value decomposition:
$$
R_{r \times c}=U_{r \times w} D_{w \times w} V_{w \times c}^{T}
$$
where $r$ is the number of rows, $c$ is the number of columns and $w=\min (r, c) . U$ and $V$ are called the right and left singular vectors, respectively. $D$ is a diagonal matrix with sorted elements $d_{i}$, called singular values. Another way of writing this is:
$$
R_{i j}=\sum_{k=1}^{w} U_{i k} d_{k} V_{j k}
$$
As with eigendecompositions, it is not uncommon for the first few singular values to be much larger than the rest. Suppose that the first two dominate so that:
$$
R_{i j} \approx U_{i 1} d_{1} V_{j 1}+U_{i 2} d_{2} V_{j 2}
$$

We usually absorb the $d$ s into $U$ and $V$ for plotting purposes so that we can assess the relative contribution of the components. Thus:
$$
\begin{aligned}
R_{i j} & \approx\left(U_{i 1} \sqrt{d_{1}}\right) \times\left(V_{j 1} \sqrt{d_{1}}\right)+\left(U_{i 2} \sqrt{d_{2}}\right) \times\left(V_{j 2} \sqrt{d_{2}}\right) \
& \equiv U_{i 1} V_{j 1}+U_{i 2} V_{j 2}
\end{aligned}
$$
where in the latter expression we have redefined the $U \mathrm{~s}$ and $V \mathrm{~s}$ to include the $\sqrt{d}$.

统计代写|广义线性模型代写generalized linear model代考|Matched Pairs

\begin{aligned}
&\text { In the typical two-way contingency tables, we display accumulated information } \
&\text { about two categorical measures on the same object. In matched pairs, we observe } \
&\text { one measure on two matched objects. } \
&\text { In Stuart (1955), data on the vision of a sample of women is presented. The left } \
&\text { and right eye performance is graded into four categories: } \
&\text { data (eyegrade) } \
&\text { (ct c- xtabs }(y \sim \text { right+left, eyegrade)) } \
&\text { right best second third worst } \
&\text { best } 1520 \quad 266 \quad 124 \quad 66 \
&\text { second } 234 \text { left } 1512 \quad 432 \quad 78 \
&\text { third } 117 \quad 362 \quad 1772 \quad 205 \
&\text { worst } 36 \quad 82 \quad 179 \quad 492
\end{aligned}

If we check for independence:
summary (et)
Call: xtabs (formula – y right + left, data – eyegrade)
Number of cases in table: 7477
Number of factors: 2
Test for independence of all factors:
Chisq – 8097, df – 9, p-value $=0$
We are not surprised to find strong evidence of dependence. Most people’s eyes are similar. A more interesting hypothesis for such matched pair data is symmetry. Is $p_{i j}=p_{j i}$ ? We can fit such a model by defining a factor where the levels represent the symmetric pairs for the off-diagonal elements. There is only one observation for each level down the diagonal:
(symfac <- factor (apply (eyegrade $[, 2: 3], 1$, function (x) paste (sort $(x)$,
$\rightarrow$ collapse=” ” “))))
[1] best-best best-second best-third best-worst
[5] best-second second-second second-third second-worst
[9] best-third second-third third-third third-worst
10 Levels: best-best best-second best-third … worst-worst
We now fit this model:
mods <- glm(y symfac, eyegrade, familympoisson)
c (deviance (mods), df . residual (mods))
[1] $19.2496 .000$
pchisq (deviance (mods), df . residual (mods), lower=F)
[1] $0.0037629$
Here, we see evidence of a lack of symmetry. It is worth checking the residuals:
round (xtabs (residuals (mods) right+left, eyegrade), 3)
round (xtabs (residuals (mods) righ left
We see that the residuals above the diagonal are mostly positive, while they are mostly negative below the diagonal. So there are generally more poor left, good right eye combinations than the reverse. Furthermore, we can compute the marginals:
margin table $(c t, 1)$
right
$\begin{array}{lrr}\text { best second third } & \text { worst } \ 1976 & 2256 & 2456\end{array}$

统计代写|广义线性模型代写generalized linear model代考|Ordinal Variables

Some variables have a natural order. We can use the methods for nominal variables described earlier in this chapter, but more information can be extracted by taking advantage of the structure of the data. Sometimes we might identify a particular ordinal variable as the response. In such cases, the methods of Section $7.4$ can be used. However, sometimes we are interested in modeling the association between ordinal variables. Here the use of scores can be helpful.

Consider a two-way table where both variables are ordinal. We may assign scores $u_{i}$ and $v_{j}$ to the rows and columns such that $u_{1} \leq u_{2} \leq \cdots \leq u_{I}$ and $v_{1} \leq v_{2} \leq \cdots \leq v_{J}$. The assignment of scores requires some judgment. If you have no particular prefer-

ence, even spacing allows for the simplest interpretation. If you have an interval scale, for example, $0-10$ years old, 10-20 years old, $20-40$ years old and so on, midpoints are often used. It is a good idea to check that the inference is robust to the assignment of scores by trying some reasonable alternative choices. If your qualitative conclusions are changed, this is an indication that you cannot make any strong finding.
Now fit the linear-by-linear association model:
$$
\log E Y_{i j}=\log \mu_{i j}=\log n p_{i j}=\log n+\alpha_{i}+\beta_{j}+\gamma u_{i} v_{j}
$$
So $\gamma=0$ means independence while $\gamma$ represents the amount of association and can be positive or negative. $\gamma$ is rather like an (unscaled) correlation coefficient. Consider underlying (latent) continuous variables which are discretized by the cutpoints $u_{i}$ and $v_{j}$. We can then identify $\gamma$ with the correlation coefficient of the latent variables.
Consider an example drawn from a subset of the 1996 American National Election Study (Rosenstone et al. (1997)). Using just the data on party affiliation and level of education, we can construct a two-way table:
data (nes96)
xtabs ( PID + educ, nes96)
\begin{tabular}{lrrrrrrrr}
\multicolumn{8}{c}{ educ } \
PID & MS & HSdrop HS Coll cCdeg & BAdeg MAdeg \
strDem & 5 & 19 & 59 & 38 & 17 & 40 & 22 \
weakDem & 4 & 10 & 49 & 36 & 17 & 41 & 23 \
indDem & 1 & 4 & 28 & 15 & 13 & 27 & 20 \
indind & 0 & 3 & 12 & 9 & 3 & 6 & 4 \
indRep & 2 & 7 & 23 & 16 & 8 & 22 & 16 \
weakRep & 0 & 5 & 35 & 40 & 15 & 38 & 17 \
strRep & 1 & 4 & 42 & 33 & 17 & 53 & 25
\end{tabular}
Both variables are ordinal in this example. We need to convert this to a dataframe with one count per line to enable model fitting.

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|Correspondence Analysis

上一节中对头发-眼睛颜色数据的分析揭示了头发和眼睛颜色是如何相互依赖的。但这并没有告诉我们他们是如何依赖的。为了研究这一点，我们可以对列联表使用一种称为对应分析的残差分析。

计算皮尔逊残差r磷并将它们写成矩阵形式R一世j，在哪里一世=1,…,r和j=1,…,C，根据数据结构。执行奇异值分解：

Rr×C=在r×在D在×在在在×C吨
在哪里r是行数，C是列数和在=分钟(r,C).在和在分别称为右奇异向量和左奇异向量。D是具有排序元素的对角矩阵d一世，称为奇异值。另一种写法是：

R一世j=∑ķ=1在在一世ķdķ在jķ
与特征分解一样，前几个奇异值远大于其余奇异值的情况并不少见。假设前两个占主导地位，因此：

R一世j≈在一世1d1在j1+在一世2d2在j2

我们通常吸收d进入在和在用于绘图目的，以便我们可以评估组件的相对贡献。因此：

R一世j≈(在一世1d1)×(在j1d1)+(在一世2d2)×(在j2d2) ≡在一世1在j1+在一世2在j2
在后一个表达式中，我们重新定义了在 s和在 s包括d.

统计代写|广义线性模型代写generalized linear model代考|Matched Pairs

在典型的双向列联表中，我们显示累积信息关于同一对象的两个分类度量。在配对中，我们观察到对两个匹配的对象进行一次测量。在 Stuart (1955) 中，提供了关于女性样本视力的数据。左边右眼表现分为四类：数据（眼级） (ct c-xtabs (是∼ 右+左，眼级））对最好第二第三最差最好的 152026612466 第二 234 剩下 151243278 第三 1173621772205 最坏的 3682179492

如果我们检查独立性：
summary (et)
Call: xtabs (formula – y right + left, data – eyegrade)
表中的案例数：7477
因素数：2
测试所有因素的独立性：
Chisq – 8097, df – 9、p值=0
我们对发现依赖的有力证据并不感到惊讶。大多数人的眼睛都是相似的。这种配对数据的一个更有趣的假设是对称性。是p一世j=pj一世? 我们可以通过定义一个因子来拟合这样的模型，其中水平表示非对角元素的对称对。对角线下的每一层只有一个观察值：
(symfac <- factor (apply (eyegrade[,2:3],1, 函数 (x) 粘贴（排序(X),
→collapse=” ” “)))))
[1] 最佳-最佳-最佳-第二-最佳-第三–最差
[5] 最佳-第二-第二-第二-第三-第二-最差
[9] 最佳-第三-第二-第三-第三-第三第三最差
10 个级别：最好最好最好第二最好第三…最差
我们现在拟合这个模型：
mods <- glm(y symfac, eyegrade, familympoisson)
c (deviance (mods), df .residual (mods ))
[1]19.2496.000
pchisq (deviance (mods), df .residual (mods), lower=F)
[1]0.0037629
在这里，我们看到了缺乏对称性的证据。值得检查残差：
round (xtabs(residuals (mods) right+left, eyegrade), 3)
round (xtabs(residuals (mods) right left
我们看到对角线以上的残差大多为正，而大部分为正对角线下方的负数。因此，通常左眼和右眼的组合比相反的差。此外，我们可以计算边际：
margin table(C吨,1)
正确的
最好的第二第三最坏的 197622562456

统计代写|广义线性模型代写generalized linear model代考|Ordinal Variables

一些变量具有自然顺序。我们可以使用本章前面描述的名义变量的方法，但是可以通过利用数据的结构来提取更多信息。有时我们可能会确定一个特定的序数变量作为响应。在这种情况下，本节的方法7.4可以使用。然而，有时我们有兴趣对序数变量之间的关联进行建模。在这里，使用分数可能会有所帮助。

考虑一个双向表，其中两个变量都是有序的。我们可以分配分数在一世和在j到行和列，使得在1≤在2≤⋯≤在我和在1≤在2≤⋯≤在Ĵ. 分数的分配需要一些判断。如果你没有特别的偏好——

因此，即使间距允许最简单的解释。例如，如果您有一个区间尺度，0−10岁，10-20岁，20−40岁等，经常使用中点。通过尝试一些合理的替代选择来检查推理对分数分配是否稳健是一个好主意。如果您的定性结论发生变化，这表明您无法做出任何强有力的发现。
现在拟合线性关联模型：

日志⁡和是一世j=日志⁡μ一世j=日志⁡np一世j=日志⁡n+一个一世+bj+C在一世在j
所以C=0意味着独立，而C表示关联的量，可以是正数或负数。C更像是一个（未缩放的）相关系数。考虑由切点离散化的潜在（潜在）连续变量在一世和在j. 然后我们可以识别C与潜在变量的相关系数。
考虑一个取自 1996 年美国全国选举研究（Rosenstone 等人（1997））子集的例子。仅使用党派和教育程度的数据，我们可以构建一个双向表：
data (nes96)
xtabs ( PID + educ, nes96)

\begin{tabular}{lrrrrrrrr} \multicolumn{8}{c}{ educ} \ PID & MS & HSdrop HS Coll cCdeg & BAdeg MAdeg \ strDem & 5 & 19 & 59 & 38 & 17 & 40 & 22 \weakDem & 4 & 10 & 49 & 36 & 17 & 41 & 23 \ indDem & 1 & 4 & 28 & 15 & 13 & 27 & 20 \ indind & 0 & 3 & 12 & 9 & 3 & 6 & 4 \ indRep & 2 & 7 & 23 & 16 & 8 & 22 & 16 \weakRep & 0 & 5 & 35 & 40 & 15 & 38 & 17 \ strRep & 1 & 4 & 42 & 33 & 17 & 53 & 25 \end{tabular}\begin{tabular}{lrrrrrrrr} \multicolumn{8}{c}{ educ} \ PID & MS & HSdrop HS Coll cCdeg & BAdeg MAdeg \ strDem & 5 & 19 & 59 & 38 & 17 & 40 & 22 \weakDem & 4 & 10 & 49 & 36 & 17 & 41 & 23 \ indDem & 1 & 4 & 28 & 15 & 13 & 27 & 20 \ indind & 0 & 3 & 12 & 9 & 3 & 6 & 4 \ indRep & 2 & 7 & 23 & 16 & 8 & 22 & 16 \weakRep & 0 & 5 & 35 & 40 & 15 & 38 & 17 \ strRep & 1 & 4 & 42 & 33 & 17 & 53 & 25 \end{tabular}
在此示例中，这两个变量都是有序的。我们需要将其转换为每行一个计数的数据帧以启用模型拟合。

