这是一份northeastern东北大学(美国) MATH 3181作业代写的成功案
In this random intercept and trend model, the Cholesky factor equals
$$
\boldsymbol{T}=\left[\begin{array}{cc}
\sigma_{v_{0}} & 0 \
\frac{\sigma_{v_{01}}}{\sigma_{v_{0}}} & \left(\sigma_{v_{1}}^{2}-\frac{\sigma_{v v_{1}}^{2}}{\sigma_{v_{0}}^{2}}\right)^{1 / 2}
\end{array}\right]
$$
thus, the relationship between the unstandardized and standardized random effects is
$$
\begin{aligned}
v_{0 i} &=\sigma_{\nu_{0}} \theta_{0 i} \
v_{1 i} &=\frac{\sigma_{v_{01}}}{\sigma_{v_{0}}} \theta_{0 i}+\left(\sigma_{v_{1}}^{2}-\frac{\sigma_{v_{01}}^{2}}{\sigma_{v_{0}}^{2}}\right)^{1 / 2} \theta_{1 i} .
\end{aligned}
$$
The results of this analysis are listed in Table 10.3. Comparing this model to the previous one via a likelihood ratio test to assess whether the random trends are significant (i.e., $H_{0}: \sigma_{v_{1}}^{2}=\sigma_{v_{p} v_{1}}=0$ ) yields $X_{2}^{2}=77.90, p<.001$. Clearly there is strong evidence of subject heterogeneity in the the time trends.
MATH 3181COURSE NOTES ：
$$
p_{i j c}=\frac{\exp \left(\mathrm{z}{i j c}\right)}{\sum{h=1}^{C} \exp \left(\mathrm{z}{i j h}\right)} \text { for } c=1,2, \ldots, C, $$ where now $$ \mathrm{z}{i j c}=\boldsymbol{x}{i j}^{\prime} \boldsymbol{\Gamma} \boldsymbol{d}{c}+\left(\boldsymbol{z}{i j}^{\prime} \otimes \boldsymbol{\theta}{i}^{\prime}\right) \mathrm{J}{\mathrm{r}^{*}}^{\prime} \boldsymbol{\Lambda} \boldsymbol{d}{e}
$$