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澳洲代写｜ECOS3007｜ International Macroeconomics国际宏观经济学悉尼大学

Posted on 2023年10月13日2023年10月13日 by statistics-lab

statistics-lab^TM为您悉尼大学（英语：The University of Sydney），简称悉大、USYD，简称“NCL”Stochastic Processes随机过程代写代考和辅导服务！

课程介绍：

This unit studies macroeconomic theory and policy in a global trading world. The microfoundations of the various sectors are examined in the context of an open economy. The evolution of international money and capital markets is described, the operation of the foreign exchange market is examined, showing how its microstructure affects its macro performance. Theories and tests of the efficiency of international capital markets are surveyed, as well as core theories and tests of exchange rate and asset price determination. The unit develops the macroeconomic implications of monetary and fiscal policies for small and large open economies for different regimes.

澳洲代写｜ECOS3007｜ International Macroeconomics国际宏观经济学悉尼大学

Attribute	Detail
Course Code	ECOS3007
Course Title	International Macroeconomics
Academic Unit	Economics
Session, Year	Semester 1, 2023
Location	Camperdown/Darlington, Sydney
Credit Points	6
Coordinator/Lecturer	Not Specified
Pre-Requisites	Not Specified

International Macroeconomics国际宏观经济学相关

Assume that there exists free international capital mobility and that the world interest rate, $r^$, is $10 \%$ per period (i.e., $r^=0.1$ ). Finally,

assume that the economy’s initial net foreign asset position is zero $\left(B_0^*=0\right)$.
(a) Compute the firm’s optimal levels of period-1 investment and period-2 profits.
(b) State the maximization problem of the representative household and solve for the optimal levels of consumption in periods 1 and 2.
(c) Find the country’s net foreign asset position at the end of period 1 , the trade balance in periods 1 and 2 , and the current account in periods 1 and 2 .
(d) Now consider an investment surge. Specifically, assume that as a result of a technological improvement, the production technology becomes $Q_2=2 \sqrt{T_1}$. Find the equilibrium levels of savings, investment, the trade balance, the current account, and the country’s net foreign asset position in period 1. Compare your results with those obtained in items (a)-(c) providing interpretation and intuition.

假设存在自由的国际资本流动性，并且世界利率 $r^$ 为每期 $10 \%$ （即 $r^=0.1$ ）。最后，

假设经济体的初始净外国资产头寸为零$\left(B_0^*=0\right)$。
(a) 计算公司第一期投资和第二期利润的最佳水平。
(b) 陈述代表性家庭的最大化问题并求解第 1 期和第 2 期的最优消费水平。
(c) 求该国在第 1 期末的外国净资产头寸、第 1 期和第 2 期的贸易差额以及第 1 期和第 2 期的经常账户。
(d) 现在考虑投资激增。具体来说，假设由于技术改进，生产技术变为$Q_2=2 \sqrt{T_1}$。求第 1 期储蓄、投资、贸易平衡、经常账户和国家净外国资产头寸的均衡水平。将您的结果与 (a)-(c) 项中获得的结果进行比较，以提供解释和直觉。

An investment surge
Suppose that in period 1 agents learn that in period 2 the productivity of capital will increase. For example, suppose that the production function in period 2 was initially given by $F\left(K_2\right)=\sqrt{K_2}$ and that due to a technological advancement it changes to $\tilde{F}\left(K_2\right)=2 \sqrt{K_2}$. Another example of an investment surge is given by an expected increase in the price of exports. In Norway, for instance, the oil price increase of 1973 unleashed an investment boom of around $10 \%$ of GDP. In response to this news, firms will choose to increase investment in period 1 for any given level of the interest rate. This scenario is illustrated in figure 6.4. Initially, the investment schedule is $I^0\left(r_1\right)$ and the saving schedule is $S^0\left(r_1, Q_1\right)$. Given the world interest rate $r^*$, investment is $I_1^0$ and savings is $S_1^0$. As shown in panel (b), the current account schedule is $C A^0\left(r_1, Q_1\right)$, and the equilibrium current account balance is $C A_1^0$. The news of the future productivity increase shifts the investment schedule to the right to $I^1\left(r_1\right)$, and the new equilibrium level of investment is $I_1^1$, which is higher than $I_1^0$. The expected increase in productivity might also affect current saving through its effect on expected future income. Specifically, in period 2, firms will generate higher profits which represent a positive income effect for households who are the owners of such firms. Households will take advantage of the expected increase in profits by increasing consumption in period 1 , thus cutting savings. Therefore, the savings schedule shifts to the left to $S^1\left(r_1, Q_1\right)$ and the equilibrium level of savings falls from $S_1^0$ to $S_1^1$. With this shifts in the investment and savings schedules it follows that, for any given interest rate, the current account is lower. That is, the current account schedule shifts to the left to $C A^1\left(r_1, Q_1\right)$. Given the world interest rate $r^*$, the current account deteriorates from $C A_1^0$ to $C A_1^1$. Note that if the economy was closed, the investment surge would trigger a rise in the domestic interest rate from $r_c^0$ to $r_c^1$ and thus investment would increase by less than in the open economy.

投资激增
假设在第 1 阶段，代理人得知在第 2 阶段，资本生产率将会提高。例如，假设第 2 期的生产函数最初由 $F\left(K_2\right)=\sqrt{K_2}$ 给出，并且由于技术进步，它变为 $\tilde{F}\left( K_2\right)=2\sqrt{K_2}$。投资激增的另一个例子是出口价格的预期上涨。例如，在挪威，1973 年石油价格上涨引发了投资热潮，投资额约占 GDP 的 10%$。针对这一消息，企业将选择在任何给定利率水平下增加第一阶段的投资。这种情况如图 6.4 所示。最初，投资计划为 $I^0\left(r_1\right)$，储蓄计划为 $S^0\left(r_1, Q_1\right)$。给定世界利率 $r^$，投资为 $I_1^0$，储蓄为 $S_1^0$。如图(b)所示，经常账户表为$C A^0\left(r_1, Q_1\right)$，平衡经常账户余额为$C A_1^0$。未来生产率提高的消息使投资计划右移为$I^1\left(r_1\right)$，新的投资均衡水平为$I_1^1$，高于$I_1^0 $。生产率的预期增长也可能通过对预期未来收入的影响而影响当前储蓄。具体来说，在第二阶段，企业将产生更高的利润，这对这些企业所有者的家庭产生积极的收入影响。家庭将通过增加第一阶段的消费来利用预期的利润增长，从而减少储蓄。因此，储蓄计划向左移动到$S^1\left(r_1, Q_1\right)$，储蓄均衡水平从$S_1^0$下降到$S_1^1$。随着投资和储蓄计划的这种变化，对于任何给定的利率，经常账户都会较低。即当前账户表左移为$C A^1\left(r_1, Q_1\right)$。考虑到世界利率 $r^$，经常账户从 $C A_1^0$ 恶化到 $C A_1^1$。请注意，如果经济封闭，投资激增将引发国内利率从 $r_c^0$ 上升至 $r_c^1$，因此投资增幅将低于开放经济。

统计代写请认准statistics-lab™. statistics-lab™为您的留学生涯保驾护航。

金融工程代写

金融工程是使用数学技术来解决金融问题。金融工程使用计算机科学、统计学、经济学和应用数学领域的工具和知识来解决当前的金融问题，以及设计新的和创新的金融产品。

非参数统计代写

非参数统计指的是一种统计方法，其中不假设数据来自于由少数参数决定的规定模型；这种模型的例子包括正态分布模型和线性回归模型。

广义线性模型代考

广义线性模型（GLM）归属统计学领域，是一种应用灵活的线性回归模型。该模型允许因变量的偏差分布有除了正态分布之外的其它分布。

术语广义线性模型（GLM）通常是指给定连续和/或分类预测因素的连续响应变量的常规线性回归模型。它包括多元线性回归，以及方差分析和方差分析（仅含固定效应）。

有限元方法代写

有限元方法（FEM）是一种流行的方法，用于数值解决工程和数学建模中出现的微分方程。典型的问题领域包括结构分析、传热、流体流动、质量运输和电磁势等传统领域。

有限元是一种通用的数值方法，用于解决两个或三个空间变量的偏微分方程（即一些边界值问题）。为了解决一个问题，有限元将一个大系统细分为更小、更简单的部分，称为有限元。这是通过在空间维度上的特定空间离散化来实现的，它是通过构建对象的网格来实现的：用于求解的数值域，它有有限数量的点。边界值问题的有限元方法表述最终导致一个代数方程组。该方法在域上对未知函数进行逼近。[1] 然后将模拟这些有限元的简单方程组合成一个更大的方程系统，以模拟整个问题。然后，有限元通过变化微积分使相关的误差函数最小化来逼近一个解决方案。

