## 统计代写|高等概率论作业代写Advanced Probability Theory代考| Levy formula

• 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

1. $L(x)$ is defined on $\mathcal{R}-{0}$. It is easy to see that
$$\begin{array}{cl} L(x)=C_{1}+\int_{-\infty}^{x} \frac{1+y^{2}}{y^{2}} d G(y), & \text { if } x<0, \ C_{2}-\int_{x}^{\infty} \frac{1+y^{2}}{y^{2}} d G(y), & \text { if } x>0, \end{array}$$
for any constants $C_{1}$ and $C_{2}$. (Verify that the integrals are well-defined!)
Furthermore, it is easy to see that $L(x)$ is non-decreasing on $(-\infty, 0)$ and $(0, \infty)$, respectively, and satisfies
$$\lim {x \rightarrow-\infty} L(x)=C{1}, \quad \lim {x \rightarrow \infty} L(x)=C{2} .$$
2. Note that, for every finite $\delta>0$, we have
$$\int_{0<|x|<\delta} x^{2} d L(x)=\int_{0<|x|<\delta}\left(1+x^{2}\right) d G(x) \leq\left(1+\delta^{2}\right) \int_{0<|x|<\delta} d G(x)<\infty$$
3. In view of (5.4), the following are equivalent:
$$\int_{0<|x|<\delta} x^{2} d L(x)<\infty, \quad \Longleftrightarrow \int_{|x|>0}\left(x^{2} \wedge 1\right) d L(x)<\infty, \quad \Longleftrightarrow \quad \int_{|x|>0} \frac{x^{2}}{1+x^{2}} d L(x)<\infty$$
4. $L(x)$ is finite for $x \neq 0$. But at $x=0$, they might not be well-defined. Namely, as $x>0$ or $x \searrow 0$, we might have $|L(x)| \rightarrow \infty$ and/or $\left|L^{\prime}(x)\right|=\infty$ (if $L^{\prime}$ exists).
On the other hand, it is easy to see that, for every finite $\epsilon>0$, we have
$$L\left((-\epsilon, \epsilon)^{\kappa}\right)-\int_{|x|>\epsilon} d L(x)-\int_{|x|>c} \frac{1+x^{2}}{x^{2}} d G(x) \leq\left(1+\epsilon^{-2}\right) \int_{|x|>c} d G(x)<\infty$$
5. $L(x)$ is often called “Levy measure”, a very important concept in the studies of Levy processes.
We summarize everything in the next theorem.
THEOREM 12.5.1 (Levy formula) A function $\psi(t)$ is an i.d.c.f. if and only if it admits the following (unique) representation
$$\psi(t)=\exp \left{i t \gamma-\frac{1}{2} \sigma^{2} t^{2}+\int_{|x|>0}\left(e^{i t x}-1-\frac{i t x}{1+x^{2}}\right) d L(x)\right}$$
where $\gamma$ is a real constant, $\sigma^{2}$ is a non-negative constant, and the function $L$ is non-decreasing on the intervals $(-\infty, 0)$ and $(0, \infty)$, and satisfies
$$\int_{0<|x|<\delta} x^{2} d L(x)<\infty, \quad \text { for every finite } \delta>0 \text {. }$$
REMARK 12.5.1 An alternative Levy formula takes the following form:
\begin{aligned} \psi(t) &=\exp \left{i t \gamma-\frac{1}{2} \sigma^{2} t^{2}+\int_{|x|>0}\left(e^{i t x}-1-i t x I{|x|<1}\right) d L(x)\right} \ &=\exp \left{i t \gamma-\frac{1}{2} \sigma^{2} t^{2}+\int_{0<|x|<1}\left(e^{i t x}-1-i t x\right) d L(x)+\int_{|x| \geq 1}\left(e^{i t x}-1\right) d L(x)\right} . \end{aligned}
This corresponds to the decomposition of a Levy process, which can be written as the sum of a Brownian motion, small jump component (a martingale), and a compound Poisson process.

In the special case where the r.v. $X$ has finite second moment $E X^{2}$, we know that $\psi(t)$ is twice differentiable. In this case, we have the following simpler representation.

THEOREM 12.6.1 (Komogorov formula) A function $\psi(t)$ is an i.d.c.f. with a finite variance if and only if it admits the following (unique) representation
$$\psi(t)=\exp \left{i t \gamma+\int_{-\infty}^{\infty}\left(e^{i t x}-1-i t x\right) \frac{1}{x^{2}} d K(x)\right}$$
where $\gamma$ is a real constant, the function $K$ is a bounded non-decreasing function.

## 统计代写|高等概率论作业代写Advanced Probability Theory代考|Relationship between the sum of independent r.v. and i.d.

What do the limiting d.f. of sums of independent r.v.s look like? We start with some simple examples:

• If $E\left|X_{1}\right|<\infty$, then $\bar{X} \rightarrow d \mu$, whose c.f is $e^{i t \mu}$.
• If $E\left|X_{1}\right|<\infty$, then $\sqrt{n}(\bar{X}-\mu) / \sigma \rightarrow d N(0,1)$, whose è.f is $e^{-t^{2} / 2}$
• If $X_{n k}$ are i.i.d. $\operatorname{Bin}(n, p=\lambda / n)$, then $\sum X_{n k} \rightarrow_{d} \operatorname{Poisson}(\lambda)$, whose c.f. is $e^{\lambda\left(e^{i t}-1\right)}$.
• If $X_{k}$ are i.i.d. Cauchy $(0,1)$, then $\bar{X} \rightarrow_{d} X_{1}$, whose c.f. $e^{-|t|}$.
Note that the c.f.s of the limiting distributions are all of the form $e^{\eta(t)}$. In fact, all the above limiting distributions are i.d., this is no accident. See the next theorem.
THEOREM 12.7.1 Let $\sum X_{n k}$ be independent r.v.s satisfying the following infinitesimal condition:
$$\max {1 \leq k \leq n} P\left(\left|X{n k}\right|>\epsilon\right) \rightarrow 0, \quad \text { as } n \rightarrow \infty$$
Then,
$$\left{\text { all limiting d.f.s of } \sum X_{n k}\right}={\text { all i.d. d.f.s }}$$

## 高等概率论代写

1. 一世(X)定义在R−0. 很容易看出
一世(X)=C1+∫−∞X1+和2和2dG(和), 如果 X<0, C2−∫X∞1+和2和2dG(和), 如果 X>0,
对于任何常数C1和C2. （验证积分是明确定义的！）
此外，很容易看出一世(X)不减少(−∞,0)和(0,∞), 分别满足
$$\lim {x \rightarrow-\infty} L(x)=C {1}, \quad \lim {x \rightarrow \infty} L(x)=C {2} 。$$
2. 注意，对于每一个有限d>0， 我们有
∫0<|X|<dX2d一世(X)=∫0<|X|<d(1+X2)dG(X)≤(1+d2)∫0<|X|<ddG(X)<∞
3. 鉴于（5.4），以下是等价的：
∫0<|X|<dX2d一世(X)<∞,⟺∫|X|>0(X2∧1)d一世(X)<∞,⟺∫|X|>0X21+X2d一世(X)<∞
4. 一世(X)是有限的X≠0. 但在X=0，它们可能没有明确定义。即，作为X>0要么X0，我们可能有|一世(X)|→∞和/或|一世′(X)|=∞（如果一世′存在）。
另一方面，很容易看出，对于每一个有限ε>0， 我们有
一世((−ε,ε)ķ)−∫|X|>εd一世(X)−∫|X|>C1+X2X2dG(X)≤(1+ε−2)∫|X|>CdG(X)<∞
5. 一世(X)常被称为“Levy 度量”，是 Levy 过程研究中的一个非常重要的概念。
我们在下一个定理中总结了所有内容。
定理 12.5.1（Levy 公式）函数ψ(吨)是一个 idcf 当且仅当它承认以下（唯一）表示
\psi(t)=\exp \left{it \gamma-\frac{1}{2} \sigma^{2} t^{2}+\int_{|x|>0}\left(e^{ itx}-1-\frac{itx}{1+x^{2}}\right) d L(x)\right}\psi(t)=\exp \left{it \gamma-\frac{1}{2} \sigma^{2} t^{2}+\int_{|x|>0}\left(e^{ itx}-1-\frac{itx}{1+x^{2}}\right) d L(x)\right}
在哪里C是一个实常数，σ2是一个非负常数，函数一世在区间上不减少(−∞,0)和(0,∞)，并且满足
∫0<|X|<dX2d一世(X)<∞, 对于每一个有限 d>0.
备注 12.5.1 另一种征费公式采用以下形式：
\begin{对齐} \psi(t) &=\exp \left{it \gamma-\frac{1}{2} \sigma^{2} t^{2}+\int_{|x|>0} \left(e^{itx}-1-itx I{|x|<1}\right) d L(x)\right} \ &=\exp \left{it \gamma-\frac{1}{2 } \sigma^{2} t^{2}+\int_{0<|x|<1}\left(e^{itx}-1-itx\right) d L(x)+\int_{|x | \geq 1}\left(e^{i t x}-1\right) d L(x)\right} 。\end{对齐}\begin{对齐} \psi(t) &=\exp \left{it \gamma-\frac{1}{2} \sigma^{2} t^{2}+\int_{|x|>0} \left(e^{itx}-1-itx I{|x|<1}\right) d L(x)\right} \ &=\exp \left{it \gamma-\frac{1}{2 } \sigma^{2} t^{2}+\int_{0<|x|<1}\left(e^{itx}-1-itx\right) d L(x)+\int_{|x | \geq 1}\left(e^{i t x}-1\right) d L(x)\right} 。\end{对齐}
这对应于 Levy 过程的分解，可以写成布朗运动、小跳跃分量（鞅）和复合泊松过程的总和。

\psi(t)=\exp \left{it \gamma+\int_{-\infty}^{\infty}\left(e^{itx}-1-itx\right) \frac{1}{x^{ 2}} d K(x)\right}\psi(t)=\exp \left{it \gamma+\int_{-\infty}^{\infty}\left(e^{itx}-1-itx\right) \frac{1}{x^{ 2}} d K(x)\right}

## 统计代写|高等概率论作业代写Advanced Probability Theory代考|Relationship between the sum of independent r.v. and i.d.

• 如果和|X1|<∞， 然后X¯→dμ, 其 cf 是和一世吨μ.
• 如果和|X1|<∞， 然后n(X¯−μ)/σ→dñ(0,1)，其 è.f 是和−吨2/2
• 如果Xn到是独立同居是⁡(n,p=λ/n)， 然后∑Xn到→d鱼⁡(λ), 其 cf 是和λ(和一世吨−1).
• 如果X到是 iid 柯西吗(0,1)， 然后X¯→dX1，其cf和−|吨|.
请注意，限制分布的 cfs 都是形式和这(吨). 其实上面所有的限制分布都是id，这绝非偶然。参见下一个定理。
定理 12.7.1 让∑Xn到是满足以下无穷小条件的独立 rvs：
最大限度1≤到≤n磷(|Xn到|>ε)→0, 作为 n→∞
然后，
\left{\text { 所有限制 dfs } \sum X_{n k}\right}={\text { all id dfs }}

## 广义线性模型代考

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## MATLAB代写

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## 统计代写|高等概率论作业代写Advanced Probability Theory代考| Proof of necessity

• 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|高等概率论作业代写Advanced Probability Theory代考|Proof of necessity

THEOREM 12.4.2 If $\psi(t)$ is i.d.c.f., then $\psi(t)$ has the representation (t.1).
Proof. We start with $\psi(t)=\left(\psi_{n}(t)\right)^{n}$. Let $F_{n}$ denote the d.f. corresponding to the c.f. $\psi_{n}$. Now as $0<|\psi(t)| \leq 1, \ln \psi(t)$ exists and is finite. Furthermore,
$$\psi^{1 / n}(t)=\exp \left(\frac{1}{n} \ln \psi(t)\right)=1+\frac{1}{n} \ln \psi(t)+O\left(\frac{1}{n^{2}}\right)=: 1+\frac{1}{n} \eta(t)+O\left(\frac{1}{n^{2}}\right),$$
from which wẻ gèt
\begin{aligned} \eta(t) &=\lim {n}\left(\psi^{1 / n}(t)-1-O\left(\frac{1}{n^{2}}\right)\right) \ &=\lim {n \rightarrow \infty} n\left(\psi^{1 / n}(t)-1\right) \ &=\lim {n \rightarrow \infty} n\left(\psi{n}(t)-1\right)=\lim {n \rightarrow \infty} \int{-\infty}^{\infty} n\left(e^{i t x}-1\right) d F_{n}(x) \ &=\lim {n \rightarrow \infty} \int{-\infty}^{\infty} n\left(e^{i t x}-1-\frac{i t x}{1+x^{2}}+\frac{i t x}{1+x^{2}}\right) d F_{n}(x) \ &=\lim {n \rightarrow \infty}\left{i t \int{-\infty}^{\infty} \frac{n x}{1+x^{2}} d F_{n}(x)+\int_{-\infty}^{\infty} n\left(e^{i t x}-1-\frac{i t x}{1+x^{2}}\right) d F_{n}(x)\right} \ &=\lim {n \rightarrow \infty}\left{i t \int{-\infty}^{\infty} \frac{n x}{1+x^{2}} d F_{n}(x)+\int_{-\infty}^{\infty} n\left(e^{i t x}-1-\frac{i t x}{1+x^{2}}\right) \frac{1+x^{2}}{x^{2}} \frac{x^{2}}{1+x^{2}} d F_{n}(x)\right} \ &=\lim {n \rightarrow \infty}\left{i t \int{-\infty}^{\infty} \frac{n x}{1+x^{2}} d F_{n}(x)+\int_{-\infty}^{\infty}\left(e^{i t x}-1-\frac{i t x}{1+x^{2}}\right) \frac{1+x^{2}}{x^{2}} d \int_{-\infty}^{x} \frac{n y^{2}}{1+y^{2}} d F_{n}(y)\right} \ &=\lim {n \rightarrow \infty}\left{i t \gamma{n}+\int_{-\infty}^{\infty}\left(e^{i t x}-1-\frac{i t x}{1+x^{2}}\right) \frac{1+x^{2}}{x^{2}} d G_{n}(x)\right} \ &=\lim {n \rightarrow \infty} \eta{n}(t), \end{aligned}
where
$$\eta_{n}(t)=: i t \gamma_{n}+\int_{-\infty}^{\infty}\left(e^{i t x}-1-\frac{i t x}{1+x^{2}}\right) \frac{1+x^{2}}{x^{2}} d G_{n}(x)$$
and
$$\gamma_{n}=\int_{-\infty}^{\infty} \frac{n x}{1+x^{2}} d F_{n}(x), \quad G_{n}(x)=\int_{-\infty}^{x} \frac{n y^{2}}{1+y^{2}} d F_{n}(y)$$
We have shown that
$$\lim {n \rightarrow \infty} \eta{n}(t)=\eta(t),$$
where $\eta(t)$ is continuous at 0 . From Lemma 12.4.3 below, we have that $\gamma_{n} \rightarrow \gamma$ and $G_{n} \Rightarrow G$ for some constant $\gamma$ and some “nice” function $G$ as described in Theorem 12.4.1.

