金融数学代考

分类：金融数学代考
金融数学描述了应用数学和数学模型来解决金融问题。它有时被称为定量金融，金融工程，和计算金融。这门学科结合了统计学、概率和随机过程的工具，并与经济理论相结合。

金融数学代写，金融数学代考请认准statistics-lab

STAT0013｜Discrete Mathematics 离散数学伦敦大学学院

Posted on 2023年9月22日2023年10月6日 by statistics-lab

statistics-lab^TM为您提供伦敦大学学院 London’s Global University STAT0013 Mathematical Analysis数学分析英国代写代考和辅导服务！

课程介绍：

STAT0013 is specified as a formal option for third and fourth year undergraduates from the Department of Mathematics. Any such student who has taken one of the standard prerequisites (simultaneous attendance of STAT0005 is also acceptable) should simply register for STAT0013 on Portico and await a decision. Mathematics students who do not satisfy the standard prerequisites must additionally consult a member of staff in the Department of Statistical Science.

STAT0013｜Discrete Mathematics (Advanced)进阶离散数学伦敦大学学院

Discrete Mathematics 离散数学作业案例

问题 1.

Let
$$
B=\left(\begin{array}{rrr}
-7 & 4 & -3 \
2 & -5 & 3 \
1 & -3 & -2
\end{array}\right)
$$
Calculate det $B$.

Solution: We calculate that
$$
\begin{aligned}
\operatorname{det} B= & (-7) \cdot \operatorname{det}\left(\begin{array}{rr}
-5 & 3 \
-3 & -2
\end{array}\right) \
& -4 \cdot \operatorname{det}\left(\begin{array}{rr}
2 & 3 \
1 & -2
\end{array}\right) \
& +(-3) \cdot \operatorname{det}\left(\begin{array}{rr}
2 & -5 \
1 & -3
\end{array}\right) \
= & (-7) \cdot 19-4 \cdot(-7)+(-3) \cdot(-1) \
= & -102 \
\neq & 0 .
\end{aligned}
$$
It is worth summarizing the main rule for inverses for which we have given an indication of the justification:

Rule for inverses: A square matrix has an inverse if and only if the matrix has nonzero determinant.

This still does not tell us how to find the inverse, but we shall get to that momentarily.
There are in fact a number of ways to calculate the inverse of a matrix, and we shall indicate two of them here. The first is the method of Gaussian elimination, which is a powerful technique that can be used for many purposes. The idea is to take the given square matrix
$$
A=\left(\begin{array}{ll}
a & b \
c & d
\end{array}\right)
$$
and to augment it by adjoining the identity matrix as in the display below:
$$
\left(\begin{array}{ll|ll}
a & b & 1 & 0 \
c & d & 0 & 1
\end{array}\right)
$$
Now the method of gaussian elimination allows us to perform certain operations on the rows of this augmented matrix, the goal being to reduce the square matrix on the left to the identity matrix. Whatever matrix results on the right will be the inverse.

问题 2.

The multiplicative identity for a group is unique.

Proof: Let $G$ be a group. Let $e$ and $e^{\prime}$ both be elements of $G$ that satisfy Axiom 2 . Then
$$
e=e \cdot e^{\prime}=e^{\prime}
$$
Thus $e$ and $e^{\prime}$ must be the same group element.

STAT0013｜Discrete Mathematics 离散数学伦敦大学学院

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

Math1904｜Discrete Mathematics (Advanced)进阶离散数学悉尼大学

Posted on 2023年9月22日2023年10月6日 by statistics-lab

statistics-lab^TM为您提供悉尼大学University of Sydney Math1904 Mathematical Analysis数学分析澳洲代写代考和辅导服务！

课程介绍：

This unit of study provides an introduction to programming and numerical methods. Topics covered include computer arithmetic and computational errors, systems of linear equations, interpolation and approximation, solution of nonlinear equations, quadrature, initial value problems for ordinary differential equations and boundary value problems, and optimisation.

Math1904｜Discrete Mathematics (Advanced)进阶离散数学悉尼大学

Discrete Mathematics (Advanced)进阶离散数学作业案例

问题 1.

Let $g:[1,+\infty) \rightarrow[8,+\infty)$ be given by $g(x)=x^2+3 x+4$. Then $g(1)=8$, and $g$ is strictly increasing without bound. So $g$ is one-to-one and onto. We conclude that $g$ has an inverse. One may use the quadratic formula to solve for that inverse, and find that $g^{-1}(x)=(-3+\sqrt{-7+4 x}) / 2$.

The idea of inverse function lends itself particularly well to the notation of ordered pairs. For $f: S \rightarrow T$ is inverse to $g: T \rightarrow S$ (and vice versa) provided that for every ordered pair $(s, t) \in f$ there is an ordered pair $(t, s) \in g$ and conversely.
Not every function has an inverse. For instance, let $f: S \rightarrow T$. Suppose that $f(s)=t$ and also that $f\left(s^{\prime}\right)=t$ with $s \neq s^{\prime}$ (in other words, suppose that $f$ is not one-to-one). If $g: T \rightarrow S$ then $g(f(s))=g(t)=g\left(f\left(s^{\prime}\right)\right)$ so it cannot be that both $g(f(s))=s$ and $g\left(f\left(s^{\prime}\right)\right)=s^{\prime}$. In other words, $f$ cannot have an irver se. We conclude that a function that does have an inverse must be one-to-one.

On the other hand, suppose that $t \in T$ has the property that there is no $s \in S$ with $f(s)=t$ (in other words, suppose that $f$ is not onto). Then, in particular, it could not be that $f(g(t))=t$ for any function $g: T \rightarrow S$. So $f$ could not be invertible. We conclude that a function that does have an inverse must be onto.

问题 2.

For each $m$, the set $\mathcal{Q}(m)$ is well ordered.

