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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

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｜MAST30021｜Complex Analysis复杂分析墨尔本大学

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：

Complex analysis is a core subject in pure and applied mathematics, as well as the physical and engineering sciences. While it is true that physical phenomena are given in terms of real numbers and real variables, it is often too difficult and sometimes not possible, to solve the algebraic and differential equations used to model these phenomena without introducing complex numbers and complex variables and applying the powerful techniques of complex analysis.

Topics include:the topology of the complex plane; convergence of complex sequences and series; holomorphic functions, the Cauchy-Riemann equations, harmonic functions and applications; contour integrals and the Cauchy Integral Theorem; singularities, Laurent series, the Residue Theorem, evaluation of integrals using contour integration, conformal mapping; and aspects of the gamma function.

澳洲代写｜MAST30021｜Complex Analysis复杂分析墨尔本大学

Complex Analysis复杂分析问题集

问题 1.

By writing $z$ in the form $z=a+b \mathrm{i}$, find all solutions $z$ of the following equations:
(i) $z^2=-5+12 \mathrm{i}$
(ii) $z^2=2+\mathrm{i}$
(iii) $(7+24 \mathrm{i}) z=375$
(iv) $z^2-(3+\mathrm{i}) z+(2+2 \mathrm{i})=0$
(v) $z^2-3 z+1+\mathrm{i}=0$

问题 2.

If $\lambda$ is a positive real number, show that
$$
{z \in \mathbb{C}:|z|=\lambda|z-1|}
$$
is a circle, unless $\lambda$ takes one particular value (which?)

问题 3.

Draw the set of points
$$
{z \in \mathbb{C}: \operatorname{re}(z+1)=|z-1|}
$$
by substituting $z=x+\mathrm{i} y$ and computing the real equation relating $x$ and $y$.
Now note that re $(z+1)$ is the distance from $z$ to the line $y=-1$, and $|z-1|$ is the distance between $z$ and 1. Compare with the classical ‘focus-directrix’ definition of a parabola: the locus of a point equidistant from a fixed line (here $y=-1$ ) and a fixed point (here $(x, y)=(1,0)$ ).

问题 4.

Let $r, s, \theta, \phi$ be real. Let
$$
\begin{aligned}
z & =r(\cos \theta+\mathrm{i} \sin \theta) \
w & =s(\cos \phi+\mathrm{i} \sin \phi)
\end{aligned}
$$
Form the product $z w$ and use the standard formulas for $\cos (\theta+\phi), \sin (\theta+\phi)$ to show that $\arg (z w)=\arg (z)+\arg (w)$ (for any values of $\arg$ on the right, and some value of arg on the left).
By induction on $n$, derive De Moivre’s Theorem
$$
(\cos \theta+\mathrm{i} \sin \theta)^n=\cos n \theta+\mathrm{i} \sin n \theta
$$
for all natural numbers $n$.

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｜MAST90125｜Bayesian Statistical Learning贝叶斯统计学习墨尔本大学

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”）Bayesian Statistical Learning贝叶斯统计学习要素澳洲代写代考和辅导服务！

课程介绍：

Bayesian inference treats all unknowns as random variables, and the core task is to update the probability distribution for each unknown as new data is observed. After introducing Bayes’ Theorem to transform prior probabilities into posterior probabilities, the first part of this subject introduces theory and methodological aspects underlying Bayesian statistical learning including credible regions, comparisons of means and proportions, multi-model inference and model selection. The second part of the subject focuses on advanced supervised and unsupervised Bayesian machine learning methods in the context of Gaussian processes and Dirichlet processes. The subject will also cover practical implementations of Bayesian methods through Markov Chain Monte Carlo computing and real data applications.

澳洲代写｜MAST90125｜Bayesian Statistical Learning贝叶斯统计学习墨尔本大学

Bayesian Statistical Learning贝叶斯统计学习问题集

Example 3: Given the condition $C$ of a symmetrical die the probability is $2 / 6=1 / 3$ to throw a two or a three according to the classical definition (2.24) of probability.
$\Delta$
Example 4: A card is taken from a deck of 52 cards under the condition $C$ that no card is marked. What is the probability that it will be an ace or a diamond? If $A$ denotes the statement of drawing a diamond and $B$ the one of drawing an ace, $P(A \mid C)=13 / 52$ and $P(B \mid C)=4 / 52$ follow from (2.24). The probability of drawing the ace of diamonds is $P(A B \mid C)=1 / 52$. Using (2.17) the probability of an ace or diamond is then $P(A+B \mid C)=$ $13 / 52+4 / 52-1 / 52=4 / 13$.

Example 5: Let the condition $C$ be true that an urn contains 15 red and 5 black balls of equal size and weight. Two balls are drawn without being replaced. What is the probability that the first ball is red and the second one black? Let $A$ be the statement to draw a red ball and $B$ the statement to draw a black one. With $(2.24)$ we obtain $P(A \mid C)=15 / 20=3 / 4$. The probability $P(B \mid A C)$ of drawing a black ball under the condition that a red one has been drawn is $P(B \mid A C)=5 / 19$ according to (2.24). The probability of drawing without replacement a red ball and then a black one is therefore $P(A B \mid C)=(3 / 4)(5 / 19)=15 / 76$ according to the product rule (2.12).