## 统计代写|高等概率论作业代写Advanced Probability Theory代考|Appendix: Several useful lemmas

Let
$$A(y)=:\left(1-\frac{\sin y}{y}\right) \frac{1+y^{2}}{y^{2}}$$
Note that for $y=o(1)$, we have
$$A(y)=\left(1-\frac{1}{y}\left(y-\frac{y^{3}}{3 !}+\frac{y^{5}}{5 !}-\ldots\right)\right) \frac{1+y^{2}}{y^{2}}=\left(\frac{1}{3 !}-\frac{y^{2}}{5 !}-\ldots\right)\left(1+y^{2}\right)$$
so if we define $A(y)$ to be $1 / 3 !$ at $y=0$, so that $A(y)$ is a nonnegative bounded continuous function. Furthermore, it can be shown easily that
$$0<c_{1} \leq A(y) \leq c s<\infty . \quad \text { for all } y$$
232
Now define
$$\Lambda(x)=\int_{-\infty}^{x} A(y) d G(y), \quad \text { and } \quad \Lambda_{n}(x)=\int_{-\infty}^{x} A(y) d G_{n}(y)$$
Therefore, $\Lambda(x)$ and $\Lambda_{n}(x)$ are bounded and non-decreasing, $\Lambda(-\infty)=0$ and $\Lambda(\infty)<\infty$ as $G(x)$ is bounded and non-decreasing. Furthermore, we can easily work out their Fourier transforms.
LEMMA 12.4.1 We have
\begin{aligned} &\int_{-\infty}^{\infty} e^{i t x} d \Lambda(x)=\eta(t)-\frac{1}{2} \int_{0}^{1}[\eta(t+h)+\eta(t-h)] d h=: \lambda(t) \ &\int_{-\infty}^{\infty} e^{i t x} d \Lambda_{n}(x)=\eta_{n}(t)-\frac{1}{2} \int_{0}^{1}\left[\eta_{n}(t+h)+\eta_{n}(t-h)\right] d h=: \lambda_{n}(t) . \end{aligned}
Proof.
\begin{aligned} \int_{-\infty}^{\infty} e^{i t x} d \Lambda(x) &=\int_{-\infty}^{\infty} e^{i t x}\left(1-\frac{\sin x}{x}\right) \frac{1+x^{2}}{x^{2}} d G(x) \ &=\int_{-\infty}^{\infty} \int_{0}^{1} e^{i t x}(1-\cos h x) \frac{1+x^{2}}{x^{2}} d h d G(x) \end{aligned}
(by Fubini’s theorem since the integrand is bounded
\begin{aligned} &\text { contintuous in }[0,1] \times(-\infty, \infty)) \ =& \int_{0}^{1} \int_{-\infty}^{\infty} e^{i t x}(1-\cos h x) \frac{1+x^{2}}{x^{2}} d G(x) d h \ =& \int_{0}^{1}\left(\eta(t)-\frac{1}{2}[\eta(t+h)+\eta(t-h)]\right) d h \ =& \eta(t)-\frac{1}{2} \int_{0}^{1}[\eta(t+h)+\eta(t-h)] d h \ =& \lambda(t) . \end{aligned}

## 统计代写|高等概率论作业代写Advanced Probability Theory代考|Proof of sufficiency

THEOREM 12.4.3 If $\psi(t)$ has the representation $(4.1)$, then $\psi(t)$ is i.d.c.f.
Proof. Denote the integral in $(4.1)$ by $I(t)$, which can be written as
\begin{aligned} I(t) &=\int_{-\infty}^{\infty} g(t, x) d G(x) \ &=\int_{{x>0}}+\int_{{x<0}}+\int_{{x=0}} g(t, x) d G(x) \ &=I_{+}(t)+I_{-}(t)+g(t, 0)[G(0+)-G(0-)] \ &=I_{+}(t)+I_{-}(t)-\frac{t^{2}}{2}[G(0+)-G(0-)] \end{aligned}
Then, we have
Therefore, it suffices to show that $\exp \left{I_{+}(t)\right}$ and $\exp \left{I_{-}(t)\right}$ are i.d.c.f. We will look at the first one nnly sinee the sernd ean he done similarly Note that
$$\exp \left{I_{+}(t)\right}=\lim {m} \exp \left{I{+}^{1 / m}(t)\right}$$
where
\begin{aligned} I_{+}^{\epsilon}(t)=& \int_{\epsilon}^{1 / e} g(t, x) d G(x) \ =& \lim {n} \sum{k=0}^{n-1}\left(e^{i t \xi_{k}}-1-\frac{i t \xi_{k}}{1+\xi_{k}^{2}}\right) \frac{1+\xi_{k}^{2}}{\xi_{k}^{2}}\left[G\left(x_{k+1}\right)-G\left(x_{k+1}\right)\right] \ =&\left(\epsilon=x_{0}<x_{1}<\ldots<x_{n}=1 / \epsilon_{0} \quad x_{k} \leq \xi_{k}<x_{k+1}\right) \ =& \lim {n} \sum{k=0}^{n-1}\left(e^{i+\xi_{0}}-1-\frac{i t \xi_{k}}{1+\xi_{k}^{2}}\right) \lambda_{n k} \end{aligned}
234
\begin{aligned} &=: \lim {n} \sum{k=0}^{n-1}\left(e^{i t \xi_{k}}-1\right) \lambda_{n k}-i t \frac{\xi_{k} \lambda_{n k}}{1+\xi_{k}^{2}} \ &=: \quad \lim {n} \sum{k=0}^{n-1}\left(i t a_{n k}+\left(e^{i t \xi_{k}}-1\right) \lambda_{n k}\right) \ &=: \quad \lim {n} \sum{k=0}^{n-1} T_{n k} . \end{aligned}

## 统计代写|高等概率论作业代写Advanced Probability Theory代考|Proof of necessity

ψ1/n(吨)=经验⁡(1nln⁡ψ(吨))=1+1nln⁡ψ(吨)+○(1n2)=:1+1n这(吨)+○(1n2),

\begin{对齐} \eta(t) &=\lim {n}\left(\psi^{1 / n}(t)-1-O\left(\frac{1}{n^{2}} \right)\right) \ &=\lim {n \rightarrow \infty} n\left(\psi^{1 / n}(t)-1\right) \ &=\lim {n \rightarrow \infty} n\left(\psi{n}(t)-1\right)=\lim {n \rightarrow \infty} \int{-\infty}^{\infty} n\left(e^{itx}-1 \right) d F_{n}(x) \ &=\lim {n \rightarrow \infty} \int{-\infty}^{\infty} n\left(e^{itx}-1-\frac{ itx}{1+x^{2}}+\frac{itx}{1+x^{2}}\right) d F_{n}(x) \ &=\lim {n \rightarrow \infty}\左{it \int{-\infty}^{\infty} \frac{nx}{1+x^{2}} d F_{n}(x)+\int_{-\infty}^{\infty} n\left(e^{itx}-1-\frac{itx}{1+x^{2}}\right) d F_{n}(x)\right} \ &=\lim {n \rightarrow \infty}\left{it \int{-\infty}^{\infty} \frac{nx}{1+x^{2}} d F_{n}(x)+\ int_{-\infty}^{\infty} n\left(e^{itx}-1-\frac{itx}{1+x^{2}}\right) \frac{1+x^{2} }{x^{2}} \frac{x^{2}}{1+x^{2}} d F_{n}(x)\right} \ &=\lim {n \rightarrow \infty}\左{it \int{-\infty}^{\infty} \frac{nx}{1+x^{2}} d F_{n}(x)+\int_{-\infty}^{\infty} \left(e^{itx}-1-\frac{itx}{1+x^{2}}\right) \frac{1+x^{2}}{x^{2}} d \int_{ -\infty}^{x} \frac{ny^{2}}{1+y^{2}} d F_{n}(y)\right} \ &=\lim {n \rightarrow \infty}\左{it \gamma{n}+\int_{-\infty}^{\infty}\left(e^{itx}-1-\frac{itx}{1+x^{2}}\right) \ frac{1+x^{2}}{x^{2}} d G_{n}(x)\right} \ &=\lim {n \rightarrow \infty} \eta{n}(t), \结束{对齐}=\lim {n \rightarrow \infty}\left{it \int{-\infty}^{\infty} \frac{nx}{1+x^{2}} d F_{n}(x)+\ int_{-\infty}^{\infty}\left(e^{itx}-1-\frac{itx}{1+x^{2}}\right) \frac{1+x^{2}} {x^{2}} d \int_{-\infty}^{x} \frac{ny^{2}}{1+y^{2}} d F_{n}(y)\right} \ & =\lim {n \rightarrow \infty}\left{it \gamma{n}+\int_{-\infty}^{\infty}\left(e^{itx}-1-\frac{itx}{1 +x^{2}}\right) \frac{1+x^{2}}{x^{2}} d G_{n}(x)\right} \ &=\lim {n \rightarrow \infty } \eta{n}(t), \end{对齐}=\lim {n \rightarrow \infty}\left{it \int{-\infty}^{\infty} \frac{nx}{1+x^{2}} d F_{n}(x)+\ int_{-\infty}^{\infty}\left(e^{itx}-1-\frac{itx}{1+x^{2}}\right) \frac{1+x^{2}} {x^{2}} d \int_{-\infty}^{x} \frac{ny^{2}}{1+y^{2}} d F_{n}(y)\right} \ & =\lim {n \rightarrow \infty}\left{it \gamma{n}+\int_{-\infty}^{\infty}\left(e^{itx}-1-\frac{itx}{1 +x^{2}}\right) \frac{1+x^{2}}{x^{2}} d G_{n}(x)\right} \ &=\lim {n \rightarrow \infty } \eta{n}(t), \end{对齐}\begin{aligned} \eta(t) &=\lim {n}\left(\psi^{1 / n}(t)-1-O\left(\frac{1}{n^{2}}\right)\right) \ &=\lim {n \rightarrow \infty} n\left(\psi^{1 / n}(t)-1\right) \ &=\lim {n \rightarrow \infty} n\left(\psi{n}(t)-1\right)=\lim {n \rightarrow \infty} \int{-\infty}^{\infty} n\left(e^{i t x}-1\right) d F_{n}(x) \ &=\lim {n \rightarrow \infty} \int{-\infty}^{\infty} n\left(e^{i t x}-1-\frac{i t x}{1+x^{2}}+\frac{i t x}{1+x^{2}}\right) d F_{n}(x) \ &=\lim {n \rightarrow \infty}\left{i t \int{-\infty}^{\infty} \frac{n x}{1+x^{2}} d F_{n}(x)+\int_{-\infty}^{\infty} n\left(e^{i t x}-1-\frac{i t x}{1+x^{2}}\right) d F_{n}(x)\right} \ &=\lim {n \rightarrow \infty}\left{i t \int{-\infty}^{\infty} \frac{n x}{1+x^{2}} d F_{n}(x)+\int_{-\infty}^{\infty} n\left(e^{i t x}-1-\frac{i t x}{1+x^{2}}\right) \frac{1+x^{2}}{x^{2}} \frac{x^{2}}{1+x^{2}} d F_{n}(x)\right} \ &=\lim {n \rightarrow \infty}\left{i t \int{-\infty}^{\infty} \frac{n x}{1+x^{2}} d F_{n}(x)+\int_{-\infty}^{\infty}\left(e^{i t x}-1-\frac{i t x}{1+x^{2}}\right) \frac{1+x^{2}}{x^{2}} d \int_{-\infty}^{x} \frac{n y^{2}}{1+y^{2}} d F_{n}(y)\right} \ &=\lim {n \rightarrow \infty}\left{i t \gamma{n}+\int_{-\infty}^{\infty}\left(e^{i t x}-1-\frac{i t x}{1+x^{2}}\right) \frac{1+x^{2}}{x^{2}} d G_{n}(x)\right} \ &=\lim {n \rightarrow \infty} \eta{n}(t), \end{aligned}

Cn=∫−∞∞nX1+X2dFn(X),Gn(X)=∫−∞Xn和21+和2dFn(和)

## 统计代写|高等概率论作业代写Advanced Probability Theory代考|Appendix: Several useful lemmas

0<C1≤一种(和)≤Cs<∞. 对所有人 和
232

Λ(X)=∫−∞X一种(和)dG(和), 和 Λn(X)=∫−∞X一种(和)dGn(和)

∫−∞∞和一世吨XdΛ(X)=这(吨)−12∫01[这(吨+H)+这(吨−H)]dH=:λ(吨) ∫−∞∞和一世吨XdΛn(X)=这n(吨)−12∫01[这n(吨+H)+这n(吨−H)]dH=:λn(吨).

∫−∞∞和一世吨XdΛ(X)=∫−∞∞和一世吨X(1−没有⁡XX)1+X2X2dG(X) =∫−∞∞∫01和一世吨X(1−某物⁡HX)1+X2X2dHdG(X)
（由 Fubini 定理，因为被积函数是有界的
连续在 [0,1]×(−∞,∞)) =∫01∫−∞∞和一世吨X(1−某物⁡HX)1+X2X2dG(X)dH =∫01(这(吨)−12[这(吨+H)+这(吨−H)])dH =这(吨)−12∫01[这(吨+H)+这(吨−H)]dH =λ(吨).

## 统计代写|高等概率论作业代写Advanced Probability Theory代考|Proof of sufficiency

$$\exp \left{I_{+}(t)\right}=\lim {m} \exp \left{I {+ }^{1 / m}(t)\right} 在H和r和 一世+ε(吨)=∫ε1/和G(吨,X)dG(X) =林n∑到=0n−1(和一世吨X到−1−一世吨X到1+X到2)1+X到2X到2[G(X到+1)−G(X到+1)] =(ε=X0<X1<…<Xn=1/ε0X到≤X到<X到+1) =林n∑到=0n−1(和一世+X0−1−一世吨X到1+X到2)λn到 234 =:林n∑到=0n−1(和一世吨X到−1)λn到−一世吨X到λn到1+X到2 =:林n∑到=0n−1(一世吨一种n到+(和一世吨X到−1)λn到) =:林n∑到=0n−1吨n到.$$

## 广义线性模型代考

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## MATLAB代写

MATLAB 是一种用于技术计算的高性能语言。它将计算、可视化和编程集成在一个易于使用的环境中，其中问题和解决方案以熟悉的数学符号表示。典型用途包括：数学和计算算法开发建模、仿真和原型制作数据分析、探索和可视化科学和工程图形应用程序开发，包括图形用户界面构建MATLAB 是一个交互式系统，其基本数据元素是一个不需要维度的数组。这使您可以解决许多技术计算问题，尤其是那些具有矩阵和向量公式的问题，而只需用 C 或 Fortran 等标量非交互式语言编写程序所需的时间的一小部分。MATLAB 名称代表矩阵实验室。MATLAB 最初的编写目的是提供对由 LINPACK 和 EISPACK 项目开发的矩阵软件的轻松访问，这两个项目共同代表了矩阵计算软件的最新技术。MATLAB 经过多年的发展，得到了许多用户的投入。在大学环境中，它是数学、工程和科学入门和高级课程的标准教学工具。在工业领域，MATLAB 是高效研究、开发和分析的首选工具。MATLAB 具有一系列称为工具箱的特定于应用程序的解决方案。对于大多数 MATLAB 用户来说非常重要，工具箱允许您学习应用专业技术。工具箱是 MATLAB 函数（M 文件）的综合集合，可扩展 MATLAB 环境以解决特定类别的问题。可用工具箱的领域包括信号处理、控制系统、神经网络、模糊逻辑、小波、仿真等。

## 统计代写|高等概率论作业代写Advanced Probability Theory代考| Infinitely divisible distributions

• 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

EXAMPLE 12.2.1
1 Degenemte df (ie $P(X=C)=1)$
$$\psi(t)=e^{i+C}=\left(e^{i t(C / n)}\right)^{n}=\left(\psi_{n}(t)\right)^{n}$$

1. Normal d.f.
$$\psi(t)=\exp \left{i \mu t-\sigma^{2} t^{2} / 2\right}=\left(\exp \left{i(\mu / n) t-\left(\sigma^{2} / n\right) t^{2} / 2\right}\right)^{n}=\left(\psi_{n}(t)\right)^{n}$$
2. Poisson d.f.
$$\psi(t)=\exp \left{\lambda\left(\varepsilon^{\text {it }}-1\right)\right}=\left(\exp \left{(\lambda / n)\left(\varepsilon^{\text {it }}-1\right)\right}\right)^{n}=\left(\psi_{n}(t)\right)^{n} .$$
3. Compound Poisson d.f.: $S_{N}=X_{1}+\ldots+X_{N}$, where $X \sim F$ and $N \sim$ Poisson $(\lambda)$. So
$$\psi(t)=e^{\lambda\left(\varphi_{x}(t)-1\right)}=\left(e^{(\lambda / n)\left(\varphi_{x}(t)-1\right)}\right)^{n}=\left(\psi_{n}(t)\right)^{n} .$$
Remark: if $P(X=1)=1$, this reduces to Poisson d.f.
229
4. Cauchy d.f. with p.d.f. $f(x)=\frac{a}{\pi} \frac{1}{a^{2}+x^{2}}$ :
$$\dot{\psi}(t)=\exp {-a|t|}=(\exp {-(a / n)|t|})^{n}=\left(\psi_{n}(t)\right)^{n}$$
5. – stable d.f. (including Cauchy d.f.)
6. Gamma d.f. with p.d.f. $f(x)=\frac{\beta^{\alpha}}{\Gamma(\alpha)} x^{\alpha-1} e^{-\beta x}$ :
$$\psi(t)=\left(1-\frac{i t}{\beta}\right)^{-\alpha}=\left(\left(1-\frac{i t}{\beta}\right)^{-\alpha / n}\right)^{n}=\left(\psi_{n}(t)\right)^{n}$$
Two special cases:
(a) The $\chi^{2}$-distribution is $i d$. since it is a special case of Gamma d.f.
(b) The $t$-distribution is i.d. since it is a special case of the $\chi^{2}$-distribution (with degree of freedom 1).