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。

随机分析代写

随机微积分是数学的一个分支，对随机过程进行操作。它允许为随机过程的积分定义一个关于随机过程的一致的积分理论。这个领域是由日本数学家伊藤清在第二次世界大战期间创建并开始的。

时间序列分析代写

随机过程，是依赖于参数的一组随机变量的全体，参数通常是时间。随机变量是随机现象的数量表现，其时间序列是一组按照时间发生先后顺序进行排列的数据点序列。通常一组时间序列的时间间隔为一恒定值（如1秒，5分钟，12小时，7天，1年），因此时间序列可以作为离散时间数据进行分析处理。研究时间序列数据的意义在于现实中，往往需要研究某个事物其随时间发展变化的规律。这就需要通过研究该事物过去发展的历史记录，以得到其自身发展的规律。

回归分析代写

多元回归分析渐进（Multiple Regression Analysis Asymptotics）属于计量经济学领域，主要是一种数学上的统计分析方法，可以分析复杂情况下各影响因素的数学关系，在自然科学、社会和经济学等多个领域内应用广泛。

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，工具箱允许您学习和应用专业技术。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。

R语言代写	问卷设计与分析代写
PYTHON代写	回归分析与线性模型代写
MATLAB代写	方差分析与试验设计代写
STATA代写	机器学习/统计学习代写
SPSS代写	计量经济学代写
EVIEWS代写	时间序列分析代写
EXCEL代写	深度学习代写
SQL代写	各种数据建模与可视化代写

澳洲代写｜MATH1510｜Stochastic Processes随机过程悉尼大学

Posted on 2023年10月13日2023年10月13日 by statistics-lab

statistics-lab^TM为您悉尼大学（英语：The University of Sydney），简称悉大、USYD，简称“NCL”Stochastic Processes随机过程代写代考和辅导服务！

课程介绍：

A stochastic process is a mathematical model of time-dependent random phenomena and is employed in numerous fields of application, including economics, finance, insurance, physics, biology, chemistry and computer science. This unit will establish basic properties of discrete-time Markov chains including random walks and branching processes. This unit will derive key results of Poisson processes and simple continuous-time Markov chains. This unit will investigate simple queuing theory. This unit will also introduce basic concepts of Brownian motion and martingales. Throughout the unit, various illustrative examples are provided in modelling and analysing problems of practical interest. By completing this unit, you will develop an essential basis for further studies stochastic analysis, stochastic differential equations, stochastic control, financial mathematics and statistical inference.

澳洲代写｜MATH1510｜Stochastic Processes随机过程悉尼大学

Attribute	Detail
Course Code	STAT3021
Course Title	Stochastic Processes
Academic Unit	Mathematics and Statistics Academic Operations
Session, Year	Semester 1, 2023
Location	Camperdown/Darlington, Sydney
Credit Points	6
Coordinator/Lecturer	Not Specified
Pre-Requisites	Not Specified

Stochastic Processes随机过程代写代考作业问题集

问题 1.

Consider the unit interval $I:=[0,1]$. Moreover, for every $n$ and $\left(i_1, \ldots, i_n\right) \in{0,1,2}^n$ we consider the interval $I_{i_1 \ldots i_n} \subset I$ which is the set of those numbers whose base 3 expansion starts with $\left(i_1 \ldots i_n\right)$. That is
$$
I_{i_1 \ldots i_n}:=\left[\sum_{k=1}^n \frac{i_k}{3^k}, \frac{1}{3^n}+\sum_{k=1}^n \frac{i_k}{3^k}\right] .
$$
Let $X_0, X_1, X_2$ be independent $\operatorname{Bernoulli}\left(p_0\right)$, Bernoulli $\left(p_1\right)$ and $\operatorname{Bernoulli}\left(p_2\right)$ random variables respectively. That is $\mathbb{P}\left(X_i=1\right)=p_i$ and $\mathbb{P}\left(X_i=0\right)=1-p_i$ for $i=0,1,2$. Moreover for every $n$ and $\left(i_1, \ldots, i_n\right) \in{0,1,2}^n$ we are given the random variables $X_{i_1 \ldots i_n}$ such that on the one hand $\left{X_{i_1 \ldots i_n}\right}_{n \geq 1,\left(i_1, \ldots, i_n\right) \in{0,1,2}^n}$ are independent and on the other hand: $X_{i_1 \ldots i_n} \stackrel{d}{=} X_{i_n}$. For every $n \geq 1$ we define the set $E_n \subset[0,1]$ by
$$
E_n:=\bigcup_{X_{i_1} \cdot X_{i_1, i_2} \cdots X_{i_1, i_2, \ldots, i_n}=1} I_{i_1 \ldots i_n} .
$$
Finally, we define the set $E:=\bigcap_{n=1}^{\infty} E_n$. Assume that $p_0=\frac{2}{3}, p_1=\frac{3}{4}$ and $p_2=\frac{1}{2}$. Question: Is it true that $\mathbb{P}(E \neq \emptyset)>0$

问题 2.

Given a branching process with the following offspring distribution, determine the extinction probabilities $q$ :
(a) $p_0=0.25, p_1=0.4, p_2=0.35, p_n=0$ if $n \geq 3$,
(b) $p_0=0.5, p_1=0.1, p_2=0, p_3=0.4, p_n=0$ if $n \geq 4$

问题 3.

Consider the branching process with offspring distribution as in the previous exercise part (b). What is the probability that the population is extinct in the second generation $X_2=0$, given that it did not die out in the first generation?

英国代写｜MATH1510｜Discrete Mathematics离散数学悉尼大学

统计代写请认准statistics-lab™. statistics-lab™为您的留学生涯保驾护航。

金融工程代写

非参数统计代写

非参数统计指的是一种统计方法，其中不假设数据来自于由少数参数决定的规定模型；这种模型的例子包括正态分布模型和线性回归模型。

广义线性模型代考

广义线性模型（GLM）归属统计学领域，是一种应用灵活的线性回归模型。该模型允许因变量的偏差分布有除了正态分布之外的其它分布。

有限元方法代写

随机分析代写

时间序列分析代写

回归分析代写

MATLAB代写

R语言代写	问卷设计与分析代写
PYTHON代写	回归分析与线性模型代写
MATLAB代写	方差分析与试验设计代写
STATA代写	机器学习/统计学习代写
SPSS代写	计量经济学代写
EVIEWS代写	时间序列分析代写
EXCEL代写	深度学习代写
SQL代写	各种数据建模与可视化代写

英国代写｜MATH1510｜Discrete Mathematics离散数学纽卡斯尔大学

Posted on 2023年10月12日2023年10月13日 by statistics-lab

statistics-lab^TM为您纽卡斯尔大学（Newcastle University），简称“NCL”Discrete Mathematics离散数学英国代写代考和辅导服务！

课程介绍：

Discrete mathematics is the study of mathematical structures that are discrete, separated or distinct; in contrast with calculus which deals with continuous change. It is an important area of pure and applied mathematics, as well as providing the mathematical basis for the understanding of computers and modern computation. Discrete Mathematics is important in the sciences, where it has increasing application in many areas, an exemplar of which is the understanding of DNA sequences in molecular biology. The Discrete Mathematics course introduces first year students to the basic concepts of discrete mathematics, covering topics such as sets, logic, enumeration methods, probability, recurrence relations, induction and graph theory. The course provides important background for students pursuing a BMath degree. It covers much of the mathematics essential for students majoring in Computer Science or Software Engineering, and is a compulsory course in those degree programs.

