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澳洲代写｜STAT7203｜Probability Models & Data Analysis概率模型与数据分析昆士兰大学

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

statistics-lab^TM为您提供昆士兰大学（The University of Queensland）Probability Models & Data Analysis概率模型与数据分析澳洲代写代考和辅导服务！

课程介绍：

Probability and Statistics for Data Science: basic probability theory, distributions and properties, sampling methods, EDA, estimation, hypothesis tests, regression, experimental design, transform methods, model construction, reliability, and joint distributions.

Detail	Information
Course Code	STAT7203
Prerequisite	MATH7501 or MATH7051
Academic Unit	School of Mathematics and Physics
Course Convener	Dr Fred Roosta-Khorasani
Units	2

Probability and Statistics 概率与统计相关

问题 1.

Polygraph analogy. In a experiment on the accuracy of polygraph tests, 140 people were instructed to tell the truth and 140 people were instructed to lie. Testers use a polygraph to guess whether or not each person is lying. By analogy, let’s say $H_0$ corresponds to the testee telling the truth and $H_A$ corresponds to the testee lying.
(a) Describe the meaning of type I and type II errors in this context, and estimate their probabilities based on the table.
\begin{tabular}{l|c|c|}
& Testee is truthful & Testee is lying \
\hline Tester thinks testee is truthful & 131 & 15 \
\hline Tester thinks tested is lying & 9 & 125 \
\hline
\end{tabular}

(a) Type I error is rejecting the null-hypothesis when it is indeed true. This corresponds to thinking someone is lying when they are in fact being truthful. The experiment had $\frac{9}{140}$ type I errors. This is our estimate of the probability of a type I error.

Type II error is not rejecting the null-hypothesis when it is indeed false. This corresponds to thinking someone is telling the truth when they are in fact lying. Based on the data our estimate of the probability of a Type II error is $\frac{15}{140}$.

问题 2.

(b) In NHST, what relationships exist between the terms significance level, power, type 1 error, and type 2 error?

(b) Significance $=P($ type I error $)=P\left(\right.$ reject $\left.H_0 \mid H_0\right)$.
Power $=1-P($ type II error $)=P\left(\right.$ reject $\left.H_0 \mid H_A\right)$.

问题 3.

We perform a $t$-test for the null hypothesis $H_0: \mu=10$ at significance level $\alpha=0.05$ by means of a dataset consisting of $n=16$ elements with sample mean 11 and sample variance 4 .
(a) Should we reject the null hypothesis in favor of $H_A: \mu \neq 10$ ?
(b) What if we test against $H_A^{\prime}: \mu>10$ ?

(a) This is a two-sided alternative. The $t$-statistic is
$$
\frac{\bar{x}-\mu}{s / \sqrt{n}}=\frac{1}{2 / 4}=2 \text {. }
$$
Since we have $n=16$ our $t$ statistic has 15 degrees of freedom.
We have the two-sided $p$-value
$$
p=P\left(|t|>2 \mid H_0\right)=2 *(1-\mathrm{pt}(2,15))=0.063945 .
$$
Since $p>\alpha=0.05$ we don’t reject the null hypothesis.
Alternatively we could have done the problem in terms of rejection regions. We are given $\bar{x}=11, s^2=4$, and $n=16$. The null hypothesis is $\mu=10$. Using $\bar{x}$ as our test statistic the rejection region is
$$
\left(-\infty, 10-t_{15,0.025} \frac{s}{\sqrt{n}}\right] \cup\left[10+t_{15,0.025} \frac{s}{\sqrt{n}}, \infty\right)=(-\infty, 8.93] \cup[11.07, \infty)
$$
Here $t_{15,0.025}$ means a critical value, i.e. the value with right tail probability 0.025 : for $T \sim t(15)$ we have $P\left(t>t_{15,0.025}\right)=0.025$.
Since 11 lies outside the rejection region, we should not reject the null-hypothesis.
(b) This is a one-sided alternative. The $t$-statistic is the same
$$
\frac{\bar{x}-\mu}{s / \sqrt{n}}=\frac{1}{2 / 4}=2 .
$$

So we have the one-sided $p$-value
$$
p=P\left(t>2 \mid H_0\right)=2 *(1-\mathrm{pt}(2,15))=0.031973 .
$$
Since $p<\alpha=0.05$ we reject the null hypothesis in favor of the alternative. Again looking at rejection regions. We use the critical value $t_{15,0.05} \approx 1.753$. The rejection region for $\bar{x}$ is $$ \left[10+t_{15,0.05} \frac{s}{\sqrt{n}}, \infty\right)=[10.876, \infty) . $$ Since 11 lies inside the rejection region, we should reject the null-hypothesis in favor of $H_1: \mu>10$

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

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：

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.

