国际经济学代写 - 统计代写答疑辅导

分类：国际经济学代写

STAT0021｜Introductory Statistical Methods and Computing 统计方法和计算入门伦敦大学学院

Posted on 2023年9月22日2023年10月6日 by statistics-lab

statistics-lab^TM为您提供澳洲国立大学Australian National University STAT0021 Introductory Statistical Methods and Computing 统计方法英国代写代考和辅导服务！

课程介绍：

This module aims to provide an introduction to statistical methods and interpretation of data, along with associated computing, and to provide some expertise in applying quantitative methods in the Life and Physical Sciences. The statistical methods covered are useful in the routine analysis of scientific methods, as might be encountered in other modules.

STAT0021｜Introductory Statistical Methods and Computing 统计方法和计算入门伦敦大学学院

Introductory Statistical Methods and Computing 统计方法和计算入门作业案例

问题 1.

Suppose an economist is organizing a survey of American minimum wage workers, and is interested in understanding how many workers that earn the minimum wage are teenagers. ${ }^1$ Suppose further that one out of every four minimum wage workers is a teenager. If the economist finds 80 minimum wage workers for his survey, what’s the probability that he interviews exactly 14 teenagers? 35 teenagers? What’s the probability that he gets at least 5 teenagers in his survey?

Solution to 2: In this case, we have $p=0.25$ again and $n=80$. We again use the binomial formula to obtain the solution to the first question:
$$
P(K=14)=\left(\begin{array}{c}
80 \
14
\end{array}\right) \cdot 25^{14}(1-.25)^{80-14}=0.0319 .
$$
The second uses the same formula:
$$
P(K=35)=\left(\begin{array}{l}
80 \
35
\end{array}\right) \cdot 25^{35}(1-.25)^{80-35}=0.00011704
$$

问题 2.

Suppose you flip a weighted coin (probability of heads is $p$ and probability of tails is $q=1-p$ ) $n$ times.

What is the probability that you get a particular ordering of $k$ heads and $n-k$ tails?

Solution to 2: From part 1, the probability of getting any particular ordering of $k$ heads and $n-k$ tails would be $p^k q^{n-k}$, but we have to take into account the number of permutations that exist for getting $k$ heads and $n-k$ tails. We can think about this as the “ALGEBRA” and “CALCULUS” problem, by simply labeling the heads as ” $\mathrm{H}$ ” and tails as “T” and then figure out how many different words we can write. This is pretty simple: $\frac{n !}{k !(n-k) !}$ which just so happens to be the same as the binomial coefficient, $\left(\begin{array}{l}n \ k\end{array}\right)$. So, it turns out that we just have the binomial formula for this problem:
$$
P(k \text { heads })=\left(\begin{array}{c}
n \
k
\end{array}\right) p^k q^{n-k} .
$$

问题 3.

Let $X \equiv$ the number of heads in $n$ flips. What is the probability density function of $X$ ?

Solution to 3 : Well, I just have to rewrite the answer to 2 slightly differently to change the notation to refer to the random variable, $X$ and its PDF, $f(x ; n)$ :
$$
P(X=x)=f(x ; n)=\left(\begin{array}{c}
n \
x
\end{array}\right) p^x(1-p)^{n-x} .
$$

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

Econ8009｜International Monetary Economics 国际货币经济学澳洲国立大学

Posted on 2023年9月22日2023年10月6日 by statistics-lab

statistics-lab^TM为您提供澳洲国立大学Australian National University Econ8009 International Monetary Economics 国际货币经济学澳洲代写代考和辅导服务！

课程介绍：

This course will provide students with models that can be used to analyse issues in international monetary economics. The course introduces students to so-called global imbalances and the basic stylized facts about current accounts with cross-section and the time-series of selected countries such as the US and China. Issues of current account sustainability will also be addressed. The underlying intertemporal modelling framework will be applied to the current account and be used to understand the macroeconomic effects of external shocks, such as changes in terms of trade and the world interest rate.

Econ8009｜International Monetary Economics 国际货币经济学澳洲国立大学

International Monetary Economics 国际货币经济学作业案例

问题 1.

A delegation of three is to be chosen from the untenured faculty of the MIT Economics Department (numbering ten) to represent the department in an Institute-wide committee. In how many ways
a) can the delegation be chosen?

Solution to (a): Since there are 10 untenured faculty members, there are “10 choose 3” $\left(\begin{array}{c}10 \ 3\end{array}\right)=\frac{10 !}{3 ! 7 !}=\frac{10 \cdot 9 \cdot 8}{3 \cdot 2 \cdot 1}=120$ different ways to choose the three faculty members. Further, if the three faculty members are selected to unique positions such as the President, VicePresident, and Secretary of the delegation, then there would be $\frac{10 !}{7 !}=10 \cdot 9 \cdot 8=720$ total ways since each of the 120 different combinations of faculty members can be mixed into the 3 positions in 6 different ways.

问题 2.

Does a monkey have a better chance of rearranging “ACCLLUUS” to spell “CALCULUS,”
or of rearranging “AABEGLR” to spell “ALGEBRA?” (2 points.)

Solution: How many letters is “CALCULUS”? Eight. How many letters is “ALGEBRA”? Seven. How many unique permutations of the letters “ACCLLUUS” are there? We have $\frac{8 !}{2 ! 212 !}=7$ ! different permutations since there are two C’s, L’s, and U’s. How many unique permutations of the letters “AABEGLR” are there? We start out with at most 7!, but we have duplicate A’s, so we know that there must be $7 ! / 2$ ! different permutations, which is less than 7 !. Thus, we know that “ALGEBRA” is twice as likely to come up randomly on a monkey’s Shakespearean typewriter.

问题 3.

can it be chosen, if two people must be chosen from MIT assistant faculty (6 professors) and one person must be chosen from visiting assistant faculty

Solution to (d): This should be easy if you’ve already figured out parts (a)-(c), not because you’ve computed the answer yet, but because the methods are similar. We just break the problem into its two parts: choosing MIT assistant faculty and choosing visiting assistant faculty. This give us $\underbrace{\frac{6 \cdot 5}{2}}{\text {MIT }} \cdot \underbrace{4}{\text {Visiting }}=60$. Does this answer make sense? It is less than if there were just two people who hated each other, as this is equivalent to the problem where four people refuse to go together, but we have to have one of them. So, we would expect the count to be less than what we found in (b).

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

STAT0013｜Discrete Mathematics 离散数学伦敦大学学院

Posted on 2023年9月22日2023年10月6日 by statistics-lab

statistics-lab^TM为您提供伦敦大学学院 London’s Global University STAT0013 Mathematical Analysis数学分析英国代写代考和辅导服务！

课程介绍：

STAT0013 is specified as a formal option for third and fourth year undergraduates from the Department of Mathematics. Any such student who has taken one of the standard prerequisites (simultaneous attendance of STAT0005 is also acceptable) should simply register for STAT0013 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.