Attribute	Detail
Course Code	MATH1510
Course Title	Discrete Mathematics
Units	10 units
Level	1000 level
Pre-Requisites	HSC Advanced Mathematics (Bands 5 or 6), or equivalent
Course Coordinator/Lecturer	Not explicitly mentioned in the provided text

Discrete Mathematics离散数学的案例

问题 1.

Let $\phi: G \rightarrow H$ be a group isomorphism. If $g \in G$ then the group inverse, in the group $H$, of $\phi(g)$ is $\phi\left(g^{-1}\right)$.

Proof: We may check that
$$
\phi(g) \cdot \phi\left(g^{-1}\right)=\phi\left(g \cdot g^{-1}\right)=\phi\left(e_G\right)=e_H
$$
(by Proposition 9.5)
Also
$$
\phi\left(g^{-1}\right) \cdot \phi(g)=\phi\left(g^{-1} \cdot g\right)=\phi\left(e_G\right)=e_H
$$
(by Proposition 9.5)
Thus $\phi\left(g^{-1}\right)$ possesses the defining properties of the group inverse of $\phi(g)$. Since the group inverse of any group element is unique, our result follows.

The theory of groups has become a large and essential part of modern mathematics. It is also used in physics (in quantum mechanics, for instance), in engineering, and in theoretical computer science (for example, data compression theory uses group theory).

It is a classical result of basic group theory that all finite abelian groups have been classified. Indeed, it can be shown that any such group is a product (in the sense of set theory) of cyclic groups. One of the triumphs of twentieth century mathematics is that all groups of finite order have been classified. This result is the product of the work of hundreds of mathematicians and will ultimately produce a book of several thousand pages.

问题 2.

If $n$ is a positive integer and $k$ is relatively prime to $n$ then
$$
k^{\varphi(n)}=1 \bmod n
$$

Proof: The proof of this result is easy. For the collection $\mathcal{P}(n)$ of numbers relatively prime to $n$ forms a group under multiplication. That is, if $a$ is relatively prime to $n$ and $b$ is relatively prime to $n$ then logic dictates that $a \cdot b$ is relatively prime to $n$. Now it is a fundamental fact-we cannot prove it here, but see [BMS] – that if a group has $m$ elements and $g$ is an element of the group then $g^m$ is the group identity. Thus any element of the group, raised to the power $\varphi(n)$ (the number of elements in the group) will equal 1 modulo $n$.

For later use, it is worth noting that if $p, q$ are prime numbers and $n=p \cdot q$ then $\varphi(n)=(p-1) \cdot(q-1)$.

The reason is that the only numbers less than or equal to $n$ that are not relatively prime to $n$ are $p, 2 p, 3 p, \ldots q \cdot p$ and $q, 2 q, 3 q, \cdots(p-1) q$.

There are $q$ numbers in the first list and $p-1$ numbers in the second list. The set $\mathcal{P}(n)$ of numbers relatively prime to $n$ is the complement of these two lists, and it therefore has
$$
p q-q-(p-1)=p q-q-p+1=(p-1) \cdot(q-1) \equiv \varphi(n)
$$
elements.

问题 3.

A quantity of radioactive material decays. At the beginning of each week there is half as much as there was the previous week. The initial quantity is 5 grams. Use sequence notation to express the amount of material at the beginning of the $j$ th week.

Solution: The amount of radioactive material at the beginning of the second week is $5 / 2$ (half as much as the initial amount at the beginning of the first week). The amount at the beginning of the third week is 5/4. The amount at the beginning of the fourth week is $5 / 8$. And so forth.

As a result, according to the description, the amount of material at the start of the $j$ th week is
$$
a_j=5 \cdot\left(\frac{1}{2}\right)^{j-1}
$$
The sequence exhibits in an elegant way the process of radioactive decay: the first several values are
$$
5, \frac{5}{2}, \frac{5}{4}, \frac{5}{8}
$$
It is easy to see intuitively, or with a calculator, that the amount of radioactive material tends to 0 as time tends to $\infty$.

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金融工程代写

非参数统计代写

非参数统计指的是一种统计方法，其中不假设数据来自于由少数参数决定的规定模型；这种模型的例子包括正态分布模型和线性回归模型。

广义线性模型代考

广义线性模型（GLM）归属统计学领域，是一种应用灵活的线性回归模型。该模型允许因变量的偏差分布有除了正态分布之外的其它分布。

有限元方法代写

随机分析代写

时间序列分析代写

回归分析代写

MATLAB代写

R语言代写	问卷设计与分析代写
PYTHON代写	回归分析与线性模型代写
MATLAB代写	方差分析与试验设计代写
STATA代写	机器学习/统计学习代写
SPSS代写	计量经济学代写
EVIEWS代写	时间序列分析代写
EXCEL代写	深度学习代写
SQL代写	各种数据建模与可视化代写

英国代写｜MATH0094 Market Risk and Portfolio Theory市场风险和投资组合理论｜伦敦大学学院

Posted on 2023年10月9日2023年10月9日 by statistics-lab

statistics-lab^TM为您提供伦敦大学学院（University College London，简称：UCL）Differential and Integral Calculus微分和积分微积分英国代写代考和辅导服务！

课程介绍：

Risk is an intrinsic element in financial markets. Its quantitative modelling and
understanding is a cornerstone of modern financial theory, as it is essential to
many activities like choosing investment strategies, calculating capital requirements and creating new financial products.
This module aims to study quantitatively (by using several mathematical tools
from probability, optimisation, linear algebra,…) the effects of market risk under some modelling assumptions. We will pay particular attention to the effects
associated to decision making for investors and regulators. Important aspects
related to the implementation of these concepts will be highlighted.

英国代写｜MATH0094 Market Risk and Portfolio Theory市场风险和投资组合理论｜伦敦大学学院

Attribute	Detail
Year	2022-2023
Code	MATH0094
Value	15 UCL credits (= 7.5 ECTS)
Term	1
Structure	On Campus
Assessment	In-class test with programming component (20%) and a final examination (80%)
Pre-requisites	None
Lecturers	Dr C Garcia Trillos

MATH0094 Market Risk and Portfolio Theory市场风险和投资组合理论课程内容（摘要）

C. The Simplest Notion of Equilibrium—The Standard Capital Asset Pricing Model
The most widely used model to explain asset prices is known as the standard capital
asset pricing model (CAPM). Developed independently by three well-known financial
economists, it is often referred to as the Sharpe-Lintner-Mossin capital asset pricing
model in honor of its discoverers. While alternative models that explain why expected
10
returns differ across assets have been produced in recent years (discussed in later
sections of this appendix), the standard CAPM still plays a central role in the financial
community.

Underlying Assumptions
The CAPM is a model that describes how investors should behave and how prices and
returns at which markets clear are set. Before discussing the model in more detail, it is
worthwhile specifying the assumptions underlying the CAPM.
The first assumption behind the CAPM is that there are no transaction costs.
There is no cost (friction) of buying or selling any asset. If transaction costs were
present, the expected return from owning any asset and the desirability of owning it
would be function of whether or not the investor already owned it. Thus, to include
transaction costs in the model add a great deal of complexity. While transaction costs
may be large enough to play a role in the decision-making process of the small investor
or in affecting the alternative investments that are close substitutes (as discussed in the
previous and later sections of this appendix), they are sufficiently small for enough
(large) investors that they should have only a minor effect on equilibrium prices.
The second assumption behind the CAPM is that assets are infantile divisible.
This means that investors could take any position in an investment, regardless of the
size of their wealth. For example, they can buy $1 of IBM stock. While this is not
strictly true for the small investor, given the opportunity to hold commingled funds
and mutual funds it is not an unrealistic assumption.
The third assumption is the absence of personal income tax. This means, for
example, that the individual is indifferent to the form (dividends or capital gains) in
which the return on the investment is received. This is a realistic assumption for
pension and other tax-exempt funds.
The fourth assumption is that an individual’s or institution’s purchases or sales
cannot affect the price of a stock. This is analogous to the assumption of perfect