1. For every c.f. $\psi(t)$, it is continuous and $\psi(0)=1$. Thus, given $\epsilon \in(0,1)$, we have $|\psi(t)|=$ $|\psi(0)-[\psi(0)-\psi(t)]| \geq|\psi(0)|-|\psi(0)-\psi(t)|>1-\epsilon$ for $|t| \leq \delta$. Since $\psi(t)=\left(\psi_{n}(t)\right)^{n}$ for every $n \geq 1$, we have
$$\left|\gamma_{n}(t)\right|=\left|q_{1}(t)\right|^{1 / n}>(1-\epsilon)^{1 / n} \rightarrow 1, \quad \text { as } n \rightarrow \infty \text {, }$$
which implies that, for $n \geq N_{0}$, we have $\left|\psi_{n}(t)\right|>1-\epsilon / 8>0$ for $|t| \leq \delta$ and $n \geq N_{0}$.
Next we will show that $\left|\psi_{n}(t)\right|>0$ for $|t|<2 \delta$ and $n>N_{0}$. To do that, we use Lemma ??????? to get
$$1-\left|\psi_{n}(2 t)\right| \leq 8\left(1-\left|\psi_{n}(t)\right|\right)<\epsilon .$$ so $\left|\psi_{n}(2 t)\right| \geq 1-\epsilon>0$ for $|t| \leq \delta$ and $n \geq N_{0}$. That is, $\left|\psi_{n}(t)\right|>0$ for $|t| \leq 2 \delta$ and $n \geq N_{0}$. Continuing like this, we get $\left|\psi_{n}(t)\right|>0$ for all $t \in R$ and $n \geq N_{0}$. Therefore,
$$|\psi(t)|=\left|\left(\psi_{N_{0}}(t)\right)^{N_{0}}\right|=\left|\psi_{N_{0}}(t)\right|^{N_{0}}>0 .$$
2. Take $m=2$ for instance. Since $\psi_{i}(t)=\left(\psi_{i n}(t)\right)^{n}, i=1,2$, we have $\psi_{1}(t) \psi_{2}(t)=\left(\psi_{1 n}(t) \psi_{2 n}(t)\right)^{n}$.
230
3. If $\psi(t)$ is i.d., $\psi(t)=\left(\psi_{2 n}(t)\right)^{2 n}$ for all $n \geq 1$. Hence,
$$|\psi(t)|^{2}=\left|\left(\psi_{2 n}(t)\right)^{2 n}\right|^{2}=\left(\left|\psi_{2 n}(t)\right|^{2}\right)^{2 n}$$
therefore,
$$|\psi(t)|=\left(\left|\psi_{2 n}(t)\right|^{2}\right)^{n}=:\left(\psi_{n}(t)\right)^{n}$$
where $\psi_{n}(t)=\left|\psi_{2 n}(t)\right|^{2}$ is a c.f.
4. We have $\psi^{(m)}(t)=\left(\psi_{n}^{(m)}(t)\right)^{n}$ for all $m$ and $n$. From the assumption, we have
$$\lim {m \rightarrow \infty} \psi^{(m)}(t)=\lim {m \rightarrow \infty}\left(\psi_{n}^{(m)}(t)\right)^{n}=\left(\lim {m \rightarrow \infty} \psi{n}^{(m)}(t)\right)^{n}=\psi(t)$$
for each fixed $n$, namely,
$$\lim {m \rightarrow \infty} \psi{n}^{(m)}(t)=(\psi(t))^{1 / n} .$$
Since $\left{\psi_{n}^{(m)}(t), m \geq 1\right}$ are c.f.s, and $(\psi(t))^{1 / n}$ is continuous at 0 , then by Levy continuity theorem. $(\psi(t))^{1 / n}$ is a c.f. as well. Therefore, $\psi(t)=\left(\psi^{1 / n}(t)\right)^{n}$ is i.d.
EXAMPLE 12.3.1 We can use Theorem 12.3.1 (part 1) to judge whether a c.f. is NOT i.d.
(a) Let $X \sim$ Uniform $[-1,1]$ with c.f. $f(t)=(\sin t) / t$. Since $f(k \pi)=0$, then, $f(t)$ is $N O T$ i. $d$.
(b) Let $P(X=\pm 1)$ with c.f. $f(t)=\cos t$. Similarly, it is $N O T$ i.d.

## 统计代写|高等概率论作业代写Advanced Probability Theory代考|Levy-Khintchine representation of infinitely divisible c.f.s

The key theorem in this section is the “Levy-Khintchine” representation.
THEOREM 12.4.1 (“Levy-Khintchine” representation) A function $\psi(t)$ is an i.d.c.f. if and only if it admits the representation
$$\psi(t)=\exp \left{i t \gamma+\int_{-\infty}^{\infty}\left(e^{i c x}-1-\frac{i t x}{1+x^{2}}\right) \frac{1+x^{2}}{x^{2}} d G(x)\right}=: e^{x(t)}$$
where $\gamma$ is a real constant, and $G$ is a bounded non-decreasing function.
(W.L.O.G., we assume that $G$ is left-contenuous and $G(-\infty)=0$; see the remark below.)
REMARK 12.4.1

1. Denote the function under the integral sign by
$$g(t, x)=\left(e^{i t x}-1-\frac{i t x}{1+x^{2}}\right) \frac{1+x^{2}}{x^{2}}$$
It is easy to see that $\lim _{x \rightarrow 0} g(t, x)=-\frac{t^{2}}{2}$, so we can define $g(t, 0)=-\frac{t^{2}}{2}$.
2. First, we note that the values of $G(x)$ at points of discontinuity do not influence the value of the integral on the RHS of (4.1) since $g(t, x)$ is continuous in $x$. Secondly, adding any constant, $C$, to $G(x)$ does not influence the value of the integral on the RHS of (4.1) either. For the purpose of definiteness, we may assume that it is left-continuous and $G(-\infty)=0$ from now on.

## 高等概率论代写

1 Degenemte df（即磷(X=C)=1)
ψ(吨)=和一世+C=(和一世吨(C/n))n=(ψn(吨))n

1. 正常 df
\psi(t)=\exp \left{i \mu t-\sigma^{2} t^{2} / 2\right}=\left(\exp \left{i(\mu / n) t- \left(\sigma^{2} / n\right) t^{2} / 2\right}\right)^{n}=\left(\psi_{n}(t)\right)^{n}\psi(t)=\exp \left{i \mu t-\sigma^{2} t^{2} / 2\right}=\left(\exp \left{i(\mu / n) t- \left(\sigma^{2} / n\right) t^{2} / 2\right}\right)^{n}=\left(\psi_{n}(t)\right)^{n}
2. df鱼
\psi(t)=\exp \left{\lambda\left(\varepsilon^{\text {it }}-1\right)\right}=\left(\exp \left{(\lambda / n)\左(\varepsilon^{\text {it }}-1\right)\right}\right)^{n}=\left(\psi_{n}(t)\right)^{n} 。\psi(t)=\exp \left{\lambda\left(\varepsilon^{\text {it }}-1\right)\right}=\left(\exp \left{(\lambda / n)\左(\varepsilon^{\text {it }}-1\right)\right}\right)^{n}=\left(\psi_{n}(t)\right)^{n} 。
3. 复合泊松df：小号ñ=X1+…+Xñ， 在哪里X∼F和ñ∼鱼(λ). 所以
ψ(吨)=和λ(披X(吨)−1)=(和(λ/n)(披X(吨)−1))n=(ψn(吨))n.
备注：如果磷(X=1)=1, 这减少到泊松 df
229
4. 带有 pdf 的柯西 dfF(X)=一种圆周率1一种2+X2:
ψ˙(吨)=经验⁡−一种|吨|=(经验⁡−(一种/n)|吨|)n=(ψn(吨))n
5. – 稳定 df（包括 Cauchy df）
6. 带有 pdf 的伽玛 dfF(X)=b一种Γ(一种)X一种−1和−bX:
ψ(吨)=(1−一世吨b)−一种=((1−一世吨b)−一种/n)n=(ψn(吨))n
两种特殊情况：
(a)χ2-分布是一世d.
因为它是 Gamma df (b) 的一个特例吨-distribution 是 id，因为它是χ2-分布（自由度为 1）。

1. 对于每一个cfψ(吨), 它是连续的并且ψ(0)=1. 因此，给定ε∈(0,1)， 我们有|ψ(吨)|= |ψ(0)−[ψ(0)−ψ(吨)]|≥|ψ(0)|−|ψ(0)−ψ(吨)|>1−ε为了|吨|≤d. 自从ψ(吨)=(ψn(吨))n对于每个n≥1， 我们有
|Cn(吨)|=|q1(吨)|1/n>(1−ε)1/n→1, 作为 n→∞,
这意味着，对于n≥ñ0， 我们有|ψn(吨)|>1−ε/8>0为了|吨|≤d和n≥ñ0.
接下来我们将展示|ψn(吨)|>0为了|吨|<2d和n>ñ0. 为此，我们使用引理 ??????? 要得到
1−|ψn(2吨)|≤8(1−|ψn(吨)|)<ε.所以|ψn(2吨)|≥1−ε>0为了|吨|≤d和n≥ñ0. 那是，|ψn(吨)|>0为了|吨|≤2d和n≥ñ0. 像这样继续下去，我们得到|ψn(吨)|>0对所有人吨∈R和n≥ñ0. 所以，
|ψ(吨)|=|(ψñ0(吨))ñ0|=|ψñ0(吨)|ñ0>0.
2. 拿米=2例如。自从ψ一世(吨)=(ψ一世n(吨))n,一世=1,2， 我们有ψ1(吨)ψ2(吨)=(ψ1n(吨)ψ2n(吨))n.
230
3. 如果ψ(吨)是身份证，ψ(吨)=(ψ2n(吨))2n对所有人n≥1. 因此，
|ψ(吨)|2=|(ψ2n(吨))2n|2=(|ψ2n(吨)|2)2n
所以，
|ψ(吨)|=(|ψ2n(吨)|2)n=:(ψn(吨))n
在哪里ψn(吨)=|ψ2n(吨)|2是一个cf
4. 我们有ψ(米)(吨)=(ψn(米)(吨))n对所有人米和n. 根据假设，我们有
林米→∞ψ(米)(吨)=林米→∞(ψn(米)(吨))n=(林米→∞ψn(米)(吨))n=ψ(吨)
对于每个固定n，即，
林米→∞ψn(米)(吨)=(ψ(吨))1/n.
自从\left{\psi_{n}^{(m)}(t), m \geq 1\right}\left{\psi_{n}^{(m)}(t), m \geq 1\right}是cfs，并且(ψ(吨))1/n在 0 处是连续的，则由 Levy 连续性定理。(ψ(吨))1/n也是一个cf。所以，ψ(吨)=(ψ1/n(吨))nis id
示例 12.3.1 我们可以使用定理 12.3.1（第 1 部分）来判断一个 cf 是否不是 id
(a) 让X∼制服[−1,1]与cfF(吨)=(没有⁡吨)/吨. 自从F(到圆周率)=0， 然后，F(吨)是ñ○吨一世。d.
(b) 让磷(X=±1)与cfF(吨)=某物⁡吨. 同样，它是ñ○吨ID

## 统计代写|高等概率论作业代写Advanced Probability Theory代考|Levy-Khintchine representation of infinitely divisible c.f.s

\psi(t)=\exp \left{it \gamma+\int_{-\infty}^{\infty}\left(e^{icx}-1-\frac{itx}{1+x^{2} }\right) \frac{1+x^{2}}{x^{2}} d G(x)\right}=: e^{x(t)}\psi(t)=\exp \left{it \gamma+\int_{-\infty}^{\infty}\left(e^{icx}-1-\frac{itx}{1+x^{2} }\right) \frac{1+x^{2}}{x^{2}} d G(x)\right}=: e^{x(t)}

（WLOG，我们假设G是左连续的并且G(−∞)=0; 见下面的备注。）

1. 将积分符号下的函数表示为
G(吨,X)=(和一世吨X−1−一世吨X1+X2)1+X2X2
很容易看出林X→0G(吨,X)=−吨22，所以我们可以定义G(吨,0)=−吨22.
2. 首先，我们注意到G(X)在不连续点处不影响 (4.1) 的 RHS 上的积分值，因为G(吨,X)是连续的X. 其次，添加任何常数，C， 到G(X)也不影响 (4.1) 的 RHS 上的积分值。为了明确起见，我们可以假设它是左连续的G(−∞)=0从现在开始。

## 广义线性模型代考

statistics-lab作为专业的留学生服务机构，多年来已为美国、英国、加拿大、澳洲等留学热门地的学生提供专业的学术服务，包括但不限于Essay代写，Assignment代写，Dissertation代写，Report代写，小组作业代写，Proposal代写，Paper代写，Presentation代写，计算机作业代写，论文修改和润色，网课代做，exam代考等等。写作范围涵盖高中，本科，研究生等海外留学全阶段，辐射金融，经济学，会计学，审计学，管理学等全球99%专业科目。写作团队既有专业英语母语作者，也有海外名校硕博留学生，每位写作老师都拥有过硬的语言能力，专业的学科背景和学术写作经验。我们承诺100%原创，100%专业，100%准时，100%满意。

## MATLAB代写

MATLAB 是一种用于技术计算的高性能语言。它将计算、可视化和编程集成在一个易于使用的环境中，其中问题和解决方案以熟悉的数学符号表示。典型用途包括：数学和计算算法开发建模、仿真和原型制作数据分析、探索和可视化科学和工程图形应用程序开发，包括图形用户界面构建MATLAB 是一个交互式系统，其基本数据元素是一个不需要维度的数组。这使您可以解决许多技术计算问题，尤其是那些具有矩阵和向量公式的问题，而只需用 C 或 Fortran 等标量非交互式语言编写程序所需的时间的一小部分。MATLAB 名称代表矩阵实验室。MATLAB 最初的编写目的是提供对由 LINPACK 和 EISPACK 项目开发的矩阵软件的轻松访问，这两个项目共同代表了矩阵计算软件的最新技术。MATLAB 经过多年的发展，得到了许多用户的投入。在大学环境中，它是数学、工程和科学入门和高级课程的标准教学工具。在工业领域，MATLAB 是高效研究、开发和分析的首选工具。MATLAB 具有一系列称为工具箱的特定于应用程序的解决方案。对于大多数 MATLAB 用户来说非常重要，工具箱允许您学习应用专业技术。工具箱是 MATLAB 函数（M 文件）的综合集合，可扩展 MATLAB 环境以解决特定类别的问题。可用工具箱的领域包括信号处理、控制系统、神经网络、模糊逻辑、小波、仿真等。

## 统计代写|高等概率论作业代写Advanced Probability Theory代考| Non-Uniform Berry-Esseen Bounds

• 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|高等概率论作业代写Advanced Probability Theory代考|A generalization of Berry-Esseen bounds

Here we shall give Berry-Esseen bounds assuming only second-order moments.
THEOREM 11.2.2 Let $X_{1}, \ldots, X_{n}$ be independent $r . v$.’s such that $E X_{j}=0$ and $E\left|X_{j}\right|^{2+\delta}<\infty(j=$ $1, \ldots, n)$. Put \begin{aligned} &\Lambda_{n}(\epsilon)=B_{n}^{-2} \sum_{j=1}^{n} E\left|X_{j}\right|^{2} I\left{\left|X_{j}\right|>\epsilon B_{n}\right} \ &\lambda_{n}(\epsilon)=B_{n}^{-3} \sum_{j=1}^{n} E\left|X_{j}\right|^{3} I\left{\left|X_{j}\right| \leq \epsilon B_{n}\right} \end{aligned}
Then for all $n$ and $\epsilon>0$,
$$\sup {x \in R}\left|F{n}(x)-\Phi(x)\right| \leq A\left(\Lambda_{n}(\epsilon)+\lambda_{n}(\epsilon)\right)$$
Proof. Omitted.
Remarks.

