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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｜MATH4512｜Stochastic Analysis随机分析悉尼大学

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

statistics-lab^TM为您悉尼大学（英语：The University of Sydney）Linear Algebra (Advanced)进阶实分析澳洲代写代考和辅导服务！

课程介绍：

Capturing random phenomena is a challenging problem in many disciplines from biology, chemistry and physics through engineering to economics and finance. There is a wide spectrum of problems in these fields, which are described using random processes that evolve with time. Hence it is of crucial importance that applied mathematicians are equipped with tools used to analyse and quantify random phenomena. This unit will introduce an important class of stochastic processes, using the theory of martingales. You will study concepts such as the Ito stochastic integral with respect to a continuous martingale and related stochastic differential equations. Special attention will be given to the classical notion of the Brownian motion, which is the most celebrated and widely used example of a continuous martingale. By completing this unit, you will learn how to rigorously describe and tackle the evolution of random phenomena using continuous time stochastic processes. You will also gain a deep knowledge about stochastic integration, which is an indispensable tool to study problems arising, for example, in Financial Mathematics.

澳洲代写｜MATH4512｜Stochastic Analysis随机分析悉尼大学

Attribute	Detail
Course Code	MATH4512
Course Title	Stochastic Analysis
Academic Unit	Mathematics and Statistics Academic Operations
Session	Semester 2, 2023
Credit Points	6
Pre-Requisites	Not explicitly mentioned in the provided text
Course Coordinator/Lecturer	Not explicitly mentioned in the provided text

Linear Algebra (Advanced)进阶实分析的案例

问题 1.

Problem 1. The following identity was used in the proof of Theorem 1 in Lecture 4: $\sup {\theta>0: M(\theta)<\exp (C \theta)}=\inf _{t>0} t I\left(C+\frac{1}{t}\right)$ (see the proof for details). Establish this identity.

Proof. The lecture note 4 has shown that ${\theta>0: M(\theta)<\exp (C \theta)}$ is nonempty. Let $$ \theta^:=\sup {\theta>0: M(\theta)<\exp (C \theta)} $$ If $\theta^=\infty$, which implies that for all $\theta>0, M(\theta)<\exp (C \theta)$ holds, we have $$ \inf {t>0} t I\left(C+\frac{1}{t}\right)=\inf {t>0} \sup {\theta \in \mathbb{R}}{t(C \theta-\log M(\theta))+\theta}=\infty=\theta^* $$ Consider the case in which $\theta^$ is finite. According to the definition of $I\left(C+\frac{1}{t}\right)$, we have $$ \begin{aligned} I\left(C+\frac{1}{t}\right) & \geq \theta^\left(C+\frac{1}{t}\right)-\log M\left(\theta^\right) \ \Rightarrow \inf {t>0} t I\left(C+\frac{1}{t}\right) & \geq \inf {t>0} t\left(\theta^\left(C+\frac{1}{t}\right)-\log M\left(\theta^\right)\right) \ & =\inf {t>0} t\left(\theta^ C-\log M\left(\theta^\right)\right)+\theta^ \
& \geq \theta^*
\end{aligned}
$$
Next, we will establish the convexity of $\log M(\theta)$ on ${\theta \in \mathbb{R}: M(\theta)<\infty}$. For two $\theta_1, \theta_2 \in{\theta \in \mathbb{R}: M(\theta)<\infty}$ and $0<\alpha<1$, Hölder’s inequality gives
$$
\mathbb{E}\left[\exp \left(\left(\alpha \theta_1+(1-\alpha) \theta_2\right) X\right)\right] \leq \mathbb{E}\left[\left(\exp \left(\alpha \theta_1 X\right)\right)^{\frac{1}{\alpha}}\right]^\alpha \mathbb{E}\left[\left(\exp \left((1-\alpha) \theta_1 X\right)\right)^{\frac{1}{1-\alpha}}\right]^{1-\alpha}
$$
Taking the log operations on both sides gives
$$
\log M\left(\alpha \theta_1+(1-\alpha) \theta_2\right) \leq \alpha \log M\left(\theta_1\right)+(1-\alpha) M\left(\theta_2\right)
$$
By the convexity of $\log M(\theta)$, we have
$$
\begin{aligned}
\left(C+\frac{1}{t}\right) \theta-\log M(\theta) & \leq\left(C+\frac{1}{t}\right) \theta-\theta^* C-\frac{\dot{M}\left(\theta^\right)}{M\left(\theta^\right)}\left(\theta-\theta^\right) \ & =\left(C-\frac{\dot{M}\left(\theta^\right)}{M\left(\theta^\right)}+\frac{1}{t}\right)\left(\theta-\theta^\right)+\frac{\theta^*}{t}
\end{aligned}
$$