C. 最简单的均衡概念–标准资本资产定价模型
解释资产价格最广泛使用的模型是标准资本资产定价模型（CAPM）。
资产定价模型（CAPM）。该模型由三位著名的金融
通常被称为夏普-林特纳-莫辛资本资产定价模型，以纪念其发现者。
以纪念其发现者。虽然有其他模型可以解释为什么预期
10
尽管近年来出现了解释不同资产预期收益不同原因的替代模型（将在本附录后面的章节中讨论
的讨论），但标准 CAPM 仍然在金融界发挥着核心作用。
的核心作用。

基本假设
CAPM 模型描述了投资者应如何行为，以及市场清算的价格和收益如何设定。
回报率。在对该模型进行更详细的讨论之前，有必要说明一下 CAPM 的基本假设。
CAPM 的基本假设。
CAPM 的第一个假设是不存在交易成本。
买卖任何资产都没有成本（摩擦）。如果存在交易成本
存在，那么拥有任何资产的预期收益以及拥有该资产的可取性
将是投资者是否已经拥有该资产的函数。因此
交易成本会增加模型的复杂性。虽然交易成本
虽然交易成本可能足够大，足以在小投资者的决策过程中发挥作用
或影响作为近似替代品的替代投资（如本附录前后部分所述）。
对足够多的（大）投资者来说，交易成本已经小到足以影响他们的决策。
(大）投资者而言，它们的影响足够小，因此对均衡价格的影响应该很小。
CAPM 背后的第二个假设是资产是可分割的。
这意味着投资者可以在一项投资中持有任何头寸，无论其财富大小。
他们的财富大小。例如，他们可以购买 1 美元的 IBM 股票。虽然这
虽然这对小投资者来说并非完全正确，但如果有机会持有混合基金和共同基金
和共同基金的机会，这并非不切实际的假设。
第三个假设是没有个人所得税。这意味着
例如，个人对投资收益的形式（股息或资本利得）并不关心。
投资收益的形式。对于
对于养老金和其他免税基金来说，这是一个现实的假设。
第四个假设是个人或机构的购买或出售
不会影响股票价格。这类似于完美投资假设。

知识点： financial markets in discrete time离散时间金融市场

Proposition 2.1. Constant repayments. Assuming that the loan is completely repaid at the beginning of year $N+1$, the amount $m$ refunded every year is given by
$$
m=\frac{r(1+r)^N A}{(1+r)^N-1}=\frac{r}{1-(1+r)^{-N}} A .
$$
Proof. Denoting by $A_k$ the amount owed by the borrower at the beginning of year $\mathrm{n}^o k=1,2, \ldots, N$ with $A_1=A$, the amount $m$ refunded at the end of the first year can be decomposed as
$$
m=r A_1+\left(m-r A_1\right),
$$
into $r A_1$ paid in interest and $m-r A_1$ in principal repayment, i.e. there remains
$$
\begin{aligned}
A_2 & =A_1-\left(m-r A_1\right) \
& =(1+r) A_1-m,
\end{aligned}
$$
to be refunded. Similarly, the amount $m$ refunded at the end of the second year can be decomposed as
$$
m=r A_2+\left(m-r A_2\right),
$$

into $r A_2$ paid in interest and $m-r A_2$ in principal repayment, i.e. there remains
$$
\begin{aligned}
A_3 & =A_2-\left(m-r A_2\right) \
& =(1+r) A_2-m \
& =(1+r)\left((1+r) A_1-m\right)-m \
& =(1+r)^2 A_1-m-(1+r) m
\end{aligned}
$$
to be refunded. After repeating the argument we find that at the beginning of year $k$ there remains
$$
\begin{aligned}
A_k & =(1+r)^{k-1} A_1-m-(1+r) m-\cdots-(1+r)^{k-2} m \
& =(1+r)^{k-1} A_1-m \sum_{i=0}^{k-2}(1+r)^i \
& =(1+r)^{k-1} A_1+m \frac{1-(1+r)^{k-1}}{r}
\end{aligned}
$$
to be refunded, i.e.

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金融工程代写

非参数统计代写

非参数统计指的是一种统计方法，其中不假设数据来自于由少数参数决定的规定模型；这种模型的例子包括正态分布模型和线性回归模型。

广义线性模型代考

广义线性模型（GLM）归属统计学领域，是一种应用灵活的线性回归模型。该模型允许因变量的偏差分布有除了正态分布之外的其它分布。

有限元方法代写

随机分析代写

时间序列分析代写

回归分析代写

MATLAB代写

R语言代写	问卷设计与分析代写
PYTHON代写	回归分析与线性模型代写
MATLAB代写	方差分析与试验设计代写
STATA代写	机器学习/统计学习代写
SPSS代写	计量经济学代写
EVIEWS代写	时间序列分析代写
EXCEL代写	深度学习代写
SQL代写	各种数据建模与可视化代写

PX281-15｜Computational Physics计算物理学沃里克大学

Posted on 2023年9月28日2023年10月6日 by statistics-lab

statistics-lab^TM为您沃里克大学（英语：University of Warwick）Computational Physics计算物理学英国代写代考和辅导服务！

课程介绍：

This is an indicative module outline only to give an indication of the sort of topics that may be covered. Actual sessions held may differ.

Vectorized programming in Python using Numpy
Handling, processing and analysing physics data: plotting distributions, data fitting, hypothesis testing
Monte Carlo simulation for physics modelling: Different types of random numbers, generation of random numbers according to specific distributions. Brownian motion and diffusion
Digital Signal Processing: the Fourier transform and convolution method, digital filters
Numerical solutions of ordinary differential equations: the Verlet algorithm for many coupled ODE’s
Speeding up Python: why, when and what again is a compiler

PX281-15｜Computational Physics计算物理学沃里克大学

Computational Physics计算物理学计案例

问题 1. Let $u \in L^2(\mathbb{R})$, and let $v$ be the grid function defined on $h \mathbb{Z}$ by $v_j=u\left(x_j\right)$ (Note that $h \mathbb{Z}$ has an infinite number of discrete points separated by the distance $h$, going from $-\infty$ to $+\infty$.). Then for all $k \in[-\pi / h, \pi / h]$

(a) If $u$ has $p-1$ continuous derivatives in $L^2(\mathbb{R})$, and a pth derivative of bounded value, $(p \geq 1)$ then
$$
|\hat{v}(k)-\hat{u}(k)|=O\left(h^{p+1}\right) \text { as } h \rightarrow 0 .
$$
(b) If $u$ has $\infty$ many continuous derivatives in $L^2(\mathbb{R})$, then
$$
|\hat{v}(k)-\hat{u}(k)|=O\left(h^m\right) \text { as } h \rightarrow 0 \text { for every } m>0,
$$
i.e., the expansion converges super-algebraically.

问题 2. Let $u \in L^2(\mathbb{R})$ have a vth derivative $(v \geq 1)$ of bounded variation, and let $w$ be the $v$ ‘ the spectral derivative of $u$ on the grid $h \mathbb{Z}$. Hence, for all $x \in h \mathbb{Z}$ the following holds uniformly:

(a) If $u$ has $p-1$ continuous derivatives in $L^2(\mathbb{R})$ for some $p \geq v+1$, and a pth derivative of bounded value, $(p \geq 1)$ then
$$
\left|w_j-u^{(v)}\left(x_j\right)\right|=O\left(h^{p-v}\right) \text { as } h \rightarrow 0 .
$$
(b) If $u$ has $\infty$ many continuous derivatives in $L^2(\mathbb{R})$, then
$$
\left|w_j-u^{(v)}\left(x_j\right)\right|=O\left(h^m\right) \text { as } h \rightarrow 0 \text { for every } m>0
$$

统计代写请认准statistics-lab™. statistics-lab™为您的留学生涯保驾护航。

金融工程代写

非参数统计代写

非参数统计指的是一种统计方法，其中不假设数据来自于由少数参数决定的规定模型；这种模型的例子包括正态分布模型和线性回归模型。

广义线性模型代考

广义线性模型（GLM）归属统计学领域，是一种应用灵活的线性回归模型。该模型允许因变量的偏差分布有除了正态分布之外的其它分布。

有限元方法代写

随机分析代写

时间序列分析代写

回归分析代写

MATLAB代写

R语言代写	问卷设计与分析代写
PYTHON代写	回归分析与线性模型代写
MATLAB代写	方差分析与试验设计代写
STATA代写	机器学习/统计学习代写
SPSS代写	计量经济学代写
EVIEWS代写	时间序列分析代写
EXCEL代写	深度学习代写
SQL代写	各种数据建模与可视化代写

MATH0039｜Differential and Integral Calculus微分和积分微积分伦敦大学学院

Posted on 2023年9月27日2023年10月6日 by statistics-lab

statistics-lab^TM为您提供伦敦大学学院（University College London，简称：UCL）Differential and Integral Calculus微分和积分微积分英国代写代考和辅导服务！

课程介绍：

This course provides a fairly rapid introduction to calculus. Calculus underlies almost all areas
of mathematics and a great deal of science and engineering. The aim of the course will be to
provide a solid grounding in this fundamental branch of mathematics for students who have a
limited mathematical background.