统计代写|广义线性模型代写generalized linear model代考请认准statistics-lab™

统计代写请认准statistics-lab™. statistics-lab™为您的留学生涯保驾护航。

金融工程代写

金融工程是使用数学技术来解决金融问题。金融工程使用计算机科学、统计学、经济学和应用数学领域的工具和知识来解决当前的金融问题，以及设计新的和创新的金融产品。

非参数统计代写

非参数统计指的是一种统计方法，其中不假设数据来自于由少数参数决定的规定模型；这种模型的例子包括正态分布模型和线性回归模型。

广义线性模型代考

广义线性模型（GLM）归属统计学领域，是一种应用灵活的线性回归模型。该模型允许因变量的偏差分布有除了正态分布之外的其它分布。

术语广义线性模型（GLM）通常是指给定连续和/或分类预测因素的连续响应变量的常规线性回归模型。它包括多元线性回归，以及方差分析和方差分析（仅含固定效应）。

有限元方法代写

有限元方法（FEM）是一种流行的方法，用于数值解决工程和数学建模中出现的微分方程。典型的问题领域包括结构分析、传热、流体流动、质量运输和电磁势等传统领域。

有限元是一种通用的数值方法，用于解决两个或三个空间变量的偏微分方程（即一些边界值问题）。为了解决一个问题，有限元将一个大系统细分为更小、更简单的部分，称为有限元。这是通过在空间维度上的特定空间离散化来实现的，它是通过构建对象的网格来实现的：用于求解的数值域，它有有限数量的点。边界值问题的有限元方法表述最终导致一个代数方程组。该方法在域上对未知函数进行逼近。[1] 然后将模拟这些有限元的简单方程组合成一个更大的方程系统，以模拟整个问题。然后，有限元通过变化微积分使相关的误差函数最小化来逼近一个解决方案。

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。

随机分析代写

随机微积分是数学的一个分支，对随机过程进行操作。它允许为随机过程的积分定义一个关于随机过程的一致的积分理论。这个领域是由日本数学家伊藤清在第二次世界大战期间创建并开始的。

时间序列分析代写

随机过程，是依赖于参数的一组随机变量的全体，参数通常是时间。随机变量是随机现象的数量表现，其时间序列是一组按照时间发生先后顺序进行排列的数据点序列。通常一组时间序列的时间间隔为一恒定值（如1秒，5分钟，12小时，7天，1年），因此时间序列可以作为离散时间数据进行分析处理。研究时间序列数据的意义在于现实中，往往需要研究某个事物其随时间发展变化的规律。这就需要通过研究该事物过去发展的历史记录，以得到其自身发展的规律。

回归分析代写

多元回归分析渐进（Multiple Regression Analysis Asymptotics）属于计量经济学领域，主要是一种数学上的统计分析方法，可以分析复杂情况下各影响因素的数学关系，在自然科学、社会和经济学等多个领域内应用广泛。

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，工具箱允许您学习和应用专业技术。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。

R语言代写	问卷设计与分析代写
PYTHON代写	回归分析与线性模型代写
MATLAB代写	方差分析与试验设计代写
STATA代写	机器学习/统计学习代写
SPSS代写	计量经济学代写
EVIEWS代写	时间序列分析代写
EXCEL代写	深度学习代写
SQL代写	各种数据建模与可视化代写

统计代写|广义线性模型代写generalized linear model代考|STAT3015

Posted on 2022年6月10日2022年6月10日 by statistics-lab

如果你也在怎样代写广义线性模型generalized linear model这个学科遇到相关的难题，请随时右上角联系我们的24/7代写客服。

我们提供的广义线性模型generalized linear model及其相关学科的代写，服务范围广, 其中包括但不限于:

Statistical Inference 统计推断
Statistical Computing 统计计算
Advanced Probability Theory 高等概率论
Advanced Mathematical Statistics 高等数理统计学
(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|STAT3015

统计代写|广义线性模型代写generalized linear model代考|Negative Binomial

Given a series of independent trials, each with probability of success $p$, let $Z$ be the number of trials until the $k^{h t h}$ success. Then:
$$
P(Z=z)=\left(\begin{array}{l}
z-1 \
k-1
\end{array}\right) p^{k}(1-p)^{z-k} \quad z=k, k+1, \ldots
$$
The negative binomial can arise naturally in several ways. Imagine a system that can withstand $k$ hits before failing. The probability of a hit in a given time period is $p$ and we count the number of time periods until failure. The negative binomial also arises from a generalization of the Poisson where the parameter $\lambda$ is gamma distributed. The negative binomial also comes up as a limiting distribution for urn schemes that can be used to model contagion.

We get a more convenient parameterization if we let $Y=Z-k$ and $p=(1+\alpha)^{-1}$ so that:
$$
P(Y=y)=\left(\begin{array}{c}
y+k-1 \
k-1
\end{array}\right) \frac{\alpha^{y}}{(1+\alpha)^{y+k}}, \quad y=0,1,2, \ldots
$$
then $E Y=\mu=k \alpha$ and var $Y=k \alpha+k \alpha^{2}=\mu+\mu^{2} / k$
The log-likelihood is then:
$$
\sum_{i=1}^{n}\left(y_{i} \log \frac{\alpha}{1+\alpha}-k \log (1+\alpha)+\sum_{j=0}^{y_{i}-1} \log (j+k)-\log \left(y_{i} !\right)\right)
$$
The most convenient way to link the mean response $\mu$ to a linear combination of the predictors $X$ is:
$$
\eta=x^{T} \beta=\log \frac{\alpha}{1+\alpha}=\log \frac{\mu}{\mu+k}
$$
We can regard $k$ as fixed and determined by the application or as anditional parameter to be estimated. More on regression models for negative binomial responses may be found in Cameron and Trivedi (1998) and Lawless (1987).

统计代写|广义线性模型代写generalized linear model代考|Zero Inflated Count Models

Sometimes we see count response data where the number of zeroes appearing is significantly greater than the Poisson or negative binomial models would predict. Consider the number of arrests for criminal offenses incurred by individuals. A large number of people have never been arrested by the police while a smaller number have been detained on multiple occasions. Modifying the Poisson by adding a dispersion parameter does not adequately model this divergence from the standard count distributions.

We consider a sample of 915 biochemistry graduate students as analyzed by Long $(1990)$. The response is the number of articles produced during the last three years of the PhD. We are interested in how this is related to the gender, marital status, number of children, prestige of the department and productivity of the advisor of the student. The dataset may be found in the pscl package of Zeileis et al. (2008) which also provides the new model fitting functions needed in this section. We start by fitting a Poisson regression model:
$n=915 p-6$
Deviance $=1634.371$ Null Deviance $=1817.405$ (Difference $=183.034$ )
We can see that deviance is significantly larger than the degrees of freedom. Some experimentation reveals that this cannot be solved by using a richer linear predictor or by eliminating some outliers. We might consider a dispersed Poisson model or negative binomial but some thought suggests that there are good reasons why a student might produce no articles at all. We count and predict how many students produce between zero and seven articles. Very few students produce more than seven articles so we ignore these. The predprob function produces the predicted probabilities for each case. By summing these, we get the expected number for each article count.

统计代写|广义线性模型代写generalized linear model代考|Two-by-Two Tables

The data shown in Table $6.1$ were collected as part of a quality improvement study at a semiconductor factory. A sample of wafers was drawn and cross-classified according to whether a particle was found on the die that produced the wafer and whether the wafer was good or bad. More details on the study may be found in Hall (1994). The data might have arisen under several possible sampling schemes:

We observed the manufacturing process for a certain period of time and observed 450 wafers. The data were then cross-classified. We could use a Poisson model.
We decided to sample 450 wafers. The data were then cross-classified. We could use a multinomial model.
We selected 400 wafers without particles and 50 wafers with particles and then recorded the good or bad outcome. We could use a binomial model.
We selected 400 wafers without particles and 50 wafers with particles that also included, by design, 334 good wafers and 116 bad ones. We could use a hypergeometric model.

The first three sampling schemes are all plausible. The fourth scheme seems less likely in this example, but we include it for completeness. Such a scheme is more attractive when one level of each variable is relatively rare and we choose to oversample both levels to ensure some representation.

The main question of interest concerning these data is whether the presence of particles on the wafer affects the quality outcome. We shall see that all four sampling schemes lead to exactly the same conclusion. First, let’s set up the data in a convenient form for analysis:
$y<-\mathrm{c}(320,14,80,36)$ particle <- gl $(2,1,4$, labelsmc (“no”, “yes”) quality $<-\mathrm{g}(2,2$, labelsmc (“good”, “bad”)) (wafer <- data. frame (y, particle, quality)) y particle quality $\begin{array}{llll}1 & 320 & \text { no } & \text { good } \ 2 & 14 & \text { yes } & \text { good } \ 3 & 80 & \text { no } & \text { bad } \ 4 & 36 & \text { yes } & \text { bad }\end{array}$.

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|Negative Binomial

给定一系列独立试验，每个试验都有成功的概率p，让从是试验次数，直到ķH吨H成功。然后：

磷(从=和)=(和−1 ķ−1)pķ(1−p)和−ķ和=ķ,ķ+1,…
负二项式可以通过多种方式自然产生。想象一个可以承受的系统ķ在失败之前命中。给定时间段内命中的概率为p我们计算直到失败的时间段数。负二项式也源于泊松的推广，其中参数λ是伽马分布的。负二项式也作为可用于模拟传染的瓮方案的限制分布。

如果我们让我们得到一个更方便的参数化是=从−ķ和p=(1+一个)−1以便：

磷(是=是)=(是+ķ−1 ķ−1)一个是(1+一个)是+ķ,是=0,1,2,…
然后和是=μ=ķ一个和 var是=ķ一个+ķ一个2=μ+μ2/ķ
那么对数似然是：

∑一世=1n(是一世日志⁡一个1+一个−ķ日志⁡(1+一个)+∑j=0是一世−1日志⁡(j+ķ)−日志⁡(是一世!))
链接平均响应的最方便方法μ预测变量的线性组合X是：

这=X吨b=日志⁡一个1+一个=日志⁡μμ+ķ
我们可以认为ķ由应用程序固定和确定或作为要估计的附加参数。有关负二项式响应的回归模型的更多信息，请参见 Cameron 和 Trivedi (1998) 和 Lawless (1987)。

统计代写|广义线性模型代写generalized linear model代考|Zero Inflated Count Models

有时我们会看到计数响应数据，其中出现的零数量明显大于泊松或负二项式模型预测的数量。考虑个人因刑事犯罪而被捕的人数。许多人从未被警方逮捕，少数人多次被拘留。通过添加分散参数来修改泊松并不能充分模拟这种与标准计数分布的差异。

我们考虑由 Long 分析的 915 名生物化学研究生样本(1990). 答案是博士最后三年发表的文章数量。我们感兴趣的是这与性别、婚姻状况、孩子数量、部门声望和学生顾问的生产力有何关系。该数据集可以在 Zeileis 等人的 pscl 包中找到。（2008），它还提供了本节所需的新模型拟合函数。我们首先拟合泊松回归模型：
n=915p−6
偏差=1634.371零偏差=1817.405（区别=183.034)
我们可以看到偏差明显大于自由度。一些实验表明，这不能通过使用更丰富的线性预测器或消除一些异常值来解决。我们可能会考虑分散泊松模型或负二项式，但一些想法表明，学生可能根本不写文章是有充分理由的。我们计算并预测有多少学生发表了零到七篇文章。很少有学生发表超过七篇文章，所以我们忽略了这些。predprob 函数为每个案例生成预测概率。通过将这些相加，我们得到每个文章计数的预期数量。

统计代写|广义线性模型代写generalized linear model代考|Two-by-Two Tables

数据见表6.1作为半导体工厂质量改进研究的一部分收集。根据是否在生产晶圆的芯片上发现颗粒以及晶圆的好坏，抽取晶圆样本并进行交叉分类。有关该研究的更多详细信息，请参见 Hall (1994)。数据可能是在几种可能的抽样方案下产生的：

我们观察了一段时间的制造过程，观察了450个晶圆。然后对数据进行交叉分类。我们可以使用泊松模型。
我们决定对 450 个晶圆进行采样。然后对数据进行交叉分类。我们可以使用多项式模型。
我们选择了 400 个没有颗粒的晶圆和 50 个有颗粒的晶圆，然后记录了结果的好坏。我们可以使用二项式模型。
我们选择了 400 个没有颗粒的晶圆和 50 个有颗粒的晶圆，按照设计，还包括 334 个好晶圆和 116 个坏晶圆。我们可以使用超几何模型。

前三个抽样方案都是合理的。在此示例中，第四种方案似乎不太可能，但为了完整起见，我们将其包括在内。当每个变量的一个级别相对罕见并且我们选择对两个级别进行过采样以确保具有某种代表性时，这种方案更具吸引力。

与这些数据有关的主要问题是晶片上颗粒的存在是否会影响质量结果。我们将看到所有四种抽样方案都得出完全相同的结论。首先，让我们以方便分析的形式设置数据：
是<−C(320,14,80,36)粒子<-gl(2,1,4，labelsmc（“否”，“是”）质量<−G(2,2, labelsmc (“good”, “bad”)) (wafer <- data.frame (y,particle, quality)) y 粒子质量1320 不好的 214 是的好的 380 不坏的 436 是的坏的 .

