### 统计代写|多元统计分析代写Multivariate Statistical Analysis代考|Left Truncation

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• Statistical Inference 统计推断
• Statistical Computing 统计计算
• (Generalized) Linear Models 广义线性模型
• Statistical Machine Learning 统计机器学习
• Longitudinal Data Analysis 纵向数据分析
• Foundations of Data Science 数据科学基础

## 统计代写|多元统计分析代写Multivariate Statistical Analysis代考|Left Truncation

Suppose that a unit comes under scrutiny sometime after switching on, meaning the time at which its inexorable downward spiral toward its demise begins. Specifically, suppose that observation on the unit starts at random time $Z$ after switch-on and its lifetime is $T$ : let $S=T-Z$. If $S<0$, the unit will fail before it comes under observation; otherwise, it will be under observation for time $S \geq 0$. Let $\bar{F}(t \mid z)=\mathrm{P}(T>t \mid Z=z)$ and let $Z$ have density function $g(z)$ on $(0, \infty)$. Then,
$$\mathrm{P}(S>s \mid Z=z)=\mathrm{P}(T>s+z \mid Z=z)= \begin{cases}\bar{F}(s+z \mid z) & \text { if } s+z \geq 0 \ 1 & \text { if } s+z<0\end{cases}$$ Hence, $$\mathrm{P}(S>s)=\int_{-s}^{\infty} F(s+z \mid z) g(z) d z+\int_{0}^{-s} g(z) d z:$$
for $s \geq 0$ the second integral is zero, and for $s<0$ it is equal to $\mathrm{P}(Z \leq-s)$. Assume now that $T$ has an exponential distribution with mean $\xi$, and that $Z$ has an exponential distribution with mean $v$. Then, for $s \geq 0$, $$\mathrm{P}(S>s)=\int_{0}^{\infty} \mathrm{e}^{-(s+z) / \xi} v^{-1} \mathrm{e}^{-z / v} d z=\left(\frac{\xi}{\xi+v}\right) \mathrm{e}^{-s / \xi},$$
and for $s<0$, $$\mathrm{P}(S>s)=\left(\frac{\xi}{\xi+v}\right) \mathrm{e}^{s / v}+\left(1-\mathrm{e}^{s / v}\right)=1-\left(\frac{v}{\xi+v}\right) \mathrm{e}^{\mathrm{s} / v}$$
The proportion of units that survive to become observed is $\mathrm{P}(S>0)=\frac{\xi}{\xi+v}$, which is greater than $50 \%$ if $\xi>v$.

Lawless (2003, Section 2.4) gave a general treatment of this sort of situation from a slightly different standpoint.

## 统计代写|多元统计分析代写Multivariate Statistical Analysis代考|Probabilities of Observation versus Censoring

Consider the situation where a unit is liable to failure, at time $T^{f}$, or censoring (being lost to observation), at time $T^{c}$. Assume that $T^{f}$ and $T^{c}$ are independent with probability functions $\left(f^{f}, \bar{F} f\right)$ and $\left(f^{c}, \bar{F}^{c}\right)$, respectively. The probabilities of observed failure and censoring at time $t$ are
$$\mathrm{P}\left(T^{f}=t, T^{c}>t\right)=f^{f}(t) \bar{F}^{c}(t) \text { and } \mathrm{P}\left(T^{c}=t, T^{f}>t\right)=f^{c}(t) \bar{F}^{f}(t)$$

The overall probabilities of observed failure and of censorship are obtained by integration over $t$. The likelihood function for a random sample is
\begin{aligned} L &=\prod_{\text {obs }}\left{f^{f}(t) \bar{F}^{c}(t)\right} \times \prod_{\text {cens }}\left{f^{c}(t) \bar{F}^{f}(t)\right} \ &=\left{\prod_{\text {obs }} f^{f}(t)\right}\left{\prod_{\text {cens }} \bar{F}^{f}(t)\right} \times\left{\prod_{\text {obs }} f^{c}(t)\right}\left{\prod_{\text {cens }} \bar{F}^{c}(t)\right} . \end{aligned}
The first product here contains the probabilities determining the $T f$ distribution, which is usually the one of primary interest. Suppose that the $T^{c}$ probabilities are not linked in any way to those of $T f$, through having parameters in common, for example. Then we may focus upon the first term, which is the likelihood function shown at the beginning of this section. Of course, we can likewise use the second product to estimate the $T^{c}$ probabilities if we want to.

If the censoring process is not independent of the failure process, we have dependent competing risks, which is covered in Part III.