MATH0039｜Differential and Integral Calculus微分和积分微积分伦敦大学学院

Differential and Integral Calculus微分和积分微积分案例

问题 1.

On a toll road a driver takes a time stamped toll-card from the starting booth and drives directly to the end of the toll section. After paying the required toll, the driver is surprised to receive a speeding ticket along with the toll receipt. Which of the following best describes the situation?

(a) The booth attendant does not have enough information to prove that the driver was speeding.
(b) The booth attendant can prove that the driver was speeding during his trip.
(c) The driver will get a ticket for a lower speed than his actual maximum speed.
(d) Both (b) and (c).
Be prepared to justify your answer.

Answer: (d). First answer this question in the easiest form, discussing instantaneous and average velocity – then go back and try to answer again in a way that allows the application of the mean value theorem easily. The fact that (c) is also true, can be shown by using a graph. This is probably the hardest part of the problem.

问题 2.

Find two nonnegative numbers whose sum is 9 and so that the product of one number and the square of the other number is a maximum.

Let variables $x$ and $y$ represent two nonnegative numbers. The sum of the two numbers is given to be
$$
9=x+y,
$$
so that
$$
y=9-x
$$
We wish to MAXIMIZE the PRODUCT
$$
P=x y^2
$$
However, before we differentiate the right-hand side, we will write it as a function of $x$ only. Substitute for $y$ getting
$$
\begin{gathered}
P=x y^2 \
=x(9-x)^2 .
\end{gathered}
$$
Now differentiate this equation using the product rule and chain rule, getting
$$
\begin{gathered}
P^{\prime}=x(2)(9-x)(-1)+(1)(9-x)^2 \
=(9-x)[-2 x+(9-x)] \
=(9-x)[9-3 x] \
=(9-x)(3)[3-x] \
=0
\end{gathered}
$$
for
$$
x=9 \text { or } x=3 \text {. }
$$

问题 3.

2A-5 Making reasonable assumptions, if a person 5 feet tall weighs on the average $120 \mathrm{lbs}$., approximately how much does a person $5^{\prime} 1^{\prime \prime}$ tall weigh?

2A-5 A reasonable assumption is that $w$ is propotional to volume $v$, which is in turn proportional to the cube of a linear dimension, i.e., a given person remains similar to him/herself, for small weight changes.) Thus $w=C h^3$; since 5 feet $=60$ inches, we get
$$
\frac{w(60+\epsilon)}{w(60)}=\frac{C(60+\epsilon)^3}{C(60)^3}=\left(1+\frac{\epsilon}{60}\right)^3 \Rightarrow w(60+\epsilon) \approx w(60) \cdot\left(1+\frac{3 \epsilon}{60}\right) \approx 120 \cdot\left(1+\frac{1}{20}\right) \approx 126 .
$$
[Or you can calculate the linearization of $w(h)$ arround $h=60$ using derivatives, and using the value $w(60)$ to determine $C$. getting $w(h) \approx 120+6(h-60)$

统计代写请认准statistics-lab™. statistics-lab™为您的留学生涯保驾护航。

金融工程代写

非参数统计代写

非参数统计指的是一种统计方法，其中不假设数据来自于由少数参数决定的规定模型；这种模型的例子包括正态分布模型和线性回归模型。

广义线性模型代考

广义线性模型（GLM）归属统计学领域，是一种应用灵活的线性回归模型。该模型允许因变量的偏差分布有除了正态分布之外的其它分布。

有限元方法代写

随机分析代写

时间序列分析代写

回归分析代写

MATLAB代写

R语言代写	问卷设计与分析代写
PYTHON代写	回归分析与线性模型代写
MATLAB代写	方差分析与试验设计代写
STATA代写	机器学习/统计学习代写
SPSS代写	计量经济学代写
EVIEWS代写	时间序列分析代写
EXCEL代写	深度学习代写
SQL代写	各种数据建模与可视化代写

MATH0030｜Mathematical Ecology数学生态学伦敦大学学院

Posted on 2023年9月27日2023年10月6日 by statistics-lab

statistics-lab^TM为您提供伦敦大学学院（University College London，简称：UCL）Mathematical Ecology数学生态学英国代写代考和辅导服务！

课程介绍：

Mathematical models are used extensively in many areas of the Biological Sciences. This course
aims to give a sample of the construction and mathematical analysis of such models in Population Ecology. The fundamental question to be addressed is: what natural (or human) factors
control the abundance and distribution of the various populations of animals and plants that
we see in Nature?
No special knowledge of Ecology is required or assumed. However, an interest in, and willingness
to learn about, concepts and problems in this area are essential. Mathematical techniques used
include calculus, mathematical methods and linear algebra, and those developed include the
important qualitative technique of phase plane analysis which the course uses extensively.

MATH0030｜Mathematical Ecology数学生态学伦敦大学学院

Mathematical Ecology数学生态学案例

问题 1.

The general deterministic model for the dynamics of a single, unstructured population reads:
$$
\dot{N}(t)=f(N, t)=f(N(t), t)
$$
$N(t) \in \mathbf{R}$ is usually interpreted as population density rather than the total size.

The most basic model is exponential growth with
$$
\dot{N}(t)=r N(t)
$$
and explicit solution $N(t)=N_0 \exp (r t)$. The net growth rate $r$ is also called the Malthusian parameter (Thomas Malthus, 1798: Essay on the Principle of Population). Exponential growth with $r>0$ leads to population explosion and is unsustainable in Nature. With $r<0$, there is a trivial stable equilibrium at $N=0$.

The simplest model with a stable equilibrium population derives from a dynamics with immigration and death (negative growth),
$$
\dot{N}(t)=c-d N(t)
$$
with $c, d>0$ and explicit solution
$$
N(t)=\frac{c}{d}+\left(N_0-\frac{c}{d}\right) \exp [-d t] .
$$
The model describes the dynamics in a population sink. It also describes pure migration if immigration is constant and emigration occurs with a constant per-capita rate.

An alternative approach assumes that larvae (or juveniles) do not compete primarily among themselves, but with their adult conspecifics. This will be true, in particular, if adult fish eat eggs and larvae of their own species (which is indeed true for many species, where predation is purely size-dependent).