Thus, we have
$$
\inf {t>0} t \sup {\theta \in \mathbb{R}}\left[\left(C+\frac{1}{t}\right) \theta-\log M(\theta)\right] \leq \inf {t>0} t \sup {\theta \in \mathbb{R}}\left[\left(C-\frac{\dot{M}\left(\theta^\right)}{M\left(\theta^\right)}+\frac{1}{t}\right)\left(\theta-\theta^\right)\right]+\theta^
$$
Then we will establish the fact that $\frac{\dot{M}\left(\theta^\right)}{M\left(\theta^\right)} \geq C$. If not, then there exists a sufficiently small $h>0$ such that
$$
\frac{\log M\left(\theta^-h\right)-\log M\left(\theta^\right)}{-h}\log M\left(\theta^\right)-C h \ \Rightarrow \log M\left(\theta^-h\right) & >C\left(\theta^-h\right) \Rightarrow M\left(\theta^-h\right) \geq \exp \left(C\left(\theta^-h\right)\right) \end{aligned} $$ which contradicts the definition of $\theta^$. By the facts that
$$
\inf {t>0} t \sup {\theta \in \mathbb{R}}\left[\left(C-\frac{\dot{M}\left(\theta^\right)}{M\left(\theta^\right)}+\frac{1}{t}\right)\left(\theta-\theta^\right)\right] \geq 0,\left(\text { when } \theta=\theta^\right)
$$
and $\frac{\dot{M}\left(\theta^\right)}{M\left(\theta^\right)} \geq C$, we have that
$$
\inf {t>0} t \sup {\theta \in \mathbb{R}}\left[\left(C-\frac{\dot{M}\left(\theta^\right)}{M\left(\theta^\right)}+\frac{1}{t}\right)\left(\theta-\theta^\right)\right]=0 $$ and the infimum is obtained at $t^>0$ such that $C+\frac{1}{t^}-\frac{\dot{M}\left(\theta^\right)}{M\left(\theta^\right)}=0$. From (2), we have $$ \begin{aligned} & \inf {t>0} t \sup {\theta \in \mathbb{R}}\left[\left(C+\frac{1}{t}\right) \theta-\log M(\theta)\right] \leq \theta^ \
\Rightarrow & \inf {t>0} t I\left(C+\frac{1}{t}\right) \leq \theta^* \end{aligned} $$ From (1) and (3), we have the result $\inf {t>0} t I\left(C+\frac{1}{t}\right)=\theta^*$.

问题 2.

Problem 2. This problem concerns the rate of convergence to the limits for the large deviations bounds. Namely, how quickly does $n^{-1} \log \mathbb{P}\left(n^{-1} S_n \in A\right)$ converge to $-\inf _{x \in A} I(x)$, where $S_n$ is the sum of $n$ i.i.d. random variables? Of course the question is relevant only to the cases when this convergence takes place.
(a) Let $S_n$ be the sum of $n$ i.i.d. random variables $X_i, 1 \leq i \leq n$ taking values in $\mathbb{R}$. Suppose the moment generating function $M(\theta)=\mathbb{E}[\exp (\theta X)]$ is finite everywhere. Let $a \geq \mu=\mathbb{E}[X]$. Recall that we have established in class that in this case the convergence $\lim _n n^{-1} \log \mathbb{P}\left(n^{-1} S_n \geq a\right)=$ $-I(a)$ takes place. Show that in fact there exists a constant $C$ such that
$$
\left|n^{-1} \log \mathbb{P}\left(n^{-1} S_n \geq a\right)+I(a)\right| \leq \frac{C}{n},
$$
for all sufficiently large $n$. Namely, the rate of convergence is at least as fast as $O(1 / n)$.

(a). Let $\theta_0$ be the one satisfying $I(a)=\theta_0 a-\log M\left(\theta_0\right)$ and $\delta$ be a small positive number. Following the proof of the lower bound of Cramer’s theorem, we have
$$
\begin{aligned}
n^{-1} \log \mathbb{P}\left(n^{-1} S_n \geq a\right) & \geq n^{-1} \log \mathbb{P}\left(n^{-1} S_n \in[a, a+\delta)\right) \
& \geq-I(a)-\theta_0 \delta-n^{-1} \log \mathbb{P}\left(n^{-1} \tilde{S}n-a \in[0, \delta)\right) \end{aligned} $$ where $\tilde{S}_n=Y_1+\ldots+Y_n$ and $Y_i(1 \leq i \leq n)$ is i.i.d. random variable following the distribution $\mathbb{P}\left(Y_i \leq z\right)=M\left(\theta_0\right)^{-1} \int{-\infty}^z \exp \left(\theta_0 x\right) d P(x)$. Recall that
$$
\mathbb{P}\left(n^{-1} \tilde{S}n-a \in[0, \delta)\right)=\mathbb{P}\left(\frac{\sum{i=1}^n\left(Y_i-a\right)}{\sqrt{n}} \in[0, \sqrt{n} \delta)\right)
$$
By the CLT, setting $\delta=O\left(n^{-1 / 2}\right)$ gives
$$
\mathbb{P}\left(n^{-1} \tilde{S}_n-a \in[0, \delta)\right)=O(1)
$$
Thus, we have
$$
\begin{aligned}
n^{-1} \log \mathbb{P}\left(n^{-1} S_n \geq a\right)+I(a) & \geq-\theta_0 \delta-n^{-1} \log \mathbb{P}\left(n^{-1} \tilde{S}_n-a \in[0, \delta)\right) \
& =-O\left(n^{-1 / 2}\right)
\end{aligned}
$$
Combining the result from the upper bound $n^{-1} \log \mathbb{P}\left(n^{-1} S_n \geq a\right) \leq-I(a)$, we have
$$
\left|n^{-1} \log \mathbb{P}\left(n^{-1} S_n \geq a\right)+I(a)\right| \leq \frac{C}{\sqrt{n}}
$$

问题 3.

(b) Show that the rate $O(1 / n)$ cannot be improved.
Hint: Consider the case $a=\mu$.