STAT0013｜Discrete Mathematics (Advanced)进阶离散数学伦敦大学学院

Discrete Mathematics 离散数学作业案例

问题 1.

Let
$$
B=\left(\begin{array}{rrr}
-7 & 4 & -3 \
2 & -5 & 3 \
1 & -3 & -2
\end{array}\right)
$$
Calculate det $B$.

Solution: We calculate that
$$
\begin{aligned}
\operatorname{det} B= & (-7) \cdot \operatorname{det}\left(\begin{array}{rr}
-5 & 3 \
-3 & -2
\end{array}\right) \
& -4 \cdot \operatorname{det}\left(\begin{array}{rr}
2 & 3 \
1 & -2
\end{array}\right) \
& +(-3) \cdot \operatorname{det}\left(\begin{array}{rr}
2 & -5 \
1 & -3
\end{array}\right) \
= & (-7) \cdot 19-4 \cdot(-7)+(-3) \cdot(-1) \
= & -102 \
\neq & 0 .
\end{aligned}
$$
It is worth summarizing the main rule for inverses for which we have given an indication of the justification:

Rule for inverses: A square matrix has an inverse if and only if the matrix has nonzero determinant.

This still does not tell us how to find the inverse, but we shall get to that momentarily.
There are in fact a number of ways to calculate the inverse of a matrix, and we shall indicate two of them here. The first is the method of Gaussian elimination, which is a powerful technique that can be used for many purposes. The idea is to take the given square matrix
$$
A=\left(\begin{array}{ll}
a & b \
c & d
\end{array}\right)
$$
and to augment it by adjoining the identity matrix as in the display below:
$$
\left(\begin{array}{ll|ll}
a & b & 1 & 0 \
c & d & 0 & 1
\end{array}\right)
$$
Now the method of gaussian elimination allows us to perform certain operations on the rows of this augmented matrix, the goal being to reduce the square matrix on the left to the identity matrix. Whatever matrix results on the right will be the inverse.

问题 2.

The multiplicative identity for a group is unique.

Proof: Let $G$ be a group. Let $e$ and $e^{\prime}$ both be elements of $G$ that satisfy Axiom 2 . Then
$$
e=e \cdot e^{\prime}=e^{\prime}
$$
Thus $e$ and $e^{\prime}$ must be the same group element.

STAT0013｜Discrete Mathematics 离散数学伦敦大学学院

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

Math1904｜Discrete Mathematics (Advanced)进阶离散数学悉尼大学

Posted on 2023年9月22日2023年10月6日 by statistics-lab

statistics-lab^TM为您提供悉尼大学University of Sydney Math1904 Mathematical Analysis数学分析澳洲代写代考和辅导服务！

课程介绍：

This unit of study provides an introduction to programming and numerical methods. Topics covered include computer arithmetic and computational errors, systems of linear equations, interpolation and approximation, solution of nonlinear equations, quadrature, initial value problems for ordinary differential equations and boundary value problems, and optimisation.

Math1904｜Discrete Mathematics (Advanced)进阶离散数学悉尼大学

Discrete Mathematics (Advanced)进阶离散数学作业案例

问题 1.

Let $g:[1,+\infty) \rightarrow[8,+\infty)$ be given by $g(x)=x^2+3 x+4$. Then $g(1)=8$, and $g$ is strictly increasing without bound. So $g$ is one-to-one and onto. We conclude that $g$ has an inverse. One may use the quadratic formula to solve for that inverse, and find that $g^{-1}(x)=(-3+\sqrt{-7+4 x}) / 2$.

The idea of inverse function lends itself particularly well to the notation of ordered pairs. For $f: S \rightarrow T$ is inverse to $g: T \rightarrow S$ (and vice versa) provided that for every ordered pair $(s, t) \in f$ there is an ordered pair $(t, s) \in g$ and conversely.
Not every function has an inverse. For instance, let $f: S \rightarrow T$. Suppose that $f(s)=t$ and also that $f\left(s^{\prime}\right)=t$ with $s \neq s^{\prime}$ (in other words, suppose that $f$ is not one-to-one). If $g: T \rightarrow S$ then $g(f(s))=g(t)=g\left(f\left(s^{\prime}\right)\right)$ so it cannot be that both $g(f(s))=s$ and $g\left(f\left(s^{\prime}\right)\right)=s^{\prime}$. In other words, $f$ cannot have an irver se. We conclude that a function that does have an inverse must be one-to-one.

On the other hand, suppose that $t \in T$ has the property that there is no $s \in S$ with $f(s)=t$ (in other words, suppose that $f$ is not onto). Then, in particular, it could not be that $f(g(t))=t$ for any function $g: T \rightarrow S$. So $f$ could not be invertible. We conclude that a function that does have an inverse must be onto.

问题 2.

For each $m$, the set $\mathcal{Q}(m)$ is well ordered.

Proof: The proof is by induction on $m$.
When $m=1$ there is nothing to prove.
Assume that the assertion has been proved for $m=k$. Now let $U$ be a subset of $\mathcal{Q}\left(k^{\prime}\right)$. There are now three possibilities:

If in fact $U \subset \mathcal{Q}(k)$ then $U$ has a least element by the inductive hypothesis.
If $U={k}$ then $U$ has but one element and that element, namely $k$, is the least element that we seek.
The last possibility is that $U$ contains $k$ and some other natural numbers as well. But then $U \backslash{k} \subset \mathcal{Q}(k)$. Hence $U \backslash{k}$ has a least element $s$ by the inductive hypothesis. Since $s$ is automatically less than $k$, it follows that $s$ is a least element for the entire set $U$.
Now we have all our tools in place and we can prove the full result:

问题 3.

The natural numbers $\mathrm{N}$ are well ordered.

Proof: Let $\emptyset \neq S \subset \mathbb{N}$. Select an element $m \in S$. There are now two possibilities:

If $\mathcal{Q}(m) \cap S=\emptyset$ then $m$ is the least element of $S$ that we seek.
If $T=\mathcal{Q}(m) \cap S \neq \emptyset$ then notice that if $x \in T$ and $y \in S \backslash T$ then $x<y$. So it suffices for us to find a least element of $T$. But such an element exists by applying the preceding proposition to $U=\mathcal{Q}(m) \cap S$.
The proof is complete.

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

Math3076｜Mathematical Computing数学计算悉尼大学

Posted on 2023年9月22日2023年10月6日 by statistics-lab

statistics-lab^TM为您提供悉尼大学University of Sydney MATH0048 Mathematical Analysis数学分析澳洲代写代考和辅导服务！

课程介绍：

Mathematical Computing数学计算作业案例

问题 1.

(b) (0 points) Are these system specifications consistent? _________Prove it!

Solution. There are several ways to approach this problem. One is to construct a truth table with sixteen lines-one for each way of assigning truth values to the four variables $M, N, K$, and $I$.