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

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｜MATH3102｜Methods & Models of Applied Mathematics应用数学方法与模型昆士兰大学

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

statistics-lab^TM为您提供昆士兰大学（The University of Queensland）Methods & Models of Applied Mathematics应用数学方法与模型澳洲代写代考和辅导服务！

课程介绍：

Classical methods and models in applied mathematics: dimensional analysis, perturbation methods, traffic models (continuity equation, shocks, Rankine-Hugoniot condition), introduction to continuum mechanics (elasticity, visco-elasticity, fluid dynamics, Navier-Stokes equation).

澳洲代写｜MATH3102｜Methods & Models of Applied Mathematics应用数学方法与模型昆士兰大学

Methods & Models of Applied Mathematics应用数学方法与模型相关

问题 1.

In each case compute $y^{\prime}=\frac{d y}{d p}$ as a function of $y$ and $p$, given that $y=y(p)$ satisfies:

$p^3+p y+2=0$.
$y=\sin (y+p)$.
$\ln (y)=p$.
$\cos ^2(y)=p$, for $p>0$.
$y=f(c-y p)$.
$y=f(p-c y)$.
Note: in (5) and (6) $f$ is an arbitrary function, and $c$ is a constant.

问题 2. Two variable dependence implicit differentiation.
In each case compute $u_x=\frac{\partial u}{\partial x}$ and $u_y=\frac{\partial u}{\partial y}$ (as functions of $u, x$, and $y$ ), given that $u=u(x, y)$ satisfies:

$x^3 y+2 x u+y u^3=0$.
$y=f(y+u x)$.
$\ln (1+y)=u e^{x u}$.
Note: In (2) $f$ is an arbitrary function of a single variable, $f=f(\zeta)$. Assume that $f^{\prime} \neq 0$.

问题 3.

In each case compute $u_x=\frac{\partial u}{\partial x}$ and $u_p=\frac{\partial u}{\partial p}$ (as functions of $u, x$, and $p$ ), given that $u=u(x, p)$ satisfies:

$\cos \left(p^2 u\right)=p e^{-x^2}$.
$p=\cos (x+u)$.
$u=p f(x+u)$.
Note: In (3) $f$ is an arbitrary function of a single variable, $f=f(\zeta)$.

问题 4.

Consider the wave equation for $u=u(x, t)$, where $c>0$ is a constant,
$$
u_{t t}-c^2 u_{x x}=0 .
$$
Introduce the new independent variables $\boldsymbol{\eta}=\boldsymbol{x}-\boldsymbol{c t}$, and $\boldsymbol{\xi}=\boldsymbol{x}+\boldsymbol{c t}$. Change variables to write the equation for $u$ as a function of these new variables: $\boldsymbol{u}=\boldsymbol{u}(\boldsymbol{\eta}, \boldsymbol{\xi})$. Using this transformed form of the equation, integrate it twice to show that it must be
$$
u=f(\eta)+g(\xi),
$$
for some arbitrary functions $f$ and $g$. This shows that any solution of the wave equation (1.1) must have the form $u=f(x-c t)+g(x+c t)$.

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｜STAT3007｜Deep Learning深度学习昆士兰大学

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

statistics-lab^TM为您提供昆士兰大学（The University of Queensland）Deep Learning深度学习澳洲代写代考和辅导服务！

课程介绍：

Deep learning has become a much sought-after game-changing technology that has enabled breakthroughs in applications such as intelligent virtual assistants, medical diagnosis, recommender systems, and autonomous driving. This course provides a comprehensive and rigorous coverage of deep learning from both applied and theoretical perspectives. Students taking this course will understand how, why and when the algorithms work, and be able to effectively apply deep learning methods to practical problems. This course begins with the basics of machine learning, followed by a broad coverage of deep neural networks, including some major deep neural network architectures, optimization of network parameters, and applications in classification, regression and reinforcement learning. This course is suitable for both students who want to build data-driven enabling applications with deep learning, and students who want to develop a solid foundation for doing research in deep learning in particular, and machine learning or artificial intelligence more broadly. To maximise the learning outcomes, students are expected to have a solid foundation in statistics, calculus, linear algebra, and programming. Python will be used for this course.