Proof: The proof is by induction on $m$.
When $m=1$ there is nothing to prove.
Assume that the assertion has been proved for $m=k$. Now let $U$ be a subset of $\mathcal{Q}\left(k^{\prime}\right)$. There are now three possibilities:

If in fact $U \subset \mathcal{Q}(k)$ then $U$ has a least element by the inductive hypothesis.
If $U={k}$ then $U$ has but one element and that element, namely $k$, is the least element that we seek.
The last possibility is that $U$ contains $k$ and some other natural numbers as well. But then $U \backslash{k} \subset \mathcal{Q}(k)$. Hence $U \backslash{k}$ has a least element $s$ by the inductive hypothesis. Since $s$ is automatically less than $k$, it follows that $s$ is a least element for the entire set $U$.
Now we have all our tools in place and we can prove the full result:

问题 3.

The natural numbers $\mathrm{N}$ are well ordered.

Proof: Let $\emptyset \neq S \subset \mathbb{N}$. Select an element $m \in S$. There are now two possibilities:

If $\mathcal{Q}(m) \cap S=\emptyset$ then $m$ is the least element of $S$ that we seek.
If $T=\mathcal{Q}(m) \cap S \neq \emptyset$ then notice that if $x \in T$ and $y \in S \backslash T$ then $x<y$. So it suffices for us to find a least element of $T$. But such an element exists by applying the preceding proposition to $U=\mathcal{Q}(m) \cap S$.
The proof is complete.

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

Math3076｜Mathematical Computing数学计算悉尼大学

Posted on 2023年9月22日2023年10月6日 by statistics-lab

statistics-lab^TM为您提供悉尼大学University of Sydney MATH0048 Mathematical Analysis数学分析澳洲代写代考和辅导服务！

课程介绍：

Mathematical Computing数学计算作业案例

问题 1.

(b) (0 points) Are these system specifications consistent? _________Prove it!

Solution. There are several ways to approach this problem. One is to construct a truth table with sixteen lines-one for each way of assigning truth values to the four variables $M, N, K$, and $I$.

We can avoid the cumbersome truthtable if we reason by cases. Case 1: $I$ is true. Then the last formula (9) is false, and the whole specification is false. Case 2: $I$ is false. Now formula (8) can be true only if $\neg M$ is false, that is, only if $M$ is true. Likewise, formula (7) can be true only if $\neg K$ is true, that is, $K$ is false.

Since $K$ is false, formula (6) can be true only if $N$ is false. Thus, we have deduced that in order to be consistent in this case, we must have
$$
\begin{aligned}
I & =\text { false } \
M & =\text { true } \
K & =\text { false } \
N & =\text { false }
\end{aligned}
$$
But now formula (5) is false, so it is impossible for all the formulas to be true: the system is inconsistent.

问题 2.

Problem 2 (20 points). For each of the following logical formulas with domain of discourse the natural numbers, $\mathbb{N}$, indicate whether it is a possible formulation of
I: the Induction Axiom,
S: the Strong Induction Axiom,
L: the Least Number Principle (also known as Well-ordering), or
$\mathrm{N}$ : None of these.
For example, the ordinary Induction Axiom could be expressed by the following formula, so it gets labelled “I”.
$$
(P(0) \wedge[\forall k P(k) \longrightarrow P(k+1)]) \longrightarrow \forall k P(k) \quad \longrightarrow
$$
This is a multiple choice problem: do not explain your answer.

(a) (0 points) $(P(b) \wedge[\forall k \geq b P(k) \longrightarrow P(k+1)]) \longrightarrow \forall k \geq b P(k)$

Solution. I. This is a perfect formulation of the Induction Axiom. $b$ is used for the base case; $P(k) \longrightarrow P(k+1)$ is the inductive case.
(b)$(P(b) \wedge[\forall k \leq b P(k) \longrightarrow P(k+1)]) \longrightarrow \forall k \leq b P(k)$
Solution. N. The two occurences of $k<=b$ should have been $k>=b$
(c) $[\forall b(\forall k<b P(k)) \longrightarrow P(b)] \longrightarrow \forall k P(k)$
Solution. S. Since you are assuming $P(k)$ for all $k<b$, this is strong induction.
(d) $\quad(\exists n P(n)) \longrightarrow \exists n \forall k<n \overline{P(k)}$
Solution. N. This statement is in fact always true; when $n=0, \forall k<n \overline{P(k)}$. It should say $P(n) \wedge \forall k<n ; \overline{P(k)}$

问题 3.

Problem :To encourage collaborative study, the 6.042 staff is considering assigning each student to a study group with two or three other students. Prove that as long as the enrollment is large enough, the class can always be divided into such study groups.

Solution. Proof. The proof is by strong induction. The induction hypothesis is that a class with $n \geq 6$ students can be divided into teams of 3 or 4 . More precisely
$$
P(n)::=n \geq 6 \longrightarrow \exists x, y \in \mathbb{N} 3 x+4 y=n .
$$
For any $n \geq 0$, we may assume $P(6), \ldots, P(n-1)$ to prove $P(n)$.
Case 1: $(n<6) . P(n)$ holds because the hypothesis $n \geq 6$ is false.
Case 2: $(n=6,7$, or 8$) P(6)$ is true because there could be two teams of $3, P(7)$ is true because there could be a team of 3 and a team of 4 , and $P(8)$ is true because there could be two teams of 4 .

Case 3: $(n \geq 9)$. Of course $n>n-3$ so $P(n-3)$ holds by the strong induction hypothesis. But $n-3 \geq 6$, so $P(n-3)$ implies $3 x^{\prime}+4 y^{\prime}=n-3$ for some $x^{\prime}, y^{\prime} \in \mathbb{N}$, and therefore $3 x+4 y=n$ where $x::=x^{\prime}+1$ and $y::=y^{\prime}$. So $P(n)$ holds, as required.

