### 统计代写|工程统计代写engineering statistics代考|STAT 2201

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• Statistical Inference 统计推断
• Statistical Computing 统计计算
• Advanced Probability Theory 高等概率论
• Advanced Mathematical Statistics 高等数理统计学
• (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|工程统计代写engineering statistics代考|RELATION TO OTHER DISTRIBUTIONS

It is a special case of gamma distribution with $m=1$ (Chapter 6), and Weibull distribution. The Rosin-Rammler-Bennett (RRB) distribution used in mineral engineering is a special case of one-parameter exponential distribution. If $X \sim \operatorname{EXP}(\lambda)$, and $b$ is a constant, then $Y=X^{1 / b} \sim \operatorname{WEIB}(\lambda, b)$. The sum of $n$ IID exponential variates with the same parameter is gamma (also called Erlang) distributed, and with different parameters has a hyper-exponential distribution. Similarly, if $X_{1}, X_{2}, \ldots, X_{n}$ are IID $\operatorname{EXP}(\lambda)$ and $S_{n}=X_{1}+X_{2}+\cdots+X_{n}$, then $\operatorname{Pr}\left[S_{n}<t<S_{n+1}\right]$ has a Poisson distribution with parameter $\lambda t$. It is also related to the $\mathrm{U}(0,1)$ distribution, and power-law distribution, which is another discrete analogue of this distribution [34]. The extreme value distribution can be considered as a nonlinear generalization of $\operatorname{EXP}(\lambda)$. Its relationship with zero-truncated Poisson (ZTP) distribution is used to generate random numbers [98]. If $X_{1}$ and $X_{2}$ are IID $\operatorname{EXP}(\lambda)$, then $Y=X_{1} /\left(X_{1}+X_{2}\right) \sim \mathrm{U}(0,1)$. This has the implication that “if two random numbers between 0 and 1 are chosen from $\mathrm{U}(0,1)$, the ratio of one of them to their sum is more likely to be close to half; but when the two numbers are chosen from an exponential distribution, the same ratio is uniform between 0 and 1 .” Putting $Y=1 / X$ results in the inverse exponential distribution (Figure 3.2) with PDF
$$f(y ; \lambda)=\left(\lambda / y^{2}\right) \exp (-\lambda / y)$$
If $X_{1}$ and $X_{2}$ are IID EXP(1) random variables, $Y=X_{1}-X_{2}$ has standard Laplace distribution. A mixture of exponential distributions with gamma mixing weights gives rise to Lomax distribution.

## 统计代写|工程统计代写engineering statistics代考|PROPERTIES OF EXPONENTIAL DISTRIBUTION

This distribution has a single parameter, which is positive. It is a reverse-J shaped distribution which is always positively skewed (see Figure 3.1; page 28). Variance of this distribution is the square of the mean, as shown below. This means that when $\lambda \rightarrow 0$, the variance and kurtosis increases without limit (see Figure 3.1; page 28). The CDF is given by
$$F(x ; \lambda)=\left{\begin{array}{cl} 1-\exp (-\lambda x), x \geq 0 & \ 0 & \text { otherwise. } \end{array}\right.$$
The SF is $S(x ; \lambda)=1-\mathrm{CDF}=\exp (-\lambda x)$, so that $f(x ; \lambda)=\lambda S(x ; \lambda)$. From this the hazard function is obtained as
$$h(x)=f(x) /[1-F(x)]=f(x) / S(x)=\lambda,$$

which is constant. When a device or an equipment approximately exhibits constant hazard rate, it is an indication that the $\operatorname{EXP}(\lambda)$ may be a good choice to model the lifetime. If $x_{\alpha}$ is the $\alpha^{\text {th }}$ percentage point, $\alpha=1-\exp \left(-\lambda x_{\alpha}\right)$ from which $x_{\alpha}=-\log (1-\alpha) / \lambda$

Problem 3.1 A left-truncated exponential distribution with truncation point $c$ has PDF $f(x ; \lambda)=\lambda e^{-\lambda x} /\left[1-e^{-c \lambda}\right]$ for $x>c$. Obtain the mean and variance.

Problem 3.2 Prove that the exponential distribution is the continuous-time analog of the geometric distribution.

