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澳洲代写｜CIVL3612｜Fluid Mechanics流体力学悉尼大学

Posted on 2023年10月16日2023年10月16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”Business Entrepreneurship: Business Models企业创业：商业模式澳洲代写代考和辅导服务！

课程介绍：

This unit of study aims to provide an understanding of the conservation of mass and momentum in differential forms for viscous fluid flows. It provides the foundation for advanced study of turbulence, flow around immersed bodies, open channel flow, pipe flow and pump design.

澳洲代写｜OLET1201｜Business Entrepreneurship: Business Models企业创业：商业模式悉尼大学

Key Information	Details
Course Number	CIVL3612
Pre-requisites	(Not mentioned in the text)
Academic Unit (Major)	Civil Engineering
Instructor	(Not mentioned in the text)
Credit Points	6

Business Models商业模式问题集

问题 1.

A common observation in big rivers or other fast-flowing bodies of water (e.g. during floods) is shown in the figures and sketch below. A fast moving stream of water that is steadily flowing along suddenly decelerates and the position of the free surface ‘jumps’ upwards. After a lot of local turbulent motion, the flow settles down again but is now steadily moving at a significantly slower speed.

We will represent the free surface height as $h(x)$ and the velocity by the function $u(x)$. The fluid has constant density $\rho$ and we will treat the problem as one-dimensional. You can assume that viscous stresses along the control surfaces of the volume shown above are negligibly small, and neglect the density of air.
PART I:
a) consider a streamline drawn (line $\mathrm{AB}$ in the figure) just above the smooth flat lower surface of the channel. How is the static pressure in the fluid along this line related to the height of the river? How does the static pressure vary along the line DEA?

(a):
The pressure distribution on line $\mathrm{AB}$ follows the hydrostatic rule. It is true that the flow is not static but by picking an arbitrary control volume at any point on line AB (green dashed control volume in Figure 1) one can see that the balance of forces in the $y$-direction will tell us that the difference between the pressure at the bottom and the ambient pressure should balance the weight of the liquid inside the control volume. This simply implies that the static pressure on line $\mathrm{AB}$ should be equal
The pressure distribution on line DEA also follows the hydrostatic change merely due to the fact that there is no curvature in the streamlines as one integrates the Euler equation normal to them and thus the only change in pressure when one moves from $\mathrm{E}$ to A will be the hydrostatic part. Ignoring the density of air one can see that the pressure is constant from D to E and then start to grow linearly with height as we move from $\mathrm{E}$ to $\mathrm{A}$. The result is shown in Figure

问题 2.

b) Using the control volume shown in the sketch develop two expressions that relate the velocity and height of the stream at station 1 and the velocity and height of the stream at station 2. Developing a table of relevant quantities along each face of the control volume ABCDEA is highly recommended!
c) [2 points] Combine your expressions from (a) and (b) together to show that the speed of the river can be simply evaluated from simple measurements of the river height (e.g. using marked yardsticks attached to the channel floor):
$$
u_1=\sqrt{\frac{g h_2}{2 h_1}\left(h_1+h_2\right)}
$$

}(b) and (c): The selected control volume is shown in Figure 3 (dashed green line). One can subtract the ambient pressure from the entire problem and knowing that the net effect of uniform $P_a$ acting on the control volume is zero then there will be no change in the problem analysis if we only deal with gauge pressures $\left(P(x, y)-P_a\right)$Table 1 summarizes all the important parameters acting on different control surfaces for the selected control volume:

Now we can start by writing the conservation rules using the RTT. It is important to notice that due to the turbulent mixing happening in the region of the hydraulic jump, energy will not be conserved and thus either applying the conservation of energy or the Bernoulli equation will not be the right approach. If we write the conservation of mass for the selected control volume then we will have:

$$
\text { C.O.Mass: } 0=\frac{d}{d t} \int_{\text {c.v. }} \rho d V+\int_{\text {c.s. }} \rho\left(v-v_c\right) \cdot n d A
$$

Knowing that the problem is steady state and using the tabulated quantities, conservation of mass can be simplified to:
$\rho u_1 h_1=\rho u_2 h_2 \Rightarrow u_1 h_1=u_2 h_2$
The conservation of linear momentum in the $x$ direction can also be written in the RTT form:

$$
\text { C.O.Momentum: } \frac{1}{W} \sum F_x=\frac{d}{d t} \int_{c . v .} \rho v_x d V+\int_{\text {c.s. }} \rho v_x\left(v-v_c\right) \cdot n d A
$$

where $W$ is the width into the page.
The net of external forces acting in the $x$-direction on the control volume neglecting the wall shear effect is a result of pressure forces acting on the (AD) and (BC) control surfaces:

$$
\frac{1}{W} \sum F_x=\int_{A D}\left(P-P_a\right) d y-\int_{B C}\left(P-P_a\right) d y=\int_0^{h_1} \rho g y d y-\int_0^{h_2} \rho g y d y=\rho g\left(\frac{h_1^2}{2}-\frac{h_2^2}{2}\right)
$$

The right hand side of the RTT for the conservation of linear momentum can also be simplified to (knowing that the problem is steady and using the tabulated identities):

$$
\text { R.H.S. of RTT for C.O. Momentum }=\rho u_2^2 h_2-\rho u_1^2 h_1
$$

thus the conservation of linear momentum implies that:

$$
\rho g\left(\frac{h_1^2}{2}-\frac{h_2^2}{2}\right)=\rho u_2^2 h_2-\rho u_1^2 h_1 \Rightarrow \frac{g}{2}\left(h_1^2-h_2^2\right)=u_2^2 h_2-u_1^2 h_1
$$

using the result from conservation of mass (equation (1)) one can eliminate $u_2$ from equation (2) to give:

$$
\frac{g}{2}\left(h_1^2-h_2^2\right)=h_1 u_1^2\left(h_1 / h_2-1\right) \Rightarrow u_1=\sqrt{\frac{g h_2}{2 h_1}\left(h_1+h_2\right)}
$$

where we have used the identity $h_1^2-h_2^2=\left(h_1-h_2\right)\left(h_1+h_2\right)$.

问题 3.

A deeper question to answer is why is the water moving so fast locally to begin with. To answer this we must consider the topography of the river bed that is upstream of station 1 , as shown in the drawing below. We denote the height of the fluid stream above the river bed as $h(x)$ and the height of the riverbed by $b(x)$ :
d) [1 point] Consider a slice of river $d x$ and show that conservation of mass can be written in the form:
$$
u(x) \frac{d h(x)}{d x}+h(x) \frac{d u(x)}{d x}=0
$$

(d):
For the selected control volume (Figure 4 ) one can easily write the conservation of mass using Taylor series to obtain expressions for $u(x+\Delta x)$ and $h(x+\Delta x)$ :
$$
u(x) h(x)=u(x+\Delta x) h(x+\Delta x) \rightarrow u(x) h(x)=\left(u(x)+\frac{d u}{d x} \Delta x\right)\left(h(x)+\frac{d h}{d x} \Delta x\right)
$$
which after ignoring the second order terms such $\left(\Delta x^2\right)$ it can be rewritten as:
$$
\Delta x\left(u(x) \frac{d h}{d x}+h(x) \frac{d u}{d x}\right)=0 \Rightarrow u \frac{d h}{d x}+h \frac{d u}{d x}=0
$$
Another way to reach the same result is to say that since the flow is incompressible then the volumetric flow rate should remain unchanged thus $d(u h) / d x=0$ which will lead to the same result we just derived in equation (4).