统计代写请认准statistics-lab™. statistics-lab™为您的留学生涯保驾护航。

金融工程代写

非参数统计代写

非参数统计指的是一种统计方法，其中不假设数据来自于由少数参数决定的规定模型；这种模型的例子包括正态分布模型和线性回归模型。

广义线性模型代考

广义线性模型（GLM）归属统计学领域，是一种应用灵活的线性回归模型。该模型允许因变量的偏差分布有除了正态分布之外的其它分布。

有限元方法代写

随机分析代写

时间序列分析代写

回归分析代写

MATLAB代写

R语言代写	问卷设计与分析代写
PYTHON代写	回归分析与线性模型代写
MATLAB代写	方差分析与试验设计代写
STATA代写	机器学习/统计学习代写
SPSS代写	计量经济学代写
EVIEWS代写	时间序列分析代写
EXCEL代写	深度学习代写
SQL代写	各种数据建模与可视化代写

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

Posted on 2022年6月10日2022年6月10日 by statistics-lab

如果你也在怎样代写广义线性模型generalized linear model这个学科遇到相关的难题，请随时右上角联系我们的24/7代写客服。

我们提供的广义线性模型generalized linear model及其相关学科的代写，服务范围广, 其中包括但不限于:

Statistical Inference 统计推断
Statistical Computing 统计计算
Advanced Probability Theory 高等概率论
Advanced Mathematical Statistics 高等数理统计学
(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|MAST30025

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:

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：

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=
表达式 (帽子 (mu)), ylab = 表达式((是−H一个吨(米在))∧2))
下线(0,1)
我们看到方差与均值成正比，但大于均值。当泊松回归模型的方差假设被打破但链接函数和预测变量的选择正确时，b是一致的，但标准错误

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。

泊松分布只有一个参数，因此对于经验拟合目的不是很灵活。我们可以通过给自己一个色散参数来概括。在 Poisson 模型中，过度或欠分散可能以各种方式发生。例如，假设泊松响应是有率λ它本身就是一个随机变量。一台机器的故障趋势可能因单元而异，即使它们是同一型号。我们可以通过让λ是伽马分布的和λ=μ和 varλ=μ/φ. 现在是是负二项式，均值和是=μ. 均值与泊松相同，但方差 var是=μ(1+φ)/φ这不等于μ. 在这种情况下，会发生过度分散，并且可以使用负二项式模型进行建模，如第 5.4 节所示。

如果我们知道具体机制，如上例所示，我们可以将响应建模为负二项式或其他更灵活的分布。如果机制未知，我们可以引入色散参数φ这样 var是=φ和是=φμ. φ=1是常规泊松回归情况，而φ>1是过度分散和φ<1是欠分散的。
可以使用以下方法估计色散参数：

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

观察到的事件数量可能取决于确定事件发生机会的数量的大小变量。例如，如果我们记录不同城市报告的入室盗窃数量，观察到的数量将取决于这些城市的家庭数量。在其他情况下，大小变量可能是时间。例如，如果我们记录销售人员服务的客户数量，我们必须考虑不同的工作时间。

有时，可以使用二项式响应模型分析此类数据。对于上面的入室盗窃示例，我们可以根据家庭数量对入室盗窃数量进行建模。但是，如果比例很小，泊松近似

对二项式的化验是有效的。此外，在某些示例中，可能无法准确知道潜在病例的总数。罕见疾病的建模说明了这个问题，因为我们可能知道病例数但没有精确的人口数据。有时，二项式模型根本无法使用。在入室盗窃的例子中，一些家庭可能不止一次被抢劫。在客户服务示例中，大小变量不是计数。另一种方法是对比率进行建模。然而，当采用这种方法时，通常存在正态性和不等方差的困难，特别是在计数很小的情况下。

在 Purott 和 Reeder (1976) 中，一些数据来自一项实验，以确定伽马辐射对观察到的染色体异常 (ca) 数量的影响。每次运行中暴露的数百个细胞中的数量（细胞）不同。应用剂量的剂量 (doseamt) 和速率 (doserate) 是感兴趣的预测因子。我们可以像这样格式化观察数据：
data (dicentric, package=”faraway”)
round (xtabs (ca/cells doseamt+doserate, dicentric),2)

统计代写请认准statistics-lab™. statistics-lab™为您的留学生涯保驾护航。

金融工程代写

非参数统计代写

非参数统计指的是一种统计方法，其中不假设数据来自于由少数参数决定的规定模型；这种模型的例子包括正态分布模型和线性回归模型。

广义线性模型代考

广义线性模型（GLM）归属统计学领域，是一种应用灵活的线性回归模型。该模型允许因变量的偏差分布有除了正态分布之外的其它分布。

有限元方法代写

随机分析代写

时间序列分析代写

回归分析代写

MATLAB代写

R语言代写	问卷设计与分析代写
PYTHON代写	回归分析与线性模型代写
MATLAB代写	方差分析与试验设计代写
STATA代写	机器学习/统计学习代写
SPSS代写	计量经济学代写
EVIEWS代写	时间序列分析代写
EXCEL代写	深度学习代写
SQL代写	各种数据建模与可视化代写

统计代写|广义线性模型代写generalized linear model代考|MAST90084

Posted on 2022年6月10日2022年6月10日 by statistics-lab

如果你也在怎样代写广义线性模型generalized linear model这个学科遇到相关的难题，请随时右上角联系我们的24/7代写客服。

我们提供的广义线性模型generalized linear model及其相关学科的代写，服务范围广, 其中包括但不限于:

Statistical Inference 统计推断
Statistical Computing 统计计算
Advanced Probability Theory 高等概率论
Advanced Mathematical Statistics 高等数理统计学
(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|MAST90084

统计代写|广义线性模型代写generalized linear model代考|Prospective and Retrospective Sampling

Consider the data shown in Table $4.1$ from a study on infant respiratory disease which shows the proportions of children developing bronchitis or pneumonia in their first year of life by type of feeding and sex, which may be found in Payne (1987):

\begin{tabular}{llll}
& Bottle Only & Some Breast with Supplement & Breast Only \
\hline Boys & $77 / 458$ & $19 / 147$ & $47 / 494$ \
Girls & $48 / 384$ & $16 / 127$ & $31 / 464$
\end{tabular}
Table $4.1$ Incidence of respiratory disease in infants to the age of 1 year.
We can recover the layout above with the proportions as follows:
data (babyfood, package=” faraway”)
xtabs (disease/ (disease + nondisease) $\sim$ sex + food, babyfood)
food
sex $\quad$ Bottle Breast Suppl
G1rl $0.125000 .066810 \quad 0.12598$
In prospective sampling, the predictors are fixed and then the outcome is observed. This is also called a cohort study. In the infant respiratory disease example shown in Table 4.1, we would select a sample of newborn girls and boys whose parents had chosen a particular method of feeding and then monitor them for their first year.

In retrospective sampling, the outcome is fixed and then the predictors are observed. This is also called a case-control study. Typically, we would find infants coming to a doctor with a respiratory disease in the first year and then record their sex and method of feeding. We would also obtain a sample of respiratory diseasefree infants and record their information. The method for obtaining the samples is important – we require that the probability of inclusion in the study is independent of the predictor values.

统计代写|广义线性模型代写generalized linear model代考|Prediction and Effective Doses

Sometimes we wish to predict the outcome for given values of the covariates. For binomial data this will mean estimating the probability of success. Given covariates $x_{0}$, the predicted response on the link scale is $\hat{\eta}=x_{0} \hat{\beta}$ with variance given by $x_{0}^{T}\left(X^{T} W X\right)^{-1} x_{0}$. Approximate confidence intervals may be obtained using a normal approximation. To get an answer in the probability scale, it will be necessary to transform back using the inverse of the link function. We predict the response for the insect data:
data (bliss, packagem” faraway”)
$1 \mathrm{mod}<-$ glm(cbind (dead, alive) conc, familymbinomial, data=bliss)
lmodsum <- summary (lmod)
We show how to predict the response at a dose of $2.5$ :
$x 0<-c(1,2.5)$
eta0 $<-\operatorname{sum}(x 0 * \operatorname{coe}(1 \mathrm{mod}))$
$1 \log i t(\mathrm{eta0})$
[1) $0.64129$
A $64 \%$ predicted chance of death at this dose – now compute a $95 \%$ confidence interval (CI) for this probability. First, extract the variance matrix of the coefficients:
(cm \&- lmodsum\$cov. unscaled) (Intercept) $\begin{array}{rr}\text { (Intercept) } & \text { conc } \ \text { conc } & -0.065823\end{array}$
se <- sqrt $(t(x 0)$ 왛은 $\mathrm{cm}$ 화화 $x 0)$
so the CI on the probability scale is:
ilogit (c (eta0 $-1.96 *$ se, eta0 $1.96 * \mathrm{se})$ )
[1) $0.534300 .73585$
A more direct way of obtaining the same result is:
predict (lmod, newdata data. frame (conc=2.5), se= $T$ )
[1] $0.58095$
\$se.fit
[1] $0.2263$
1logit (c (0.58095-1.960.2263,0.58095+1.960.2263))
[1] $0.534300 .73585$
Note that in contrast to the linear regression situation, there is no distinction possible between confidence intervals for a future observation and those for the mean response. Now we try predicting the response probability at the low dose of $-5$ :
$x 0<-c(1,-5)$
se $<-\operatorname{sqrt}(t(x 0)$ 왛의 $\mathrm{cm} \mathrm{~ ㅇ}$
eta0 <- sum $(x 0 * 1 \mathrm{mod}$ scoef $)$
ilogit (c (eta0 -1.96*se, eta0 $0+1.96 * s e)$ )
[1) $2.3577 \mathrm{e}-053.6429 \mathrm{e}-03$

统计代写|广义线性模型代写generalized linear model代考|Matched Case-Control Studies

In a case-control study, we try to determine the effect of certain risk factors on the outcome. We understand that there are other confounding variables that may affect the outcome. One approach to dealing with these is to measure or record them, include them in the logistic regression model as appropriate and thereby control for

their effect. But this method requires that we model these confounding variables with the correct functional form. This may be difficult. Also, making an appropriate adjustment is problematic when the distribution of the confounding variables is quite different in the cases and controls. So we might consider an alternative where the confounding variables are explicitly adjusted for in the design.

In a matched case-control study, we match each case (diseased person, defective object, success, etc.) with one or more controls that have the same or similar values of some set of potential confounding variables. For example, if we have a 56-year-old, Hispanic male case, we try to match him with some number of controls who are also 56-year-old Hispanic males. This group would be called a matched set. Obviously, the more confounding variables one specifies, the more difficult it will be to make the matches. Loosening the matching requirements, for example, accepting controls who are 50-60 years old, might be necessary. Matching also gives us the possibility of adjusting for confounders that are difficult to measure. For example, suppose we suspect an environmental effect on the outcome. However, it is difficult to measure exposure, particularly when we may not know which substances are relevant. We could match subjects based on their place of residence or work. This would go some way to adjusting for the environmental effects.

Matched case-control studies also have some disadvantages apart from the difficulties of forming the matched sets. One loses the possibility of discovering the effects of the variables used to determine the matches. For example, if we match on sex, we will not be able to investigate a sex effect. Furthermore, the data will likely be far from a random sample of the population of interest. So although relative effects may be found, it may be difficult to generalize to the population.

Sometimes, cases are rare but controls are readily available. A $1: M$ design has $M$ controls for each case. $M$ is typically small and can even vary in size from matched set to matched set due to difficulties in finding matching controls and missing values. Each additional control yields a diminished return in terms of increased efficiency in estimating risk factors – it is usually not worth exceeding $M=5$.