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

Math0048｜Mathematical Analysis数学分析伦敦大学学院

Posted on 2023年9月22日2023年10月6日 by statistics-lab

statistics-lab^TM为您提供伦敦大学学院 London’s Global University MATH0048 Mathematical Analysis数学分析英国代写代考和辅导服务！

课程介绍：

This module is an introduction to mathematical analysis, one of the most important and welldeveloped strands of pure mathematics with many elegant and beautiful theorems, and also
with applications to many areas of mathematics, theoretical statistics, econometrics, and optimisation.
The aim is to introduce students to the ideas of formal definitions and rigorous proofs (one
of the fundamental features of modern mathematics, and something that is not familiar from
A-level), and to develop their powers of logical thinking.
This module is a prerequisite for Complex Analysis, MATH0013 and provides a useful foundation for courses such as Logic, MATH0050.
The module is intended for second or third year students in departments outside Mathematics,
particularly in Economics or Statistics. Students taking this module should be mathematically
able and will normally have demonstrated this by achieving a strong result in a module such as
MATH0047 or having an A* in Further Mathematics A-level.

Math0048｜Mathematical Analysis数学分析伦敦大学学院

Mathematical Analysis数学分析作业案例

问题 1.

Let $A: X \rightarrow Y$ be a linear operator between two Euclidean or two Hermitian spaces and let $A^: Y \rightarrow X$ be its adjoint. Then $\operatorname{Rank} A^=\operatorname{Rank} A$. Moreover,
$$
\begin{array}{ll}
(\operatorname{Im} A)^{\perp}=\operatorname{ker} A^, & \operatorname{Im} A=\left(\operatorname{ker} A^\right)^{\perp}, \
\left(\operatorname{Im} A^\right)^{\perp}=\operatorname{ker} A, & \operatorname{Im} A^=(\operatorname{ker} A)^{\perp} .
\end{array}
$$

Proof. Fix two orthonormal bases on $X$ and $Y$, and let $\mathbf{A}$ be the matrix associated to $A$ using these bases. Then, see (3.8), the matrix associated to $A^$ is $\mathbf{A}^T$, hence $$ \operatorname{Rank} A^=\operatorname{Rank} \mathbf{A}^T=\operatorname{Rank} \mathbf{A}=\operatorname{Rank} A,
$$
and
$$
\operatorname{dim}\left(\operatorname{ker} A^\right)^{\perp}=\operatorname{dim} Y-\operatorname{dim} \operatorname{ker} A^=\operatorname{Rank} A^=\operatorname{Rank} A=\operatorname{dim} \operatorname{Im} A . $$ On the other hand, $\operatorname{Im} A \subset\left(\operatorname{ker} A^\right)^{\perp}$ since, if $y=A(x)$ and $A^(v)=0$, then $(y \mid v)=$ $(A(x) \mid v)=\left(x \mid A^(v)\right)=0$. We then conclude that $\left(\operatorname{ker} A^\right)^{\perp}=\operatorname{Im} A$. The other claims easily follow. In fact, they are all equivalent to $\operatorname{Im} A=\left(\operatorname{ker} A^\right)^{\perp}$.

问题 2.

Let $X$ be a finite-dimensional vector space and let $\left(e_1, e_2, \ldots, e_n\right)$ be a $g$-orthogonal basis for a metric $g$ on $X$. Denote by $n_{+}, n_{-}$and $n_0$ the numbers of elements in the basis such that respectively, we have $g\left(e_i, e_i\right)>0, g\left(e_i, e_i\right)<0, g\left(e_i, e_i\right)=0$. Then $n_{+}=i_{+}(g)$, $n_{-}=i_{-}(g)$ and $n_0=i_0(g)$. In particular, $n_{+}, n_{-}, n_0$ do not depend on the chosen g-orthogonal basis,
$$
i_{+}(g)+i_{-}(g)=r(g) \quad \text { and } \quad i_{+}(g)+i_{-}(g)+i_0(g)=n .
$$

Proof. Suppose that $g\left(e_i, e_i\right)>0$ for $i=1, \ldots, n_{+}$. For each $v=\sum_{i=1}^{n_{+}} v^i e_i$, we have
$$
g(v, v)=\sum_{i=1}^{n_{+}}\left|v^i\right|^2 g\left(e_i, e_i\right)>0,
$$
hence $\operatorname{dim} \operatorname{Span}\left{e_1, e_2, \ldots, e_{n_{+}}\right} \leq i_{+}(g)$. On the other hand, if $W \subset X$ is a subspace of dimension $i_{+}(g)$ such that $g(v, v)>0 \forall v \in W$, we have
$$
W \cap \operatorname{Span}\left{e_{n_{+}+1}, \ldots, e_n\right}={0}
$$
since $g(v, v) \leq 0$ for all $v \in \operatorname{Span}\left{e_{n_{+}+1}, \ldots, e_n\right}$. Therefore we also have $i_{+}(g) \leq$ $n-\left(n-n_{+}\right)=n_{+}$.

Similarly, one proves that $n_{-}=i_{-}(g)$. Finally, since $\mathbf{G}:=\left[g\left(e_i, e_j\right)\right]$ is the matrix associated to $g$ in the basis $\left(e_1, e_2, \ldots, e_n\right)$, we have $i_0(g)=\operatorname{dim} \operatorname{rad}(g)=\operatorname{dim} \operatorname{ker} \mathbf{G}$, and, since $\mathbf{G}$ is diagonal, $\operatorname{dim} \operatorname{ker} G=n_0$.

问题 3.

Let $g$ be a metric on a finite-dimensional real vector space $X$. Then $g$ has a $g$-orthogonal basis.

Proof. Let $r$ be the rank of $g, r:=n-\operatorname{dim} \operatorname{rad}(g)$, and let $\left(w_1, w_2, \ldots, w_{n-r}\right)$ be a basis of $\operatorname{rad}(g)$. If $V$ denotes a supplementary subspace of $\operatorname{rad}(g)$, then $V$ is $g$-orthogonal to $\operatorname{rad} g$ and $\operatorname{dim} V=r$. Moreover, for every $v \in V$ there is $z \in X$ such that $g(v, z) \neq 0$. Decomposing $z$ as $z=w+t, w \in V, t \in \operatorname{rad}(g)$, we then have $g(v, w)=g(v, w)+$ $g(v, t)=g(v, z) \neq 0$, i.e., $g$ is nondegenerate on $V$. Since trivially, $\left(w_1, w_2, \ldots, w_{n-r}\right)$ is $g$-orthogonal and $V$ is $g$-orthogonal to $\left(w_1, w_2, \ldots, w_{n-r}\right)$, in order to conclude it suffices to complete the basis $\left(w_1, w_2, \ldots, w_{n-r}\right)$ with a $g$-orthogonal basis of $V$; in other words, it suffices to prove the claim under the further assumption that $g$ be nondegenerate.

