分类：计算机代写

澳洲代写｜PHYC30018｜Quantum Physics量子计算墨尔本大学

Posted on 2023年10月16日2024年1月7日 by statistics-lab

statistics-lab^TM为您提供墨尔本大学（The University of Melbourne，简称UniMelb，中文简称“墨大”）Quantum Physics量子计算澳洲代写代考和辅导服务！

课程介绍：

Quantum mechanics plays a central role in our understanding of fundamental phenomena, primarily in the microscopic domain. It lays the foundation for an understanding of atomic, molecular, condensed matter, nuclear and particle physics.

Quantum Physics量子计算问题集

问题 1. Consider a two-dimensional strip of material of length $L$ and width $W$, placed in a magnetic field perpendicular to the strip and with an electric field established in the direction of the length $L$.
(a) Suppose that the resistivity matrix is given by the classical result
$$
\rho=\left(\begin{array}{cc}
\rho_0 & -\rho_H \
\rho_H & \rho_0
\end{array}\right)
$$
where $\rho_H=B /$ nec is the Hall resistivity and $\rho_0$ is the usual Ohmic resistivity. Find the conductivity matrix, $\sigma=\rho^{-1}$. Write it in the form:
$$
\sigma=\left(\begin{array}{cc}
\sigma_0 & \sigma_H \
-\sigma_H & \sigma_0
\end{array}\right) .
$$
What are $\sigma_0$ and $\sigma_H$ ?
(b) Suppose $B=0$, so the Hall resistivity is zero. Notice that the Ohmic conductivity, $\sigma_0$, is just $1 / \rho_0$. In particular, note that $\sigma_0 \rightarrow \infty$ as $\rho_0 \rightarrow 0$. Now suppose $\rho_H \neq 0$. Show that $\sigma_0 \rightarrow 0$ as $\rho_0 \rightarrow 0$, so it is possible to have both $\sigma_0$ and $\rho_0$ equal to zero

问题 2.

This problem asks you to give a complete presentation of a calculation that is almost the same as one you saw in lecture.

Consider a constant electric field, $\vec{E}=\left(0, E_0, 0\right)$ and a constant magnetic field, $\vec{B}=\left(0,0, B_0\right)$.
(a) Choose an electrostatic potential $\phi$ and a vector potential $\vec{A}$ which describe the $\vec{E}$ and $\vec{B}$ fields, and write the Hamiltonian for a charged particle of mass $m$ and charge $q$ in these fields. Assume that the particle is restricted to move in the $x y$-plane.
(b) What are the allowed energies as a function of $B_0$ and $E_0$ ? Draw a figure to show how the Landau levels (energy levels when $E_0=0$ ) change as $E_0$ increases.

问题 3.

You will see the “standard presentation” of the Aharonov-Bohm effect in lecture, on the day that this problem set is due. The standard presentation has its advantages, and in particular is more general than the presentation you will work through in this problem. However, students often come away from the standard presentation of the Aharonov-Bohm effect thinking that the only way to detect this effect is to do an interference experiment. This is not true, and the purpose of this problem is to disabuse you of this misimpression before you form it.

As Griffiths explains on pages 385-387 (344-345 in 1st Ed.), the Aharonov-Bohm effect modifies the energy eigenvalues of suitably chosen quantum mechanical systems. In this problem, I ask you to work through the same physical example that Griffiths uses, but in a different fashion which makes more use of gauge invariance.

Imagine a particle constrained to move on a circle of radius $b$ (a bead on a wire ring, if you like.) Along the axis of the circle runs a solenoid of radius $a<b$, carrying a magnetic field $\vec{B}=\left(0,0, B_0\right)$. The field inside the solenoid is uniform

and the field outside the solenoid is zero. The setup is depicted in Griffiths’ Fig. 10.10. (10.12 in 1st Ed.)
(a) Construct a vector potential $\vec{A}$ which describes the magnetic field (both inside and outside the solenoid) and which has the form $A_r=A_z=0$ and $A_\phi=\alpha(r)$ for some function $\alpha(r)$. I am using cylindrical coordinates $z, r$, $\phi$.
(b) Since $\vec{\nabla} \times \vec{A}=0$ for $r>a$, it must be possible to write $\vec{A}=\vec{\nabla} f$ in any simply connected region in $r>a$. [This is a theorem in vector calculus.] Show that if we find such an $f$ in the region
$$
r>a \text { and }-\pi+\epsilon<\phi<\pi-\epsilon,
$$
then
$$
f(r, \pi-\epsilon)-f(r,-\pi+\epsilon) \rightarrow \Phi \text { as } \epsilon \rightarrow 0 .
$$
Here, the total magnetic flux is $\Phi=\pi a^2 B_0$. Now find an explicit form for $f$, which is a function only of $\phi$.
(c) Now consider the motion of a “bead on a ring”: write the Schrödinger equation for the particle constrained to move on the circle $r=b$, using the $\vec{A}$ you found in (a). Hint: the answer is given in Griffiths.
(d) Use the $f(\phi)$ found in (b) to gauge transform the Schrödinger equation for $\psi(\phi)$ within the angular region $-\pi+\epsilon<\phi<\pi-\epsilon$ to a Schrödinger equation for a free particle within this angular region. Call the original wave function $\psi(\phi)$ and the gauge-transformed wave function $\psi^{\prime}(\phi)$.
(e) The original wave function $\psi$ must be single-valued for all $\phi$, in particular at $\phi=\pi$. That is, $\psi(\pi-\epsilon)-\psi(-\pi+\epsilon) \rightarrow 0$ and $\frac{\partial \psi}{\partial \phi}(\pi-\epsilon)-\frac{\partial \psi}{\partial \phi}(-\pi+\epsilon) \rightarrow 0$ as $\epsilon \rightarrow 0$. What does this say about the gauge-transformed wave function? I.e., how must $\psi^{\prime}(\pi-\epsilon)$ and $\psi^{\prime}(-\pi+\epsilon)$ be related as $\epsilon \rightarrow 0$ ?
[Hint: because the $f(\phi)$ is not single valued at $\phi=\pi$, the gauge transformed wave function $\psi^{\prime}(\phi)$ is not single valued there either.]
(f) Solve the Schrödinger equation for $\psi^{\prime}$, and find energy eigenstates which satisfy the boundary conditions you derived in (e). What are the allowed energy eigenvalues?
(g) Undo the gauge transformation, and find the energy eigenstates $\psi(\phi)$ in the original gauge. Do the energy eigenvalues in the two gauges differ?
(h) Plot the energy eigenvalues as a function of the enclosed flux, $\Phi$. Show that the energy eigenvalues are periodic functions of $\Phi$ with period $\Phi_0$, where you must determine $\Phi_0$. For what values of $\Phi$ does the enclosed magnetic field have no effect on the spectrum of a particle on a ring? Show that the

Aharonov-Bohm effect can only be used to determine the fractional part of $\Phi / \Phi_0$.
[Note: you have shown that even though the bead on a ring is everywhere in a region in which $\vec{B}=0$, the presence of a nonzero $\vec{A}$ affects the energy eigenvalue spectrum. However, the effect on the energy eigenvalues is determined by $\Phi$, and is therefore gauge invariant. To confirm the gauge invariance of your result, you can compare your answer for the energy eigenvalues to Griffiths’ result, obtained using a different gauge.]

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｜COMP30027｜Machine Learning机器学习墨尔本大学

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

statistics-lab^TM为您提供墨尔本大学（The University of Melbourne，简称UniMelb，中文简称“墨大”）Complex Analysis复杂分析澳洲代写代考和辅导服务！

课程介绍：

Machine Learning, a core discipline in data science, is prevalent across Science, Technology, the Social Sciences, and Medicine; it drives many of the products we use daily such as banner ad selection, email spam filtering, and social media newsfeeds. Machine Learning is concerned with making accurate, computationally efficient, interpretable and robust inferences from data. Originally borne out of Artificial Intelligence, Machine Learning has historically been the first to explore more complex prediction models and to emphasise computation, while in the past two decades Machine Learning has grown closer to Statistics gaining firm theoretical footing.

澳洲代写｜COMP30027｜Machine Learning机器学习墨尔本大学

Machine Learning机器学习问题集

问题 1.