Example 6: The grey value $g$ of a picture element, also called pixel, of a digital image takes on the values $0 \leq g \leq 255$. If 100 pixels of a digital image with $512 \times 512$ pixels have the gray value $g=0$, then the relative frequency of this value equals $100 / 512^2$ according to (2.24). The distribution of the relative frequencies of the gray values $g=0, g=1, \ldots, g=255$ is called a histogram.
$\Delta$

Axioms of Probability
Probabilities of random events are introduced by axioms for the probability theory of traditional statistics, see for instance KосH (1999, p.78). Starting from the set $S$ of elementary events of a random experiment, a special system $Z$ of subsets of $S$ known as $\sigma$-algebra is introduced to define the random events. $Z$ contains as elements subsets of $S$ and in addition as elements the
10
2 Probability
empty set and the set $S$ itself. $Z$ is closed under complements and countable unions. Let $A$ with $A \in Z$ be a random event, then the following axioms are presupposed,

Axiom 1: A real number $P(A) \geq 0$ is assigned to every event $A$ of $Z . P(A)$ is called the probability of $A$.
Axiom 2: The probability of the sure event is equal to one, $P(S)=1$.
Axiom 3: If $A_1, A_2, \ldots$ is a sequence of a finite or infinite but countable number of events of $Z$ which are mutually exclusive, that is $A_i \cap A_j=\emptyset$ for $i \neq j$, then
$$
P\left(A_1 \cup A_2 \cup \ldots\right)=P\left(A_1\right)+P\left(A_2\right)+\ldots .
$$
The axioms introduce the probability as a measure for the sets which are the elements of the system $Z$ of random events. Since $Z$ is a $\sigma$-algebra, it may contain a finite or infinite number of elements, whereas the rules given in Chapter 2.1.4 and 2.1.5 are valid only for a finite number of statements.
If the system $Z$ of random events contains a finite number of elements, the $\sigma$-algebra becomes a set algebra and therefore a Boolean algebra, as already mentioned at the end of Chapter 2.1.2. Axiom 1 is then equivalent to the requirement 1 of Chapter 2.1.4, which was formulated with respect to the plausibility. Axiom 2 is identical with (2.13) and Axiom 3 with (2.21), if the condition $C$ in (2.13) and (2.21) is not considered. We may proceed to an infinite number of statements, if a well defined limiting process exists. This is a limitation of the generality, but is is compensated by the fact that the probabilities (2.12) to (2.14) have been derived as rules by consistent and logical reasoning. This is of particular interest for the product rule (2.12). It is equivalent in the form
$$
P(A \mid B C)=\frac{P(A B \mid C)}{P(B \mid C)} \text { with } \quad P(B \mid C)>0,
$$
if the condition $C$ is not considered, to the definition of the conditional probability of traditional statistics. This definition is often interpreted by relative frequencies which in contrast to a derivation is less obvious.

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｜ECOM30004｜Time Series Analysis and Forecasting时间序列分析与预测墨尔本大学

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”）Time Series Analysis and Forecasting时间序列分析与预测要素澳洲代写代考和辅导服务！

课程介绍：

Normally topics will include current techniques used in forecasting in finance, accounting and economics such as regression models, Box-Jenkins, ARIMA models, vector autoregression, causality analysis, cointegration and forecast evaluation, and ARCH models. The computer software used is Eviews.

澳洲代写｜ECOM30004｜Time Series Analysis and Forecasting时间序列分析与预测墨尔本大学

Time Series Analysis and Forecasting时间序列分析与预测问题集

问题 1.

Transforming $A R(p)$ to $M A$. If a $p$-order autoregressive process $\phi(L) y_t=\varepsilon_t$ is stationary, with moving average representation $y_t=\psi(L) \varepsilon_t$, show that
$$
0=\sum_{j=0}^p \phi_j \psi_{k-j}=\phi(L) \psi_k, \quad k=p, p+1, \ldots
$$
i.e., show that the moving average coefficients satisfy the autoregressive difference equation.

问题 2.

Sims’ formula for spectrum. Assume that we have a sample $\left{y_t, x_t\right}_{t=1}^T$ from infinite distributed lag model $y_t=B(L) x_t+e_t, B(L)=\sum_{j=1}^{\infty} b_j L^j$ with absolutely summable coefficients $\sum\left|b_j\right|<\infty$ (here $e_t$ is a white-noise, $x_t$ is stationary and weakly exogenous). Assume that one estimates (misspecified) model with $q$ lags, that is, he regresses $y_t$ on to $x_{t-1}, \ldots, x_{t-q-1}$ and obtains $\widehat{a}1, \ldots, \widehat{a}_q$. As the sample size increases to infinity (but $q$ is kept constant), the estimated coefficients converge to some non-random limits: $\widehat{a}_j \stackrel{p}{\rightarrow} a_j$. Let $A(L)=a_1 L+\ldots+a_p L^p$. Show that $A(\cdot)$ is a solution to the following problem: $$ \min {a_1, \ldots, a_q} \frac{1}{2 \pi} \int_{-\pi}^\pi\left(A\left(e^{-i \omega}\right)-B\left(e^{-i \omega}\right)\right) S_X(\omega)\left(A\left(e^{i \omega}\right)-B\left(e^{i \omega}\right)\right),
$$
where $S_X(\cdot)$ is the spectrum of the process $x_t$. That is, one minimizes the quadratic form in the differences between true and estimated polynomial, assigning the greatest weights to the frequencies for which spectral density is the greatest.

问题 3.

Spectrum and filters. This is your first empirical problem. Choose a software package you feel comfortable using (I would recommend MatLab).You may use any users-written codes you find on Internet. Always make sure that the code is doing what you think it is doing. Please, do not forget to cite whatever you are using.
(i) Download quarterly values of Real GDP for the US from Mark W. Watson personal web-site (you may use any other aggregate macro time series from any other source if you wish. Economic Database (FRED II) maintained by the Federal Reserve Bank of St. Louis is a fantastic source ).
(ii) Define the growth rate for real GDP. Estimate and plot spectrum for the growth rate. Discuss the graph. Find which peak in the spectrum corresponds to business cycles.
(iii) Use the following three cycle removing devices: a) run the OLS to detrend the series ; b) use Prescott-Hodrick filter; c) apply Baxter-King filter.
(iv) Re-estimate spectrum for all series after applying each of the three procedures. Draw spectrum functions. Discuss the differences.

Note. As in real life empirical research, you will need to make a lot of choices while performing the task, such as choosing lag length, kernel function, and so on. Try to be reasonable, always check whether you results are sensitive to these choices. Also check original papers for suggestions.

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｜MAST20026｜Real Analysis实分析墨尔本大学

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”）Real Analysis实分析要素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 emphasized and used to motivate rigorous notions of convergence and divergence of sequences, including the Cauchy criterion. These ideas are extended to cover the theory of infinite series, including common tests for convergence and divergence. 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. The definitions and properties of the Riemann integral allow rigorous 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.