We proceed by induction on the dimension of $X$. Let $\left(f_1, f_2, \ldots, f_n\right)$ be a basis of $X$. We claim that there exists $e_1 \in X$ with $g\left(e_1, e_1\right) \neq 0$. In fact, if for some $f_i$ we have $g\left(f_i, f_i\right) \neq 0$, we simply choose $e_1:=f_i$, otherwise, if $g\left(f_i, f_i\right)=0$ for all $i$, for some $k \neq 0$ we must have $g\left(f_1, f_k\right) \neq 0$, since by assumption $\operatorname{rad}(g)={0}$. In this case, we choose $e_1:=f_1+f_k$ as
$$
g\left(f_1+f_k, f_1+f_k\right)=g\left(f_1, f_1\right)+2 g\left(f_1, f_k\right)+g\left(f_k, f_k\right)=0+2 g\left(f_1, f_k\right)+0 \neq 0 .
$$
Now it is easily seen that the subspace
$$
V_1:=\left{v \in X \mid g\left(e_1, v\right)=0\right}
$$
supplements Span $\left{e_1\right}$, and we find a basis $\left(v_2, \ldots, v_n\right)$ of $V_1$ such that $g\left(v_j, e_1\right)=0$ for all $j=2, \ldots, n$ by setting
$$
v_j:=f_j-\frac{g\left(f_j, e_1\right)}{g\left(e_1, e_1\right)} e_1 .
$$
Since $g$ is nondegenerate on $V_1$, by the induction assumption we find a $g$-orthogonal basis $\left(e_2, \ldots, e_n\right)$ of $V_1$, and the vectors $\left(e_1, e_2, \ldots, e_n\right)$ form a $g$-orthogonal basis of $X$.

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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

Math0033｜Numerical Methods 数值方法伦敦大学学院

Posted on 2023年9月21日2023年10月7日 by statistics-lab

statistics-lab^TM为您提供悉尼大学 London’s Global University MATH0048 Mathematical Analysis数学分析英国代写代考和辅导服务！

课程介绍：

Many phenomena in engineering and the physical and biological sciences can be described
using mathematical models. Frequently the resulting models cannot be solved analytically,
in which case a common approach is to use a numerical method to find an approximate
solution. The aim of this course is to introduce the basic ideas underpinning computational mathematics, study a series of numerical methods to solve different problems, and
carry out a rigorous mathematical analysis of their accuracy and stability

Numerical Methods 数值方法作业案例

问题 1.

we derived the an equation to describe a mass that is attached to a spring that would break when its elongation reached $0.03 \mathrm{~m}$ during resonant vibration of the springmass system. We need to determine the time $\mathrm{t}_{\mathrm{f}}$ at which the spring breaks from Equation (a):
$$
\left(0.05-\frac{t_f}{20}\right) \cos 10 t_f+\frac{1}{200} \sin 10 t_f-0.03=0
$$

We will use the Newton-Raphson’s method to solve the unknown quantity $t_f$ in Equation (a) by first assuming a solution on $t_f=0.75$. We made this assumed solution based on a crude approximated value of $\mathrm{t}{\mathrm{f}}=0.7$ in Example 8.9. Again, let us replace the unknown quantity $t_f$ in Equation (a) by conventional unknown symbol $\mathrm{x}$ in the following alternative form: $$ \begin{gathered} \left(0.05-\frac{x}{20}\right) \cos 10 x+\frac{1}{200} \sin 10 x-0.03=0 \ f(x)=\left(0.05-\frac{x}{20}\right) \cos 10 x+\frac{1}{200} \sin 10 x-0.03 \ f^{\prime}(x)=-(0.05-0.5 x) \sin 10 x \end{gathered} $$ Thus, the estimated root $x{i+1}$ after the previously estimated root $x_i$ may be computed by using the expression, as will be shown in the next slide.

Solution $\mathrm{x}$ from the equation: $\left(0.05-\frac{x}{20}\right) \cos 10 x+\frac{1}{200} \sin 10 x-0.03=0$

问题 2.

Evaluate the following integral by using the Gaussian quadrature $$
I=\int_0^\pi \cos x d x
$$

We have the function $y(x)=\cos x$ over the integration limits $x_a=0$ and $x_b=\pi$. The transformation of coordinates makes use of the relationship $x=\frac{\pi}{2} \xi+\frac{\pi}{2}$ from Equation (10.13), from which we get:
$$
y(x)=\cos x=F(\xi)=\cos \left(\frac{\pi}{2} \xi+\frac{\pi}{2}\right)=\sin \left(\frac{\pi}{2} \xi\right)
$$
Also, from Equation (10.14) with the use of the trigonometric relationships such as:
$$
\sin \left(\frac{\pi}{2}+\theta\right)=\cos \theta \text { and } \cos \left(\frac{\pi}{2}+\theta\right)=\sin \theta
$$
We may arrive at the following expression for integrating I in Equation (a) using Gaussian quadrature:
$$
I=\int_0^\pi \cos x d x=\int_{-1}^1\left[\sin \left(\frac{\pi}{2} \xi\right)\left(\frac{\pi}{2} d \xi\right)\right]=\frac{\pi}{2} \int_{-1}^1 \sin \frac{\pi}{2} \xi d \xi=\frac{\pi}{2} \sum_{i=1}^n H_i \sin \left(\frac{\pi}{2} a_i\right)
$$
Let us take, for example, 3 sampling points, i.e., $\mathrm{n}=3$ from Table 10.3 on P.357 with:
$$
\begin{array}{lll}
\mathrm{a}_1=0 & \mathrm{a}_2=+0.77459 & \mathrm{a}_3=-0.77459 \
\mathrm{H}_1=0.88888 & \mathrm{H}_2=0.55555 & \mathrm{H}_3=0.55555
\end{array}
$$
Substituting the above numbers into Equation (b) will lead to the solution:
$$
\begin{aligned}
I & =\frac{\pi}{2}\left[0.88888 \sin (0)+0.55555 \sin \left(\frac{\pi}{2} \times 0.77459\right)+0.55555 \sin \left(-\frac{\pi}{2} \times 0.77459\right)\right] \
& =\frac{\pi}{2}[0.55555 \sin (1.2167)-0.55555 \sin (1.2167)]=0
\end{aligned}
$$
25

问题 3.