The data and scripts for this problem are available in hw2/prob1. You can load the data using the MATLAB script load_al_data. This script should load the matrices y_noisy, y_true, X_in. The $y$ vectors are $n \times 1$ while $\mathrm{X}{-}$in is a $n \times 3$ matrix with each row corresponding to a point in $\mathcal{R}^3$. The $y{\text {true }}$ vectors correspond to the ideal $y$ values, generated directly from the “true” model (whatever it may be) without any noise. In contrast, the $y_{\text {noisy }}$ vectors are the actual, noisy observations, generated by adding Gaussian noise to the $y_{\text {true }}$ vectors. You should use $y_{n o i s y}$ for any estimation. $y_{\text {true }}$ is provided only to make it easier to evaluate the error in your predictions (simulate an infinite test data). You would not have $y_{\text {true }}$ in any real task.
(a) Write MATLAB functions theta = linear_regress $(\mathrm{y}, \mathrm{X})$ and $\mathrm{y}$ hat $=$ linear_pred(theta, X_test). Note that we are not explicitly including the offset parameter but instead rely on the feature vectors to provide a constant component. See part (b).
(b) The feature mapping can substantially affect the regression results. We will consider two possible feature mappings:
$$
\begin{aligned}
& \phi_1\left(x_1, x_2, x_3\right)=\left[1, x_1, x_2, x_3\right]^T \
& \phi_2\left(x_1, x_2, x_3\right)=\left[1, \log x_1^2, \log x_2^2, \log x_3^2\right]^T
\end{aligned}
$$
Use the provided MATLAB function feature mapping to transform the input data matrix into a matrix
$$
X=\left[\begin{array}{c}
\phi\left(\mathbf{x}1\right)^T \ \phi\left(\mathbf{x}_2\right)^T \ \cdots \ \phi\left(\mathbf{x}{\mathbf{n}}\right)^T
\end{array}\right]
$$
For example, $\mathrm{X}$ = feature_mapping ( $\mathrm{X}_{-}$in, 1 ) would get you the first feature representation. Using your completed linear regression functions, compute the mean squared prediction error for each feature mapping (2 numbers).

问题 2.

(c) The selection of points to query in an active learning framework might depend on the feature representation. We will use the same selection criterion as in the lectures, the expected squared error in the parameters, proportional to $\operatorname{Tr}\left[\left(X^T X\right)^{-1}\right]$. Write a MATLAB function $\mathrm{idx}=\operatorname{active_ learn}(\mathrm{X}, \mathrm{k} 1, \mathrm{k} 2)$. Your function should assume that the top $k_1$ rows in $X$ have been queried and your goal is to sequentially find the indices of the next $k_2$ points to query. The final set of $k_1+k_2$ indices should be returned in idx. The latter may contain repeated entries. For each feature mapping, and $k_1=5$ and $k_2=10$, compute the set of points that should be queried (i.e., $\mathrm{X}(:, \mathrm{idx})$ ). For each set of points, use the feature mapping $\phi_2$ to perform regression and compute the resulting mean squared prediction errors (MSE) over the entire data set (again, using $\phi_2$ ).

问题 3.

(d) Let us repeat the steps of part (c) with randomly selected additional points to query. We have provided a MATLAB function $i d x=\operatorname{randomly}(\operatorname{select}(\mathrm{X}, \mathrm{k} 1, \mathrm{k} 2)$ which is essentially the same as active_learn except that it selects the $k_2$ points uniformly at random from $X$. Repeat the regression steps as in previous part, and compute the resulting mean squared prediction error again. To get a reasonable comparison you should repeat this process 50 times, and use the median MSE. Compare the resulting errors with the active learning strategies. What conclusions can you draw?

问题 4.

(e) Let us now compare the two sets of points chosen by active learning due to the different feature representations. We have provided a function plot_points(X,idx_r,idx_b) which will plot each row of $X$ as a point in $\mathbf{R}^3$. The points indexed by $i d x _r$ will be circled in red and those marked by idx_b will be circled (larger) in blue (some of the points indexed by idx_r and idx_b might be common). Plot the original data points using the indexes of the actively selected points based on the two feature representations. Also plot the same indexes using $\mathrm{X}$ from the second feature representation with its first constant column removed. In class, we saw an example where the active learning strategy chose points at the extrema of the available space. Can you see evidence of this in the two plots?

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｜MAST90104｜A First Course In Statistical Learning 统计学习第一门课程墨尔本大学

Posted on 2023年10月11日2023年10月11日 by statistics-lab

statistics-lab^TM为您提供墨尔本大学（The University of Melbourne，简称UniMelb，中文简称“墨大”）A First Course In Statistical Learning 统计学习第一门课程澳洲代写代考和辅导服务！

课程介绍：

Supervised statistical learning is based on the widely used linear models that model a response as a linear combination of explanatory variables. Initially this subject develops an elegant unified theory for a quantitative response that includes the estimation of model parameters, hypothesis testing using analysis of variance, model selection, diagnostics on model assumptions, and prediction. Some classification methods for qualitative responses are then developed. This subject then considers computational techniques, including the EM algorithm. Bayes methods and Monte-Carlo methods are considered. The subject concludes by considering some unsupervised learning techniques.

澳洲代写｜MAST90104｜A First Course In Statistical Learning 统计学习第一门课程墨尔本大学

A First Course In Statistical Learning 统计学习第一门课程案例

问题 1.

Describe the unethical behavior in each example and describe how it could impact the reliability of the resulting data. Explain how the problem should be corrected.
A researcher is collecting data in a community.
a. She selects a block where she is comfortable walking because she knows many of the people living on the street.
b. No one seems to be home at four houses on her route. She does not record the addresses and does not return at a later time to try to find residents at home.
c. She skips four houses on her route because she is running late for an appointment. When she gets home, she fills in the forms by selecting random answers from other residents in the neighborhood.

a. By selecting a convenient sample, the researcher is intentionally selecting a sample that could be biased. Claiming that this sample represents the community is misleading. The researcher needs to select areas in the community at random.
b. Intentionally omitting relevant data will create bias in the sample. Suppose the researcher is gathering information about jobs and child care. By ignoring people who are not home, she may be missing data from working families that are relevant to her study. She needs to make every effort to interview all members of the target sample.
c. It is never acceptable to fake data. Even though the responses she uses are “real” responses provided by other participants, the duplication is fraudulent and can create bias in the data. She needs to work diligently to interview everyone on her route.

问题 2.

Listed are 29 ages for Academy Award winning best actors in order from smallest to largest.
$$
18 ; 21 ; 22 ; 25 ; 26 ; 27 ; 29 ; 30 ; 31 ; 33 ; 36 ; 37 ; 41 ; 42 ; 47 ; 52 ; 55 ; 57 ; 58 ; 62 ; 64 ; 67 ; 69 ; 71 ; 72 ; 73 ; 74 ; 76 ; 77
$$
a. Find the percentile for 58 .
b. Find the percentile for 25 .

a. Counting from the bottom of the list, there are 18 data values less than 58 . There is one value of 58 . $x=18$ and $y=1 . \frac{x+0.5 y}{n}(100)=\frac{18+0.5(1)}{29}(100)=63.80 .58$ is the $64^{\text {th }}$ percentile.
b. Counting from the bottom of the list, there are three data values less than 25 . There is one value of 25 . $x=3$ and $y=1 . \frac{x+0.5 y}{n}(100)=\frac{3+0.5(1)}{29}(100)=12.07$. Twenty-five is the $12^{\text {th }}$ percentile.

问题 3.

Suppose that in a small town of 50 people, one person earns $\$ 5,000,000$ per year and the other 49 each earn $\$ 30,000$. Which is the better measure of the “center”: the mean or the median?

$$
\begin{aligned}
\bar{x} & =\frac{5,000,000+49(30,000)}{50}=129,400 \
M & =30,000
\end{aligned}
$$
(There are 49 people who earn $\$ 30,000$ and one person who earns $\$ 5,000,000$.)
The median is a better measure of the “center” than the mean because 49 of the values are 30,000 and one is $5,000,000$. The $5,000,000$ is an outlier. The 30,000 gives us a better sense of the middle of the data.

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｜MAST30020｜Probability for Inference推理概率墨尔本大学

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

statistics-lab^TM为您提供墨尔本大学（The University of Melbourne，简称UniMelb，中文简称“墨大”）Probability for Inference推理概率要素澳洲代写代考和辅导服务！

课程介绍：

This subject introduces a measured-theoretic approach to probability theory and presents its fundamentals concepts and results.

Topics covered include: probability spaces and random variables, expectation, conditional expectation and distributions, elements of multivariate distribution theory, modes of convergence in probabilty theory, characteristics functions and their application in key limit theorems.