Consider a random sample of Weibull-distributed lifetimes. The survivor function has form $\bar{F}(t ; \theta)=\exp \left{-(t / \xi)^{v}\right}$, with $\theta=(\xi, v)$. This can be recast as
$$\log {-\log \bar{F}(t ; \theta)}=v \log t-v \log \xi$$
Given a random sample of uncensored times, $\left(t_{1}, \ldots, t_{n}\right)$, we can order them as $t_{(1)}<t_{(2)}<\ldots<t_{(n)}$; the $t_{(i)}$ are the order statistics. An estimate of $F\left(t_{(i)} ; \theta\right)$ can then be extracted as $(n-i) / n$, the sample proportion of times beyond $t_{(i)}(i=1, \ldots, n)$; to avoid the extreme and unlikely value 1 for $F\left(t_{(n)} ; \theta\right)$, in practice we use a slightly modified version such as $a_{i}=(n-i+1) /(n+1)$. Now, consider plotting the points ( $\log t_{(i)}, \log \left(-\log a_{i}\right)$ ). According to the equation above, with Weibull-distributed times, the points plotted should lie near a straight line with slope $v$ and intercept $-v \log \xi$. This is the basic Weibull probability plot, widely used by engineers, for example. When there is right censoring a more sophisticated approach to estimating the survivor function is needed-see Section 4.1.

For random samples that may include right-censored observations, we can write down the likelihood function as follows, using the Weibull form of $\bar{F}(t ; \theta)$ given above together with the corresponding hazard function $h(t)=$ $(v / \xi)(t / \xi)^{v-1}$.
\begin{aligned} \log L(\xi, v) &=\log \prod_{i=1}^{n}\left[\left{(v / \xi)\left(t_{i} / \xi\right)^{v-1}\right}^{k_{i}} \exp \left{-\left(t_{i} / \xi\right)^{v}\right}\right] \ &=n_{c}(\log v-v \log \xi)+(v-1) \sum_{c e n s} \log t_{i}-\sum_{i=1}^{n}\left(t_{i} / \xi\right)^{v} \end{aligned}where $n_{c}$ is the number of right-censored times. Setting $\partial \log L / \partial \xi$ to zero produces $\xi^{v}=\sum_{i=1}^{t t} t_{i}^{v} / n_{c}$, which yields the $m l e \hat{\xi}{v}$ in terms of $v$. So, $\xi$ in $L$ can be replaced by $\hat{\xi}{v}$ to produce the profile likelihood $L\left(\hat{\xi}_{v}, v\right)$ for $v$. This can then be maximised to compute $\hat{v}$ using a simple one-dimensional search.

## 统计代写|多元统计分析代写Multivariate Statistical Analysis代考|Left Truncation

Lawless（2003 年，第 2.4 节）从稍微不同的角度对这种情况进行了一般性处理。

## 统计代写|多元统计分析代写Multivariate Statistical Analysis代考|Probabilities of Observation versus Censoring

\begin{aligned} L &=\prod_{\text {obs }}\left{f^{f}(t) \bar{F}^{c}(t)\right} \times \prod_{\text {cens }}\left{f^{c}(t) \bar{F}^{f}(t)\right} \ &=\left{\prod_{\text {obs }} f^{f} (t)\right}\left{\prod_{\text {cens }} \bar{F}^{f}(t)\right} \times\left{\prod_{\text {obs }} f^{ c}(t)\right}\left{\prod_{\text {cens }} \bar{F}^{c}(t)\right} 。\end{对齐}\begin{aligned} L &=\prod_{\text {obs }}\left{f^{f}(t) \bar{F}^{c}(t)\right} \times \prod_{\text {cens }}\left{f^{c}(t) \bar{F}^{f}(t)\right} \ &=\left{\prod_{\text {obs }} f^{f} (t)\right}\left{\prod_{\text {cens }} \bar{F}^{f}(t)\right} \times\left{\prod_{\text {obs }} f^{ c}(t)\right}\left{\prod_{\text {cens }} \bar{F}^{c}(t)\right} 。\end{对齐}

\begin{对齐} \log L(\xi, v) &=\log \prod_{i=1}^{n}\left[\left{(v / \xi)\left(t_{i} / \ xi\right)^{v-1}\right}^{k_{i}} \exp \left{-\left(t_{i} / \xi\right)^{v}\right}\right] \ &=n_{c}(\log vv \log \xi)+(v-1) \sum_{c e n s} \log t_{i}-\sum_{i=1}^{n}\left(t_{i } / \xi\right)^{v} \end{对齐}\begin{对齐} \log L(\xi, v) &=\log \prod_{i=1}^{n}\left[\left{(v / \xi)\left(t_{i} / \ xi\right)^{v-1}\right}^{k_{i}} \exp \left{-\left(t_{i} / \xi\right)^{v}\right}\right] \ &=n_{c}(\log vv \log \xi)+(v-1) \sum_{c e n s} \log t_{i}-\sum_{i=1}^{n}\left(t_{i } / \xi\right)^{v} \end{对齐}在哪里nC是右删失次数。环境∂日志⁡大号/∂X归零产生X在=∑一世=1吨吨吨一世在/nC，这产生米l和X^在按照在. 所以，X在大号可以替换为X^在产生轮廓似然性大号(X^在,在)为了在. 然后可以将其最大化以计算在^使用简单的一维搜索。

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