## 统计代写|概率论代写Probability theory代考|MATHS7103

2022年10月17日

couryes-lab™ 为您的留学生涯保驾护航 在代写概率论Probability theory方面已经树立了自己的口碑, 保证靠谱, 高质且原创的统计Statistics代写服务。我们的专家在代写概率论Probability theory代写方面经验极为丰富，各种代写概率论Probability theory相关的作业也就用不着说。

• 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
couryes™为您提供可以保分的包课服务

## 统计代写|概率论代写Probability theory代考|Supplement: Signed Measures

In this section, we show the decomposition theorems for signed measures (Hahn, Jordan) and deliver an alternative proof for Lebesgue’s decomposition theorem. We owe some of the proofs to [89].

Definition 7.35 Let $\mu$ and $v$ be two measures on $(\Omega, \mathcal{A}) . v$ is called totally continuous with respect to $\mu$ if, for any $\varepsilon>0$, there exists a $\delta>0$ such that for all $A \in \mathcal{A}$
$$\mu(A)<\delta \quad \text { implies } \quad v(A)<\varepsilon .$$
Remark 7.36 The definition of total continuity is similar to that of uniform integrability (see Theorem 6.24(iii)), at least for finite $\mu$. We will come back to this connection in the framework of the martingale convergence theorem that will provide an alternative proof of the Radon-Nikodym theorem (Corollary 7.34). $\diamond$
Theorem $7.37$ Let $\mu$ and $v$ be measures on $(\Omega, \mathcal{A})$. If $\nu$ is totally continuous with respect to $\mu$, then $v \ll \mu$. If $v(\Omega)<\infty$, then the converse also holds.

Proof $” \Longrightarrow “$ Let $v$ be totally continuous with respect to $\mu$. Let $A \in \mathcal{A}$ with $\mu(A)=0$. For all $\varepsilon>0$, by assumption, $v(A)<\varepsilon$; hence $v(A)=0$ and thus $v \ll \mu$. exist an $\varepsilon>0$ and sets $A_n \in \mathcal{A}$ with $\mu\left(A_n\right)<2^{-n}$ but $v\left(A_n\right) \geq \varepsilon$ for all $n \in \mathbb{N}$. Define $A:=\limsup {n \rightarrow \infty} A_n=\bigcap{n=1}^{\infty} \bigcup_{k=n}^{\infty} A_k$. Then
$$\mu(A)=\lim {n \rightarrow \infty} \mu\left(\bigcup{k=n}^{\infty} A_k\right) \leq \lim {n \rightarrow \infty} \sum{k=n}^{\infty} \mu\left(A_k\right) \leq \lim {n \rightarrow \infty} \sum{k=n}^{\infty} 2^{-k}=0$$

Since $\nu$ is finite and upper semicontinuous (Theorem 1.36), we have
$$v(A)=\lim {n \rightarrow \infty} v\left(\bigcup{k=n}^{\infty} A_k\right) \geq \inf {n \in \mathbb{N}} v\left(A_n\right) \geq \varepsilon>0 .$$ Thus $v \nless \mu$. Example 7.38 In the converse implication of the theorem, the assumption of finiteness is essential. For example, let $\mu=\mathcal{N}{0,1}$ be the standard normal distribution on $\mathbb{R}$ and let $v$ be the Lebesgue measure on $\mathbb{R}$. Then $v$ has the density $f(x)=$ $\sqrt{2 \pi} e^{x^2 / 2}$ with respect to $\mu$. In particular, we have $v \ll \mu$. On the other hand, $\mu([n, \infty)) \stackrel{n \rightarrow \infty}{\longrightarrow} 0$ and $v([n, \infty))=\infty$ for any $n \in \mathbb{N}$. Hence $v$ is not totally continuous with respect to $\mu$. $\diamond$

## 统计代写|概率论代写Probability theory代考|Supplement: Dual Spaces

By the Riesz-Fréchet theorem (Theorem 7.26), every continuous linear functional $F: L^2(\mu) \rightarrow \mathbb{R}$ has a representation $F(g)=\langle f, g\rangle$ for some $f \in L^2(\mu)$. On the other hand, for any $f \in L^2(\mu)$, the map $L^2(\mu) \rightarrow \mathbb{R}, g \mapsto\langle f, g\rangle$ is continuous and linear. Hence $L^2(\mu)$ is canonically isomorphic to its topological dual space $\left(L^2(\mu)\right)^{\prime}$. This dual space is defined as follows.