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|Prospective and Retrospective Sampling

考虑表中显示的数据4.1来自一项关于婴儿呼吸道疾病的研究，该研究显示了儿童在出生后第一年发生支气管炎或肺炎的比例，按喂养类型和性别分类，这可以在 Payne (1987) 中找到：

\begin{tabular}{llll} & 瓶装 & 一些含补充剂的乳房 & 仅乳房 \ \hline 男孩 & $77 / 458$ & $19 / 147$ & $47 / 494$ \ 女孩 & $48 / 384$ & $16 / 127$ & $31 / 464$ \end{表格}\begin{tabular}{llll} & 瓶装 & 一些含补充剂的乳房 & 仅乳房 \ \hline 男孩 & $77 / 458$ & $19 / 147$ & $47 / 494$ \ 女孩 & $48 / 384$ & $16 / 127$ & $31 / 464$ \end{表格}
桌子4.11岁以下婴儿呼吸道疾病的发病率。
我们可以按照如下比例恢复上面的布局：
data (babyfood, package=”faraway”)
xtabs (disease/ (disease + nondisease)∼性+食物，婴儿食品）
食物
性奶瓶
补充 G1rl0.125000.0668100.12598
在前瞻性抽样中，预测变量是固定的，然后观察结果。这也称为队列研究。在表 4.1 所示的婴儿呼吸道疾病示例中，我们将选择一个新生儿女孩和男孩的样本，他们的父母选择了一种特定的喂养方法，然后对他们的第一年进行监测。

在回顾性抽样中，结果是固定的，然后观察预测变量。这也称为病例对照研究。通常，我们会发现婴儿在第一年就患有呼吸道疾病就医，然后记录他们的性别和喂养方式。我们还将获得无呼吸道疾病婴儿的样本并记录他们的信息。获取样本的方法很重要——我们要求纳入研究的概率与预测值无关。

统计代写|广义线性模型代写generalized linear model代考|Prediction and Effective Doses

有时我们希望预测给定协变量值的结果。对于二项式数据，这将意味着估计成功的概率。给定协变量X0，链接尺度上的预测响应为这^=X0b^方差由X0吨(X吨在X)−1X0. 可以使用正态近似来获得近似置信区间。为了在概率尺度上得到答案，有必要使用链接函数的逆函数进行转换。我们预测昆虫数据的反应：
数据（幸福，包装“遥远”）
1米○d<−glm(cbind (dead, alive) conc, familymbinomial, data=bliss)
lmodsum <- summary (lmod)
我们展示了如何预测剂量为2.5 :
X0<−C(1,2.5)
和0<−和⁡(X0∗心电图⁡(1米○d))
1日志⁡一世吨(和吨一个0)
[1) 0.64129
一个64%在这个剂量下预测的死亡机会——现在计算一个95%此概率的置信区间 (CI)。首先，提取系数的方差矩阵：
(cm \&- lmodsum $ cov. unscaled) (Intercept) （截距）浓浓 −0.065823
se <- sqrt(吨(X0)哇C米华华X0)
所以概率尺度上的CI为：
ilogit (c (eta0−1.96∗硒, eta01.96∗s和) )
[1) 0.534300.73585
获得相同结果的更直接的方法是：
predict (lmod, newdata data.frame (conc=2.5), se=吨 )
[1] 0.58095
$ se.fit
[1]0.2263
1logit (c (0.58095-1.960.2263,0.58095+1.960.2263))
[1]0.534300.73585
请注意，与线性回归情况相比，未来观察的置信区间和平均响应的置信区间之间没有区别。现在我们尝试预测低剂量的反应概率−5 :
X0<−C(1,−5)
瑟<−平方⁡(吨(X0)谁的ㅇC米是的
eta0 <- 总和(X0∗1米○d斯科夫)
ilogit (c (eta0 -1.96*se, eta00+1.96∗s和) )
[1) 2.3577和−053.6429和−03

统计代写|广义线性模型代写generalized linear model代考|Matched Case-Control Studies

在病例对照研究中，我们试图确定某些风险因素对结果的影响。我们知道还有其他混杂变量可能会影响结果。处理这些问题的一种方法是测量或记录它们，酌情将它们包括在逻辑回归模型中，从而控制

他们的影响。但是这种方法要求我们用正确的函数形式对这些混杂变量进行建模。这可能很困难。此外，当混杂变量的分布在案例和对照中完全不同时，进行适当的调整是有问题的。因此，我们可能会考虑在设计中明确调整混杂变量的替代方案。

在匹配的病例对照研究中，我们将每个病例（患病的人、有缺陷的对象、成功等）与一个或多个具有相同或相似值的一组潜在混杂变量的对照进行匹配。例如，如果我们有一个 56 岁的西班牙裔男性病例，我们会尝试将他与一些同样是 56 岁的西班牙裔男性的对照匹配。该组将被称为匹配集。显然，指定的混杂变量越多，匹配就越困难。放宽匹配要求，例如，接受 50-60 岁的控制，可能是必要的。匹配还为我们提供了调整难以衡量的混杂因素的可能性。例如，假设我们怀疑环境对结果的影响。但是，很难测量曝光，特别是当我们可能不知道哪些物质是相关的时。我们可以根据他们的居住地或工作地点来匹配主题。这将在某种程度上调整环境影响。

除了难以形成匹配集外，匹配病例对照研究也有一些缺点。人们失去了发现用于确定匹配的变量的影响的可能性。例如，如果我们根据性别进行匹配，我们将无法调查性别效应。此外，数据可能与感兴趣人群的随机样本相去甚远。因此，尽管可能会发现相对影响，但可能难以推广到人群。

有时，病例很少见，但控制措施很容易获得。一个1:米设计有米控制每种情况。米由于难以找到匹配控件和缺失值，它通常很小，甚至在匹配集之间的大小也可能不同。就风险因素估计效率的提高而言，每个额外的控制都会产生减少的回报——通常不值得超过米=5.

统计代写请认准statistics-lab™. statistics-lab™为您的留学生涯保驾护航。

金融工程代写

非参数统计代写

非参数统计指的是一种统计方法，其中不假设数据来自于由少数参数决定的规定模型；这种模型的例子包括正态分布模型和线性回归模型。

广义线性模型代考

广义线性模型（GLM）归属统计学领域，是一种应用灵活的线性回归模型。该模型允许因变量的偏差分布有除了正态分布之外的其它分布。

有限元方法代写

随机分析代写

时间序列分析代写

回归分析代写

MATLAB代写

R语言代写	问卷设计与分析代写
PYTHON代写	回归分析与线性模型代写
MATLAB代写	方差分析与试验设计代写
STATA代写	机器学习/统计学习代写
SPSS代写	计量经济学代写
EVIEWS代写	时间序列分析代写
EXCEL代写	深度学习代写
SQL代写	各种数据建模与可视化代写

统计代写|广义线性模型代写generalized linear model代考|STAT8111

Posted on 2022年6月10日2022年6月10日 by statistics-lab

如果你也在怎样代写广义线性模型generalized linear model这个学科遇到相关的难题，请随时右上角联系我们的24/7代写客服。

我们提供的广义线性模型generalized linear model及其相关学科的代写，服务范围广, 其中包括但不限于:

Statistical Inference 统计推断
Statistical Computing 统计计算
Advanced Probability Theory 高等概率论
Advanced Mathematical Statistics 高等数理统计学
(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|STAT8111

统计代写|广义线性模型代写generalized linear model代考|Beta Regression

Beta regression is useful for responses that are bounded in $(0,1)$ such as proportions. It could also be used for variables that are bounded in some other finite interval simply by rescaling to $(0,1)$. A Beta-distributed random variable $Y$ has density:
$$
f(y \mid a, b)=\frac{\Gamma(a+b)}{\Gamma(a) \Gamma(b)} y^{a-1}(1-y)^{b-1}
$$
for parameters $a, b$ and Gamma function $\Gamma()$. It is more convenient to transform the parameters so $\mu=a /(a+b)$ and $\phi=a+b$ so that $E Y=\mu$ and $\operatorname{var} Y=\mu(1-\mu) /(1+$ $\emptyset)$. We can then link the linear predictor $\eta$ using $\eta=g(\mu)$ using a link function $g$ where any of the choices used for the binomial model would be suitable.

An implementation of the Beta regression model can be found in the macv package of Wood (2006). We can apply this to the mammalsleep also used in the previous section.

The default choice of link is the logit function. The estimated value of $\phi$ is $8.927$. A comparison of the fitted values of this model and the quasi-binomial model fitted earlier reveals no substantial difference. The advantage of the Beta-based model is the full distributional model which would allow the construction of full predictive distributions rather than just a point estimate and standard error.

统计代写|广义线性模型代写generalized linear model代考|Latent Variables

Suppose that students answer questions on a test and that a specific student has an aptitude $T$. A particular question might have difficulty $d$ and the student will get the answer correct only if $T>d$. Now if we consider $d$ fixed and $T$ as a random variable with density $f$ and distribution function $F$, then the probability that the student will get the answer wrong is:
$$
p=P(T \leq d)=F(d)
$$
$T$ is called a latent variable. Suppose that the distribution of $T$ is logistic:
$$
F(y)=\frac{\exp (y-\mu) / \sigma}{1+\exp (y-\mu) / \sigma}
$$
$\mathrm{SO}$
$$
\operatorname{logit}(p)=-\mu / \sigma+d / \sigma
$$
If we set $\beta_{0}=-\mu / \sigma$ and $\beta_{1}=1 / \sigma$, we now have a logistic regression model. We can illustrate this in the following example where we set $d=1$ and let $T$ have mean $-1$ and $\sigma=1$ :
$x<-\operatorname{seq}(-6,4,0.1)$
$y<-d \log _{\text {is }}(x$, location $=-1)$
plot (x,y, type=”1″, ylab=”density”, $x l a b=” t “)$
i1 $<-(x<1)$
polygon (c $(x[11], 1,-6), c(y[11], 0,0)$, col=’ gray’)
The plot in Figure $4.1$ shows a logistically distributed latent variable. We can see that this distribution is apparently very similar to the normal distribution. The shaded area represents the probability of getting an answer wrong. As the mean aptitude of this student is somewhat less than the difficulty of the question, this probability is substantially greater than one half.

This idea also arises in a bioassay where we might treat an animal, plant or person with some concentration of a treatment and observe the outcome. For example, suppose we are interested in the concentration of insecticide to be used in exterminating a pest. Insects will have varying tolerances for the toxin and will survive if their tolerance is greater than the dose. In this context, the term tolerance distribution for $T$ is used. Applications in several other areas exist where we observe only a binary outcome but believe this to be generated by some continuous but unobserved variable.

统计代写|广义线性模型代写generalized linear model代考|Link Functions

Until now we have used logit link function to connect the probability and the linear predictor. But other choices of link function are reasonable. We need a function that bounds the probability between zero and one. We also expect the link function to be monotone. It is conceivable that the success probability may go up and down as the linear predictor increases but this circumstance is best modeled by adding nonlinear components to the linear predictors such as quadratic terms rather than modifying the link function. The latent variable formulation suggests some other possibilities for the link function. Here are some choices which are implemented in the glm () function:

Probit: $\eta=\Phi^{-1}(p)$ where $\Phi$ is the normal cumulative distribution function. This arises from a normally distributed latent variable.
Complementary $\log -\log : \eta=\log (-\log (1-p))$. A Gumbel-distributed latent variable will lead to this.
Cauchit: $\eta=\tan ^{-1}(\pi(p-1 / 2))$ which is motivated by a Cauchy-distributed latent variable.

We can illustrate the choices using some data from Bliss (1935) on the numbers of insects dying at different levels of insecticide concentration. We fit all four link functions:

These are not very different, but now look at a wider range from $[-4,8]$. We apply the predict function to each of the four models forming a matrix of predicted values. We label the columns and add the information about the dose. The tidyr package is useful for reformatting data from a wide format of multiple measured values per row to a long format where there is only one response value per row. This is accomplished using the gather () function. This format, where each row is an observation and each column is a variable, is the most convenient form for many $R$ analyses. Finally, the ggplot 2 package is useful for completing and well-labeled plot.

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|Beta Regression

Beta 回归对于有界的响应很有用(0,1)比如比例。它也可以用于限制在其他有限区间内的变量，只需重新缩放到(0,1). 一个 Beta 分布的随机变量是有密度：

F(是∣一个,b)=Γ(一个+b)Γ(一个)Γ(b)是一个−1(1−是)b−1
对于参数一个,b和 Gamma 函数Γ(). 转换参数更方便，所以μ=一个/(一个+b)和φ=一个+b以便和是=μ和曾是⁡是=μ(1−μ)/(1+ ∅). 然后我们可以链接线性预测器这使用这=G(μ)使用链接功能G用于二项式模型的任何选择都是合适的。

Beta 回归模型的实现可以在 Wood (2006) 的 macv 包中找到。我们可以将其应用于上一节中使用的哺乳动物睡眠。

链接的默认选择是 logit 函数。的估计值φ是8.927. 该模型的拟合值与之前拟合的准二项式模型的比较显示没有实质性差异。基于 Beta 的模型的优点是完整的分布模型，它允许构建完整的预测分布，而不仅仅是点估计和标准误差。

统计代写|广义线性模型代写generalized linear model代考|Latent Variables

假设学生在测试中回答问题并且特定学生具有能力吨. 一个特定的问题可能有困难d并且学生只有在以下情况下才能得到正确的答案吨>d. 现在如果我们考虑d固定和吨作为具有密度的随机变量F和分布函数F，则学生答错的概率为：

p=磷(吨≤d)=F(d)
吨称为潜变量。假设分布吨是逻辑：

F(是)=经验⁡(是−μ)/σ1+经验⁡(是−μ)/σ
小号○

罗吉特⁡(p)=−μ/σ+d/σ
如果我们设置b0=−μ/σ和b1=1/σ，我们现在有一个逻辑回归模型。我们可以在下面设置的示例中说明这一点d=1然后让吨有意思−1和σ=1 :
X<−序列⁡(−6,4,0.1)
是<−d日志是 ⁡(X，地点=−1)
绘图（x，y，类型=“1”，ylab=“密度”，Xl一个b=”吨“)
i1<−(X<1)
多边形（c(X[11],1,−6),C(是[11],0,0), col=’gray’)
图中的情节4.1显示了一个逻辑分布的潜变量。我们可以看到，这种分布显然与正态分布非常相似。阴影区域代表回答错误的概率。由于该学生的平均能力略低于问题的难度，因此该概率大大大于二分之一。