Detail	Information
Course Code	STAT3007
Prerequisite	(STAT2004 or STAT2203 or equivalent) + programming experience (e.g., MATH2504 or CSSE2002 or equivalent)
School	Mathematics & Physics School
Course Enquiries	Dr Nan Ye
Units	2

Multiplying Matrices and Vectors矩阵和向量相乘相关

Matrix product operations have many useful properties that make mathematical analysis of matrices more convenient. For example, matrix multiplication is distributive:
$$
A(B+C)=A B+A C .
$$
It is also associative:
$$
\boldsymbol{A}(\boldsymbol{B C})=(\boldsymbol{A B}) \boldsymbol{C} .
$$
Matrix multiplication is not commutative (the condition $\boldsymbol{A B}=\boldsymbol{B A}$ does not always hold), unlike scalar multiplication. However, the dot product between two vectors is commutative:
$$
\boldsymbol{x}^{\top} \boldsymbol{y}=\boldsymbol{y}^{\top} \boldsymbol{x} .
$$
The transpose of a matrix product has a simple form:
$$
(\boldsymbol{A B})^{\top}=\boldsymbol{B}^{\top} \boldsymbol{A}^{\top} \text {. }
$$
This allows us to demonstrate equation 2.8 , by exploiting the fact that the value of such a product is a scalar and therefore equal to its own transpose:
$$
\boldsymbol{x}^{\top} \boldsymbol{y}=\left(\boldsymbol{x}^{\top} \boldsymbol{y}\right)^{\top}=\boldsymbol{y}^{\top} \boldsymbol{x} .
$$
Since the focus of this textbook is not linear algebra, we do not attempt to develop a comprehensive list of useful properties of the matrix product here, but the reader should be aware that many more exist.

We now know enough linear algebra notation to write down a system of linear equations:
$$
A x=b
$$
where $\boldsymbol{A} \in \mathbb{R}^{m \times n}$ is a known matrix, $\boldsymbol{b} \in \mathbb{R}^m$ is a known vector, and $\boldsymbol{x} \in \mathbb{R}^n$ is a vector of unknown variables we would like to solve for. Each element $x_i$ of $\boldsymbol{x}$ is one of these unknown variables. Each row of $\boldsymbol{A}$ and each element of $\boldsymbol{b}$ provide another constraint. We can rewrite equation 2.11 as:
$$
\begin{gathered}
\boldsymbol{A}{1,:} \boldsymbol{x}=b_1 \ \boldsymbol{A}{2,:} \boldsymbol{x}=b_2 \
\ldots \
\boldsymbol{A}{m, x} \boldsymbol{x}=b_m \end{gathered} $$ or, even more explicitly, as: $$ \boldsymbol{A}{1,1} x_1+\boldsymbol{A}{1,2} x_2+\cdots+\boldsymbol{A}{1, n} x_n=b_1
$$

Identity and Inverse Matrices恒等矩阵和逆矩阵相关

Linear algebra offers a powerful tool called matrix inversion that allows us to analytically solve equation 2.11 for many values of $\boldsymbol{A}$.

To describe matrix inversion, we first need to define the concept of an identity matrix. An identity matrix is a matrix that does not change any vector when we multiply that vector by that matrix. We denote the identity matrix that preserves $n$-dimensional vectors as $\boldsymbol{I}_n$. Formally, $\boldsymbol{I}_n \in \mathbb{R}^{n \times n}$, and
$$
\forall x \in \mathbb{R}^n, I_n x=x .
$$
The structure of the identity matrix is simple: all of the entries along the main diagonal are 1 , while all of the other entries are zero. See figure 2.2 for an example.
The matrix inverse of $\boldsymbol{A}$ is denoted as $\boldsymbol{A}^{-1}$, and it is defined as the matrix such that
$$
\boldsymbol{A}^{-1} \boldsymbol{A}=\boldsymbol{I}_n
$$
We can now solve equation 2.11 by the following steps:
$$
\begin{gathered}
A x=b \
A^{-1} A x=A^{-1} b \
I_n x=A^{-1} b
\end{gathered}
$$
$$
\boldsymbol{x}=\boldsymbol{A}^{-1} \boldsymbol{b} .
$$
Of course, this process depends on it being possible to find $\boldsymbol{A}^{-1}$. We discuss the conditions for the existence of $\boldsymbol{A}^{-1}$ in the following section.

