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## 统计代写|生存模型代考Survival Models代写|Anniversary-to-Anniversary Studies

When insuring ages are used, a natural choice for the observation period is one that opens on the policy anniversary in a designated year for each insured person involved in the study. Similarly, the observation period would close on the policy anniversary in a later year. For example, an observation period might be defined as running from policy anniversaries in 1994 to those in 1998. Note that each person involved in the study has his or her own observation period.

Observation periods that run from a fixed date to a later fixed date can be used with insuring ages, but there are definite advantages to the anniversary-to-anniversary study when insuring ages are used. In this text we emphasize anniversary-to-anniversary studies with insuring ages.

The major convenient consequence of anniversary-to-anniversary studies with insuring ages is that all persons enter the study at an integral age $y_i$. This is true whether the person enters the study by joining the group via policy issue during the O.P. (at an integral insuring age), or by already being in the group when the O.P. opens. In the latter case, entry is at a policy anniversary which is always the attainment of an integral (insuring) age. Since $y_i$ is an integer, it follows that $r_i=0$ for any estimation interval $(x, x+1)$, since $x$ is integral.

Similarly, with the O.P. ending on a policy anniversary, all scheduled ending ages $z_i$ are integers, from which it follows that $s_i=1$ for all estimation intervals $(x, x+1]$. Hence all vectors $\boldsymbol{u}_{i, x}$ are of the convenient form $\left[0,1, \iota_i, \kappa_i\right]$, and all anniversary-to-anniversary insuring age studies belong to our Special Case A.

If there are no withdrawals, then the estimate $\hat{q}x$ is found from (6.7) simply by counting the number of vectors $\mathbf{u}{i, x}$ with $r_i=0$, which gives $n_x$, and the number with $\iota_i \neq 0$, which gives $d_x$, for $(x, x+1]$. If there are withdrawals, the number of them in $(x, x+1], w_x$, is the number of vectors $u_{i, x}$ with $\kappa_i \neq 0$. With $n_x, d_x$ and $w_x$ available, $q_x^{\prime(d)}$ and $q_x^{\prime(w)}$ can be estimated by (6.34) or (6.37). Of course any of the exposure-based estimators could be used as well. Note that since all $s_i=1$, the moment estimator and the actuarial estimator are the same.

## 统计代写|生存模型代考Survival Models代写|Select Studies

For the purpose of estimating $S(t ; x)$, as defined in Section 1.2, we consider only those policies issued at $I A=x$. Again assuming an anniversary-toanniversary observation period, the data processing is quite similar to that described for insuring age studies, with the following parallel features:
(1) The vector $\boldsymbol{v}i$ now represents the policy durations at entry, scheduled exit, death or withdrawal, rather than the insuring ages of the insured at such events. Note that for policies issued during the O.P., the duration at entry to the study is 0 . (2) The vector $\mathbf{u}{i, t}$ represents the location of these events within estimation interval $(t, t+1]$.
(3) Withdrawals are grouped by calendar duration, instead of calendar insuring age. Note that calendar duration is defined by
$$C D=C Y W-C Y I .$$
An example will show the similarity of select studies to insuring age studies.

# 生存模型代考

## 统计代写|生存模型代考Survival Models代写|Select Studies

(1)向量$\boldsymbol{v}i$现在表示进入、计划退出、死亡或退出时的保单期限，而不是被保险人在此类事件中的保险年龄。注意，对于在op期间发布的策略，进入研究的持续时间为0。(2)向量$\mathbf{u}{i, t}$表示这些事件在估计区间$(t, t+1]$内的位置。
(3)提款按日历期限分组，而不是按日历投保年龄分组。注意，日历持续时间由
$$C D=C Y W-C Y I .$$

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In this model the total hazard $\lambda(t ; \mathbf{z})$ is made up of an underlying hazard which is a function of time, $\lambda(t)$, and additional hazards arising from the concomitant variables, all added together. If, in general,
$$\lambda\left(t ; z_j\right)=h_j(t) \cdot g_j\left(z_j\right)$$
represents the extra hazard at time $t$ due to the $j^{t h}$ concomitant variable, then the total hazard is
$$\lambda(t ; \mathbf{z})=\lambda(t)+\sum_{j=1}^s h_j(t) \cdot g_j\left(z_j\right)$$
Now (8.68) is written to suggest the most general form for $\lambda\left(t ; z_j\right)$, expressed as the product of a function of time and a function of $z_j$. (By subscripting the functions $h$ and $g$ with $j$, we allow them to vary with $j$.) As a special case, we might have $g_j\left(z_j\right)=z_j$ for all $j$. Similarly we might let $h_j(t)=a_j$, say. If these two simplifications are combined, we then have
$$\lambda\left(t ; z_j\right)=a_j z_j$$
so that the additive model becomes
$$\lambda(t ; \mathbf{z})=\lambda(t)+\sum_{j=1}^s a_j z_j$$

## 统计代写|生存模型代考Survival Models代写|The Multiplicative Model

In this model the underlying hazard and the additional hazards from the concomitant variables are multiplied together, rather than added.

The most popular model in this family is the Cox model (see Cox [21]), where
$$\lambda\left(t ; z_j\right)=\exp \left(a_j z_j\right)$$
so the entire hazard is
$$\lambda(t ; \mathbf{z})=\lambda(t) \cdot \prod_{j=1}^s \lambda\left(t ; z_j\right)=\lambda(t) \cdot \exp \left(\sum_{j=1}^s a_j z_j\right) .$$
Furthermore, if the underlying hazard is constant over time, then we can let $\lambda(t)=\lambda=e^{a_0}$, so that
$$\lambda(t ; \mathbf{z})=\exp \left(\sum_{j=0}^s a_j z_j\right)=e^{\mathbf{a}^{\prime} \mathbf{z}},$$
where $\mathbf{a}^{\prime}=\left[a_0, a_1, \cdots, a_s\right], \mathbf{z}^{\prime}=\left[z_0, z_1, \cdots, z_s\right]$, and $z_0=1$, necessarily.
The hazard rate is again constant, so the survival model is exponential. Furthermore, in $\lambda(t ; \mathbf{z})$ is linear in $\mathbf{z}$, so (8.79) is sometimes called a log-linear exponential model.

# 生存模型代考

$$\lambda\left(t ; z_j\right)=h_j(t) \cdot g_j\left(z_j\right)$$

$$\lambda(t ; \mathbf{z})=\lambda(t)+\sum_{j=1}^s h_j(t) \cdot g_j\left(z_j\right)$$

$$\lambda\left(t ; z_j\right)=a_j z_j$$

$$\lambda(t ; \mathbf{z})=\lambda(t)+\sum_{j=1}^s a_j z_j$$

## 统计代写|生存模型代考Survival Models代写|The Multiplicative Model

$$\lambda\left(t ; z_j\right)=\exp \left(a_j z_j\right)$$

$$\lambda(t ; \mathbf{z})=\lambda(t) \cdot \prod_{j=1}^s \lambda\left(t ; z_j\right)=\lambda(t) \cdot \exp \left(\sum_{j=1}^s a_j z_j\right) .$$

$$\lambda(t ; \mathbf{z})=\exp \left(\sum_{j=0}^s a_j z_j\right)=e^{\mathbf{a}^{\prime} \mathbf{z}},$$

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## 统计代写|生存模型代考Survival Models代写|The Nelson-Aalen Estimator

Recall Equation (2.13) which states that $S(t)=e^{-\Lambda(t)}$, where $\Lambda(t)$ is the cumulative hazard function. It would be reasonable to estimate $S(t)$ by first estimating $\Lambda(t)$ and then defining
$$\hat{S}(t)=e^{-\hat{A}(t)}$$
From (7.75) we have the relationship
$$\hat{\Lambda}(t)=-\ln \hat{S}(t)$$
Consider again Equation (7.70), the product-limit estimator of $S(t)$, for $t_m \leq t<t_{m+1}$. Substituting (7.70) into (7.76) we have
\begin{aligned} \hat{\Lambda}(t) & =-\ln \left[\prod_{j=1}^m\left(\frac{r_j-d_j}{r_j}\right)\right] \ & =-\sum_{j=1}^m \ln \left(1-\frac{d_j}{r_j}\right), \quad t_m \leq \ell<t_{m+1} . \end{aligned}

## 统计代写|生存模型代考Survival Models代写|Exact Times of Death

Suppose a sample of $n$ lives, all existing at $t=0$, produces observed times of death $t_1, t_2, \cdots, t_n$, assumed to be independent. We wish to use this data to estimate $S(t)$ as a parametric survival model. In other words, we adopt a particular mathematical function of the variable $t$, depending on one or more parameters as well, and then use the sample data to estimate these unknown parameters, using a particular estimation procedure.