We can avoid the cumbersome truthtable if we reason by cases. Case 1: $I$ is true. Then the last formula (9) is false, and the whole specification is false. Case 2: $I$ is false. Now formula (8) can be true only if $\neg M$ is false, that is, only if $M$ is true. Likewise, formula (7) can be true only if $\neg K$ is true, that is, $K$ is false.

Since $K$ is false, formula (6) can be true only if $N$ is false. Thus, we have deduced that in order to be consistent in this case, we must have
$$
\begin{aligned}
I & =\text { false } \
M & =\text { true } \
K & =\text { false } \
N & =\text { false }
\end{aligned}
$$
But now formula (5) is false, so it is impossible for all the formulas to be true: the system is inconsistent.

问题 2.

Problem 2 (20 points). For each of the following logical formulas with domain of discourse the natural numbers, $\mathbb{N}$, indicate whether it is a possible formulation of
I: the Induction Axiom,
S: the Strong Induction Axiom,
L: the Least Number Principle (also known as Well-ordering), or
$\mathrm{N}$ : None of these.
For example, the ordinary Induction Axiom could be expressed by the following formula, so it gets labelled “I”.
$$
(P(0) \wedge[\forall k P(k) \longrightarrow P(k+1)]) \longrightarrow \forall k P(k) \quad \longrightarrow
$$
This is a multiple choice problem: do not explain your answer.

(a) (0 points) $(P(b) \wedge[\forall k \geq b P(k) \longrightarrow P(k+1)]) \longrightarrow \forall k \geq b P(k)$

Solution. I. This is a perfect formulation of the Induction Axiom. $b$ is used for the base case; $P(k) \longrightarrow P(k+1)$ is the inductive case.
(b)$(P(b) \wedge[\forall k \leq b P(k) \longrightarrow P(k+1)]) \longrightarrow \forall k \leq b P(k)$
Solution. N. The two occurences of $k<=b$ should have been $k>=b$
(c) $[\forall b(\forall k<b P(k)) \longrightarrow P(b)] \longrightarrow \forall k P(k)$
Solution. S. Since you are assuming $P(k)$ for all $k<b$, this is strong induction.
(d) $\quad(\exists n P(n)) \longrightarrow \exists n \forall k<n \overline{P(k)}$
Solution. N. This statement is in fact always true; when $n=0, \forall k<n \overline{P(k)}$. It should say $P(n) \wedge \forall k<n ; \overline{P(k)}$

问题 3.

Problem :To encourage collaborative study, the 6.042 staff is considering assigning each student to a study group with two or three other students. Prove that as long as the enrollment is large enough, the class can always be divided into such study groups.

Solution. Proof. The proof is by strong induction. The induction hypothesis is that a class with $n \geq 6$ students can be divided into teams of 3 or 4 . More precisely
$$
P(n)::=n \geq 6 \longrightarrow \exists x, y \in \mathbb{N} 3 x+4 y=n .
$$
For any $n \geq 0$, we may assume $P(6), \ldots, P(n-1)$ to prove $P(n)$.
Case 1: $(n<6) . P(n)$ holds because the hypothesis $n \geq 6$ is false.
Case 2: $(n=6,7$, or 8$) P(6)$ is true because there could be two teams of $3, P(7)$ is true because there could be a team of 3 and a team of 4 , and $P(8)$ is true because there could be two teams of 4 .

Case 3: $(n \geq 9)$. Of course $n>n-3$ so $P(n-3)$ holds by the strong induction hypothesis. But $n-3 \geq 6$, so $P(n-3)$ implies $3 x^{\prime}+4 y^{\prime}=n-3$ for some $x^{\prime}, y^{\prime} \in \mathbb{N}$, and therefore $3 x+4 y=n$ where $x::=x^{\prime}+1$ and $y::=y^{\prime}$. So $P(n)$ holds, as required.

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

Math0048｜Mathematical Analysis数学分析伦敦大学学院

Posted on 2023年9月22日2023年10月6日 by statistics-lab

statistics-lab^TM为您提供伦敦大学学院 London’s Global University MATH0048 Mathematical Analysis数学分析英国代写代考和辅导服务！

课程介绍：

This module is an introduction to mathematical analysis, one of the most important and welldeveloped strands of pure mathematics with many elegant and beautiful theorems, and also
with applications to many areas of mathematics, theoretical statistics, econometrics, and optimisation.
The aim is to introduce students to the ideas of formal definitions and rigorous proofs (one
of the fundamental features of modern mathematics, and something that is not familiar from
A-level), and to develop their powers of logical thinking.
This module is a prerequisite for Complex Analysis, MATH0013 and provides a useful foundation for courses such as Logic, MATH0050.
The module is intended for second or third year students in departments outside Mathematics,
particularly in Economics or Statistics. Students taking this module should be mathematically
able and will normally have demonstrated this by achieving a strong result in a module such as
MATH0047 or having an A* in Further Mathematics A-level.

Math0048｜Mathematical Analysis数学分析伦敦大学学院

Mathematical Analysis数学分析作业案例

问题 1.

Let $A: X \rightarrow Y$ be a linear operator between two Euclidean or two Hermitian spaces and let $A^: Y \rightarrow X$ be its adjoint. Then $\operatorname{Rank} A^=\operatorname{Rank} A$. Moreover,
$$
\begin{array}{ll}
(\operatorname{Im} A)^{\perp}=\operatorname{ker} A^, & \operatorname{Im} A=\left(\operatorname{ker} A^\right)^{\perp}, \
\left(\operatorname{Im} A^\right)^{\perp}=\operatorname{ker} A, & \operatorname{Im} A^=(\operatorname{ker} A)^{\perp} .
\end{array}
$$

Proof. Fix two orthonormal bases on $X$ and $Y$, and let $\mathbf{A}$ be the matrix associated to $A$ using these bases. Then, see (3.8), the matrix associated to $A^$ is $\mathbf{A}^T$, hence $$ \operatorname{Rank} A^=\operatorname{Rank} \mathbf{A}^T=\operatorname{Rank} \mathbf{A}=\operatorname{Rank} A,
$$
and
$$
\operatorname{dim}\left(\operatorname{ker} A^\right)^{\perp}=\operatorname{dim} Y-\operatorname{dim} \operatorname{ker} A^=\operatorname{Rank} A^=\operatorname{Rank} A=\operatorname{dim} \operatorname{Im} A . $$ On the other hand, $\operatorname{Im} A \subset\left(\operatorname{ker} A^\right)^{\perp}$ since, if $y=A(x)$ and $A^(v)=0$, then $(y \mid v)=$ $(A(x) \mid v)=\left(x \mid A^(v)\right)=0$. We then conclude that $\left(\operatorname{ker} A^\right)^{\perp}=\operatorname{Im} A$. The other claims easily follow. In fact, they are all equivalent to $\operatorname{Im} A=\left(\operatorname{ker} A^\right)^{\perp}$.

问题 2.