Real Analysis实分析案例

Let us prove that for all $n \in \mathbb{N}$,
$$
2^{n-1} \leq n ! \quad(\text { recall } n !=1 \cdot 2 \cdot 3 \cdots n) .
$$
We let $P(n)$ be the statement that $2^{n-1} \leq n !$ is true. Plug in $n=1$ to see that $P(1)$ is true.
Suppose $P(n)$ is true. That is, suppose $2^{n-1} \leq n$ ! holds. Multiply both sides by 2 to obtain
$$
2^n \leq 2(n !)
$$
As $2 \leq(n+1)$ when $n \in \mathbb{N}$, we have $2(n !) \leq(n+1)(n !)=(n+1) !$. That is,
$$
2^n \leq 2(n !) \leq(n+1) !
$$
and hence $P(n+1)$ is true. By the principle of induction, $P(n)$ is true for all $n \in \mathbb{N}$. In other words, $2^{n-1} \leq n$ ! is true for all $n \in \mathbb{N}$.
We claim that for all $c \neq 1$,
$$
1+c+c^2+\cdots+c^n=\frac{1-c^{n+1}}{1-c} .
$$
Proof: It is easy to check that the equation holds with $n=1$. Suppose it is true for $n$. Then
$$
\begin{aligned}
1+c+c^2+\cdots+c^n+c^{n+1} & =\left(1+c+c^2+\cdots+c^n\right)+c^{n+1} \
& =\frac{1-c^{n+1}}{1-c}+c^{n+1} \
& =\frac{1-c^{n+1}+(1-c) c^{n+1}}{1-c} \
& =\frac{1-c^{n+2}}{1-c} .
\end{aligned}
$$
Sometimes, it is easier to use in the inductive step that $P(k)$ is true for all $k=1,2, \ldots, n$, not just for $k=n$. This principle is called strong induction and is equivalent to the normal induction above. The proof of that equivalence is left as an exercise.

Let $P(n)$ be a statement depending on a natural number $n$. Suppose that
(i) (basis statement) $P(1)$ is true.
(ii) (induction step) If $P(k)$ is true for all $k=1,2, \ldots, n$, then $P(n+1)$ is true.
Then $P(n)$ is true for all $n \in \mathbb{N}$.

Real Analysis实分析案例2

Let $A, B \subset \mathbb{R}$ be nonempty sets such that $x \leq y$ whenever $x \in A$ and $y \in B$. Then $A$ is bounded above, $B$ is bounded below, and sup $A \leq \inf B$.

Proof. Any $x \in A$ is a lower bound for $B$. Therefore $x \leq \inf B$ for all $x \in A$, so inf $B$ is an upper bound for $A$. Hence, $\sup A \leq \inf B$.

We must be careful about strict inequalities and taking suprema and infima. Note that $x<y$ whenever $x \in A$ and $y \in B$ still only implies sup $A \leq \inf B$, and not a strict inequality. For example, take $A:={0}$ and $B:={1 / n: n \in \mathbb{N}}$. Then $0<1 / n$ for all $n \in \mathbb{N}$. However, sup $A=0$ and inf $B=0$. This important subtle point comes up often.

The proof of the following often used fact is left to the reader. A similar result holds for infima.

If $S \subset \mathbb{R}$ is nonempty and bounded above, then for every $\epsilon>0$ there exists an $x \in S$ such that (sup $S)-\epsilon<x \leq \sup S$.

To make using suprema and infima even easier, we may want to write sup $A$ and $\inf A$ without worrying about $A$ being bounded and nonempty. We make the following natural definitions.

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｜ECON20001｜Intermediate Macroeconomics中级宏观经济学墨尔本大学

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”）Probability for Inference推理概率要素澳洲代写代考和辅导服务！

课程介绍：

Intermediate macroeconomic analysis develops the tools, skills and knowledge base necessary to operate as a practicing macroeconomist. These may include: models of long run economic growth; an assessment of the evidence on economic growth and its implications; the flexible-price macroeconomic model in which markets continuously clear; an assessment of the evidence regarding whether prices and wages are flexible or sticky; the sticky price macroeconomic model in which markets do not always clear; assessment of the flexible and sticky price models; the analysis of macroeconomic policy making.

澳洲代写｜ECON20001｜Intermediate Macroeconomics中级宏观经济学墨尔本大学

Intermediate Macroeconomics中级宏观经济学案例

To an economist, a model is a simplified representation of the economy; it is essentially a representation of the economy in which only the main ingredients are being accounted for. Since we are interested in analyzing the direction of relationships (e.g. does investment go up or down when interest rates increase?) and the quantitative impact of those (e.g. how much does investment change after a one percentage point increase in interest rates?), in economics, a model is composed of a set of mathematical relationships. Through these mathematical relationships, the economist determines how variables (like an interest rate) affect each other (e.g. investment). Models are not the only way to study human behavior. Indeed, in the natural sciences, scientists typically follow a different approach.

Imagine a chemist wants to examine the effectiveness of a certain new medicine in addressing a specific illness. After testing the effects of drugs on guinea pigs the chemist decides to perform experiments on humans. How would she go about it? Well, she will select a group of people willing to participate – providing the right incentives as, we know from
45
principles, incentives can affect behavior – and, among these, she randomly divides members in two groups: a control and a treatment group. The control group will be given a placebo (something that resembles the medicine to be given but has no physical effect in the person who takes it). The treatment group is composed by the individuals that were selected to take the real medicine. As you may suspect, the effects of the medicine on humans will be based on the differences between the treatment and control group. As these individuals were randomly selected, any difference to which the illness is affecting them can be attributed to the medicine. In other words, the experiment provides a way of measuring the extent to which that particular drug is effective in diminishing the effects of the disease.

Ideally, we would like to perform the same type of experiments with respect to economic policies. Variations of lab experiments have proven to be a useful approach in some areas of economics that focus on very specific markets or group of agents. In macroeconomics, however, things are different.