Evaluate the following integral in Example 10.8 using Gaussian quadrature method

Solution:
$$
I=\int_{0.5}^{3.5} y(x) d x=\int_{0.5}^{3.5} x \sqrt{\left(16-x^2\right)^3} d x
$$

We notice the function $y(x)$ in the integral in Equation (a) is identical to what we had in previous Examples 10.5 (p.349), 6 (p.350) and 8 (p.355).
We will first derive the expression of the function $F(\xi)$ for the function $y(x)$ in Equation (a) from the integration limits of 0.5 and 3.5 to the limit -1 and +1 , as required in Gaussian quadrature.
The transformation of coordinate systems as illustrated in Figure 10.15 begins with the transformation of variable from $x$ to $\xi$ using the relationship in Equation (10.13), leading to the following relationship between the variables $x$ and $\xi$ as shown below:
$x=\frac{1}{2}\left(x_b-x_a\right) \xi+\frac{1}{2}\left(x_b+x_a\right)=1.5 \xi+2$, from which we obtain: $\mathrm{dx}=1.5 \mathrm{~d} \xi$
The integral in Equation (a) in the $y(x)$ vs. $x$ coordinates can thus be transformed to the $F(\xi)$ vs. $\xi$ coordinates by using Equation (10.14), yielding the following expression:
$$
\begin{aligned}
I=\int_{0.5}^{3.5} y(x) d x=\int_{0.5}^{3.5} x \sqrt{\left(16-x^2\right)^3} d x & =\frac{x_b-x_a}{2} \int_{-1}^1 F(\xi) d \xi \
& =\left.\frac{3.5-0.5}{2} \int_{-1}^1 y(x)\right|{x=1.5 \xi+2} d \xi \end{aligned} $$ We may thus evaluate the integral with the expression in Equation (d) as: $$ I=1.5 \int{-1}^1(1.5 \xi+2) \sqrt{\left[16-(1.5 \xi+2)^2\right]^3} d \xi=1.5 \int_{-1}^1(1.5 \xi+2) \sqrt{\left(-2.25 \xi^2-6 \xi+12\right)^3} d \xi

We have arrived at the following integral:
$$
I=1.5 \int^1(1.5 \xi+2) \sqrt{\left[16-(1.5 \xi+2)^2\right]^3} d \xi=1.5 \int^1(1.5 \xi+2) \sqrt{\left(-2.25 \xi^2-6 \xi+12\right)^3} d \xi
$$

Math0033｜Numerical Methods 数值方法伦敦大学学院请认准statistics-lab™

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

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

数学代写|金融数学代写Intro to Mathematics of Finance代考|TFIN101

Posted on 2022年11月5日2022年11月5日 by statistics-lab

如果你也在怎样代写金融数学Intro to Mathematics of Finance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。

金融数学是将数学方法应用于金融问题。(有时使用的同等名称是定量金融、金融工程、数学金融和计算金融）。它借鉴了概率、统计、随机过程和经济理论的工具。传统上，投资银行、商业银行、对冲基金、保险公司、公司财务部和监管机构将金融数学的方法应用于诸如衍生证券估值、投资组合结构、风险管理和情景模拟等问题。依赖商品的行业（如能源、制造业）也使用金融数学。定量分析为金融市场和投资过程带来了效率和严谨性，在监管方面也变得越来越重要。

statistics-lab™ 为您的留学生涯保驾护航在代写金融数学Intro to Mathematics of Finance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。我们的专家在代写金融数学Intro to Mathematics of Finance代写方面经验极为丰富，各种代写金融数学Intro to Mathematics of Finance相关的作业也就用不着说。

我们提供的金融数学Intro to Mathematics of Finance及其相关学科的代写，服务范围广, 其中包括但不限于:

Statistical Inference 统计推断
Statistical Computing 统计计算
Advanced Probability Theory 高等概率论
Advanced Mathematical Statistics 高等数理统计学
(Generalized) Linear Models 广义线性模型
Statistical Machine Learning 统计机器学习
Longitudinal Data Analysis 纵向数据分析
Foundations of Data Science 数据科学基础

数学代写|金融数学代写Intro to Mathematics of Finance代考|TFIN101

数学代写|金融数学代写Intro to Mathematics of Finance代考|Approximations Using Taylor Series

If $f(x)$ is a function which has derivatives of all orders $\left(f^{\prime}, f^{\prime \prime}, f^{\prime \prime \prime}\right.$ etc., all exist) it can be shown that (under certain restrictions) $f(x)$ can be computed as an infinite sum of terms involving its derivatives.
$$
f(x)=f(a)+f^{\prime}(a)(x-a)^2+\frac{f^{(2)}(a)}{2 !}(x-a)^2+\cdots+\frac{f^{(n)}(a)}{n !}(x-a)^n+\cdots
$$
In the expression above $f^{(n)}(a)$ refers to the $n^{\text {th }}$ derivative of $f$ evaluated at $a$. We can compute approximate values of $f(x)$ near a known value $f(a)$ by using the first few terms in $1.12$.