澳洲代写｜MAST30020｜Probability for Inference推理概率墨尔本大学

Probability for Inference推理概率案例

A survey was taken of a group’s viewing habits of sporting events on TV during the last year. Let $A={$ watched football $}, B={$ watched basketball $}, C={$ watched baseball $}$. The results indicate that if a person is selected at random from the surveyed group, then $P(A)=0.43, P(B)=0.40, P(C)=0.32, P(A \cap B)=0.29$, $P(A \cap C)=0.22, P(B \cap C)=0.20$, and $P(A \cap B \cap C)=0.15$. It then follows that
$$
\begin{aligned}
P(A \cup B \cup C)= & P(A)+P(B)+P(C)-P(A \cap B)-P(A \cap C) \
& -P(B \cap C)+P(A \cap B \cap C) \
= & 0.43+0.40+0.32-0.29-0.22-0.20+0.15 \
= & 0.59
\end{aligned}
$$
is the probability that this person watched at least one of these sports.
Let a probability set function be defined on a sample space $S$. Let $S=\left{e_1, e_2, \ldots, e_m\right}$, where each $e_i$ is a possible outcome of the experiment. The integer $m$ is called the total number of ways in which the random experiment can terminate. If each of these outcomes has the same probability of occurring, we say that the $m$ outcomes are equally likely. That is,
$$
P\left(\left{e_i\right}\right)=\frac{1}{m}, \quad i=1,2, \ldots, m .
$$
If the number of outcomes in an event $A$ is $h$, then the integer $h$ is called the number of ways that are favorable to the event $A$. In this case, $P(A)$ is equal to the number of ways favorable to the event $A$ divided by the total number of ways in which the experiment can terminate. That is, under this assumption of equally likely outcomes, we have
$$
P(A)=\frac{h}{m}=\frac{N(A)}{N(S)},
$$
where $h=N(A)$ is the number of ways $A$ can occur and $m=N(S)$ is the number of ways $S$ can occur. Exercise 1.1-15 considers this assignment of probability in a more theoretical manner.

It should be emphasized that in order to assign the probability $h / m$ to the event $A$, we must assume that each of the outcomes $e_1, e_2, \ldots, e_m$ has the same probability $1 / \mathrm{m}$. This assumption is then an important part of our probability model; if it is not realistic in an application, then the probability of the event $A$ cannot be computed in this way. Actually, we have used this result in the simple case given in Example 1.1-3 because it seemed realistic to assume that each of the possible outcomes in $S={H H, H T, T H, T T}$ had the same chance of being observed.

我们对一个群体去年在电视上观看体育赛事的习惯进行了调查。让 $A={$看橄榄球的$/}，B={$看篮球的$/}，C={$看棒球的$/}$。结果表明，如果从调查组中随机抽取一个人，那么 $P(A)=0.43，P(B)=0.40，P(C)=0.32，P(A \cap B)=0.29$，$P(A \cap C)=0.22，P(B \cap C)=0.20$，$P(A \cap B \cap C)=0.15$。由此得出
$$
\begin{aligned}
P(A\cap B\cap C)= & P(A)+P(B)+P(C)-P(A\cap B)-P(A\cap C)\
& -P(B \cap C)+P(A \cap B \cap C) \
= & 0.43+0.40+0.32-0.29-0.22-0.20+0.15 \
= & 0.59
\end{aligned}
$$
是这个人至少观看了其中一项运动的概率。
在样本空间 $S$ 上定义一个概率集函数。让 $S=\left{e_1, e_2, \ldots, e_m\right}$, 其中每个 $e_i$ 是实验的一个可能结果。整数 $m$ 称为随机实验终止的方式总数。如果每个结果出现的概率相同，我们就说 $m$ 结果的可能性相同。也就是说
$$
P\left(\left{e_i\right}\right)=\frac{1}{m}, \quad i=1,2, \ldots, m .
$$
如果一个事件 $A$ 的结果数为 $h$，那么整数 $h$ 称为对事件 $A$ 有利的方式数。在这种情况下，$P(A)$ 等于对事件 $A$ 有利的方式数除以实验可能终止的方式总数。也就是说，在结果可能性相同的假设下，我们有
$$
P(A)=\frac{h}{m}=\frac{N(A)}{N(S)},
$$
其中，$h=N(A)$ 是 $A$ 发生的方式数，$m=N(S)$ 是 $S$ 发生的方式数。练习 1.1-15 以更理论化的方式考虑了概率的分配。

需要强调的是，为了给事件 $A$ 分配概率 $h /m$，我们必须假设每个结果 $e_1、e_2、\ldots、e_m$ 都有相同的概率 $1 /\mathrm{m}$。这个假设是我们的概率模型的重要组成部分；如果它在应用中不现实，那么事件 $A$ 的概率就不能用这种方法计算。事实上，我们在例 1.1-3 中给出的简单案例中使用了这一结果，因为假设 $S={H H, H T, T H, T}$ 中的每个可能结果都有相同的被观察到的机会似乎是现实的。

Probability for Inference推理概率案例2

Let the random experiment be the cast of a die. Then the outcome space associated with this experiment is $S={1,2,3,4,5,6}$, with the elements of $S$ indicating the number of spots on the side facing up. For each $s \in S$, let $X(s)=s$. The space of the random variable $X$ is then ${1,2,3,4,5,6}$.

If we associate a probability of $1 / 6$ with each outcome, then, for example, $P(X=5)=1 / 6, P(2 \leq X \leq 5)=4 / 6$, and $P(X \leq 2)=2 / 6$ seem to be reasonable assignments, where, in this example, ${2 \leq X \leq 5}$ means ${X=2,3,4$, or 5$}$ and ${X \leq 2}$ means ${X=1$ or 2$}$.
The student will no doubt recognize two major difficulties here:

In many practical situations, the probabilities assigned to the events are unknown.
Since there are many ways of defining a function $X$ on $S$, which function do we want to use?

As a matter of fact, the solutions to these problems in particular cases are major concerns in applied statistics. In considering (2), statisticians try to determine what measurement (or measurements) should be taken on an outcome; that is, how best do we “mathematize” the outcome? These measurement problems are most difficult and can be answered only by getting involved in a practical project. For (1), we often need to estimate these probabilities or percentages through repeated observations (called sampling). For example, what percentage of newborn girls in the University of Iowa Hospital weigh less than 7 pounds? Here a newborn baby girl is the outcome, and we have measured her one way (by weight), but obviously there are many other ways of measuring her. If we let $X$ be the weight in pounds, we are interested in the probability $P(X<7)$, and we can estimate this probability only by repeated observations. One obvious way of estimating it is by the use of the relative frequency of ${X<7}$ after a number of observations. If it is reasonable to make additional assumptions, we will study other ways of estimating that probability. It is this latter aspect with which the field of mathematical statistics is concerned. That is, if we assume certain models, we find that the theory of statistics can explain how best to draw conclusions or make predictions.

In many instances, it is clear exactly what function $X$ the experimenter wants to define on the outcome space. For example, the caster in the dice game called craps is concerned about the sum of the spots (say $X$ ) that are facing upward on the pair of dice. Hence, we go directly to the space of $X$, which we shall denote by the same letter $S$. After all, in the dice game the caster is directly concerned only with the probabilities associated with $X$. Thus, for convenience, in many instances the reader can think of the space of $X$ as being the outcome space.

Let $X$ denote a random variable with space $S$. Suppose that we know how the probability is distributed over the various subsets $A$ of $S$; that is, we can compute $P(X \in A)$. In this sense, we speak of the distribution of the random variable $X$, meaning, of course, the distribution of probability associated with the space $S$ of $X$.

假设随机实验是掷骰子。那么与此实验相关的结果空间为 $S={1,2,3,4,5,6}$，$S$ 中的元素表示朝上一面的点数。对于 S$ 中的每个 $s，让 $X(s)=s$。随机变量 $X$ 的空间为 ${1,2,3,4,5,6}$。

如果我们为每个结果设定一个 1 / 6$ 的概率，那么，举例来说，$P(X=5)=1 / 6$，P(2 \leq X \leq 5)=4 / 6$、在这个例子中，${2 \leq X \leq 5}$表示${X=2,3,4$或5$/}$，而${X \leq 2}$表示${X=1$或2$/}$。
毫无疑问，学生在这里会遇到两个主要困难：

在许多实际情况中，事件的概率是未知的。
既然有许多方法可以定义 $S$ 上的函数 $X$，那么我们要使用哪个函数呢？

事实上，如何在特定情况下解决这些问题是应用统计学的主要关注点。在考虑第(2)点时，统计学家试图确定应该对结果进行哪些（或哪些）测量；也就是说，我们如何才能最好地将结果 “数学化”？这些测量问题最为棘手，只有通过参与实际项目才能找到答案。对于 (1)，我们通常需要通过重复观察（称为抽样）来估计这些概率或百分比。例如，爱荷华大学医院的新生女婴体重不足 7 磅的比例是多少？这里的结果是一个新生女婴，我们用一种方法（按体重）对她进行了测量，但显然还有很多其他测量方法。如果让 $X$ 作为体重（以磅为单位），我们感兴趣的是概率 $P(X<7)$，而我们只能通过重复观察来估计这个概率。一个显而易见的估计方法是使用多次观察后 ${X<7}$ 的相对频率。如果做出额外的假设是合理的，我们将研究估计该概率的其他方法。数理统计领域关注的正是后一方面。也就是说，如果我们假设了某些模型，我们就会发现统计理论可以解释如何最好地得出结论或进行预测。

