这是一份kcl伦敦大学学院 7CCMCS06作业代写的成功案
This expression shows that independent variables act multiplicatively on the baseline hazard, $\lambda_{0}$. With no loss of generality, the baseline hazard term $\left(\lambda_{0}\right)$ can be absorbed into the intercept term as
$$
\begin{aligned}
\lambda_{i} &=\exp \left(\sum_{k} \beta_{k} x_{i k}\right) \quad k=0, \ldots, K \
&=\exp \left(\mathbf{x}_{i}^{\prime} \boldsymbol{\beta}\right),
\end{aligned}
$$
When the independent variables are categorical (or can be treated as such), the data can be grouped as in Table 5.2. For a contingency table with $J$ cells, the data likelihood can be expressed in terms of the cell-specific occurrences $\left(D_{j}\right)$ and exposures $\left(E_{j}\right)(j=1, \ldots, J)$ as
$$
L=\prod_{j=1}^{J} \lambda_{j}^{D_{j}} \exp \left(-E_{j} \lambda_{j}\right)
$$
7CCMCS06 COURSE NOTES ：
$$
\mu_{i}=t_{i} \lambda_{i}=t_{i} \exp \left(\mathbf{x}{i}^{\prime} \boldsymbol{\beta}\right) . $$ The likelihood is a product of the individual Poisson probabilities in Eq. $5.11$ and is proportional to $$ L=\prod{i=1}^{n}\left(t_{i} \lambda_{i}\right)^{d_{i}} \exp \left(-t_{i} \lambda_{i}\right) .
$$
To show the equivalence between Poisson regression and the exponential rate model, note that exponential likelihood in Eq. $5.9$ can also be written as
$$
L=\prod_{i=1}^{n}\left(t_{i} \lambda_{i}\right)^{d_{i}} \exp \left(-t_{i} \lambda_{i}\right) / \prod_{i=1}^{n} t_{i}^{d_{i}}
$$