Example 1.6: Use the first four terms of Equation $1.12$ and $a=0$ to approximate $\sin (x)$
Solution:
$$
\begin{aligned}
\sin (x) &=\sin (0)+\sin ^{\prime}(0)(x-0)+\frac{\sin ^{\prime \prime}(0)}{2 !}(x-0)^2+\cdots+\frac{\sin ^{\prime \prime \prime}(0)}{3 !}(x-0)^3 \
&=0+\cos (0)(x-0)+\frac{-\sin (0)}{2 !}(x-0)^2+\frac{-\cos (0)}{3 !}(x-0)^2 \
&=x-\frac{x^3}{6}
\end{aligned}
$$
As the sketch below illustrates this approximation is quite good for values of $x$ close to 0 (Figure $1.3$ ).

数学代写|金融数学代写Intro to Mathematics of Finance代考|Exponents and Logarithms

For convenience we state some of the basic properties of the exponential and logarithmic functions. Unless otherwise stated, we will use the logarithm to base $e$, indicated as $\ln (x)$ and often referred to as the natural logarithm The TI BA II Plus and TI-30XS both have a button devoted to $\ln$ – the 2ND function for this button is $e^x$.
BASIC IDENTITIES
$$
\begin{aligned}
\ln (a b) &=\ln (a)+\ln (b) \
\ln \left(a^r\right) &=r \ln (a) \
\ln \left(\frac{a}{b}\right) &=\ln (a)-\ln (b) \
\frac{d \ln (x)}{d x} &=\frac{1}{x}, \quad \frac{d \ln (1+i)}{d i}=\frac{1}{1+i} \
\ln \left(e^x\right) &=e^{\ln (x)}=x \
\frac{d e^x}{d x} &=e^x \
\int e^u d u &=e^u+c \
\int \frac{1}{u} d u &=\ln (|u|)+c \
\int
\end{aligned}
$$
Example 1.9: Solve $(1.05)^n=2$.
Solution: We take $\ln$ of both sides to obtain $n \cdot \ln (1.05)=\ln (2)$. Thus, $n=\frac{\ln (2)}{\ln (1.05)} \approx 14.21$.
Example 1.10: Solve for $i$ :
$$
(1+i)^3=1+3 \cdot(.05)=1.15 .
$$
Solution: We take $\ln$ of both sides to obtain $3 \ln (1+i)=\ln (1.15)$. This gives us $\ln (1+i)=\frac{\ln (1.15)}{3}=0.04658731412$. As a result, $1+i=e^{0.04658731412}=$ 1.047689553. Hence $i=.047686553$. We could also solve this problem by taking the cube root of both sides of the equation. $(1+i)=\sqrt[3]{1.15}=(1.15)^{\frac{1}{3}}=$ $1.047689553$.

数学代写|金融数学代写Intro to Mathematics of Finance代考|Approximations Using Taylor Series

如果 $f(x)$ 是一个具有所有阶导数的函数 $\left(f^{\prime}, f^{\prime \prime}, f^{\prime \prime \prime}\right.$ 等等，都存在) 可以证明（在某些限制下) $f(x)$ 可以计算为涉及其导数的项的无限总和。
$$
f(x)=f(a)+f^{\prime}(a)(x-a)^2+\frac{f^{(2)}(a)}{2 !}(x-a)^2+\cdots+\frac{f^{(n)}(a)}{n !}(x-a)^n+\cdots
$$
在上面的表达式中 $f^{(n)}(a)$ 指的是 $n^{\text {th }}$ 的导数 $f$ 评价为 $a$. 我们可以计算近似值 $f(x)$ 接近已知值 $f(a)$ 通过使用前几个术语 $1.12$.
例 1.6：使用公式的前四项 $1.12$ 和 $a=0$ 近似 $\sin (x)$
解决方案:
$$
\sin (x)=\sin (0)+\sin ^{\prime}(0)(x-0)+\frac{\sin ^{\prime \prime}(0)}{2 !}(x-0)^2+\cdots+\frac{\sin ^{\prime \prime \prime}(0)}{3 !}(x-0)^3=0+\cos (0)(x
$$
正如下面的草图所示，这个近似值对于 $x$ 接近于 0 (图 $1.3$ ).

数学代写|金融数学代写Intro to Mathematics of Finance代考|Exponents and Logarithms

为方便起见，我们陈述了指数函数和对数函数的一些基本性质。除非另有说明，否则我们将以对数为底 $e$ ，表示为 $\ln (x)$ 并且通常被称为自然对数 TI BA II Plus 和 TI-30XS 都有一个专门用于ln- 此按钮的 2ND 功能是 $e^x$. 基本身份
$$
\ln (a b)=\ln (a)+\ln (b) \ln \left(a^r\right) \quad=r \ln (a) \ln \left(\frac{a}{b}\right)=\ln (a)-\ln (b) \frac{d \ln (x)}{d x} \quad=\frac{1}{x}, \quad \frac{d \ln (1+i)}{d i}
$$
例 1.9: 求解 $(1.05)^n=2$.
解决方案: 我们采取ln双方获得 $n \cdot \ln (1.05)=\ln (2)$. 因此， $n=\frac{\ln (2)}{\ln (1.05)} \approx 14.21$.
例 1.10: 求解 $i$ :
$$
(1+i)^3=1+3 \cdot(.05)=1.15 .
$$
解决方案: 我们采取 $\ln$ 双方获得 $3 \ln (1+i)=\ln (1.15)$. 这给了我们 $\ln (1+i)=\frac{\ln (1.15)}{3}=0.04658731412$. 因此， $1+i=e^{0.04658731412}=1.047689553$ 。因此 $i=.047686553$. 我们也可以通过对等式两边取立方根来解决这个问题。 $(1+i)=\sqrt[3]{1.15}=(1.15)^{\frac{1}{3}}=$ $1.047689553$.