Let $X$ be a finite-dimensional vector space and let $\left(e_1, e_2, \ldots, e_n\right)$ be a $g$-orthogonal basis for a metric $g$ on $X$. Denote by $n_{+}, n_{-}$and $n_0$ the numbers of elements in the basis such that respectively, we have $g\left(e_i, e_i\right)>0, g\left(e_i, e_i\right)<0, g\left(e_i, e_i\right)=0$. Then $n_{+}=i_{+}(g)$, $n_{-}=i_{-}(g)$ and $n_0=i_0(g)$. In particular, $n_{+}, n_{-}, n_0$ do not depend on the chosen g-orthogonal basis,
$$
i_{+}(g)+i_{-}(g)=r(g) \quad \text { and } \quad i_{+}(g)+i_{-}(g)+i_0(g)=n .
$$

Proof. Suppose that $g\left(e_i, e_i\right)>0$ for $i=1, \ldots, n_{+}$. For each $v=\sum_{i=1}^{n_{+}} v^i e_i$, we have
$$
g(v, v)=\sum_{i=1}^{n_{+}}\left|v^i\right|^2 g\left(e_i, e_i\right)>0,
$$
hence $\operatorname{dim} \operatorname{Span}\left{e_1, e_2, \ldots, e_{n_{+}}\right} \leq i_{+}(g)$. On the other hand, if $W \subset X$ is a subspace of dimension $i_{+}(g)$ such that $g(v, v)>0 \forall v \in W$, we have
$$
W \cap \operatorname{Span}\left{e_{n_{+}+1}, \ldots, e_n\right}={0}
$$
since $g(v, v) \leq 0$ for all $v \in \operatorname{Span}\left{e_{n_{+}+1}, \ldots, e_n\right}$. Therefore we also have $i_{+}(g) \leq$ $n-\left(n-n_{+}\right)=n_{+}$.

Similarly, one proves that $n_{-}=i_{-}(g)$. Finally, since $\mathbf{G}:=\left[g\left(e_i, e_j\right)\right]$ is the matrix associated to $g$ in the basis $\left(e_1, e_2, \ldots, e_n\right)$, we have $i_0(g)=\operatorname{dim} \operatorname{rad}(g)=\operatorname{dim} \operatorname{ker} \mathbf{G}$, and, since $\mathbf{G}$ is diagonal, $\operatorname{dim} \operatorname{ker} G=n_0$.

问题 3.

Let $g$ be a metric on a finite-dimensional real vector space $X$. Then $g$ has a $g$-orthogonal basis.

Proof. Let $r$ be the rank of $g, r:=n-\operatorname{dim} \operatorname{rad}(g)$, and let $\left(w_1, w_2, \ldots, w_{n-r}\right)$ be a basis of $\operatorname{rad}(g)$. If $V$ denotes a supplementary subspace of $\operatorname{rad}(g)$, then $V$ is $g$-orthogonal to $\operatorname{rad} g$ and $\operatorname{dim} V=r$. Moreover, for every $v \in V$ there is $z \in X$ such that $g(v, z) \neq 0$. Decomposing $z$ as $z=w+t, w \in V, t \in \operatorname{rad}(g)$, we then have $g(v, w)=g(v, w)+$ $g(v, t)=g(v, z) \neq 0$, i.e., $g$ is nondegenerate on $V$. Since trivially, $\left(w_1, w_2, \ldots, w_{n-r}\right)$ is $g$-orthogonal and $V$ is $g$-orthogonal to $\left(w_1, w_2, \ldots, w_{n-r}\right)$, in order to conclude it suffices to complete the basis $\left(w_1, w_2, \ldots, w_{n-r}\right)$ with a $g$-orthogonal basis of $V$; in other words, it suffices to prove the claim under the further assumption that $g$ be nondegenerate.

We proceed by induction on the dimension of $X$. Let $\left(f_1, f_2, \ldots, f_n\right)$ be a basis of $X$. We claim that there exists $e_1 \in X$ with $g\left(e_1, e_1\right) \neq 0$. In fact, if for some $f_i$ we have $g\left(f_i, f_i\right) \neq 0$, we simply choose $e_1:=f_i$, otherwise, if $g\left(f_i, f_i\right)=0$ for all $i$, for some $k \neq 0$ we must have $g\left(f_1, f_k\right) \neq 0$, since by assumption $\operatorname{rad}(g)={0}$. In this case, we choose $e_1:=f_1+f_k$ as
$$
g\left(f_1+f_k, f_1+f_k\right)=g\left(f_1, f_1\right)+2 g\left(f_1, f_k\right)+g\left(f_k, f_k\right)=0+2 g\left(f_1, f_k\right)+0 \neq 0 .
$$
Now it is easily seen that the subspace
$$
V_1:=\left{v \in X \mid g\left(e_1, v\right)=0\right}
$$
supplements Span $\left{e_1\right}$, and we find a basis $\left(v_2, \ldots, v_n\right)$ of $V_1$ such that $g\left(v_j, e_1\right)=0$ for all $j=2, \ldots, n$ by setting
$$
v_j:=f_j-\frac{g\left(f_j, e_1\right)}{g\left(e_1, e_1\right)} e_1 .
$$
Since $g$ is nondegenerate on $V_1$, by the induction assumption we find a $g$-orthogonal basis $\left(e_2, \ldots, e_n\right)$ of $V_1$, and the vectors $\left(e_1, e_2, \ldots, e_n\right)$ form a $g$-orthogonal basis of $X$.

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

Math0033｜Numerical Methods 数值方法伦敦大学学院

Posted on 2023年9月21日2023年10月7日 by statistics-lab

statistics-lab^TM为您提供悉尼大学 London’s Global University MATH0048 Mathematical Analysis数学分析英国代写代考和辅导服务！

课程介绍：

Many phenomena in engineering and the physical and biological sciences can be described
using mathematical models. Frequently the resulting models cannot be solved analytically,
in which case a common approach is to use a numerical method to find an approximate
solution. The aim of this course is to introduce the basic ideas underpinning computational mathematics, study a series of numerical methods to solve different problems, and
carry out a rigorous mathematical analysis of their accuracy and stability

Numerical Methods 数值方法作业案例

问题 1.

we derived the an equation to describe a mass that is attached to a spring that would break when its elongation reached $0.03 \mathrm{~m}$ during resonant vibration of the springmass system. We need to determine the time $\mathrm{t}_{\mathrm{f}}$ at which the spring breaks from Equation (a):
$$
\left(0.05-\frac{t_f}{20}\right) \cos 10 t_f+\frac{1}{200} \sin 10 t_f-0.03=0
$$

We will use the Newton-Raphson’s method to solve the unknown quantity $t_f$ in Equation (a) by first assuming a solution on $t_f=0.75$. We made this assumed solution based on a crude approximated value of $\mathrm{t}{\mathrm{f}}=0.7$ in Example 8.9. Again, let us replace the unknown quantity $t_f$ in Equation (a) by conventional unknown symbol $\mathrm{x}$ in the following alternative form: $$ \begin{gathered} \left(0.05-\frac{x}{20}\right) \cos 10 x+\frac{1}{200} \sin 10 x-0.03=0 \ f(x)=\left(0.05-\frac{x}{20}\right) \cos 10 x+\frac{1}{200} \sin 10 x-0.03 \ f^{\prime}(x)=-(0.05-0.5 x) \sin 10 x \end{gathered} $$ Thus, the estimated root $x{i+1}$ after the previously estimated root $x_i$ may be computed by using the expression, as will be shown in the next slide.