• From the definition, we have, for all $\epsilon>0$,
$$\lambda_{n}(\epsilon) \leq \epsilon B_{n}^{-2} \sum_{j=1}^{n} E\left|X_{j}\right|^{2} I\left{\left|X_{j}\right| \leq \epsilon B_{n}\right} \leq \epsilon$$
Therefore, we have the inequality,
$$\sup {x \in R}\left|F{n}(x)-\Phi(x)\right| \leq A\left(\Lambda_{n}(\epsilon)+\epsilon\right),$$
which, in turn, implies the Lindeberg CLT since $\Lambda_{n}(\epsilon) \rightarrow 0$ as $\epsilon \rightarrow 0$.
• On the other hand, we note that, for every $\delta \in(0,1]$.
$$\Lambda_{n}(\epsilon)+\lambda_{n}(\epsilon) \leq \frac{\sum_{j=1}^{n} E\left|X_{j}\right|^{2+\delta}}{B_{n}^{2+\delta}}$$
Then we derive the Berry-Esseen bounds
$$\sup {x \in R}\left|F{n}(x)-\Phi(x)\right| \leq \frac{A \sum_{j-1}^{n} E\left|X_{j}\right|^{2+\delta}}{B_{n}^{2+\delta}},$$
which, in turn, implies the Lynapnov CLT.

## 统计代写|高等概率论作业代写Advanced Probability Theory代考|Non-Uniform Berry-Esseen Bounds

So a more informative way to describe rates of convergence to normality is via the Non-uniform Berry-Esseen bounds, which involve both $n$ and $x$.
THEOREM 11.3.1 Let $X_{1}, \ldots, X_{n}$ be i.i.d. r.v.’s. Let
$$E X_{1}=0, \quad E X_{1}^{2}=\sigma^{2}>0, \quad E\left|X_{1}\right|^{3}<\infty, \quad \rho=E\left|X_{1}\right|^{3} / \sigma^{3} .$$
and
$$F_{n}(x)=P\left(\frac{1}{\sigma \sqrt{n}} \sum_{j=1}^{n} X_{i} \leq x\right)$$
Then for all $x$ and $n$,
$$\left|F_{n}(x)-\Phi(x)\right| \leq \frac{A \rho}{\sqrt{n}} \frac{1}{\left(1+|x|^{3}\right)}$$
For the independent case, we have
THEOREM 11.3.2 Let $X_{1}, \ldots, X_{n}$ be independent r.v.’s such that $E X_{j}=0$ and $E\left|X_{j}\right|^{3}<\infty(j=1, \ldots, n)$. Put
\begin{aligned} E X_{j}^{2}=\sigma_{j}^{2}, \quad B_{n}^{2} &=\sum_{j=1}^{n} \sigma_{j}^{2}, \quad L_{n}=B_{n}^{-3 / 2} \ F_{n}(x) &=P\left(B_{n}^{-1} \sum_{j=1}^{n} X_{j} \leq x\right) \end{aligned}
Then for all $n$ and $x$,
$$\left|F_{n}(x)-\Phi(x)\right| \leq \frac{A L_{n}}{1+|x|^{3}}$$
Tineonem 11.3.3 Let $X_{1}, \ldots, X_{n}$ be i.i.d. with $E\left|X_{1}\right|^{r}<\infty, r \geq 3$. Then for all $x$ and $n$,
$$\left|F_{n}(x)-\Phi(x)\right| \leq C_{r}\left(\frac{E\left|X_{1}\right|^{3} / \sigma^{3}}{\sqrt{n}}+\frac{E\left|X_{1}\right|^{r} / \sigma^{r}}{n^{(r-2) / 2}}\right) \frac{1}{(1+|x|)^{r}} .$$
THEOREM 11.3.4 Let $X_{1}, \ldots, X_{n}$ be i.i.d. with $E\left|X_{1}\right|^{2+\delta}<\infty, \delta \in(0,1]$. Then for all $x$ and $n$,
$$\left|F_{n}(x)-\Phi(x)\right| \leq \frac{C_{\delta} E\left|X_{1}\right|^{2+\delta}}{\sigma^{2+\delta} n^{\delta / 2}} \frac{1}{1+|x|^{2+\delta}}$$

## 统计代写|高等概率论作业代写Advanced Probability Theory代考|Heuristic argument for informal Edgeworth expansions

For simplicity, assume that $X, X_{1}, X_{2}, \ldots$ are i.i.d. r.v.s with $E X=0, E X^{2}=1$, and $\rho=E X^{3}$. Let
$$F_{n}(x)=P(\sqrt{n}(\bar{X}-0) / 1 \leq x)=P(\sqrt{n} \bar{X} \leq x) .$$
It is known that $F_{n}(x) \Longrightarrow \Phi(x)$. We will use the c.f. approach to derive a more accurate approximation. Note that $\psi_{X}(t)=1+i t E X+(i t)^{2} E X^{2}+\frac{1}{6}(i t)^{3} E X^{3}+o\left(t^{3}\right)=1-t^{2}+\frac{1}{6}(i t)^{3} \rho+o\left(t^{3}\right)$. Hence,
$$\psi_{\sqrt{n} \mathcal{X}}(t)=\psi_{X}^{n}(t / \sqrt{n})=\left(1-\frac{1}{2 n} t^{2}+\frac{1}{6 n^{3 / 2}}(i t)^{3} \rho+n^{-3 / 2} o\left(t^{3}\right)\right)^{n}$$
Hence, using $\ln (1+x)=x-\frac{x^{2}}{2}+\frac{x^{3}}{3}+\ldots$, we have
\begin{aligned} \ln \psi_{\sqrt{n} \mathcal{X}}(t) &=n \ln \left(1-\frac{1}{2 n} t^{2}+\frac{1}{6 n^{3 / 2}}(i t)^{3} \rho+n^{-3 / 2} o\left(t^{3}\right)\right) \ &=n\left(-\frac{1}{2 n} t^{2}+\frac{1}{6 n^{3 / 2}}(i t)^{3} \rho+n^{-3 / 2} o\left(t^{3}\right)+\ldots\right) \ &=-\frac{t^{2}}{2}+\frac{1}{6 n^{1 / 2}}(i t)^{3} \rho+n^{-1 / 2} o\left(t^{3}\right)+\ldots \end{aligned}
Thus,
\begin{aligned} \psi_{\sqrt{n} \mathcal{X}}(t)=\psi_{X}^{n}(t / \sqrt{n}) &=e^{-t^{2} / 2} \exp \left{\frac{1}{6 n^{1 / 2}}(i t)^{3} \rho+n^{-1 / 2} o(t\right.\ &=e^{-t^{2} / 2}\left(1+\frac{1}{6 n^{1 / 2}}(i t)^{3} \rho+\ldots\right) \ &=e^{-t^{2} / 2}+\frac{\rho}{6 n^{1 / 2}}(i t)^{3} e^{-t^{2} / 2}+\ldots \end{aligned}
Therefore, the “formal density” of $\sqrt{n} \bar{X}$ is
\begin{aligned} f_{\sqrt{n} \bar{X}}(x) &=\frac{1}{2 \pi} \int e^{-i t x} \psi_{\sqrt{n}}(t) d t \ &=\frac{1}{2 \pi} \int e^{-i t x} e^{-t^{2} / 2} d t+\frac{\rho}{6 n^{1 / 2}} \frac{1}{2 \pi} \int e^{-i t x}(i t)^{\frac{3}{3}} e^{-t^{2} / 2} d t+\ldots \ &=\phi(x)+\frac{\rho}{6 n^{1 / 2}} H_{3}(x) \phi(x)+\ldots \end{aligned}
Finally, we integrate the “density” to get the d.f.
\begin{aligned} P(\sqrt{n} \bar{X} \leq x) &=\int_{-\infty}^{x}\left(\phi(x)+\frac{\rho}{6 n^{1 / 2}} H_{3}(x) \phi(x)+\ldots\right) d x \ &=\Phi(x)-\frac{\rho}{6 n^{1 / 2}} H_{2}(x) \phi(x)+\ldots \end{aligned}
where the last line follows since
$$\begin{gathered} \frac{d}{d x} H_{2}(x) \phi(x)=H_{2}^{\prime}(x) \phi(x)+H_{2}(x)(-x) \phi(x)=2 x \phi(x)+\left(-x^{3}+x\right) \phi(x) \ =-\left(x^{3}-3 x\right) \phi(x)=-H_{3}(x) \phi(x) \end{gathered}$$

## 统计代写|高等概率论作业代写Advanced Probability Theory代考|A generalization of Berry-Esseen bounds

• 根据定义，我们有，对于所有人ε>0,
\lambda_{n}(\epsilon) \leq \epsilon B_{n}^{-2} \sum_{j=1}^{n} E\left|X_{j}\right|^{2} I\左{\left|X_{j}\right| \leq \epsilon B_{n}\right} \leq \epsilon\lambda_{n}(\epsilon) \leq \epsilon B_{n}^{-2} \sum_{j=1}^{n} E\left|X_{j}\right|^{2} I\左{\left|X_{j}\right| \leq \epsilon B_{n}\right} \leq \epsilon
因此，我们有不等式，
支持X∈R|Fn(X)−披(X)|≤一种(Λn(ε)+ε),
反过来，这意味着林德伯格 CLT，因为Λn(ε)→0作为ε→0.
• 另一方面，我们注意到，对于每个d∈(0,1].
Λn(ε)+λn(ε)≤∑j=1n和|Xj|2+d乙n2+d
然后我们推导出 Berry-Esseen 边界
支持X∈R|Fn(X)−披(X)|≤一种∑j−1n和|Xj|2+d乙n2+d,
反过来，这意味着 Lynapnov CLT。

## 统计代写|高等概率论作业代写Advanced Probability Theory代考|Non-Uniform Berry-Esseen Bounds

Fn(X)=磷(1σn∑j=1nX一世≤X)

|Fn(X)−披(X)|≤一种ρn1(1+|X|3)

|Fn(X)−披(X)|≤一种一世n1+|X|3
Tineon 11.3.3 让X1,…,Xn同住和|X1|r<∞,r≥3. 那么对于所有人X和n,
|Fn(X)−披(X)|≤Cr(和|X1|3/σ3n+和|X1|r/σrn(r−2)/2)1(1+|X|)r.

|Fn(X)−披(X)|≤Cd和|X1|2+dσ2+dnd/211+|X|2+d

## 统计代写|高等概率论作业代写Advanced Probability Theory代考|Heuristic argument for informal Edgeworth expansions

Fn(X)=磷(n(X¯−0)/1≤X)=磷(nX¯≤X).

ψnX(吨)=ψXn(吨/n)=(1−12n吨2+16n3/2(一世吨)3ρ+n−3/2○(吨3))n

ln⁡ψnX(吨)=nln⁡(1−12n吨2+16n3/2(一世吨)3ρ+n−3/2○(吨3)) =n(−12n吨2+16n3/2(一世吨)3ρ+n−3/2○(吨3)+…) =−吨22+16n1/2(一世吨)3ρ+n−1/2○(吨3)+…

\begin{aligned} \psi_{\sqrt{n} \mathcal{X}}(t)=\psi_{X}^{n}(t / \sqrt{n}) &=e^{ -t^{2} / 2} \exp \left{\frac{1}{6 n^{1 / 2}}(it)^{3} \rho+n^{-1 / 2} o(t \right.\ &=e^{-t^{2} / 2}\left(1+\frac{1}{6 n^{1 / 2}}(it)^{3} \rho+\ldots\对) \ &=e^{-t^{2} / 2}+\frac{\rho}{6 n^{1 / 2}}(it)^{3} e^{-t^{2} / 2}+\ldots \end{对齐} 吨H和r和F○r和,吨H和“F○r米一种一世d和ns一世吨和”○FnX¯一世s FnX¯(X)=12圆周率∫和−一世吨Xψn(吨)d吨 =12圆周率∫和−一世吨X和−吨2/2d吨+ρ6n1/212圆周率∫和−一世吨X(一世吨)33和−吨2/2d吨+… =φ(X)+ρ6n1/2H3(X)φ(X)+… F一世n一种一世一世和,在和一世n吨和Gr一种吨和吨H和“d和ns一世吨和”吨○G和吨吨H和d.F. 磷(nX¯≤X)=∫−∞X(φ(X)+ρ6n1/2H3(X)φ(X)+…)dX =披(X)−ρ6n1/2H2(X)φ(X)+… 在H和r和吨H和一世一种s吨一世一世n和F○一世一世○在ss一世nC和 ddXH2(X)φ(X)=H2′(X)φ(X)+H2(X)(−X)φ(X)=2Xφ(X)+(−X3+X)φ(X) =−(X3−3X)φ(X)=−H3(X)φ(X)

## 广义线性模型代考

statistics-lab作为专业的留学生服务机构，多年来已为美国、英国、加拿大、澳洲等留学热门地的学生提供专业的学术服务，包括但不限于Essay代写，Assignment代写，Dissertation代写，Report代写，小组作业代写，Proposal代写，Paper代写，Presentation代写，计算机作业代写，论文修改和润色，网课代做，exam代考等等。写作范围涵盖高中，本科，研究生等海外留学全阶段，辐射金融，经济学，会计学，审计学，管理学等全球99%专业科目。写作团队既有专业英语母语作者，也有海外名校硕博留学生，每位写作老师都拥有过硬的语言能力，专业的学科背景和学术写作经验。我们承诺100%原创，100%专业，100%准时，100%满意。

## MATLAB代写

MATLAB 是一种用于技术计算的高性能语言。它将计算、可视化和编程集成在一个易于使用的环境中，其中问题和解决方案以熟悉的数学符号表示。典型用途包括：数学和计算算法开发建模、仿真和原型制作数据分析、探索和可视化科学和工程图形应用程序开发，包括图形用户界面构建MATLAB 是一个交互式系统，其基本数据元素是一个不需要维度的数组。这使您可以解决许多技术计算问题，尤其是那些具有矩阵和向量公式的问题，而只需用 C 或 Fortran 等标量非交互式语言编写程序所需的时间的一小部分。MATLAB 名称代表矩阵实验室。MATLAB 最初的编写目的是提供对由 LINPACK 和 EISPACK 项目开发的矩阵软件的轻松访问，这两个项目共同代表了矩阵计算软件的最新技术。MATLAB 经过多年的发展，得到了许多用户的投入。在大学环境中，它是数学、工程和科学入门和高级课程的标准教学工具。在工业领域，MATLAB 是高效研究、开发和分析的首选工具。MATLAB 具有一系列称为工具箱的特定于应用程序的解决方案。对于大多数 MATLAB 用户来说非常重要，工具箱允许您学习应用专业技术。工具箱是 MATLAB 函数（M 文件）的综合集合，可扩展 MATLAB 环境以解决特定类别的问题。可用工具箱的领域包括信号处理、控制系统、神经网络、模糊逻辑、小波、仿真等。

## 统计代写|高等概率论作业代写Advanced Probability Theory代考|Uniform Berry-Esseen Bounds

• 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|高等概率论作业代写Advanced Probability Theory代考|Central Limit Theorems with infinite variances

So far, we have discussed the CLT under the second moment condition. In fact, CLT can hold under slightly weaker condition.
THEOREM 11.1.4 If $X,\left{X_{n}, n \geq 1\right}$ are i.i.d. r.v.s with non-degenerate d.f. $F$, then
$$\lim {n \rightarrow \infty} P\left(\frac{1}{B{n}} \sum_{i=1}^{n} X_{i}-A_{n} \leq x\right)=\Phi(x)$$
for some $B_{n}>0$ and $A_{n}$ iff
$$\lim {C \rightarrow \infty} \frac{P(|X|>C)}{C^{-2} E X^{2} I{|X| \leq C}}=0$$ Moreover, $A{n}, B_{n}$ may be chosen as
\begin{aligned} B_{n} &=\sup \left{C: C^{-2} E X^{2} I{|X| \leq C} \geq \frac{1}{n}\right}, \ A_{n} &=\frac{n}{B_{n}} E X I\left{|X|<B_{n}\right} . \end{aligned}
Proof. See Theorem 4, Chow and Teicher, p323.