数学代写|金融数学代写Intro to Mathematics of Finance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。

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

数学代写|金融数学代写Intro to Mathematics of Finance代考|MATH3090

Posted on 2022年11月5日2022年11月5日 by statistics-lab

如果你也在怎样代写金融数学Intro to Mathematics of Finance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。

我们提供的金融数学Intro to Mathematics of Finance及其相关学科的代写，服务范围广, 其中包括但不限于:

Statistical Inference 统计推断
Statistical Computing 统计计算
Advanced Probability Theory 高等概率论
Advanced Mathematical Statistics 高等数理统计学
(Generalized) Linear Models 广义线性模型
Statistical Machine Learning 统计机器学习
Longitudinal Data Analysis 纵向数据分析
Foundations of Data Science 数据科学基础

数学代写|金融数学代写Intro to Mathematics of Finance代考|Approximation Techniques

In many cases, we will need to solve equations for which no direct method applies. You are probably familiar with the quadratic formula: The solutions to $a x^2+b x+c=0$ are
$$
x=\frac{-b \pm \sqrt{b^2-4 a c}}{2 a}
$$
There are similar equations for polynomials of degrees 3 and 4 , but no such formula exists for polynomials of degree 5 or higher. In some cases, we can reduce a higher degree polynomial to a quadratic, but these techniques won’t always work. As a result, we will utilize approximating techniques to solve such equations. We will use four methods.
a) Excel’s financial functions.
b) Newton’s Method (not used much anymore, provided as an historical note).
c) MAPLE (very powerful tool, but requires interpretation of results); MAPLE seems little used by financial folk.
d) TI Calculator internal calculation. Along with Excel, this will be the tool you will use most often in “the real world.”

数学代写|金融数学代写Intro to Mathematics of Finance代考|Newton’s Method

Isaac Newton (1643-1727), an English philosopher and mathematician, did important work in both physics and calculus. His method for approximating roots to polynomials is a very nice application of the tangent line. Joseph Raphson (1648-1715), also English, was made a member of the Royal Society prior to his graduation from Cambridge. See more about these two at the MacTutor History of Mathematics site: http://www-history.mes.standrews.ac.uk/index.html

Newton’s Method solves the equation $f(x)=0$ using an iteration technique. An iteration technique involves three stages:
1) Determining an initial guess (or approximation) called $x_0$,
2) Constructing an algorithm to compute $x_{i+1}$ in terms of $x_i$,
3) A proof that the sequence $x_n$ converges to the required value, in our case a solution of the equation $f(x)=0$.

The process starts with the initial approximation $x_0$ and then computes $x_1$, $x_2$, etc., until a desired degree of accuracy is attained. We will discuss how to make an educated guess (the $x_0$ ) in the context of specific problems ${ }^4$. At this point, we are interested only in describing how Newton’s Method generates the iteration sequence in 2). A proof that the method works is beyond the scope of this text – consult an Advanced Calculus text, if you would like to see a proof.

To create the sequence of approximations using the Newton-Raphson Method, we start with a reasonable first approximation, $x_0$. Often this is done by using a graphing calculator to graph the function and then reading off an estimate from the graph. To find $x_1$, we first construct the tangent line to the graph of $f$ at the point $\left(x_0, f\left(x_0\right)\right)$. The second estimate, $x_1$, is the $x$-intercept of this tangent line.

数学代写|金融数学代写Intro to Mathematics of Finance代考|MATH3090

金融数学代考

数学代写|金融数学代写Intro to Mathematics of Finance代考|Approximation Techniques

在许多情况下，我们需要求解没有直接方法适用的方程。您可能熟悉二次公式：一个X2+bX+C=0是

X=−b±b2−4一个C2一个
对于 3 次和 4 次多项式有类似的方程，但对于 5 次或更高次的多项式不存在这样的公式。在某些情况下，我们可以将更高次多项式简化为二次，但这些技术并不总是有效。因此，我们将利用近似技术来求解这些方程。我们将使用四种方法。
a) Excel 的财务功能。
b) 牛顿法（不再使用太多，作为历史记录提供）。
c) MAPLE（非常强大的工具，但需要解释结果）；MAPLE 似乎很少被金融界人士使用。
d) TI 计算器内部计算。与 Excel 一起，这将是您在“现实世界”中最常使用的工具。

数学代写|金融数学代写Intro to Mathematics of Finance代考|Newton’s Method

英国哲学家和数学家艾萨克·牛顿（Isaac Newton，1643-1727）在物理学和微积分方面都做出了重要贡献。他将根近似为多项式的方法是切线的一个很好的应用。Joseph Raphson (1648-1715) 也是英国人，在他从剑桥大学毕业之前就成为了英国皇家学会的成员。在 MacTutor 数学史网站上查看更多关于这两个的信息：http://www-history.mes.standrews.ac.uk/index.html

牛顿法求解方程F(X)=0使用迭代技术。迭代技术涉及三个阶段：
1) 确定初始猜测（或近似值），称为X0,
2) 构造一个算法来计算X一世+1按照X一世,
3) 证明该序列Xn收敛到所需的值，在我们的例子中是方程的解F(X)=0.

该过程从初始近似值开始X0然后计算X1, X2等，直到达到所需的准确度。我们将讨论如何做出有根据的猜测（X0) 在具体问题的背景下4. 此时，我们只对描述牛顿法如何生成 2) 中的迭代序列感兴趣。该方法有效的证明超出了本文的范围——如果您想查看证明，请参阅高级微积分文本。

为了使用 Newton-Raphson 方法创建近似序列，我们从一个合理的第一近似开始，X0. 通常这是通过使用图形计算器绘制函数图，然后从图中读取估计值来完成的。寻找X1，我们首先构造图的切线F在这一点上(X0,F(X0)). 第二个估计，X1，是个X-这条切线的截距。

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

数学代写|金融数学代写Intro to Mathematics of Finance代考|ACTL20001

Posted on 2022年11月5日2022年11月5日 by statistics-lab

如果你也在怎样代写金融数学Intro to Mathematics of Finance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。

我们提供的金融数学Intro to Mathematics of Finance及其相关学科的代写，服务范围广, 其中包括但不限于:

Statistical Inference 统计推断
Statistical Computing 统计计算
Advanced Probability Theory 高等概率论
Advanced Mathematical Statistics 高等数理统计学
(Generalized) Linear Models 广义线性模型
Statistical Machine Learning 统计机器学习
Longitudinal Data Analysis 纵向数据分析
Foundations of Data Science 数据科学基础

数学代写|金融数学代写Intro to Mathematics of Finance代考|ACTL20001

数学代写|金融数学代写Intro to Mathematics of Finance代考|Sequences and Series

A sequence of payments over time is known as an annuity. We will often need to compute the value of an annuity at a particular point in time. To do so we compute the value of each payment in the sequence (which will depend on the time that payment will be made) and then add those values to obtain the total value. The sequence of sums obtained by adding the terms of a sequence is called a series. For example, if our sequence of terms (payments, usually) is $100,200,300,400$ the series of sums is $100,100+200,100+200+300,100+$ $200+300+400$.