Suppose we are interested in studying the effects of training programs in improving the chances unemployed workers find jobs. Clearly the best way to do this would be to split the pool of unemployed workers in two groups, whose members are randomly selected. Here is where the problem with experiments of this sort becomes clear. Given the cost associated with unemployment, would it be morally acceptable to prevent some workers from joining a program that could potentially reduce the time without a job? What if we are trying to understand the effects a sudden reduction in income has on consumption for groups with different levels of savings? Would it be morally acceptable to suddenly confiscate income from a group? Most would agree not. As such, economists develop models, and in these models we run experiments. A model provides us a fictitious economy in which these issues can be analyzed and the economic mechanisms can be understood.

All models are not created equal and some models are better to answer one particular question but not another one. Given this, you may wonder how to judge when a model is appropriate. This is a difficult question. The soft consensus, however, is that a model should be able to capture features of the data that it was not artificially constructed to capture. Any simplified representation of reality will not have the ability to explain every aspect of that reality. In the same way, a simplified version of the economy will not be able to account for all the data that an economy generates. A model that can be useful to study how unemployed workers and firms find each other will not necessarily be able to account for the behavior of important economic variables such as the interest rate. What is expected is that the model matches relevant features of the process through which workers and firms meet.

对经济学家而言，模型是经济的简化表示；它本质上是只考虑主要成分的经济表示。由于我们感兴趣的是分析各种关系的方向（例如，利率上升时投资是上升还是下降？）以及这些关系的定量影响（例如，利率上升一个百分点后投资会发生多大变化？通过这些数学关系，经济学家可以确定变量（如利率）如何相互影响（如投资）。模型并不是研究人类行为的唯一方法。事实上，在自然科学领域，科学家通常采用不同的方法。

想象一下，一位化学家想研究某种新药在治疗特定疾病方面的效果。在对豚鼠进行药物效果测试后，化学家决定在人类身上进行实验。她会怎么做呢？她会选择一群愿意参与实验的人，并提供适当的激励措施。
45
在这些人中，她会随机将成员分成两组：对照组和治疗组。对照组将服用安慰剂（类似于要服用的药物，但对服用者没有任何生理作用）。治疗组由被选中服用真药的人组成。正如您所猜测的那样，药物对人体的影响将基于治疗组和对照组之间的差异。由于这些人是随机抽取的，因此疾病对他们造成的任何影响都可以归因于药物。换句话说，实验提供了一种方法来衡量特定药物对减轻疾病影响的有效程度。

在理想情况下，我们也希望在经济政策方面进行同样的实验。在经济学的某些领域，实验室实验的变体已被证明是一种有用的方法，这些领域关注的是非常具体的市场或代理群体。但在宏观经济学中，情况有所不同。

假设我们有兴趣研究培训计划在提高失业工人找到工作机会方面的效果。显然，最好的办法是将失业工人分成两组，随机抽取其成员。在这里，此类实验的问题就很明显了。考虑到与失业相关的成本，阻止一些工人参加一个有可能缩短失业时间的项目，在道德上是否可以接受？如果我们试图了解收入突然减少对不同储蓄水平群体的消费有何影响？突然没收一个群体的收入在道义上可以接受吗？大多数人都认为不会。因此，经济学家建立模型，并在这些模型中进行实验。模型为我们提供了一个虚构的经济环境，我们可以在其中分析这些问题，了解经济机制。

并非所有的模型都是一样的，有些模型可以更好地回答某个问题，但却不能更好地回答另一个问题。有鉴于此，您可能会问，如何判断一个模型是否合适。这是一个难题。不过，一个软性共识是，模型应该能够捕捉数据的特征，而不是人为地构建模型来捕捉这些特征。任何对现实的简化表示都无法解释现实的方方面面。同样，经济的简化版本也无法解释经济产生的所有数据。一个可以用来研究失业工人和企业如何互相找到对方的模型，并不一定能够解释利率等重要经济变量的行为。我们所期望的是，模型能够与工人和企业相遇过程的相关特征相匹配。

Mathematical Diversion数学发散思维定义

Referring back to the assumed mathematical properties of the production function, we assumed that the production function has constant returns to scale. In words, this means that doubling both inputs results in a doubling of output. A fancier term for constant returns to scale is to say that the function is homogeneous of degree 1. More generally, a function is homogeneous of degree $\rho$ if:
$$
F\left(\gamma K_t, \gamma N_t\right)=\gamma^\rho F\left(K_t, N_t\right)
$$
where $\gamma=1$ corresponds to the case of constant returns to scale. $\gamma<1$ is what is called decreasing returns to scale (meaning that doubling both inputs results in a less than doubling of output), while $\gamma>1$ is increasing returns to

scale (doubling both inputs results in a more than doubling of output). Euler’s theorem for homogeneous functions states (see Mathworld (2016)) if a function is homogeneous of degree $\rho$, then:
$$
\rho F\left(K_t, N_t\right)=F_K\left(K_t, N_t\right) K_t+F_N\left(K_t, N_t\right) N_t
$$
If $\rho=1$ (as we have assumed), this says that the function can be written as the sum of partial derivatives times the factor being differentiated with respect to. To see this in action for the Cobb-Douglas production function, note:
$$
\begin{aligned}
K_t^\alpha N_t^{1-\alpha} & =\alpha K_t^{\alpha-1} N_t^{1-\alpha} K_t+(1-\alpha) K_t^\alpha N_t^{-\alpha} N_t \
& =\alpha K_t^\alpha N_t^{1-\alpha}+(1-\alpha) K_t^\alpha N_t^{1-\alpha} \
& =K_t^\alpha N_t^{1-\alpha}(\alpha+1-\alpha)=K_t^\alpha N_t^{1-\alpha} .
\end{aligned}
$$
Euler’s theorem also states that, if a function is homogeneous of degree $\rho$, then its first partial derivatives are homogeneous of degree $\rho-1$. This has the implication, for example, that:
$$
F_K\left(\gamma K_t, \gamma N_t\right)=\gamma^{\rho-1} F_K\left(K_t, N_t\right)
$$

