# 统计代写|广义线性模型代写generalized linear model代考|STAT6175

## 统计代写|广义线性模型代写generalized linear model代考|Checking the missing data mechanism via a graphical procedure

As shown in Table $7.1$, the Beat the Blues Data set includes a lot of missing data due to dropouts. As described in Chapter 6, if the missing data can be assumed to be MCAR or MAR, we can apply the likelihood-based ignorable analysis by simply ignoring the records that include missing data. Although we cannot judge from the observed data at hand whether the missing data are MCAR, MAR or MNAR, we can apply some ad hoc graphical procedures for assessing the missing data mechanism to a certain extent in longitudinal study. So, let us apply here the procedure suggested by Carpenter et al. (2002). It involves just plotting the repeated measurements at each time point, differentiating between two groups of subjects; those who come in (o) and those who do not come in $(\times)$ to their next scheduled visit. Figure $7.1$ shows such a plot for all the subjects combined. This figure indicates that there is no obvious difference between the distributions of the observed BDI scores for these two groups (o vs. $x$ ) at each visit. This observation can be expressed as the following equations:
$$\operatorname{Pr}\left{B D I_j \mid r_{B D I_{j+1}}=0\right}=\operatorname{Pr}\left{B D I_j \mid r_{B D I_{j+1}}=1\right}$$
Namely, we have
$$\frac{\operatorname{Pr}\left{r_{B D I_{j+1}}=0 \mid B D I_j\right}}{\operatorname{Pr}\left{r_{B D I_{j+1}}=0\right}}=\frac{\operatorname{Pr}\left{r_{B D I_{j+1}}=1 \mid B D I_j\right}}{\operatorname{Pr}\left{r_{B D I_{j+1}}=1\right}} .$$
This means that, given the value of $B D I_j$, the conditional probability of the observation at the next scheduled visit being missing, the probability of an observation being missing does not depend on the observed data at a previous scheduled visit, i.e.,
$$\operatorname{Pr}\left{r_{B D I_{j+1}}=\delta \mid B D I_j\right}=\operatorname{Pr}\left{r_{B D I_{j+1}}=\delta\right},(\delta=0,1)$$

## 统计代写|广义线性模型代写generalized linear model代考|Model IV: Random intercept model

The random intercept model with a random intercept $b_{0 i}$ is expressed as
\begin{aligned} &E\left(y_{i 0} \mid b_{0 i}\right) \sim\left{\begin{array}{l} \beta_0+b_{0 i}+\boldsymbol{w}i^t \boldsymbol{\xi}, i=1, \ldots, n_1 \text { (TAU group) } \ \beta_0+b{0 i}+\beta_1+\boldsymbol{w}i^t \boldsymbol{\xi}, i=n_1+1, \ldots, N \text { (BtheB group) } \end{array}\right. \ &E\left(y{i j} \mid b_{0 i}\right) \sim\left{\begin{array}{l} \beta_0+b_{0 i}+\beta_2+\boldsymbol{w}i^t \boldsymbol{\xi}, i=1, \ldots, n_1(\text { TAU group) } \ \beta_0+b{0 i}+\beta_1+\beta_2+\beta_3+\boldsymbol{w}i^t \boldsymbol{\xi}, i=n_1+1, \ldots, N \end{array}\right. \ &j=1, \ldots, 4, \ &\text { (BtheB group) } \end{aligned} which can be re-expressed as \begin{aligned} y{i j} \mid b_{0 i}=& \beta_0+b_{0 i}+\beta_1 x_{1 i}+\beta_2 x_{2 i j} \ \quad+\beta_3 x_{1 i} x_{2 i j}+\boldsymbol{w}i^t \boldsymbol{\xi}+\epsilon{i j} \ & i=1, \ldots, N ; j=0,1, \ldots, 4 \ b_{0 i} \sim N\left(0, \sigma_{B 0}^2\right), \quad \epsilon_{i j} \sim N\left(0, \sigma_E^2\right) \end{aligned}
where the $b_{0 i}$ and $\epsilon_{i j}$ are assumed to be independent of each other and the meanings of the fixed-effects parameters are as follows:

1. $\beta_0$ denotes the mean $\mathrm{BDI}$ score at the baseline period in the TAU group.
2. $\beta_0+\beta_1$ denotes the mean $\mathrm{BDI}$ score at the baseline period in the BtheB group.
3. $\beta_0+\beta_2$ denotes the mean BDI score during the evaluation period in the TAU group.
4. $\beta_0+\beta_1+\beta_2+\beta_3$ denotes the mean $\mathrm{BDI}$ score during the evaluation period in the BtheB group.
5. $\beta_2$ denotes the mean change from the baseline period to the evaluation period in the TAU group.
6. $\beta_2+\beta_3$ denotes the same quantity in the BtheB group.

## 统计代写|广义线性模型代写generalized linear model代考|Model IV: Random intercept model

$\$ \\backslash begin{aligned \&E \left(y_{ 0} \backslash mid b_{0 i}\right) \backslash sim \left{ \beta_0+b_{0 i}+\boldsymbol{w} i^t \boldsymbol{\xi}, i=1, \ldots, n_1 (TAU group) \beta_0+b 0 i+\beta_1+\boldsymbol{w} i^t \boldsymbol{\xi}, i=n_1+1, \ldots, N (BtheB group) \beta_0+b_{0 i}+\beta_2+\boldsymbol{w} i^t \boldsymbol{\xi}, i=1, \ldots, n_1 (TAU group) \beta_0+b 0 i+\beta_1+\beta_2+\beta_3+\boldsymbol{w} i^t \boldsymbol{\xi}, i=n_1+1, \ldots, N 正确的。 \backslash \& j=1, \Idots, 4 ＼backslash \& \backslash text {( BtheB 组 )} \backslash end { 对齐 } whichcanbere – expressedas y i j \mid b_{0 i}=\beta_0+b_{0 i}+\beta_1 x_{1 i}+\beta_2 x_{2 i j} \quad+\beta_3 x_{1 i} x_{2 i j}+\boldsymbol{w} i^t \boldsymbol{\xi}+\epsilon i j \quad i=1, \ldots, N ; j=0,1, \ldots, 4 b_{0 i} \sim N \ \

1. $\beta_0$ 表示平均值BDI在 $\mathrm{TAU}$ 组的基线期得分。
2. $\beta_0+\beta_1$ 表示平均值BDIBtheB 组的基线期得分。
3. $\beta_0+\beta_2$ 表示 $\mathrm{TAU}$ 组评估期间的平均 BDI 分数。
4. $\beta_0+\beta_1+\beta_2+\beta_3$ 表示平均值BDIBtheB 组在评估期间的得分。
5. $\beta_2$ 表示 $\mathrm{TAU}$ 组中从基线期到评估期的平均变化。
6. $\beta_2+\beta_3$ 表示 BtheB 组中的相同数量。

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