\end{prob}

We then get

$$
\frac{\partial L_t(\tau)}{\partial \tau}=-m_1 L_t(\tau)-m_2 N_t L_t(\tau)
$$
which simply results in
$$
L_t(\Delta)=b N_t \exp \left[-\left(m_1+m_2 N_t\right) \Delta\right] .
$$
We then get a so-called Ricker model for stock recruitment
$$
N_{t+1}=s N_t+N_t \exp \left[r\left(1-N_t / K\right)\right]
$$
with constants
$$
r=\log [b]-m_1 \Delta \quad, \quad K=\frac{\log [b]-m_1 \Delta}{m_2 \Delta}
$$

统计代写请认准statistics-lab™. statistics-lab™为您的留学生涯保驾护航。

金融工程代写

非参数统计代写

非参数统计指的是一种统计方法，其中不假设数据来自于由少数参数决定的规定模型；这种模型的例子包括正态分布模型和线性回归模型。

广义线性模型代考

广义线性模型（GLM）归属统计学领域，是一种应用灵活的线性回归模型。该模型允许因变量的偏差分布有除了正态分布之外的其它分布。

有限元方法代写

随机分析代写

时间序列分析代写

回归分析代写

MATLAB代写

R语言代写	问卷设计与分析代写
PYTHON代写	回归分析与线性模型代写
MATLAB代写	方差分析与试验设计代写
STATA代写	机器学习/统计学习代写
SPSS代写	计量经济学代写
EVIEWS代写	时间序列分析代写
EXCEL代写	深度学习代写
SQL代写	各种数据建模与可视化代写

数学代写Mathematics代考2023

Posted on 2023年9月26日2023年10月6日 by statistics-lab

如果你也在数学 Mathematics这个学科遇到相关的难题，请随时添加vx号联系我们的代写客服。我们会为你提供专业的服务。

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statistics-lab™ 为您的留学生涯保驾护航在代写代考数学 Mathematics方面已经树立了自己的口碑, 保证靠谱, 高质量且原创的统计Statistics和数学Math代写服务。我们的专家在代考数学 Mathematics代写方面经验极为丰富，各种代写数学 Mathematics相关的作业也就用不着说。

数学代写Mathematics代考

数学是研究数字、数量和图形的学科，是科学的一个分支。它被认为是 “算术、代数、几何、分析、微分和积分的总称”。

数学既可归类为自然科学的一种，也可归类为不属于自然科学的 “抽象或理论科学”（”the abstract or theoretical sciences”）的一种。

数学包含几个不同的主题，列举如下：

代数学algebra代写代考

代数（algebra）是数学的一个分支，主要研究用字母代替数字解方程。近代代数大大扩展了其研究范围，成为研究半群、群、环、多群（代数）、体和束（抽象代数）等代数系统的学科。代数的思想也渗透到分析、几何等领域，为数学的各个分支提供了共同语言。

下面列出的各代数分支名称中出现的半群、群、环、多群（代数）、体和束都是典型的代数结构。

群、环、代数和体的理论起源于伽罗瓦等人对代数方程解的研究，而束理论则起源于布尔等人对逻辑的数学研究。

在现代日本大学中，学生在一、二年级学习微积分的同时也学习线性代数，线性代数是代数学的一个分支，研究的代数系统称为线性空间。

几何学Geometry代写代考

几何学（古希腊语：γεωμετρία）是研究图形和空间性质的数学分支。

它最初诞生于埃及，是出于测量的需要，后来在古希腊得到了独特的发展，特别是成为研究人类可以识别的各种图形性质的数学领域，公元前 300 年左右，欧几里得将这些研究的主要成果总结为欧几里得理论。从中世纪开始，欧洲出现了以欧几里得几何学为首的各种几何学。

几何 “一词通常指处理具体平面和空间图形的几何学，如欧几里得几何学，为大众所熟悉，但几何学有许多不同类型，其对象、方法和公理体系也各不相同，在现代已发展成高度抽象的理论，如微分几何学、代数几何学和拓扑几何学。

数学分析mathematical analysis代写代考

分析（英语：analysis，mathematical analysis）是数学的一个分支，涉及极限和收敛等概念。它与代数和几何并称为数学的三大分支。

分析作为一个数学术语不同于元素还原论，通常被称为使用初等微积分和数列来研究函数变化量等性质的领域。这是由于分析最初使用泰勒级数和傅里叶级数来研究函数的性质。

例如，当函数的变量轻微移动时，函数值如何变化以及变化多少的问题被视为一个分析问题。

其他相关科目课程代写：

statistics统计学
set theory集合论

数学Mathematics历史

数学的起源在很大程度上与人类农业的开始有关”。数学的前三个必要条件是：用于作物分配管理和商业交易的计算，用于农田管理的测量，以及用于确定农活时间的历法的天文现象周期性的阐明。可以说，这三种需求大致对应于数学的三大范畴：结构、空间和变化的研究。例如，根据土木工程的经验，人们知道边长比为 3 : 4 : 5 的三角形是直角三角形，但一般人并不知道直角三角形的边长比是 c2 = a2 + b2（c、b、a 是边长）（毕达哥拉斯定理）。在数学纯粹是实用数学而不是一门独立学科的时代，把这些关系当作自然科学中的数据来处理，并通过列举大量实例来证明它们的正确性，是不成问题的。然而，由于数的数量是无限的，因此不可能通过研究大量的数来证明一个完整的陈述。自从数学作为一门学科被研究以来，利用逻辑判断真假的 “数学证明 “就发展起来了。数学证明在现代数学中也受到高度重视。

The origin of mathematics is to a large extent connected with the beginning of human agriculture”. The first three requisites of mathematics were: calculations for the management of crop distribution and commercial transactions, measurements for the management of agricultural land, and the elucidation of the periodicity of astronomical phenomena for the calendars used in determining the time of agricultural activity. It can be argued that these three imperatives correspond roughly to the three main areas of mathematics: the study of structure, space and change. For example, from experience in civil engineering, it is known that a triangle with a side ratio of 3 : 4 : 5 is a right-angled triangle, but it is not generally known that the side ratio of a right-angled triangle is c2 = a2 + b2 (with c, b, and a being the side ratios) (Pythagoras’ theorem). In the days when mathematics was purely practical rather than a separate discipline, it was not a problem to treat these relationships as if they were data in the natural sciences and to prove them correct by citing a large number of examples. However, since the number of numbers is infinite, it is impossible to prove a complete statement by studying a large number of numbers. Since mathematics has been studied as a discipline, “mathematical proofs”, which use logic to determine truth or falsehood, have been developed. Mathematical proofs are also highly valued in modern mathematics.

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数学Mathematics的相关课后作业范例

这是一篇关于数学Mathematics的作业

问题 1.

Let $u$ and $v$ be arbitrary vectors of an inner product space. Then
$$
|(\boldsymbol{u}, \boldsymbol{v})| \leq(\boldsymbol{u}, \boldsymbol{u})^{\frac{1}{2}}(\boldsymbol{v}, \boldsymbol{v})^{\frac{1}{2}}
$$

PROOF If $v=0$, the inequality is obviously satisfied. Suppose $v \neq 0$. Then for an arbitrary scalar $\alpha \in \mathbb{C}(\mathbb{R})$
$$
0 \leq(\boldsymbol{u}-\alpha \boldsymbol{v}, \boldsymbol{u}-\alpha \boldsymbol{v})=(\boldsymbol{u}, \boldsymbol{u})-\alpha(\boldsymbol{v}, \boldsymbol{u})-\bar{\alpha}(\boldsymbol{u}, \boldsymbol{v})+\alpha \bar{\alpha}(\boldsymbol{v}, \boldsymbol{v})
$$
Take $\alpha=\overline{(\boldsymbol{v}, \boldsymbol{u})} /(\boldsymbol{v}, \boldsymbol{v})$. Then $\bar{\alpha}=\overline{(\boldsymbol{u}, \boldsymbol{v})} /(\boldsymbol{v}, \boldsymbol{v})$ and
$$
(\boldsymbol{u}, \boldsymbol{u})-\frac{|(\boldsymbol{v}, \boldsymbol{u})|^2}{(\boldsymbol{v}, \boldsymbol{v})}-\frac{|(\boldsymbol{u}, \boldsymbol{v})|^2}{(\boldsymbol{v}, \boldsymbol{v})}+\frac{|(\boldsymbol{u}, \boldsymbol{v})|^2}{(\boldsymbol{v}, \boldsymbol{v})} \geq 0
$$
or
$$
(\boldsymbol{u}, \boldsymbol{u})(\boldsymbol{v}, \boldsymbol{v})-|(\boldsymbol{u}, \boldsymbol{v})|^2 \geq 0
$$
from which the assertion follows.
The Cauchy-Schwarz inequality is a useful tool in many proofs in analysis. For brevity, we shall follow common practice and refer to it as simply the Schwarz inequality.