这个想法也出现在生物测定中，我们可能会对动物、植物或人进行一定程度的治疗并观察结果。例如，假设我们对用于消灭害虫的杀虫剂浓度感兴趣。昆虫对毒素有不同的耐受性，如果它们的耐受性大于剂量，它们就会存活下来。在这种情况下，术语公差分布吨用来。在其他几个领域的应用中，我们只观察到二元结果，但认为这是由一些连续但未观察到的变量产生的。

统计代写|广义线性模型代写generalized linear model代考|Link Functions

到目前为止，我们已经使用 logit 链接函数来连接概率和线性预测器。但链接功能的其他选择是合理的。我们需要一个将概率限制在零和一之间的函数。我们还期望链接函数是单调的。可以想象，随着线性预测器的增加，成功概率可能会上下波动，但这种情况最好通过向线性预测器添加非线性分量（例如二次项）而不是修改链接函数来建模。潜变量公式为链接函数提出了一些其他的可能性。以下是在 glm () 函数中实现的一些选项：

概率：这=披−1(p)在哪里披是正态累积分布函数。这是由一个正态分布的潜变量引起的。
补充日志−日志:这=日志⁡(−日志⁡(1−p)). 一个 Gumbel 分布的潜在变量将导致这一点。
考希特：这=棕褐色−1⁡(圆周率(p−1/2))这是由柯西分布的潜在变量驱动的。

我们可以使用 Bliss (1935) 中关于在不同杀虫剂浓度水平下死亡的昆虫数量的一些数据来说明这些选择。我们适合所有四个链接功能：

这些并没有太大的不同，但现在从更广泛的范围来看[−4,8]. 我们将预测函数应用于形成预测值矩阵的四个模型中的每一个。我们标记列并添加有关剂量的信息。tidyr 包可用于将数据从每行多个测量值的宽格式重新格式化为每行只有一个响应值的长格式。这是使用gather () 函数完成的。这种格式，其中每一行是一个观察值，每一列是一个变量，对于许多人来说是最方便的形式R分析。最后，ggplot 2 包对于完成和标记良好的绘图很有用。

统计代写请认准statistics-lab™. statistics-lab™为您的留学生涯保驾护航。

金融工程代写

非参数统计代写

非参数统计指的是一种统计方法，其中不假设数据来自于由少数参数决定的规定模型；这种模型的例子包括正态分布模型和线性回归模型。

广义线性模型代考

广义线性模型（GLM）归属统计学领域，是一种应用灵活的线性回归模型。该模型允许因变量的偏差分布有除了正态分布之外的其它分布。

有限元方法代写

随机分析代写

时间序列分析代写

回归分析代写

MATLAB代写

R语言代写	问卷设计与分析代写
PYTHON代写	回归分析与线性模型代写
MATLAB代写	方差分析与试验设计代写
STATA代写	机器学习/统计学习代写
SPSS代写	计量经济学代写
EVIEWS代写	时间序列分析代写
EXCEL代写	深度学习代写
SQL代写	各种数据建模与可视化代写

统计代写|广义线性模型代写generalized linear model代考|OLET5608

Posted on 2022年6月10日2022年6月10日 by statistics-lab

如果你也在怎样代写广义线性模型generalized linear model这个学科遇到相关的难题，请随时右上角联系我们的24/7代写客服。

我们提供的广义线性模型generalized linear model及其相关学科的代写，服务范围广, 其中包括但不限于:

Statistical Inference 统计推断
Statistical Computing 统计计算
Advanced Probability Theory 高等概率论
Advanced Mathematical Statistics 高等数理统计学
(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|OLET5608

统计代写|广义线性模型代写generalized linear model代考|Pearson’s χ 2 Statistic

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 count and $E_{i}$ is the expected count for case $i$. For a binomial response, we count the number of successes for which $O_{i}=y_{i}$ while $E_{i}=n_{i} \hat{p}{i}$ and failures for which $O{i}=n_{i}-y_{i}$ and $E_{i}=n_{i}\left(1-\hat{p}{i}\right)$, 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:
[1] $28.067$
Compare this to:
deviance (lmod)
[1] $16.912$
In this case there is more than the typical small difference between $X^{2}$ and the deviance. However, a test for model fit:
1 -pchisq $(28.067,21)$
[1] $0.13826$
results in a moderate sized $p$-value which would not reject this model which agrees with decision based on the deviance statistic.

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

If the binomial model specification is correct, we expect that the residual deviance will be approximately distributed $\chi^{2}$ with the appropriate degrees of freedom. Sometimes, we observe a deviance that is much larger than would be expected if the model were correct. We must then determine which aspect of the model specification is incorrect.

The most common explanation is that we have the wrong structural form for the model. We have not included the right predictors or we have not transformed or combined them in the correct way. We have a number of ways of determining the importance of potential additional predictors and diagnostics for determining better transformations – see Section 8.4. Suppose, however, that we are able to exclude this explanation. This is difficult to achieve, but when we have only one or two predictors, it is feasible to explore the model space quite thoroughly and be sure that there is not a plausible superior model formula.

Another common explanation for a large deviance is the presence of a small

number of outliers. Fortunately, these are easily checked using diagnostic methods. When larger numbers of points are identified as outliers, they become unexceptional, and we might more reasonably conclude that there is something amiss with the error distribution.

Sparse data can also lead to large deviances. In the extreme case of a binary response, the deviance is not even approximately $\chi^{2}$. In situations where the group sizes are simply small, the approximation is poor. Because we cannot judge the fit using the deviance, we shall exclude this case from further consideration in this section.
Having excluded these other possibilities, we might explain a large deviance by deficiencies in the random part of the model. A binomial distribution for $Y$ arises when the probability of success $p$ is independent and identical for each trial within the group. If the group size is $m$, then var $Y=m p(1-p)$ if the binomial assumptions are correct. However, if the assumptions are broken, the variance may be greater. This is overdispersion. In rarer cases, the variance is less and underdispersion results.

统计代写|广义线性模型代写generalized linear model代考|Quasi-Binomial

In the previous section, we have demonstrated ways to model data where the supposedly binomial response is more variable than should be expected. A quasi-binomial model is another way to allow for extra-binomial variation. We will explain the method in greater generality than immediately necessary because the idea can be used across a wider range of response types.

The idea is to specify only how the mean and variance of the response are connected to the linear predictor. The method of weighted least squares, as used for standard linear models, would be a simple example of this. An examination of the fitting of the binomial model reveals that this only requires the mean and variance information and does not use any additional information about the binomial distribution. Hence, we can obtain the parameter estimates $\hat{\beta}$ and standard errors without making the full binomial assumption.

The problem arises when we attempt to do inference. To construct a confidence interval or perform an hypothesis test, we need some distributional assumptions. Previously we have used the deviance, but for this we need a likelihood and to compute a likelihood we need a distribution. Now we need a suitable substitute for a likelihood that can be computed without assuming a distribution.

Let $Y_{i}$ have mean $\mu_{i}$ and variance $\phi V\left(\mu_{i}\right)$. We assume that $Y_{i}$ are independent. We

define a score, $U_{i}$ :
$$
U_{i}=\frac{Y_{i}-\mu_{i}}{\phi V\left(\mu_{i}\right)}
$$
Now:
$$
\begin{gathered}
E U_{i}=0 \
\operatorname{var} U_{i}=\frac{1}{\phi V\left(\mu_{i}\right)} \
-E \frac{\partial U_{i}}{\partial \mu_{i}}=-E \frac{-\phi V\left(\mu_{i}\right)-\left(Y_{i}-\mu_{i}\right) \phi V^{\prime}\left(\mu_{i}\right)}{\left[\phi V\left(\mu_{i}\right)\right]^{2}}=\frac{1}{\phi V\left(\mu_{i}\right)}
\end{gathered}
$$
These properties are shared by the derivative of the log-likelihood, $l^{\prime}$. This suggests that we can use $U$ in place of $l^{\prime}$. So we define:
$$
Q_{i}=\int_{y_{i}}^{\mu_{i}} \frac{y_{i}-t}{\phi V(t)} d t
$$
The intent is that $Q$ should behave like the log-likelihood. We then define the log quasi-likelihood for all $n$ observations as:
$$
Q=\sum_{i=1}^{n} Q_{i}
$$

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|Pearson’s χ 2 Statistic

偏差是衡量模型与数据拟合程度的一种衡量标准，但还有其他选择。皮尔逊的X2统计量采用一般形式：

X2=∑一世=1n(○一世−和一世)2和一世

在哪里○一世是观察到的计数和和一世是案例的预期计数一世. 对于二项式响应，我们计算成功的次数○一世=是一世尽管和一世=n一世p^一世和失败的原因○一世=n一世−是一世和和一世=n一世(1−p^一世)，这导致：

X2=∑一世=1n(是一世−n一世p^一世)2n一世p^一世(1−p^一世)
如果我们将 Pearson 残差定义为：

r一世磷=(是一世−n一世p^一世)/曾是⁡是^一世
可以看作是一种标准化残差，那么X2=∑一世=1n(r一世磷)2. 所以皮尔逊X2类似于正常线性模型中使用的残差平方和。

皮尔逊X2通常会在大小上接近偏差，并且可以以相同的方式使用。上述假设检验的替代版本可能使用X2代替具有相同近似零分布的偏差。但是，需要注意一些，因为该模型适合最小化偏差，而不是 Pearson’sX2. 这意味着尽管不太可能，但有可能X2可能会随着向模型中添加预测变量而增加。X2可以这样计算：
[1]28.067
将此与：
偏差 (lmod)
[1]进行比较16.912
在这种情况下，两者之间的差异不仅仅是典型的小差异X2和偏差。但是，模型拟合测试：
1 -pchisq(28.067,21)
[1] 0.13826
导致中等大小p- 不会拒绝该模型的值，该模型与基于偏差统计的决策一致。

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

如果二项式模型规范是正确的，我们预计残差将近似分布χ2具有适当的自由度。有时，如果模型正确，我们会观察到比预期大得多的偏差。然后我们必须确定模型规范的哪个方面是不正确的。

最常见的解释是我们的模型结构形式错误。我们没有包含正确的预测变量，或者我们没有以正确的方式转换或组合它们。我们有多种方法来确定潜在的附加预测变量和诊断对确定更好的转换的重要性——参见第 8.4 节。然而，假设我们能够排除这种解释。这很难实现，但是当我们只有一两个预测变量时，可以相当彻底地探索模型空间并确保没有一个看似合理的优越模型公式。

对大偏差的另一个常见解释是存在小偏差

异常值的数量。幸运的是，这些都可以使用诊断方法轻松检查。当大量点被识别为异常值时，它们就变得不例外，我们可能会更合理地得出错误分布存在问题的结论。

稀疏数据也可能导致较大的偏差。在二元响应的极端情况下，偏差甚至不是近似的χ2. 在组大小很小的情况下，近似值很差。因为我们不能使用偏差来判断拟合，我们将在本节进一步考虑这个案例。
排除了这些其他可能性后，我们可能会通过模型随机部分的缺陷来解释大偏差。二项分布是当成功的概率出现时p对于组内的每个试验是独立且相同的。如果组大小是米, 然后 var是=米p(1−p)如果二项式假设是正确的。但是，如果假设被打破，方差可能会更大。这是过度分散。在极少数情况下，方差较小且分散不足。

统计代写|广义线性模型代写generalized linear model代考|Quasi-Binomial

在上一节中，我们已经展示了对假设二项式响应比预期的变化更大的数据进行建模的方法。准二项式模型是允许额外二项式变化的另一种方式。我们将比立即必要的更广泛地解释该方法，因为该想法可以用于更广泛的响应类型。

这个想法是仅指定响应的均值和方差如何连接到线性预测变量。用于标准线性模型的加权最小二乘法就是一个简单的例子。对二项式模型拟合的检查表明，这仅需要均值和方差信息，而无需使用有关二项式分布的任何附加信息。因此，我们可以得到参数估计b^和标准误差，而不做完整的二项式假设。

当我们尝试进行推理时，问题就出现了。为了构建置信区间或执行假设检验，我们需要一些分布假设。以前我们使用过偏差，但为此我们需要一个可能性，并且为了计算一个可能性，我们需要一个分布。现在我们需要一个合适的替代品来代替可以在不假设分布的情况下计算的可能性。

让是一世有意思μ一世和方差φ在(μ一世). 我们假设是一世是独立的。我们

定义分数，在一世 :

在一世=是一世−μ一世φ在(μ一世)
现在：

和在一世=0 曾是⁡在一世=1φ在(μ一世) −和∂在一世∂μ一世=−和−φ在(μ一世)−(是一世−μ一世)φ在′(μ一世)[φ在(μ一世)]2=1φ在(μ一世)
这些属性由对数似然的导数共享，l′. 这表明我们可以使用在代替l′. 所以我们定义：