回到生产函数的假定数学特性，我们假定生产函数的规模收益不变。换句话说，这意味着投入增加一倍，产出就会增加一倍。规模收益恒定的一个更高级的说法是，该函数是1度均质的。更一般地说，如果出现以下情况，函数就是$\rho$度均质的：
$$
F\left(\gamma K_t, \gamma N_t\right)=\gamma^\rho F\left(K_t, N_t\right)
$$
其中，$\gamma=1$对应于规模收益不变的情况。$\gamma<1$是所谓的规模收益递减（即两种投入加倍导致产出增加不到一倍），而$\gamma>1$是规模收益递增（两种投入加倍导致产出增加不到一倍）。

而 $/gamma>1$ 则是规模收益递增（两种投入加倍会导致产出增加一倍以上）。同质函数的欧拉定理指出（见 Mathworld (2016)），如果一个函数的同质度为 $\rho$，那么：
$$
\rho F\left(K_t, N_t\right)=F_K\left(K_t, N_t\right) K_t+F_N\left(K_t, N_t\right) N_t
$$
如果 $\rho=1$（就像我们假设的那样），这说明函数可以写成偏导数乘以被微分因子的和。要在柯布-道格拉斯生产函数中看到这一点，请注意：
$$
\begin{aligned}
K_t^\alpha N_t^{1-\alpha} & =\alpha K_t^{\alpha-1} N_t^{1-\alpha} K_t+(1-\alpha) K_t^\alpha N_t^{-\alpha} N_t \
& =\alpha K_t^\alpha N_t^{1-\alpha}+(1-\alpha) K_t^\alpha N_t^{1-\alpha} \
& =K_t^\alpha N_t^{1-\alpha}(\alpha+1-\alpha)=K_t^\alpha N_t^{1-\alpha} .
\end{aligned}
$$
欧拉定理还指出，如果一个函数是阶为 $\rho$ 的同调函数，那么它的第一个偏导数就是阶为 $\rho-1$ 的同调函数。例如，这意味着
$$
F_K\left(\gamma K_t, \gamma N_t\right)=\gamma^{\rho-1} F_K\left(K_t, N_t\right)
$$

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

MATH2222｜Introduction to Mathematical Thinking: Problem-Solving and Proofs数学思维导论：解决问题与证明澳洲国立大学

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

statistics-lab^TM为您提供澳大利亚国立大学（The Australian National University）Introduction to Mathematical Thinking: Problem-Solving and Proofs数学思维导论：解决问题与证明澳洲代写代考和辅导服务！

课程介绍：

This course focuses on the language of mathematical arguments. Rather than attacking advanced topics, we will use simple mathematics to develop an understanding of how results are established. We begin with clearly stated and plausible assumptions or axioms and then develop a more and more complex theory from them. The course, and the lecturer, will have succeeded if you finish the course able to construct valid arguments of your own and to criticise those that are presented to you.

MATH2222｜Introduction to Mathematical Thinking: Problem-Solving and Proofs数学思维导论：解决问题与证明澳洲国立大学

Introduction to Mathematical Thinking: Problem-Solving and Proofs数学思维导论：解决问题与证明案例

问题 1.

If $x$ and $y$ are real numbers, then $|x+y| \leq|x|+|y|$.

Proof: We start with the inequality $2 x y \leq 2|x||y|$. By adding $x^2+y^2$ to both sides and using $z^2=|z|^2$, we obtain
$$
x^2+2 x y+y^2 \leq x^2+2|x||y|+y^2=|x|^2+2|x||y|+|y|^2 .
$$
we may take the positive square root of both sides and preserve the inequality. Thus $|x+y| \leq|x|+|y|$, as desired.

In order to prove a statement, we derive it from known facts. Before we find a proof, we may not know which known facts to use. To discover a proof, it may be helpful to ask what is needed to make the conclusion true. In this approach, we try to “reduce” the desired conclusion to a statement known to be true. The written proof must be a rigorous justification of the conclusion from known facts.

The next proposition illustrates this. Manipulating the desired inequality leads to a known inequality, but the proof starts with the known inequality and derives the desired one from it. The arithmetic mean (or “average”) of $x$ and $y$ is $(x+y) / 2$. The geometric mean of nonnegative numbers $x$ and $y$ is $\sqrt{x y}$. The term AGM Inequality stands for Arithmetic Mean-Geometric Mean Inequality; it states that the arithmetic mean of two nonnegative numbers is always at least their geometric mean.

问题 2.

If $x$ and $y$ are real numbers, then $2 x y \leq x^2+y^2$ and $x y \leq\left(\frac{x+y}{2}\right)^2$. If $x$ and $y$ are also nonnegative, then $\sqrt{x y} \leq(x+y) / 2$. Equality holds in each only when $x=y$.

Proof: We begin with $0 \leq(x-y)^2=x^2-2 x y+y^2$ and observe that equality holds only when $x=y$. Adding $2 x y$ yields $2 x y \leq x^2+y^2$. Adding another $2 x y$ yields $4 x y \leq x^2+2 x y+y^2=(x+y)^2$, which we divide by 4 to obtain $x y \leq\left(\frac{x+y}{2}\right)^2$.

If $x \geq 0$ and $y \geq 0$, then also $x y \geq 0$, and we can take positive square roots in $x y \leq\left(\frac{x+y}{2}\right)^2$. Proposition 1.1 yields $\sqrt{x y} \leq(x+y) / 2$.

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

金融工程代写

非参数统计代写

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

广义线性模型代考

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

有限元方法代写

随机分析代写

时间序列分析代写

回归分析代写

MATLAB代写

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

EMET8012｜Business and Economic Forecasting商业和经济预测澳洲国立大学

Posted on 2023年9月27日2023年10月9日 by statistics-lab

statistics-lab^TM为您提供澳大利亚国立大学（The Australian National University）Business and Economic Forecasting商业和经济预测澳洲代写代考和辅导服务！

课程介绍：

Accurate forecasting of future events and their outcomes is a crucial input into a successful business or economic planning process. This course provides an introduction to the application of various forecasting techniques. The methods include trend curve extrapolation, smoothing, autoregressions, regression modelling, leading indicators. The course also looks at techniques for the evaluation of performance of forecasting methods and examines the role of forecasts in the decision making process. Students will learn how to use the various techniques in real world forecasting applications.