最后的总结：

通过对数学Mathematics各方面的介绍，想必您对这门学科有了初步的认识。如果你仍然不确定或对这方面感到困难，你仍然可以依靠我们的代写和辅导服务。我们拥有各个领域、具有丰富经验的专家。他们将保证你的 essay、assignment或者作业都完全符合要求、100%原创、无抄袭、并一定能获得高分。需要如何学术帮助的话，随时联系我们的客服。

R语言代写	问卷设计与分析代写
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MATLAB代写	方差分析与试验设计代写
STATA代写	机器学习/统计学习代写
SPSS代写	计量经济学代写
EVIEWS代写	时间序列分析代写
EXCEL代写	深度学习代写
SQL代写	各种数据建模与可视化代写

STAT0008｜Statistical Inference统计推断伦敦大学学院

Posted on 2023年9月25日2023年10月6日 by statistics-lab

statistics-lab^TM为您提供伦敦大学学院（University College London，简称：UCL）Statistical Inference统计推理英国代写代考和辅导服务！

课程介绍：

STAT0008 is specified as a formal option for fourth year undergraduates from the Department of Mathematics. Any such student who has taken the standard prerequisite should simply register for STAT0008 on Portico and await a decision. Mathematics students who do not satisfy the standard prerequisites must additionally consult a member of staff in the Department of Statistical Science.

STAT0008｜Statistical Inference统计推理伦敦大学学院

Statistical Inference统计推断案例

问题 1.

(Binomial Theorem) For any real numbers $x$ and $y$ and integer $n \geq 0$,
$$
(x+y)^n=\sum_{i=0}^n\left(\begin{array}{l}
n \
i
\end{array}\right) x^i y^{n-i} .
$$

Proof: Write
$$
(x+y)^n=(x+y)(x+y) \cdots(x+y),
$$
and consider how the right-hand side would be calculated. From each factor $(x+y)$ we choose either an $x$ or $y$, and multiply together the $n$ choices. For each $i=0,1, \ldots, n$, the number of such terms in which $x$ appears exactly $i$ times is $\left(\begin{array}{l}n \ i\end{array}\right)$. Therefore, this term is of the form $\left(\begin{array}{c}n \ i\end{array}\right) x^i y^{n-i}$ and the result follows.
If we take $x=p$ and $y=1-p$ in (3.2.4), we get
$$
1=(p+(1-p))^n=\sum_{i=0}^n\left(\begin{array}{c}
n \
i
\end{array}\right) p^i(1-p)^{n-i},
$$
and we see that each term in the sum is a binomial probability. As another special case, take $x=y=1$ in Theorem 3.2.2 and get the identity
$$
2^n=\sum_{i=0}^n\left(\begin{array}{l}
n \
i
\end{array}\right) \text {. }
$$

The mean and variance of the binomial distribution have already been derived in Examples 2.2.3 and 2.3.5, so we will not repeat the derivations here. For completeness, we state them. If $X \sim \operatorname{binomial}(n, p)$, then
$$
\mathrm{E} X=n p, \quad \operatorname{Var} X=n p(1-p) .
$$
The mg f of the binomial distribution was calculated in Example 2.3.9. It is
$$
M_X(t)=\left[p e^t+(1-p)\right]^n
$$

问题 2.

Let $f(x)$ be any pdf and let $\mu$ and $\sigma>0$ be any given constants. Then the function
$$
g(x \mid \mu, \sigma)=\frac{1}{\sigma} f\left(\frac{x-\mu}{\sigma}\right)
$$
is a $p d f$.

Proof: To verify that the transformation has produced a legitimate pdf, we need to check that $(1 / \sigma) f((x-\mu) / \sigma)$, as a function of $x$, is a pdf for every value of $\mu$ and $\sigma$ we might substitute into the formula. That is, we must check that $(1 / \sigma) f((x-\mu) / \sigma)$ is nonnegative and integrates to 1 . Since $f(x)$ is a pdf, $f(x) \geq 0$ for all values of $x$. So, $(1 / \sigma) f((x-\mu) / \sigma) \geq 0$ for all values of $x, \mu$, and $\sigma$. Next we note that
$$
\begin{aligned}
\int_{-\infty}^{\infty} \frac{1}{\sigma} f\left(\frac{x-\mu}{\sigma}\right) d x & =\int_{-\infty}^{\infty} f(y) d y & \left(\text { substitute } y=\frac{x-\mu}{\sigma}\right) \
& =1, & \text { (since } f(y) \text { is a pdf) }
\end{aligned}
$$
as was to be verified.
We now turn to the first of our constructions, that of location families.

问题 3.

Let $f(\cdot)$ be any pdf. Let $\mu$ be any real number, and let $\sigma$ be any positive real number. Then $X$ is a random variable with $p d f(1 / \sigma) f((x-\mu) / \sigma)$ if and only if there exists a random variable $Z$ with pdf $f(z)$ and $X=\sigma Z+\mu$.

Proof: To prove the “if” part, define $g(z)=\sigma z+\mu$. Then $X=g(Z), g$ is a monotone function, $g^{-1}(x)=(x-\mu) / \sigma$, and $\left|(d / d x) g^{-1}(x)\right|=1 / \sigma$. Thus by Theorem 2.1.5, the pdf of $X$ is
$$
f_X(x)=f_Z\left(g^{-1}(x)\right)\left|\frac{d}{d x} g^{-1}(x)\right|=f\left(\frac{x-\mu}{\sigma}\right) \frac{1}{\sigma} .
$$
To prove the “only if” part, define $g(x)=(x-\mu) / \sigma$ and let $Z=g(X)$. Theorem 2.1.5 again applies: $g^{-1}(z)=\sigma z+\mu,\left|(d / d z) g^{-1}(z)\right|=\sigma$, and the pdf of $Z$ is
$$
f_Z(z)=f_X\left(g^{-1}(z)\right)\left|\frac{d}{d z} g^{-1}(z)\right|=\frac{1}{\sigma} f\left(\frac{(\sigma z+\mu)-\mu}{\sigma}\right) \sigma=f(z) .
$$
Also,
$$
\sigma Z+\mu=\sigma g(X)+\mu=\sigma\left(\frac{X-\mu}{\sigma}\right)+\mu=X .
$$

An important fact to extract from Theorem 3.5 .6 is that the random variable $Z=(X-\mu) / \sigma$ has pdf
$$
f_Z(z)=\frac{1}{1} f\left(\frac{z-0}{1}\right)=f(z) .
$$
That is, the distribution of $Z$ is that member of the location-scale family corresponding to $\mu=0, \sigma=1$. This was already proved for the special case of the normal family in Section 3.3.

Often, calculations can be carried out for the “standard” random variable $Z$ with pdf $f(z)$ and then the corresponding result for the random variable $X$ with pdf $(1 / \sigma) f((x-\mu) / \sigma)$ can be easily derived. An example is given in the following, which is a generalization of a computation done in Section 3.3 for the normal family.

STAT0008｜Statistical Inference统计推断伦敦大学学院

统计代写请认准statistics-lab™. statistics-lab™为您的留学生涯保驾护航。

金融工程代写

非参数统计代写

非参数统计指的是一种统计方法，其中不假设数据来自于由少数参数决定的规定模型；这种模型的例子包括正态分布模型和线性回归模型。

广义线性模型代考

广义线性模型（GLM）归属统计学领域，是一种应用灵活的线性回归模型。该模型允许因变量的偏差分布有除了正态分布之外的其它分布。

有限元方法代写

随机分析代写

时间序列分析代写

回归分析代写

MATLAB代写

R语言代写	问卷设计与分析代写
PYTHON代写	回归分析与线性模型代写
MATLAB代写	方差分析与试验设计代写
STATA代写	机器学习/统计学习代写
SPSS代写	计量经济学代写
EVIEWS代写	时间序列分析代写
EXCEL代写	深度学习代写
SQL代写	各种数据建模与可视化代写

ENG2005｜Advanced engineering mathematics高级工程数学伦敦大学学院

Posted on 2023年9月25日2023年10月6日 by statistics-lab

statistics-lab^TM为您提供伦敦大学学院（University College London，简称：UCL）ENG2005高级工程数学英国代写代考和辅导服务！

课程介绍：

This unit introduces advanced mathematical concepts required for all disciplines of engineering that extend on the concepts in ENG1005 Engineering mathematics.