(b). Take $a=\mu$. It is obvious, $\mathbb{P}\left(n^{-1} S_n \geq \mu\right) \rightarrow \frac{1}{2}$ as $n \rightarrow \infty$. Recalling that $I(\mu)=0$, we have
$$
\left|n^{-1} \log \mathbb{P}\left(n^{-1} S_n \geq \mu\right)+I(\mu)\right| \sim \frac{C}{n}
$$
Namely, this bound can not be improved.

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｜MATH1902｜Linear Algebra (Advanced)进阶实分析悉尼大学

Posted on 2023年10月12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）Linear Algebra (Advanced)进阶实分析澳洲代写代考和辅导服务！

课程介绍：

This unit is designed to provide a thorough preparation for further study in mathematics and statistics. It is a core unit of study providing three of the twelve credit points required by the Faculty of Science as well as a foundations requirement in the Faculty of Engineering. It parallels the normal unit MATH1002 but goes more deeply into the subject matter and requires more mathematical sophistication.

澳洲代写｜MATH1902｜ Linear Algebra (Advanced)进阶实分析悉尼大学

Attribute	Detail
Course Code	MATH1902
Course Title	Linear Algebra (Advanced)
Academic Unit	Mathematics and Statistics Academic Operations
Session	Semester 1, 2023
Number of Units	3
Pre-Requisites	It parallels the normal unit MATH1002
Course Coordinator	Not explicitly mentioned in the provided text
Lecturer	Not explicitly mentioned in the provided text

Linear Algebra (Advanced)进阶实分析的案例

问题 1.

Suppose $A$ is a positive definite symmetric $n$ by $n$ matrix.
(a) How do you know that $A^{-1}$ is also positive definite? (We know $A^{-1}$ is symmetric. I just had an e-mail from the International Monetary Fund with this question.)

Solution Since a matrix is positive-definite if and only if all its eigenvalues are positive
(5 points), and since the eigenvalues of $A^{-1}$ are simply the inverses of the eigenvalues of $A, A^{-1}$ is also positive definite (the inverse of a positive number is positive)

问题 2.

(b) Suppose $Q$ is any orthogonal $n$ by $n$ matrix. How do you know that $Q A Q^T=Q A Q^{-1}$ is positive definite? Write down which test you are using.

Solution Using the energy text $\left(x^T A x>0\right.$ for nonzero $x$ ), we find that $x^T Q A Q^T x=$ $\left(Q^T x\right)^T A\left(Q^T x\right)>0$ for all nonzero $x$ as well (since $Q$ is invertible). Using the positiv eigenvalue test, since $A$ is similar to $Q A Q^{-1}$ and similar matrices have the sam eigenvalues, $Q A Q^{-1}$ also has all positive eigenvalues.

问题 3.

Show that the block matrix
$$
B=\left[\begin{array}{ll}
A & A \
A & A
\end{array}\right]
$$
is positive semidefinite. How do you know $B$ is not positive definite?

Solution First, since $B$ is singular, it cannot be positive definite (it has eigenvalues of 0 ). However, the pivots of $B$ are the pivots of $A$ in the first $n$ rows followed by 0 s in the remaining rows, so by the pivot test, $B$ is still semi-definite. Similarly, the first $n$ upper-left determinants of $B$ are the same as those of $A$, while the remaining ones are 0s, giving another proof. Finally, given a nonzero vector
$$
u=\left[\begin{array}{l}
x \
y
\end{array}\right]
$$
where $x$ and $y$ are vectors in $\mathbf{R}^n$, one has $u^T B u=(x+y)^T A(x+y)$ which is nonnegative (and zero when $x+y=0$ ).

Linear Algebra (Advanced)进阶实分析的案例2

问题 4.

(a) $p=A \widehat{x}$ is the vector in $C(A)$ nearest to a given vector $b$. If $A$ has independent columns, what equation determines $\widehat{x}$ ? What are all the vectors perpendicular to the error $e=b-A \widehat{x}$ ? What goes wrong if the columns of $A$ are dependent?

Solution $\widehat{x}$ is determined by the equation $\widehat{x}=\left(A^T A\right)^{-1} A^T b$ (since $A$ has independent columns, $A^T A$ is invertible whether or not $A$ is square). The vectors perpendicular to an arbitrary error vector are the elements of the column space of $A$. If the columns of $A$ are dependent, $A^T A$ is no longer invertible, and there is no unique nearest vector (i.e. there are multiple solutions).

问题 5.

(b) Suppose $A=Q R$ where $Q$ has orthonormal columns and $R$ is upper triangular invertible. Find $\widehat{x}$ and $p$ in terms of $Q$ and $R$ and $b(\operatorname{not} A)$.

Solution Since $Q^T Q=I$ and $R$ is invertible, we obtain
$$
\begin{aligned}
\widehat{x} & =\left(A^T A\right)^{-1} A^T b=\left((Q R)^T(Q R)\right)^{-1}(Q R)^T b \
& =\left(R^T Q^T Q R\right)^{-1} R^T Q^T b=R^{-1}\left(R^T\right)^{-1} R^T Q^T b=R^{-1} Q^T b \
p & =(Q R) \widehat{x}=Q Q^T b
\end{aligned}
$$
Note that $Q Q^T$ is not the identity matrix in general.

问题 6.

(c) If $q_1$ and $q_2$ are any orthonormal vectors in $R^5$, give a formula for the projection $p$ of any vector $b$ onto the plane spanned by $q_1$ and $q_2$ (write $p$ as a combination of $q_1$ and $q_2$ ).