## 统计代写|高等概率论作业代写Advanced Probability Theory代考|Some useful lemmas

In this section, we write $X_{n, k}=X_{k} / B_{n}, \psi_{n, k}(t)=E e^{i t X_{n, k}}$ and $\psi_{n}(t)=E e^{i t \sum_{k=1}^{n} X_{n, k}}$.
LEMMA 11.2.1
$$\left|\psi_{n}(t)-e^{-t^{2} / 2}\right| \leq 3 L_{n, \delta}|t|^{2+\delta} e^{-t^{2} / 2} \quad \text { for }|t|<\frac{1}{2} L_{n, \delta}^{-1 /(2+\delta)} .$$
Remark: In the i.i.d. case, $L_{n, \delta}^{-1 /(2+\delta)}=C n^{\delta /[2(2+\delta)]}$.
Proof. Note that
\begin{aligned} \left|\psi_{n}(t)-e^{-t^{2} / 2}\right| &=\left|\prod_{k=1}^{n} \psi_{n, k}(t)-e^{-t^{2} / 2}\right| \ &=\left|\exp \left{\sum_{k=1}^{n} \ln \psi_{n, k}(t)\right}-e^{-t^{2} / 2}\right| \ &=e^{-t^{2} / 2}\left|\exp \left{\sum_{k=1}^{n} \ln \psi_{n, k}(t)+\frac{t^{2}}{2}\right}-1\right| \end{aligned}
Using Theorem 10.5.3, we have
\begin{aligned} \psi_{n, k}(t) &=1+i t E X_{n, k}+\frac{1}{2}(i t)^{2} E X_{n, k}^{2}+\theta|t|^{2+\delta} E\left|X_{n, k}\right|^{2+\delta} \ &=1-\frac{1}{2} t^{2} \sigma_{n, k}^{2}+\theta|t|^{2+\delta} E\left|X_{n, k}\right|^{2+\delta} \end{aligned}
where $\theta$ is a complex number with $|\theta| \leq 1$. Noting that
$$|t| \sigma_{n, k} \leq|t|\left(E\left|X_{n, k}\right|^{2+\delta}\right)^{1 /(2+\delta)} \leq|t| L_{n, \delta}^{1 /(2+\delta)}<\frac{1}{2}$$
Thus,
\begin{aligned} \left|\psi_{n, k}(t)-1\right| & \leq \frac{1}{2} \times \frac{1}{4}+\frac{1}{2^{2+\delta}}<\frac{3}{8} \ \left|\psi_{n, k}(t)-1\right|^{2} & \leq 2\left[\left(\frac{1}{2} t^{2} \sigma_{n, k}^{2}\right)^{2}+\left(|t|^{2+\delta} E\left|X_{n, k}\right|^{2+\delta}\right)^{2}\right] \ &=2\left(\frac{1}{8}\left|t \sigma_{n, k}\right|^{2+\delta}\left|t \sigma_{n, k}\right|^{2-\delta}+\left(|t|^{2+\delta} E\left|X_{n, k}\right|^{2+\delta}\right)\left(|t|^{2+\delta} E\left|X_{n, k}\right|^{2+\delta}\right)\right) \ & \leq 2\left(\frac{1}{8}+\frac{1}{2^{2+\delta}}\right)|t|^{2+\delta} E\left|X_{n, k}\right|^{2+\delta} \ &=\frac{3}{4}|t|^{2+\delta} E\left|X_{k}\right|^{2+\delta} . \end{aligned}
By Taylor expansion of $\log (1+z)$, we can easily find that (see Appendix)
$$\log (1+z)=z+\frac{4}{5} \theta|z|^{2}, \quad \text { where }|\theta| \leq 1 \text { and }|z|<\frac{3}{8} \text {. }$$

THEOREM 11.2.1 (Berry-Esseen bounds for independent r.v.’s) Let $X_{1}, \ldots, X_{n}$ be independent r.v.’s such that $E X_{2}=0$ and $E\left|X_{j}\right|^{2+\delta}<\infty(j=1, \ldots, n)$ for some $0<\delta \leq 1$. Put $$L_{n, \sigma}=B_{n}^{+(2+s)} \sum_{j=1}^{n} E\left|X_{j}\right|^{2+s} .$$ Then for all $n$, $$\sup {x \in R}\left|F{\mathrm{n}}(x)-\Phi(x)\right| \leq A L_{n, \delta^{-}}$$ Proof. In the Smoothing Lemma we set $T=\left(36 L_{n, \delta}\right)^{-1 / \delta}$. Note that $\Phi^{\prime}(x)=\phi(x)=(2 \pi)^{-1 / 2} e^{-x^{2} / 2} \leq$ $(2 \pi)^{-1 / 2}$. Then we have \begin{aligned} \sup {x}\left|F{n}(x)-\Phi(x)\right| & \leq \frac{2}{\pi} \int_{0}^{T}|t|^{-1}\left|\psi_{n}(t)-e^{-t^{2} / 2}\right| d t+\frac{24 \lambda}{\pi T} \ & \leq \pi^{-1} \int_{0}^{T} 16 L_{n, \delta} t^{1+\delta} e^{-t^{2} / 3} d t+\frac{24 \lambda}{\pi}\left(36 L_{n, \delta}\right)^{1 / \delta} \ & \leq C_{\delta} L_{n, \delta}+C_{\delta} L_{n, \delta}^{1 / \delta} . \end{aligned} If $L_{n, \delta}^{1 / \delta} \leq 1$, then $\sup {x}\left|F{n}(x)-\Phi(x)\right| \leq 2 C_{\delta} L_{n, \delta}$ with $A=2 C_{\delta}$. On the other hand, if $L_{n, \delta}^{1 / \delta}>1$, then we can simply take $A=1$.
In the i.i.d. case, Theorem $11.2 .1$ reduces to the following corollary.
COROLLARY 11.2.3 (Berry-Esseen bounds for i.i.d. r.v.’s) Let $X_{1}, \ldots, X_{n}$ be i.i.d. r.v.’s. Let $\delta \in$ $(0,1]$, and
$$E X_{1}=u, \quad E X_{1}^{2}=\sigma^{2}>u, \quad E\left|X_{1}\right|^{2+s}<\infty, \quad \rho_{\delta}=E\left|X_{1}\right|^{2+s} / \sigma^{2+s} .$$
Then for all $n$,
$$\sup {x \in R}\left|F{n}(x)-\Phi(x)\right| \leq \frac{\mathcal{A}_{5}}{n^{\delta / 2}}$$
In particular, the case $\delta=1$ gives the most familiar Berry-Esseen bound.

## 统计代写|高等概率论作业代写Advanced Probability Theory代考|Central Limit Theorems with infinite variances

\begin{aligned} B_{n} &=\sup \left{C: C^{-2} E X^{2} I{|X| \leq C} \geq \frac{1}{n}\right}, \ A_{n} &=\frac{n}{B_{n}} EXI\left{|X|<B_{n}\right } 。\end{对齐}\begin{aligned} B_{n} &=\sup \left{C: C^{-2} E X^{2} I{|X| \leq C} \geq \frac{1}{n}\right}, \ A_{n} &=\frac{n}{B_{n}} EXI\left{|X|<B_{n}\right } 。\end{对齐}

## 统计代写|高等概率论作业代写Advanced Probability Theory代考|Some useful lemmas

|ψn(吨)−和−吨2/2|≤3一世n,d|吨|2+d和−吨2/2 为了 |吨|<12一世n,d−1/(2+d).

\begin{对齐} \left|\psi_{n}(t)-e^{-t^{2} / 2}\right| &=\left|\prod_{k=1}^{n} \psi_{n, k}(t)-e^{-t^{2} / 2}\right| \ &=\left|\exp \left{\sum_{k=1}^{n} \ln \psi_{n, k}(t)\right}-e^{-t^{2} / 2} \对| \ &=e^{-t^{2} / 2}\left|\exp \left{\sum_{k=1}^{n} \ln \psi_{n, k}(t)+\frac{ t^{2}}{2}\right}-1\right| \end{对齐}\begin{对齐} \left|\psi_{n}(t)-e^{-t^{2} / 2}\right| &=\left|\prod_{k=1}^{n} \psi_{n, k}(t)-e^{-t^{2} / 2}\right| \ &=\left|\exp \left{\sum_{k=1}^{n} \ln \psi_{n, k}(t)\right}-e^{-t^{2} / 2} \对| \ &=e^{-t^{2} / 2}\left|\exp \left{\sum_{k=1}^{n} \ln \psi_{n, k}(t)+\frac{ t^{2}}{2}\right}-1\right| \end{对齐}

ψn,到(吨)=1+一世吨和Xn,到+12(一世吨)2和Xn,到2+θ|吨|2+d和|Xn,到|2+d =1−12吨2σn,到2+θ|吨|2+d和|Xn,到|2+d

|吨|σn,到≤|吨|(和|Xn,到|2+d)1/(2+d)≤|吨|一世n,d1/(2+d)<12

|ψn,到(吨)−1|≤12×14+122+d<38 |ψn,到(吨)−1|2≤2[(12吨2σn,到2)2+(|吨|2+d和|Xn,到|2+d)2] =2(18|吨σn,到|2+d|吨σn,到|2−d+(|吨|2+d和|Xn,到|2+d)(|吨|2+d和|Xn,到|2+d)) ≤2(18+122+d)|吨|2+d和|Xn,到|2+d =34|吨|2+d和|X到|2+d.

## 广义线性模型代考

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## MATLAB代写

MATLAB 是一种用于技术计算的高性能语言。它将计算、可视化和编程集成在一个易于使用的环境中，其中问题和解决方案以熟悉的数学符号表示。典型用途包括：数学和计算算法开发建模、仿真和原型制作数据分析、探索和可视化科学和工程图形应用程序开发，包括图形用户界面构建MATLAB 是一个交互式系统，其基本数据元素是一个不需要维度的数组。这使您可以解决许多技术计算问题，尤其是那些具有矩阵和向量公式的问题，而只需用 C 或 Fortran 等标量非交互式语言编写程序所需的时间的一小部分。MATLAB 名称代表矩阵实验室。MATLAB 最初的编写目的是提供对由 LINPACK 和 EISPACK 项目开发的矩阵软件的轻松访问，这两个项目共同代表了矩阵计算软件的最新技术。MATLAB 经过多年的发展，得到了许多用户的投入。在大学环境中，它是数学、工程和科学入门和高级课程的标准教学工具。在工业领域，MATLAB 是高效研究、开发和分析的首选工具。MATLAB 具有一系列称为工具箱的特定于应用程序的解决方案。对于大多数 MATLAB 用户来说非常重要，工具箱允许您学习应用专业技术。工具箱是 MATLAB 函数（M 文件）的综合集合，可扩展 MATLAB 环境以解决特定类别的问题。可用工具箱的领域包括信号处理、控制系统、神经网络、模糊逻辑、小波、仿真等。

## 统计代写|高等概率论作业代写Advanced Probability Theory代考|Central Limit Theorems

• 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|高等概率论作业代写Advanced Probability Theory代考|Central Limit Theorems

11.1.1 CLT for i.i.d. r.v.s
When $X_{1}, \ldots, X_{n}$ are i.i.d., we get the following simple CLT.
THEOREM 11.1.1 (Levy theorem) Let $X_{1}, \ldots, X_{n}$ be i.i.d. r.v.’s with $E X_{1}=0$, and $\sigma^{2}=E X_{1}^{2}<\infty$. Let $F_{n}(x)=P(\sqrt{n} \bar{X} / \sigma \leq x)$. Then
$$\sup {x \in R}\left|F{n}(x)-\Phi(x)\right| \rightarrow 0$$
Proof. We provide two methods.
Method 1. Since $E X_{1}^{2}<\infty, \psi(t)$ is twice differentiable and has the following Taylor expansion,
$$\psi(t)=\psi(0)+\psi^{\prime}(0) t+\frac{1}{2} \psi^{\prime \prime}(0) t^{2}+o\left(t^{2}\right)=1-\frac{\sigma^{2} t^{2}}{2}+o\left(t^{2}\right)$$
Then the c.f. of $F_{n}$ is
$$\psi \sqrt{n} \Omega / \sigma(t)=\psi^{n}\left(\frac{t}{\sqrt{n} \sigma}\right)=\left(1-\frac{t^{2}}{2 n}+o\left(\frac{1}{n}\right)\right)^{n} \rightarrow e^{-t^{2} / 2}$$
Method 2. The Lindeberg condition in Corollary 11.1.2 holds since
$$\frac{1}{B_{n}^{2}} \sum_{k=1}^{n} E X_{k}^{2} I\left{\left|X_{k}\right| \geq \epsilon B_{n}\right}=\frac{E X_{1}^{2} I\left{\left|X_{1}\right| \geq \epsilon \sqrt{n E X_{1}^{2}}\right}}{E X_{1}^{2}} \rightarrow 0$$