The simplest parametric models to use are those which depend on only one parameter, such as the exponential distribution with $S(t)=e^{-\lambda t}$. How can we estimate $\lambda$ from our sample data?
Let us calculate
$$\bar{t}=\frac{1}{n} \cdot \sum_{i=1}^n t_i$$
which is the mean time of death of the sample. We know from (2.29) that the mean of $T$ is $E[T]=\frac{1}{\lambda}$ if $T$ has an exponential distribution, so we can estimate $\lambda$ by equating these two means. Thus
$$\hat{\lambda}=\frac{1}{\bar{t}}=\sum_{i=1}^n t_i .$$
This estimation procedure is called the method of moments, so Estimator (8.2) is the moment estimator of $\lambda$.

# 生存模型代考

## 统计代写|生存模型代考Survival Models代写|The Nelson-Aalen Estimator

$$\hat{S}(t)=e^{-\hat{A}(t)}$$

$$\hat{\Lambda}(t)=-\ln \hat{S}(t)$$

\begin{aligned} \hat{\Lambda}(t) & =-\ln \left[\prod_{j=1}^m\left(\frac{r_j-d_j}{r_j}\right)\right] \ & =-\sum_{j=1}^m \ln \left(1-\frac{d_j}{r_j}\right), \quad t_m \leq \ell<t_{m+1} . \end{aligned}

## 统计代写|生存模型代考Survival Models代写|Exact Times of Death

$$\bar{t}=\frac{1}{n} \cdot \sum_{i=1}^n t_i$$

$$\hat{\lambda}=\frac{1}{\bar{t}}=\sum_{i=1}^n t_i .$$

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## 统计代写|生存模型代考Survival Models代写|Special Case $C$ with Random Censoring

As an alternative to recording, and using, exact or average values of $s_i$, for the $c_x$ persons with $0<s_i<1$, it is sometimes assumed that the values of $s_i$ are randomly distributed over $(x, x+1]$. Then scheduled ending time is a random variable, $S$, and we let $g(s)$ denote its PDF. In survival data analysis, where termination of observation due to the ending of the observation period is called censoring, this structure is referred to as a random censoring mechanism.

To construct the likelihood for Special Case $C$ with partial data, we need the probability that a person in $c_x$ will die before the scheduled ending age. If a precise scheduled ending age is not used, and a random distribution of scheduled ending ages is assumed instead, then we must first calculate the marginal probability of death before scheduled ending age. If we let $\bar{q}_x$ denote this marginal probability, then
$$\bar{q}_x=1-\bar{p}_x=1-\int_0^1 g(s) \cdot{ }_s p_x d s .$$
The integral in (7.47) shows that the marginal probability $\bar{p}_x$ is a weighted average value of ${ }_s p_x$, where the sum of the weights is $\int_0^1 g(s) d s=1$.
To evaluate (7.47), we must make a distribution assumption with respect to mortality, and specify the distribution of $S$ as well. If we make the linear assumption for mortality, and the uniform distribution for $S$ (so that $g(s)=1)$, then $(7.47)$ evaluates to
$$\bar{q}_x=\frac{1}{2} \cdot q_x$$
The Special Case $\mathrm{C}$ likelihood is
$$L=\left(q_x\right)^{d^{\prime \prime}} \cdot\left(1-q_x\right)^{n-c-d^{\prime \prime}} \cdot\left(\bar{q}_x\right)^{d^{\prime}} \cdot\left(1-\bar{q}_x\right)^{c-d^{\prime}}$$
under random censoring, as opposed to (7.35) for the likelihood when an average (fixed) value of $s$ is assumed. Of course if (7.48) is used for $\bar{q}_x$, then (7.49) becomes the same as (7.35) evaluated under the uniform assumption with $s=\frac{1}{2}$, and the resulting MLE is given by Estimator (7.32) with $s=\frac{1}{2}$.
The evaluation of (7.47) assuming other mortality distributions, and the resulting MLE’s, is pursued in the exercises.

## 统计代写|生存模型代考Survival Models代写|Summary of Single-Decrement MLE ‘s

We have derived a general, full data, exponential distribution MLE, given by Estimator (7.23), which, of course, is applicable to Special Cases A, B, and $C$ as well.
Similarly, Estimator (7.27) is the general, full data, uniform distribution MLE, with Special Cases A, B, and C versions given by (7.7), (7.30) and (7.32), respectively.
For partial data, we have developed results for Special Cases A and $\mathrm{C}$ only, under each of the exponential and uniform distributions. The partial data MLE’s for Special Case B are derived analogously to those for Special Case $C$, and are left to the exercises. The general case, partial data, MLE’s are considerably more complex, and are not pursued in this text.

If both death and withdrawal are random events operating in the estimation interval $(x, x+1]$, then we seek to estimate $q_x=q_x^{\prime(d)}$ within a doubledecrement environment. The basic mathematics and notation for doubledecrement theory was presented in Section 5.3, and employed in Section 6.3 in connection with moment estimation. Now we will explore maximum likelihood estimation in a double-decrement environment.

As before, let $x+r_i$ be the age at which person $i$ enters $(x, x+1], 0 \leq r_i<1$, and let $n_x$ be the total number of persons in the sample. Let $x+t_i$ be the age at which person $i$ leaves $(x, x+1], 0<t_i \leq 1$, whether as an interval survivor $\left(t_i=1\right)$, as an observed ender $\left(t_i<1\right)$, or as a result of one of the random events death or withdrawal.

Thus each person is under observation from age $x+r_i$ to age $x+t_i$, and the probability of this is $t_i-r_i p_{x+r_i}^{(\tau)}$. For interval survivors and enders, this “total survival” (i.e., neither dying nor withdrawing) is the contribution to the likelihood. For each death and withdrawal, the respective density functions, given by $(5.11 \mathrm{a})$ and $(5.11 \mathrm{~b})$, are needed, so $t_{t,-r_i} p_{x+r_i}^{(\tau)}$ must be multiplied by the appropriate force. Thus the overall likelihood is
\begin{aligned} L & =\prod_{i=1}^n t_t-r_i p_{x+r_i}^{(\tau)} \cdot \prod_{\mathcal{D}} \mu_{x+t_i}^{(d)} \cdot \prod_W \mu_{x+t_i}^{(w)} \ & =\prod_{i=1}^n \frac{t_i-r_i}{} p_{x+r_i}^{\prime(d)} \cdot \prod_{i=1}^n t_t-r_i p_{x+r_i}^{\prime(w)} \cdot \prod_{\mathcal{D}} \mu_{x+t_i}^{(d)} \cdot \prod_W \mu_{x+t_i \cdot}^{(w)} \end{aligned}

# 生存模型代考

## 统计代写|生存模型代考Survival Models代写|Special Case $C$ with Random Censoring

$$\bar{q}_x=1-\bar{p}_x=1-\int_0^1 g(s) \cdot{ }_s p_x d s .$$

$$\bar{q}_x=\frac{1}{2} \cdot q_x$$

$$L=\left(q_x\right)^{d^{\prime \prime}} \cdot\left(1-q_x\right)^{n-c-d^{\prime \prime}} \cdot\left(\bar{q}_x\right)^{d^{\prime}} \cdot\left(1-\bar{q}_x\right)^{c-d^{\prime}}$$