EMET8012｜Business and Economic Forecasting商业和经济预测澳洲国立大学

Attribute	Detail
Course Code	EMET8012
Co-taught Course	EMET3007, EMET4312
Subject/Area of Interest	Econometrics
Course Convener	Dr James Taylor
Unit Value (学分)	6 units

Business and Economic Forecasting商业和经济预测案例

问题 1.

We begin by assuming that the risk sharing unit is the household. Assume a unitarian household model in which allocations are decided as a result of an efficient social-planner-like decision rule with weights $\mu_{i, h}$ on individual utility functions.
(1) Write down the program that a household $\mathrm{h}$ solves when deciding consumption and labor allocations for its members.

Solution: The household takes the time path of transfers $\left{\tau_h\left(s^t\right)\right}_t$ as given. Hence, total household income after history $s^t$ is given by
$$
X_h\left(s^t\right)=\tau_h\left(s^t\right)+\sum_{i=1}^{I_h} y_{i, h}\left(s_t\right)
$$

The problem of the family is given by
$$
\begin{array}{r}
\max {\left[\left{c^{i, h}\left(s^t\right), l^{i, h}\left(s^t\right)\right}_i^{I_h}\right]{s^t}} \sum_{i=1}^{I^h} \mu^{i, h} \sum_{t, s^t} \beta^t \pi\left(s^t\right) u^{i, h}\left(l^{i, h}\left(s^t\right), c^{i, h}\left(s^t\right)\right) \
\text { s.t. } \sum_{i=1}^{I^h} c^{i, h}\left(s^t\right)+\sum_{i=1}^{I^h} w^{i, h}\left(s^t\right) l^{i, h}\left(s^t\right)=\sum_{i=1}^{I^h} w^{i, h}\left(s^t\right) T^{i, h}+X^h\left(s^t\right) \text { for all } s^t .
\end{array}
$$
As usual we can solve (1) as a static problem, i.e. for each $s^t,\left{c^{i, h}\left(s^t\right), l^{i, h}\left(s^t\right)\right}_i^{I_h}$ solve
$$
\begin{array}{r}
\max {\left{c^{i, h}, l^{i, h}\right}_i^{I_h}} \sum{i=1}^{I^h} \mu^{i, h} u^{i, h}\left(l^{i, h}, c^{i, h}\right) \
\text { s.t. } \sum_{i=1}^{I^h} c^{i, h}+\sum_{i=1}^{I^h} w^{i, h} l^{i, h}=\sum_{i=1}^{I^h} w^{i, h} T^{i, h}+X^h,
\end{array}
$$
where I now suppressed the explicit dependence on $s^t$.

问题 2.

(2) Characterize the solution to the allocation problem. Please be explicit on which variables $c^{i, h}$ and $l^{i, h}$ depend. Provide a precise intuition why the solution depends on those variables (and why not on some others).

Let $\lambda$ be the multiplier on the constraint.

The necessary conditions for this problem are
$$
\begin{aligned}
& \mu^{i, h} u_c^{i, h}\left(l^{i, h}, c^{i, h}\right)=\lambda \
& \mu^{i, h} u_l^{i, h}\left(l^{i, h}, c^{i, h}\right)=\lambda w^{i, h} .
\end{aligned}
$$
Hence, the allocation between consumption and leisure is given by
$$
\frac{u_l^{i, h}\left(l^{i, h}, c^{i, h}\right)}{u_c^{i, h}\left(l^{i, h}, c^{i, h}\right)}=w^{i, h}
$$
and the consumption allocation across households is given by
$$
\mu^{i, h} u_c^{i, h}\left(l^{i, h}, c^{i, h}\right)=\mu^{g, h} u_c^{g, h}\left(l^{g, h}, c^{g, h}\right) .
$$
From (7) we can express leisure as an individual-specific function of consumption and the wage rate, i.e.
$$
l^{i, h}=\phi^{i, h}\left(c^{i, h}, w^{i, h}\right) .
$$
Hence, we can express (4) as
$$
\mu^{i, h} u_c^{i, h}\left(\phi^{i, h}\left(c^{i, h}, w^{i, h}\right), c^{i, h}\right)=\mu^{g, h} u_c^{g, h}\left(\phi^{g, h}\left(c^{g, h}, w^{g, h}\right), c^{g, h}\right) .
$$
In particular, consider $i=1$ so that (6) determines $c^{g, h}$ as a function of $c^{1, h}$ and the wage rates $w^{g, h}$ and $w^{i, h}$, i.e.
$$
c^{g, h}=\chi^{g, h}\left(c^{1, h}, w^{1, h}, w^{g, h}\right) .
$$

Hence,
$$
\begin{aligned}
l^{g, h} & =\phi^{g, h}\left(c^{g, h}, w^{g, h}\right) \
& =\phi^{g, h}\left(\chi^{g, h}\left(c^{1, h}, w^{1, h}, w^{g, h}\right), w^{g, h}\right) \
& \equiv \kappa^{g, h}\left(c^{1, h}, w^{1, h}, w^{g, h}\right),
\end{aligned}
$$
where $\kappa^{g, h}$ is some function specific to individual $g$.
From the budget constraint of (2) we therefore get that
$$
\begin{aligned}
\sum_{i=1}^{I^h} w^{i, h} T^{i, h}+X^h= & \sum_{i=1}^{I^h} c^{i, h}+\sum_{i=1}^{I^h} w^{i, h} l^{i, h} \
= & c^{1, h}+w^{1, h} l^{1, h}+\sum_{g=2}^{I^h} c^{g, h}+\sum_{g=2}^{I^h} w^{g, h} l^{g, h} \
= & c^{1, h}+w^{1, h} \phi^{1, h}\left(c^{1, h}, w^{1, h}\right)+ \
& \sum_{g=2}^{I^h} \chi^{g, h}\left(c^{1, h}, w^{1, h}, w^{g, h}\right)+\sum_{g=2}^{I^h} \kappa^{g, h}\left(c^{1, h}, w^{1, h}, w^{g, h}\right) .
\end{aligned}
$$
This is an equation which determines $c^{1, h}$ as a function of a bunch of things, in particular
$$
c^{1, h}=f^{1, h}\left(w^{1, h}, w^{2, h}, \ldots, w^{3, h}, \sum_{i=1}^{I^h} w^{i, h} T^{i, h}+X^h\right),
$$
i.e. consumption depends on all wage rates $\left[w^{1, h}, \ldots, w^{I_h, h}\right]$ and total income $\sum_{i=1}^{I^h} w^{i, h} T^{i, h}+X^h$. That consumption only depends on total income and not on its individual components is the usual result. However, now consumption depends on all wage rates of household members. This is due to the leisure-labor choice encapsulated in (7).