问一世=∫是一世μ一世是一世−吨φ在(吨)d吨
意图是问应该表现得像对数似然。然后我们定义所有的对数准似然n观察结果为：

问=∑一世=1n问一世

统计代写请认准statistics-lab™. statistics-lab™为您的留学生涯保驾护航。

金融工程代写

非参数统计代写

非参数统计指的是一种统计方法，其中不假设数据来自于由少数参数决定的规定模型；这种模型的例子包括正态分布模型和线性回归模型。

广义线性模型代考

广义线性模型（GLM）归属统计学领域，是一种应用灵活的线性回归模型。该模型允许因变量的偏差分布有除了正态分布之外的其它分布。

有限元方法代写

随机分析代写

时间序列分析代写

回归分析代写

MATLAB代写

R语言代写	问卷设计与分析代写
PYTHON代写	回归分析与线性模型代写
MATLAB代写	方差分析与试验设计代写
STATA代写	机器学习/统计学习代写
SPSS代写	计量经济学代写
EVIEWS代写	时间序列分析代写
EXCEL代写	深度学习代写
SQL代写	各种数据建模与可视化代写

统计代写|广义线性模型代写generalized linear model代考|STA441H5

Posted on 2022年6月10日2022年6月10日 by statistics-lab

如果你也在怎样代写广义线性模型generalized linear model这个学科遇到相关的难题，请随时右上角联系我们的24/7代写客服。

我们提供的广义线性模型generalized linear model及其相关学科的代写，服务范围广, 其中包括但不限于:

Statistical Inference 统计推断
Statistical Computing 统计计算
Advanced Probability Theory 高等概率论
Advanced Mathematical Statistics 高等数理统计学
(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|STA441H5

统计代写|广义线性模型代写generalized linear model代考|Estimation Problems

Estimation of the logistic regression model using the Fisher scoring algorithm, described in Section 8.2, is usually fast. However, difficulties can sometimes arise. When convergence fails, it is sometimes due to a problem exhibited by the following dataset. We take a subset of the famous Fisher Iris data to consider only two of the three species of Iris and use only two of the potential predictors:
ibrary (dplyr)
irisr <- filter(iris, Species ! “virginica”) to ? select (Sepal. Width,
$\rightarrow$ Sepal. Length, Species)
We plot the data using a different shape of plotting symbol for the two species:
(p <- ggplot(irisr, aes (x=Sepa1. Width, y=Sepal. Length, shape=Species))
$\hookrightarrow$ tgeom point ())
We now fit a logistic regression model to see if the species can be predicted from the two sepal dimensions.

$n-100 p-3$ Deviance $-0.000$ Null Deviance $-138.629$ (Difference $-138.629$ ) Notice that the residual deviance is zero indicating a perfect fit and yet none of the predictors are significant due to the high standard errors. A look at the data reveals the reason for this. We see that the two groups are linearly separable so that a perfect fit is possible. We suffer from an embarrassment of riches in this example – we can fit the data perfectly. Unfortunately, this results in unstable estimates of the parameters and their standard errors and would (probably falsely) suggest that perfect predictions can be made. An alternative fitting approach might be considered in such cases called exact logistic regression. See Cox (1970) or Mehta and Patel (1995). Implementations can be found in the elrm and logistix packages in R.

统计代写|广义线性模型代写generalized linear model代考|Binomial Regression Model

Suppose the response variable $Y_{i}$ for $i=1, \ldots, n$ is binomially distributed $B\left(m_{i}, p_{i}\right)$ so that:
$$
P\left(Y_{i}=y_{i}\right)=\left(\begin{array}{c}
m_{i} \
y_{i}
\end{array}\right) p_{i}^{y_{i}}\left(1-p_{i}\right)^{m_{i}-y_{i}}
$$
We further assume that the $Y_{i}$ are independent. The individual outcomes or trials that compose the response $Y_{i}$ are all subject to the same $q$ predictors $\left(x_{i 1}, \ldots, x_{i q}\right)$. The group of trials is known as a covariate class. For example, we might record whether customers of a particular type make a purchase or not. Conventionally, one outcome is labeled a success (say, making purchase in this example) and the other outcome is labeled as a failure. No emotional meaning should be attached to success and failure in this context. For example, success might be the label given to a patient death with survival being called a failure. Because we need to have multiple trials for each covariate class, data for binomial regression models is more likely to result from designed experiments with a few predictors at chosen values rather than observational data which is likely to be more sparse.
As in the binary case, we construct a linear predictor:
$$
\eta_{i}=\beta_{0}+\beta_{1} x_{i 1}+\cdots+\beta_{q} x_{i q}
$$
We can use a logistic link function $\eta_{i}=\log \left(p_{i} /\left(1-p_{i}\right)\right)$. The log-likelihood is then given by:
$$
l(\beta)=\sum_{i=1}^{n}\left[y_{i} \eta_{i}-m_{i} \log \left(1+e_{i}^{\eta}\right)+\log \left(\begin{array}{c}
m_{i} \
y_{i}
\end{array}\right)\right]
$$
Let’s work through an example to see how the analysis differs from the binary response case.

In January 1986, the space shuttle Challenger exploded shortly after launch. An investigation was launched into the cause of the crash and attention focused on the rubber O-ring seals in the rocket boosters. At lower temperatures, rubber becomes more brittle and is a less effective sealant. At the time of the launch, the temperature was $31^{\circ} \mathrm{F}$. Could the failure of the O-rings have been predicted? In the 23 previous shuttle missions for which data exists, some evidence of damage due to blow by and erosion was recorded on some O-rings. Each shuttle had two boosters, each with three O-rings. For each mission, we know the number of $\mathrm{O}$-rings out of six showing some damage and the launch temperature. This is a simplification of the problem see Dalal et al. (1989) for more details.

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

We use the same likelihood-based methods as in Section $2.3$ to derive the binomial deviance:
$$
D=2 \sum_{i=1}^{n}\left{y_{i} \log y_{i} / \hat{y}{i}+\left(m{i}-y_{i}\right) \log \left(m_{i}-y_{i}\right) /\left(m_{i}-\hat{y}{i}\right)\right} $$ where $\hat{y}{i}$ are the fitted values from the model.
Provided that $Y$ is truly binomial and that the $m_{i}$ are relatively large, the deviance is approximately $\chi^{2}$ distributed with $n-q-1$ degrees of freedom if the model is correct. Thus we can use the deviance to test whether the model is an adequate fit. For the logit model of the Challenger data, we may compute:
pchisq (deviance (1mod), df . residual (1mod), lower FALSE)
[1] $0.71641$
Since this $p$-value is well in excess of $0.05$, we conclude that this model fits sufficiently well. Of course, this does not mean that this model is correct or that a simpler model might not also fit adequately. Even so, for the null model:
pchisq $(38.9,22$, lower FALSE)
[1] $0.014489$
We see that the fit is inadequate, so we cannot ascribe the response to simple variation not dependent on any predictor. Note that a $\chi_{d}^{2}$ variable has mean $d$ and standard deviation $\sqrt{2 d}$ so that it is often possible to quickly judge whether a deviance is large or small without explicitly computing the $p$-value. If the deviance is far in excess of the degrees of freedom, the null hypothesis can be rejected.

The $\chi^{2}$ distribution is only an approximation that becomes more accurate as the $m_{i}$ increase. The approximation is very poor for small $m_{i}$ and fails entirely in binary cases where $m_{i}=1$. Although it is not possible to say exactly how large $m_{i}$ should be for an adequate approximation, $m_{i} \geq 5 \forall i$ has often been suggested. Permutation or bootstrap methods might be considered as an alternative.

We can also use the deviance to compare two models, with smaller model $S$ representing a subspace (usually a subset) of a larger model $L$. The likelihood ratio test statistic becomes $D_{S}-D_{L}$. This test statistic is asymptotically distributed $\chi_{l-s}^{2}$, assuming that the smaller model is correct and the distributional assumptions hold. We can use this to test the significance of temperature by computing the difference in the deviances between the model with and without temperature. The model without temperature is just the null model and the difference in degrees of freedom or parameters is one:
pchisq (38.9-16.9,1, lower=FALSE)
[1] $2.7265 \mathrm{e}-06$

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|Estimation Problems

使用第 8.2 节中描述的 Fisher 评分算法估计逻辑回归模型通常很快。然而，有时也会出现困难。当收敛失败时，有时是由于以下数据集表现出的问题。我们采用著名的 Fisher Iris 数据的子集，仅考虑三种鸢尾花中的两种，并仅使用其中两种潜在的预测变量：
ibrary (dplyr)
irisr <- filter(iris, Species ! “virginica”) to ? 选择（萼片。宽度，
→萼片。长度，物种）
我们使用两个物种的不同形状的绘图符号绘制数据：
（p <- ggplot(irisr, aes (x=Sepa1. Width, y=Sepal.Length, shape=Species)）
tgeom point ())
我们现在拟合一个逻辑回归模型，看看是否可以从两个萼片维度预测物种。

n−100p−3偏差−0.000零偏差−138.629（区别−138.629) 请注意，残差为零表示完美拟合，但由于高标准误差，所有预测变量均不显着。看一下数据就可以发现其中的原因。我们看到这两组是线性可分的，因此可以完美拟合。在这个例子中，我们遭受了财富的尴尬——我们可以完美地拟合数据。不幸的是，这会导致参数及其标准误差的估计不稳定，并且会（可能错误地）表明可以做出完美的预测。在这种情况下，可以考虑另一种拟合方法，称为精确逻辑回归。参见 Cox (1970) 或 Mehta 和 Patel (1995)。可以在 R 的 elrm 和 logistix 包中找到实现。

统计代写|广义线性模型代写generalized linear model代考|Binomial Regression Model

假设响应变量是一世为了一世=1,…,n是二项分布的乙(米一世,p一世)以便：

磷(是一世=是一世)=(米一世是一世)p一世是一世(1−p一世)米一世−是一世
我们进一步假设是一世是独立的。构成响应的单个结果或试验是一世都受同样的约束q预测因子(X一世1,…,X一世q). 这组试验称为协变量类。例如，我们可能会记录特定类型的客户是否购买。按照惯例，一个结果被标记为成功（例如，在此示例中进行购买），而另一个结果被标记为失败。在这种情况下，不应该对成功和失败附加任何情感意义。例如，成功可能是患者死亡的标签，而生存则称为失败。因为我们需要对每个协变量类进行多次试验，所以二项式回归模型的数据更有可能来自设计的实验，其中一些预测变量处于选定的值，而不是可能更稀疏的观察数据。
与二进制情况一样，我们构造一个线性预测器：

这一世=b0+b1X一世1+⋯+bqX一世q
我们可以使用逻辑链接功能这一世=日志⁡(p一世/(1−p一世)). 对数似然由下式给出：

l(b)=∑一世=1n[是一世这一世−米一世日志⁡(1+和一世这)+日志⁡(米一世是一世)]
让我们通过一个示例来了解分析与二元响应案例的不同之处。

1986 年 1 月，挑战者号航天飞机在发射后不久爆炸。对坠机原因进行了调查，并将注意力集中在火箭助推器中的橡胶 O 形密封圈上。在较低温度下，橡胶变得更脆，并且是一种不太有效的密封剂。发射时气温为31∘F. O 形圈的故障是否可以预测？在有数据的之前的 23 次航天飞机任务中，一些 O 形环上记录了一些因吹漏和腐蚀而损坏的证据。每个航天飞机有两个助推器，每个助推器有三个 O 形环。对于每个任务，我们知道○- 六环显示一些损坏和发射温度。这是对问题的简化，请参见 Dalal 等人。(1989) 了解更多详情。

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

我们使用与第 1 节中相同的基于可能性的方法2.3导出二项式偏差：

D=2 \sum_{i=1}^{n}\left{y_{i} \log y_{i} / \hat{y}{i}+\left(m{i}-y_{i}\右) \log \left(m_{i}-y_{i}\right) /\left(m_{i}-\hat{y}{i}\right)\right}D=2 \sum_{i=1}^{n}\left{y_{i} \log y_{i} / \hat{y}{i}+\left(m{i}-y_{i}\右) \log \left(m_{i}-y_{i}\right) /\left(m_{i}-\hat{y}{i}\right)\right}在哪里是^一世是模型的拟合值。
前提是是是真正的二项式，并且米一世都比较大，偏差大约是χ2分布于n−q−1模型正确时的自由度。因此，我们可以使用偏差来测试模型是否合适。对于 Challenger 数据的 logit 模型，我们可以计算：
pchisq (deviance (1mod), df .residual (1mod), lower FALSE)
[1]0.71641
从此p-价值远远超过0.05，我们得出结论，该模型非常适合。当然，这并不意味着这个模型是正确的，或者更简单的模型也可能不适合。即便如此，对于空模型：
pchisq(38.9,22, 下 FALSE)
[1]0.014489
我们看到拟合不足，因此我们不能将响应归因于不依赖于任何预测变量的简单变化。请注意，一个χd2变量有均值d和标准差2d因此，通常可以快速判断偏差是大还是小，而无需明确计算p-价值。如果偏差远远超过自由度，则可以拒绝原假设。