(7.47)假设其他死亡率分布，以及由此产生的MLE，是在练习中进行评价的。

## 统计代写|生存模型代考Survival Models代写|Summary of Single-Decrement MLE ‘s

\begin{aligned} L & =\prod_{i=1}^n t_t-r_i p_{x+r_i}^{(\tau)} \cdot \prod_{\mathcal{D}} \mu_{x+t_i}^{(d)} \cdot \prod_W \mu_{x+t_i}^{(w)} \ & =\prod_{i=1}^n \frac{t_i-r_i}{} p_{x+r_i}^{\prime(d)} \cdot \prod_{i=1}^n t_t-r_i p_{x+r_i}^{\prime(w)} \cdot \prod_{\mathcal{D}} \mu_{x+t_i}^{(d)} \cdot \prod_W \mu_{x+t_i \cdot}^{(w)} \end{aligned}

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## MATLAB代写

MATLAB 是一种用于技术计算的高性能语言。它将计算、可视化和编程集成在一个易于使用的环境中，其中问题和解决方案以熟悉的数学符号表示。典型用途包括：数学和计算算法开发建模、仿真和原型制作数据分析、探索和可视化科学和工程图形应用程序开发，包括图形用户界面构建MATLAB 是一个交互式系统，其基本数据元素是一个不需要维度的数组。这使您可以解决许多技术计算问题，尤其是那些具有矩阵和向量公式的问题，而只需用 C 或 Fortran 等标量非交互式语言编写程序所需的时间的一小部分。MATLAB 名称代表矩阵实验室。MATLAB 最初的编写目的是提供对由 LINPACK 和 EISPACK 项目开发的矩阵软件的轻松访问，这两个项目共同代表了矩阵计算软件的最新技术。MATLAB 经过多年的发展，得到了许多用户的投入。在大学环境中，它是数学、工程和科学入门和高级课程的标准教学工具。在工业领域，MATLAB 是高效研究、开发和分析的首选工具。MATLAB 具有一系列称为工具箱的特定于应用程序的解决方案。对于大多数 MATLAB 用户来说非常重要，工具箱允许您学习应用专业技术。工具箱是 MATLAB 函数（M 文件）的综合集合，可扩展 MATLAB 环境以解决特定类别的问题。可用工具箱的领域包括信号处理、控制系统、神经网络、模糊逻辑、小波、仿真等。

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## 统计代写|生存模型代考Survival Models代写|Partial Data

As defined in Section 6.2.2, this situation simply states that, of $n_x$ lives exactly age $x, d_x$ of them die in $(x, x+1]$, and $n_x-d_x$ survive to age $x+1$. We recognize this as a binomial model, so the likelihood is simply the binomial probability of obtaining the sample result actually obtained. That is,
$$L\left(q_x \mid n_x, d_x\right)=\frac{n_{x} !}{d_{x} !\left(n_x-d_x\right) !}\left(q_x\right)^{d_x}\left(1-q_x\right)^{n_x-d_x} .$$
One of the basic properties of MLE is that any multiplicative constants can be ignored, and the same estimate of $q_x$ will still result. When this is done, the likelihood is no longer the probability of the sample per se, but rather is proportional to it. Thus many writers prefer to write
$$L\left(q_x \mid n_x, d_x\right) \propto\left(q_x\right)^{d_x} \cdot\left(1-q_x\right)^{n_x-d_x},$$
where $\alpha$ is read “is proportional to.” We wish to take the point of view that it is just as reasonable to call the right side of (7.2) the likelihood itself, as to call it something to which the likelihood is proportional. Thus we would write simply
$$L\left(q_x \mid n_x, d_x\right)=\left(q_x\right)^{d_x} \cdot\left(1-q_x\right)^{n_x-d_x} .$$
The notation $L\left(q_x \mid n_x, d_x\right)$ reminds us that the likelihood is a function of the unknown $q_x$, and that $n_x$ and $d_x$ are given values, namely those observed in the sample upon which our estimate of $q_x$ is to be based. When there is no doubt as to the unknown and the given values, we will simply use $L$ instead of $L\left(q_x \mid n_x, d_x\right)$. Finally, for convenience we will frequently suppress the subscript $x$. Thus we will write the likelihood for the Special Case A partial data situation as
$$L=q^d \cdot(1-q)^{n-d}$$

## 统计代写|生存模型代考Survival Models代写|Full Data

Now ive assume that we have the precise age at death for each of the $d_x$ deaths in the interval. Since this age is different for each death, we consider them individually, and take the product of each death’s contribution to the likelihood function.

The likelihood for the $i^{\text {th }}$ death is given by the probability density function (PDF) for death at that particular age, given alive at age $x$. That is, for death at age $x_i$,
$$L_i=f\left(x_i \mid X>x\right)=\frac{f\left(x_i\right)}{S(x)}=\frac{S\left(x_i\right) \cdot \lambda\left(x_i\right)}{S(x)}$$
is the contribution to $L$ of the $i^{t h}$ death. If we let $s_i=x_i-x$ be the time of the $i^{t h}$ death within $(x, x+1]$, where $0<s_i \leq 1$, then we have
$$L_i=\frac{S\left(x+s_i\right) \cdot \lambda\left(x+s_i\right)}{S(x)}=s_{s_i} p_x \mu_{x+s_i}$$
in standard actuarial notation. The contribution to $L$ for all deaths combined is
$$\prod_{i=1}^d s_i p_x \mu_{x+s_i},$$
which is commonly written as $\prod_D s_s p_x \mu_{x+s_1}$, and read as “multiplied over all deaths.”

Of course the $n_x-d_x$ survivors contribute $\left(p_x\right)^{n_x-d_x}=\left(1-q_x\right)^{n_x-d_x}$ to $L$, so we have the total likelihood
$$L=\left(1-q_x\right)^{n_x-d_s} \cdot \prod_{\mathcal{D}} p_i p_x \mu_{x+s_i}$$
for our Special Case A full data situation.
To solve (7.11) for $\hat{q}x$ it is necessary to make a distribution assumption which will express $s_i p_x \mu{x+s_t}$ in terms of $q_x$. We will consider two such assumptions.

# 生存模型代考

## 统计代写|生存模型代考Survival Models代写|Partial Data

$$L\left(q_x \mid n_x, d_x\right)=\frac{n_{x} !}{d_{x} !\left(n_x-d_x\right) !}\left(q_x\right)^{d_x}\left(1-q_x\right)^{n_x-d_x} .$$
MLE的一个基本特性是可以忽略任何乘法常数，并且仍然会得到相同的$q_x$估计。当这样做时，可能性不再是样本本身的概率，而是与之成正比。因此，许多作家更喜欢写作
$$L\left(q_x \mid n_x, d_x\right) \propto\left(q_x\right)^{d_x} \cdot\left(1-q_x\right)^{n_x-d_x},$$
$\alpha$的意思是“与…成正比”。我们认为，把式7.2的右边称为似然本身，正如把它称为与似然成正比的某种东西一样，都是合理的。因此，我们可以简单地写
$$L\left(q_x \mid n_x, d_x\right)=\left(q_x\right)^{d_x} \cdot\left(1-q_x\right)^{n_x-d_x} .$$

$$L=q^d \cdot(1-q)^{n-d}$$

## 统计代写|生存模型代考Survival Models代写|Full Data

$i^{\text {th }}$死亡的可能性由该特定年龄的死亡概率密度函数(PDF)给出，给定其存活年龄为$x$。也就是说，在$x_i$岁时死亡
$$L_i=f\left(x_i \mid X>x\right)=\frac{f\left(x_i\right)}{S(x)}=\frac{S\left(x_i\right) \cdot \lambda\left(x_i\right)}{S(x)}$$

$$L_i=\frac{S\left(x+s_i\right) \cdot \lambda\left(x+s_i\right)}{S(x)}=s_{s_i} p_x \mu_{x+s_i}$$

$$\prod_{i=1}^d s_i p_x \mu_{x+s_i},$$

$$L=\left(1-q_x\right)^{n_x-d_s} \cdot \prod_{\mathcal{D}} p_i p_x \mu_{x+s_i}$$

avatest.org 为您提供可靠及专业的论文代写服务以便帮助您完成您学术上的需求，让您重新掌握您的人生。我们将尽力给您提供完美的论文，并且保证质量以及准时交稿。除了承诺的奉献精神，我们的专业写手、研究人员和校对员都经过非常严格的招聘流程。所有写手都必须证明自己的分析和沟通能力以及英文水平，并通过由我们的资深研究人员和校对员组织的面试。