Figure 4: An arbitrary control volume selected to derive the conservation of mass in the differential form.

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｜OLET1201｜Business Entrepreneurship: Business Models企业创业：商业模式悉尼大学

Posted on 2023年10月16日2023年10月16日 by statistics-lab

课程介绍：

Where the 0 credit point Business Entrepreneurship unit provides students with a theoretical perspective on business entrepreneurship, this for-credit upgrade provides an opportunity for students to apply this knowledge, and to refine their understanding. To this aim, students are presented with entrepreneurial challenges and are assisted to develop viable prototypes of services or products that address the challenges. With the help of research-based entrepreneurship literature, students analyse the market potential of the prototypes, formulate a suitable value proposition for their prototypes, and develop a business model that enables them to progress from idea to venture. Through this experiential exercise and the accompanying literature on business models and prototyping, students develop relevant prototyping and analytical skills, an understanding of the role and nature of business models, and learn how to combine both toward the goal of venture growth.

Business Models商业模式问题集

问题 1.

The Key Activities element includes the most important things a company must do to make its business model work.
In order to be successful, a company must carry out key actions that are primarily dictated by its business model.

When planning the key activities, it is necessary to know answers to the following questions:

What kinds of activities are crucial to our business?
What kinds of activities are crucial to our distribution channels?
What kinds of activities are important if we want to maintain our customer_relationships?
What kinds of activities are fundamental for our revenue streams?
Some typical key activities that are commonly practiced by most organizations are listed below:

Research \& Development,
Production,
Marketing, and
$\quad$ Sales \& Customer Services.

关键活动要素包括公司为使其商业模式行之有效而必须做的最重要的事情。
为了取得成功，公司必须开展主要由其商业模式决定的关键行动。

在规划关键活动时，有必要了解以下问题的答案：

哪些活动对我们的业务至关重要？
哪些活动对我们的分销渠道至关重要？
如果我们要维护客户关系，哪些活动是重要的？
哪些活动对我们的收入流至关重要？
下面列出了大多数组织通常开展的一些典型的关键活动：

研究与开发、
生产
市场营销
销售和客户服务。

问题 2.

Cost structure covers all expenses, which are important in the company activity.
Having in mind the financial aspect, we should answer the following questions:

What are the main costs that are generated in our company?
Which key resources are the most expensive?
Which key actions require a major financial investment?
In several business models, it is particularly important to maintain low costs. Therefore, it is worth distinguishing between the two categories of structure:

The structure focused on costs – The maintenance of a low-cost structure needs reducing costs whenever it is possible. It can be ensured by lowering the costs of value proposition, and introducing maximum automation in production and outsourcing.
Structure focused on value – Some companies pay more attention to the quality of the products.
The cost structure may concern the following:

成本结构包括公司活动中的所有重要开支。
考虑到财务问题，我们应该回答以下问题：

我们公司产生的主要成本是什么？
哪些关键资源最昂贵？
哪些关键行动需要大量资金投入？
在一些商业模式中，保持低成本尤为重要。因此，值得区分两类结构：

以成本为中心的结构 – 要保持低成本结构，就必须尽可能降低成本。可以通过降低价值主张的成本、在生产和外包过程中引入最大程度的自动化来确保这一点。
注重价值的结构 – 有些公司更注重产品质量。
成本结构可能涉及以下方面

the residuals $r_t=X_t-\hat{m}_t-\hat{S}_t$. Does it look like a white noise sequence? If not, can you make any suggestions?

Fixed cost – These are the costs that the company bears even in the period in which the production is at zero level. These costs are incurred every month on operating activities, such as media, accounting, etc. Fixed costs are major cost components for many businesses, especially service providers, including restaurants, cinemas, theatres, and hotels.
Variable costs – These change in proportion to the quantity of goods produced or services provided. For this type of costs, it is possible to include costs associated with renting variable factors of production, for example, work, or raw materials. For example, companies have signed contracts with employees and suppliers of raw materials, and they may use quite a lot of flexibility through work in a timeless or part-time, employment of seasonal workers or the purchase of raw materials in the market.

固定成本 – 这些是公司在生产处于零水平期间也要承担的成本。这些成本每月都会在媒体、会计等运营活动中产生。固定成本是许多企业的主要成本构成，尤其是服务提供商，包括餐馆、电影院、剧院和酒店。
可变成本 – 这些成本的变化与生产的商品或提供的服务数量成正比。对于这类成本，可以包括与租赁可变生产要素（如工作或原材料）相关的成本。例如，公司与雇员和原材料供应商签订合同，可以通过定时工作或非全时工作、雇用季节性工人或在市场上购买原材料等方式使用相当大的灵活性。

问题 3.

The methods that can be used are the following (Osterwalder \& Pigneur (2010):
A. Asset sale
This kind of sale refers to the transfer of ownership rights of a physical product from the seller to the buyer.
B. Usage fee
This kind of fee is usually charged by service providers to customers for the use of the service. A doctor may charge the patient according to the number and nature of treatments the patient undergoes while under his care.
C. Subscription fees
When a user requires long-term or continuous access to the products of a company, they pay a subscription fee. For example, a gym may sell a yearly membership subscription to its customer.
D. Lending/renting/leasing
Some organizations provide their customers with exclusive rights to their product for a limited amount of time for a set fee. Upon the end of this period, the organization regains ownership of the product. The company enjoys recurring revenue from the customer for the mentioned period, while the customer has exclusive access to the product for the time he/ she require it without having to make a hefty investment.
E. Licensing
Licensing is generally used when we are talking about products, services, or ideas that fall under the parameter of intellectual property. It is common in the technology industry for patent holders to license the use of patents to other companies and to charge a licensing fee for it.

A. 资产销售
这种销售是指将实物产品的所有权从卖方转移给买方。
B. 使用费
这种费用通常是服务提供商向客户收取的服务使用费。医生可根据病人接受治疗的次数和性质向病人收费。
C. 订购费
当用户需要长期或持续使用某公司的产品时，他们需要支付订购费。例如，一家健身房可能会向其客户出售一年的会员订阅费。
D. 出借/出租/租赁
一些机构向客户提供在有限时间内使用其产品的专有权，并收取一定费用。期限结束后，该组织重新获得产品所有权。公司可在上述期间从客户那里获得经常性收入，而客户则可在其需要的时间内独家使用产品，而无需进行巨额投资。
E. 许可证
当我们谈论属于知识产权范畴的产品、服务或创意时，通常会用到许可。在技术行业，专利持有者向其他公司许可使用专利并收取许可费的做法很常见。

澳洲代写｜ECMT2130｜Financial Econometrics金融计量经济学悉尼大学

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

澳洲代写｜QBUS6830｜Financial Time Series and Forecasting金融时间序列和预测悉尼大学

Posted on 2023年10月13日2024年1月7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”Financial Time Series and Forecasting金融时间序列和预测澳洲代写代考和辅导服务！

课程介绍：

Time series and statistical modelling is a fundamental component of the theory and practice of modern financial asset pricing as well as financial risk measurement and management. Further, forecasting is a required component of financial and investment decision making. This unit provides an introduction to the time series models used for the analysis of data arising in financial markets. It then considers methods for forecasting, testing and sensitivity analyses, in the context of these models. Topics include: the properties of financial return data; the Capital Asset Pricing Model (CAPM); financial return factor models, with known and unknown factors, in panel data settings; modelling and forecasting conditional volatility, via ARCH and GARCH; forecasting market risk measures such as Value at Risk. Emphasis is placed on applications involving the analysis of many real market datasets. Students are encouraged to undertake hands-on analysis using an appropriate computing package.