$p=\left(q_1^T b\right) q_1+\left(q_2^T b\right) q^2$

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｜PHYS3034｜ Quantum, Statistical and Comp Physics量子、统计和复合物理悉尼大学

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

statistics-lab^TM为您悉尼大学（英语：The University of Sydney） Quantum, Statistical and Comp Physics量子、统计和复合物理澳洲代写代考和辅导服务！

课程介绍：

The dynamics of complex systems are often described in terms of how they process information and self-organise; for example regarding how genes store and utilise information, how information is transferred between neurons in undertaking cognitive tasks, and how swarms process information in order to collectively change direction in response to predators. The language of information also underpins many of the central concepts of complex adaptive systems, including order and randomness, self-organisation and emergence. Shannon information theory, which was originally founded to solve problems of data compression and communication, has found contemporary application in how to formalise such notions of information in the world around us and how these notions can be used to understand and guide the dynamics of complex systems. This unit of study introduces information theory in this context of analysis of complex systems, foregrounding empirical analysis using modern software toolkits, and applications in time-series analysis, nonlinear dynamical systems and data science. Students will be introduced to the fundamental measures of entropy and mutual information, as well as dynamical measures for time series analysis and information flow such as transfer entropy, building to higher-level applications such as feature selection in machine learning and network inference. They will gain experience in empirical analysis of complex systems using comprehensive software toolkits, and learn to construct their own analyses to dissect and design the dynamics of self-organisation in applications such as neural imaging analysis, natural and robotic swarm behaviour, characterisation of risk factors for and diagnosis of diseases, and financial market dynamics.

澳洲代写｜PHYS3034｜ Quantum, Statistical and Comp Physics量子、统计和复合物理悉尼大学

Attribute	Detail
Course Code	PHYS3034
Course Title	Quantum, Statistical and Comp Physics
Academic Unit	Physics Academic Operations
Session	Semester 1, 2023
Number of Units	6
Pre-Requisites	2000-level physics (from the context provided)
Course Coordinator	Not explicitly mentioned in the provided text
Lecturer	Not explicitly mentioned in the provided text

Statistical Physics统计物理的案例

问题 1.

For a particular model of a gene in a cell, the probability density that said gene produces a concentration $x$ of proteins during the cell cycle is given by
$$
p(x)=A\left(\frac{x}{b}\right)^N e^{-x / b},
$$
where $b$ is a biological constant with units of concentration, $N$ is a physical constant, and $A$ is a normalization parameter.
(a) (5 points) The concentration of proteins that can be produced ranges from zero to infinite. What must $A$ be in order for Eq.(1) to be normalized?
(b) (5 points) What is the mean of the normalized probability density?
(c) (Removed in Final Version) For this probability distribution, compute the average
$$
\left\langle e^{x / a}\right\rangle
$$
and write the result in terms of an infinite sum over a binomial coefficient. (Note that $e^x=$ $\sum_{j=0}^{\infty} x^j / j$ !.)

(a) Given the range of possible protein production, $x$ can go from 0 to $\infty$. Therefore, for $p(x)$ to be normalized, we must obtain 1 when we integrate the function over this entire domain:
$$
\int_0^{\infty} d x p(x)=1 .
$$
From the definition of the probability density we have
$$
\begin{aligned}
1 & =\int_0^{\infty} d x A\left(\frac{x}{b}\right)^N e^{-x / b} \
& =A \int_0^{\infty} d x\left(\frac{x}{b}\right)^N e^{-x / b} \
& =A \int_0^{\infty} d u b u^N e^{-u} \
& =A b \int_0^{\infty} d u u^N e^{-u},
\end{aligned}
$$
where we changed variables with $u=x / b$ in the third line, and factored the $u$-independent constant out of the integral in the final line. By the integral definition of factorial, we have
$$
N !=\int_0^{\infty} d u u^N e^{-u} .
$$
Therefore, the final line of Eq.(4) becomes
$$
1=A b N !
$$
and we can conclude
$$
A=\frac{1}{b N !}
$$

The normalized probability density is therefore
$$
p(x)=\frac{1}{b N !}\left(\frac{x}{b}\right)^N e^{-x / b} .
$$
(b) The mean of a random variable defined by the probability density $p(x)$ (which has a nonzero domain for $x \in[0, \infty))$ is
$$
\langle x\rangle=\int_0^{\infty} d x x p(x) .
$$
Using Eq.(8) to compute this value, we obtain
$$
\begin{aligned}
\langle x\rangle & =\int_0^{\infty} d x x \frac{1}{b N !}\left(\frac{x}{b}\right)^N e^{-x / b} \
& =\frac{1}{N !} \int_0^{\infty} d x\left(\frac{x}{b}\right)^{N+1} e^{-x / b} \
& =\frac{1}{N !} \int_0^{\infty} d u b u^{N+1} e^{-u} \
& =b \frac{(N+1) !}{N !},
\end{aligned}
$$
where in the third line we performed a change of variables with $u=x / b$ and in the final line we used Eq.(5). By the definition of factorial, we ultimately find
$$
\langle x\rangle=b(N+1)
$$
(c) (Not part of final version of exam) We now seek to compute the average of $e^{x / a}$. Noting the Taylor series definition of the exponential
$$
e^{x / a}=\sum_{j=0}^{\infty} \frac{(x / a)^j}{j !},
$$
we have
$$
\begin{aligned}
\left\langle e^{x / a}\right\rangle & =\left\langle\sum_{j=0}^{\infty} \frac{(x / a)^j}{j !}\right\rangle \
& =\sum_{j=0}^{\infty} \frac{1}{j !} \frac{1}{a^j}\left\langle x^j\right\rangle .
\end{aligned}
$$
Computing $\left\langle x^j\right\rangle$ yields