## MATLAB代写

MATLAB 是一种用于技术计算的高性能语言。它将计算、可视化和编程集成在一个易于使用的环境中，其中问题和解决方案以熟悉的数学符号表示。典型用途包括：数学和计算算法开发建模、仿真和原型制作数据分析、探索和可视化科学和工程图形应用程序开发，包括图形用户界面构建MATLAB 是一个交互式系统，其基本数据元素是一个不需要维度的数组。这使您可以解决许多技术计算问题，尤其是那些具有矩阵和向量公式的问题，而只需用 C 或 Fortran 等标量非交互式语言编写程序所需的时间的一小部分。MATLAB 名称代表矩阵实验室。MATLAB 最初的编写目的是提供对由 LINPACK 和 EISPACK 项目开发的矩阵软件的轻松访问，这两个项目共同代表了矩阵计算软件的最新技术。MATLAB 经过多年的发展，得到了许多用户的投入。在大学环境中，它是数学、工程和科学入门和高级课程的标准教学工具。在工业领域，MATLAB 是高效研究、开发和分析的首选工具。MATLAB 具有一系列称为工具箱的特定于应用程序的解决方案。对于大多数 MATLAB 用户来说非常重要，工具箱允许您学习应用专业技术。工具箱是 MATLAB 函数（M 文件）的综合集合，可扩展 MATLAB 环境以解决特定类别的问题。可用工具箱的领域包括信号处理、控制系统、神经网络、模糊逻辑、小波、仿真等。

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## 统计代写|生存模型代考Survival Models代写|The Basic Moment Relationships

As explained in Section 5.3, we now consider both death and withdrawal to be random events. Thus, analogous to the moment Equation (6.2) in the single-decrement case, we now write the pair of moment equations
$$E\left[D_x\right]=\sum_{i=1}^n s_i-r_1 q_{x+r_1}^{(d)}=d_x$$
and
$$E\left[W_x\right]=\sum_{i=1}^n s_i-r_i q_{x+r_i}^{(w)}=w_x$$
where $W_x$ is the random variable for withdrawals in $(x, x+1]$, and $w_x$ denotes the observed number of withdrawals in the sample.

A simple way to solve Equations (6.29a) and (6.29b) for estimates of the double-decrement probabilities $q_x^{(d)}$ and $q_x^{(w)}$ would be to use approximations analogous to (6.3), namely
$$s_i-r_i q_{x+r_i}^{(d)} \approx\left(s_i-r_i\right) \cdot q_x^{(d)}$$
and
$$s_i-r_i q_{x+r_i}^{(w)} \approx\left(s_i-r_i\right) \cdot q_x^{(w)}$$
producing
$$\hat{q}x^{(d)}=\frac{d_x}{\sum{i=1}^n\left(s_i-r_i\right)}$$
and
$$\hat{q}x^{(w)}=\frac{w_x}{\sum{i=1}^n\left(s_i-r_i\right)}$$
However, our objective is usually to estimate the single-decrement probabilities $q_x^{(d)}$ and $q_x^{(w)}$, in this case from data that have been collected in a double-decrement environment. To accomplish this, we could calculate estimates of ${q_x^{\prime}}^{(d)}$ and ${q_x^{\prime}}^{(w)}$ from the estimates $\hat{q}_x^{(d)}$ and $\hat{q}_x^{(w)}$, produced by (6.31a) and (6.31b), using one of the approximate relationships between singledecrement and double-decrement probabilities presented in Section 5.3.

Alternatively, we could calculate estimates of $q_x^{\prime(d)}$ and $q_x^{\prime(w)}$ directly from our sample data by expressing $s_i-r_i q_{x+r_i}^{(d)}$ and $s_i-r_i q_{x+r_i}^{(w)}$ directly in terms of $q_x^{\prime}(d)$ and $q_x^{\prime(w)}$, under a chosen assumption, without using the preliminary approximations (6.30a) and (6.30b). This approach normally results in equations which must be solved numerically for $\hat{q}_x^{{ }^{(d)}}$ and $\hat{q}_x^{(w)}$

## 统计代写|生存模型代考Survival Models代写|Uniform Distribution Assumptions

If the random events, death and withdrawal, are each assumed to have uniform distributions over $(x, x+1)$ in the single-decrement context, we know from Section 3.5 that ${ }{u-r} q{x+r}^{\prime(d)}=\frac{(u-r) \cdot q_x^{\prime(d)}}{1-r \cdot q_x^{\prime(d)}}, 0 \leq r<u \leq 1$, and also that $\mu_{x+u}^{(d)}=\frac{q_x^{\prime(d)}}{1-u \cdot q_x^{\prime(d)}}$, so that $\left(1-u-r q_{x+r)}^{\prime(d)} \mu_{x+u}^{(d)}=\frac{q_x^{\prime(d)}}{1-r \cdot q_x^{\prime(\sigma)}}\right.$. Then Equation (5.14a) becomes
\begin{aligned} s-r q_{x+r}^{(d)} & =\int_r^s\left[1-u_{-r} q_{x+r}^{\prime(w)}\right]\left[1-u_{u-r} q_{x+r}^{(d)}\right] \mu_{x+u}^{(d)} d u \ & =\int_r^s \frac{q_x^{\prime(d)}}{1-r \cdot q_x^{\prime(d)}} \cdot \frac{1-u \cdot q_x^{\prime(w)}}{1-r \cdot q_x^{\prime(w)}} d u \ & =\frac{q_x^{\prime(d)}}{\left[1-r \cdot q_x^{\prime(d)}\right]\left[1-r \cdot q_x^{(w)}\right]} \int_r^s\left[1-u \cdot q_x^{\prime(w)}\right] d u \ & =\frac{q_x^{\prime(d)}\left[(s-r)-\frac{1}{2}\left(s^2-r^2\right) \cdot q_x^{\prime(w)}\right]}{\left[1-r \cdot q_x^{\prime(d)}\right]\left[1-r \cdot q_x^{\prime(w)}\right]} \end{aligned}
Similarly, Equation (5.14b) would lead to
$${ }{s-r} q{x+r}^{(w)}=\frac{q_x^{\prime(w)}\left[(s-r)-\frac{1}{2}\left(s^2-r^2\right) \cdot q_x^{\prime(d)}\right]}{\left[1-r \cdot q_x^{\prime(d)}\right]\left[1-r \cdot q_x^{\prime(w)}\right]}$$
Substitution of (6.32a) and (6.32b) into the basic moment equations (6.29a) and $(6.29 \mathrm{~b})$, respectively, results in
$$E\left[D_x\right]=\sum_{i=1}^n \frac{q_x^{(d)}\left[\left(s_i-r_i\right)-\frac{1}{2}\left(s_i^2-r_i^2\right) \cdot q_x^{\prime(w)}\right]}{\left[1-r_i \cdot q_x^{\prime(d)}\right]\left[1-r_i \cdot q_x^{\prime(w)}\right]}=d_x$$
and
$$E\left[W_x\right]=\sum_{i=1}^n \frac{q_x^{\prime(w)}\left[\left(s_i-r_i\right)-\frac{1}{2}\left(s_i^2-r_i^2\right) \cdot q_x^{\prime(d)}\right]}{\left[1-r_i \cdot q_x^{\prime(d)}\right]\left[1-r_i \cdot q_x^{\prime(w)}\right]}=w_x$$
The complexity of Equations (6.33a) and (6.33b), necessitating a numerical solution, is obvious.