## 统计代写|高等概率论作业代写Advanced Probability Theory代考|CLT for triangular arrays with finite variances

REMARK 11.1.1 Glearly, the Lindeberg condition implies that
$$\forall \epsilon>0: \quad \lim {n \rightarrow \infty} \sum{m=1}^{n} P\left(\left|X_{n, k}\right| \geq \epsilon\right)=0$$
which further implies that
$$\forall \epsilon>0: \sup {n} P\left(\left|X{n, k}\right| \geq \epsilon\right) \longrightarrow 0 .$$
That is, all the individual terms $X_{n, k}$ are uniformly small.
Proof. $”($ i $\Longrightarrow$ (ii)”. We first prove (a) of (ii). For $1 \leq k \leq n$, we have
\begin{aligned} \sigma_{n, k}^{2} &=E X_{n, k}^{2} I\left{\left|X_{n, k}\right|<\epsilon\right}+E X_{n, k}^{2} I\left{\left|X_{n, k}\right| \geq \epsilon\right} \ & \leq \epsilon^{2}+E X_{n, k}^{2} I\left{\left|X_{n, k}\right| \geq \epsilon\right} . \end{aligned}
Thus $\max {1 \leq k \leq n} \sigma{n, k}^{2} \leq \epsilon^{2}+\sum_{k=1}^{n} E X_{n, k}^{2} I\left{\left|X_{n, k}\right| \geq \epsilon\right}$. Letting $n \rightarrow \infty$, the Lindeberg condition implies that $\max {1 \leq k \leq n} \sigma{n, k}^{2} \leq 2 \epsilon^{2}$. Since $\epsilon$ can be chosen to be arbitrarily small, we have
$$\max {1 \leq m \leq n} \sigma{n, k}^{2} \rightarrow 0$$
We now prove (b) of (ii). Write $\psi_{n, k}(t)=E e^{i t X_{n, k}}$. It suffices to show that, $\forall t \in R$,
$$\prod_{k=1}^{n} \psi_{n, k}(t) \longrightarrow e^{-t^{2} / 2} \Longleftrightarrow \sum_{k=1}^{n} \ln \psi_{n, k}(t)+t^{2} / 2 \longrightarrow 0$$
It suffices to show that, as $n \rightarrow \infty, \forall t \in R$,
$$\begin{array}{r} \sum_{k=1}^{n} \ln \psi_{n, k}(t)-\sum_{k=1}^{n}\left(\psi_{n, k}(t)-1\right) \quad \rightarrow \quad 0, \ \sum_{k=1}^{n}\left(\psi_{n, k}(t)-1\right)+\frac{t^{x}}{2} \rightarrow 0 . \end{array}$$
Let us prove (1.2) first. From the inequality $\left|e^{i t}-1-i t\right| \leq t^{2} / 2$ for any real $t$, we have
$$\left|\psi_{n, k}(t)-1\right|=\left|E e^{i t X_{n, k}}-1-i t E X_{n, k}\right| \leq E\left|e^{i t X_{n, k}}-1-i t X_{n, k}\right| \leq \frac{1}{2} t^{2} E X_{n, k}^{2}=\frac{1}{2} t^{2} \sigma_{n, k}^{2} .$$
Thus, as $n \rightarrow \infty$
$$\max {1 \leq k \leq n}\left|\psi{n, k}(t)-1\right| \leq \frac{1}{2} t^{2} \max {1 \leq k \leq n} \sigma{n, k}^{2}=o(1), \quad \text { and } \quad \sum_{k=1}^{n}\left|\psi_{n, k}(t)-1\right| \leq \frac{t^{2}}{2}$$

Hence, from Theorem $11.8 .1:|\ln (1+z)-z| \leq|z|^{2}$ for $|z| \leq 1 / 2$, (1.2) follows from
$$\sum_{k=1}^{n}\left|\ln \psi_{n, k}(t)-\left(\psi_{n, k}(t)-1\right)\right| \leq \sum_{k=1}^{n}\left|\psi_{n, k}(t)-1\right|^{2} \leq o(1) \sum_{k=1}^{n}\left|\psi_{n, k}(t)-1\right|=o(1)$$
Next let us prove (1.3). By using the inequality $\left|e^{i t}-1-i t-\frac{1}{2}(i t)^{2}\right| \leq \min \left{t^{2}, \frac{1}{6}|t|^{3}\right}$ for any real t. we have
\begin{aligned} \left|\sum_{k=1}^{n}\left(\psi_{n, k}(t)-1\right)+\frac{t^{2}}{2}\right| &=\left|\sum_{k=1}^{n} E\left(e^{i t X_{n, k}}-1-i t X_{n, k}-\frac{1}{2}\left(i t X_{k}\right)^{2}\right)\right| \ & \leq \sum_{k=1}^{n} E \min \left{t^{2} X_{n, k}^{2}, \frac{1}{6}\left|t X_{n, k}\right|^{3}\right} \ & \leq t^{2} \sum_{k=1}^{n} E X_{n, k}^{2} I\left{\left|X_{k}\right| \geq \epsilon\right}+\frac{|t|^{3} \epsilon}{6} \sum_{k=1}^{n} E\left|X_{n, k}\right|^{2} I\left{\left|X_{n, k}\right|<\epsilon\right} \ & \leq t^{2} \sum_{k=1}^{n} E X_{n, k}^{2} I\left{\left|X_{k}\right| \geq \epsilon\right}+\frac{1}{6}|t|^{3} \epsilon . \end{aligned} Then (1.3) follows from this, the Lindeberg condition and by choosing $\epsilon$ arbitrarily small. $”$ (ii) $\Longrightarrow$ (i)”. Assume that (ii) holds. First, part (b) of (ii) implies (1.1). Secondly, from the proceeding proof, we can see that that (1.2) is implied from part (a) of (ii). Putting these two together, we see that (1.3) still holds. In particular, the real part of the left hand side in (1.3) should tend to 0 , i.e., $$0 \longleftarrow \operatorname{Re}\left(\sum_{k=1}^{n}\left(\psi_{n, k}(t)-1\right)+\frac{t^{2}}{2}\right)$$ \begin{aligned} &=\sum_{k=1}^{n} E\left(\cos \left(t X_{n, k}\right)-1+\frac{1}{2} t^{2} X_{n, k}^{2}\right) \ &\geq \sum_{k=1}^{n} E\left(\cos \left(t X_{n, k}\right)-1+\frac{1}{2} t^{2} X_{n, k}^{2}\right) I\left{\left|X_{n, k}\right| \geq \epsilon\right} \ &\geq \quad\left(\text { as } \cos (y)-1+\frac{1}{2} y^{2} \geq 0\right) \ &\geq \quad \sum_{k=1}^{n} E\left(\frac{1}{2} t^{2} X_{n, k}^{2}-2\right) I\left{\left|X_{n, k}\right| \geq \epsilon\right} \quad(\text { as } \cos (y) \geq-1) \ &=\sum_{k=1}^{n} E X_{n, k}^{2}\left(\frac{1}{2} t^{2}-\frac{2}{X_{n, k}^{2}}\right) I\left{\left|X_{n, k}\right| \geq \epsilon\right} \ &\geq \quad\left(\frac{t^{2}}{2}-\frac{2}{\epsilon^{2}}\right) \sum_{k=1}^{n} E X_{n, k}^{2} I\left{\left|X_{n, k}\right| \geq \epsilon\right} \end{aligned} so long as $t$ is chosen so that $t^{2} / 2-2 / \epsilon^{2}>0$, i.e., $t^{2}>4 / \epsilon^{2}$. Thus the right side tends to zero. Hence the Lindeberg condition holds.

## 统计代写|高等概率论作业代写Advanced Probability Theory代考|Central Limit Theorems

11.1.1 用于 iidrv 的 CLT

$$\sup {x \in R}\left|F {n}(x)-\Phi(x)\right| \右箭头 0 磷r○○F.在和pr○v一世d和吨在○米和吨H○ds.米和吨H○d1.小号一世nC和和X12<∞,ψ(吨)一世s吨在一世C和d一世FF和r和n吨一世一种b一世和一种ndH一种s吨H和F○一世一世○在一世nG吨一种和一世○r和Xp一种ns一世○n, \psi(t)=\psi(0)+\psi^{\prime}(0) t+\frac{1}{2} \psi^{\prime \prime}(0) t^{2}+o \left(t^{2}\right)=1-\frac{\sigma^{2} t^{2}}{2}+o\left(t^{2}\right) 吨H和n吨H和C.F.○FFn一世s \psi \sqrt{n} \Omega / \sigma(t)=\psi^{n}\left(\frac{t}{\sqrt{n} \sigma}\right)=\left(1-\frac {t^{2}}{2 n}+o\left(\frac{1}{n}\right)\right)^{n} \rightarrow e^{-t^{2} / 2} 米和吨H○d2.吨H和一世一世nd和b和rGC○nd一世吨一世○n一世nC○r○一世一世一种r和11.1.2H○一世dss一世nC和 \frac{1}{B_{n}^{2}} \sum_{k=1}^{n} E X_{k}^{2} I\left{\left|X_{k}\right| \geq \epsilon B_{n}\right}=\frac{E X_{1}^{2} I\left{\left|X_{1}\right| \geq \epsilon \sqrt{n E X_{1}^{2}}\right}}{E X_{1}^{2}} \rightarrow 0$$

## 统计代写|高等概率论作业代写Advanced Probability Theory代考|CLT for triangular arrays with finite variances

∀ε>0:林n→∞∑米=1n磷(|Xn,到|≥ε)=0

∀ε>0:支持n磷(|Xn,到|≥ε)⟶0.

\begin{对齐} \sigma_{n, k}^{2} &=E X_{n, k}^{2} I\left{\left|X_{n, k}\right|<\epsilon\right }+E X_{n, k}^{2} I\left{\left|X_{n, k}\right| \geq \epsilon\right} \ & \leq \epsilon^{2}+E X_{n, k}^{2} I\left{\left|X_{n, k}\right| \geq \epsilon\right} 。\end{对齐}\begin{对齐} \sigma_{n, k}^{2} &=E X_{n, k}^{2} I\left{\left|X_{n, k}\right|<\epsilon\right }+E X_{n, k}^{2} I\left{\left|X_{n, k}\right| \geq \epsilon\right} \ & \leq \epsilon^{2}+E X_{n, k}^{2} I\left{\left|X_{n, k}\right| \geq \epsilon\right} 。\end{对齐}

∏到=1nψn,到(吨)⟶和−吨2/2⟺∑到=1nln⁡ψn,到(吨)+吨2/2⟶0

∑到=1nln⁡ψn,到(吨)−∑到=1n(ψn,到(吨)−1)→0, ∑到=1n(ψn,到(吨)−1)+吨X2→0.

|ψn,到(吨)−1|=|和和一世吨Xn,到−1−一世吨和Xn,到|≤和|和一世吨Xn,到−1−一世吨Xn,到|≤12吨2和Xn,到2=12吨2σn,到2.

∑到=1n|ln⁡ψn,到(吨)−(ψn,到(吨)−1)|≤∑到=1n|ψn,到(吨)−1|2≤○(1)∑到=1n|ψn,到(吨)−1|=○(1)

\begin{对齐} \left|\sum_{k=1}^{n}\left(\psi_{n, k}(t)-1\right)+\frac{t^{2}}{2} \对| &=\left|\sum_{k=1}^{n} E\left(e^{it X_{n, k}}-1-it X_{n, k}-\frac{1}{2} \left(it X_{k}\right)^{2}\right)\right| \ & \leq \sum_{k=1}^{n} E \min \left{t^{2} X_{n, k}^{2}, \frac{1}{6}\left|t X_ {n, k}\right|^{3}\right} \ & \leq t^{2} \sum_{k=1}^{n} E X_{n, k}^{2} I\left{ \left|X_{k}\right| \geq \epsilon\right}+\frac{|t|^{3} \epsilon}{6} \sum_{k=1}^{n} E\left|X_{n, k}\right|^{ 2} I\left{\left|X_{n, k}\right|<\epsilon\right} \ & \leq t^{2} \sum_{k=1}^{n} E X_{n, k }^{2} I\left{\left|X_{k}\right| \geq \epsilon\right}+\frac{1}{6}|t|^{3} \epsilon 。\end{对齐}\begin{对齐} \left|\sum_{k=1}^{n}\left(\psi_{n, k}(t)-1\right)+\frac{t^{2}}{2} \对| &=\left|\sum_{k=1}^{n} E\left(e^{it X_{n, k}}-1-it X_{n, k}-\frac{1}{2} \left(it X_{k}\right)^{2}\right)\right| \ & \leq \sum_{k=1}^{n} E \min \left{t^{2} X_{n, k}^{2}, \frac{1}{6}\left|t X_ {n, k}\right|^{3}\right} \ & \leq t^{2} \sum_{k=1}^{n} E X_{n, k}^{2} I\left{ \left|X_{k}\right| \geq \epsilon\right}+\frac{|t|^{3} \epsilon}{6} \sum_{k=1}^{n} E\left|X_{n, k}\right|^{ 2} I\left{\left|X_{n, k}\right|<\epsilon\right} \ & \leq t^{2} \sum_{k=1}^{n} E X_{n, k }^{2} I\left{\left|X_{k}\right| \geq \epsilon\right}+\frac{1}{6}|t|^{3} \epsilon 。\end{对齐}然后（1.3）由此得出林德伯格条件并通过选择ε任意小。”(二)⟹（一世）”。假设 (ii) 成立。首先，（ii）的（b）部分暗示（1.1）。其次，从前面的证明中，我们可以看出（1.2）是从（ii）的（a）部分隐含的。将这两者放在一起，我们看到（1.3）仍然成立。特别是，（1.3）中左侧的实部应该趋向于 0 ，即0⟵Re⁡(∑到=1n(ψn,到(吨)−1)+吨22)\begin{aligned} &=\sum_{k=1}^{n} E\left(\cos \left(t X_{n, k}\right)-1+\frac{1}{2} t^ {2} X_{n, k}^{2}\right) \ &\geq \sum_{k=1}^{n} E\left(\cos \left(t X_{n, k}\right) -1+\frac{1}{2} t^{2} X_{n, k}^{2}\right) I\left{\left|X_{n, k}\right| \geq \epsilon\right} \ &\geq \quad\left(\text { as } \cos (y)-1+\frac{1}{2} y^{2} \geq 0\right) \ & \geq \quad \sum_{k=1}^{n} E\left(\frac{1}{2} t^{2} X_{n, k}^{2}-2\right) I\left {\left|X_{n, k}\right| \geq \epsilon\right} \quad(\text { as } \cos (y) \geq-1) \ &=\sum_{k=1}^{n} E X_{n, k}^{2} \left(\frac{1}{2} t^{2}-\frac{2}{X_{n, k}^{2}}\right) I\left{\left|X_{n, k} \对| \geq \epsilon\right} \ &\geq \quad\left(\frac{t^{2}}{2}-\frac{2}{\epsilon^{2}}\right) \sum_{k= 1}^{n} E X_{n, k}^{2} I\left{\left|X_{n, k}\right| \geq \epsilon\right} \end{对齐}\begin{aligned} &=\sum_{k=1}^{n} E\left(\cos \left(t X_{n, k}\right)-1+\frac{1}{2} t^ {2} X_{n, k}^{2}\right) \ &\geq \sum_{k=1}^{n} E\left(\cos \left(t X_{n, k}\right) -1+\frac{1}{2} t^{2} X_{n, k}^{2}\right) I\left{\left|X_{n, k}\right| \geq \epsilon\right} \ &\geq \quad\left(\text { as } \cos (y)-1+\frac{1}{2} y^{2} \geq 0\right) \ & \geq \quad \sum_{k=1}^{n} E\left(\frac{1}{2} t^{2} X_{n, k}^{2}-2\right) I\left {\left|X_{n, k}\right| \geq \epsilon\right} \quad(\text { as } \cos (y) \geq-1) \ &=\sum_{k=1}^{n} E X_{n, k}^{2} \left(\frac{1}{2} t^{2}-\frac{2}{X_{n, k}^{2}}\right) I\left{\left|X_{n, k} \对| \geq \epsilon\right} \ &\geq \quad\left(\frac{t^{2}}{2}-\frac{2}{\epsilon^{2}}\right) \sum_{k= 1}^{n} E X_{n, k}^{2} I\left{\left|X_{n, k}\right| \geq \epsilon\right} \end{对齐}只要吨被选择使得吨2/2−2/ε2>0， IE，吨2>4/ε2. 因此右侧趋于零。因此，林德伯格条件成立。

## 广义线性模型代考

statistics-lab作为专业的留学生服务机构，多年来已为美国、英国、加拿大、澳洲等留学热门地的学生提供专业的学术服务，包括但不限于Essay代写，Assignment代写，Dissertation代写，Report代写，小组作业代写，Proposal代写，Paper代写，Presentation代写，计算机作业代写，论文修改和润色，网课代做，exam代考等等。写作范围涵盖高中，本科，研究生等海外留学全阶段，辐射金融，经济学，会计学，审计学，管理学等全球99%专业科目。写作团队既有专业英语母语作者，也有海外名校硕博留学生，每位写作老师都拥有过硬的语言能力，专业的学科背景和学术写作经验。我们承诺100%原创，100%专业，100%准时，100%满意。