\begin{aligned}
\left\langle x^j\right\rangle & =\int_0^{\infty} d x x^j \frac{1}{b N !}\left(\frac{x}{b}\right)^N e^{-x / b} \
& =\frac{b^j}{b N !} \int_0^{\infty} d x\left(\frac{x}{b}\right)^{N+j} e^{-x / b} \
& =\frac{b^j}{b N !} \int_0^{\infty} d u b u^{N+j} e^{-u}
\end{aligned}

$$
=b^j \frac{(N+j) !}{N !}
$$
where in the second line we multiplied the numerator and the denominator by $b^j$, in the third line we performed a change of variables $u=x / b$, and in the final line we used Eq.(5). Inserting this result into Eq.(13), we find
$$
\left\langle e^{x / a}\right\rangle=\sum \sum_{j=0}^{\infty} \frac{1}{j !} \frac{1}{a^j} b^j \frac{(N+j) !}{N !}=\sum_{j=0}^{\infty} \frac{(N+j) !}{j ! N !}\left(\frac{b}{a}\right)^j,
$$
or
$$
\left\langle e^{x / a}\right\rangle=\sum_{j=0}^{\infty}\left(\begin{array}{c}
N+j \
j
\end{array}\right)\left(\frac{b}{a}\right)^j .
$$
We note that we could evaluate $\left\langle e^{x / a}\right\rangle$ directly using a change of variables in the argument of the exponential of the distribution. The result would be
$$
\begin{aligned}
\left\langle e^{x / a}\right\rangle & =\frac{1}{b N !} \int_0^{\infty} d x e^{x / a}\left(\frac{x}{b}\right)^N e^{-x / b} \
& =\frac{1}{b N !} \int_0^{\infty} d x\left(\frac{x}{b}\right)^N e^{-(1 / b-1 / a) x} \
& =\frac{1}{b N !} \int_0^{\infty} d u \frac{a b}{a-b}\left(\frac{a u}{a-b}\right)^N e^{-u} \
& =\frac{1}{N !}\left(\frac{a}{a-b}\right)^{N+1} \int_0^{\infty} d u u^N e^{-u} \
& =\frac{1}{(1-b / a)^{N+1}},
\end{aligned}
$$
where in the second line we made the change of variables $u=x(a-b) / a b$. Considering Eq.(16), the result Eq.(17) implies
$$
\sum_{j=0}^{\infty}\left(\begin{array}{c}
N+j \
j
\end{array}\right) q^j=\frac{1}{(1-q)^{N+1}}
$$

Quantum Physics量子物理问题

问题 2.

True or false questions [ 20 points] No explanations required. Just indicate $\mathrm{T}$ or $\mathrm{F}$ for true or false, respectively.
(1) The operators $\sigma_1 \otimes \sigma_1$ and $\sigma_3 \otimes \sigma_1$ commute.
(2) The operators $\sigma_1 \otimes \sigma_1$ and $\sigma_3 \otimes \sigma_3$ commute.
(3) Let $T \otimes S$ be a linear operator on $V \otimes W$. Then $(T \otimes S)^{\dagger}=S^{\dagger} \otimes T^{\dagger}$.
(4) A linear operator on a finite-dimensional vector space is invertible if it is injective.
(5) Angular momentum conservation prevents a particle with spin $1 / 2$ from decaying into two spin- $1 / 2$ particles.
(6) $\left[\mathbf{L}^2, \hat{x}_i\right]=0$. Here $\hat{x}_i$ is the position operator in any of the three directions.
(7) $\mathbf{r} \cdot \mathbf{p}=\mathbf{p} \cdot \mathbf{r}-3 i \hbar$.
(8) $\mathbf{A} \cdot \mathbf{L}=\mathbf{L} \cdot \mathbf{A}$, when $\mathbf{A}$ is a vector under rotations.
(9) In the hydrogen atom the Runge-Lenz (RL) vector $\mathbf{R}$ satisfies the algebra of angular momentum.
(10) Both classically and quantum mechanically the $R L$ vector $\mathbf{R}$ satisfies $\mathbf{R} \cdot \mathbf{L}=0$.

问题 3.

Expectation value on a generalized squeezed state [10 points] Consider the general squeezed state $|\alpha, \gamma\rangle=D(\alpha) S(\gamma)|0\rangle$ of the harmonic oscillator at time equal zero (here $\alpha \in \mathbb{C}, \gamma \in \mathbb{R}$ ). Find the expectation value of the number operator $\hat{N}$ in this state. As we let time change, does this expectation value exhibit time dependence?

问题 4.

$3 \mathrm{D}$ bound state $[10$ points]
A particle of mass $m$ is in a potential $V(r)$ that represents a finite depth spherical well of radius $a$ :
$$
V(r)=\left{\begin{aligned}
-V_0 & \text { for } ra,
\end{aligned}\right.
$$
Here $V_0>0$. For the potential to have bound states it should be deep enough. Derive the inequality that $V_0$ must satisfy in order that the potential have a bound state.