When $\boldsymbol{A}^{-1}$ exists, several different algorithms exist for finding it in closed form. In theory, the same inverse matrix can then be used to solve the equation many times for different values of $\boldsymbol{b}$. However, $\boldsymbol{A}^{-1}$ is primarily useful as a theoretical tool, and should not actually be used in practice for most software applications. Because $\boldsymbol{A}^{-1}$ can be represented with only limited precision on a digital computer, algorithms that make use of the value of $\boldsymbol{b}$ can usually obtain more accurate estimates of $\boldsymbol{x}$.

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｜ECON2070｜Introduction to Strategic Thinking战略思维导论昆士兰大学

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

statistics-lab^TM为您提供昆士兰大学（The University of Queensland）Introduction to Strategic Thinking战略思维导论澳洲代写代考和辅导服务！

课程介绍：

The way that economists think about strategic situations is through the application of game theory. One aim of the course is to teach you some strategic considerations to take into account when making your own choices. A second aim is to predict how other people or organizations behave when they are in strategic settings. We will see that these aims are closely related. We will learn new concepts, methods and terminology. A third aim is to apply these tools to settings from economics and other disciplines. The course will emphasize examples.

澳洲代写｜ECON2070｜Introduction to Strategic Thinking战略思维导论昆士兰大学

Detail	Information
Course Code	ECON2070
Course Title	Introduction to Strategic Thinking
Coordinating Unit	School of Economics
Pre-Requisites	ECON1010 or 1011
Course Coordinator	Dr Metin Uyanik
Number of Units	2

Introduction to Strategic Thinking战略思维导论相关

Later GM leaders’ preoccupation with financial strategy had exacerbated a lack of attention to other business policies, such as labor relations, safety, and product quality. GM leaders’ antagonistic attitude toward labor fostered continuing labor troubles for GM (that greatly contributed to the deterioration of GM’s capability for efficient, high-quality, and low-cost production). GM leaders resisted providing leadership in automobile safety and environment (eventually fostering U.S. government intervention to mandate safety and environmental standards for the auto industry). GM’s different automobile brands lost distinctiveness, quality, competitiveness, and performance – particularly compared to imported foreign-produced automobiles. By the 1970s, GM had failed to compete effectively in the low-end automobile market. It also lost quality leadership in the middle and high end. German automobile makers dominated high-end sector quality; whereas Japanese automobile makers dominated lower and middle sector quality. The holding company strategy as a focus on finance had not encouraged the portfolio companies to maintain quality and brand distinctiveness.

As we saw in the GM decline and bankruptcy, the leadership of a diversified firm needs to attend to providing adequate resources to its portfolio businesses to keep them competitive. Otherwise market share of its portfolio firms decline. Profits decline, and eventually the whole firm can go bankrupt.

The decision to maintain a portfolio business or to buy or sell a business are strategic decisions by a CEO of a diversified firm – which have major impacts upon the presidents of the businesses in the firm’s business portfolio. There are six critical factors for successful strategic management that need to be recognized:
(1) The need for relationships of trust between levels of management;
(2) The impact of unequal power relationships between a holding company and the businesses of its portfolio;
(3) The effect of long-term and short-term differences of control over finances between the firm and its portfolio businesses;
(4) The possible results of differing incentives and rewards for levels of management;
(5) The inherent conflicts of interest of different levels of management;
(6) The influence of external forces on business valuation.

后来通用汽车领导人对财务战略的关注加剧了对其他业务政策的缺乏关注，例如劳动关系、安全和产品质量。通用汽车领导人对劳工的敌对态度，导致通用汽车公司持续陷入劳工困境（这极大地导致了通用汽车公司高效、高质量、低成本生产能力的恶化）。通用汽车领导人拒绝在汽车安全和环境方面发挥领导作用（最终促成美国政府干预，强制汽车行业制定安全和环境标准）。通用汽车的不同汽车品牌失去了独特性、质量、竞争力和性能——特别是与进口的外国生产的汽车相比。到了 20 世纪 70 年代，通用汽车未能在低端汽车市场进行有效竞争。它还失去了中高端的质量领先地位。德国汽车制造商主导高端领域品质；而日本汽车制造商在中低质量行业占据主导地位。以金融为重点的控股公司战略并没有鼓励投资组合公司保持质量和品牌独特性。