在许多情况下，实验者想要在结果空间上定义什么函数 $X$是很清楚的。例如，在掷骰子游戏中，掷骰子者关心的是一对骰子上朝上的点数总和（即 X$）。因此，我们直接进入 $X$ 的空间，用同一个字母 $S$ 表示。毕竟，在掷骰子游戏中，掷骰者直接关心的只是与 $X$ 相关的概率。因此，为了方便起见，在许多情况下，读者可以将 $X$ 的空间视为结果空间。

让 $X$ 表示空间为 $S$ 的随机变量。假设我们知道概率在 $S$ 的各个子集 $A$ 上是如何分布的；也就是说，我们可以计算 $P(X\inA)$。在这个意义上，我们所说的随机变量 $X$ 的分布，当然是指与 $X$ 的空间 $S$ 相关的概率分布。

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｜MAST20018｜Discrete Maths and Operations Research离散数学和运筹学墨尔本大学

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

statistics-lab^TM为您提供墨尔本大学（The University of Melbourne，简称UniMelb，中文简称“墨大”）Elements of Quantum Computing量子计算要素澳洲代写代考和辅导服务！

课程介绍：

This subject introduces the essential features of Operations Research methods, and also introduces the area of Discrete Mathematics with a focus on applications. Operations Research topics include mathematical modelling, linear programming, simplex methods, and duality theory. Discrete Mathematics topics include scheduling, graph theory, assignments and matchings, and fair division. The subject material has a common theme of applications of mathematics in realistic settings encountered in the business world, industry and day-to-day life.

澳洲代写｜MAST20018｜Discrete Maths and Operations Research离散数学和运筹学墨尔本大学

Discrete Maths and Operations Research离散数学和运筹学案例

For a discrete random variable $X$, the cdf is a step function. Why? Because, the edf $F$ ‘jumps’ at the very values the random variable takes on. For continuous random variables, the range is real-valued(uncountable). In this case, $F$ is often continuous, even differentiable. For continuous random variables, we are interested in intervals, $X \in[a, b]$, the set of all reals between $a$ and $b$. Now,
$$
P(X \in[a, b])=P(X \in(-\infty, b])-P(X \in(-\infty, a])=P(X \leq b)-P(X \leq a)=F(b)-F(a) .
$$
where $F$ is the cdf. So, given the cdf, one can compute $P(X \in[a, b])$. Next, if $F$ is differentiable, then there exists a function $f$ such that,

$F(b)-F(a)=\int_a^b f(x) d x$.
$F^{\prime}(x)=f(x)$.
Thus,
$$
P(X \in[a, b])=\int_a^b f(x) d x .
$$
$f(x)$ is called the probability density function(pdf). Note:
For a cdf $F(y)$,
(a) $F(y)$ is an increasing function, $y_1 \geq y_2, F\left(y_1\right) \geq F\left(y_2\right)$.
(b) $F(y) \rightarrow 1$ as $y \rightarrow \infty$.
(c) $F(y) \rightarrow 0$ as $y \rightarrow-\infty$.
$P(X=a)=0$ for an continuous random variable $X$ and where $a$ is a point of continuity in $F(y)$. We will mostly deal with differentiable functions $F(y)$. So, $P(X=a)=0$ will always be true for every $a$.
$P(X \leq a)=P(X<a)+P(X=a)=P(X<a)+0=P(X<a)$. So we do not need to distinguish between less-than and less-than-and-equal.
$\int_{-\infty}^{\infty} f(x) d x=1$.
$\int_a^b f(x) d x=P(X \in[a, b])$ is the area under the curve $f(x)$ in the interval $[a, b]$.
$f(x)$ is not a probability. It is only a device used in computing probabilities.
$F(y)=\int_{-\infty}^{\infty} f(x) d x$.

Discrete Maths and Operations Research离散数学和运筹学案例2

7.3.3 Chebyshev’s Theorem
For any $k>1$,
$$
P\left(\frac{|x-\mu|}{\sigma} \geq k\right) \leq \frac{1}{k^2}=P\left(\frac{|x-\mu|}{\sigma} \leq k\right) \geq 1-\frac{1}{k^2}=P(\mu-k \sigma<x<\mu+k \sigma) \geq 1-\frac{1}{k^2} .
$$
$k$ is the standard deviation. The distribution does not need to be known for the theorem to work. The above expression can also be thought of as a confidence interval of $X$. The confidence interval of $\mu$ if $\sigma$ is known and the distribution of $X$ is unknown is,
$$
\begin{aligned}
& E(\bar{X})=\mu, \operatorname{Var}(\bar{X})=\frac{\sigma^2}{n} . \
& P\left(\mu-\frac{2 \sigma}{\sqrt{n}}<\bar{X}<\mu+\frac{2 \sigma}{\sqrt{n}}\right) \geq 0.75=P\left(\bar{X}-\frac{2 \sigma}{\sqrt{n}}<\mu<\bar{X}+\frac{2 \sigma}{\sqrt{n}}\right) \geq 0.75 .
\end{aligned}
$$
Example: Police are interested in the age profile of serial killers. Based on 15 solved cases of serial killers in the US, the average age is 28.5 years old with a standard deviation of 2 years. Give a $75 \%$ confidence interval for the age of the serial killer. $X$ is the age of the serial killer.
$$
\begin{aligned}
& P(\mu-2 \sigma<X<\mu+2 \sigma) \geq 0.75 . \
& P(28.5 \pm 2(2))=[24.5,32.5] .
\end{aligned}
$$

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｜MULT90063｜Introduction to Quantum Computing 量子计算入门墨尔本大学

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

课程介绍：

This subject will introduce students to the world of quantum information technology, focusing on the fast developing area of quantum computing. The subject will cover basic principles of quantum logic operations in both digital and analogue approaches to quantum processors, through to quantum error correction and the implementation of quantum algorithms for real-world problems. In lab-based classes students will learn to use state-of-the-art quantum computer programing and simulation environments to complete a range of projects.

澳洲代写｜MULT90063｜Introduction to Quantum Computing 量子计算入门墨尔本大学

Introduction to Quantum Computing 量子计算入门定义

Two vectors are said to be orthogonal if their inner product is zero. The norm of a vector $|\psi\rangle$, denoted $||\psi\rangle |$, is the square root of the inner product of $|\psi\rangle$ with itself. That is,
$$
||\psi\rangle | \equiv \sqrt{\langle\psi \mid \psi\rangle} .
$$
The quantity $||\psi\rangle |$ is called the Euclidean norm of $|\psi\rangle$. A vector is called a unit vector if it has norm 1 . A set of unit vectors that are mutually orthogonal is called an orthonormal set.

The Kronecker delta function, $\delta_{i, j}$, is defined to be equal to 1 whenever $i=j$, and 0 otherwise. We use the Kronecker delta function in our definition of an orthonormal basis.

Definition 2.2.3 Consider a Hilbert space $\mathcal{H}$ of dimension $2^n$. A set of $2^n$ vectors $B=\left{\left|b_m\right\rangle\right} \subseteq \mathcal{H}$ is called an orthonormal basis for $\mathcal{H}$ if
$$
\left\langle b_n \mid b_m\right\rangle=\delta_{n, m} \quad \forall b_m, b_n \in B
$$
and every $|\psi\rangle \in \mathcal{H}$ can be written as
$$
|\psi\rangle=\sum_{b_n \in B} \psi_n\left|b_n\right\rangle, \text { for some } \psi_n \in \mathbb{C} \text {. }
$$
The values of $\psi_n$ satisfy $\psi_n=\left\langle b_n \mid \psi\right\rangle$, and are called the ‘coefficients of $|\psi\rangle$ with respect to basis $\left{\left|b_n\right\rangle\right}$ ‘.

如果两个向量的内积为零，则称它们正交。向量 $|\psi\rangle$ 的范数，表示为 $||\psi\rangle |$，是 $|\psi\rangle$ 与其自身的内积的平方根。那是，
$$
||\psi\rangle | \equiv \sqrt{\langle\psi \mid \psi\rangle} 。
$$
数量 $||\psi\rangle |$ 称为 $|\psi\rangle$ 的欧几里得范数。如果向量的范数为 1 ，则该向量称为单位向量。一组相互正交的单位向量称为正交集。

克罗内克德尔塔函数 $\delta_{i, j}$ 定义为只要 $i=j$ 就等于 1，否则等于 0。我们在标准正交基的定义中使用克罗内克 delta 函数。

定义2.2.3 考虑维度为$2^n$的希尔伯特空间$\mathcal{H}$。一组 $2^n$ 个向量 $B=\left{\left|b_m\right\rangle\right} \subseteq \mathcal{H}$ 称为 $\mathcal{H}$ 的标准正交基，如果
$$
\left\langle b_n \mid b_m\right\rangle=\delta_{n, m} \quad \forall b_m, b_n \in B
$$
并且 \mathcal{H}$ 中的每个 $|\psi\rangle \ 可以写为
$$
|\psi\rangle=\sum_{b_n \in B} \psi_n\left|b_n\right\rangle, \text { 对于某些 } \psi_n \in \mathbb{C} \text {. }
$$
$\psi_n$ 的值满足 $\psi_n=\left\langle b_n \mid \psi\right\rangle$，称为“$|\psi\rangle$ 相对于基 $\left{ 的系数” \left|b_n\right\rangle\right}$ ‘.

