### 统计代写|工程统计代写engineering statistics代考|STATS 7053

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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|工程统计代写engineering statistics代考|MOMENTS AND GENERATING FUNCTIONS

The moments are easy to find using the MGF. The mean is directly obtained as $\mu=[1 /(b-$ $a)] \int_{a}^{b} x d x=[1 /(b-a)] \frac{x^{2}}{2} |_{a}^{b}=\left(b^{2}-a^{2}\right) /[2(b-a)]=(a+b) / 2$. Higher-order moments are found using the MGF. Thus,
$$E\left(X^{n}\right)=(1 /(n+1)) \sum_{k=0}^{n} a^{k} b^{n-k}$$
The MGF is
$$M_{x}(t)=E\left(e^{t x}\right)=\int_{x=a}^{b}[1 /(b-a)] e^{t x} d x=[1 /(b-a)] e^{t x} /\left.t\right|{a} ^{b}=\left(e^{b t}-e^{a t}\right) /[(b-a) t]$$ The characteristic function $(\mathrm{ChF})$ is $$\phi{x}(t)=(\exp (i b t)-\exp (i a t)) /[(b-a) i t] \quad \text { for } \quad t \neq 0 .$$
This reduces to $\sinh (a t) / a t$ for $\operatorname{CUNI}(-a,+a)$.
Moments can be found from the MGF as follows. Consider $e^{b t} / t=1 / t+b+b^{2} t / 2 !+$ $\cdots+b^{k} t^{k-1} / k !+\cdots .$ As $(1 / t)$ is common in both $e^{b t} / t$ and $e^{a t} / t$, it cancels out. The second term is $(b-a) /(b-a)=1$. Thus,
$$\left(e^{b t}-e^{a t}\right) /[(b-a) t]=1+\frac{1}{b-a}\left(\sum_{k=2}^{\infty}\left[\left(b^{k}-a^{k}\right) /(b-a)\right] t^{k-1} / k !\right) \text {. }$$
If we differentiate $(2.13)(k-1)$ times w.r.t. $t$, all terms below the $(k-1)^{t h}$ term will vanish (as they are derivatives of constants independent of $t$ ‘s) and all terms beyond the $k^{t h}$ term will contain powers of $t$. Only the $(k-1)^{t h}$ term is a constant with a $(k-1)$ ! in the numerator, which cancels out with the $k !$ giving a $k$ in the denominator. By taking the limit as $t \rightarrow 0$, we get
$$\mu_{k-1}^{\prime}=\left.\left(\partial^{k-1} / \partial t^{k-1}\right) M_{x}(t)\right|_{t=0}=\left(b^{k}-a^{k}\right) /[(b-a) k]$$

## 统计代写|工程统计代写engineering statistics代考|TRUNCATED UNIFORM DISTRIBUTIONS

Truncation in the left-tail or right-tail results in the same distribution with a reduced range. Suppose $X \sim \operatorname{CUNI}(a, b)$. If truncation occurs in the left-tail at $x=c$ where $a<c<b$, the PDF is given by
$$g(x ; a, b, c)=f(x ; a, b) /(1-F(c))=(1 /(b-a))[1 /(1-(c-a) /(b-a))]=1 /(b-c) .$$
If truncation occurs at $c$ in the left-tail and $d$ in the right-tail, the PDF is given by
$$g(x ; a, b, c, d)=f(x ; a, b) /(F(d)-F(c))=1 /(d-c)$$
This shows that truncation results in rectangular distributions.
Example 2.34 Even moments of rectangular distribution Prove that the $k^{\text {th }}$ central moment is zero for $k$ odd, and is given by $\mu_{k}=(b-a)^{k} /\left[2^{k}(k+1)\right]$ for $k$ even.

Solution 2.35 By definition, $\mu_{k}=\frac{1}{b-a} \int_{a}^{b}\left(x-\frac{a+b}{2}\right)^{k} d x$. Make the change of variable $y=$ $x-(a+b) / 2$. For $x=a$, we get $y=a-(a+b) / 2=(a-b) / 2=-(b-a) / 2$. Similarly for $x=b$, we get $y=b-(a+b) / 2=(b-a) / 2$. As the Jacobian is $\partial y / \partial x=1$, the integral becomes $\mu_{k}=\frac{1}{b-a} \int_{-(b-a) / 2}^{(b-a) / 2} y^{k} d y$. When $k$ is odd, this is an integral of an odd function in symmetric range, which is identically zero. For $k$ even, we have $\mu_{k}=\frac{2}{b-a} \int_{0}^{(b-a) / 2} y^{k} d y$ $=\left.\frac{2}{b-a}\left[y^{k+1} /(k+1)\right]\right|_{0} ^{(b-a) / 2}=(b-a)^{k} /\left[2^{k}(k+1)\right]$, as the constant $2 /(b-a)$ cancels out.
Example 2.36 Ratio of independent uniform distributions If $X$ and $Y$ are IID CUNI$(0, b)$ variates, find the distribution of $U=X / Y$.