正如我们在通用汽车的衰落和破产中看到的那样，多元化公司的领导层需要为其投资组合业务提供足够的资源，以保持其竞争力。否则，其投资组合公司的市场份额会下降。利润下降，最终整个公司可能破产。

维持投资组合业务或购买或出售业务的决定是多元化公司首席执行官的战略决策，这对公司业务组合中的业务总裁产生重大影响。成功的战略管理需要认识到六个关键因素：
(1) 管理层之间需要建立信任关系；
(2) 控股公司与其投资组合业务之间不平等权力关系的影响；
(3) 公司及其投资组合业务之间长期和短期财务控制权差异的影响；
（四）对管理层实行不同的激励和奖励可能产生的结果；
（五）各级管理人员固有的利益冲突；
(6)外部力量对企业估值的影响。

External Forces on Business Valuation
Stocks create a value of a return from investment either (1) through dividends paid out annually by the company or (2) by any increase in the price of the stock. Accordingly, earnings of a corporation can be used for paying out dividends, investments for improving businesses, or acquisitions. Earnings used for dividends provide an immediate return to the shareholder, while earnings used for improvement or acquisitions may provide future capital accumulation to raise share value. There is an important trade-off in optimizing shareholder value of a company, between how earnings are used for (1) immediate returnon-investment or (2) future return-on-investment. Economically, this trade-off should be
made to balance appropriately short-and long-term shareholder value and short- and longterm competitiveness of the company. However, external forces can make an important influence on this balancing, particularly when a government’s tax policies unwittingly bias this balance. And in the U.S. Corporate world of the twentieth century, Federal government tax policy had biased corporate strategy strongly against returning investments in stocks via dividends.

Government policies can often make an important impact upon the environments of a business. In the United States at the close of the twentieth century, Federal government income tax policy had a major external impact upon business policies. The Federal government taxed returns-on-investment from stocks very unequally as dividends or stock appreciation. Wealthy individuals who owned stock would have any dividends from their stocks taxed at a top income tax rate of $36 \%$ in the year 2000 . In contrast, any gains on sales of their stocks held at least one year would be taxed at a lower capital gains tax rate of $28 \%$. This tax rate difference of $36-28=8 \%$ had in effect created a $22 \%$ tax penalty on received stock returns by dividends rather that appreciation. One can see how government policy of the United States in the last part of the twentieth century encouraged corporations to pursue strategies which aimed at continually increasing stock prices, as opposed to traditional business practice of sharing earnings with investors through dividends. Thus earnings were often used to buy growth through acquisitions, even when a company could not properly manage acquired businesses.

The tax policies of the United States biased twentieth-century corporate strategy away from traditional dividend strategies toward quick capital-gain strategies – making it difficult to properly run companies in mature industries with little growth but steady earnings – as all successful companies eventually become. When the twentieth century began, the corporate situation in the United States was one where the U.S. Federal government unwittingly biased the economic playing field through tax policy wherein:
(1) only companies in new and growing industries (such as Cisco) could be properly run and rewarded with high price/earnings $(P / E)$ ratios by the financial structure, and
(2) companies in mature industries (such as Sunbeam) could not be properly run and at the same time rewarded by reasonable $P / E$ ratios by the financial structure.

企业估值的外部力量
股票通过（1）通过公司每年支付的股息或（2）通过股票价格的上涨创造投资回报价值。因此，公司的收益可用于支付股息、改善业务的投资或收购。用于股息的收益可以立即为股东带来回报，而用于改进或收购的收益则可以提供未来的资本积累以提高股票价值。在优化公司股东价值时，需要在收益用于（1）即时投资回报或（2）未来投资回报之间进行重要权衡。从经济角度来看，这种权衡应该是
适当平衡短期和长期股东价值以及公司短期和长期竞争力。然而，外部力量可以对这种平衡产生重要影响，特别是当政府的税收政策无意中使这种平衡产生偏差时。在二十世纪的美国企业界，联邦政府的税收政策强烈偏向企业战略，反对通过股息返还股票投资。