澳洲代写｜QBUS6830｜Financial Time Series and Forecasting金融时间序列和预测悉尼大学

Financial Econometrics金融计量经济学问题集

问题 1.

(a) Show that a linear filter $\left{a_j\right}$ passes an arbitrary polynomial of degree $k$ without distortion, that is,
$$
m_t=\sum_j a_j m_{t-j}
$$
for all $k$ th-degree polynomials $m_t=c_0+c_1 t+\cdots+c_k t^k$ if and only if
$$
\sum_j a_j=1 \text {, and } \sum_j j^r a_j=0 \text { for } r=1, \ldots, k \text {. }
$$
(b) Show that the Spencer 15-point moving average filter does not distort a cubic trend.

问题 2.

In Splus, get hold of the yearly airline passenger data set by assigning it to an object. You can use the command
x_rts (scan(‘airline.dat’), freq=12, start=1949)
The data are now stored in the object $x$, which forms the time series $\left{X_t\right}$. This data set consists of monthly totals (in thousands) of international airline passengers from January 1949 to December 1960 [details can be found in Brockwell and Davis (1991)]. It is stored under the file airline.dat on the Web page for this book.
(a) Do a time series plot of this data set. Are there any obvious trends?
(b) Is it necessary to transform the data? If a transformation is needed, what would you suggest?
(c) Do a yearly running median for this data set. Sketch the box plots for each year to detect any other trends.
(d) Find a trend estimate by using a moving average filter. Plot this trend.
(e) Estimate the seasonal component $S_k$, if any.
(f) Consider the deseasonalized data $d_t=X_t-\hat{S}_t, t=1, \ldots, n$. Reestimate a trend from $\left{d_t\right}$ by applying a moving average filter to $\left{d_t\right}$; call it $\hat{m}_t$, say.
(g) Plot the residuals $r_t=X_t-\hat{m}_t-\hat{S}_t$. Does it look like a white noise sequence? If not, can you make any suggestions?

问题 3.

If $\left{X_t=A \cos t \omega: t=1, \ldots, n\right}$ where $A$ is a fixed constant and $\omega$ is a constant in $(0, \pi)$, show that $r_k \rightarrow \cos k \omega$ as $n \rightarrow \infty$. Hint: You need to use the double-angle and summation formulas for a trigonometric function.
Let $Z_t \sim \mathrm{N}(0,1)$ i.i.d. Match each of the following time series with its corresponding correlogram in Figure 2.1.
(a) $X_t=Z_t$.
(b) $X_t=-0.3 X_{t-1}+Z_t$.
(c) $X_t=\sin (\pi / 3) t+Z_t$.
(d) $X_t=Z_t-0.3 Z_{t-1}$.

问题 4.

Determine which of the following processes are causal and/or invertible:
(a) $Y_t+0.2 Y_{t-1}-0.48 Y_{t-2}=Z_t$.
(b) $Y_t+1.9 Y_{t-1}+0.88 Y_{t-2}=Z_t+0.2 Z_{t-1}+0.7 Z_{t-2}$.
(c) $Y_t+0.6 Y_{t-2}=Z_t+1.2 Z_{t-1}$.
(d) $Y_t+1.8 Y_{t-1}+0.81 Y_{t-2}=Z_t$.
(e) $Y_t+1.6 Y_{t-1}=Z_t-0.4 Z_{t-1}+0.04 Z_{t-2}$.
Let $\left{Y_t: t=0, \pm 1, \ldots\right}$ be the stationary solution of the noncausal $\operatorname{AR}(1)$ equation
$$
Y_t=\phi Y_{t-1}+Z_t, \quad|\phi|>1, \quad\left{Z_t\right} \sim \mathrm{WN}\left(0, \sigma^2\right) .
$$
Show that $\left{Y_t\right}$ also satisfies the causal AR(1) equation
$$
Y_t=\phi^{-1} Y_{t-1}+W_t, \quad\left{W_t\right} \sim \mathrm{WN}\left(0, \tilde{\sigma}^2\right)
$$
for a suitably chosen white noise process $\left{W_t\right}$. Determine $\tilde{\sigma}^2$.
Show that for an MA(2) process with moving average polynomial $\theta(z)=$ $1-\theta_1 z-\theta_2 z^2$ to be invertible, the parameters $\left(\theta_1, \theta_2\right)$ must lie in the triangular region determined by the intersection of the three regions
$$
\begin{aligned}
& \theta_2+\theta_1<1, \
& \theta_2-\theta_1<1, \
& \left|\theta_2\right|<1 .
\end{aligned}
$$

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

澳洲代写｜ECMT2130｜Financial Econometrics金融计量经济学悉尼大学

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”Financial Econometrics金融计量经济学澳洲代写代考和辅导服务！

课程介绍：

This unit focuses on the financial models and econometric methods necessary to critically evaluate the risk and return characteristics of various fund-management strategies. Asset-pricing models and market efficiency are tested using econometric models that are popular in banking and finance, using industry-standard software. A core learning outcome is competency with that software. Students work with real and simulated data to specify, estimate, and test the linear regression models and the univariate time-series models that are at the core of the unit. The unit equips students with the conceptual framework and applied skills relevant to quantitative careers in finance and policy.

Financial Econometrics金融计量经济学问题集

问题 1.

When the regressors in a multiple regression are highly correlated, then we have a practical problem: the standard errors of individual coefficients tend to be large.
As a simple example, consider the regression
$$
y_t=\beta_1 x_{1 t}+\beta_2 x_{2 t}+u_t
$$
where (for simplicity) the dependent variable and the regressors have zero means. In this case, the variance of
$$
\operatorname{Var}\left(\hat{\beta}2\right)=\frac{1}{1-\operatorname{Corr}\left(x{1 t}, x_{2 t}\right)^2} \frac{1}{\operatorname{Var}\left(x_{2 t}\right)} \frac{\sigma^2}{T},
$$
where the new term is the (squared) correlation. If the regressors are highly correlated, then the uncertainty about the slope coefficient is high. The basic reason is that we see that the variables have an effect on $y_t$, but it is hard to tell if that effect is from regressor one or two.