Solution 2.37 Let $U=X / Y, V=Y$ so that the inverse mapping is $Y=V, X=U V$. The Jacobian is $|J|=v$. The joint PDF is $f(x, y)=1 / b^{2}$. Hence, $f(u, v)=v / b^{2}$. The PDF of $u$ is obtained by integrating out $v$. The region of interest is a rectangle of sides $1 \times b$ at the left,and a curve $u v=b$ to its right. Integrating out $v$, we obtain $f(u)=\int_{0}^{b} \frac{v}{b^{2}} d v$ for $0<u \leq 1$, and $f(u)=\int_{0}^{b / u} v / b^{2} d v=1 /\left(2 u^{2}\right)$ for $1<u<\infty .$
$$f(u)=\left{\begin{array}{rll} 1 / 2 & \text { for } \quad 0<u<1 \ 1 /\left(2 u^{2}\right) & \text { for } \quad 1<u<\infty \end{array}\right.$$
which is independent of the parameter $b$.

## 统计代写|工程统计代写engineering statistics代考|APPLICATIONS

This distribution finds applications in many fields. For instance, heat conductivity, thermal diffusion, and voltage fluctuations in a short time period (temporal dimension) are assumed to be uniform distributed. Some of these properties can also be extended to spatial dimensions (areas that are unit distance away from the source). It is used in nonparametric tests like KolmogorovSmirnov test. The rounding errors resulting from grouping data into classes uses a $U(0,1)$ to obtain a correction factor known as Sheppard’s correction. Quantization errors in audio-coding (compression) use this distribution. It is also used in stratified sampling, non-random clustering, etc. Random numbers from other distributions are easy to generate using $\mathrm{U}[0,1]$. These are discussed in subsequent chapters.

Example 2.38 Estimating proportions A jar contains a mixture of two liquids, $\mathrm{L}{1}$ and $\mathrm{L}{2}$, that mixes well in each other (as water and wine, or acid and water). All that is known is that “there is at most three times as much of one as the other.” Find the probability that (i) $\mathrm{L}{1} / \mathrm{L}{2} \leq 2$ and (ii) $\mathrm{L}{1} / \mathrm{L}{2} \geq 1$.

Solution $2.39$ The given condition is $\frac{1}{3} \leq \mathrm{L}{1} / \mathrm{L}{2} \leq 3$. Let $\mathrm{U}=\mathrm{L}{1} / \mathrm{L}{2}$. Assume that $U$ is uniformly distributed in $[1 / 3,3]$. As $3-1 / 3=8 / 3$, we take the density function as $\mathrm{f}(\mathrm{x})=3 / 8$, $\frac{1}{3} \leq x \leq 3$. The required answer for (i) is $\mathrm{P}[\mathrm{U} \leq 2]=\int_{1 / 3}^{2} f(x) d x=\left.(3 / 8) * x\right|{1 / 3} ^{2}=(3 / 8) *$ $(2-1 / 3)=5 / 8$; and (ii) $\mathrm{L}{1} / \mathrm{L}{2} \geq 1=\int{1}^{3} \mathrm{f}(\mathrm{x}) \mathrm{dx}=\left.(3 / 8) * x\right|_{1} ^{3}=6 / 8=0.75$

## 统计代写|工程统计代写engineering statistics代考|MOMENTS AND GENERATING FUNCTIONS

MGF 是

φX(吨)=(经验⁡(一世b吨)−经验⁡(一世一个吨))/[(b−一个)一世吨] 为了 吨≠0.

(和b吨−和一个吨)/[(b−一个)吨]=1+1b−一个(∑ķ=2∞[(bķ−一个ķ)/(b−一个)]吨ķ−1/ķ!).

μķ−1′=(∂ķ−1/∂吨ķ−1)米X(吨)|吨=0=(bķ−一个ķ)/[(b−一个)ķ]

## 统计代写|工程统计代写engineering statistics代考|TRUNCATED UNIFORM DISTRIBUTIONS

G(X;一个,b,C)=F(X;一个,b)/(1−F(C))=(1/(b−一个))[1/(1−(C−一个)/(b−一个))]=1/(b−C).

G(X;一个,b,C,d)=F(X;一个,b)/(F(d)−F(C))=1/(d−C)

$$f(u)=\左{ 1/2 为了 0<在<1 1/(2在2) 为了 1<在<∞\正确的。$$

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## MATLAB代写

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