政府政策通常会对企业环境产生重要影响。在二十世纪末的美国，联邦政府的所得税政策对商业政策产生了重大的外部影响。联邦政府对股票投资回报的征税与股息或股票增值的征税非常不平等。 2000 年，持有股票的富人将按照 36%$ 的最高所得税税率对股票股息征税。相比之下，出售持有至少一年的股票所获得的任何收益都将按较低的资本利得税率征税，即 28 \%$。 $36-28=8 \%$ 的税率差异实际上对通过股息而不是升值获得的股票回报产生了 $22 \%$ 的税收罚款。人们可以看到，二十世纪后半叶美国政府的政策如何鼓励企业采取旨在不断提高股价的战略，而不是通过股息与投资者分享收益的传统商业做法。因此，即使公司无法正确管理所收购的业务，收益也经常被用来通过收购来购买增长。

美国的税收政策使二十世纪的企业战略从传统的股利战略转向快速的资本收益战略——这使得在成长缓慢但盈利稳定的成熟行业中正确经营公司变得困难——就像所有成功的公司最终都会遇到的那样。二十世纪初，美国的企业状况是美国联邦政府无意中通过税收政策扭曲了经济竞争环境，其中：
(1) 只有新兴行业和成长型行业的公司（例如思科）才能正常运营并通过财务结构获得高市盈率$(P / E)$，并且
(2)成熟行业的公司(如Sunbeam)无法通过财务结构得到合理的运营并同时获得合理的$P/E$比率奖励。

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｜FINM3407｜Behavioural Finance行为金融学昆士兰大学

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

statistics-lab^TM为您提供昆士兰大学（The University of Queensland）Behavioural Finance行为金融学澳洲代写代考和辅导服务！

课程介绍：

This course provides an introduction to behavioural finance, a discipline that integrates insights from psychology into the study of finance. There will be a focus on understanding the psychological underpinnings of financial decision-making as well as the institutional frictions that may allow these psychological mechanisms to influence economic outcomes. Applications include the pricing of assets relative to fundamental value, trading strategies, managerial behaviour, and household savings and investment decisions.

澳洲代写｜MATH3404｜Optimisation Theory最优化理论昆士兰大学

Attribute	Detail
Course Code	FINM3407
Course Title	Behavioural Finance
Coordinating Unit	School of Business
Pre-Requisites	(FINM2415 + 2416) or (FINM2411 + 3412) or (ECON1310 + 2010)
Course Coordinator/Lecturer	Associate Professor Kelvin Tan
Number of Units	2

Risk and Return of a Portfolio投资组合的风险和回报

Risk and Return of a Portfolio
A portfolio comprises of two or more assets. Smart investors know that combining several assets in a portfolio usually leads to risk reduction, thanks to the benefit of diversification. Remember the old adage which says, “don’t put all your eggs in the same basket.”

To understand the quantitative impact of diversification on risk (variability), let us consider a portfolio of two assets. As long as the returns on the two assets do not move in perfect lockstep, diversification reduces risk. Covariance and correlation are statistical measures of how random variables are related. If the two variables tend to move in the same (opposite) direction, the covariance and correlation are positive (negative).
The correlation of a sample including $n$ returns for assets 1 and 2 is:
$$
\rho_{12}=\frac{\sigma\left(R_1, R_2\right)}{\sigma_1 \sigma_2}
$$
$\sigma\left(R_1, R_2\right)$ is the covariance between returns on assets 1 and 2 , and $\sigma_1$ and $\sigma_2$ are standard deviations of returns on assets 1 and 2 . While the covariance can take any positive or negative value, the correlation always lies between -1.0 and +1.0 .

Given information on how the returns for the two assets are correlated, we can compute the portfolio mean return and portfolio variance for two asset portfolios as follows:
$$
\begin{gathered}
\bar{R}p=w_1 \bar{R}_1+w_2 \bar{R}_2 \ \sigma_p^2=w_1^2 \sigma_1^2+w_2^2 \sigma_2^2+2 w_1 w_2 \rho{12} \sigma_1 \sigma_2
\end{gathered}
$$
where $\bar{R}\rho$ is portfolio mean return, $w_1$ and $w_2$ are the weights associated with assets 1 and $2\left(w_1+w_2=1\right), \bar{R}_1$ and $\bar{R}_2$ are the mean returns for assets 1 and $2, \sigma\rho^2$ is the variance of the portfolio return, $\sigma_1{ }^2$ and $\sigma_2{ }^2$ are the variance of the returns for assets 1 and 2 , and $\rho_{12}$ is the coefficient of correlation between the returns on assets 1 and 2 .

