统计代写|多元统计分析代写Multivariate Statistical Analysis代考| Reliability Topics

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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|Reliability of Systems

Consider a system, comprising components $c_{1}, \ldots, c_{n}$, and let $p_{j}$ be the reliability of component $c_{j}$, that is, its probability of being $u p$ (in working order). We wish to calculate the reliability $R$ of the whole system from the component reliabilities. It is assumed throughout that the components operate independently.
Example

1. Series system: this is shown in Figure 2.1a. Since the system can operate only if both components are up, the reliability is
$$R=\mathrm{P}\left(c_{1} u p \text { and } c_{2} u p\right)=\mathrm{P}\left(c_{1} u p\right) \times \mathrm{P}\left(c_{2} u p\right)=p_{1} p_{2}$$
using the independence of $c_{1}$ and $c_{2}$.
2. Parallel system: this is shown in Figure $2.1 \mathrm{~b}$. The system can operate if either $c_{1}$ or $c_{2}$ is up, so
\begin{aligned} R &=1-\mathrm{P}\left(c_{1} \text { down and } c_{2} \text { down }\right) \ &=1-\mathrm{P}\left(c_{1} \text { down }\right) \times \mathrm{P}\left(c_{2} \text { down }\right)=1-\left(1-p_{1}\right)\left(1-p_{2}\right) \end{aligned}
This can easily be extended to $n$ components: for a series system $R=$ $p_{1} p_{2} \ldots p_{n}$, and for a parallel system $R=1-\left(1-p_{1}\right)\left(1-p_{2}\right) \ldots\left(1-p_{n}\right)$. Further, components can be replaced by subsystems in more complex systems and networks.

统计代写|多元统计分析代写Multivariate Statistical Analysis代考|Stress and Strength

Suppose that a system (electronic, mechanical, chemical, biological) has strength $X$ on some scale that measures its resistance to failure. The system will have been constructed or assembled to some nominal strength specification, say $\mu_{X} ;$ but in practice, $X$ will be randomly distributed around $\mu_{X}$. Once in operation, suppose that the system will be exposed to stress $Y$ randomly distributed around some level $\mu_{Y}$. The system will fail if $X<Y$, and we are interested in the probability of this event.
Let $f(x, y)$ be the joint density of $X$ and $Y$, then
$$\mathrm{P}(\text { failure })=\mathrm{P}(X<Y)=\int_{x<y} f(x, y) d x d y=\int_{0}^{\infty} d x \int_{x}^{\infty} d y{f(x, y)}$$
If, as is usual, $X$ and $Y$ are independent, $f(x, y)=f_{X}(x) f_{Y}(y)$, and then
$$\mathrm{P}(X<Y)=\int_{0}^{\infty} d x \int_{x}^{\infty} d y\left{f_{X}(x) f_{Y}(y)\right}=\int_{0}^{\infty} f_{X}(x)\left{1-F_{Y}(x)\right} d x$$
where $F_{Y}$ is the distribution function of $Y$. Equivalently,
$$\mathrm{P}(X<Y)=\int_{0}^{\infty} d y \int_{0}^{y} d x{f(x, y)}=\int_{0}^{\infty} f_{Y}(y) F_{X}(y) d y .$$
In the survival version of the situation, $X$ and $Y$ may vary over time, and then we have $X(t)$ and $Y(t)$. The system will fail at time $s$ if $X(t) \geq Y(t)$ for $0 \leq t<s$, and $X(s)<Y(s)$. The forms taken by $X(t)$ and $Y(t)$ can vary widely between different applications; for example, $X(t)$ might be constant or steadily decreasing over time, and $Y(t)$ might be constant, steadily increasing, randomly fluctuating, or result from intermittent shocks to the system. The analysis in such cases can be quite difficult: $T$ is the first time at which the stochastic process $X(t)-Y(t)$ becomes negative.

统计代写|多元统计分析代写Multivariate Statistical Analysis代考|Survival Distributions

1. Derive the density $f(t)=\xi^{-1} \mathrm{e}^{-t / \xi}$ of the exponential distribution and show that its mean is $\xi$ and that its variance is $\xi^{2}$. Prove the lack-of-memory property. What is the distribution of $(T / \xi)^{v}$ for $v>0$ ? What about $v<0$ ?
2. Derive the mean and variance of the Weibull distribution. Verify that the upper $q$ th quantile is $t_{q}=\xi(-\log q)^{1 / v}$.
3. Suppose that $Y$ has distribution $N\left(\mu, \sigma^{2}\right)$ and that $T=\mathrm{e}^{Y}$ : then $T$ has a log-normal distribution. Confirm that $\mathrm{E}\left(T^{r}\right)=\mathrm{e}^{r \mu+\sigma^{2} / 2}$. Derive the mean and variance of $T$ as $\mu_{T}=\mathrm{E}(T)=\mathrm{e}^{\mu+\sigma^{2} / 2}$ and

$\sigma_{T}^{2}=\operatorname{var}(T)=\left(\mathrm{e}^{\sigma^{2}}-1\right) \mu_{T}^{2}$. Note that the survivor and hazard functions are not explicit, only expressible in terms of $\Phi$, the standard normal distribution function.