An Overview of Statistical Inference 统计推断基础理论

The Basics of Point Estimation
The problem here, briefly stated, is as follows. Let $X$ be a r.v. with a p.d.f. $f$ which, however, involves a parameter. This is the case, for instance, in the Binomial distribution $B(1, p)$, the Poisson distribution $P(\lambda)$, the Negative Exponential $f(x)=\lambda e^{-\lambda x}, x>0$ distribution, the Uniform distribution $U(0, \alpha)$, and the Normal distribution $N\left(\mu, \sigma^2\right)$ with one of the quantities $\mu$ and $\sigma^2$ known. The parameter is usually denoted by $\theta$, and the set of its possible values is denoted by $\Omega$ and is called the parameter space. In order to emphasize the fact that the p.d.f. depends on $\theta$, we write $f(\cdot ; \theta)$. Thus, in the distributions mentioned above, we have for the respective p.d.f.’s and the parameter spaces:
$$
f(x ; \theta)=\theta^x(1-\theta)^{1-x}, \quad x=0,1, \quad \theta \in \Omega=(0,1) .
$$
The situations described in Examples 5, 6, 8, 9, and 10 of Chapter 1 may be described by a Binomial distribution.
$$
f(x ; \theta)=\frac{e^{-\theta} \theta^x}{x !}, \quad x=0,1, \ldots, \quad \theta \in \Omega=(0, \infty) .
$$
The Poisson distribution can be used appropriately in the case described in Example 12 of Chapter 1.
$$
\begin{aligned}
& f(x ; \theta)=\theta e^{-\theta x}, \quad x>0, \quad \theta \in \Omega=(0, \infty) . \
& f(x ; \theta)=\left{\begin{array}{ll}
\frac{1}{\theta}, & 0<x<\theta \
0, & \text { otherwise, }
\end{array} \quad \theta \in \Omega=(0, \infty) .\right. \
& f(x ; \theta)=\frac{1}{\sqrt{2 \pi} \sigma} e^{-\frac{(x-\theta)^2}{2 \sigma^2}}, \quad x \in \Re, \quad \theta \in \Omega=\Re, \quad \sigma^2 \text { known, }
\end{aligned}
$$
and
$$
f(x ; \theta)=\frac{1}{\sqrt{2 \pi \theta}} e^{-\frac{(x-\mu)^2}{2 \vartheta}}, \quad x \in \Re, \quad \theta \in \Omega=(0, \infty), \quad \mu \text { known. }
$$
Normal distributions are suitable for modeling the situations described in Examples 16 and 17 of Chapter 1.

Our objective is to draw a random sample of size $n, X_1, \ldots, X_n$, from the underlying distribution, and on the basis of it to construct a point estimate (or estimator) for $\theta$, that is, a statistic $\hat{\theta}=\hat{\theta}\left(X_1, \ldots, X_n\right)$, which is used for

Finally, another relatively popular method (in particular, in the context of certain models) is the method of Least Squares (LS). The method of LS leads to the construction of an estimate for $\theta$, the Least Squares Estimate (LSE) of $\theta$, through a minimization (with respect to $\theta$ ) of the sum of certain squares. This sum of squares represents squared deviations between what we actually observe after experimentation is completed and what we would expect to have on the basis of an assumed model. Once again, details will be presented later on, more specifically, in Chapter 13.

In all of the preceding discussion, it was assumed that the underlying p.d.f. depended on a single parameter, which was denoted by $\theta$. It may very well be the case that there are two or more parameters involved. This may happen, for instance, in the Uniform distribution $U(\alpha, \beta),-\infty<\alpha<\beta<\infty$, where both $\alpha$ and $\beta$ are unknown; the Normal distribution, $N\left(\mu, \sigma^2\right)$, where both $\mu$ and $\sigma^2$ are unknown; and it does happen in the Multinomial distribution, where the number of parameters is $k, p_1, \ldots, p_k$ (or more precisely, $k-1$, since the $k$ th parameter, for example, $\left.p_k=1-p_1-\cdots-p_{k-1}\right)$. For instance, Examples 20 and 21 of Chapter 1 refer to situations where a Multinomial distribution is appropriate. In such multiparameter cases, one simply applies to each parameter separately what was said above for a single parameter. The alternative option to use the vector notation for the parameters involved does simplify things in a certain way but also introduces some complications in other ways.

点估计的基础知识
这里的问题简单地说如下。令 $X$ 为 r.v. 带有 pdf。然而，$f$ 涉及一个参数。例如，二项式分布 $B(1, p)$、泊松分布 $P(\lambda)$、负指数 $f(x)=\lambda e^{-\lambda x }、x>0$ 分布、均匀分布 $U(0, \alpha)$ 和正态分布 $N\left(\mu, \sigma^2\right)$ ，其中数量为 $\mu$ 之一和 $\sigma^2$ 已知。参数通常用$\theta$表示，其可能值的集合用$\Omega$表示，称为参数空间。为了强调这个事实，pdf。取决于$\theta$，我们写为$f(\cdot ; \theta)$。因此，在上述分布中，我们有各自的 p.d.f. 和参数空间：
$$
f(x ; \theta)=\theta^x(1-\theta)^{1-x}, \quad x=0,1, \quad \theta \in \Omega=(0,1) 。
$$
第 1 章示例 5、6、8、9 和 10 中描述的情况可以通过二项式分布来描述。
$$
f(x ; \theta)=\frac{e^{-\theta} \theta^x}{x !}, \quad x=0,1, \ldots, \quad \theta \in \Omega=(0 ，\infty) 。
$$
在第 1 章示例 12 中描述的情况下可以适当使用泊松分布。
$$
\开始{对齐}
& f(x ; \theta)=\theta e^{-\theta x}, \quad x>0, \quad \theta \in \Omega=(0, \infty) 。 \
& f(x ; \theta)=\left{\begin{array}{ll}
\frac{1}{\theta}, & 0<x<\theta \
0, & \text { 否则, }
\end{array} \quad \theta \in \Omega=(0, \infty) .\right。 \
& f(x ; \theta)=\frac{1}{\sqrt{2 \pi} \sigma} e^{-\frac{(x-\theta)^2}{2 \sigma^2}}, \quad x \in \Re、\quad \theta \in \Omega=\Re、\quad \sigma^2 \text { 已知， }
\结束{对齐}
$$
和
$$
f(x ; \theta)=\frac{1}{\sqrt{2 \pi \theta}} e^{-\frac{(x-\mu)^2}{2 \vartheta}}, \quad x \in \Re, \quad \theta \in \Omega=(0, \infty), \quad \mu \text { 已知。 }
$$
正态分布适合对第 1 章示例 16 和 17 中描述的情况进行建模。

我们的目标是从底层分布中抽取大小为 $n, X_1, \ldots, X_n$ 的随机样本，并在此基础上构建 $\theta$ 的点估计（或估计器），即统计量 $\hat{\theta}=\hat{\theta}\left(X_1, \ldots, X_n\right)$，用于

最后，另一种相对流行的方法（特别是在某些模型的背景下）是最小二乘法（LS）。 LS 方法通过某些平方和的最小化（相对于 $\theta$ ）来构建 $\theta$ 的估计，即 $\theta$ 的最小二乘估计 (LSE)。这个平方和代表实验完成后我们实际观察到的结果与我们根据假设模型期望得到的结果之间的平方偏差。稍后将再次介绍详细信息，更具体地说，将在第 13 章中介绍。

在前面的所有讨论中，都假设基础 p.d.f. 依赖于一个参数，用 $\theta$ 表示。很可能涉及两个或多个参数。例如，这种情况可能发生在均匀分布 $U(\alpha, \beta),-\infty<\alpha<\beta<\infty$ 中，其中 $\alpha$ 和 $\beta$ 都是未知的；正态分布，$N\left(\mu, \sigma^2\right)$，其中 $\mu$ 和 $\sigma^2$ 都是未知的；它确实发生在多项式分布中，其中参数的数量为 $k, p_1, \ldots, p_k$ （或更准确地说，$k-1$，因为第 $k$ 个参数，例如 $\left .p_k=1-p_1-\cdots-p_{k-1}\right)$. 例如，第一章的示例 20 和 21 提到了适合多项式分布的情况。在这种多参数情况下，只需将上面针对单个参数所述的内容分别应用于每个参数即可。对所涉及的参数使用向量表示法的替代选项确实以某种方式简化了事情，但也以其他方式引入了一些复杂性。

澳洲代写｜MULT20015｜Elements of Quantum Computing量子计算要素墨尔本大学

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｜MULT20015｜Elements of Quantum Computing量子计算要素墨尔本大学

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

课程介绍：

This subject provides an introduction to quantum computing for students from a range of backgrounds that will position students for further studies in quantum computing and programming. The syllabus will cover basic principles of quantum bits (qubits) and quantum logic operations, through to applications in communication, security, optimisation, finance and machine learning. In computer-based tutorials, students will learn quantum computer programming using a state-of-the-art simulation environment, and access online quantum computer systems where available. This subject will be taught by a multi-disciplinary teaching team from physics, mathematics, computer science and finance.