Proof. (of 2.21). Recall that for a $2 \times 2$ matrix we have
$$
\left[\begin{array}{ll}
a & b \
c & d
\end{array}\right]^{-1}=\frac{1}{a d-b c}\left[\begin{array}{cc}
d & -b \
-c & a
\end{array}\right] .
$$
For the regression (2.20) we get
$$
\begin{aligned}
& {\left[\begin{array}{cc}
\sum_{t=1}^T x_{1 t}^2 & \sum_{t=1}^T x_{1 t} x_{2 t} \
\sum_{t=1}^T x_{1 t} x_{2 t} & \sum_{t=1}^T x_{2 t}^2
\end{array}\right]^{-1}=} \
& \quad \frac{1}{\sum_{t=1}^T x_{1 t}^2 \sum_{t=1}^T x_{2 t}^2-\left(\sum_{t=1}^T x_{1 t} x_{2 t}\right)^2}\left[\begin{array}{cc}
\sum_{t=1}^T x_{2 t}^2 & -\sum_{t=1}^T x_{1 t} x_{2 t} \
-\sum_{t=1}^T x_{1 t} x_{2 t} & \sum_{t=1}^T x_{1 t}^2
\end{array}\right] .
\end{aligned}
$$
The variance of the second slope coefficient is $\sigma^2$ time the lower right element of this

matrix. Multiply and divide by $T$ to get
$$
\begin{aligned}
\operatorname{Var}\left(\hat{\beta}2\right) & =\frac{\sigma^2}{T} \frac{\sum{t=1}^T x_{1 t}^2 / T}{\sum_{t=1}^T \frac{1}{T} x_{1 t}^2 \sum_{t=1}^T \frac{1}{T} x_{2 t}^2-\left(\sum_{t=1}^T \frac{1}{T} x_{1 t} x_{2 t}\right)^2} \
& =\frac{\sigma^2}{T} \frac{\operatorname{Var}\left(x_{1 t}\right)}{\operatorname{Var}\left(x_{1 t}\right) \operatorname{Var}\left(x_{2 t}\right)-\operatorname{Cov}\left(x_{1 t}, x_{2 t}\right)^2} \
& =\frac{\sigma^2}{T} \frac{1 / \operatorname{Var}\left(x_{2 t}\right)}{1-\frac{\operatorname{Cov}\left(x_{1 t}, x_{2 t}\right)^2}{\operatorname{Var}\left(x_{1 t}\right) \operatorname{Var}\left(x_{2 t}\right)}}
\end{aligned}
$$

问题 2.

Suppose we have monthly data with $\widehat{\alpha}i=0.2 \%$ (that is, $0.2 \% \times 12=2.4 \%$ per year), Std $\left(\varepsilon{i t}\right)=3 \%$ (that is, $3 \% \times \sqrt{12} \approx 10 \%$ per year) and a market Sharpe ratio of 0.15 (that is, $0.15 \times \sqrt{12} \approx 0.5$ per year). (This corresponds well to US CAPM regressions for industry portfolios.) A significance level of $10 \%$ requires a $t$-statistic (6.4) of at least 1.65 , so
$$
\frac{0.2}{\sqrt{1+0.15^2} 3 / \sqrt{T}} \geq 1.65 \text { or } T \geq 626 .
$$
We need a sample of at least 626 months (52 years)! With a sample of only 26 years (312 months), the alpha needs to be almost $0.3 \%$ per month (3.6\% per year) or the standard deviation of the residual just $2 \%$ (7\% per year). Notice that cumulating a $0.3 \%$ return over 25 years means almost 2.5 times the initial value.

Proof. (*Proof of (6.8)) Consider the regression equation $y_t=x_t^{\prime} b+\varepsilon_t$. With iid errors that are independent of all regressors (also across observations), the LS estimator, $\hat{b}{L s}$, is asymptotically distributed as $$ \sqrt{T}\left(\hat{b}{L s}-b\right) \stackrel{d}{\rightarrow} N\left(\mathbf{0}, \sigma^2 \Sigma_{x x}^{-1}\right) \text {, where } \sigma^2=\operatorname{Var}\left(\varepsilon_t\right) \text { and } \Sigma_{x x}=\operatorname{plim} \Sigma_{t=1}^T x_t x_t^{\prime} / T .
$$
When the regressors are just a constant (equal to one) and one variable regressor, $f_t$, so $x_t=\left[1, f_t\right]^{\prime}$, then we have
$$
\begin{aligned}
\Sigma_{x x} & =\mathrm{E} \sum_{t=1}^T x_t x_t^{\prime} / T=\mathrm{E} \frac{1}{T} \sum_{t=1}^T\left[\begin{array}{cc}
1 & f_t \
f_t & f_t^2
\end{array}\right]=\left[\begin{array}{cc}
1 & \mathrm{E} f_t \
\mathrm{E} f_t & \mathrm{E} f_t^2
\end{array}\right], \text { so } \
\sigma^2 \Sigma_{x x}^{-1} & =\frac{\sigma^2}{\mathrm{E} f_t^2-\left(\mathrm{E} f_t\right)^2}\left[\begin{array}{cc}
\mathrm{E} f_t^2 & -\mathrm{E} f_t \
-\mathrm{E} f_t & 1
\end{array}\right]=\frac{\sigma^2}{\operatorname{Var}\left(f_t\right)}\left[\begin{array}{cc}
\operatorname{Var}\left(f_t\right)+\left(\mathrm{E} f_t\right)^2 & -\mathrm{E} f_t \
-\mathrm{E} f_t & 1
\end{array}\right] .
\end{aligned}
$$
(In the last line we use $\operatorname{Var}\left(f_t\right)=\mathrm{E} f_t^2-\left(\mathrm{E} f_t\right)^2$.)

问题 3.

It is then straightfoward to show that the VaR for a portfortfolio
$$
R_p=w_1 R_1+w_2 R_2,
$$

where $w_1+w_2=1$ can be written
$$
\operatorname{VaR}p=\left(\left[\begin{array}{ll} w_1 \operatorname{Var}_1 & w_2 \operatorname{Var}_2 \end{array}\right]\left[\begin{array}{cc} 1 & \rho{12} \
\rho_{12} & 1
\end{array}\right]\left[\begin{array}{l}
w_1 \operatorname{Var}1 \ w_2 \operatorname{Var}_2 \end{array}\right]\right)^{1 / 2}, $$ where $\rho{12}$ is the correlation of $R_1$ and $R_2$. The extension to $n$ (instead of 2) assets is straightforward.

This expression highlights the importance of both the individual $\mathrm{VaR}i$ values and the correlation. Clearly, a worst case scenario is when the portfolio is long in all assets $\left(w_i>\right.$ $0)$ and the correlation turns out to be perfect $\left(\rho{12}=1\right)$.

Proof. (of (11.8)) Recall that $\mathrm{VaR}p=1.64 \sigma_p$, and that $$ \sigma_p^2=w_1^2 \sigma{11}+w_2^2 \sigma_{22}+2 w_1 w_2 \rho_{12} \sigma_1 \sigma_2
$$
Use (11.6) to substitute as $\sigma_i=\operatorname{VaR}i / 1.64$ $$ \sigma_p^2=w_1^2 \operatorname{VaR}_1^2 / 1.64^2+w_2^2 \operatorname{VaR}_2^2 / 1.64^2+2 w_1 w_2 \rho{12} \times \operatorname{VaR}_1 \times \mathrm{VaR}_2 / 1.64^2 .
$$
Multiply both sides by $1.64^2$ and take the square root to get (11.8).