这χ2分布只是一个近似值，随着米一世增加。对于小型的近似值很差米一世并且在二进制情况下完全失败米一世=1. 虽然无法确切地说出多大米一世应该是一个足够的近似值，米一世≥5∀一世经常被建议。置换或引导方法可能被视为一种替代方法。

我们也可以使用偏差来比较两个模型，较小的模型小号表示较大模型的子空间（通常是子集）大号. 似然比检验统计量变为D小号−D大号. 该检验统计量是渐近分布的χl−s2，假设较小的模型是正确的并且分布假设成立。我们可以通过计算有温度和没有温度的模型之间的偏差差异来使用它来测试温度的显着性。没有温度的模型只是空模型，自由度或参数的差异为一：
pchisq (38.9-16.9,1, lower=FALSE)
[1]2.7265和−06

统计代写请认准statistics-lab™. statistics-lab™为您的留学生涯保驾护航。

金融工程代写

非参数统计代写

非参数统计指的是一种统计方法，其中不假设数据来自于由少数参数决定的规定模型；这种模型的例子包括正态分布模型和线性回归模型。

广义线性模型代考

广义线性模型（GLM）归属统计学领域，是一种应用灵活的线性回归模型。该模型允许因变量的偏差分布有除了正态分布之外的其它分布。

有限元方法代写

随机分析代写

时间序列分析代写

回归分析代写

MATLAB代写

R语言代写	问卷设计与分析代写
PYTHON代写	回归分析与线性模型代写
MATLAB代写	方差分析与试验设计代写
STATA代写	机器学习/统计学习代写
SPSS代写	计量经济学代写
EVIEWS代写	时间序列分析代写
EXCEL代写	深度学习代写
SQL代写	各种数据建模与可视化代写

统计代写|广义线性模型代写generalized linear model代考|ST411

Posted on 2022年6月10日2022年6月10日 by statistics-lab

如果你也在怎样代写广义线性模型generalized linear model这个学科遇到相关的难题，请随时右上角联系我们的24/7代写客服。

我们提供的广义线性模型generalized linear model及其相关学科的代写，服务范围广, 其中包括但不限于:

Statistical Inference 统计推断
Statistical Computing 统计计算
Advanced Probability Theory 高等概率论
Advanced Mathematical Statistics 高等数理统计学
(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|ST411

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

Regression diagnostics are useful in checking the assumptions of the model and in identifying any unusual points. As with linear models, residuals are the most important means of determining how well the data fits the model and where any changes or

improvements might be advisable. We can compute residuals as a difference between observed and fitted values. There are two kinds of fitted (or predicted) values:
linpred <- prediet (lmod)
predprob <- predict (lmod, type=” response”)
The former is the predicted value in the linear predictor scale, $\eta$, while the latter is the
predicted probability $p=\operatorname{logit}^{-1}(\eta)$. Here are the first few values and a confirmation
of the relationship between them:
head (linpred)
These can also be obtained as residuals (lmod, type $=$ “response”). Following
the standard practice for diagnostics in linear models, we plot the residuals against
the fitted values:

plot (rawres linpred, xlab” “linear predictor”, ylab=” residuals”)
The plot, as seen in the first panel of Figure 2.6, is not very helpful. Because $y=0$ or 1 , the residual can take only two values given a fixed linear predictor. The upper line in the plot corresponds to $y=1$ and the lower line to $y=0$. We gain no insight into the fit of the model. We have chosen to plot the linear predictor rather than the predicted probability on the horizontal axis because the former provides a better spacing of the points in this direction.

统计代写|广义线性模型代写generalized linear model代考|Model Selection

The analysis thus far has used only two of the predictors available but we might construct a better model for the response if we used some of the other predictors. We might find that not all these predictors are helpful in explaining the response. We would like to identify a subset of the predictors that model the response well without including any superfluous predictors.

We could use the inferential methods to construct hypothesis tests to compare various candidate models and use this as a mechanism for choosing a model. Back-

ward elimination is one such method which is relatively easy to implement. The method proceeds sequentially:

Start with the full model including all the available predictors. We can add derived predictors formed from transformations or interactions between two or more predictors.
Compare this model with all the models consisting of one less predictor. Compute the $p$-value corresponding to each dropped predictor. The dropl function in $\mathrm{R}$ can be used for this purpose.
Eliminate the term with largest $p$-value that is greater than some preset critical value, say $0.05$. Return to the previous step. If no such term meets this criterion, stop and use the current model.

Thus predictors are sequentially eliminated until a final model is settled upon. Unfortunately, this is an inferior procedure. Although the algorithm is simple to use, it is hard to identify the problem to which it provides a solution. It does not identify the best set of predictors for predicting future responses. It is not a reliable indication of which predictors are the best explanation for the response. Even if one believes the fiction that there is a true model, this procedure would not be best for identifying such a model.

The Akaike information criterion (AIC) is a popular way of choosing a model see Section A.3 for more. The criterion for a model with likelihood $L$ and number of parameters $q$ is defined by
$$
A I C=-2 \log L+2 q
$$
We select the model with the smallest value of AIC among those under consideration. Any constant terms in the definition of log-likelihood can be ignored when comparing different models that will have the same constants. For this reason we can use $\mathrm{AIC}=$ deviance $+2 q$

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

As mentioned earlier, we cannot use the deviance for a binary response GLM as a measure of fit. We can use diagnostic plots of the binned residuals to help us identify inadequacies in the model but these cannot tell us whether the model fits or not. Even so the process of binning can help us develop a test for this purpose. We divide the observations up into $J$ bins based on the linear predictor. Let the mean response in the $j^{t h}$ bin be $y_{j}$ and the mean predicted probability be $\hat{p}{j}$ with $m{j}$ observations within the bin. We compute these values:
wogsm $<-$ na.omit (wogs)
wegsm \&- mutate (wcgsm, predprob”predict (lmod, type=” response”))
gdf <- group by (wcgsm, cut (linpred, breaksmunique (quantile (linpred,
$\hookrightarrow(1: 100) / 101))))$
hldf <- summarise (gde, y=sum (y), ppredmean (predprob), count=n()) There are a few missing values in the data. The default method is to ignore these cases. The na. omit command drops these cases from the data frame for the purposes of this calculation. We use the same method of binning the data as for the residuals but now we need to compute the number of observed cases of heart disease and total observations within each bin. We also need the mean predicted probability within
each bin. When we make a prediction with probability $p$, we would hope that the event oc-
When we make a prediction with probability $p$, we would hope that the event occurs in practice with that proportion. We can check that by plotting the observed proportions against the predicted probabilities as seen in Figure 2.9. For a wellcalibrated prediction model, the observed proportions and predicted probabilities should be close.

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|Diagnostics

回归诊断可用于检查模型的假设和识别任何异常点。与线性模型一样，残差是确定数据与模型的拟合程度以及任何更改或

改进可能是可取的。我们可以将残差计算为观察值和拟合值之间的差异。拟合（或预测）值有两种：
linpred <- prediet (lmod)
predprob <- predict (lmod, type=”response”)
前者是线性预测量表中的预测值，这，而后者是
预测概率p=罗吉特−1⁡(这). 以下是前几个值以及
它们之间关系的确认：
head (linpred)
这些也可以作为残差获得（lmod, type=“回复”）。按照
线性模型诊断的标准做法，我们将残差与
拟合值作图：

plot (rawres linpred, xlab” “linear predictor”, ylab=”residuals”)
如图 2.6 的第一幅图所示，该图不是很有帮助。因为是=0或 1 ，给定一个固定的线性预测器，残差只能取两个值。图中的上一行对应于是=1和下线是=0. 我们无法深入了解模型的拟合。我们选择在水平轴上绘制线性预测变量而不是预测概率，因为前者在这个方向上提供了更好的点间距。

统计代写|广义线性模型代写generalized linear model代考|Model Selection

到目前为止的分析只使用了两个可用的预测变量，但如果我们使用其他一些预测变量，我们可能会为响应构建一个更好的模型。我们可能会发现并非所有这些预测变量都有助于解释响应。我们想确定一个预测变量的子集，可以很好地模拟响应，而不包括任何多余的预测变量。

我们可以使用推理方法构建假设检验来比较各种候选模型，并将其用作选择模型的机制。后退-

病房消除是一种相对容易实施的方法。该方法按顺序进行：

从包含所有可用预测变量的完整模型开始。我们可以添加由两个或多个预测变量之间的转换或交互形成的派生预测变量。
将此模型与包含一个较少预测变量的所有模型进行比较。计算p- 对应于每个丢弃的预测器的值。中的 dropl 函数R可用于此目的。
消除最大的项p- 大于某个预设临界值的值，例如0.05. 返回上一步。如果没有此类术语符合此标准，请停止并使用当前模型。

因此，预测变量被依次消除，直到最终模型确定。不幸的是，这是一个较差的程序。尽管该算法易于使用，但很难确定它提供解决方案的问题。它没有确定预测未来响应的最佳预测变量集。对于哪些预测变量是响应的最佳解释，这并不是一个可靠的指示。即使人们相信存在一个真实模型的虚构，这个过程也不是识别这种模型的最佳选择。

Akaike 信息标准 (AIC) 是一种流行的选择模型的方法，请参阅第 A.3 节了解更多信息。具有似然性的模型的标准大号和参数数量q定义为

一个我C=−2日志⁡大号+2q
我们在考虑的模型中选择 AIC 值最小的模型。在比较具有相同常数的不同模型时，可以忽略对数似然定义中的任何常数项。出于这个原因，我们可以使用一个我C=越轨+2q

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

如前所述，我们不能使用二元响应 GLM 的偏差作为拟合度量。我们可以使用分箱残差的诊断图来帮助我们识别模型中的不足之处，但这些不能告诉我们模型是否适合。即便如此，分箱过程可以帮助我们为此目的开发测试。我们将观察结果分为Ĵ基于线性预测器的箱。让平均响应在j吨H是吗是j平均预测概率为 $\hat{p} {j}在一世吨H米{j}○bs和r在一个吨一世○ns在一世吨H一世n吨H和b一世n.在和C○米p在吨和吨H和s和在一个l在和s:在○Gs米<-n一个.○米一世吨(在○Gs)在和Gs米&−米在吨一个吨和(在CGs米,pr和dpr○b”pr和d一世C吨(l米○d,吨是p和=”r和sp○ns和”))GdF<−Gr○在pb是(在CGs米,C在吨(l一世npr和d,br和一个ķs米在n一世q在和(q在一个n吨一世l和(l一世npr和d,\hookrightarrow(1: 100) / 101))))HldF<−s在米米一个r一世s和(Gd和,是=s在米(是),ppr和d米和一个n(pr和dpr○b),C○在n吨=n())吨H和r和一个r和一个F和在米一世ss一世nG在一个l在和s一世n吨H和d一个吨一个.吨H和d和F一个在l吨米和吨H○d一世s吨○一世Gn○r和吨H和s和C一个s和s.吨H和n一个.○米一世吨C○米米一个nddr○ps吨H和s和C一个s和sFr○米吨H和d一个吨一个Fr一个米和F○r吨H和p在rp○s和s○F吨H一世sC一个lC在l一个吨一世○n.在和在s和吨H和s一个米和米和吨H○d○Fb一世nn一世nG吨H和d一个吨一个一个sF○r吨H和r和s一世d在一个lsb在吨n○在在和n和和d吨○C○米p在吨和吨H和n在米b和r○F○bs和r在和dC一个s和s○FH和一个r吨d一世s和一个s和一个nd吨○吨一个l○bs和r在一个吨一世○ns在一世吨H一世n和一个CHb一世n.在和一个ls○n和和d吨H和米和一个npr和d一世C吨和dpr○b一个b一世l一世吨是在一世吨H一世n和一个CHb一世n.在H和n在和米一个ķ和一个pr和d一世C吨一世○n在一世吨Hpr○b一个b一世l一世吨是p,在和在○在ldH○p和吨H一个吨吨H和和在和n吨○C−在H和n在和米一个ķ和一个pr和d一世C吨一世○n在一世吨Hpr○b一个b一世l一世吨是p$，我们希望事件在实践中以该比例发生。我们可以通过将观察到的比例与预测概率作图来检查这一点，如图 2.9 所示。对于校准良好的预测模型，观察到的比例和预测概率应该接近。

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金融工程代写

非参数统计代写

非参数统计指的是一种统计方法，其中不假设数据来自于由少数参数决定的规定模型；这种模型的例子包括正态分布模型和线性回归模型。

广义线性模型代考

广义线性模型（GLM）归属统计学领域，是一种应用灵活的线性回归模型。该模型允许因变量的偏差分布有除了正态分布之外的其它分布。

有限元方法代写

随机分析代写

时间序列分析代写

回归分析代写

MATLAB代写

R语言代写	问卷设计与分析代写
PYTHON代写	回归分析与线性模型代写
MATLAB代写	方差分析与试验设计代写
STATA代写	机器学习/统计学习代写
SPSS代写	计量经济学代写
EVIEWS代写	时间序列分析代写
EXCEL代写	深度学习代写
SQL代写	各种数据建模与可视化代写