1. Another generalisation of the exponential distribution is the gamma, which has density $f(t)=\Gamma(v)^{-1} \xi^{-v} t^{v-1} \mathrm{e}^{t / \xi}$. The parameters are $\xi>0$ (scale) and $v>0$ (shape); the exponential is recovered with $v=1$. Derive the survivor function as $\bar{F}(t)=1-\Gamma^{(r)}(v ; t / \xi)$, where $\Gamma^{(r)}(v ; z)=\Gamma(v)^{-1} \int_{0}^{z} y^{v-1} \mathrm{e}^{-y} d y$ is the incomplete gamma function ratio.
2. Calculate the mean, variance, and upper $100 q \%$ quantile of the Pareto distribution. Take care regarding the size of $\gamma$.
3. As a generalisation of the Pareto distribution, step up the Burr: this has survivor function $\bar{F}(t)=\left{1+(t / \alpha)^{\rho}\right}^{-\gamma}$, just replacing $t / \alpha$ by $(t / \alpha)^{\rho}$ with $\rho>0$. Derive its hazard function: is it IFR, DFR, or what? You might suspect that, by analogy with the derivation of the Pareto given above, the Burr can be mocked up as some sort of Weibullgamma combination. Can it? The special case $\gamma=1$ gives the loglogistic distribution.
4. Prove the following:
a. If $\int_{0}^{\infty} h(t) d t<\infty$, then $\mathrm{P}(T=\infty)>0$.
b. If $h(t)=h_{1}(t)+h_{2}(t)$, where $h_{1}(t)$ and $h_{2}(t)$ are the hazard functions of independent failure time variates $T_{1}$ and $T_{2}$, then $T$ has the same distribution as $\min \left(T_{1}, T_{2}\right)$.
5. Calculate $\mathrm{P}\left(T>t_{0}\right)$ and $\mathrm{P}\left(T>2 t_{0}\right)$ when $T$ has hazard function $h(t)=a$ for $0<t<t_{0}, h(t)=b$ for $t \geq t_{0}$.
6. Show that, for continuous $T, \mathrm{E}(T)=\int_{0}^{\infty} \bar{F}(t) d t$, provided that $t \bar{F}(t) \rightarrow$ 0 as $t \rightarrow \infty$. For discrete $T$, taking values $0,1, \ldots$ with probabilities $p_{0}, p_{1}, \ldots$, let $q_{j}=\mathrm{P}(T>j)$ : show that $\mathrm{E}(T)=\sum_{j=1}^{\infty} q_{j}$.
7. Negative binomial distribution: verify that the probabilities $p_{r}=$ $\left(\begin{array}{c}\kappa+r-1 \ \kappa-1\end{array}\right) \rho^{r}(1-\rho)^{\kappa}(r=0,1,2, \ldots)$ sum to $1 .$
8. Sometimes survival data are reduced to binary outcomes. Thus, all that is recorded is whether $T>t^{}$ or not, where $t^{}$ is some threshold, for example, five-year survival after cancer treatment. For Weibull lifetimes $p^{}=\mathrm{P}\left(T>t^{}\right)=\exp \left{-\left(t^{} / \xi\right)^{v}\right}$, solog $\left(-\log p^{}\right)=v \log t^{}-$ $v \log \xi .$ A log-linear regression model for $\xi$ then gives a complementary $\log -\log$ form for $p^{}: \log \left(-\log p^{*}\right)=\beta_{0}+\mathbf{x}^{T} \beta$. What lifetime distribution corresponds to a logit-linear model for binary data?