## MATLAB代写

MATLAB 是一种用于技术计算的高性能语言。它将计算、可视化和编程集成在一个易于使用的环境中，其中问题和解决方案以熟悉的数学符号表示。典型用途包括：数学和计算算法开发建模、仿真和原型制作数据分析、探索和可视化科学和工程图形应用程序开发，包括图形用户界面构建MATLAB 是一个交互式系统，其基本数据元素是一个不需要维度的数组。这使您可以解决许多技术计算问题，尤其是那些具有矩阵和向量公式的问题，而只需用 C 或 Fortran 等标量非交互式语言编写程序所需的时间的一小部分。MATLAB 名称代表矩阵实验室。MATLAB 最初的编写目的是提供对由 LINPACK 和 EISPACK 项目开发的矩阵软件的轻松访问，这两个项目共同代表了矩阵计算软件的最新技术。MATLAB 经过多年的发展，得到了许多用户的投入。在大学环境中，它是数学、工程和科学入门和高级课程的标准教学工具。在工业领域，MATLAB 是高效研究、开发和分析的首选工具。MATLAB 具有一系列称为工具箱的特定于应用程序的解决方案。对于大多数 MATLAB 用户来说非常重要，工具箱允许您学习应用专业技术。工具箱是 MATLAB 函数（M 文件）的综合集合，可扩展 MATLAB 环境以解决特定类别的问题。可用工具箱的领域包括信号处理、控制系统、神经网络、模糊逻辑、小波、仿真等。

## 统计代写|高等概率论作业代写Advanced Probability Theory代考| Application to Weak Law of Large Numbers

• 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|高等概率论作业代写Advanced Probability Theory代考|Appendix: Several useful lemmas

LEMMA 10.5.1 For $n=0,1,2, \ldots$ and any real $t$,
$$\left|e^{i t}-1-i t-\frac{(i t)^{2}}{2 !}-\ldots-\frac{(i t)^{n}}{n !}\right| \leq \min \left{\frac{|t|^{n+1}}{(n+1) !}, \frac{2|t|^{n}}{n !}\right}$$
Proof. By integration by parts, we have, for any $m \geq 0$,
\begin{aligned} \int_{0}^{t}(t-s)^{m} e^{i s} d s &=\frac{-1}{m+1} \int_{0}^{t} e^{i s} d(t-s)^{m+1} \ &=\frac{t^{m+1}}{m+1}+\frac{i}{m+1} \int_{0}^{t}(t-s)^{m+1} e^{i s} d s . \end{aligned}
Therefore, by iteration we get
\begin{aligned} e^{i t} &=1+\left(e^{i t}-1\right)=1+i \int_{0}^{t} e^{i s} d s \ &=1+i t+i^{2} \int_{0}^{t}(t-s) e^{i s} d s \ &=\ldots \ldots \ &=1+i t+\frac{(i t)^{2}}{2 !}+\ldots+\frac{(i t)^{n}}{n !}+\frac{i^{-n+1}}{n !} \int_{0}^{t}(t-s)^{n} e^{i s} d s \end{aligned}
Note that
$$\left|\int_{0}^{t}(t-s)^{n} e^{i s} d s\right| \leq \int_{0}^{|t|}|t-s|^{n} d s \leq \frac{|t|^{n+1}}{n+1}$$
By integration by parts,
\begin{aligned} \int_{0}^{t}(t-s)^{n} e^{i s} d s &=(-i) \int_{0}^{t}(t-s)^{n} d e^{i s} \ &=-i t^{n}+i n \int_{0}^{t}(t-s)^{n-1} e^{i s} d s \ &=i n \int_{0}^{t}(t-s)^{n-1}\left[e^{i s}-1\right] d s \end{aligned}
and hence
$$\left|\int_{0}^{t}(t-s)^{n} e^{i s} d s\right| \leq 2 n \int_{0}^{|t|}|t-s|^{n-1} d s=2|t|^{n}$$
189
Using these relationships, we get
$$\left|e^{i t}-1-i t-\frac{(i t)^{2}}{2 !}-\ldots-\frac{(i t)^{n}}{n !}\right| \leq \min \left{\frac{|t|^{n+1}}{(n+1) !}, \frac{2|t|^{n}}{n !}\right}$$
The proof is complete.

## 统计代写|高等概率论作业代写Advanced Probability Theory代考|Esseen’s Smoothing Lemma

Often we are interested in difference of two functions. For instance, if a r.v. $T_{n}$ has an asymptotic normal distribution, as in the central limit theorem, then
$$\sup {x}\left|P\left(T{n} \leq x\right)-\Phi(x)\right| \rightarrow 0 \quad \text { as } n \rightarrow \infty$$
The natural question is then how fast this limit goes to zero. In other words, we are interested in the rates of convergence to normality. One fundamental tool in studying the difference in two functions is the “smoothing lemma”.

The word “smoothing” is derived from the fact that: any r.v. $X$ perturbed by an independent continuous r.v. $Y$, will also be a continuous r.v.. That is, if $X$ and $Y$ are independent and $Y$ is a continuous r.v., then $X+Y$ is a continuous r.v. for all $X$. Furthermore, the degree of smoothness for $X+Y$ also depends on the degree of smoothness for $Y$. This follows from the following identity:
$$F_{X+Y}(t)=\int_{-\infty}^{\infty} F_{Y}(t-y) d F_{X}(y)$$
Let $V_{T}$ be the d.f. with a p.d.f. (i.e. inverse triangular d.f.)
$$v_{T}(x)=\frac{1-\cos (T x)}{\pi T x^{2}}$$
which is the p.d.f of the sum of two independent $U[-1 /(2 T), 1 /(2 T)]$ (try to plot it!) The corresponding c.f. is given by
$$\omega_{T}(t)=\left(1-\frac{t}{T}\right) I{|t| \leq T}$$
The explicit form of $\omega_{T}(t)$ is of no importance. What matters is that $\omega_{T}(t)$ vanishes for $|t| \geq T$, since this eliminates all questions of convergence.
For any function $\Delta(x)$, we denote its convolution with $V_{T}(x)$ by
$$\Delta^{T}(t) \equiv \Delta \star V_{T}(t):=\int_{-\infty}^{\infty} \Delta(t-x) v_{T}(x) d x$$
Our objective is to estimate the maximum of $|\Delta|$ in terms of the maximum of $\left|\Delta^{T}\right|$.

## 统计代写|高等概率论作业代写Advanced Probability Theory代考|Characteristic functions and smoothness condition

DEFINITION 10.8.1 If all points of increase of $F$ are among $b, b \pm h, b \pm 2 h, \ldots .$, then we say that $F$ is $a$ lattice d.f. with span $h$.
The following two theorems give a characterization of lattice distribution.
THEOREM 10.8.1 If $\lambda \neq 0$, the following three statements are equivalent:
(a) $\psi(\lambda)=1$.
(b) $\psi(t)$ has period $\lambda$, i.e., $\psi(t \neq n \lambda)=\psi(t)$ for all $t$ and $n$.
(c) All points of increase of $F$ are among $0, \pm h, \pm 2 h, \ldots .$ where $h=2 \pi / \lambda$.
Proof. We shall show that $(c) \rightarrow(b) \rightarrow(a) \rightarrow(c)$.
If (c) is true and $F$ attributes weight $p_{k}$ to $k h, k=0, \pm 1, \pm 2, \ldots$, then $\psi(t)=\sum_{k=-\infty}^{\infty} p_{k} e^{i k h t}$, which has period $2 \pi / h-\lambda$. So (c) implies (b).
If (b) is true, by taking $n=1$ and $t=0$, we get $\psi(\lambda)=\psi(0)=1$, which proves (a).
If $(\mathrm{a})$ is true, $\psi(\lambda)=E \cos (\lambda X)+i E \sin (\lambda X)=1$, then $\int_{-\infty}^{\infty}[1-\cos (\lambda x)] d F(x)=0$. Note that the integrand is nonnegative. So at every point $x$ of increase for $F$, we must have $1-\cos (\lambda x)=0$. Thus $F$ is concentrated on the multiplesi of $2 \pi / \lambda$, and hence (c) is true.
It is easy to deduce the next corollary by applying the last lemma to $X-b$.
COROLLARY 10.8.1 If $\lambda \neq 0$, the following three statements are equivalent:
(a) $\psi(\lambda)=e^{i b \lambda}$
(b) $\psi(t)$ satisfies $\psi(t+n \lambda)=\psi(t) e^{i n \lambda b}$ for all $t$ and $n$.
(c) All points of increase of $F$ are among $b, b \pm h, b \pm 2 h, \ldots .$ where $h=2 \pi / \lambda$.
The following result is thus immediate. It states that any distribution is either lattice, or nonlattice, or degenerate.
THEOREM 10.8.2 There exist only the following three possibilities:

1. $|\psi(t)| \equiv 1$ for all t. In this case, $\psi(t)=e^{i b t}$ (degenerate at b).
2. $|\psi(\lambda)|=1$ and $|\psi(t)|<1$ for $0<t<\lambda$ (lattice with span $h=2 \pi / \lambda$.)
3. $|\psi(t)|<1$ for all $t \neq 0$ (non-lattice distribution).

## 统计代写|高等概率论作业代写Advanced Probability Theory代考|Appendix: Several useful lemmas

\left|e^{it}-1-i t-\frac{(it)^{2}}{2 !}-\ldots-\frac{(it)^{n}}{n !}\right | \leq \min \left{\frac{|t|^{n+1}}{(n+1) !}, \frac{2|t|^{n}}{n !}\right}\left|e^{it}-1-i t-\frac{(it)^{2}}{2 !}-\ldots-\frac{(it)^{n}}{n !}\right | \leq \min \left{\frac{|t|^{n+1}}{(n+1) !}, \frac{2|t|^{n}}{n !}\right}

∫0吨(吨−s)米和一世sds=−1米+1∫0吨和一世sd(吨−s)米+1 =吨米+1米+1+一世米+1∫0吨(吨−s)米+1和一世sds.

|∫0吨(吨−s)n和一世sds|≤∫0|吨||吨−s|nds≤|吨|n+1n+1

∫0吨(吨−s)n和一世sds=(−一世)∫0吨(吨−s)nd和一世s =−一世吨n+一世n∫0吨(吨−s)n−1和一世sds =一世n∫0吨(吨−s)n−1[和一世s−1]ds

|∫0吨(吨−s)n和一世sds|≤2n∫0|吨||吨−s|n−1ds=2|吨|n
189

\left|e^{it}-1-i t-\frac{(it)^{2}}{2 !}-\ldots-\frac{(it)^{n}}{n !}\right | \leq \min \left{\frac{|t|^{n+1}}{(n+1) !}, \frac{2|t|^{n}}{n !}\right}\left|e^{it}-1-i t-\frac{(it)^{2}}{2 !}-\ldots-\frac{(it)^{n}}{n !}\right | \leq \min \left{\frac{|t|^{n+1}}{(n+1) !}, \frac{2|t|^{n}}{n !}\right}

## 统计代写|高等概率论作业代写Advanced Probability Theory代考|Esseen’s Smoothing Lemma

“平滑”一词源于以下事实：任何 rvX受独立连续 rv 扰动和, 也将是一个连续的 rv。也就是说，如果X和和是独立的并且和是一个连续的 rv，那么X+和对所有人来说是一个连续的 rvX. 此外，平滑度为X+和还取决于平滑度和. 这来自以下身份：
FX+和(吨)=∫−∞∞F和(吨−和)dFX(和)

v吨(X)=1−某物⁡(吨X)圆周率吨X2

ω吨(吨)=(1−吨吨)一世|吨|≤吨

Δ吨(吨)≡Δ⋆五吨(吨):=∫−∞∞Δ(吨−X)v吨(X)dX

## 统计代写|高等概率论作业代写Advanced Probability Theory代考|Characteristic functions and smoothness condition

(a)ψ(λ)=1.
(二)ψ(吨)有期λ， IE，ψ(吨≠nλ)=ψ(吨)对所有人吨和n.
(c) 所有增加点F是其中0,±H,±2H,….在哪里H=2圆周率/λ.

(a)ψ(λ)=和一世bλ
(二)ψ(吨)满足ψ(吨+nλ)=ψ(吨)和一世nλb对所有人吨和n.
(c) 所有增加点F是其中b,b±H,b±2H,….在哪里H=2圆周率/λ.

1. |ψ(吨)|≡1对于所有 t。在这种情况下，ψ(吨)=和一世b吨（在 b 处退化）。
2. |ψ(λ)|=1和|ψ(吨)|<1为了0<吨<λ（带跨度的格子H=2圆周率/λ.)
3. |ψ(吨)|<1对所有人吨≠0（非晶格分布）。

## 广义线性模型代考

statistics-lab作为专业的留学生服务机构，多年来已为美国、英国、加拿大、澳洲等留学热门地的学生提供专业的学术服务，包括但不限于Essay代写，Assignment代写，Dissertation代写，Report代写，小组作业代写，Proposal代写，Paper代写，Presentation代写，计算机作业代写，论文修改和润色，网课代做，exam代考等等。写作范围涵盖高中，本科，研究生等海外留学全阶段，辐射金融，经济学，会计学，审计学，管理学等全球99%专业科目。写作团队既有专业英语母语作者，也有海外名校硕博留学生，每位写作老师都拥有过硬的语言能力，专业的学科背景和学术写作经验。我们承诺100%原创，100%专业，100%准时，100%满意。

## MATLAB代写

MATLAB 是一种用于技术计算的高性能语言。它将计算、可视化和编程集成在一个易于使用的环境中，其中问题和解决方案以熟悉的数学符号表示。典型用途包括：数学和计算算法开发建模、仿真和原型制作数据分析、探索和可视化科学和工程图形应用程序开发，包括图形用户界面构建MATLAB 是一个交互式系统，其基本数据元素是一个不需要维度的数组。这使您可以解决许多技术计算问题，尤其是那些具有矩阵和向量公式的问题，而只需用 C 或 Fortran 等标量非交互式语言编写程序所需的时间的一小部分。MATLAB 名称代表矩阵实验室。MATLAB 最初的编写目的是提供对由 LINPACK 和 EISPACK 项目开发的矩阵软件的轻松访问，这两个项目共同代表了矩阵计算软件的最新技术。MATLAB 经过多年的发展，得到了许多用户的投入。在大学环境中，它是数学、工程和科学入门和高级课程的标准教学工具。在工业领域，MATLAB 是高效研究、开发和分析的首选工具。MATLAB 具有一系列称为工具箱的特定于应用程序的解决方案。对于大多数 MATLAB 用户来说非常重要，工具箱允许您学习应用专业技术。工具箱是 MATLAB 函数（M 文件）的综合集合，可扩展 MATLAB 环境以解决特定类别的问题。可用工具箱的领域包括信号处理、控制系统、神经网络、模糊逻辑、小波、仿真等。

## 统计代写|高等概率论作业代写Advanced Probability Theory代考| Levy Continuity Theorem

• 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|高等概率论作业代写Advanced Probability Theory代考|Levy Continuity Theorem

Instead of studying d.f.s directly, we could study their corresponding c.f.s. This can be done due to Levy continuity theorem.
LEMMA 10.4.1 For any $a>0$, we have
$$P\left(|X|>\frac{2}{a}\right) \leq \frac{1}{a} \int_{-a}^{a}(1-\psi(t)) d t .$$
Proof.
\begin{aligned} \int_{-a}^{a}(1-\psi(t)) d t &=2 a-\int_{-a}^{a} E e^{i t X} d t \ &=2 a-E\left(\int_{-a}^{a} e^{i t X} d t\right) \quad \text { (by Fubini theorem } \ &=2 a-E\left(\int_{-a}^{a} \cos (t X) d t\right) \ &=2 a-E\left(\frac{2 \sin (a X)}{X}\right) \ &=2 a E\left(1-\frac{\sin (a X)}{a X}\right) \ & \geq 2 a E\left(1-\frac{\sin (a X)}{a X}\right) I{|a X|>2} \ & \geq 2 a E\left(1-\frac{1}{2}\right) I{|a X|>2} \ & \geq a P(|a X|>2) \end{aligned}
REMARK 10.4.1 When a is chosen to be very small, Lemma 10.4.1 shows that the tail probability behavior of a r.v. $X$ is actually determined by the behavior of its c.f. around the origin.

