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

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• 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|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:
$$Q=\sum_{i=1}^{n} Q_{i}$$

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

X2=∑一世=1n(○一世−和一世)2和一世

X2=∑一世=1n(是一世−n一世p^一世)2n一世p^一世(1−p^一世)

r一世磷=(是一世−n一世p^一世)/曾是⁡是^一世

[1]28.067

[1]进行比较16.912

1 -pchisq(28.067,21)
[1] 0.13826

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