Elements of Quantum Computing量子计算要素进阶实分析案例

问题 1.

Suppose we have a qubit in the state $|0\rangle$, and we measure the observable $X\left[\sigma_x\right]$. What is the average value of $X$ ? What is the standard deviation of $X$ ?

If we have a qubit in the state $|0\rangle$, and we measure the observable $\sigma_x$, we project onto the two eigenvectors of $\sigma_x$, which are $|+\rangle=\frac{1}{\sqrt{2}}(|0\rangle+|1\rangle)$, with eigenvalue 1 , and $|-\rangle=\frac{1}{\sqrt{2}}(|0\rangle-|1\rangle)$, with eigenvalue -1 . It is easy to check that $|0\rangle=\frac{1}{\sqrt{2}}(|+\rangle+|-\rangle)$, so the probability of seeing $|+\rangle$ and $|-\rangle$ is each $\frac{1}{2}$. We thus observe +1 with probability $\frac{1}{2}$ and -1 with probability $\frac{1}{2}$. The expectation is 0 and the standard deviation is 1 .

问题 2.

Show that $\vec{v} \cdot \vec{\sigma}$ has eigenvalues \pm 1 , and that the projectors onto the corresponding eigenspaces are given by $P_{ \pm} \equiv(I \pm \vec{v} \cdot \vec{\sigma}) / 2$.

Solution： We have
$$
\vec{v} \cdot \vec{\sigma}=v_x \sigma_x+v_y \sigma_y+v_z \sigma_z .
$$
Consider
$$
(\vec{v} \cdot \vec{\sigma})^2
$$
By using the relations $\sigma_x^2=\sigma_y^2=\sigma_z^2=I$, and the fact that any two distinct Pauli matrices anticommute, i.e., $\sigma_x \sigma_y=-\sigma_y \sigma_x$, we can see that
$$
(\vec{v} \cdot \vec{\sigma})^2=\left(v_x^2+v_y^2+v_z^2\right) I=I .
$$
Therefore, its eigenvalues must be \pm 1 . It is easy to check that $\vec{v} \cdot \vec{\sigma}$ is not $\pm I$, so one eigenvalue must be +1 and the other must be -1 . Now, if we let the eigenvectors be $\left|\phi_{+}\right\rangle$and $\left|\phi_{-}\right\rangle$, we have that
$$
\vec{v} \cdot \vec{\sigma}=\left|\phi_{+}\right\rangle\left\langle\phi_{+}|-| \phi_{-}\right\rangle\left\langle\phi_{-}\right|
$$
and so
$$
(I+\vec{v} \cdot \vec{\sigma}) / 2=\left|\phi_{+}\right\rangle\left\langle\phi_{+}\right|=P_{+},
$$
and similarly for the -1 eigenvector.

问题 3.

Calculate the probability of obtaining the result +1 for a measurement of $\vec{v} \cdot \vec{\sigma}$, given that the state prior to measurement is $|0\rangle$. What is the state of the system after the measurement if +1 is obtained?

The expectation of the observable $\vec{v} \cdot \vec{\sigma}$ when $|0\rangle$ is measured is
$$
\langle 0|\vec{v} \cdot \vec{\sigma}| 0\rangle=v_x\left\langle 0\left|\sigma_x\right| 0\right\rangle+v_y\left\langle 0\left|\sigma_y\right| 0\right\rangle+v_z\left\langle 0\left|\sigma_z\right| 0\right\rangle .
$$
This is easily seen to be equal to $v_z$. Thus,
$$
v_z=\operatorname{Prob}(+1)-\operatorname{Prob}(-1)
$$
and using the fact that the two probabilities add to 1 gives
$$
\operatorname{Prob}(+1)=\left(v_z+1\right) / 2 .
$$Problems 1-3 deal with what is known as the GHZ state (after Greenberger, Horne and Zeilinger), and the proof of non-locality using this state which was discovered by Greenberger, Horne, Shimony and Zeilinger.

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金融工程代写

非参数统计代写

非参数统计指的是一种统计方法，其中不假设数据来自于由少数参数决定的规定模型；这种模型的例子包括正态分布模型和线性回归模型。

广义线性模型代考

广义线性模型（GLM）归属统计学领域，是一种应用灵活的线性回归模型。该模型允许因变量的偏差分布有除了正态分布之外的其它分布。

有限元方法代写

随机分析代写

时间序列分析代写

回归分析代写

MATLAB代写

R语言代写	问卷设计与分析代写
PYTHON代写	回归分析与线性模型代写
MATLAB代写	方差分析与试验设计代写
STATA代写	机器学习/统计学习代写
SPSS代写	计量经济学代写
EVIEWS代写	时间序列分析代写
EXCEL代写	深度学习代写
SQL代写	各种数据建模与可视化代写

澳洲代写｜MAST20033｜Real Analysis: Advanced 进阶实分析墨尔本大学

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

statistics-lab^TM为您提供墨尔本大学（The University of Melbourne，简称UniMelb，中文简称“墨大”）Real Analysis: Advanced 进阶实分析澳洲代写代考和辅导服务！

课程介绍：

This subject introduces the field of mathematical analysis both with a careful theoretical framework as well as selected applications. Many of the important results are proved rigorously and students are introduced to methods of proof such as mathematical induction and proof by contradiction.

The important distinction between the real numbers and the rational numbers is emphasised and used to motivate rigorous notions of convergence and divergence of sequences, including the Cauchy criterion. Various constructions of the real numbers, for example using Dedekind cuts or by completion, are discussed and shown to be equivalent. These ideas are extended to cover the theory of infinite series, including common tests for convergence and divergence. Compactness of the unit interval is established and various consequences of compactness, such as the Extreme Value Theorem, are discussed. A similar treatment of continuity and differentiability of functions of a single variable leads to applications such as the Mean Value Theorem and Taylor’s theorem. We define and compare both the Lebesgue and Riemann integral, establish basic properties of both, and dis- cuss the proof of the Fundamental Theorem of Calculus. The convergence properties of sequences and series are explored, with applications to power series representations of elementary functions and their generation by Taylor series. Fourier series are introduced as a way to represent periodic functions. Further topics may include: uniform continuity, equicontinuity, the Arzela-Ascoli theorem, and the Stone-Weierstrass theorem.

澳洲代写｜MAST20033｜Real Analysis: Advanced 进阶实分析墨尔本大学

Real Analysis: Advanced 进阶实分析案例

问题 1.

Let $K$ be any field, and $x, y \in K$ elements such that $x^2=y^2$. Prove, using only the axioms of a field, that $x=y$ or $x=-y$. Be careful to explain which axiom you use at each step.

Suppose $F$ is a field, and $x, y \in F$ satisfy $x^2=y^2$. We need to show that either $x=y$ or $x=-y$. For the sake of sparing the proliferation of parenthesis, we will adopt the standard conventions about multiplication and addition, i.e. $x+y \cdot z$ is $x+(y \cdot z)$ rather than $(x+y) \cdot z$.

Now consider the quantity $(x+(-y)) \cdot(x+y)$. Using the distributive property, this is equal to $(x+(-y)) x+(x+(-y)) y$. Using distributivity again, we get $\left(x^2+(-y) \cdot x\right)+(x \cdot y+(-y) \cdot y)$. Using associativity of addition and commutativity of multiplication, we can rearrange this to $x^2+((x \cdot y+x \cdot(-y))+(-y) \cdot y)$. Now, we have
$$
x \cdot y+x \cdot(-y)=x \cdot(y+(-y))=x \cdot 0=0
$$
Where the first step follows from distributivity, the second from the definition of $-y$, and the third from noting that that $x \cdot 0=x \cdot(0+0)=x \cdot 0+x \cdot 0$ and adding $-(x \cdot 0)$ to both sides. Similarly, $y \cdot(-y)+y^2=0$, and adding $-y^2$ on the right to both sides gives $(-y) \cdot y=-y^2$. Going back, we have
$$
\begin{gathered}
(x-y) \cdot(x+y)=x^2+((x \cdot y+x \cdot(-y))+(-y) \cdot y) \
=x^2+\left((0)+\left(-y^2\right)\right)=x^2+\left(-y^2\right)
\end{gathered}
$$
But by assumption $x^2=y^2$, so this is equal to $y^2+\left(-y^2\right)=0$. In other words, $(x-y) \cdot(x+y)=0$.