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｜CSYS5030｜Information Theory and Self-Organisation信息论和自组织悉尼大学

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

statistics-lab^TM为您悉尼大学（英语：The University of Sydney）澳洲代写｜ECMT3150｜The Econometrics of Financial Markets金融市场计量经济学悉尼大学澳洲代写代考和辅导服务！

课程介绍：

澳洲代写｜CSYS5030｜Information Theory and Self-Organisation信息论和自组织悉尼大学

Attribute	Detail
Course Code	CSYS5030
Course Title	Information Theory and Self-Organisation
Academic Unit	Computer Science
Session	Semester 2, 2023
Number of Units	6
Pre-Requisites	MATH2000 or MATH2001
Course Coordinator	Associate Professor Min-Chun Hong (as per the previous course)
Lecturer	Not explicitly mentioned in the provided text

Information Theory 信息论的问题

问题 1.

Prove or disprove the following:
a) $I(X+Y ; Y \mid X)=0$.
b) $H(X \mid Z)-H(X \mid Y, Z)=H(Y \mid Z)-H(Y \mid X, Z)$.
c) $I\left(\left(X_1, \ldots X_n\right) ;\left(Y_1, \ldots Y_n\right)\right)=\sum_{i=1}^n H\left(Y_i \mid Y_1 \ldots Y_{i-1}\right)-H\left(Y_i, X_i \ldots X_n \mid Y_1 \ldots Y_{i-1}, X_1 \ldots X_{i-1}\right)+$ $H\left(X_i \ldots X_n \mid Y_1 \ldots Y_{i-1}, X_1 \ldots X_{i-1}\right)$.
d) $I\left(\left(X_1, \ldots X_n\right) ;\left(Y_1, \ldots Y_n\right)\right) \geq \sum_{i=1}^n H\left(Y_i \mid Y_1 \ldots Y_{i-1}\right)-H\left(Y_i \mid Y_1 \ldots Y_{i-2}, X_1 \ldots X_n\right.$.
Let $Z=X+Y$ in the following:
e) $H(Z)=H(X)+H(Y)$.
f) $H(Z, X)=H(X)+H(Y)$.

问题 2.

Problem 1 Two semi-working street lamps turn on and off independently as follows: within each one-minute interval, a lamp that is on turns off with probability $p$, and a lamp that is off turns on with probability $p$. At time $t=0,1,2 \ldots$ minutes, an observer records the number $N_t$ of street lamps that are on, as well as the change $D_t=N_t-N_{t-1}$ from the previous recorded number.
(a) Do $N_0, N_1, \ldots$ form a Markov process? What is the entropy rate of this sequence?
(b) Do $D_0, D_1, \ldots$ form a Markov process? What is the entropy rate of this sequence?

问题 3.

Problem 3 Consider a sequence of IID binary r.v.s $A_0, A_1, \ldots$ such that $A_i=0$ with probability $\xi$ and $A_i=1$ with probability $1-\xi$ for some $0<\xi<1$. Consider another sequence of IID quaternary r.v.s $\Xi_0, \Xi_1, \ldots$ such that $\Xi_i=0$ with probability $\frac{1-\theta}{3}, \Xi_i=1$ with probability $\frac{1-\theta}{3}, \Xi_i=2$ with probability $\frac{1-\theta}{3}, \Xi_i=3$ with probability $\ddot{\theta}$ for some $0<\theta<1$. The $\Xi_i \mathrm{~s}$ and the $A_i \mathrm{~s}$ are all mutually independent. Consider a sequence of quaternary r.v.s $Z_0, Z_1, \ldots$ such that $\forall i>0$
$$
Z_i=A_i\left(\Xi_{i-1} \oplus Z_{i-1}\right) \oplus \overline{A_i} \Xi_{i-1}
$$
and $Z_0, \Xi_0$ are IID, where $\oplus$ denotes addition $\bmod 4$.
a) What is $H\left(Z_i \mid Z_{i-1}\right)$ ?
b) What is $H\left(Z_i \mid Z_{i-j}\right)$ ?
c) Can you find some form of the AEP that holds for the r.v.s $Z_0, Z_1, \ldots$ ?

Information storage信息存储知识点

Publications of the CRG:

there are a number of bibliographic and bibliometric studies of the CRG
joint publications of the Group are relatively few
regular (although not frequent) Bulletins were published in the Journal of Documentation
three of these contained bibliographies of members’ publications
Vickery is by far the most prolific author, as he continued to be throughout his life.
The new general classification scheme:
in the late 1950 s and 1960 s the main focus of CRG work was the proposed new general scheme of classification
several papers were written, and a conference held, on this topic
a grant from NATO subsidised the original work which never produced a classification, but did result in the PRECIS indexing system
throughout the 1970 s and later the objective was pursued through the revision of Bliss’s Bibliographic Classification.
Divergence of classification and IR:
during the 1960 s ‘classification’ and ‘information retrieval’ begin to develop as distinct and separate fields
this happens in both the US and UK
it’s at this time that Vickery ceases to contribute to the CRG’s activities and his name disappears from the record.
Factors in the ‘split’:
there are distinct ‘library’ and ‘information science’ groupings within the CRG
bibliometric analysis of the CRG publications show quite clear associations of scholars
at some stage the ideas of ‘classification’ and ‘information retrieval’ were uncoupled
the main group continue with work on a classification scheme and on classification for organization
Vickery’s agenda is somewhat different and he turns in other directions.
Faceted classification today:
over the last twenty years facet analysis has become increasingly important as a methodological approach for all kinds of organizational and retrieval tools
it features in classification, subject heading lists, thesauri, search interfaces, in taxonomies, ontologies, and semantic web applications
a number of researchers are attempting to model faceted structures using representation languages and mathematical logic
it looks as if ‘classification’ and ‘information retrieval’ have been reconciled and re-united.
An evaluation of Vickery’s contribution:
a driving force in uniting and stimulating the study and understanding of classification
two intellectual achievements:

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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

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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｜ECMT3150｜The Econometrics of Financial Markets金融市场计量经济学悉尼大学

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

课程介绍：

This unit provides students with an understanding of the economic foundations of financial theory and the economic framework upon which that theory is based. Much of the work covered is an application of both microeconomic and macroeconomic theory to the special problems encountered in the study of the financial side of an economy. The relevance of these foundations is illustrated with empirical research using Australian and international data.

澳洲代写｜ECMT3150｜The Econometrics of Financial Markets金融市场计量经济学悉尼大学

Attribute	Detail
Course Code	ECMT3150
Course Title	The Econometrics of Financial Markets
Academic Unit	Economics
Session	Semester 1, 2023
Number of Units	6
Pre-Requisites	ECMT2130
Course Coordinator	Not explicitly mentioned in the provided text
Lecturer	Not explicitly mentioned in the provided text

Linear time series analysis线性时间序列分析的问题

问题 1.

Let $\mathbf{V}$ be an $n \times n$ positive definite matrix, and let $\mathbf{U}$ and $\mathbf{T}$ be $n \times k$ and $n \times(n-k)$ matrices, respectively, such that, if $\mathbf{W}=[\mathbf{U}, \mathbf{T}]$, then $\mathbf{W}^{\prime} \mathbf{W}=\mathbf{W} \mathbf{W}^{\prime}=\mathbf{I}_n$. Then
$$
\mathbf{V}^{-1}-\mathbf{V}^{-1} \mathbf{U}\left(\mathbf{U}^{\prime} \mathbf{V}^{-1} \mathbf{U}\right)^{-1} \mathbf{U}^{\prime} \mathbf{V}^{-1}=\mathbf{T}\left(\mathbf{T}^{\prime} \mathbf{V T}\right)^{-1} \mathbf{T}^{\prime}
$$

Let $\mathbf{P}=\mathbf{P}{\mathbf{X}}$ be the usual projection matrix on the column space of $\mathbf{X}$ from let $\mathbf{M}=\mathbf{I}_T-\mathbf{P}$, and let $\mathbf{G}$ and $\mathbf{H}$ be matrices such that $\mathbf{M}=\mathbf{G}^{\prime} \mathbf{G}$ and $\mathbf{P}=\mathbf{H}^{\prime} \mathbf{H}$, in which case $\mathbf{W}=\left[\mathbf{H}^{\prime}, \mathbf{G}^{\prime}\right]$ satisfies $\mathbf{W}^{\prime} \mathbf{W}=\mathbf{W} \mathbf{W}^{\prime}=\mathbf{I}_T$. Theorem 1.8 For the regression model given with $\widehat{\epsilon}=\mathbf{M}{\mathbf{\Sigma}} \mathbf{Y}$ from (1.62),
$$
\widehat{\epsilon}^{\prime} \Sigma^{-1} \widehat{\epsilon}=\epsilon^{\prime} \mathbf{G}^{\prime}\left(\mathbf{G} \Sigma \mathbf{G}^{\prime}\right)^{-1} \mathbf{G} \epsilon \text {. }
$$
Proof: $\mathbf{T}=\mathbf{G}^{\prime}, \mathbf{U}=\mathbf{H}^{\prime}$, and $\mathbf{V}=\mathbf{\Sigma}$, and the fact that $\mathbf{H}^{\prime}$ can be written as $\mathbf{X K}$, where $\mathbf{K}$ is a $k \times k$ full rank transformation matrix, we have
$$
\begin{aligned}
& \epsilon^{\prime} \mathbf{G}^{\prime}\left(\mathbf{G} \boldsymbol{\Sigma} \mathbf{G}^{\prime}\right)^{-1} \mathbf{G} \boldsymbol{\epsilon}=\mathbf{U}^{\prime}\left(\mathbf{\Sigma}^{-1}-\mathbf{\Sigma}^{-1} \mathbf{H}^{\prime}\left(\mathbf{H} \boldsymbol{\Sigma}^{-1} \mathbf{H}^{\prime}\right)^{-1} \mathbf{H} \boldsymbol{\Sigma}^{-1}\right) \mathbf{U} \
& =\mathbf{U}^{\prime}\left(\mathbf{\Sigma}^{-1}-\boldsymbol{\Sigma}^{-1} \mathbf{X K}\left(\mathbf{K}^{\prime} \mathbf{X}^{\prime} \mathbf{\Sigma}^{-1} \mathbf{X K}\right)^{-1} \mathbf{K}^{\prime} \mathbf{X}^{\prime} \mathbf{\Sigma}^{-1}\right) \mathbf{U} \
& =\mathbf{U}^{\prime}\left(\boldsymbol{\Sigma}^{-1}-\boldsymbol{\Sigma}^{-1} \mathbf{X}\left(\mathbf{X}^{\prime} \boldsymbol{\Sigma}^{-1} \mathbf{X}\right)^{-1} \mathbf{X}^{\prime} \boldsymbol{\Sigma}^{-1}\right) \mathbf{U}=\widehat{\boldsymbol{\epsilon}}^{\prime} \boldsymbol{\Sigma}^{-1} \widehat{\boldsymbol{\epsilon}}, \
&
\end{aligned}
$$

问题 2.