Solution $\mathrm{x}$ from the equation: $\left(0.05-\frac{x}{20}\right) \cos 10 x+\frac{1}{200} \sin 10 x-0.03=0$

问题 2.

Evaluate the following integral by using the Gaussian quadrature $$
I=\int_0^\pi \cos x d x
$$

We have the function $y(x)=\cos x$ over the integration limits $x_a=0$ and $x_b=\pi$. The transformation of coordinates makes use of the relationship $x=\frac{\pi}{2} \xi+\frac{\pi}{2}$ from Equation (10.13), from which we get:
$$
y(x)=\cos x=F(\xi)=\cos \left(\frac{\pi}{2} \xi+\frac{\pi}{2}\right)=\sin \left(\frac{\pi}{2} \xi\right)
$$
Also, from Equation (10.14) with the use of the trigonometric relationships such as:
$$
\sin \left(\frac{\pi}{2}+\theta\right)=\cos \theta \text { and } \cos \left(\frac{\pi}{2}+\theta\right)=\sin \theta
$$
We may arrive at the following expression for integrating I in Equation (a) using Gaussian quadrature:
$$
I=\int_0^\pi \cos x d x=\int_{-1}^1\left[\sin \left(\frac{\pi}{2} \xi\right)\left(\frac{\pi}{2} d \xi\right)\right]=\frac{\pi}{2} \int_{-1}^1 \sin \frac{\pi}{2} \xi d \xi=\frac{\pi}{2} \sum_{i=1}^n H_i \sin \left(\frac{\pi}{2} a_i\right)
$$
Let us take, for example, 3 sampling points, i.e., $\mathrm{n}=3$ from Table 10.3 on P.357 with:
$$
\begin{array}{lll}
\mathrm{a}_1=0 & \mathrm{a}_2=+0.77459 & \mathrm{a}_3=-0.77459 \
\mathrm{H}_1=0.88888 & \mathrm{H}_2=0.55555 & \mathrm{H}_3=0.55555
\end{array}
$$
Substituting the above numbers into Equation (b) will lead to the solution:
$$
\begin{aligned}
I & =\frac{\pi}{2}\left[0.88888 \sin (0)+0.55555 \sin \left(\frac{\pi}{2} \times 0.77459\right)+0.55555 \sin \left(-\frac{\pi}{2} \times 0.77459\right)\right] \
& =\frac{\pi}{2}[0.55555 \sin (1.2167)-0.55555 \sin (1.2167)]=0
\end{aligned}
$$
25

问题 3.

Evaluate the following integral in Example 10.8 using Gaussian quadrature method

Solution:
$$
I=\int_{0.5}^{3.5} y(x) d x=\int_{0.5}^{3.5} x \sqrt{\left(16-x^2\right)^3} d x
$$

We notice the function $y(x)$ in the integral in Equation (a) is identical to what we had in previous Examples 10.5 (p.349), 6 (p.350) and 8 (p.355).
We will first derive the expression of the function $F(\xi)$ for the function $y(x)$ in Equation (a) from the integration limits of 0.5 and 3.5 to the limit -1 and +1 , as required in Gaussian quadrature.
The transformation of coordinate systems as illustrated in Figure 10.15 begins with the transformation of variable from $x$ to $\xi$ using the relationship in Equation (10.13), leading to the following relationship between the variables $x$ and $\xi$ as shown below:
$x=\frac{1}{2}\left(x_b-x_a\right) \xi+\frac{1}{2}\left(x_b+x_a\right)=1.5 \xi+2$, from which we obtain: $\mathrm{dx}=1.5 \mathrm{~d} \xi$
The integral in Equation (a) in the $y(x)$ vs. $x$ coordinates can thus be transformed to the $F(\xi)$ vs. $\xi$ coordinates by using Equation (10.14), yielding the following expression:
$$
\begin{aligned}
I=\int_{0.5}^{3.5} y(x) d x=\int_{0.5}^{3.5} x \sqrt{\left(16-x^2\right)^3} d x & =\frac{x_b-x_a}{2} \int_{-1}^1 F(\xi) d \xi \
& =\left.\frac{3.5-0.5}{2} \int_{-1}^1 y(x)\right|{x=1.5 \xi+2} d \xi \end{aligned} $$ We may thus evaluate the integral with the expression in Equation (d) as: $$ I=1.5 \int{-1}^1(1.5 \xi+2) \sqrt{\left[16-(1.5 \xi+2)^2\right]^3} d \xi=1.5 \int_{-1}^1(1.5 \xi+2) \sqrt{\left(-2.25 \xi^2-6 \xi+12\right)^3} d \xi

We have arrived at the following integral:
$$
I=1.5 \int^1(1.5 \xi+2) \sqrt{\left[16-(1.5 \xi+2)^2\right]^3} d \xi=1.5 \int^1(1.5 \xi+2) \sqrt{\left(-2.25 \xi^2-6 \xi+12\right)^3} d \xi
$$

Math0033｜Numerical Methods 数值方法伦敦大学学院请认准statistics-lab™

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

Econ8009｜International Monetary Economics 国际货币经济学澳洲国立大学

Posted on 2023年9月21日2023年10月7日 by statistics-lab

statistics-lab^TM为您提供澳洲国立大学 Australian National University Econ8009 Mathematical Analysis数学分析澳洲代写代考和辅导服务！

课程介绍：This course will provide students with models that can be used to analyse issues in international monetary economics. The course introduces students to so-called global imbalances and the basic stylized facts about current accounts with cross-section and the time-series of selected countries such as the US and China. Issues of current account sustainability will also be addressed.

International Monetary Economics 国际货币经济学作业案例

问题 1.

2) Jo and Maureen (aka Mo) are the only two people who drive cabs from the Leominster Airport to the Leominster Inn. The two cabs have the following identical total cost functions.
$$
\mathrm{TC}{\mathrm{Jo}}=\$ 3.00 \mathrm{xQ}{\mathrm{Jo}}, \quad \mathrm{TC}{\mathrm{Mo}}=\$ 3.00 \mathrm{x} \mathrm{Q}{\mathrm{Mo}} \text { where } \mathrm{QX}_{\mathrm{X}}=\text { number of trips per day made by driver } \mathrm{X} .
$$
Demand for trips is given by the demand function:
$$
\text { Price per Trip }=\$ 6.00-.05 \mathrm{xQ}
$$
a) Write down the average and marginal cost functions for this cab “industry”.