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

ENG2005｜Advanced engineering mathematics高级工程数学伦敦大学学院

Posted on 2023年9月25日2023年10月6日 by statistics-lab

statistics-lab^TM为您提供伦敦大学学院（University College London，简称：UCL）ENG2005高级工程数学英国代写代考和辅导服务！

课程介绍：

This unit introduces advanced mathematical concepts required for all disciplines of engineering that extend on the concepts in ENG1005 Engineering mathematics.

Major concepts taught using engineering contexts include multivariable and vector calculus, Fourier series and partial differential equations. Specific topics include partial derivatives, double and triple integrals with change of coordinates, vector functions and operators, parametric curves and surfaces, periodic functions, Laplace transforms, ODE boundary value problems, partial differential equation solution techniques and applications.

ENG2005｜Advanced engineering mathematics高级工程数学伦敦大学学院

Advanced engineering mathematics高级工程数学案例

问题 1.

Establish the following results:
(a) $\operatorname{Re}\left(z_1+z_2\right)=\operatorname{Re}\left(z_1\right)+\operatorname{Re}\left(z_2\right)$, but $\operatorname{Re}\left(z_1 z_2\right) \neq \operatorname{Re}\left(z_1\right) \operatorname{Re}\left(z_2\right)$ in general;

Solution. (a) We want to show that $\operatorname{Re}\left(z_1+z_2\right)=\operatorname{Re}\left(z_1\right)+\operatorname{Re}\left(z_2\right)$. Let $z_1=a_1+i b_1$, $z_2=a_2+i b_2$, then
$$
z_1+z_2=\left(a_1+a_2\right)+i\left(b_1+b_2\right) \text {. }
$$
Hence
$$
\operatorname{Re}\left(z_1+z_2\right)=a_1+a_2
$$
and clearly
$$
\operatorname{Re}\left(z_1\right)+\operatorname{Re}\left(z_2\right)=a_1+a_2 .
$$
Let us show that in general $\operatorname{Re}\left(z_1 z_2\right) \neq \operatorname{Re}\left(z_1\right) \operatorname{Re}\left(z_2\right)$. We have
$$
z_1 z_2=\left(a_1+i b_1\right)\left(a_2+i b_2\right)=\left(a_1 a_2-b_1 b_2\right)+i\left(a_1 b_2+a_2 b_1\right),
$$
therefore
$$
\operatorname{Re}\left(z_1 z_2\right)=a_1 a_2-b_1 b_2 .
$$
On the other hand
$$
\operatorname{Re}\left(z_1\right) \operatorname{Re}\left(z_2\right)=a_1 a_2 \neq a_1 a_2-b_1 b_2
$$
in general.

\end{prob}

问题 2.

(b) $\operatorname{Im}\left(z_1+z_2\right)=\operatorname{Im}\left(z_1\right)+\operatorname{Im}\left(z_2\right)$, but $\operatorname{Im}\left(z_1 z_2\right) \neq \operatorname{Im}\left(z_1\right) \operatorname{Im}\left(z_2\right)$ in general;

(b) We want to show that $\operatorname{Im}\left(z_1+z_2\right)=\operatorname{Im}\left(z_1\right)+\operatorname{Im}\left(z_2\right)$. From part (a) we have that
$$
z_1+z_2=\left(a_1+a_2\right)+i\left(b_1+b_2\right) .
$$
Hence
$$
\operatorname{Im}\left(z_1+z_2\right)=b_1+b_2
$$
and clearly
$$
\operatorname{Im}\left(z_1\right)+\operatorname{Im}\left(z_2\right)=b_1+b_2 .
$$
Let us show that in general $\operatorname{Im}\left(z_1 z_2\right) \neq \operatorname{Im}\left(z_1\right) \operatorname{Im}\left(z_2\right)$. We have
$$
z_1 z_2=\left(a_1+i b_1\right)\left(a_2+i b_2\right)=\left(a_1 a_2-b_1 b_2\right)+i\left(a_1 b_2+a_2 b_1\right),
$$
therefore
$$
\operatorname{Im}\left(z_1 z_2\right)=a_1 b_2+a_2 b_1 .
$$

问题 3.

