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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

Some systems operate intermittently rather than continuously. The cycles of operation may be regular or irregular, and the system lifetime is the number of cycles until failure. Regular operation is commonly encountered in such areas as manufacturing production runs, cycles of an electrical system, machines producing individual items, and orbiting satellites exposed to solar radiation each time they emerge from the earth’s shadow. Another example occurs where certain electrical and structural units on aircraft have to be inspected after each flight: the location of one or more critical faults will lead to failing the system. In such cases, the lifetime, the number of cycles to failure, is a discrete variable.

Let $T$ be a discrete failure time taking possible values $0=\tau_{0}<\tau_{1}<\ldots<\tau_{m}$; $m$ may be finite or infinite, as may $\tau_{m}$. In many situations it is sufficient to take $\tau_{l}=l(l=0,1, \ldots)$, that is, integer-valued failure times. However, when we come on to likelihood functions in later chapters, the generality is necessary to smooth the transition from discrete to continuous time.

The survivor function is $\bar{F}(t)=\mathrm{P}(T>t)$ and the corresponding discrete density function, or probability mass function, is defined as $f(t)=\mathrm{P}(T=t)$. They are related by $f(t)=\bar{F}(t-)-\bar{F}(t)$ and $\bar{F}(t)=\sum f\left(\tau_{s}\right)$, where the summation is over $\left{s: \tau_{s}>t\right}$. The notation $\bar{F}(t-)$ is a useful abbreviation for $\lim {8 \downarrow 0} F(t-\delta)$. (For continuous failure times, $F(t-)=F(t)$.) If $t$ is not equal to one of the $\tau{s}, f(t)=0$. Also, we will adopt the convention $F\left(\tau_{0}\right)=1$, that is, $f(0)=0$, so that zero lifetimes are ruled out. The discrete hazard function is defined as
$$h(t)=\mathrm{P}(T=t \mid T \geq t)=f(t) / F(t-) .$$
(For continuous failure times, the denominator is usually written equivalently as $F(t)$.) Note that $0 \leq h(t) \leq 1$ for all $t$, with $h(t)=0$ except at the points $\tau_{l}(l=1, \ldots, m)$; also, $h(0)=0$, and $h(t)=1$ only at the upper end point $\tau_{m}$ of the distribution, where $F(t-)=f(t)$. The inverse relationship, expressing $\bar{F}(t)$ in terms of $h(t)$, can be derived as
$$\bar{F}(t)=\bar{F}(t-)-f(t)=\bar{F}(t-){1-h(t)}=\prod_{s=1}^{l(t)}\left{1-h\left(\tau_{s}\right)\right}$$
where $l(t)=\max \left{l: \tau_{l} \leq t\right}$ so that the product is over $\left{s: \tau_{s} \leq t\right}$. Also,
$$f(t)=h(t) \prod_{s=1}^{l(t-)}\left{1-h\left(\tau_{s}\right)\right} .$$
Otherwise expressed, and writing $h_{l}$ for $h\left(\tau_{l}\right)$, these representations are
$$\bar{F}\left(\tau_{l}\right)=\prod_{s=1}^{l}\left(1-h_{s}\right) \text { for } l \geq 1, f\left(\tau_{l}\right)=h_{l} \prod_{s=1}^{l-1}\left(1-h_{s}\right) \text { for } l \geq 2,$$

with $\bar{F}\left(\tau_{0}\right)=1, \bar{F}\left(\tau_{m}\right)=0, f\left(\tau_{0}\right)=0=h_{0}$, and $f\left(\tau_{1}\right)=h_{1}$. If we interpret $\prod_{s=1}^{0}$ as simply contributing a factor 1 , then the product formulae here hold for all $l$.
The integrated hazard function is
$$H(t)=-\log F(t)=-\sum_{s=1}^{l(t)} \log \left(1-h_{\mathrm{s}}\right) .$$
If the $h_{s}$ in the summation are small, then $\log \left(1-h_{s}\right) \approx-h_{s}$ and $H(t) \approx$ $\sum_{s=1}^{l(t)} h_{s}$, which can justifiably be called the cumulative hazard function.

## 统计代写|多元统计分析代写Multivariate Statistical Analysis代考|The Geometric Distribution

This is just about the simplest discrete waiting-time distribution. It arises as the time to failure in a sequence of Bernoulli trials, that is, independent trials each with probability $\rho$ of success. In the present context the process carries on while it is still winning. The $\tau_{l}$ here are the non-negative integers: $\tau_{l}=l$ for $l=0,1, \ldots, m=\infty$. We have $f(0)=0$ and, for $l \geq 1$,
$$f(l)=\rho^{l-1}(1-\rho), \quad F(l)=\rho^{l}, \quad h_{l}=(1-\rho) .$$
Note that the hazard function is constant, independent of $I$, as for the exponential among continuous distributions. Further, the famed lack of memory property of the exponential is also shared by the geometric:
$$\mathrm{P}(T>t+s \mid T>s)=F(t+s) / F(s)=\rho^{t}=F(t)=\mathrm{P}(T>t) .$$
Actually, of course, this is really only another way of saying that the hazard function is constant, as can be seen by considering the general identity
$$\bar{F}(t+s) / \bar{F}(s)=\prod_{r=s+1}^{s+t}\left(1-h_{r}\right):$$
if $h_{r}$ is independent of $r$, this expression is equal to $\bar{F}(t)$; conversely, if the expression is equal to $\bar{F}(1)$ for $t=1$ and every $s$, then $\bar{F}(1)=1-h_{s+1}$, and so $h_{s+1}$ is independent of $s$.