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

澳洲代写｜ENVX2001｜Applied Statistical Method应用统计方法悉尼大学

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”Applied Statistical Method应用统计方法代写代考和辅导服务！

课程介绍：

This unit builds on introductory 1st year statistics units and is targeted towards students in the agricultural, life and environmental sciences. It consists of two parts and presents, in an applied manner, the statistical methods that students need to know for further study and their future careers. In the first part the focus is on designed studies including both surveys and formal experimental designs. Students will learn how to analyse and interpret datasets collected from designs from more than 2 treatment levels, multiple factors and different blocking designs. In the second part the focus is on finding patterns in data. In this part the students will learn to model relationships between response and predictor variables using regression, and find patterns in datasets with many variables using principal components analysis and clustering. This part provides the foundation for the analysis of big data. In the practicals the emphasis is on applying theory to analysing real datasets using the statistical software package R. A key feature of the unit is using R to develop coding skills that have become essential in science for processing and analysing datasets of ever-increasing size.

澳洲代写｜ENVX2001｜Applied Statistical Method应用统计方法悉尼大学

Detail	Information
Unit Code	ENVX2001
Unit Name	Applied Statistical Methods
Academic Unit	Life and Environmental Sciences Academic Operations
Session, Year	Semester 1, 2023
Location	Camperdown/Darlington, Sydney
Credit Points	6

Applied Statistical Method应用统计方法问题集

问题 1.

Consider the following set of numbers: $2,5,6,7,11,15,20,22$, and 23 . Find the sum of the first 3 numbers.

Solution:
This set of numbers forms an array, since they are listed in order from the smallest to the largest. To sum the first three numbers we write
$$
\sum_{i=1}^3 X_i=X_1+X_2+X_3=2+5+6=13 .
$$
The expression $i=1$, below the summation sign, is called the lower limit of the summation, and the number 3 , in this case, is called the upper limit. In general, in case we like to add all the numbers in the array, the order here does not matter. We can add them in any order they are given. There is no need to arrange them in an array.

问题 2.

Consider the $\mathrm{X}$ array as $2,4,6$, and 8 ; while the $\mathrm{Y}$ array to be given by $3,5,7$, and 9 .

$$
\begin{aligned}
& \sum_{i=1}^4 X_i Y_i=2(3)+4(5)+6(7)+8(9)=6+20+42+72=140 \
& \left(\sum_{i=1}^4 X_i\right) \cdot\left(\sum_{i=1}^4 Y_i\right)=(2+4+6+8) \cdot(3+5+7+9)=20 \cdot 24=480 .
\end{aligned}
$$
No doubt, we see that $140 \neq 480$.

问题 3.

Consider the following set of data: $5,8,12,15$, and 20. For this data, find
a) The geometric mean,
b) The harmonic mean.
c) Compare the above three means: $\bar{x}, \bar{G}$, and $\bar{H}$.

a) The geometric mean is given by $\bar{G}=\left(x_1, x_2 \ldots x_n\right)^{1 / n}$, and we have $n=5$, and $X_1=5, X_2=8$, $\mathrm{X}_3=12, \mathrm{X}_4=15$, and $\mathrm{X}_5=20$. Applying the formula for, $\bar{G}$ we see that with a graphing calculator that $\bar{G}=\left(5^{\star} 8^{\star} 12^{\star} 15^{\star} 20\right)^{1 / 5}=(144000)^{1 / 5}=10.7565$.
b) The Harmonic mean is given by $\bar{H}=n / \sum_1^n\left(1 / X_i\right)$. From the data, and by using a graphing calculator we find that $\bar{H}=5 /[1 / 5+1 / 8+1 / 12+1 / 15+1 / 20]=9.524$.
c) For the comparison, we need to calculate the arithmetic mean $\bar{x}$. It is easily found that it equals to $60 / 5=12$. Therefore we have $\bar{H}<\bar{G}<\bar{x}$.

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

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。

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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

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

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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｜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 $\infty$.

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

澳洲代写随机过程Stochastic process代考2023

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

如果你也在随机过程Stochastic process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随机过程Stochastic process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。我们的专家在代考随机过程Stochastic process相关的作业也就用不着说。

随机过程Stochastic process代考

在数学中，更确切地说，在概率论中，随机过程（或随机过程）是动态系统概念的概率版本。随机过程是某个参数（通常是时间）的实函数的有序集合，具有某些统计特性。一般来说，可以将这一过程确定为实数随机变量 $X(t)$ 的单参数族，代表从初始状态到一定时间 $t$ 之后的状态的变换。更精确地说，这是基于实数极限上的随机变量（如 $\mathbb{R}^n$，或函数空间，或实数的连续）。当时间参数也被考虑在内时，随机过程就是随机变量概念的扩展。

随机过程Stochastic process包含几个不同的主题，列举如下：

泛函分析Functional analysis代写代考

函数分析（也写作函数分析，Analyse fonctionnelle，法语。也称拓扑分析）。是数学（尤其是分析）的一个分支，起源于对傅立叶变换、微分方程和积分方程的研究。它研究函数空间的结构，为由某类函数组成的向量空间定义了某种拓扑结构，并研究由其公理化得到的线性拓扑空间。主要兴趣在于通过各种函数空间上由积分和微分定义的线性算子的行为，对积分和微分方程进行线性代数处理，这通常被视为无穷维向量空间上的线性代数。它也可被视为无穷维空间上的微分和积分微积分，因为它涉及无穷维空间上的导数（如弗雷谢特导数）。

傅立叶分析Fourier analysis代写代考

在数学分析中，傅里叶分析，也称为调和分析，是一个研究分支，始于让·巴蒂斯特·约瑟夫·傅里叶的研究，他在十九世纪初成功地用数学方法证明了如何将任何周期函数分解为无限“适当”正弦函数或分量（正弦和余弦）的总和，称为谐波。从这一观察中诞生了将复杂函数分解为一系列函数（称为傅里叶级数）的想法，使它们的分析更简单、更有利。傅里叶变换的概念以及相关的频域概念也源自傅里叶级数的数学概念。

其他相关科目课程代写：

数学模型Mathematical model
线性代数Linear algebra
概率学Probability

随机过程Stochastic process定义

随机或随机过程可以定义为由某个数学集合索引的随机变量集合，这意味着随机过程中的每个随机变量都与集合中的某个元素唯一相关。从历史上看，索引集是实线的某个子集，如自然数，这就赋予了索引集时间的解释[1]。例如，状态空间可以是整数、实线或……。n 维欧几里得空间]。增量是随机过程在两个指数值之间的变化量，通常被解释为两个时间点。由于随机性，随机过程可以有很多结果，随机过程的单一结果被称为样本函数或实现。

许多领域使用观测值作为时间的函数（或者，更罕见的是，空间变量）。在最简单的情况下，这些观察结果会产生一条明确的曲线。事实上，从地球科学到人文学科，观测或多或少都会出现不稳定的情况。因此，对这些观察结果的解释存在一定的不确定性，这可能反映在使用概率来表示它们上。

随机过程概括了概率中使用的随机变量的概念。它被定义为与所有值 t ∈ T（通常是时间）相关的一系列随机变量 X(t)。

从统计的角度来看，我们将所有可用的观测值x(t)视为过程的实现，这会带来一定的困难。第一个问题涉及以下事实：构建过程的持续时间通常是无限的，而实现则涵盖有限的持续时间。因此，不可能完美地再现现实。第二个更严重的困难是，与随机变量问题不同，有关过程的可用信息通常被简化为单个实现。

A stochastic or stochastic process can be defined as a collection of random variables indexed by some mathematical set, which means that each random variable in the stochastic process is uniquely related to some element in the set. Historically, the index set was some subset of the real lines, such as the natural numbers, which gave the index set a temporal interpretation. For example, the state space can be integers, solid lines, or… n-dimensional Euclidean space]. An increment is the amount of change of a random process between two exponential values, usually interpreted as two points in time. Due to randomness, a random process can have many outcomes, and a single outcome of a random process is called a sample function or realization.