Major concepts taught using engineering contexts include multivariable and vector calculus, Fourier series and partial differential equations. Specific topics include partial derivatives, double and triple integrals with change of coordinates, vector functions and operators, parametric curves and surfaces, periodic functions, Laplace transforms, ODE boundary value problems, partial differential equation solution techniques and applications.

ENG2005｜Advanced engineering mathematics高级工程数学伦敦大学学院

Advanced engineering mathematics高级工程数学案例

问题 1.

Establish the following results:
(a) $\operatorname{Re}\left(z_1+z_2\right)=\operatorname{Re}\left(z_1\right)+\operatorname{Re}\left(z_2\right)$, but $\operatorname{Re}\left(z_1 z_2\right) \neq \operatorname{Re}\left(z_1\right) \operatorname{Re}\left(z_2\right)$ in general;

Solution. (a) We want to show that $\operatorname{Re}\left(z_1+z_2\right)=\operatorname{Re}\left(z_1\right)+\operatorname{Re}\left(z_2\right)$. Let $z_1=a_1+i b_1$, $z_2=a_2+i b_2$, then
$$
z_1+z_2=\left(a_1+a_2\right)+i\left(b_1+b_2\right) \text {. }
$$
Hence
$$
\operatorname{Re}\left(z_1+z_2\right)=a_1+a_2
$$
and clearly
$$
\operatorname{Re}\left(z_1\right)+\operatorname{Re}\left(z_2\right)=a_1+a_2 .
$$
Let us show that in general $\operatorname{Re}\left(z_1 z_2\right) \neq \operatorname{Re}\left(z_1\right) \operatorname{Re}\left(z_2\right)$. We have
$$
z_1 z_2=\left(a_1+i b_1\right)\left(a_2+i b_2\right)=\left(a_1 a_2-b_1 b_2\right)+i\left(a_1 b_2+a_2 b_1\right),
$$
therefore
$$
\operatorname{Re}\left(z_1 z_2\right)=a_1 a_2-b_1 b_2 .
$$
On the other hand
$$
\operatorname{Re}\left(z_1\right) \operatorname{Re}\left(z_2\right)=a_1 a_2 \neq a_1 a_2-b_1 b_2
$$
in general.

\end{prob}

问题 2.

(b) $\operatorname{Im}\left(z_1+z_2\right)=\operatorname{Im}\left(z_1\right)+\operatorname{Im}\left(z_2\right)$, but $\operatorname{Im}\left(z_1 z_2\right) \neq \operatorname{Im}\left(z_1\right) \operatorname{Im}\left(z_2\right)$ in general;

(b) We want to show that $\operatorname{Im}\left(z_1+z_2\right)=\operatorname{Im}\left(z_1\right)+\operatorname{Im}\left(z_2\right)$. From part (a) we have that
$$
z_1+z_2=\left(a_1+a_2\right)+i\left(b_1+b_2\right) .
$$
Hence
$$
\operatorname{Im}\left(z_1+z_2\right)=b_1+b_2
$$
and clearly
$$
\operatorname{Im}\left(z_1\right)+\operatorname{Im}\left(z_2\right)=b_1+b_2 .
$$
Let us show that in general $\operatorname{Im}\left(z_1 z_2\right) \neq \operatorname{Im}\left(z_1\right) \operatorname{Im}\left(z_2\right)$. We have
$$
z_1 z_2=\left(a_1+i b_1\right)\left(a_2+i b_2\right)=\left(a_1 a_2-b_1 b_2\right)+i\left(a_1 b_2+a_2 b_1\right),
$$
therefore
$$
\operatorname{Im}\left(z_1 z_2\right)=a_1 b_2+a_2 b_1 .
$$

问题 3.

(c) We want to show that $\left|z_1 z_2\right|=\left|z_1\right|\left|z_2\right|$. From part (a) we have
$$
z_1 z_2=\left(a_1+i b_1\right)\left(a_2+i b_2\right)=\left(a_1 a_2-b_1 b_2\right)+i\left(a_1 b_2+a_2 b_1\right) .
$$
Hence
$$
\begin{aligned}
\left|z_1 z_2\right| & =\sqrt{\left(a_1 a_2-b_1 b_2\right)^2+\left(a_1 b_2+a_2 b_1\right)^2} \
& =\sqrt{\left(a_1^2 a_2^2-2 a_1 a_2 b_1 b_2+b_1^2 b_2^2\right)+\left(a_1^2 b_2^2+2 a_1 b_2 a_2 b_1+a_2^2 b_1^2\right)} \
& =\sqrt{a_1^2 a_2^2+b_1^2 b_2^2+a_1^2 b_2^2+a_2^2 b_1^2} .
\end{aligned}
$$
On the other hand $\left|z_1\right|=\sqrt{a_1^2+b_1^2}$ and $\left|z_2\right|=\sqrt{a_2^2+b_2^2}$. Therefore
$$
\left|z_1\right|\left|z_2\right|=\left(\sqrt{a_1^2+b_1^2}\right)\left(\sqrt{a_2^2+b_2^2}\right)=\sqrt{\left(a_1^2+b_1^2\right)\left(a_2^2+b_2^2\right)}=\sqrt{a_1^2 a_2^2+b_1^2 b_2^2+a_1^2 b_2^2+a_2^2 b_1^2}
$$
which is equal to $\left|z_1 z_2\right|$.
Let us show that $\left|z_1+z_2\right| \neq\left|z_1\right|+\left|z_2\right|$ in general. From part (a) we have
$$
z_1+z_2=\left(a_1+a_2\right)+i\left(b_1+b_2\right) \text {. }
$$
Hence
$$
\left|z_1+z_2\right|=\sqrt{\left(a_1+a_2\right)^2+\left(b_1+b_2\right)^2} .
$$
On the other hand
$$
\left|z_1\right|+\left|z_2\right|=\sqrt{\left(a_1+a_2\right)^2+\left(b_1+b_2\right)^2}
$$
and
$$
\sqrt{\left(a_1+a_2\right)^2+\left(b_1+b_2\right)^2} \neq \sqrt{\left(a_1+a_2\right)^2+\left(b_1+b_2\right)^2}
$$
in general (choose for example $a_1=1, b_1=0, a_2=0, b_2=1$ ).
(d) We want to show that $\overline{z_1+z_2}=\overline{z_1}+\overline{z_2}$. From part (a) we have
$$
z_1+z_2=\left(a_1+a_2\right)+i\left(b_1+b_2\right)
$$
then
$$
\overline{z_1+z_2}=\left(a_1+a_2\right)-i\left(b_1+b_2\right) .
$$
On the other hand
$$
\overline{z_1}+\overline{z_2}=\left(a_1-i b_1\right)+\left(a_2-i b_2\right)=\left(a_1+a_2\right)-i\left(b_1+b_2\right)
$$
which is equal to $\overline{z_1+z_2}$.
Let us show that $\overline{z_1 z_2}=\overline{z_1 z_2}$. From part (a) we have that
$$
z_1 z_2=\left(a_1 a_2-b_1 b_2\right)+i\left(a_1 b_2+a_2 b_1\right) .
$$

Hence
$$
\overline{z_1 z_2}=\left(a_1 a_2-b_1 b_2\right)-i\left(a_1 b_2+a_2 b_1\right) .
$$
On the other hand,
$$
\overline{z_1 z_2}=\left(a_1-i b_1\right)\left(a_2-i b_2\right)=\left(a_1 a_2-b_1 b_2\right)-i\left(a_1 b_2+a_2 b_1\right)
$$
which is equal to $\overline{z_1 z_2}$.

统计代写请认准statistics-lab™. statistics-lab™为您的留学生涯保驾护航。

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