(c) We want to show that $\left|z_1 z_2\right|=\left|z_1\right|\left|z_2\right|$. From part (a) we have
$$
z_1 z_2=\left(a_1+i b_1\right)\left(a_2+i b_2\right)=\left(a_1 a_2-b_1 b_2\right)+i\left(a_1 b_2+a_2 b_1\right) .
$$
Hence
$$
\begin{aligned}
\left|z_1 z_2\right| & =\sqrt{\left(a_1 a_2-b_1 b_2\right)^2+\left(a_1 b_2+a_2 b_1\right)^2} \
& =\sqrt{\left(a_1^2 a_2^2-2 a_1 a_2 b_1 b_2+b_1^2 b_2^2\right)+\left(a_1^2 b_2^2+2 a_1 b_2 a_2 b_1+a_2^2 b_1^2\right)} \
& =\sqrt{a_1^2 a_2^2+b_1^2 b_2^2+a_1^2 b_2^2+a_2^2 b_1^2} .
\end{aligned}
$$
On the other hand $\left|z_1\right|=\sqrt{a_1^2+b_1^2}$ and $\left|z_2\right|=\sqrt{a_2^2+b_2^2}$. Therefore
$$
\left|z_1\right|\left|z_2\right|=\left(\sqrt{a_1^2+b_1^2}\right)\left(\sqrt{a_2^2+b_2^2}\right)=\sqrt{\left(a_1^2+b_1^2\right)\left(a_2^2+b_2^2\right)}=\sqrt{a_1^2 a_2^2+b_1^2 b_2^2+a_1^2 b_2^2+a_2^2 b_1^2}
$$
which is equal to $\left|z_1 z_2\right|$.
Let us show that $\left|z_1+z_2\right| \neq\left|z_1\right|+\left|z_2\right|$ in general. From part (a) we have
$$
z_1+z_2=\left(a_1+a_2\right)+i\left(b_1+b_2\right) \text {. }
$$
Hence
$$
\left|z_1+z_2\right|=\sqrt{\left(a_1+a_2\right)^2+\left(b_1+b_2\right)^2} .
$$
On the other hand
$$
\left|z_1\right|+\left|z_2\right|=\sqrt{\left(a_1+a_2\right)^2+\left(b_1+b_2\right)^2}
$$
and
$$
\sqrt{\left(a_1+a_2\right)^2+\left(b_1+b_2\right)^2} \neq \sqrt{\left(a_1+a_2\right)^2+\left(b_1+b_2\right)^2}
$$
in general (choose for example $a_1=1, b_1=0, a_2=0, b_2=1$ ).
(d) We want to show that $\overline{z_1+z_2}=\overline{z_1}+\overline{z_2}$. From part (a) we have
$$
z_1+z_2=\left(a_1+a_2\right)+i\left(b_1+b_2\right)
$$
then
$$
\overline{z_1+z_2}=\left(a_1+a_2\right)-i\left(b_1+b_2\right) .
$$
On the other hand
$$
\overline{z_1}+\overline{z_2}=\left(a_1-i b_1\right)+\left(a_2-i b_2\right)=\left(a_1+a_2\right)-i\left(b_1+b_2\right)
$$
which is equal to $\overline{z_1+z_2}$.
Let us show that $\overline{z_1 z_2}=\overline{z_1 z_2}$. From part (a) we have that
$$
z_1 z_2=\left(a_1 a_2-b_1 b_2\right)+i\left(a_1 b_2+a_2 b_1\right) .
$$

Hence
$$
\overline{z_1 z_2}=\left(a_1 a_2-b_1 b_2\right)-i\left(a_1 b_2+a_2 b_1\right) .
$$
On the other hand,
$$
\overline{z_1 z_2}=\left(a_1-i b_1\right)\left(a_2-i b_2\right)=\left(a_1 a_2-b_1 b_2\right)-i\left(a_1 b_2+a_2 b_1\right)
$$
which is equal to $\overline{z_1 z_2}$.

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

金融工程代写

非参数统计代写

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

广义线性模型代考

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

有限元方法代写

随机分析代写

时间序列分析代写

回归分析代写

MATLAB代写

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

MATH0057｜Probability and Statistics概率与统计伦敦大学学院

Posted on 2023年9月25日2023年10月6日 by statistics-lab

statistics-lab^TM为您提供伦敦大学学院（University College London，简称：UCL）MATH0057Probability and Statistics概率与统计英国代写代考和辅导服务！

课程介绍：

The aim of the course is to introduce students to the theory of probability and some of the
statistical methods based upon it. Many physical processes involve random components which
can only be modelled using probabilistic methods. Statistical theory is vital for analysing
scientific data where it is necessary to distinguish genuine patterns from random fluctuations.

MATH0057｜Probability and Statistics概率与统计伦敦大学学院

Probability and Statistics概率与统计作业案例

问题 1.

Ignoring leap days, the days of the year can be numbered 1 to 365 . Assume that birthdays are equally likely to fall on any day of the year. Consider a group of $n$ people, of which you are not a member. An element of the sample space $\Omega$ will be a sequence of $n$ birthdays (one for each person).
(a) Define the probability function $P$ for $\Omega$..

answer: The sample space $\Omega$ is the set of all sequences of $n$ birthdays. That is, all sequences
$$
\omega=\left(b_1, b_2, b_3, \ldots, b_n\right),
$$

where each entry is a number between 1 and 365 .

问题 2.

(b) Consider the following events:
A: “someone in the group shares your birthday”
B: “some two people in the group share a birthday”
C: “some three people in the group share a birthday”
Carefully describe the subset of $\Omega$ that corresponds to each event.

(b) Event $A$ : Suppose my birthday is on day $b$. Then “an outcome $\omega$ is in $A$ ” is equivalent to ” $b$ is in the sequence for $\omega$ “, i.e. $b=b_k$ for some index $k$ between 1 and $n$. More symbolically,
an outcome $\omega$ is in $A$ if and only if $b_k=b$ for some index $k$ in $1, \ldots, n$.
Event B: “An outcome $\omega$ is in $B$ ” is equivalent to “two of the entries in $\omega$ are the same”. That is, an outcome $\omega$ is in $B \quad$ if and only if $\quad b_j=b_k$ for two (different) indices $j, k$ in $1, \ldots, n$.
Event $C$ : an outcome $\omega$ is in $C$ if and only if $\quad b_j=b_k=b_l$ for three (distinct) indices $j, k, l$ in $1, \ldots, n$.

问题 3.

(c) It’s easier to calculate $P\left(A^c\right)$. There are $364^n$ outcomes in $A^c$ since there are 364 choices for each birthday. So
$$
P(A)=1-P\left(A^c\right)=1-\frac{364^n}{365^n} .
$$
We can find the size of the group needed for $P(A)>.5$ by trial and error, plugging in different values of $n$. Or we can set $P(A)=.5$ and solve for $n$.
$$
1-\frac{364^n}{365^n}=.5 \Rightarrow\left(\frac{364}{365}\right)^n=.5 \Rightarrow n \cdot \ln \left(\frac{364}{365}\right)=\ln (.5) \Rightarrow n \approx 252.65
$$
So there needs to be at least 253 people for it to be more likely than not that one of them shares your birthday.

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

Machine Learning机器学习 问题集

Complex Analysis复杂分析 问题集

Bayesian Statistical Learning贝叶斯统计学习问题集

Time Series Analysis and Forecasting时间序列分析与预测问题集

Real Analysis实分析案例

Real Analysis实分析案例2

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