# 生存模型代考

## 统计代写|生存模型代考Survival Models代写|The Basic Moment Relationships

$$E\left[D_x\right]=\sum_{i=1}^n s_i-r_1 q_{x+r_1}^{(d)}=d_x$$

$$E\left[W_x\right]=\sum_{i=1}^n s_i-r_i q_{x+r_i}^{(w)}=w_x$$

$$s_i-r_i q_{x+r_i}^{(d)} \approx\left(s_i-r_i\right) \cdot q_x^{(d)}$$

$$s_i-r_i q_{x+r_i}^{(w)} \approx\left(s_i-r_i\right) \cdot q_x^{(w)}$$

$$\hat{q}x^{(d)}=\frac{d_x}{\sum{i=1}^n\left(s_i-r_i\right)}$$

$$\hat{q}x^{(w)}=\frac{w_x}{\sum{i=1}^n\left(s_i-r_i\right)}$$

## 统计代写|生存模型代考Survival Models代写|Uniform Distribution Assumptions

\begin{aligned} s-r q_{x+r}^{(d)} & =\int_r^s\left[1-u_{-r} q_{x+r}^{\prime(w)}\right]\left[1-u_{u-r} q_{x+r}^{(d)}\right] \mu_{x+u}^{(d)} d u \ & =\int_r^s \frac{q_x^{\prime(d)}}{1-r \cdot q_x^{\prime(d)}} \cdot \frac{1-u \cdot q_x^{\prime(w)}}{1-r \cdot q_x^{\prime(w)}} d u \ & =\frac{q_x^{\prime(d)}}{\left[1-r \cdot q_x^{\prime(d)}\right]\left[1-r \cdot q_x^{(w)}\right]} \int_r^s\left[1-u \cdot q_x^{\prime(w)}\right] d u \ & =\frac{q_x^{\prime(d)}\left[(s-r)-\frac{1}{2}\left(s^2-r^2\right) \cdot q_x^{\prime(w)}\right]}{\left[1-r \cdot q_x^{\prime(d)}\right]\left[1-r \cdot q_x^{\prime(w)}\right]} \end{aligned}

$${ }{s-r} q{x+r}^{(w)}=\frac{q_x^{\prime(w)}\left[(s-r)-\frac{1}{2}\left(s^2-r^2\right) \cdot q_x^{\prime(d)}\right]}{\left[1-r \cdot q_x^{\prime(d)}\right]\left[1-r \cdot q_x^{\prime(w)}\right]}$$

$$E\left[D_x\right]=\sum_{i=1}^n \frac{q_x^{(d)}\left[\left(s_i-r_i\right)-\frac{1}{2}\left(s_i^2-r_i^2\right) \cdot q_x^{\prime(w)}\right]}{\left[1-r_i \cdot q_x^{\prime(d)}\right]\left[1-r_i \cdot q_x^{\prime(w)}\right]}=d_x$$

$$E\left[W_x\right]=\sum_{i=1}^n \frac{q_x^{\prime(w)}\left[\left(s_i-r_i\right)-\frac{1}{2}\left(s_i^2-r_i^2\right) \cdot q_x^{\prime(d)}\right]}{\left[1-r_i \cdot q_x^{\prime(d)}\right]\left[1-r_i \cdot q_x^{\prime(w)}\right]}=w_x$$

avatest.org 为您提供可靠及专业的论文代写服务以便帮助您完成您学术上的需求，让您重新掌握您的人生。我们将尽力给您提供完美的论文，并且保证质量以及准时交稿。除了承诺的奉献精神，我们的专业写手、研究人员和校对员都经过非常严格的招聘流程。所有写手都必须证明自己的分析和沟通能力以及英文水平，并通过由我们的资深研究人员和校对员组织的面试。

## MATLAB代写

MATLAB 是一种用于技术计算的高性能语言。它将计算、可视化和编程集成在一个易于使用的环境中，其中问题和解决方案以熟悉的数学符号表示。典型用途包括：数学和计算算法开发建模、仿真和原型制作数据分析、探索和可视化科学和工程图形应用程序开发，包括图形用户界面构建MATLAB 是一个交互式系统，其基本数据元素是一个不需要维度的数组。这使您可以解决许多技术计算问题，尤其是那些具有矩阵和向量公式的问题，而只需用 C 或 Fortran 等标量非交互式语言编写程序所需的时间的一小部分。MATLAB 名称代表矩阵实验室。MATLAB 最初的编写目的是提供对由 LINPACK 和 EISPACK 项目开发的矩阵软件的轻松访问，这两个项目共同代表了矩阵计算软件的最新技术。MATLAB 经过多年的发展，得到了许多用户的投入。在大学环境中，它是数学、工程和科学入门和高级课程的标准教学工具。在工业领域，MATLAB 是高效研究、开发和分析的首选工具。MATLAB 具有一系列称为工具箱的特定于应用程序的解决方案。对于大多数 MATLAB 用户来说非常重要，工具箱允许您学习应用专业技术。工具箱是 MATLAB 函数（M 文件）的综合集合，可扩展 MATLAB 环境以解决特定类别的问题。可用工具箱的领域包括信号处理、控制系统、神经网络、模糊逻辑、小波、仿真等。

Posted on Categories:生存模型, 统计代写, 统计代考

## avatest™帮您通过考试

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## 统计代写|生存模型代考Survival Models代写|Uniform Distribution Assumptions

Let each decrement, death and withdrawal, have uniform distributions over $(x, x+1]$, so that
$${ }s^{\prime(d)}=s \cdot q_x^{\prime(d)}$$ and $${ }_s^{\prime(w)}=s \cdot q_x^{\prime(w)}$$ asalready established by Equation (3.54). By Equation (3.57), we now have $$\mu{x+s}^{(d)}=\frac{q_x^{\prime(d)}}{1-s \cdot q_x^{\prime(d)}}$$
and
$$\mu_{x+s}^{(w)}=\frac{q_x^{\prime(w)}}{1-s \cdot q_x^{\prime(w)}} .$$
To find $q_x^{(d)}$ in terms of $q_x^{\prime(d)}$ and $q_x^{\prime(w)}$ under the uniform assumptions, we substitute (5.16a), (5.16b), and (5.17a) into (5.12a), obtaining
\begin{aligned} q_x^{(d)} & =\int_0^1\left[1-u \cdot q_x^{\prime(d)}\right]\left[1-u \cdot q_x^{\prime(w)}\right] \frac{q_x^{\prime(d)}}{1-u \cdot q_x^{\prime(d)}} d u \ & =q_x^{\prime(d)} \int_0^1\left(1-u \cdot q_x^{\prime(w)}\right) d u \ & =q_x^{\prime(d)}\left[1-\frac{1}{2} \cdot q_x^{\prime(w)}\right] . \end{aligned}
Similarly, using (5.12b) we find
$$q_x^{(w)}=q_x^{\prime(w)}\left[1-\frac{1}{2} \cdot q_x^{\prime(d)}\right]$$
To find $q_x^{\prime(d)}$ in terms of $q_x^{(d)}$ and $q_x^{(w)}$, we solve the pair of equations given by $(5.18 \mathrm{a})$ and $(5.18 b)$. From (5.18a) we have
$$q_x^{\prime(d)}=\frac{q_x^{(d)}}{1-\frac{1}{2} \cdot q_x^{\prime(w)}}$$
and from (5.18b) we have
$$q_x^{\prime(w)}=\frac{q_x^{(w)}}{1-\frac{1}{2} \cdot q_x^{\prime(d)}}$$
$$\frac{1}{2}\left(q_x^{\prime(d)}\right)^2-\left(1-\frac{1}{2} q_x^{(w)}+\frac{1}{2} q_x^{(d)}\right) q_x^{(d)}+q_x^{(d)}=0$$
which solves for
$$q_x^{\prime(d)}=b-\sqrt{b^2-2 \cdot q_x^{(d)}}$$
where $b=1-\frac{1}{2} q_x^{(\mathrm{w})}+\frac{1}{2} q_x^{(d)}$. (Note that the positive radical is not taken since it would lead to $q_x^{\prime(d)}>1$.) By considerations of symmetry, we have
$$q_x^{\prime(w)}=c-\sqrt{c^2-2 \cdot q_x^{(w)}}$$
where $c=1-\frac{1}{2} q_x^{(d)}+\frac{1}{2} q_x^{(w)}$.