Many fields use observations as functions of time (or, more rarely, spatial variables). In the simplest case, these observations yield a well-defined curve. In fact, from the earth sciences to the humanities, observations are more or less unstable. Therefore, there is a certain uncertainty in the interpretation of these observations, which may be reflected in the use of probabilities to express them.

Stochastic processes generalize the concept of random variables used in probability. It is defined as a sequence of random variables X(t) related to all values t ∈ T (usually time).

From a statistical perspective, we treat all available observations x(t) as realizations of the process, which creates certain difficulties. The first problem concerns the fact that the duration of the build process is usually infinite, whereas the implementation covers a finite duration. Therefore, it is impossible to perfectly reproduce reality. A second, more serious difficulty is that, unlike random variable problems, the available information about the process is often reduced to a single realization.

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随机过程Stochastic process的重难点

什么是伯努利过程Bernoulli process？

每个单一随机变量 $X_{mathrm{i}}$ 只能提供两种结果：成功 (1) 或失败 $(0)$，概率分别为 $p$ 和 $q=1$ – $p$ ：

$p$ e $q=1$ – $p$ :
$$
\begin{aligned}
& P\left(X_i=1\right)=p ; \
& P\left(X_i=0\right)=q=1-p .
\end{aligned}
$$

遵循二项式定律 $B(n,p)$，概率为

$$
P\left(S_n=k\right)=\left(\begin{array}{l}
n \
k
\end{array}\right) p^k q^{n-k}
$$

等于$k$成功和$n-k$失败的序列数乘以其中任何一个出现的概率。
获得一次成功所需的投掷次数由随机变量 $N$ 给出，该随机变量遵循几何比率定律 $q$。

$$
P(N=n)=P\left(S_{n-1}=0\right) \cdot P\left(X_n=1\right)=q^n \frac{p}{q} .
$$

更一般地说，获得 $k$ 成功所需的投掷次数由随机变量 $N_k$ 给出，其规律为
$P\left(N_k=n\right)=P\left(S_{n-1}=k-1\right) \cdot P\left(X_n=1\right)=\left(\begin{array}{c}n-1 \ k-1\end{array}\right) p^k q^{n-k}$
特别是，失败的次数由帕斯卡定律（或负二项式）的随机变量 $P_k=N_k-n$ $P(p, k)$ 给出

$$
P\left(P_k=r\right)=P\left(N_k=r+k\right)=\left(\begin{array}{c}
r+k-1 \
k-1
\end{array}\right) p^k q^r=(-1)^k\left(\begin{array}{c}
-r \
k
\end{array}\right) p^k q^r
$$
Applicazioni
[ modifica | modifica wikitesto ]

什么是随机漫步Random walk？

在一维随机游走中，研究了被约束在两个允许方向上沿直线移动的点粒子的运动。每次移动时，它都会（随机）向右移动一步（以固定概率 $p$）或以概率 $1-p$ 向左移动，并且每一步长度相等且彼此独立。我们的目标是计算在 $\mathrm{N}$ 运动后粒子将返回（如果它返回的话！）到起点的概率。我们引入以下随机变量 $X(N)$，它提供 $N$ 运动后向左迈出的步数；特别是，它模拟了经过适当操纵的硬币投掷 $N$ 后出现的正面的数量。显然这是一个二项式分布的离散随机变量。我们还注意到，事件“返回原点”相当于在 $2 N$ 总步数中向左精确走了 $N$ 步；因此，所寻求的概率相当于 $P{X=N}$，其中二项式 $X$ 的参数为 $n=2 N, k=N, p$ 因此

什么是因子分析Factor analysis？

该模型试图用一组 $k$ 的公共因子 $\left(f_{i, j}\right)$ 来解释 $n$ 个体中每个个体的 $p$ 观察值，其中每个单位的因子数量少于每个单位的观察值 $(k<p)$。每个个体都有自己的 $k$ 共同因子，这些因子通过因子载荷矩阵 $(left(L)(in\mathbb{R}^{p\times k}\right)$与观测值相关，对于单个观测值，根据

$$
P{X=N}=\frac{(2 N) !}{N !(2 N-N) !} p^N(1-p)^N=\left(\begin{array}{c}
2 N \
N
\end{array}\right)\left(p-p^2\right)^N .
$$

例如，如果粒子在每一步 $(p=1 / 2)$ 向左或向右移动的机会均等，则在 $2 N$ 步骤后返回原点的概率为

$$
P{X=N}=\left(\begin{array}{c}
2 N \
N
\end{array}\right)\left(\frac{1}{2}\right)^{2 N} \approx \frac{1}{\sqrt{N \pi}},
$$我们对足够大的 $N$ 应用斯特林近似，

$$
N ! \sim \sqrt{2 \pi N}\left(\frac{N}{e}\right)^N .
$$现在记住随机变量的期望值由下式给出

$$
E[X]=\sum_{n=0}^{\infty} n P(n)
$$

随机过程Stochastic process的相关课后作业范例

这是一篇关于多元统计分析Multivariate Statistical Analysis的作业

问题 1.

Show that in successive tosses of a fair die indefinitely, the probability of obtaining no 6 is 0 .

Solution: For $n \geq 1$, let $E_n$ be the event of at least one 6 in the first $n$ tosses of the die. Clearly,
$$
E_1 \subseteq E_2 \subseteq \cdots \subseteq E_n \subseteq E_{n+1} \subseteq \cdots .
$$
Therefore, $E_n$ ‘s form an increasing sequence of events. Note that $\lim {n \rightarrow \infty} E_n=\bigcup{i=1}^{\infty} E_i$ is the event that in successive tosses of the die indefinitely, eventually a 6 will occur. By the Continuity of the Probability Function (Theorem 1.8), we have
$$
P\left(\lim {n \rightarrow \infty} E_n\right)=\lim {n \rightarrow \infty} P\left(E_n\right)=\lim {n \rightarrow \infty}\left[1-\left(\frac{5}{6}\right)^n\right]=1-\lim {n \rightarrow \infty}\left(\frac{5}{6}\right)^n=1-0=1 .
$$
This shows that, with probability 1 , eventually a 6 will occur. Therefore, the probability of no 6 ever is 0 .

