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统计代写|生存模型代考Survival Models代写|Available Data
Because of the assumption of constant forces of progression, Panjer’s model is quantified by estimating the values of $\mu_j$, for $j=1 a, 1 b, 2 a, 2 b, 3$. As demonstrated several times in Part II of this text (see Equations (7.15) and (7.23)), if we have sample data showing the exact time under observation at which sample members progressed to the next state, then we can easily calculate exact exposure and find the maximum likelihood estimate of each constant force of progression as the ratio of number of progressions to exact exposure.
In practice, however, it is not always the case that sample data exists in such an ideal form, so that the researcher is faced with the issue of how best to adapt the estimation procedure to the form of available data. This was true at the time of Panjer’s study. Panjer used data that was available from a study done at the University of Frankfurt in Germany.
The Frankfurt data recorded the following three pieces of information about each person in the study:
(1) The length of time the person was under observation.
(2) The stage the person was in at the beginning of observation (the initial stage).
(3) Whether the person remained in the initial stage throughout observation, or left that stage while under observation.
统计代写|生存模型代考Survival Models代写|Estimating Model Parameters from Sample Data
The parameters $\mu_j$, for $j=1 a, 1 b, 2 a, 2 b, 3$, are estimated separately from the data described in Section 10.2.3 by the maximum likelihood approach. For the $j^{i h}$ initial stage, recall that the Frankfurt data shows, for each of four time-under-observation groups, the total number in that group $\left(n_i\right)$ and the number (subset) of them who progressed out $\left(d_i\right)$ while under observation, where $n_i$ and $d_i$ refer to the $i^{\text {th }}$ time-under-observation group. Then $n_i-d_i$ denotes the number who remained in the initial stage throughout observation.
Let $p_i$ denote the probability of remaining in the initial stage and $q_i=1-p_i$ denote the probability of progressing out of the initial stage. It is clear that we have a binomial model here, so the likelihood for estimating $\mu_j$ is
$$
L_i\left(\mu_j\right)=\left(1-p_i\right)^{d_i} \cdot\left(p_i\right)^{n_i-d_i}
$$
for the $i^{\text {th }}$ time-under-observation group and
$$
L\left(\mu_j\right)=\prod_{i=1}^4\left(1-p_i\right)^{d_i} \cdot\left(p_i\right)^{n_i-d_i},
$$
since there are four time-under-observation groups. The log-likelihood is
$$
\ell\left(\mu_j\right)=\sum_{i=1}^4 d_i \cdot \log \left(1-p_i\right)+\left(n_i-d_i\right) \cdot \log p_i
$$
and the likelihood equation is
$$
\frac{\partial \ell\left(\mu_j\right)}{\partial \mu_j}=\sum_{i=1}^4-\frac{\partial p_i}{\partial \mu_j}\left[\frac{d_i}{1-p_i}-\frac{n_i-d_i}{p_i}\right]=0,
$$
which can be solved for $\hat{\mu}_j$ by iteration.
生存模型代考
统计代写|生存模型代考Survival Models代写|Available Data
由于假设前进力恒定,Panjer的模型通过估算$j=1 a, 1 b, 2 a, 2 b, 3$的$\mu_j$值来量化。正如本文第二部分多次展示的那样(参见公式(7.15)和(7.23)),如果我们有样本数据显示样本成员进展到下一个状态的确切观察时间,那么我们可以很容易地计算出精确的曝光,并找到每个恒定的进展力的最大似然估计,即进展数量与精确曝光的比率。
然而,在实践中,样本数据并不总是以这种理想的形式存在,因此研究人员面临的问题是如何最好地使估计过程适应可用数据的形式。这在Panjer的研究中是正确的。Panjer使用的数据来自德国法兰克福大学的一项研究。
法兰克福的数据记录了研究中每个人的以下三条信息:
(一)被观察的时间长短。
(2)观察开始时所处的阶段(初始阶段)。
(三)被观察人在整个观察过程中是否处于初始阶段,或者在观察过程中是否离开了初始阶段。
统计代写|生存模型代考Survival Models代写|Estimating Model Parameters from Sample Data
对于$j=1 a, 1 b, 2 a, 2 b, 3$,参数$\mu_j$是通过最大似然方法从第10.2.3节中描述的数据中单独估计的。对于$j^{i h}$初始阶段,回想一下法兰克福的数据显示,对于四个观察时间不足组中的每一个,该组中的总数$\left(n_i\right)$和在观察期间进展的人数$\left(d_i\right)$,其中$n_i$和$d_i$指$i^{\text {th }}$观察时间不足组。然后$n_i-d_i$表示在整个观察过程中停留在初始阶段的人数。
设$p_i$表示停留在初始阶段的概率,$q_i=1-p_i$表示走出初始阶段的概率。很明显,我们这里有一个二项模型,所以估计$\mu_j$的可能性是
$$
L_i\left(\mu_j\right)=\left(1-p_i\right)^{d_i} \cdot\left(p_i\right)^{n_i-d_i}
$$
对于$i^{\text {th }}$观察时间组和
$$
L\left(\mu_j\right)=\prod_{i=1}^4\left(1-p_i\right)^{d_i} \cdot\left(p_i\right)^{n_i-d_i},
$$
因为有四个时间不足的观察组。对数似然是
$$
\ell\left(\mu_j\right)=\sum_{i=1}^4 d_i \cdot \log \left(1-p_i\right)+\left(n_i-d_i\right) \cdot \log p_i
$$
可能性方程是
$$
\frac{\partial \ell\left(\mu_j\right)}{\partial \mu_j}=\sum_{i=1}^4-\frac{\partial p_i}{\partial \mu_j}\left[\frac{d_i}{1-p_i}-\frac{n_i-d_i}{p_i}\right]=0,
$$
可以通过迭代求解$\hat{\mu}_j$。
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