A standard textbook connection between the geometric and exponential distributions is as follows: Suppose that the continuous time scale is divided into equal intervals of length $\delta$, and that independent Bernoulli trials are performed with probability $\pi=\lambda \delta$ of a failure event within each interval. Let $M=T / \delta$, the number of intervals survived without failure. Then $M$ has the geometric survivor function $\mathrm{P}(M>m)=(1-\pi)^{m}$ for $m=1,2, \ldots$; hence, $\mathrm{P}(T / \delta>t / \delta)=(1-\lambda \delta)^{t / \delta}$. As $\delta \downarrow 0, \mathrm{P}(T>t) \rightarrow \mathrm{e}^{-\lambda t}$, that is, an exponential distribution.

A slightly extended version can be based on the discrete survivor function $\mathrm{P}(M>m)=(1-\pi)^{m^{\phi}}$, with $\phi>0 ; \pi$ is the probability of failure on the first trial, and $(1-\pi)^{m^{\natural}-(m-1)^{\natural}}$ is the probability of failure on the $m$ th, given survival that far. This leads to $\left(1-\lambda \delta^{\phi}\right)^{(t / \delta)^{\natural}}$ and thence, allowing $\delta \downarrow 0$, to a Weibull distribution with $\mathrm{P}(T>t)=\exp \left(-\lambda t^{\phi}\right)$.

## 统计代写|多元统计分析代写Multivariate Statistical Analysis代考|The Negative Binomial Distribution

The negative binomial is often introduced as a waiting-time distribution. Let $M$ be the number of independent Bernoulli trials performed up to and including the $\kappa$ th failure, where $\kappa$ is a given positive integer. The probability that there are $\kappa-1$ failures in the first $\kappa+m-1$ trials is given by the binomial expression $\left(\begin{array}{c}k+m-1 \ \kappa-1\end{array}\right)(1-\rho)^{k-1} \rho^{m}$. This is to be multiplied by $1-\rho$, the probability of failure on the $(\kappa+m)$ th trial. Thus,
$$\mathrm{P}(M=\kappa+m)=\left(\begin{array}{c} \kappa+m-1 \ \kappa-1 \end{array}\right) \rho^{m}(1-\rho)^{\kappa} \quad(m=0,1, \ldots) .$$
More generally, the expression can be taken to define a probability distribution on the non-negative integers for any positive real number $\kappa$. In the Exercises you are encouraged to verify that these probabilities sum to 1 , and also to find out why it is called negative binomial.

When $\kappa$ is an integer, $M$ can be represented as the sum of $\kappa$ consecutive waiting times to a first failure, that is, as the sum of $\kappa$ independent geometric variates. Then the limiting process described in the preceding example yields the sum of $\kappa$ independent exponential variates with rate parameter $\rho$, that is, a gamma distribution for $T$ with density $\rho^{\kappa} t^{\kappa-1} \mathrm{e}^{-\rho t} / \Gamma(\kappa)$.

## 多元统计分析代写

H(吨)=磷(吨=吨∣吨≥吨)=F(吨)/F(吨−).
（对于连续故障时间，分母通常等价地写为F(吨)。） 注意0≤H(吨)≤1对全部吨， 和H(吨)=0除了在点τl(l=1,…,米); 还，H(0)=0， 和H(吨)=1仅在上端点τ米的分布，其中F(吨−)=F(吨). 反比关系，表示F¯(吨)按照H(吨), 可以导出为
\bar{F}(t)=\bar{F}(t-)-f(t)=\bar{F}(t-){1-h(t)}=\prod_{s=1}^ {l(t)}\left{1-h\left(\tau_{s}\right)\right}\bar{F}(t)=\bar{F}(t-)-f(t)=\bar{F}(t-){1-h(t)}=\prod_{s=1}^ {l(t)}\left{1-h\left(\tau_{s}\right)\right}

f(t)=h(t) \prod_{s=1}^{l(t-)}\left{1-h\left(\tau_{s}\right)\right} 。f(t)=h(t) \prod_{s=1}^{l(t-)}\left{1-h\left(\tau_{s}\right)\right} 。

F¯(τl)=∏s=1l(1−Hs) 为了 l≥1,F(τl)=Hl∏s=1l−1(1−Hs) 为了 l≥2,

H(吨)=−日志⁡F(吨)=−∑s=1l(吨)日志⁡(1−Hs).

## 统计代写|多元统计分析代写Multivariate Statistical Analysis代考|The Geometric Distribution

F(l)=ρl−1(1−ρ),F(l)=ρl,Hl=(1−ρ).

F¯(吨+s)/F¯(s)=∏r=s+1s+吨(1−Hr):

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