If $x=y$, then we are done. So suppose $x \neq y$. Then $x-y \neq 0$, since if $x-y=0$ we have $y=0+y=(x-y)+y=x+(y-y)=x+0=x$

using, respectively, definition of 0 , substitution, associativity of addition, definition of $-y$, and definition of 0 . So there must exist a multiplicative inverse $(x-y)^{-1}$. Then we have
$$
\begin{aligned}
& 0=(x-y)^{-1} \cdot 0=(x-y)^{-1}((x-y)(x+y)= \
& \left((x-y)^{-1}(x-y)\right)(x+y)=1 \cdot(x+y)=x+y
\end{aligned}
$$
The first equality is a fact we already proved, the second is just substitution, the third is associativity of multiplication, the fourth is the definition of multiplicative inverse, and the fifth is the definition of 1 .
Now we get
$$
x=x+0=x+(y-y)=(x+y)-y=0-y=-y
$$
Using the definition of 0 , definition of $-y$, associativity of addition, substitution, and definition of 0 . In other words, if $x \neq y$, then $x=-y$, which is what we wanted to prove.

问题 2.

Let $(X, d)$ be a metric space. Recall that a subset $E \subset X$ is dense if $\bar{E}=X$. Prove: if $X$ has a dense subset which is finite, then $X$ itself is finite.

Suppose $E$ is a finite dense subset of $X$. Then $E$ contains no limit points. To see this, suppose $x \in X$ is a limit point of $E$. Then by Theorem 2.20 of Rudin, every neighbourhood of $x$ has to contain infinitely many points of $E$. But $E$ only has finitely many points, so this is obviously impossible. Since $E$ has no limit points, it is vacuously true that it contains all its limit points, so $E$ is closed and $E=\bar{E}$. But $E$ is dense in $X$, so $\bar{E}=X$. This means that $X=E$, so $X$ itself must be finite since $E$ is.

问题 3.

Recall the definition of the $p$-adic metric on the set $\mathbb{Z}$ of integers: we fix a prime number $p$. If $x \neq y$, then $d(x, y)=p^{-n}$, where $n$ is the largest integer such that $p^n$ divides $y-x$. Is $(\mathbb{Z}, d)$ compact? (With proof, of course).
Hint: look at specific examples of subsets $\left{x_1, x_2, \ldots\right}$ such that the $p$-adic distance between $x_i$ and $x_{i+1}$ becomes smaller and smaller.

We will show that $\mathbb{Z}$ is not compact in the $p$-adic topology by constructing an infinite set with no limit points. We begin with an elementary

Lemma 1: Let $m \in \mathbb{N}$ and $k$ the largest natural number with $p^k \mid m$. Let $l$ be natural number with $m<p^l$. Then $p^k$ is the largest power of $p$ dividing $m+p^l$

Proof: We have $p^k \leq m<p^l$, so $k<l$ and $p^k \mid p^l$, hence $p^k$ divides $m+p^l$. Since $k<l, k+1 \leq \bar{l}$, so $p^{k+1} \mid p^l$. But then $p^{k+1}$ cannot divide $m+p^l$, since then it would divide $\left(m+p^l\right)-p^l=m$, a contradiction.

Now for $n \in \mathbb{N}$, we introduce the finite sum $s_n=\sum_{i=0}^{i=n} p^{2 i}$, where $p$ is the prime with respect to which we define the $p$-adic metric on $\mathbb{Z}$. Let $S=\left{s_n \mid n \in \mathbb{N}\right}$ be the set of all these numbers. $S$ is obviously infinite, so if we can show that $S$ has no limit points we are done. To do so, we need another elementary

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｜MAST90083｜Computational Statistics & Data Science计算统计与数据科学墨尔本大学

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

statistics-lab^TM为您提供墨尔本大学（The University of Melbourne，简称UniMelb，中文简称“墨大”）Computational Statistics & Data Science计算统计与数据科学澳洲代写代考和辅导服务！

课程介绍：

Computing techniques and data mining methods are indispensable in modern statistical research and data science applications, where “Big Data” problems are often involved. This subject will introduce a number of recently developed methods and applications in computational statistics and data science that are scalable to large datasets and high-performance computing. The data mining methods to be introduced include general model diagnostic and assessment techniques, kernel and local polynomial nonparametric regression, basis expansion and nonparametric spline regression, generalised additive models, classification and regression trees, forward stagewise and gradient boosting models. Important statistical computing algorithms and techniques used in data science will be explained in detail. These include the bootstrap resampling and inference, cross-validation, the EM algorithm and Louis method, and Markov chain Monte Carlo methods including adaptive rejection and squeeze sampling, sequential importance sampling, slice sampling, Gibbs sampler and Metropolis-Hastings algorithm.

澳洲代写｜MAST90083｜Computational Statistics & Data Science计算统计与数据科学墨尔本大学

Computational Statistics & Data Science计算统计与数据科学案例

Say we are interested in estimating the variance of the forearm data, and we decide to use the following statistic:

Then we get the estimate of the standard error that we need for the endpoints of the interval.

% Estimate the SE using the bootstrap.
SE = std(bootreps);

Now we get the quantiles that we need for the interval given and calculate the interval.

$\%$ Get the quantiles.
k= B*alpha/2;
szval $=\operatorname{sort}(z v a l s) ;$
tlo $=\operatorname{szval}(\mathrm{k}) ;$
thi $=\operatorname{szval}(\mathrm{B}-\mathrm{k}) ;$
$\%$ Get the endpoints of the interval.
blo = thetahat - thi*SE;
bhi = thetahat - tlo*SE;
$\%$ Get the quantiles.
szval $=\operatorname{sort}($ zvals $)$;
tlo $=\operatorname{szval}(\mathrm{k})$;
thi $=\operatorname{szval}(\mathrm{B}-\mathrm{k})$;
$\%$ Get the endpoints of the interval.
blo = thetahat - thi*SE;
bhi $=$ thetahat - tlo*SE;

The bootstrap- $t$ interval for the variance of the forearm data is $(1.00,1.57)$.
Bootstrap Percentile Interval
An improved bootstrap confidence interval is based on the quantiles of the distribution of the bootstrap replicates. This technique has the benefit of being more stable than the bootstrap-t, and it also enjoys better theoretical coverage properties [Efron and Tibshirani, 1993]. The bootstrap percentile confidence interval is
$$
\left(\hat{\theta}_B^{(\alpha / 2)}, \hat{\theta}_B^{(1-\alpha / 2)}\right)
$$
where $\hat{\theta}_B^{(\alpha / 2)}$ is the $\alpha / 2$ quantile in the bootstrap distribution of $\hat{\theta}^$. For example, if $\alpha / 2=0.025$ and $B=1000$, then $\hat{\theta}_B^{(0.025)}$ is the $\hat{\theta}^{ b}$ in the 25 th position of the ordered bootstrap replicates. Similarly, $\hat{\theta}_B^{*(0.975)}$ is the replicate in position 975. As discussed previously, some other suitable estimate for the quantile can be used.

The procedure is the same as the general bootstrap method, making it easy to understand and to implement. We outline the steps next.
PROCEDURE – BOOTSTRAP PERCENTILE INTERVAL

Given a random sample, $\mathbf{x}=\left(x_1, \ldots, x_n\right)$, calculate $\hat{\boldsymbol{\theta}}$.
Sample with replacement from the original sample to get $\mathbf{x}^{* b}=\left(x_1^{* b}, \ldots, x_n^{* b}\right)$.
Calculate the same statistic using the sample in step 2 to get the bootstrap replicates, $\hat{\theta}^{* b}$.