通过对随机过程Stochastic process各方面的介绍，想必您对这门课有了初步的认识。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
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

澳洲代写多元统计分析代写Multivariate Statistical Analysis代考2023

Posted on 2023年10月10日2023年10月17日 by statistics-lab

如果你也在多元统计分析Multivariate Statistical Analysis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。

statistics-lab™ 为您的留学生涯保驾护航在代写代考多元统计分析Multivariate Statistical Analysis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。我们的专家在代考多元统计分析Multivariate Statistical Analysis相关的作业也就用不着说。

多元统计分析代写Multivariate Statistical Analysis代考

多元统计是统计学的一个分支，包括同时观察和分析一个以上的结果变量，即多元随机变量。多元统计涉及了解每种不同形式的多元分析的不同目的和背景，以及它们之间的关系。将多元统计实际应用于某一特定问题时，可能会涉及几种类型的单变量和多元分析，以便了解变量之间的关系及其与所研究问题的相关性。

此外，多元统计还涉及多元概率分布，既包括

如何使用它们来表示观测数据的分布；
如何将它们用作统计推断的一部分，特别是在同一分析涉及多个不同数量的情况下。
涉及多元数据的某些类型的问题，例如简单线性回归和多元回归，通常不被视为多元统计的特例，因为分析是通过考虑给定其他变量的单一结果变量的（单变量）条件分布来处理的。

多元统计分析Multivariate Statistical Analysis包含几个不同的主题，列举如下：

概率论Probability distribution代写代考

在概率论中，当随机变量单独成为概率讨论的主题时，就要考虑该变量的分布。例如，如果知道变量的分布，就可以计算出随机变量取某值的概率、期望值和方差等量。反之，如果考虑分布，变量 $\omega$ 与随机变量之间就失去了对应关系，与其他随机变量的关系也变得不明确。例如，如果给出随机变量 $X$ 和 $Y$ 的分布分别为 $P_X$ 和 $P_Y$，由于不知道两个变量之间的关系，因此无法计算 $X+Y$ 取某值的概率、乘积 $X Y$ 的期望值和 $X+Y$ 的方差等量。要计算这些量，需要同时得到 $X$ 和 $Y$ 的概率分布 $P_{X，Y}$。
常用的概率分布都有自己的名称，其性质也得到了很好的研究。研究成果可用于具有此类分布的随机变量。例如，如果随机变量的分布是均值为 0、方差为 1 的正态分布，那么从表中可以看出变量值为 2 或 2 以上的概率为 2.28%$。

简单线性回归Simple linear regression代写代考

在统计学中，简单线性回归是一种具有单一解释变量的线性回归模型。也就是说，它涉及具有一个自变量和一个因变量（通常是直角坐标系中的 x 坐标和 y 坐标）的二维样本点，并找到一个线性函数（非垂直直线），尽可能准确地预测因变量值与自变量的函数关系。形容词 “简单 “指的是结果变量与单一预测因子相关。

通常还规定应使用普通最小二乘法（OLS）：每个预测值的准确性由其残差平方（数据集点与拟合线之间的垂直距离）来衡量，目标是使这些平方差之和尽可能小。其他可替代普通最小二乘法的回归方法包括最小绝对偏差法（最小化残差绝对值之和）和 Theil-Sen 估计法（选择一条斜率为样本点对确定的斜率中值的直线）。戴明回归（总最小二乘法）也能找到一条与二维样本点集合拟合的直线，但（与普通最小二乘法、最小绝对偏差和中位斜率回归不同）它并不是真正的简单线性回归，因为它没有将坐标分为一个因变量和一个自变量，有可能返回一条垂直线作为拟合结果。

其他相关科目课程代写：

回归分析Regression analysis
因变量和自变量Dependent and independent variables
统计推断Statistical inference

多元统计分析Multivariate Statistical Analysis定义

多变量分析（MVA）以多变量统计原理为基础。通常，多变量分析用于处理对每个实验单元进行多次测量的情况，这些测量之间的关系及其结构非常重要：

正态和一般多元模型及分布理论
关系的研究与测量
多维区域的概率计算
数据结构和模式的探索
多变量分析可能会因希望进行物理分析以计算变量对分层 “系统体系 “的影响而变得复杂。希望使用多变量分析的研究常常因问题的维度而停滞不前。使用代用模型（基于物理的代码的高精度近似值）往往可以缓解这些问题。由于代用模型采用方程形式，因此可以非常快速地进行评估。这为大规模 MVA 研究提供了有利条件：在设计空间内进行蒙特卡罗模拟对于基于物理的代码来说非常困难，而在评估代用模型（通常采用响应面方程的形式）时则变得微不足道。

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

多元统计分析Multivariate Statistical Analysis的重难点

什么是多变量方差分析Multivariate analysis of variance？

在统计学中，多元方差分析（MANOVA）是一种比较多元样本均值的程序。作为一种多变量程序，它适用于有两个或两个以上因变量的情况，并且通常会对各个因变量分别进行显著性检验。

在与图像无关的情况下，因变量可以是在连续时间点测量的 k 个生活满意度得分和在连续时间点测量的 p 个工作满意度得分。在这种情况下，有 k+p 个因变量，它们的线性组合遵循多元正态分布、多元方差-协方差矩阵同质性和线性关系，不存在多重共线性，且每个因变量都没有异常值。

假设有 $n q$ 维观测值，其中第 $i^{prime}$ 个观测值 $y_i$ 被分配到组 $g(i) \in{1, \ldots, m}$ 中，并围绕组中心 $\mu^{(g(i))} 分布。\in \mathbb{R}^q$ 与多变量高斯噪声：
$$
y_i=\mu^{(g(i))}+\varepsilon_i \quad \varepsilon_i \stackrel{text { i.i.d. }}{\sim} \mathcal{N}_q(0, \Sigma) \quad \text { for } i=1, \ldots, n
$$
其中 $\Sigma$ 是协方差矩阵。然后，我们将零假设表述为
$$
H_0: \mu^{(1)}=\mu^{(2)}=\cdots=\mu^{(m)}
$$

什么是主成分分析Principal component analysis？

PCA 被定义为一种正交线性变换，它将数据转换到一个新的坐标系中，使数据的某个标量投影的最大方差位于第一个坐标上（称为第一个主成分），第二个最大方差位于第二个坐标上，以此类推${ }^{[12]}$
考虑一个$n \times p$ 的数据矩阵, $\mathrm{X}$,其列的经验平均值为零（每列的样本平均值已被移至零），其中每行 $n$ 表示实验的不同重复，每列 $p$ 表示一种特定的特征（例如，来自特定传感器的结果）。
在数学上，变换由一组大小为 $l$ 的 $p$ 维权重或系数向量定义 s $\mathbf{w}{(k)}=\left(w_1, \ldots, w_p\right){(k)}$、将$\mathbf{x}_{(i)}$的每个行向量 $\mathbf{X}$映射到一个新的主成分得分向量s $\mathbf{t}{(i)}=\left(t_1, \ldots, t_l\right){(i)}$。
上图是一个树枝图，用于帮助解释 PCA 并决定保留多少个成分。折线的起点（拐点）应表示保留多少个成分，因此在本例中，应保留三个因子。

$$
t_{k(i)}=\mathbf{x}{(i)} \cdot \mathbf{w}{(k)} \quad \text { for } \quad i=1, \ldots, n \quad k=1, \ldots, l
$$