统计代写|广义线性模型代写generalized linear model代考|STAT 7430

Posted on 2022年6月10日2022年6月10日 by statistics-lab

如果你也在怎样代写广义线性模型generalized linear model这个学科遇到相关的难题，请随时右上角联系我们的24/7代写客服。

我们提供的广义线性模型generalized linear model及其相关学科的代写，服务范围广, 其中包括但不限于:

Statistical Inference 统计推断
Statistical Computing 统计计算
Advanced Probability Theory 高等概率论
Advanced Mathematical Statistics 高等数理统计学
(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|STAT 7430

统计代写|广义线性模型代写generalized linear model代考|Heart Disease Example

What might affect the chance of getting heart disease? One of the earliest studies addressing this issue started in 1960 and used 3154 healthy men, aged from 39 to 59 , from the San Francisco area. At the start of the study, all were free of heart disease. Eight and a half years later, the study recorded whether these men now suffered from heart disease along with many other variables that might be related to the chance of developing this disease. We load a subset of this data from the Western Collaborative Group Study described in Rosenman et al. (1975):
data (wcgs, package=”faraway”)
We start by focusing on just three of the variables in the dataset:
We see that only 257 men developed heart disease as given by the factor variable chd. The men vary in height (in inches) and the number of cigarettes (cigs) smoked per day. We can plot these data using $R$ base graphics:
plot (height chd, wcgs)
wcgs\$y <- ifelse (wogs\$chd = “no”, 0, 1)
plot (jitter $(y, 0.1) \sim$ jitter (height), wegs, $x l a b=$ “Height”, $y$ lab=”Heart
$\hookrightarrow$ Disease”, peh=”.”)
The first panel in Figure $2.1$ shows a boxplot. This shows the similarity in the distribution of heights of the two groups of men with and without heart disease. But the heart disease is the response variable so we might prefer a plot which treats it as such. We convert the absence/presence of disease into a numerical $0 / 1$ variable and plot this in the second panel of Figure 2.1. Because heights are reported as round numbers of inches and the response can only take two values, it is sensible to add a small amount of noise to each point, called jittering, so that we can distinguish them. Again we can see the similarity in the distributions. We might think about fitting a line to this plot.

More informative plots may be obtained using the ggplot2 package of Wickham (2009). In the first panel of Figure 2.2, we see two histograms showing the distribution of heights for both those with and without heart disease. The dodge option ensures that the two histograms are interleaved. We see that the two distributions are similar. We also had to set the bin width of the histogram. It was natural to use one inch as all the height measurements are rounded to the nearest inch. In the second panel of Figure $2.2$, we see the corresponding histograms for smoking. In this case, we have shown the frequency rather than the count version of the histogram. We see that smokers are more likely to get heart disease.

统计代写|广义线性模型代写generalized linear model代考|Logistic Regression

Suppose we have a response variable $Y_{i}$ for $i=1, \ldots, n$ which takes the values zero or one with $P\left(Y_{i}=1\right)=p_{i}$. This response may be related to a set of $q$ predictors $\left(x_{i 1}, \ldots, x_{i q}\right)$. We need a model that describes the relationship of $x_{1}, \ldots, x_{q}$ to the probability $p$. Following the linear model approach, we construct a linear predictor:
$$
\eta_{i}=\beta_{0}+\beta_{1} x_{i 1}+\cdots+\beta_{q} x_{i q}
$$
Since the linear predictor can accommodate quantitative and qualitative predictors with the use of dummy variables and also allows for transformations and combinations of the original predictors, it is very flexible and yet retains interpretability. The idea that we can express the effect of the predictors on the response solely through the linear predictor is important. The idea can be extended to models for other types of response and is one of the defining features of the wider class of generalized linear models (GLMs) discussed later in Chapter 8 .

We have seen previously that the linear relation $\eta_{i}=p_{i}$ is not workable because we require $0 \leq p_{i} \leq 1$. Instead we shall use a link function $g$ such that $\eta_{i}=g\left(p_{i}\right)$. We need $g$ to be monotone and be such that $0 \leq g^{-1}(\eta) \leq 1$ for any $\eta$. The most popular choice of link function in this situation is the logit. It is defined so that:
$$
\eta=\log (p /(1-p))
$$
or equivalently:
$$
p=\frac{e^{\eta}}{1+e^{\eta}}
$$
Combining the use of the logit link with a linear predictor gives us the term logistic regression. Other choices of link function are possible but we will defer discussion of these until later. The logit and its inverse are defined as logit and ilogit in the faraway package. The relationship between $p$ and the linear predictor $\eta$ is shown in Figure 2.4.

统计代写|广义线性模型代写generalized linear model代考|Consider two models

Consider two models, a larger model with $l$ parameters and likelihood $L_{L}$ and a smaller model with $s$ parameters and likelihood $L_{S}$ where the smaller model represents a subset (or more generally a linear subspace) of the larger model. Likelihood

methods suggest the likelihood ratio statistic:
$$
2 \log \frac{L_{L}}{L_{S}}
$$
as an appropriate test statistic for comparing the two models. Now suppose we choose a saturated larger model – such a model typically has as many parameters as cases and has fitted values $\hat{p}{i}=y{i}$. The test statistic becomes:
$$
D=-2 \sum_{i=1}^{n} \hat{p}{i} \operatorname{logit}\left(\hat{p}{i}\right)+\log \left(1-\hat{p}{i}\right) $$ where $\hat{p}{i}$ are the fitted values from the smaller model. $D$ is called the deviance and is useful in making hypothesis tests to compare models.

In other examples of GLMs, the deviance is a measure of how well the model fit the data but in this case, $D$ is just a function of the fitted values $\hat{p}$ so it cannot be used for that purpose. Other methods must be used to judge goodness of fit for binary data – for example, the Hosmer-Lemeshow test described in Section 2.6.
In the summary output previously, we had:
Deviance – 1749.049 Nu11 Deviance $=1781.244$ (Difference – 32.195)
The Deviance is the deviance for the current model while the Nu1l Deviance is the deviance for a model with no predictors and just an intercept term.

We can use the deviance to compare two nested models. The test statistic in (2.1) becomes $D_{S}-D_{L}$. This test statistic is asymptotically distributed $\chi_{l-s}^{2}$, assuming that the smaller model is correct and the distributional assumptions hold. For example, we can compare the fitted model to the null model (which has no predictors) by considering the difference between the residual and null deviances. For the heart disease example, this difference is $32.2$ on two degrees of freedom (one for each predictor). Hence, the $p$-value for the test of the hypothesis that at least one of the predictors is related to the response is:
1-pchisq $(32,2,2)$
(1) $1.0183 \mathrm{e}-07$
Since this value is so small, we are confident that there is some relationship between the predictors and the response. Note that the expected value of a $\chi^{2}$-variate with $d$ degrees of freedom is simply $d$ so we knew the $p$-value would be small before even calculating it.

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|Heart Disease Example

哪些因素会影响患心脏病的几率？最早解决这一问题的研究之一始于 1960 年，使用了来自旧金山地区的 3154 名年龄在 39 至 59 岁之间的健康男性。在研究开始时，所有人都没有心脏病。八年半后，该研究记录了这些男性现在是否患有心脏病以及许多其他可能与患这种疾病的机会有关的变量。我们从 Rosenman 等人描述的西方合作小组研究中加载了这些数据的一个子集。(1975):
data (wcgs, package=”faraway”)
我们首先只关注数据集中的三个变量：
我们看到只有 257 名男性患上了由因子变量 chd 给出的心脏病。男性的身高（英寸）和每天抽的香烟（cigs）的数量各不相同。我们可以使用绘制这些数据R基本图形：
plot (height chd, wcgs)
wcgs $ y <- ifelse (wogs $ chd = “no”, 0, 1)
plot (jitter(是,0.1)∼抖动（高度），wegs，Xl一个b=“高度”，是实验室=”心脏
疾病”, peh=”.”)
图中的第一个面板2.1显示箱线图。这显示了有和没有心脏病的两组男性的身高分布相似。但是心脏病是响应变量，所以我们可能更喜欢这样对待它的情节。我们将疾病的缺席/存在转换为数字0/1变量并将其绘制在图 2.1 的第二个面板中。因为高度以英寸为整数，并且响应只能取两个值，所以在每个点上添加少量噪声（称为抖动）是明智的，这样我们就可以区分它们。我们再次可以看到分布的相似性。我们可能会考虑为该图拟合一条线。

使用 Wickham (2009) 的 ggplot2 包可以获得更多信息图。在图 2.2 的第一幅图中，我们看到了两个直方图，显示了患有和不患有心脏病的人的身高分布。闪避选项确保两个直方图交错。我们看到这两个分布是相似的。我们还必须设置直方图的 bin 宽度。使用一英寸是很自然的，因为所有高度测量值都四舍五入到最接近的英寸。在图的第二个面板中2.2，我们看到吸烟的相应直方图。在这种情况下，我们显示了直方图的频率而不是计数版本。我们看到吸烟者更容易患心脏病。

统计代写|广义线性模型代写generalized linear model代考|Logistic Regression

假设我们有一个响应变量是一世为了一世=1,…,n它采用零或一的值磷(是一世=1)=p一世. 该响应可能与一组q预测因子(X一世1,…,X一世q). 我们需要一个模型来描述X1,…,Xq到概率p. 遵循线性模型方法，我们构建了一个线性预测器：

这一世=b0+b1X一世1+⋯+bqX一世q
由于线性预测器可以通过使用虚拟变量来适应定量和定性预测器，并且还允许原始预测器的转换和组合，因此它非常灵活并且保留了可解释性。我们可以仅通过线性预测器来表达预测器对响应的影响的想法很重要。这个想法可以扩展到其他类型响应的模型，并且是第 8 章后面讨论的更广泛类别的广义线性模型 (GLM) 的定义特征之一。

我们之前已经看到线性关系这一世=p一世不可行，因为我们需要0≤p一世≤1. 相反，我们将使用链接功能G这样这一世=G(p一世). 我们需要G是单调的并且是这样的0≤G−1(这)≤1对于任何这. 在这种情况下，最流行的链接函数选择是 logit。它被定义为：

这=日志⁡(p/(1−p))
或等效地：

p=和这1+和这
将 logit 链接的使用与线性预测器相结合，我们得到了术语逻辑回归。链接功能的其他选择是可能的，但我们将推迟讨论这些直到稍后。logit 及其倒数在 faraway 包中定义为 logit 和 igit。之间的关系p和线性预测器这如图 2.4 所示。

统计代写|广义线性模型代写generalized linear model代考|Consider two models

考虑两个模型，一个更大的模型l参数和可能性大号大号和一个较小的模型s参数和可能性大号小号其中较小的模型表示较大模型的子集（或更一般地说是线性子空间）。可能性

方法建议似然比统计：

2日志⁡大号大号大号小号
作为比较两个模型的适当检验统计量。现在假设我们选择一个饱和的更大模型——这样的模型通常具有与案例一样多的参数，并且具有拟合值 $\hat{p} {i}=y {i}.吨H和吨和s吨s吨一个吨一世s吨一世Cb和C○米和s:D=−2∑一世=1np^一世罗吉特⁡(p^一世)+日志⁡(1−p^一世)在H和r和\帽子{p}{i}一个r和吨H和F一世吨吨和d在一个l在和sFr○米吨H和s米一个ll和r米○d和l.D$ 称为偏差，可用于进行假设检验以比较模型。

在 GLM 的其他示例中，偏差是衡量模型与数据拟合程度的指标，但在这种情况下，D只是拟合值的函数p^所以它不能用于那个目的。必须使用其他方法来判断二进制数据的拟合优度——例如，第 2.6 节中描述的 Hosmer-Lemeshow 检验。
在之前的摘要输出中，我们有：
Deviance – 1749.049 Nu11 Deviance=1781.244（差值 – 32.195）
偏差是当前模型的偏差，而 Nu1l 偏差是没有预测变量且只有截距项的模型的偏差。

我们可以使用偏差来比较两个嵌套模型。(2.1) 中的检验统计量变为D小号−D大号. 该检验统计量是渐近分布的χl−s2，假设较小的模型是正确的并且分布假设成立。例如，我们可以通过考虑残差和空偏差之间的差异来将拟合模型与空模型（没有预测变量）进行比较。对于心脏病示例，这种差异是32.2在两个自由度上（每个预测变量一个）。因此，p- 至少有一个预测变量与响应相关的假设检验的值为：
1-pchisq(32,2,2)
(1) 1.0183和−07
由于这个值非常小，我们确信预测变量和响应之间存在某种关系。注意 a 的期望值χ2- 与d自由度很简单d所以我们知道p-value 在计算之前会很小。

统计代写请认准statistics-lab™. statistics-lab™为您的留学生涯保驾护航。

R语言代写	问卷设计与分析代写
PYTHON代写	回归分析与线性模型代写
MATLAB代写	方差分析与试验设计代写
STATA代写	机器学习/统计学习代写
SPSS代写	计量经济学代写
EVIEWS代写	时间序列分析代写
EXCEL代写	深度学习代写
SQL代写	各种数据建模与可视化代写