Consider the simple linear regression model $Y_t=\beta_1+\beta_2 X_t+\epsilon_t, t=1, \ldots, T$.
a) By setting $\partial \mathrm{S}(\boldsymbol{\beta}) / \partial \beta_1$ to zero, show that $\hat{\beta}1=\bar{Y}-\widehat{\beta}_2 \bar{X}$. Using this with $0=\partial \mathrm{S}(\boldsymbol{\beta}) / \partial \beta_2$, show that $\hat{\beta}_2=\hat{\sigma}{X, Y} / \hat{\sigma}X^2$, where $\hat{\sigma}{X, Y}$ denotes the sample covariance between $X$ and $Y$,
$$
\hat{\sigma}{X, Y}:=\frac{1}{T-1} \sum{t=1}^T\left(X_t-\bar{X}\right)\left(Y_t-\bar{Y}\right),
$$
and $\hat{\sigma}X^2:=\hat{\sigma}{X, X}$.
b) Show that $\widehat{Y}t-\bar{Y}=\widehat{\beta}_2\left(X_t-\bar{X}\right)$. c) Define the standardized variables $x_t=\left(X_t-\bar{X}\right) / \hat{\sigma}_X$ and $y_t=\left(Y_t-\bar{Y}\right) / \hat{\sigma}_Y$, and consider the regression $y_t=\alpha_1+\alpha_2 x_t+\varepsilon_t$. Show that $\hat{\alpha}_1=0$ and $\hat{\alpha}_2=\hat{\rho}$, where $\hat{\rho}=\hat{\rho}{X, Y}$ is the sample correlation between $X$ and $Y$, with $|\hat{\rho}| \leqslant 1$. Thus, we can write
$$
\hat{Y}_t=\hat{\alpha}_1+\hat{\alpha}_2 x_t=\hat{\rho} x_t,
$$
and squaring and summing both sides yields $\hat{\rho}^2=\sum \widehat{Y}_t^2 / \sum x_t^2$. Show that the $R^2$ statistics for the two regression models are the same, namely $\hat{\rho}^2$.

Continuous time models, option pricing连续时间模型，期权定价定义

Brownian Motion
A Brownian motion is a real-valued stochastic process with continuous trajectories that have independent and stationary increments. The trajectories are denoted by $t \rightarrow W_t$ and for the standard Brownian motion:

$W_0=0$;
for any $0<t_1<\cdots<t_n$, the random variables $\left(W_{t_1}, W_{t_2}-W_{t_1}, \cdots, W_{t_n}-W_{t_{n-1}}\right)$ are independent;
for any $0 \leq s<t$, the increment $W_t-W_s$ is a centered (mean-zero) normal random variable with variance $\mathbb{E}\left{\left(W_t-W_s\right)^2\right}=t-s$. In particular $W_t$ is $\mathcal{N}(0, t)$-distributed.
$\mathcal{F}t$ denotes the $\sigma$-algebra generated by $\left(W_s\right){s \leq t}$, the information on $W$ up to time $t$.

Conditional characteristic functions
For $0 \leq s<t$ and $u \in \mathbb{R}$
$$
\mathbf{E}\left{\mathrm{e}^{\mathrm{i} \mathbf{u}\left(\mathrm{W}{\mathrm{t}}-\mathrm{W}{\mathrm{s}}\right)} \mid \mathcal{F}_{\mathrm{s}}\right}=\mathrm{e}^{-\frac{\mathrm{u}^2(\mathrm{t}-\mathrm{s})}{2}} .
$$
If $W$ is a Brownian motion, by independence of the increment $W_t-W_s$ from the past $\mathcal{F}_s$, the left-hand side of (3) is simply
$$
\mathbb{E}\left{e^{i u\left(W_t-W_s\right)}\right}
$$
which is the characteristic function of a centered normal random variable with variance $t-s$, and is equal to the right-hand side.
Conversely, if (3) holds, then the continuous process $\left(W_t\right)$ is a standard Brownian motion.

Gaussian white noise
This independence of increments makes the Brownian motion an ideal candidate to define a complete family of independent infinitesimal increments $\mathrm{dW}{\mathbf{t}}$, which are $\mathcal{N}(\mathbf{0}, \mathbf{d t})$-distributed (centered, normally distributed with variance $d t$ ) and which will serve as a model of (Gaussian white) noise. The drawback is that the trajectories of $\left(W_t\right)$ are not of bounded variation. Let $t_0=0{i-1}=t / n$. The quantity
$$
\mathbb{E}\left{\sum_{i=1}^n\left|W_{t_i}-W_{t_{i-1}}\right|\right}=n \mathbb{E}\left{\left|W_{\frac{t}{n}}\right|\right}=n \sqrt{\frac{t}{n}} \mathbb{E}\left{\left|W_1\right|\right},
$$
goes to $+\infty$ as $n \nearrow+\infty$.
The integral with respect to $d W_t$ cannot be defined in the usual way “trajectory by trajectory”.

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