Answer: You should be able to write down the answers by inspection but we can derive them mathematically.
Define $\mathrm{Q}=\mathrm{Q}{\mathrm{Joe}}+\mathrm{Q}{\mathrm{Moe}}$. Then $\mathrm{TC}{\text {Industry }}=\$ 3.00 \mathrm{xQ}$ $$ \mathrm{AC}{\text {Industry }}=\mathrm{TC}{\text {Industry }} / \mathrm{Q}=\$ 3.00 \quad \mathrm{MC}{\text {Industry }}=\frac{\mathrm{d}\left(\mathrm{TC}_{\text {Industry })}\right.}{\mathrm{dQ}}=\$ 3.00
$$
i.e. both average and marginal costs are the same horizontal line $=\$ 3.00$

问题 2.

b) Assume Jo and Mo agree to cooperate to set the price per trip and split any profits they
generate. Determine the industry equilibrium price and quantity.

Answer: They will run the industry like a monopoly setting marginal revenue $=$ marginal cost. They will then presumably equally split the profit maximizing number of trips. As we know by now, Marginal Revenue for this demand curve is $\$ 6.00-.1 \mathrm{Q}$ so:
$$
\$ 6.00-.1 \mathrm{Q}=\$ 3.00 \text { or } \mathrm{Q}=\$ 3.00 / .1=30 \text { trips }(\text { total) } \text { or } 15 \text { trips per driver. }
$$
To get the price, we plug 30 trips back into the demand curve and get $\mathrm{P}=\$ 6.00-.05 \times 30=\$ 4.50$
Profit per cab $=($ price – average cost $) \mathrm{x}$ quantity $=\$ 1.50 \mathrm{x} 15=\$ 22.50$

问题 3.

c) The Leominster Environmental Protection Agency (LEPA) has determined that each cab trip between the Leominster Airport and the Leominster Inn generate a pollution cost equal to $\$ .50$ per trip. What is the implication of this cost for your solution in (b)? Discus in as much detail as you can LEPA’s policy options for dealing with this cost.

Answer: Using the terms we developed in class, the $\$ 3.00 \mathrm{MC}$ is really the private marginal cost per trip while the $\$ .50$ is the social cost of the trip. The economically efficient outcome is that the taxi industry set:
$$
\text { Marginal revenue }=\mathrm{MC}{\text {Private }}+\mathrm{MC}{\text {social }}=\$ 3.00+\$ .50=\$ 3.50 .
$$
The $\$ .50$ per trip is then collected by the government and distributed as compensation to those harmed by the pollution.

There are two ways to reach this socially optimal outcome: by imposing a $\$ .50$ tax on each ride or by calculating the socially optimal level of trips and requiring that Jo and Mo’s trips do not exceed this level. In this second case, the government will still have to collect the $\$ .50$ per trip from the two drivers to use as compensation.

econ8009｜International Monetary Economics 国际货币经济学澳洲国立大学请认准statistics-lab™

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

经济代写|国际经济学代写International Economics代考|EC340

Posted on 2022年9月23日2022年9月23日 by statistics-lab

如果你也在怎样代写国际经济学International Economics这个学科遇到相关的难题，请随时右上角联系我们的24/7代写客服。

国际经济学关注的是生产资源和消费者偏好的国际差异对经济活动的影响，以及影响它们的国际机构。

statistics-lab™ 为您的留学生涯保驾护航在代写国际经济学International Economics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。我们的专家在代写国际经济学International Economics代写方面经验极为丰富，各种代写国际经济学International Economics相关的作业也就用不着说。

我们提供的国际经济学International Economics及其相关学科的代写，服务范围广, 其中包括但不限于:

Statistical Inference 统计推断
Statistical Computing 统计计算
Advanced Probability Theory 高等概率论
Advanced Mathematical Statistics 高等数理统计学
(Generalized) Linear Models 广义线性模型
Statistical Machine Learning 统计机器学习
Longitudinal Data Analysis 纵向数据分析
Foundations of Data Science 数据科学基础

经济代写|国际经济学代写International Economics代考|EC340

经济代写|国际经济学代写International Economics代考|Trade Indifference Curves and Alternative Derivation of Offer Curves

Trade indifference curves (TICs) are a useful tool in discussing the welfare properties of free or restricted trade equilibrium. James Meade used TICs to propose an alternative way of deriving offer curves. A TIC for a country is the locus of different combinations of its export supply and import demand that yield the same national welfare for the country. Accordingly, the country is indifferent among all such combinations. In the export-import space, each TIC is positively sloped. The reason is as follows. An increase in export supply by the country means lesser units of the export good are now available for domestic consumption. This lowers the welfare of the country. Thus, to compensate and keep the national welfare at the same level, the other good (which is imported by the country) must be available for domestic consumption in greater quantity. Hence, along a TIC, an increase in export supply must be accompanied by an increase in import demand.

The other properties of TICs are similar to that of CICs, since TICs by definition replicate CICs in the excess-demand (or offer) space. That is, each TIC is convex to the export-axis and a higher TIC, which is further away from the country’s export-axis, represents a higher national welfare. That a higher (non-intersecting) TIC represents a higher national welfare is easily comprehensible, and is left as an exercise. It is also assumed that taste patterns are such that TICs do not intersect each other so that, like the CICs, we can attach a unique utility or welfare index to each TIC and the export-import combinations on different TICs can be Pareto ranked in the sense discussed earlier. The TIC map for the home country is illustrated in Figure A4.3.
Actual offer of exports in exchange for import demand by the home country, however, should be consistent with its budget constraints at the post-trade prices. The budget constraints for different TOT are shown by TOT lines through the origin. Thus facing the TOT indicated by the ray $O a_1$, the home country makes an offer of the combination $b_1$ that maximizes its welfare. An improved TOT enables the country to improve its welfare by offering $b_2$, and so on. Note that each of these offers is consistent with balanced trade. The locus connecting this welfare-maximizing and balanced trade offers traces out the offer curve for the home country. From this alternative derivation of a country’s offer curve emerges the welfare property of offers made by it for different TOT. Any point on the offer curve represents an offer of export supply and import demand for which trade is balanced and welfare is maximized for the corresponding TOT.

Note that by construction, at the free trade international equilibrium point $E$ in Figure $4.5$, the trade indifference curves of the two countries are tangent to the TOT line $O E$ and hence are mutually tangent to each other.

经济代写|国际经济学代写International Economics代考|Measurement and Trends in Barter TOT

A country may export and import thousands of commodities, and thus it only makes sense to measure a country’s TOT by the export price index relative to the import price index. Often these price indices are measured by the unit value indices for exports and imports. These unit value indices are constructed from data on exports and imports-both value and volumereported by countries. The unit value of exports, as a proxy for the export price index, is simply the index of value of exports divided by the index of volume of exports. The unit value of imports is similarly calculated. Finally, the ratio of the unit value index of exports to the unit value index of imports gives us the net barter terms of trade of a country.