统计代写|多元统计分析代写Multivariate Statistical Analysis代考|Reliability of Systems

1. 串联系统：如图 2.1a 所示。由于系统只有在两个组件都启动时才能运行，因此可靠性是
R=磷(C1在p 和 C2在p)=磷(C1在p)×磷(C2在p)=p1p2
使用独立性C1和C2.
2. 并联系统：如图所示2.1 b. 系统可以运行，如果C1或者C2起来了，所以
R=1−磷(C1 下来和 C2 向下 ) =1−磷(C1 向下 )×磷(C2 向下 )=1−(1−p1)(1−p2)
这可以很容易地扩展到n组件：用于串联系统R= p1p2…pn, 对于并行系统R=1−(1−p1)(1−p2)…(1−pn). 此外，组件可以被更复杂的系统和网络中的子系统替换。

统计代写|多元统计分析代写Multivariate Statistical Analysis代考|Stress and Strength

\mathrm{P}(X<Y)=\int_{0}^{\infty} d x \int_{x}^{\infty} d y\left{f_{X}(x) f_{Y}(y) \right}=\int_{0}^{\infty} f_{X}(x)\left{1-F_{Y}(x)\right} d x\mathrm{P}(X<Y)=\int_{0}^{\infty} d x \int_{x}^{\infty} d y\left{f_{X}(x) f_{Y}(y) \right}=\int_{0}^{\infty} f_{X}(x)\left{1-F_{Y}(x)\right} d x

统计代写|多元统计分析代写Multivariate Statistical Analysis代考|Survival Distributions

1. 导出密度F(吨)=X−1和−吨/X的指数分布，并证明它的平均值是X并且它的方差是X2. 证明内存不足的性质。什么是分布(吨/X)在为了在>0? 关于什么在<0 ?
2. 导出 Weibull 分布的均值和方差。验证上层q第分位数是吨q=X(−日志⁡q)1/在.
3. 假设是有分布ñ(μ,σ2)然后吨=和是： 然后吨具有对数正态分布。确认这个和(吨r)=和rμ+σ2/2. 导出均值和方差吨作为μ吨=和(吨)=和μ+σ2/2和

σ吨2=曾是⁡(吨)=(和σ2−1)μ吨2. 请注意，幸存者和危险函数不是明确的，只能表示为披，标准正态分布函数。

1. 指数分布的另一个推广是 gamma，它具有密度F(吨)=Γ(在)−1X−在吨在−1和吨/X. 参数是X>0（规模）和在>0（形状）; 指数恢复为在=1. 导出幸存者函数为F¯(吨)=1−Γ(r)(在;吨/X)， 在哪里Γ(r)(在;和)=Γ(在)−1∫0和是在−1和−是d是是不完全伽马函数比。
2. 计算均值、方差和上限100q%帕累托分布的分位数。注意大小C.
3. 作为帕累托分布的推广，加强毛刺：这具有幸存者功能\bar{F}(t)=\left{1+(t / \alpha)^{\rho}\right}^{-\gamma}\bar{F}(t)=\left{1+(t / \alpha)^{\rho}\right}^{-\gamma}, 只是替换吨/一种经过(吨/一种)ρ和ρ>0. 推导出它的危险函数：是 IFR、DFR 还是什么？您可能会怀疑，通过与上面给出的 Pareto 推导类比，Burr 可以模拟为某种 Weibullgamma 组合。它可以？特殊情况C=1给出对数逻辑分布。
4. 证明以下几点
：如果∫0∞H(吨)d吨<∞， 然后磷(吨=∞)>0.
湾。如果H(吨)=H1(吨)+H2(吨)， 在哪里H1(吨)和H2(吨)是独立失效时间变量的危险函数吨1和吨2， 然后吨具有相同的分布分钟(吨1,吨2).
5. 计算磷(吨>吨0)和磷(吨>2吨0)什么时候吨具有危险功能H(吨)=一种为了0<吨<吨0,H(吨)=b为了吨≥吨0.
6. 证明，对于连续吨,和(吨)=∫0∞F¯(吨)d吨, 前提是吨F¯(吨)→0 为吨→∞. 对于离散吨, 取值0,1,…有概率p0,p1,…， 让qj=磷(吨>j)： 显示和(吨)=∑j=1∞qj.
7. 负二项分布：验证概率pr= (ķ+r−1 ķ−1)ρr(1−ρ)ķ(r=0,1,2,…)总和1.
8. 有时生存数据被简化为二元结果。因此，记录的只是是否吨>吨与否，在哪里吨是一些阈值，例如，癌症治疗后的五年生存率。威布尔一生p^{}=\mathrm{P}\left(T>t^{}\right)=\exp \left{-\left(t^{} / \xi\right)^{v}\right}p^{}=\mathrm{P}\left(T>t^{}\right)=\exp \left{-\left(t^{} / \xi\right)^{v}\right}, 独奏(−日志⁡p)=在日志⁡吨− 在日志⁡X.对数线性回归模型X然后给出一个互补的日志−日志表格p:日志⁡(−日志⁡p∗)=b0+X吨b. 什么寿命分布对应于二进制数据的 logit 线性模型？

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