这样，在数据集上考虑$t_1, \ldots, t_l$ of $\mathbf{t}$ 的单个变量 $t_1，\ldots，t_l$ 将连续地从 $\mathbf{X}$ 继承可能的最大方差，每个系数向量 $\mathbf{w}$ 被约束为一个单位向量（通常选择 $l$ 严格小于 $p$，以降低维度）。

什么是因子分析Factor analysis？

$$
x_{i, m}-\mu_i=l_{i, 1} f_{1, m}+\cdots+l_{i, k} f_{k, m}+\varepsilon_{i, m}
$$
其中

$x_{i, m}$ 是第 $m$ 个体的第 $i$ 次观测值、
$mu_i$ 是第 i 个观测值的观测平均值、
$l_{i, j}$ 是 $i$ 观测值在 $j$ 因子中的负荷、
$f_{j, m}$ 是第 $m$ 个体的第 $j$ 个因子的值，以及
$varepsilon_{i, m}$ 是第 $(i, m)$ 个未观测到的随机误差项，其均值为零，方差有限。
用矩阵符号表示
$$
X-\mathrm{M}=L F+\varepsilon
$$
其中，观测矩阵$X \in \mathbb{R}^{p \times n}$，加载矩阵$L \in \mathbb{R}^{p \times k}$，因子矩阵$F \in \mathbb{R}^{k \times n}$，误差项矩阵$\varepsilon \in \mathbb{R}^{p \times n}$，均值矩阵$\mathrm{M} \in \mathbb{R}^{p \times n}$ 个元素为 $\mathrm{M}_{i, m}=\mu_i$。
同时，我们将对 $F$ 作如下假设：
$F$ 和 $\varepsilon$ 是独立的。
$\mathrm{E}(F)=0$; 其中 $\mathrm{E}$ 是期望值。
$operatorname{Cov}(F)=I$；其中，$operatorname{Cov}$ 是协方差矩阵，以确保各因子互不相关，而 $I$ 是同一矩阵。
假设 $\operatorname{Cov}(X-\mathrm{M})=\Sigma$。那么
$$
\Sigma=\operatorname{Cov}(X-\mathrm{M})=\operatorname{Cov}(L F+\varepsilon)、
$$
因此，根据上述施加于 $F$ 的条件 1 和 2，$E[L F]=L E[F]=0$ 和 $\operatorname{Cov}(L F+\epsilon)=\operatorname{Cov}(L F)+\operatorname{Cov}(\epsilon)$ ，得出
$$
\Sigma=L \operatorname{Cov}(F) L^T+\operatorname{Cov}(\varepsilon)
$$
或者，设置 $\Psi:=\operatorname{Cov}(\varepsilon)$、
$$
\Sigma=L L^T+\Psi .
$$
请注意，对于任意正交矩阵 $Q$，如果我们设置 $L^{\prime}=L Q$ 和 $F^{\prime}=Q^T F$，因子和因子载荷的标准仍然成立。因此，一组因子和因子载荷只有在正交变换时才是唯一的。

多元统计分析Multivariate Statistical Analysis的相关课后作业范例

这是一篇关于多元统计分析Multivariate Statistical Analysis的作业

问题 1.

If $\mathbf{\Sigma}$ is positive definite, so that $\Sigma^{-1}$ exists, then
$$
\mathbf{\Sigma} \mathbf{e}=\lambda \mathbf{e} \text { implies } \mathbf{\Sigma}^{-1} \mathbf{e}=\left(\frac{1}{\lambda}\right) \mathbf{e}
$$
so $(\lambda, \mathbf{e})$ is an eigenvalue-eigenvector pair for $\mathbf{\Sigma}$ corresponding to the pair $(1 / \lambda, \mathbf{e})$ for $\mathbf{\Sigma}^{-1}$. Also, $\mathbf{\Sigma}^{-1}$ is positive definite.

Proof. For $\Sigma$ positive definite and $\mathbf{e} \neq \mathbf{0}$ an eigenvector, we have $0<\mathbf{e}^{\prime} \mathbf{\Sigma} \mathbf{e}=\mathbf{e}^{\prime}(\mathbf{\Sigma} \mathbf{e})$ $=\mathbf{e}^{\prime}(\lambda \mathbf{e})=\lambda \mathbf{e}^{\prime} \mathbf{e}=\lambda$. Moreover, $\mathbf{e}=\mathbf{\Sigma}^{-1}(\mathbf{\Sigma} \mathbf{e})=\mathbf{\Sigma}^{-1}(\lambda \mathbf{e})$, or $\mathbf{e}=\lambda \mathbf{\Sigma}^{-1} \mathbf{e}$, and division by $\lambda>0$ gives $\mathbf{\Sigma}^{-1} \mathbf{e}=(1 / \lambda)$ e. Thus, $(1 / \lambda, \mathbf{e})$ is an eigenvalue-eigenvector pair for $\mathbf{\Sigma}^{-1}$. Also, for any $p \times 1 \mathbf{x}$, by $(2-21)$
$$
\begin{aligned}
\mathbf{x}^{\prime} \mathbf{\Sigma}^{-1} \mathbf{x} & =\mathbf{x}^{\prime}\left(\sum_{i=1}^p\left(\frac{1}{\lambda_i}\right) \mathbf{e}i \mathbf{e}_i^{\prime}\right) \mathbf{x} \ & =\sum{i=1}^p\left(\frac{1}{\lambda_i}\right)\left(\mathbf{x}^{\prime} \mathbf{e}i\right)^2 \geq 0 \end{aligned} $$ since each term $\lambda_i^{-1}\left(\mathbf{x}^{\prime} \mathbf{e}_i\right)^2$ is nonnegative. In addition, $\mathbf{x}^{\prime} \mathbf{e}_i=0$ for all $i$ only if $\mathbf{x}=\mathbf{0}$. So $\mathbf{x} \neq \mathbf{0}$ implies that $\sum{i=1}^p\left(1 / \lambda_i\right)\left(\mathbf{x}^{\prime} \mathbf{e}_i\right)^2>0$, and it follows that $\mathbf{\Sigma}^{-1}$ is positive definite.
The following summarizes these concepts:
Contours of constant density for the $p$-dimensional normal distribution are ellipsoids defined by $\mathbf{x}$ such the that
$$
(\mathbf{x}-\boldsymbol{\mu})^{\prime} \mathbf{\Sigma}^{-1}(\mathbf{x}-\boldsymbol{\mu})=c^2
$$
These ellipsoids are centered at $\boldsymbol{\mu}$ and have axes $\pm c \sqrt{\lambda_i} \mathbf{e}_i$, where $\mathbf{\Sigma} \mathbf{e}_i=\lambda_i \mathbf{e}_i$ for $i=1,2, \ldots, p$.

A contour of constant density for a bivariate normal distribution with $\sigma_{11}=\sigma_{22}$ is obtained in the following example.

通过对多元统计分析Multivariate Statistical Analysis各方面的介绍，想必您对这门课有了初步的认识。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
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