Figure A4.4 illustrates movements in net barter TOT for four developed countries – Canada, Japan, UK, and USA-during 2000-19 with 2000 as the base year, as reported in UNCTAD Trade Statistics 2020. Both UK and the United States have stable and almost unchanged net barter TOT during this period. Canada experienced steady improvement in its net barter TOT till 2008 before it worsened. But, the net barter TOT for Japan deteriorated sharply before improving marginally after 2016. Figure A4.5, on the other hand, illustrates movements in net barter TOT for the developing countries like Brazil, China, India, Korea, and Mexico during the same period. Similar to Japan, Korea experienced continuous worsening of its net barter TOT till it stabilized to some extent after 2015 . The most interesting observation that can be made is movements in TOT of the other four countries. TOT of China and India, on the one hand, and of Brazil and Mexico, on the other hand, moved in tandem though almost exactly in the opposite directions. That is, while China and India experienced worsening of their respective TOTs during 2000-12, Brazil and Mexico enjoyed improving TOT. Subsequently, after 2015 in particular, the trends reversed for these countries.

There are many plausible explanations for such wide variations in TOT movement both across time and across countries as we will learn in later chapters.

国际经济学代考

经济代写|国际经济学代写国际经济学代考|贸易无差异曲线和报价曲线的可选推导

贸易无差异曲线(TICs)是讨论自由或限制贸易均衡的福利性质的有用工具。James Meade使用TICs提出了一种衍生报价曲线的替代方法。一个国家的TIC是其出口供给和进口需求的不同组合的轨迹，为该国带来相同的国民福利。因此，在所有这些组合中，国家是无所谓的。在进出口领域，每个TIC都是正倾斜的。原因如下。该国出口供应的增加意味着可供国内消费的出口商品数量减少。这降低了国家的福利。因此，为了补偿和保持国家福利在同一水平，其他商品(由国家进口)必须更多地供国内消费。因此，沿着TIC，出口供应的增加必然伴随着进口需求的增加

tic的其他属性与CICs相似，因为tic从定义上复制了过剩需求(或提供)空间中的CICs。也就是说，每个TIC都是出口轴的凸点，越高的TIC，越远离国家的出口轴，代表越高的国家福利。较高的(不交叉的)TIC代表较高的国家福利，这一点很容易理解，因此就留作练习。我们还假设，味觉模式是这样的，TIC之间不会交叉，因此，像CICs一样，我们可以给每个TIC附加一个独特的效用或福利指数，不同TIC上的进出口组合可以按照前面讨论的意义进行帕累托排序。母国的议会地图见图A4.3。但是，本国以出口换取进口需求的实际提议应符合其在贸易后价格下的预算限制。不同TOT的预算约束由通过原点的TOT线表示。因此，面对射线$O a_1$所表示的TOT，母国提出了使其福利最大化的组合$b_1$。一个改进的TOT使国家能够通过提供$b_2$来改善其福利，等等。请注意，这些提议都符合平衡贸易。连接福利最大化和平衡贸易提供的轨迹是母国提供曲线的轨迹。从这个对一国提供曲线的替代推导中，可以得出该国为不同TOT所提供的服务的福利性质。报价曲线上的任何一点都代表出口供给和进口需求的报价，对于相应的TOT，贸易是平衡的，福利是最大化的。

注意，通过构造，在图$4.5$中的自由贸易国际平衡点$E$处，两国的贸易无差异曲线与TOT线$O E$相切，因此彼此相切

经济代写|国际经济学代写国际经济学代考|物物交换TOT的测量和趋势

一个国家可能出口和进口数千种商品，因此，只有用出口价格指数相对于进口价格指数来衡量一个国家的TOT才有意义。这些价格指数通常用出口和进口的单位价值指数来衡量。这些单位价值指数是根据各国报告的出口和进口数据构建的。出口单位价值作为出口价格指数的代表，简单来说就是出口价值指数除以出口量指数。进口的单位价值也以同样的方式计算。最后，出口的单位价值指数与进口的单位价值指数之比，就得到了一个国家的净易货贸易条件

图A4.4显示了4个发达国家(加拿大、日本、英国和美国)以2000年为基准年在2000年至2019年期间的净易货TOT变动情况，见《2020年贸发会议贸易统计》报告。在此期间，英国和美国的净易货TOT都保持稳定和几乎不变。加拿大在2008年之前经历了净易货TOT的稳定改善，之后情况恶化。但是，日本的净易货TOT急剧恶化，在2016年之后略有改善。另一方面，图A4.5显示了同期巴西、中国、印度、韩国和墨西哥等发展中国家的净易货技术转移情况。与日本类似，韩国的净易货TOT也在持续恶化，直到2015年以后才有所稳定。可以作出的最有趣的观察是其他四个国家的TOT的变动。一方面，中国和印度的TOT，另一方面，巴西和墨西哥的TOT，步调一致，尽管几乎是完全相反的方向。也就是说，中国和印度在2000-12年期间经历了各自TOT的恶化，而巴西和墨西哥享受了TOT的改善。随后，特别是在2015年之后，这些国家的趋势发生了逆转

我们将在后面的章节中了解到，对于跨时间和跨国家的TOT运动如此广泛的差异，有许多合理的解释

经济代写|国际经济学代写International Economics代考请认准statistics-lab™

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

经济代写|国际经济学代写International Economics代考|ECN422

Posted on 2022年9月23日2022年9月23日 by statistics-lab

如果你也在怎样代写国际经济学International Economics这个学科遇到相关的难题，请随时右上角联系我们的24/7代写客服。

国际经济学关注的是生产资源和消费者偏好的国际差异对经济活动的影响，以及影响它们的国际机构。

我们提供的国际经济学International Economics及其相关学科的代写，服务范围广, 其中包括但不限于:

Statistical Inference 统计推断
Statistical Computing 统计计算
Advanced Probability Theory 高等概率论
Advanced Mathematical Statistics 高等数理统计学
(Generalized) Linear Models 广义线性模型
Statistical Machine Learning 统计机器学习
Longitudinal Data Analysis 纵向数据分析
Foundations of Data Science 数据科学基础

经济代写|国际经济学代写International Economics代考|ECN422

经济代写|国际经济学代写International Economics代考|Global Pareto optimality of free trade bundle

Free trade bundle, despite being sub-optimal for each large country in the case specified and illustrated above, is globally Pareto optimal. A Pareto optimal bundle is the one such that if we deviate from this bundle then at least one of the countries will be worse off. This can be verified from Figure 4.7. All the bundles along the aE segment of the foreign offer curve lie in the better set to the free trade bundle for the home country, but in the worse set to the free trade bundle for the foreign country. Thus, for all such trade bundles, achieved by the home country through a trade restriction policy, the home country will be better off but the foreign country will be worse off. Similarly, for all trade bundles along aE segment of the home offer curve, achieved by the foreign country through a trade restriction policy, the foreign country will be better off but the home country will be worse off. On the other hand, for any trade bundle that lie below $T I C_f$ below $T I C_f^$ and-such as bundles along the Oa segment of the home offer curve and along the $\mathrm{Oa}^*$ segment of the foreign offer curve-both countries are worse off relative to free trade.
Note that, free trade bundle emerges as the globally Pareto optimal bundle because of the fact that a pair of TICs of both the countries are tangent to each other for this bundle. If the TICs would have crossed each other at the free trade bundle, then some other trade bundles would have existed for which either both countries would have been better off; or one of them would have been better off with the other country being indifferent, that is, having the same welfare as for the free trade bundle.