## 统计代写|生存模型代考Survival Models代写|Exponential Distribution Assumptions

If both death and withdrawal are assumed to have exponential distributions (constant forces) over $(x, x+1]$, then
$${ }_s^{\prime(d)}=e^{-s \cdot \mu^{(d)}}$$
and
$${ }_s p_x^{\prime(w)}=e^{-s \cdot \mu^{(w)}}$$

as already established by Equation (3.64a). By the constant force property, we now have
$$\mu_{x+s}^{(d)}=\mu^{(d)}$$
and
$$\mu_{x+s}^{(w)}=\mu^{(w)}$$
To find $q_x^{(d)}$ in terms of $\mu^{(d)}$ and $\mu^{(w)}$ under the exponential assumptions, we substitute (5.21a), (5.21b), and (5.22a) into (5.12a), obtaining
\begin{aligned} q_x^{(d)} & =\int_0^1 u_x^{\prime(d)} \cdot{ }_u p_x^{\prime(w)} \cdot \mu^{(d)} d u \ & =\mu^{(d)} \int_0^1 e^{-u\left(\mu^{(d)}+\mu^{(w)}\right)} d u \ & =\frac{\mu^{(d)}}{\mu^{(d)}+\mu^{(w)}}\left[1-e^{-\left(\mu^{(d)}+\mu^{(x)}\right)}\right] \end{aligned}
Similarly, using (5.12b) we find
$$q_x^{(w)}=\frac{\mu^{(w)}}{\mu^{(d)}+\mu^{(w)}}\left[1-e^{-\left(\mu^{(\omega)}+\mu^{(w)}\right)}\right]$$

# 生存模型代考

## 统计代写|生存模型代考Survival Models代写|Uniform Distribution Assumptions

$${ }s^{\prime(d)}=s \cdot q_x^{\prime(d)}$$和$${ }s^{\prime(w)}=s \cdot q_x^{\prime(w)}$$已由式(3.54)建立。通过式(3.57)，我们现在有$$\mu{x+s}^{(d)}=\frac{q_x^{\prime(d)}}{1-s \cdot q_x^{\prime(d)}}$$ 和 $$\mu{x+s}^{(w)}=\frac{q_x^{\prime(w)}}{1-s \cdot q_x^{\prime(w)}} .$$

\begin{aligned} q_x^{(d)} & =\int_0^1\left[1-u \cdot q_x^{\prime(d)}\right]\left[1-u \cdot q_x^{\prime(w)}\right] \frac{q_x^{\prime(d)}}{1-u \cdot q_x^{\prime(d)}} d u \ & =q_x^{\prime(d)} \int_0^1\left(1-u \cdot q_x^{\prime(w)}\right) d u \ & =q_x^{\prime(d)}\left[1-\frac{1}{2} \cdot q_x^{\prime(w)}\right] . \end{aligned}

$$q_x^{(w)}=q_x^{\prime(w)}\left[1-\frac{1}{2} \cdot q_x^{\prime(d)}\right]$$

$$q_x^{\prime(d)}=\frac{q_x^{(d)}}{1-\frac{1}{2} \cdot q_x^{\prime(w)}}$$

$$q_x^{\prime(w)}=\frac{q_x^{(w)}}{1-\frac{1}{2} \cdot q_x^{\prime(d)}}$$

$$\frac{1}{2}\left(q_x^{\prime(d)}\right)^2-\left(1-\frac{1}{2} q_x^{(w)}+\frac{1}{2} q_x^{(d)}\right) q_x^{(d)}+q_x^{(d)}=0$$

$$q_x^{\prime(d)}=b-\sqrt{b^2-2 \cdot q_x^{(d)}}$$

$$q_x^{\prime(w)}=c-\sqrt{c^2-2 \cdot q_x^{(w)}}$$

## 统计代写|生存模型代考Survival Models代写|Exponential Distribution Assumptions

$${ }_s^{\prime(d)}=e^{-s \cdot \mu^{(d)}}$$

$${ }_s p_x^{\prime(w)}=e^{-s \cdot \mu^{(w)}}$$

$$\mu_{x+s}^{(d)}=\mu^{(d)}$$

$$\mu_{x+s}^{(w)}=\mu^{(w)}$$

\begin{aligned} q_x^{(d)} & =\int_0^1 u_x^{\prime(d)} \cdot{ }_u p_x^{\prime(w)} \cdot \mu^{(d)} d u \ & =\mu^{(d)} \int_0^1 e^{-u\left(\mu^{(d)}+\mu^{(w)}\right)} d u \ & =\frac{\mu^{(d)}}{\mu^{(d)}+\mu^{(w)}}\left[1-e^{-\left(\mu^{(d)}+\mu^{(x)}\right)}\right] \end{aligned}

$$q_x^{(w)}=\frac{\mu^{(w)}}{\mu^{(d)}+\mu^{(w)}}\left[1-e^{-\left(\mu^{(\omega)}+\mu^{(w)}\right)}\right]$$

avatest.org 为您提供可靠及专业的论文代写服务以便帮助您完成您学术上的需求，让您重新掌握您的人生。我们将尽力给您提供完美的论文，并且保证质量以及准时交稿。除了承诺的奉献精神，我们的专业写手、研究人员和校对员都经过非常严格的招聘流程。所有写手都必须证明自己的分析和沟通能力以及英文水平，并通过由我们的资深研究人员和校对员组织的面试。

## MATLAB代写

MATLAB 是一种用于技术计算的高性能语言。它将计算、可视化和编程集成在一个易于使用的环境中，其中问题和解决方案以熟悉的数学符号表示。典型用途包括：数学和计算算法开发建模、仿真和原型制作数据分析、探索和可视化科学和工程图形应用程序开发，包括图形用户界面构建MATLAB 是一个交互式系统，其基本数据元素是一个不需要维度的数组。这使您可以解决许多技术计算问题，尤其是那些具有矩阵和向量公式的问题，而只需用 C 或 Fortran 等标量非交互式语言编写程序所需的时间的一小部分。MATLAB 名称代表矩阵实验室。MATLAB 最初的编写目的是提供对由 LINPACK 和 EISPACK 项目开发的矩阵软件的轻松访问，这两个项目共同代表了矩阵计算软件的最新技术。MATLAB 经过多年的发展，得到了许多用户的投入。在大学环境中，它是数学、工程和科学入门和高级课程的标准教学工具。在工业领域，MATLAB 是高效研究、开发和分析的首选工具。MATLAB 具有一系列称为工具箱的特定于应用程序的解决方案。对于大多数 MATLAB 用户来说非常重要，工具箱允许您学习应用专业技术。工具箱是 MATLAB 函数（M 文件）的综合集合，可扩展 MATLAB 环境以解决特定类别的问题。可用工具箱的领域包括信号处理、控制系统、神经网络、模糊逻辑、小波、仿真等。

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## 统计代写|生存模型代考Survival Models代写|Estimation of $S(t)$ and ${ }_t \mid q_0$

It should be clear that $S(t)$ can be estimated by using the observed proportion surviving to $t$. Thus
$$\hat{S}(t)=\frac{N_t}{n}$$
is a binomial proportion random variable, with mean and variance given by (4.4a) and (4.4b), respectively, with $S^o(t)$ replaced by $\hat{S}(t)$.

Similarly, $t \mid q_0$ is naturally estimated by using the observed relative frequency of deaths between $t$ and $t+1$. Thus
$$t \hat{q}_0=\frac{D_t}{n}, t=0,1, \ldots, k-1$$
is a multinomial proportion, with mean and variance given by
$$E\left[t \mid \hat{q}_0\right]=\frac{1}{n} \cdot E\left[D_t\right]=t \mid q_0$$
and
$$\operatorname{Var}\left(t \mid \hat{q}_0\right)=\frac{1}{n^2} \cdot \operatorname{Var}\left(D_t\right)=\frac{\mid q_0 \cdot\left(1-t \mid q_0\right)}{n} .$$
We observe from (4.14) that the multinomial proportion estimator ${ }_1 \mid \hat{q}_0=\frac{D_r}{n}$ is unbiased, as defined in Appendix $\mathbf{A}$.