Solution 3.3 Let $T$ denote the lifetime of a non-repairable item (like light-bulbs, transistors, micro-batteries used in watches, etc.) that wears out over time. Divide $T$ into discrete time units of equal duration $($ say $c)$. This duration may be counted in hours for light bulbs, days for transistors, and so on. Thus, the time-clicks are counted in unit multiples of $c$ (say 1200 hours for light bulbs). Let $N$ denote the number of time-clicks until the item fails, so that $T=N c$. This equation connects the continuous lifetime with discrete time-clicks. Assume that $N$ has a geometric distribution $\mathrm{GEO}(c \lambda)$ where $c \lambda$ denotes the failure probability for each time-click. Then $\operatorname{Pr}[N=n]=(1-c \lambda)^{n-1}(c \lambda)$, where $(1-c \lambda)^{n-1}$ denotes the probability that the unit did not fail during the first $(n-1)$ time-clicks. As the $\mathrm{SF}$ of $T$ is $\operatorname{Pr}[T>k]=\operatorname{Pr}[N c>k]=$ $\operatorname{Pr}[N>\lfloor k / c\rfloor]=(1-c \lambda)^{\lfloor k / c\rfloor}$, where the integer part is taken because $N$ is discrete. This is of the form $(1-\lambda / m)^{m}$, which as $m \rightarrow \infty$ tends to $\exp (-\lambda)$ with $m=1 / c$. As $c \rightarrow 0$, the SF approaches $\exp (-k \lambda)$, which is the SF of exponential distribution.

Problem 3.4 If $X \sim E X P(\lambda)$, find (i) $\operatorname{Pr}[1 \leq X \leq 2]$ and (ii) $\operatorname{Pr}[X \geq x]$ when $\lambda=\ln (2)$.
Problem 3.5 If $\mathrm{X} \sim \mathrm{EXP}(\lambda)$ and $\mathrm{Y} \sim \operatorname{EXP}(\mu)$ prove that $\operatorname{Pr}(X<Y)=\lambda /(\lambda+\mu)$.
This distribution represents the time for a continuous process to change state (from working to non-working, from infected to recovery, detected to non-detected or vice versa). For example, the time between detection of radioactivity by a Geiger counter (absence to presence) is approximately exponentially distributed.

## 统计代写|工程统计代写engineering statistics代考|RANDOM NUMBERS

The easiest way to generate random numbers is using the inverse CDF method. As the quantiles of the distribution are given by $u=F(x)=1-\exp (-\lambda x)$, we get $x=F^{-1}(u)=-\log (1-$ $u) / \lambda$. This becomes $-\theta \log (1-u)$ for the alternate form $f(x, \theta)=(1 / \theta) \exp (-x / \theta)$. As $U(0,1)$ and $1-U$ are identically distributed (page 20 ), we could generate a random number in $[0,1$ ) and obtain $u=-\log (u) / \lambda$ as the required random number (Marsaglia (1961) [98]).

Problem $3.41$ If $X, Y$ are IID $\operatorname{EXP}(1 / 2)$, prove that $Z=(X-Y) / 2$ is Laplace distributed.
Problem 3.42 Find $k$ for the PDF $f(x)=k x^{-p} \exp (-c / x), \quad 00, p>1$.
Show that the $r^{t h}$ moment is $E\left(x^{r}\right)=c^{r} \Gamma(p-r+1) / \Gamma(p-1)$ for $r \leq(p+1)$.
Example 3.43 Mean deviation of exponential distribution Find the mean deviation of the exponential distribution $\mathrm{f}(\mathrm{x}, \lambda)=\lambda e^{-\lambda x}$.
Solution 3.44 We know that the CDF is $1-e^{-\lambda x}$. Thus, the MD is given by
$$\mathrm{MD}=2 \int_{0}^{1 / \lambda}\left(1-e^{-\lambda x}\right) d x$$
Split this into two integrals and evaluate each to get
$$\mathrm{MD}=2[1 / \lambda+(1 / \lambda) \exp (-1)-(1 / \lambda)]=2 /(e \lambda)=2 \mu_{2} * f_{m}$$
where $f_{m}=\lambda e^{-1}=\lambda / e$. Alternatively, use the $\mathrm{SF}()$ version as the exponential distribution tails off to the upper limit.

## 统计代写|工程统计代写engineering statistics代考|RELATION TO OTHER DISTRIBUTIONS

F(是;λ)=(λ/是2)经验⁡(−λ/是)

## 统计代写|工程统计代写engineering statistics代考|PROPERTIES OF EXPONENTIAL DISTRIBUTION

$$F(x ; \lambda)=\left{给出 1−经验⁡(−λX),X≥0 0 否则。 \正确的。 吨H和小号F一世s小号(X;λ)=1−CDF=经验⁡(−λX),s○吨H一个吨F(X;λ)=λ小号(X;λ).Fr○米吨H一世s吨H和H一个和一个rdF在nC吨一世○n一世s○b吨一个一世n和d一个s h(x)=f(x) /[1-F(x)]=f(x) / S(x)=\lambda,$$

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## MATLAB代写

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