This global Pareto optimality property of the free trade bundle has some far-reaching implications for trade wars, and for formation of regional trading blocs, as we will learn later.

经济代写|国际经济学代写International Economics代考|Existence, Uniqueness, and Stability of International Equilibrium

Existence of international equilibrium depicted in Figure $4.5$ requires that the two offer curves must cross each for a trade bundle containing strictly positive volumes of export offer and import demand. The assumption of non-decreasing opportunity costs and corresponding convexity of offer curves though help in this regard, but they may not be sufficient. Assuming that the offer curves are continuous, a sufficient condition for existence of international equilibrium is that at least one offer curve must be backward bending. However, though this might ensure existence, it may lead to multiple equilibria. This is evident from Figure 4.6.

Algebraically, the conditions of existence, uniqueness, and stability can be derived using the excess demand function approach. Consider the world market for the good imported by the home country and exported by the foreign country (that is, the world market for computers). Referring back to equation (4.2), define excess demand function $E\left(p^n\right)$ as:
$$
E\left(p^n\right)=M\left(p^\right)-X^\left(p^w\right)
$$
By the world-clearing condition (4.1) in the text, the (world) equilibrium price $p_c^{\text {” }}$ is such that, $E\left(p_c^w\right)=0$. Then, if this excess demand function is continuous in $p^w$, the following conditions ensure that $p_e^w>0$ (that is, an equilibrium price exists):
(a) There exists a price $p_0^w>0$ such that $E\left(p_0^w\right)>0$
(b) There exists a price $p_1^w>0$ such that $E\left(p_1^w\right)<0$ Figure A4.2 under the assumption that $p_0^w0$ will be unique. This is illustrated in Figure A4.2b. As it appears, the uniqueness is ensured by a monotonic excess demand function.

国际经济学代考

经济代写|国际经济学代写国际经济学代考|自由贸易捆绑包的全球帕累托最优性

自由贸易捆绑协议，尽管在上述指定和说明的情况下，对每个大国来说都不是最优的，但在全球范围内却是帕累托最优的。帕累托最优组合是指，如果我们偏离这个组合，那么至少有一个国家的情况会更糟。这可以从图4.7中得到验证。在外国报价曲线的aE段上，所有的捆绑包都位于较好的设置为本国的自由贸易捆绑包，但在较差的设置为外国的自由贸易捆绑包。因此，对于母国通过贸易限制政策实现的所有这类捆绑贸易，母国的境况会更好，而外国的境况会更糟。同样，对于外国通过贸易限制政策实现的本国报价曲线aE段上的所有贸易捆绑包，外国的境况会更好，而本国的境况会更糟。另一方面，对于位于$T I C_f$和$T I C_f^$以下的任何贸易捆绑包，例如沿着国内报价曲线的Oa部分和沿着国外报价曲线的$\mathrm{Oa}^*$部分的捆绑包，这两个国家相对于自由贸易而言都更糟糕。注意，自由贸易捆绑作为全球帕累托最优捆绑出现，因为两个国家的一对TICs在这个捆绑中彼此相切。如果这两个国家在自由贸易捆绑协议上相互交叉，那么就会存在一些其他的贸易捆绑协议，两国都能从中受益;或者，在另一个国家无动于衷的情况下，其中一个国家会过得更好，也就是说，享受与自由贸易捆绑协议相同的福利

自由贸易组合的这种全球帕累托最优性对贸易战和区域贸易集团的形成有着深远的影响，这一点我们稍后会讲到

经济代写|国际经济学代写国际经济学代考|国际均衡的存在性、唯一性和稳定性

图$4.5$中描述的国际均衡的存在性要求，对于一个包含严格为正的出口要约和进口需求的贸易包，两条要约曲线必须交叉。机会成本不下降的假设和相应的报价曲线的凸性虽然在这方面有帮助，但它们可能是不够的。假设报盘曲线是连续的，国际均衡存在的充分条件是至少有一条报盘曲线是向后弯曲的。然而，尽管这可能保证存在，它也可能导致多重均衡。从图4.6可以明显看出这一点

从代数上讲，存在条件、唯一性和稳定性可以用超额需求函数方法推导出来。考虑一下本国进口和外国出口商品的世界市场(即计算机的世界市场)。回到式(4.2)，将超额需求函数$E\left(p^n\right)$定义为:
$$
E\left(p^n\right)=M\left(p^\right)-X^\left(p^w\right)
$$
通过文中的世界清算条件(4.1)，(世界)均衡价格$p_c^{\text {” }}$使$E\left(p_c^w\right)=0$。那么，如果这个超额需求函数在$p^w$中是连续的，下面的条件保证了$p_e^w>0$(即均衡价格存在):
(a)存在一个价格$p_0^w>0$使得$E\left(p_0^w\right)>0$
(b)存在一个价格$p_1^w>0$使得$E\left(p_1^w\right)<0$图A4.2在假设$p_0^w0$是唯一的情况下。图A4.2b说明了这一点。正如它所显示的那样，唯一性是由一个单调的超额需求函数保证的

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

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

Introductory Statistical Methods and Computing 统计方法和计算入门作业案例

International Monetary Economics 国际货币经济学 作业案例

Discrete Mathematics 离散数学作业案例

Discrete Mathematics (Advanced)进阶离散数学作业案例

Mathematical Computing数学计算作业案例

Mathematical Analysis数学分析作业案例

Numerical Methods 数值方法作业案例

International Monetary Economics 国际货币经济学作业案例

经济代写|国际经济学代写International Economics代考|Trade Indifference Curves and Alternative Derivation of Offer Curves

经济代写|国际经济学代写International Economics代考|Measurement and Trends in Barter TOT

经济代写|国际经济学代写国际经济学代考|贸易无差异曲线和报价曲线的可选推导

经济代写|国际经济学代写国际经济学代考|物物交换TOT的测量和趋势

经济代写|国际经济学代写International Economics代考|Global Pareto optimality of free trade bundle

经济代写|国际经济学代写International Economics代考|Existence, Uniqueness, and Stability of International Equilibrium

经济代写|国际经济学代写国际经济学代考|自由贸易捆绑包的全球帕累托最优性

经济代写|国际经济学代写国际经济学代考|国际均衡的存在性、唯一性和稳定性

International Monetary Economics 国际货币经济学作业案例