## 统计代写|生存模型代考Survival Models代写|Estimation of $q_t$ and $p_t$

Recall that $q_t$ is the conditional probability of death not later than time $t+1$, given alive at time $t$. It is natural to estimate this probability by
$$\hat{q}_t=\frac{D_t}{n_t}$$

Here we recognize that, conditional on there being $n_t$ survivors at time $t, D_t$ is binonial, so $\hat{q}t=\frac{D_t}{n_t}$ is binomial proportion, with conditional mean and variance given by $$E\left[\hat{q}_t \mid n_t\right]=\frac{1}{n_t} \cdot E\left[D_t\right]=q_t$$ and $$\operatorname{Var}\left(\hat{q}_t \mid n_t\right)=\frac{1}{n_t^2} \cdot \operatorname{Var}\left(D_t\right)=\frac{q_t\left(1-q_t\right)}{n_t}$$ Furthermore, $$\hat{p}_t=1-\hat{q}_t=\frac{n_t-D_t}{n_t}=\frac{N{t+1}}{n_t}$$
is also a binomial proportion, so
$$E\left[\hat{p}_t \mid n_t\right]=p_t$$
and
$$\operatorname{Var}\left(\hat{p}_t \mid n_t\right)=\frac{p_t\left(1-p_t\right)}{n_t}$$
Since $q_t=1-p_t$, then, clearly, (4.18) and (4.21) are the same.

# 生存模型代考

## 统计代写|生存模型代考Survival Models代写|Estimation of $S(t)$ and ${ }_t \mid q_0$

$$\hat{S}(t)=\frac{N_t}{n}$$

$$t \hat{q}_0=\frac{D_t}{n}, t=0,1, \ldots, k-1$$

$$E\left[t \mid \hat{q}_0\right]=\frac{1}{n} \cdot E\left[D_t\right]=t \mid q_0$$

$$\operatorname{Var}\left(t \mid \hat{q}_0\right)=\frac{1}{n^2} \cdot \operatorname{Var}\left(D_t\right)=\frac{\mid q_0 \cdot\left(1-t \mid q_0\right)}{n} .$$

## 统计代写|生存模型代考Survival Models代写|Estimation of $q_t$ and $p_t$

$$\hat{q}_t=\frac{D_t}{n_t}$$

$$E\left[\hat{p}_t \mid n_t\right]=p_t$$

$$\operatorname{Var}\left(\hat{p}_t \mid n_t\right)=\frac{p_t\left(1-p_t\right)}{n_t}$$

avatest.org 为您提供可靠及专业的论文代写服务以便帮助您完成您学术上的需求，让您重新掌握您的人生。我们将尽力给您提供完美的论文，并且保证质量以及准时交稿。除了承诺的奉献精神，我们的专业写手、研究人员和校对员都经过非常严格的招聘流程。所有写手都必须证明自己的分析和沟通能力以及英文水平，并通过由我们的资深研究人员和校对员组织的面试。

## MATLAB代写

MATLAB 是一种用于技术计算的高性能语言。它将计算、可视化和编程集成在一个易于使用的环境中，其中问题和解决方案以熟悉的数学符号表示。典型用途包括：数学和计算算法开发建模、仿真和原型制作数据分析、探索和可视化科学和工程图形应用程序开发，包括图形用户界面构建MATLAB 是一个交互式系统，其基本数据元素是一个不需要维度的数组。这使您可以解决许多技术计算问题，尤其是那些具有矩阵和向量公式的问题，而只需用 C 或 Fortran 等标量非交互式语言编写程序所需的时间的一小部分。MATLAB 名称代表矩阵实验室。MATLAB 最初的编写目的是提供对由 LINPACK 和 EISPACK 项目开发的矩阵软件的轻松访问，这两个项目共同代表了矩阵计算软件的最新技术。MATLAB 经过多年的发展，得到了许多用户的投入。在大学环境中，它是数学、工程和科学入门和高级课程的标准教学工具。在工业领域，MATLAB 是高效研究、开发和分析的首选工具。MATLAB 具有一系列称为工具箱的特定于应用程序的解决方案。对于大多数 MATLAB 用户来说非常重要，工具箱允许您学习应用专业技术。工具箱是 MATLAB 函数（M 文件）的综合集合，可扩展 MATLAB 环境以解决特定类别的问题。可用工具箱的领域包括信号处理、控制系统、神经网络、模糊逻辑、小波、仿真等。

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## 统计代写|生存模型代考Survival Models代写|The Concept of Exposure

A survival model, expressed by the $\operatorname{SDF} S(x)$, is a probability distribution with all the properties that such distributions possess. The simple transformation of $\ell_x=\ell_0 \cdot S(x)$, where $\ell_0$ is a constant, expresses the same probability distribution. The function $\ell_x$ is the same function as is $S(x)$, except for the trivial difference that whereas $1 \geq S(x) \geq 0$, we now have $\ell_0 \geq \ell_x \geq 0$. Any function or result that can be derived from $S(x)$ can also be derived from $\ell_x$.
However, an advantage in using $\ell_x$, instead of $S(x)$, is the ability to interpret the values of $\ell_x$ as the survivors of an initial, closed, cohort of newborn lives of size of $\ell_0$. Successive values of $S(x)$ are probabilities, which are somewhat abstract, especially to non-mathematicians. But values of $\ell_x$ have a “real world” meaning, notwithstanding the fact that we are dealing with hypothetical situations.
In turn, the interpretive nature of $\ell_x$ allows for concrete (albeit hypothetical) interpretations of several functions derived from $\ell_x$. Of particular usefulness is the function $L_x$, defined by (3.37).
Recall, from (3.26), that ${ }s p_x \mu{x+s}$ is the PDF for death at age $x+s$, given alive at age $x$. If we multiply this PDF by $\ell_x$, which we interpret as the number of persons alive in a group at age $x$, we obtain $\ell_{x+s} \mu_{x+s}$, which is the rate of deaths occurring in the group at exact age $x+s$. In turn, $\ell_{x+s} \mu_{x+s} d s$ is the differential number of deaths occurring at exact age $x+s$. Then $s \cdot \ell_{x+s} \mu_{x+s} d s$ is the total number of years lived by those deaths after attaining age $x$. Finally, $\int_0^1 s \cdot \ell_{x+s} \mu_{x+s} d s$ gives the aggregate number of years lived, after age $x$, by all those who die between age $x$ and age $x+1$.
Most of the $\ell_x$ group, of course, survive to age $x+1, \ell_{x+1}$ being the number who do so. Each of these persons live one year from age $x$ to age $x+1$, so $\ell_{x+1}$ also represents the aggregate number of years lived, between ages $x$ and $x+1$, by those who survive to age $x+1$. Together,
$$\ell_{x+1}+\int_0^1 s \cdot \ell_{x+s} \mu_{x+s} d s$$
gives the aggregate number of years lived between ages $x$ and $x+1$ by the $\ell_x$ persons who comprised the group at age $x$.

## 统计代写|生存模型代考Survival Models代写|Relationship between ${ }_n q_x$ and ${ }_n m_x$

Familiarity with the concept of exposure allows us to develop a very useful formula which relates the functions ${ }n q_x$ and ${ }_n m_x$. Let us first explore this relationship with $n=1$. Since $q_x=\frac{d_x}{\ell_x}$, and $m_x=\frac{d_x}{L_x}$, then we are, in effect, exploring the relationship between $\ell_x$ and $L_x$. To do this, we first need to define a new function. From (3.40), we recall that $$\int_0^1 s \cdot \ell{x+s} \mu_{x+s} d s=\int_0^1 \ell_{x+s} d s-\ell_{x+1}=L_x-\ell_{x+1}$$
gives the aggregate number of life-years lived in $(x, x+1$ ] by those who die in that age interval, namely $d_x$. Then if (3.46) is divided by $d_x$, we obtain the average number of years lived in $(x, x+1]$ by those who die in that interval. It is clear that this average number is necessarily less than one, and could also be called the average fraction of $(x, x+1]$ lived through by those who die in that interval. We define this average fraction to be $f_x$, so that
$$f_x=\frac{L_x-\ell_{x+1}}{d_x}$$
Further, since $d_x=\ell_x-\ell_{x+1}$, then $\ell_{x+1}=\ell_x-d_x$, so we have
$$f_x \cdot d_x=L_x-\ell_x+d_x$$
or
$$L_x=\ell_x-\left(1-f_x\right) d_x$$
Then
$$m_x=\frac{d_x}{L_x}=\frac{d_x}{\ell_x-\left(1-f_x\right) d_x}=\frac{q_x}{1-\left(1-f_x\right) q_x} .$$
Alternatively,
$$\ell_x=L_x+\left(1-f_x\right) d_x$$
so
$$q_x=\frac{d_x}{\ell_x}=\frac{d_x}{L_x+\left(1-f_x\right) d_x}=\frac{m_x}{1+\left(1-f_x\right) m_x} .$$