As long as $\rho_{12}$ is less than $1, \sigma_p$ will be less than the weighted average of the standard deviations of returns for the two assets.
For a 3-asset portfolio, the portfolio mean return and portfolio variance are as follows:
$$
\begin{gathered}
\bar{R}p=w_1 \bar{R}_1+w_2 \bar{R}_2+w_3 \bar{R}_3 \ \sigma_p^2=w_1^2 \sigma_1^2+w_2^2 \sigma_2^2+w_3^2 \sigma_3^2+2 w_1 w_2 \rho{12} \sigma_1 \sigma_2+2 w_1 w_3 \rho_{13} \sigma_1 \sigma_3+2 w_2 w_3 \rho_{23} \sigma_2 \sigma_3
\end{gathered}
$$
In general, for a portfolio of $n$ assets
$$
\begin{gathered}
R_p=\sum_{i=1}^n w_i R_i \
\sigma_p^2=\sum_{i=1}^n \sum_{j=1}^n w_i w_j \sigma\left(R_i, R_j\right)
\end{gathered}
$$

Behavioural Finance行为金融学案例

问题 1.

Rakesh Gupta’s utility function for wealth is: $u(w)=w^{2 / 3}$
Suppose Rakesh Gupta has a $20 \%$ chance of wealth of ₹ 3,000,000, 30\% chance of wealth of $₹ 2,000,000$, and $50 \%$ chance of wealth of $₹ 1,000,000$.
a. What is the expected value of wealth?
b. Is Rakesh Gupta risk averse, risk neutral, or risk-seeking?
c. What is Rakesh Gupta’s certainty equivalent for the prospect?

a. The expected value of wealth is:
$$
0.2 \times 3,000,000+0.3 \times 2,000,000+0.5 \times 1,000,000=₹ 1,700,000
$$
b. Given the utility function $u(w)=w^{2 / 3}$ let us look at the utility for four levels of wealth, viz., ₹ 0 , $₹ 1,000,000$, ₹ $2,000,000$, and ₹ $3,000,000$.
$$
\begin{aligned}
& u(0)=0 \
& u(1,000,000)=10,046 \
& u(2,000,000)=15,951 \
& u(3,000,000)=20,905
\end{aligned}
$$
From these numbers, it is clear that Rakesh Gupta’s utility of wealth function is concave, implying that he is risk-averse.

c. The expected utility of the prospect is:
$$
0.2 \times 20,905+0.3 \times 15,951+0.3 \times 10,046=13,989
$$
The expected utility is 13,989 . Since the utility of wealth is $w^{2 / 3}$, the certain amount that provides an expected utility of 13,989 is:
$$
(13,989)^{3 / 2}=₹ 1,654,550
$$

问题 2.

Neha has the following utility function:
$$
\mathrm{u}(w)=\ln w \quad \text { where } w=\text { wealth }
$$
a. What is the expected utility of the following prospects:
$$
\begin{aligned}
& \text { P1 }(0.6,400,800) \
& \text { P2 }(0.8,5000,2,000) \
& \text { P3 }(0.4,6,000,3,000)
\end{aligned}
$$
b. What is the certainty equivalent of P2?

Solution
a. The expected utility of $\mathrm{P} 1$ is:
$$
0.6 U(400)+0.4 U(800)=0.6 \times 5.99+0.4 \times 6.68=6.27
$$
The expected utility of $\mathrm{P} 2$ is:
$$
0.8 \times 8.52+0.2 \times 7.60=8.34
$$
The expected utility of $\mathrm{P} 3$ is:
$$
0.4 \times 8.70+0.6 \times 8.01=8.29
$$
b. The expected utility of $P 2$ is 8.34 . Since the utility of wealth is $\ln w$, the certain wealth that provides an expected utility of 8.34 is:
$$
e^{8.34}=4,188
$$

问题 3.

The risk-free return is 7 per cent and the expected return on market portfolio is 14 per cent. If the required return on a stock is 16 per cent, what is its beta?

Solution
We have: Required return $=$ Risk-free return + Beta(Expected return on market portfolio – Risk-free return)
$$
\begin{aligned}
& 16=7+\operatorname{Beta}(14-7) \
& \text { Beta }=9 / 7=1.29
\end{aligned}
$$

澳洲代写｜FINM3407｜Behavioural Finance行为金融学昆士兰大学

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｜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.

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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。

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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:

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