## 澳洲代写｜CIVL3612｜Fluid Mechanics流体力学 悉尼大学

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.

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

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 $h_1^2-h_2^2=\left(h_1-h_2\right)\left(h_1+h_2\right)$.

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.

## 有限元方法代写

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。

## 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。

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.

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:

1. What kinds of activities are crucial to our business?
2. What kinds of activities are crucial to our distribution channels?
3. What kinds of activities are important if we want to maintain our customer_relationships?
4. 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.

1. 哪些活动对我们的业务至关重要？
2. 哪些活动对我们的分销渠道至关重要？
3. 如果我们要维护客户关系，哪些活动是重要的？
4. 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。

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

1. What are the main costs that are generated in our company?
2. Which key resources are the most expensive?
3. 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:

1. 我们公司产生的主要成本是什么？
2. 哪些关键资源最昂贵？
3. 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。

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

B. 使用费

C. 订购费

D. 出借/出租/租赁

E. 许可证

## 有限元方法代写

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。

## 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。

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

statistics-labTM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 澳洲代写代考辅导服务！

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.

## Financial Econometrics金融计量经济学问题集

(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.

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?

1. 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.
2. 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) $X_t=Z_t-0.3 Z_{t-1}$.

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}

## 有限元方法代写

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。

## 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。

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

statistics-labTM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 澳洲代写代考辅导服务！

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

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}

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 $\operatorname{Var}\left(f_t\right)=\mathrm{E} f_t^2-\left(\mathrm{E} f_t\right)^2$.)

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).

## 有限元方法代写

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。

## 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。

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

statistics-labTM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辅导服务！

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.

## Applied Statistical Method应用统计方法问题集

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.

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 $140 \neq 480$.

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 $\bar{H}<\bar{G}<\bar{x}$.

## 有限元方法代写

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。

## 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。

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

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

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

## International Macroeconomics国际宏观经济学相关

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

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

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

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

#### 简单线性回归Simple linear regression代写代考

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

## 多元统计分析Multivariate Statistical Analysis的重难点

$$y_i=\mu^{(g(i))}+\varepsilon_i \quad \varepsilon_i \stackrel{text { i.i.d. }}{\sim} \mathcal{N}_q(0, \Sigma) \quad \text { for } i=1, \ldots, n$$

$$H_0: \mu^{(1)}=\mu^{(2)}=\cdots=\mu^{(m)}$$

PCA 被定义为一种正交线性变换，它将数据转换到一个新的坐标系中，使数据的某个标量投影的最大方差位于第一个坐标上（称为第一个主成分），第二个最大方差位于第二个坐标上，以此类推${ }^{[12]}$

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

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

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

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

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

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

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