# 生存模型代考

## 统计代写|生存模型代考Survival Models代写|The Concept of Exposure

$$\ell_{x+1}+\int_0^1 s \cdot \ell_{x+s} \mu_{x+s} d s$$

## 统计代写|生存模型代考Survival Models代写|Relationship between ${ }_n q_x$ and ${ }_n m_x$

$$f_x=\frac{L_x-\ell_{x+1}}{d_x}$$

$$f_x \cdot d_x=L_x-\ell_x+d_x$$

$$L_x=\ell_x-\left(1-f_x\right) d_x$$

$$m_x=\frac{d_x}{L_x}=\frac{d_x}{\ell_x-\left(1-f_x\right) d_x}=\frac{q_x}{1-\left(1-f_x\right) q_x} .$$

$$\ell_x=L_x+\left(1-f_x\right) d_x$$

$$q_x=\frac{d_x}{\ell_x}=\frac{d_x}{L_x+\left(1-f_x\right) d_x}=\frac{m_x}{1+\left(1-f_x\right) m_x} .$$

avatest.org 为您提供可靠及专业的论文代写服务以便帮助您完成您学术上的需求，让您重新掌握您的人生。我们将尽力给您提供完美的论文，并且保证质量以及准时交稿。除了承诺的奉献精神，我们的专业写手、研究人员和校对员都经过非常严格的招聘流程。所有写手都必须证明自己的分析和沟通能力以及英文水平，并通过由我们的资深研究人员和校对员组织的面试。

## MATLAB代写

MATLAB 是一种用于技术计算的高性能语言。它将计算、可视化和编程集成在一个易于使用的环境中，其中问题和解决方案以熟悉的数学符号表示。典型用途包括：数学和计算算法开发建模、仿真和原型制作数据分析、探索和可视化科学和工程图形应用程序开发，包括图形用户界面构建MATLAB 是一个交互式系统，其基本数据元素是一个不需要维度的数组。这使您可以解决许多技术计算问题，尤其是那些具有矩阵和向量公式的问题，而只需用 C 或 Fortran 等标量非交互式语言编写程序所需的时间的一小部分。MATLAB 名称代表矩阵实验室。MATLAB 最初的编写目的是提供对由 LINPACK 和 EISPACK 项目开发的矩阵软件的轻松访问，这两个项目共同代表了矩阵计算软件的最新技术。MATLAB 经过多年的发展，得到了许多用户的投入。在大学环境中，它是数学、工程和科学入门和高级课程的标准教学工具。在工业领域，MATLAB 是高效研究、开发和分析的首选工具。MATLAB 具有一系列称为工具箱的特定于应用程序的解决方案。对于大多数 MATLAB 用户来说非常重要，工具箱允许您学习应用专业技术。工具箱是 MATLAB 函数（M 文件）的综合集合，可扩展 MATLAB 环境以解决特定类别的问题。可用工具箱的领域包括信号处理、控制系统、神经网络、模糊逻辑、小波、仿真等。

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## 统计代写|生存模型代考Survival Models代写|The Force of Mortalit

The derivative of $\ell_x$ can be interpreted as the absolute instantaneous annual rate of change of $\ell_x$. Since $\ell_x$ represents the number of persons alive at age $x$, then the derivative, which is the annual rate at which $\ell_x$ is changing, gives the annual rate at which people are dying at age $x$. This derivative is negative since $\ell_x$ is a decreasing function. To obtain the absolute magnitude of this instantaneous rate of death, we will use the negative of the derivative. Finally, since the magnitude of the derivative depends on the size of $\ell_x$ itself, we obtain the relative instantaneous rate of death by dividing the negative derivative of $\ell_x$ by $\ell_x$ itself. Thus we have
$$\mu_x=\frac{-\frac{d}{d x} \ell_x}{\ell_x}$$
which we call the force of mortality at age $x$. Now since $\ell_x=\ell_0 \cdot S(x)$, then we see that (3.8) is the same as
$$\lambda(x)=\frac{-\frac{d}{d x} S(x)}{S(x)}=\frac{f(x)}{S(x)}$$
Thus the hazard rate and the force of mortality are identical.
If we multiply both sides of Equation (2.11) by $\ell_0$, replace $t$ with $x$, and substitute $\mu_y$ for $\lambda(\mathrm{y})$, we obtain
$$\ell_x=\ell_0 \cdot S(x)=\ell_0 \cdot \exp \left[-\int_0^x \mu_y d y\right]$$

## 统计代写|生存模型代考Survival Models代写|Tlte Probability Density Function ofX

With the force of mortality, which is the same as the hazard rate, now defined, the next function to develop from $\ell_x$ is the PDF of the age-at-death random variable $X$. (Remember that we wish to show that the life table is a representation of the distribution of this random variable.)

From Equation (2.7) we have $f(x)=\lambda(x) \cdot S(x)$. In the life table context, we have $\lambda(x)=\mu_x$ and $S(x)=\ell_x / \ell_0$. Thus we have
$$f(x)=\mu_x\left(\ell_x / \ell_0\right)={ }_x p_0 \mu_x, x \geq 0$$
Since, from (3.8), $\frac{d}{d x} \ell_x=-\ell_x \mu_x$, then dividing both sides by $\ell_0$ gives
$$\frac{d}{d x} x p_0=-{ }_x p_0 \mu_x$$

# 生存模型代考

## 统计代写|生存模型代考Survival Models代写|The Force of Mortalit

$\ell_x$的导数可以解释为$\ell_x$的绝对瞬时年变化率。因为$\ell_x$代表的是活到$x$岁的人数，那么这个导数，即$\ell_x$的年变化率，就得到了人们在$x$岁时的年死亡率。这个导数是负的，因为$\ell_x$是递减函数。为了得到这个瞬时死亡率的绝对值，我们将使用导数的负值。最后，由于导数的大小取决于$\ell_x$本身的大小，我们通过将$\ell_x$的负导数除以$\ell_x$本身来获得相对瞬时死亡率。因此我们有
$$\mu_x=\frac{-\frac{d}{d x} \ell_x}{\ell_x}$$

$$\lambda(x)=\frac{-\frac{d}{d x} S(x)}{S(x)}=\frac{f(x)}{S(x)}$$

$$\ell_x=\ell_0 \cdot S(x)=\ell_0 \cdot \exp \left[-\int_0^x \mu_y d y\right]$$

## 统计代写|生存模型代考Survival Models代写|Tlte Probability Density Function ofX

$$f(x)=\mu_x\left(\ell_x / \ell_0\right)={ }_x p_0 \mu_x, x \geq 0$$

$$\frac{d}{d x} x p_0=-{ }_x p_0 \mu_x$$

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## MATLAB代写

MATLAB 是一种用于技术计算的高性能语言。它将计算、可视化和编程集成在一个易于使用的环境中，其中问题和解决方案以熟悉的数学符号表示。典型用途包括：数学和计算算法开发建模、仿真和原型制作数据分析、探索和可视化科学和工程图形应用程序开发，包括图形用户界面构建MATLAB 是一个交互式系统，其基本数据元素是一个不需要维度的数组。这使您可以解决许多技术计算问题，尤其是那些具有矩阵和向量公式的问题，而只需用 C 或 Fortran 等标量非交互式语言编写程序所需的时间的一小部分。MATLAB 名称代表矩阵实验室。MATLAB 最初的编写目的是提供对由 LINPACK 和 EISPACK 项目开发的矩阵软件的轻松访问，这两个项目共同代表了矩阵计算软件的最新技术。MATLAB 经过多年的发展，得到了许多用户的投入。在大学环境中，它是数学、工程和科学入门和高级课程的标准教学工具。在工业领域，MATLAB 是高效研究、开发和分析的首选工具。MATLAB 具有一系列称为工具箱的特定于应用程序的解决方案。对于大多数 MATLAB 用户来说非常重要，工具箱允许您学习应用专业技术。工具箱是 MATLAB 函数（M 文件）的综合集合，可扩展 MATLAB 环境以解决特定类别的问题。可用工具箱的领域包括信号处理、控制系统、神经网络、模糊逻辑、小波、仿真等。