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

金融工程代写

非参数统计代写

非参数统计指的是一种统计方法，其中不假设数据来自于由少数参数决定的规定模型；这种模型的例子包括正态分布模型和线性回归模型。

广义线性模型代考

广义线性模型（GLM）归属统计学领域，是一种应用灵活的线性回归模型。该模型允许因变量的偏差分布有除了正态分布之外的其它分布。

有限元方法代写

随机分析代写

时间序列分析代写

回归分析代写

MATLAB代写

R语言代写	问卷设计与分析代写
PYTHON代写	回归分析与线性模型代写
MATLAB代写	方差分析与试验设计代写
STATA代写	机器学习/统计学习代写
SPSS代写	计量经济学代写
EVIEWS代写	时间序列分析代写
EXCEL代写	深度学习代写
SQL代写	各种数据建模与可视化代写

计算机代写|数据库作业代写Database代考|NIT1201

Posted on 2023年8月29日2023年8月29日 by statistics-lab

如果你也在怎样代写数据库Database 这个学科遇到相关的难题，请随时右上角联系我们的24/7代写客服。数据库Database可以成为一种强大的工具，它可以做计算机程序最擅长的事情:存储、操作和显示数据。

数据库Database不仅在许多应用程序中发挥作用，而且经常发挥关键作用。如果数据没有正确存储，它可能会损坏，程序将无法有意义地使用它。如果数据组织不当，程序可能无法在合理的时间内找到所需的数据。
除非数据库安全有效地存储其数据，否则无论系统的其余部分设计得多么好，应用程序都将是无用的。

statistics-lab™ 为您的留学生涯保驾护航在代写数据库Database方面已经树立了自己的口碑, 保证靠谱, 高质且原创的统计Statistics代写服务。我们的专家在代写数据库Database代写方面经验极为丰富，各种代写数据库Database相关的作业也就用不着说。

计算机代写|数据库作业代写Database代考|Oracle PKI Integration

Oracle Advanced Security includes Oracle PKI (public key infrastructure) integration for authentication and single sign-on. You can integrate Oracle-based applications with the PKI authentication and encryption framework, using the following tools:

Oracle Wallet Manager creates an encrypted Oracle Wallet, used for digital certificates.
Oracle Enterprise Login Assistant creates the obfuscated decrypted Oracle Wallet, used by Oracle applications for Secure Sockets Layer (SSL) encryption. The Oracle Wallet is then stored on the file system or Oracle Internet Directory.
Active Directory
Oracle customers with large user populations often require enterprise-level security and schemas management. Oracle security and administration are integrated with Windows 2000 through Active Directory, Microsoft’s directory service.

Oracle9i provides native authentication and single sign-on through Windows 2000 authentication mechanisms. Native authentication uses Kerberos security protocols on Windows 2000 and allows the operating system to perform user identification for Oracle databases. With native authentication enabled, users can access Oracle applications simply by logging into Windows. Single sign-on eliminates need for multiple security credentials and simplifies administration.

Oracle native authentication also supports Oracle9 $i$ enterprise users and roles. Traditionally, administrators must create a database user on every database for each Windows user. This often equates to thousands of different database users. Oracle enterprise user mappings allow many Windows users to access a database as a single global database user. These enterprise user mappings are stored in Active Directory. For example, entire organizational units in Active Directory can be mapped to one database user.

Oracle also stores enterprise role mappings in Active Directory. With such roles, a database privilege can be managed at the domain level through directories. This is accomplished by assigning Windows 2000 users and groups to Oracle enterprise roles registered in Active Directory. Enterprise users and roles reduce administrative overhead while increasing scalability of database solutions.

计算机代写|数据库作业代写Database代考|Oracle Net Naming with Active Directory

Oracle also uses Active Directory to improve management of database connectivity information. Traditionally, users reference databases with Oracle Net-style names resolved through the tnsnames . ora configuration file. This file has to be administered on each client computer.
Oracle Net Naming with Active Directory stores and resolves names through Active Directory. By centralizing such information in a directory, Oracle Net Naming with Active Directory eliminates administrative overhead and relieves users from configuring their individual client computers.
Various tools in Windows 2000, such as Windows Explorer and Active Directory Users and Computers, have been enhanced. Users can now connect to databases and test database connectivity from these tools.
Oracle tools have also been enhanced. Database Configuration Assistant automatically registers database objects with Active Directory. Oracle Net Manager, meanwhile, registers net service objects with the directory. These enhancements further simplify administration.
ORACLEMTSRecoveryService
Microsoft Transaction Server is used in the middle tier as an application server for $\mathrm{COM} / \mathrm{COM}+$ objects and transactions in distributed environments.
ORACLEMTSRecoveryService allows Oracle $9 i$ databases to be used as resource managers in Microsoft Transaction Server-coordinated transactions, providing strong integration between Oracle solutions and Microsoft Transaction Server. ORACLEMTSRecoveryService can operate with Oracle $9 i$ databases running on any operating system.
Oracle takes advantage of a native implementation and also stores recovery information in Oracle $9 i$ database itself. ORACLEMTSRecoveryService allows development in all industry-wide data access interfaces, including Oracle Objects for OLE (OO4O), Oracle Call Interface (OCI), ActiveX Data Objects (ADO), OLE DB, and Open Database Connectivity (ODBC). The Oracle APIs, OO4O and OCI, offer greatest efficiency.

数据库代考

计算机代写|数据库作业代写Database代考|Oracle PKI Integration

Oracle Advanced Security包括Oracle PKI(公钥基础设施)集成，用于身份验证和单点登录。您可以使用以下工具将基于oracle的应用程序与PKI认证和加密框架集成:

Oracle钱包管理器创建一个加密的Oracle钱包，用于数字证书。

Oracle企业登录助手创建混淆解密的Oracle钱包，用于Oracle应用程序的安全套接字层(SSL)加密。然后将Oracle钱包存储在文件系统或Oracle Internet目录中。
活动目录
拥有大量用户的Oracle客户通常需要企业级的安全性和模式管理。Oracle的安全和管理通过Active Directory(微软的目录服务)集成到Windows 2000中。

Oracle9i通过Windows 2000认证机制提供本地认证和单点登录。本机身份验证在Windows 2000上使用Kerberos安全协议，并允许操作系统对Oracle数据库执行用户标识。启用本地身份验证后，用户只需登录Windows即可访问Oracle应用程序。单点登录消除了对多个安全凭证的需求，并简化了管理。

Oracle本地认证还支持Oracle9 $i$企业用户和角色。传统上，管理员必须为每个Windows用户在每个数据库上创建一个数据库用户。这通常相当于数千个不同的数据库用户。Oracle企业用户映射允许多个Windows用户作为单个全局数据库用户访问数据库。这些企业用户映射存储在Active Directory中。例如，活动目录中的整个组织单位可以映射到一个数据库用户。

Oracle还将企业角色映射存储在Active Directory中。使用这些角色，可以通过目录在域级别管理数据库特权。这是通过将Windows 2000用户和组分配给在Active Directory中注册的Oracle企业角色来实现的。企业用户和角色减少了管理开销，同时提高了数据库解决方案的可伸缩性。

计算机代写|数据库作业代写Database代考|Oracle Net Naming with Active Directory

Oracle还使用Active Directory来改进数据库连接信息的管理。传统上，用户使用通过tnsnames解析的Oracle net风格名称来引用数据库。Ora配置文件。必须在每台客户端计算机上管理此文件。
Oracle网络命名与活动目录存储和解析名称通过活动目录。通过将这些信息集中在一个目录中，Oracle网络命名与活动目录消除了管理开销，并减轻了用户配置其个人客户端计算机的负担。
Windows 2000中的各种工具，如Windows资源管理器和Active Directory用户和计算机，都得到了增强。用户现在可以从这些工具连接到数据库并测试数据库连接性。
Oracle工具也得到了增强。数据库配置助手自动向Active Directory注册数据库对象。同时，Oracle Net Manager将Net服务对象注册到该目录中。这些增强功能进一步简化了管理。
ORACLEMTSRecoveryService
Microsoft Transaction Server在中间层用作分布式环境中$\ mathm {COM} / \ mathm {COM}+$对象和事务的应用程序服务器。
ORACLEMTSRecoveryService允许Oracle $ 9i $数据库在Microsoft Transaction Server协调的事务中用作资源管理器，提供了Oracle解决方案和Microsoft Transaction Server之间的强大集成。ORACLEMTSRecoveryService可以在任何操作系统上运行Oracle $ 9i $数据库。
Oracle利用了本地实现，并且还将恢复信息存储在Oracle $ 9i $数据库本身中。ORACLEMTSRecoveryService允许开发所有行业范围的数据访问接口，包括Oracle object for OLE (oo40)、Oracle Call Interface (OCI)、ActiveX data Objects (ADO)、OLE DB和Open Database Connectivity (ODBC)。Oracle api oo40和OCI提供了最高的效率。

计算机代写|数据库作业代写Database代考请认准statistics-lab™

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

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

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计算机代写|数据库作业代写Database代考|Oracle PKI Integration

计算机代写|数据库作业代写Database代考|Oracle Net Naming with Active Directory

计算机代写|数据库作业代写Database代考|Oracle PKI Integration

计算机代写|数据库作业代写Database代考|Oracle Net Naming with Active Directory

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Introduction to Quantum Computing 量子计算入门定义

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