Definition 7.47 (Dual space) Let $(V,|\cdot|)$ be a Banach space. The dual space $V^{\prime}$ of $V$ is defined by
$V^{\prime}:={F: V \rightarrow \mathbb{R}$ is continuous and linear $}$.
For $F \in V^{\prime}$, we define $|F|^{\prime}:=\sup {|F(f)|:|f|=1}$.
Remark 7.48 As $F$ is continuous, for any $\delta>0$, there exists an $\varepsilon>0$ such that $|F(f)|<\delta$ for all $f \in V$ with $|f|<\varepsilon$. Hence $|F|^{\prime} \leq \delta / \varepsilon<\infty . \diamond$

We are interested in the case $V=L^p(\mu)$ for $p \in[1, \infty]$ and write $|F|_p^{\prime}$ for the norm of $F \in V^{\prime}$. In the particular case $V=L^2(\mu)$, by the Cauchy-Schwarz inequality, we have $|F|_2^{\prime}=|f|_2$. This can be generalized:
Lemma 7.49 Let $p, q \in[1, \infty]$ with $\frac{1}{p}+\frac{1}{q}=1$. The canonical map
\begin{aligned} \kappa: L^q(\mu) & \rightarrow\left(L^p(\mu)\right)^{\prime} \ \kappa(f)(g) &=\int f g d \mu \quad \text { for } f \in L^q(\mu), g \in L^p(\mu) \end{aligned}
is an isometry; that is, $|\kappa(f)|_p^{\prime}=|f|_q$.
Proof We show equality by showing the two inequalities separately.
” $\leq$ “, This follows from Hölder’s inequality.
” $\geq$ “, For any admissible pair $p, q$ and all $f \in \mathcal{L}^q(\mu), g \in \mathcal{L}^p(\mu)$, by the definition of the operator norm, $|\kappa(f)|_p^{\prime}|g|_p \geq\left|\int f g d \mu\right|$. Define the sign function $\operatorname{sign}(x)=\mathbb{1}{(0, \infty)}(x)-\mathbb{1}{(-\infty, 0)}(x)$. Replacing $g$ by $\tilde{g}:=|g| \operatorname{sign}(f)$ (note that $\left.|\tilde{g}|_p=|g|_p\right)$, we obtain
$$|\kappa(f)|_p^{\prime}|g|_p \geq\left|\int f \tilde{g} d \mu\right|=|f g|_1 .$$
First consider the case $q=1$ and $f \in \mathcal{L}^1(\mu)$. Applying (7.12) with $g \equiv 1 \in \mathcal{L}^{\infty}(\mu)$ yields $|\kappa(f)|_{\infty}^{\prime} \geq|f|_1$.
Now let $q \in(1, \infty)$. Let $g:=|f|^{q-1}$. Since $\frac{q-1}{q}=\frac{1}{p}$, we have
$$|\kappa(f)|_p^{\prime} \cdot|g|_p \geq|f g|_1=\left||f|^q\right|_1=|f|_q^q=|f|_q \cdot|f|_q^{q-1}=|f|_q \cdot|g|_p$$

# 概率论代考

## 统计代写|概率论代写概率论代考|补充:签署办法

$$\mu(A)<\delta \quad \text { implies } \quad v(A)<\varepsilon .$$总的连续性的定义类似于一致可积性的定义(见定理6.24(iii))，至少对于有限的是这样 $\mu$。我们将在鞅收敛定理的框架中回到这个联系，它将提供Radon-Nikodym定理的另一种证明(推论7.34)。 $\diamond$

$$\mu(A)=\lim {n \rightarrow \infty} \mu\left(\bigcup{k=n}^{\infty} A_k\right) \leq \lim {n \rightarrow \infty} \sum{k=n}^{\infty} \mu\left(A_k\right) \leq \lim {n \rightarrow \infty} \sum{k=n}^{\infty} 2^{-k}=0$$

## 有限元方法代写

tatistics-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。