LEMMA 10.4.2 Let $F_{n}$ be a sequence of d.f.s with c.f.s $\psi_{n}$. If $\psi_{n}(t) \rightarrow g(t)$, and $g(t)$ that is continuous at 0 , then $F_{n}$ is tight.

Proof. First note that $g(0)=\lim {n} \psi{n}(0)=1$ and $g(t)$ is continuous at 0 . Therefore, $\forall \varepsilon>0, \exists a_{0}>0$ such that $|1-g(t)|=|g(t)-g(0)|<\varepsilon / 2$ whenever |t|\frac{2}{a_{0}}\right) & \leq\left|\frac{1}{a_{0}} \int_{-a_{0}}^{a_{0}}\left(1-\psi_{n}(t)\right) d t\right| \quad(\text { Lemma 10.4.1) }\ & \rightarrow\left|\frac{1}{a_{0}} \int_{-a_{0}}^{a_{0}}(1-g(t)) d t\right| \quad \text { (dominated convergence theorem) } \ & \leq \frac{1}{a_{0}} \int_{-a_{0}}^{a_{u}}|1-g(t)| d t \leq \frac{1}{a_{0}} \int_{-a_{0}}^{a_{u}} \varepsilon d t \leq \varepsilon / 2 \end{aligned}
Thus, $\exists N_{0}>0$ such that, for all $n>N_{0}$, one has $P\left(\left|X_{n}\right|>\frac{2}{a_{0}}\right) \leq \varepsilon$.

## 统计代写|高等概率论作业代写Advanced Probability Theory代考|Moments of r.v.s and derivatives of their c.f.s

THEOREM 10.4.1 (Levy continuity theorem)
Assume that $X_{n}$ has d.f. $F_{n}$ and c.f. $\psi_{n}$ for $1 \leq n \leq \infty$.
(i) If $X_{n} \rightarrow d X_{\infty}$, (i.e., $F_{n} \Longrightarrow F_{\infty}$ ), then $\psi_{n}(t) \rightarrow \psi_{\infty}(t)$ for all $t$.
(ui) If $\psi_{n}(t) \rightarrow \psi(t)$, and $\psi(t)$ is contsnous at $v$, then there exsts a r.v. $X$ with d.f. $F$ such that $X_{n} \rightarrow$ (i.e., $F_{n} \Longrightarrow F$ ), and $\psi$ is the c.f. of $X$.
Proof.
184
(i) The proof follows from the bounded convergence theorem.
(ii). Now suppose that $\psi_{n}(t) \rightarrow \psi(t)$, and that $\psi(t)$ is continuous at 0. From Lemma 10.4.2, $F_{n}$ is tight.
Now suppose that $F_{n_{k}} \Longrightarrow_{v} F$ for some subsequence $n_{k}$ and some limit $F$. Since $F_{n}$ is tight, we have $F_{n_{k}} \Longrightarrow \bar{F}$, i.e., the limit $\bar{F}$ is a d.f. From part (i) of the current theorem, we see that $\psi_{n_{k}}(t) \rightarrow \psi_{\bar{F}}(t)$ for all $t$. On the other hand, from the assumption, we have $\psi_{n_{k}}(t) \rightarrow \psi(t)$ for all $t$. Therefore,
$$\psi_{\tilde{F}}(t)=\psi(t) .$$
Clearly, $\psi(t)$ is a c.f. Suppose that its corresponding d.f. is $F$. By the uniqueness theorem, Theorem 10.3.3, we get $\bar{F}=F$. This shows that $F$ is the only possible weak limit of the $F_{n}$. Therefore,
$$F_{n} \Longrightarrow F_{-}$$
The following corollary is immediate.
COROLLARY 10.4.1 (Levy continuity theorem)
$$X_{n} \rightarrow{ }{d} X \text { if and only if } \psi{X_{n}}(t) \rightarrow \psi_{X}(t) \text { for all } t \text {. }$$

## 统计代写|高等概率论作业代写Advanced Probability Theory代考|Relation between moments of a r.v. and derivatives of its c.f.

The smoothness of the c.f. $\psi(t)$ at $t=0$ is closely related to how many moments that $X$ possesses, and hence to the tail behavior of the d.f. of $X$.

THEOREM 10.5.1 If $E|X|^{n}<\infty$, then $\psi^{(n)}(t)$ exists and is a uniformly continuous function given by
$$\psi^{(k)}(t)=i^{k} E\left(X^{k} e^{i t X}\right)=i^{k} \int_{-\infty}^{\infty} x^{k} e^{i t x} d F(x), \quad k=0,1,2, \ldots, n .$$
In particular,
$$\psi^{(k)}(0)=i^{k} E X^{k}, \quad k=0,1, \ldots, n$$
Proof. Note that
$$\frac{\psi(t+h)-\psi(t)}{h}=\int_{-\infty}^{\infty} e^{i t x} \frac{e^{i h x}-1}{h} d F(x) .$$
Using Lemma 10.5.1, the integrand is dominated by $|x|$. So the first derivative of $\psi(t)$ exists by the dominated convergence theorem, and is given by
$$\psi^{\prime}(t)=\lim {h \rightarrow 0} \frac{\psi(t+h)-\psi(t)}{h}=i \int{-\infty}^{\infty} x e^{i t x} d F(x)$$
The uniform continuity of $\psi^{\prime}(t)$ follows from
$$\left|\psi^{\prime}(t+\delta)-\psi^{\prime}(t)\right|=\left|\int_{-\infty}^{\infty} x e^{i t x}\left(e^{i t \delta}-1\right) d F(x)\right| \rightarrow 0$$
by the dominated convergence theorem. Therefore, the assertion is true for $n=1$. The general case follows by induction.
A partial converse is given by the following theorem.
THEOREM 10.5.2 If $\psi^{(n)}(0)$ exists and is finite for some $n=1,2, \ldots$, then $E|X|^{n}<\infty$ if $n$ is even. (Consequently, $E|X|^{n-1}<\infty$ if $n$ is odd.)

## 统计代写|高等概率论作业代写Advanced Probability Theory代考|Levy Continuity Theorem

LEMMA 10.4.1 对于任何一种>0， 我们有

∫−一种一种(1−ψ(吨))d吨=2一种−∫−一种一种和和一世吨Xd吨 =2一种−和(∫−一种一种和一世吨Xd吨) （由富比尼定理  =2一种−和(∫−一种一种某物⁡(吨X)d吨) =2一种−和(2没有⁡(一种X)X) =2一种和(1−没有⁡(一种X)一种X) ≥2一种和(1−没有⁡(一种X)一种X)一世|一种X|>2 ≥2一种和(1−12)一世|一种X|>2 ≥一种磷(|一种X|>2)

## 统计代写|高等概率论作业代写Advanced Probability Theory代考|Moments of r.v.s and derivatives of their c.f.s

(一) 如果Xn→dX∞， （IE，Fn⟹F∞）， 然后ψn(吨)→ψ∞(吨)对所有人吨.
(ui) 如果ψn(吨)→ψ(吨)， 和ψ(吨)是一致的v, 那么存在一个 rvX与 dfF这样Xn→（IE，Fn⟹F）， 和ψ是 cfX.

184
(i) 证明来自有界收敛定理。
(二)。现在假设ψn(吨)→ψ(吨)， 然后ψ(吨)在 0 处连续。从引理 10.4.2，Fn很紧。

ψF~(吨)=ψ(吨).

Fn⟹F−

Xn→dX 当且仅当 ψXn(吨)→ψX(吨) 对所有人 吨.

## 统计代写|高等概率论作业代写Advanced Probability Theory代考|Relation between moments of a r.v. and derivatives of its c.f.

cf的平滑度ψ(吨)在吨=0与有多少时刻密切相关X拥有，因此对 df 的尾部行为X.

ψ(到)(吨)=一世到和(X到和一世吨X)=一世到∫−∞∞X到和一世吨XdF(X),到=0,1,2,…,n.

ψ(到)(0)=一世到和X到,到=0,1,…,n

ψ(吨+H)−ψ(吨)H=∫−∞∞和一世吨X和一世HX−1HdF(X).

ψ′(吨)=林H→0ψ(吨+H)−ψ(吨)H=一世∫−∞∞X和一世吨XdF(X)

|ψ′(吨+d)−ψ′(吨)|=|∫−∞∞X和一世吨X(和一世吨d−1)dF(X)|→0

## 广义线性模型代考

statistics-lab作为专业的留学生服务机构，多年来已为美国、英国、加拿大、澳洲等留学热门地的学生提供专业的学术服务，包括但不限于Essay代写，Assignment代写，Dissertation代写，Report代写，小组作业代写，Proposal代写，Paper代写，Presentation代写，计算机作业代写，论文修改和润色，网课代做，exam代考等等。写作范围涵盖高中，本科，研究生等海外留学全阶段，辐射金融，经济学，会计学，审计学，管理学等全球99%专业科目。写作团队既有专业英语母语作者，也有海外名校硕博留学生，每位写作老师都拥有过硬的语言能力，专业的学科背景和学术写作经验。我们承诺100%原创，100%专业，100%准时，100%满意。

## MATLAB代写

MATLAB 是一种用于技术计算的高性能语言。它将计算、可视化和编程集成在一个易于使用的环境中，其中问题和解决方案以熟悉的数学符号表示。典型用途包括：数学和计算算法开发建模、仿真和原型制作数据分析、探索和可视化科学和工程图形应用程序开发，包括图形用户界面构建MATLAB 是一个交互式系统，其基本数据元素是一个不需要维度的数组。这使您可以解决许多技术计算问题，尤其是那些具有矩阵和向量公式的问题，而只需用 C 或 Fortran 等标量非交互式语言编写程序所需的时间的一小部分。MATLAB 名称代表矩阵实验室。MATLAB 最初的编写目的是提供对由 LINPACK 和 EISPACK 项目开发的矩阵软件的轻松访问，这两个项目共同代表了矩阵计算软件的最新技术。MATLAB 经过多年的发展，得到了许多用户的投入。在大学环境中，它是数学、工程和科学入门和高级课程的标准教学工具。在工业领域，MATLAB 是高效研究、开发和分析的首选工具。MATLAB 具有一系列称为工具箱的特定于应用程序的解决方案。对于大多数 MATLAB 用户来说非常重要，工具箱允许您学习应用专业技术。工具箱是 MATLAB 函数（M 文件）的综合集合，可扩展 MATLAB 环境以解决特定类别的问题。可用工具箱的领域包括信号处理、控制系统、神经网络、模糊逻辑、小波、仿真等。

## 统计代写|高等概率论作业代写Advanced Probability Theory代考| Inversion formula

• 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|高等概率论作业代写Advanced Probability Theory代考|The inversion formula

THEOREM 10.3.1 (The inversion formula.) Let $\psi(t)=\int e^{i t x} \mu(d x)$, where $\mu$ is a probability measure. If $a<h$, then
$$\lim {T \rightarrow \infty} \frac{1}{2 \pi} \int{-T}^{T} \frac{e^{-i t a}-e^{-i t b}}{i t} \psi(t) d t=\mu(a, b)+\frac{1}{2} \mu({a, b}),$$
provided that the limit on the left hand side exists.
Proof. Let
\begin{aligned} I(T)=& \frac{1}{2 \pi} \int_{-T}^{T} \frac{e^{-i t a}-e^{-i t b}}{i t} \psi(t) d t \ =& \frac{1}{2 \pi} \int_{-T}^{T} \frac{e^{-i t a}-e^{-i t b}}{i t}\left(\int_{-\infty}^{\infty} e^{i t x} \mu(d x)\right) d t \ =& \frac{1}{2 \pi} \int_{-T}^{T}\left(\int_{-\infty}^{\infty} \frac{e^{-i t a}-e^{-i t b}}{i t} e^{i t x} \mu(d x)\right) d t \ =& \frac{1}{2 \pi} \int_{-\infty}^{\infty}\left(\int_{-T}^{T} \frac{e^{-i t a}-e^{-i t b}}{i t} e^{i t x} d t\right) \mu(d x) \ &\left(b y \text { Fubini’s theorem } \operatorname{since}\left|\frac{e^{-i t a}-e^{-i t b}}{i t} e^{i t x}\right|=\left|\int_{a}^{b} e^{-i t x} d x\right|\left|e^{i t x}\right| \leq b-a\right) \ =& \frac{1}{2 \pi} \int_{-\infty}^{\infty}\left(\int_{-T}^{T} \frac{e^{i t(x-a)}-e^{i t(x-b)}}{i t} d t\right) \mu(d x) \ =& \frac{1}{2 \pi} \int_{-\infty}^{\infty}\left(\int_{-T}^{T} \frac{1}{i t}(\cos [t(x-a)]-\cos [t(x-b)]) d t\right) \mu(d x) \ &+\frac{1}{2 \pi} \int_{-\infty}^{\infty}\left(\int_{-T}^{T} \frac{i}{i t}(\sin [t(x-a)]-\sin [t(x-b)]) d t\right) \mu(d x) \end{aligned}
180
\begin{aligned} &=\frac{1}{2 \pi} \int_{-\infty}^{\infty}\left(\int_{-T}^{T} \frac{1}{t}(\sin [t(x-a)]-\sin [t(x-b)]) d t\right) \mu(d x) \ &=\frac{1}{\pi} \int_{-\infty}^{\infty}\left(\int_{0}^{T} \frac{1}{t}(\sin [t(x-a)]-\sin [t(x-b)]) d t\right) \mu(d x) \ &=\frac{1}{\pi} \int_{-\infty}^{\infty}[I(x-a, T)-I(x-b, T)] \mu(d x) \end{aligned}

## 统计代写|高等概率论作业代写Advanced Probability Theory代考|One-to-one correspondence between d.f. and c.f.

THEOREM 10.3.3 (Uniqueness) Characteristic functions uniquely determines distribution functions. That is, there is a one-one correspondence between c.f.s and d.f.s.

## MATLAB代写

MATLAB 是一种用于技术计算的高性能语言。它将计算、可视化和编程集成在一个易于使用的环境中，其中问题和解决方案以熟悉的数学符号表示。典型用途包括：数学和计算算法开发建模、仿真和原型制作数据分析、探索和可视化科学和工程图形应用程序开发，包括图形用户界面构建MATLAB 是一个交互式系统，其基本数据元素是一个不需要维度的数组。这使您可以解决许多技术计算问题，尤其是那些具有矩阵和向量公式的问题，而只需用 C 或 Fortran 等标量非交互式语言编写程序所需的时间的一小部分。MATLAB 名称代表矩阵实验室。MATLAB 最初的编写目的是提供对由 LINPACK 和 EISPACK 项目开发的矩阵软件的轻松访问，这两个项目共同代表了矩阵计算软件的最新技术。MATLAB 经过多年的发展，得到了许多用户的投入。在大学环境中，它是数学、工程和科学入门和高级课程的标准教学工具。在工业领域，MATLAB 是高效研究、开发和分析的首选工具。MATLAB 具有一系列称为工具箱的特定于应用程序的解决方案。对于大多数 MATLAB 用户来说非常重要，工具箱允许您学习应用专业技术。工具箱是 MATLAB 函数（M 文件）的综合集合，可扩展 MATLAB 环境以解决特定类别的问题。可用工具箱的领域包括信号处理、控制系统、神经网络、模糊逻辑、小波、仿真等。