If we compute the sum of the values of the payments at the current time, the result is called the present value (PV) of the annuity. If we compute the accumulated values of the payments at some time in the future, the result is called the future value (FV) of the annuity. In either case, we will usually end up with a geometric series (the sum of a sequence where each term is a constant multiple of the preceding term) and so need the formula for the sum of such a series:
$$
\sum_{i=0}^{n-1} a v^i=a+a v+a v^2+\cdots+a v^{n-1}=a \frac{1-v^n}{1-v}
$$
Here $a$ is the initial term and $v$ is the common multiple ${ }^1$.
If $|v|<1$ then $\lim {n \rightarrow \infty} v^n=0$ and we can compute the sum of an infinite series of payments (called a perpetuity) as well: $$ \sum{i=0}^{\infty} a v^i=\lim _{n \rightarrow \infty} a \frac{1-v^n}{1-v}=\frac{a}{1-v}
$$

Using Equations $1.1$ and $1.2$ can be a bit tricky as not all series start at $i=0$. The most direct way to deal with this is to write down a few terms of the series you are dealing with and match them up with Equation $1.1$ or Equation 1.2. Note that you don’t need to figure out the last term since
$$
\begin{aligned}
&a=\text { first term } \
&v=\text { common multiple } \
&n=\text { number of terms. }
\end{aligned}
$$

数学代写|金融数学代写Intro to Mathematics of Finance代考|Arithmetic Series

An arithmetic series is created by adding the terms of a sequence where a constant (denoted by $d$ in the formula below) is added to each term to get the next term. In the case of an arithmetic series we have
$$
a+(a+d)+(a+2 d)+\cdots+(a+(n-1) d)=\frac{n(2 a+(n-1) d)}{2}
$$
Example 1.4: In the simplest case $a=d=1$ and we have the formula Carl Friederich Gauss supposedly proved at age six.
$$
\sum_{i=1}^n i=1+2+3+\cdots+n=\frac{n(n+1)}{2}
$$
In some cases, we will need to deal with a combination of an arithmetic and a geometric series:
$$
A=P v+(P+Q) v^2+(P+2 Q) v^3+\cdots+(P+(n-1) Q) v^n
$$
This situation (which we will refer to as a $P-Q$ Series) arises when we have an annuity ${ }^3$ which starts with an initial payment which is then incremented by $Q$ at the end of each subsequent period ( $Q$ can be positive or negative). In many cases $Q$ is added to account for inflation. To simplify this expression we first divide both sides by $v$, obtaining:
$$
\frac{A}{v}=P+(P+Q) v+(P+2 Q) v^2+(P+3 Q) v^3+\cdots+(P+(n-1) Q) v^{n-1}
$$
Subtracting Equation 1.5 from Equation 1.6 gives us:
$$
A\left(\frac{1}{v}-1\right)=A i=P\left(1-v^n\right)+Q\left(v+v^2+v^3+\cdots+v^n\right)-Q n v^n
$$

金融数学代考

数学代写|金融数学代写Intro to Mathematics of Finance代考|Sequences and Series

随着时间的推移，一系列付款被称为年金。我们经常需要计算特定时间点的年金价值。为此，我们计算序列中每笔付款的价值（这将取决于付款的时间），然后将这些值相加以获得总价值。将数列的各项相加得到的和的数列称为数列。例如，如果我们的条款序列（通常是付款）是 $100,200,300,400$ 总和系列是 $100,100+200,100+200+300,100+200+300+400$.
如果我们计算当前时间的付款值的总和，则结果称为年金的现值 (PV) 。如果我们计算末来某个时间支付的累计值，则结果称为年金的末来值 (FV) 。在任何一种情况下，我们通常都会得到一个几何级数（一个序列的总和，其中每一项都是前一项的常数倍数），因此需要这个数列总和的公式:
$$
\sum_{i=0}^{n-1} a v^i=a+a v+a v^2+\cdots+a v^{n-1}=a \frac{1-v^n}{1-v}
$$
这里 $a$ 是初始项，并且 $v$ 是公倍数 ${ }^1$.
如果 $|v|<1$ 然后 $\lim n \rightarrow \infty v^n=0$ 我们还可以计算无限系列支付的总和（称为永续年金)：
$$
\sum i=0^{\infty} a v^i=\lim _{n \rightarrow \infty} a \frac{1-v^n}{1-v}=\frac{a}{1-v}
$$
使用方程式 $1.1$ 和 $1.2$ 可能有点棘手，因为并非所有系列都从 $i=0$. 解决这个问题的最直接方法是写下您正在处理的系列的一些术语，并将它们与方程式匹配1.1或公式 1.2。请注意，您不需要计算上一个术语，因为
$a=$ first term $\quad v=$ common multiple $n=$ number of terms.

数学代写|金融数学代写Intro to Mathematics of Finance代考|Arithmetic Series

算术级数是通过添加一个序列的项来创建的，其中一个常数（表示为 $d$ 在下面的公式中）被添加到每个术语以获得下一个术语。在算术级数的情况下，我们有
$$
a+(a+d)+(a+2 d)+\cdots+(a+(n-1) d)=\frac{n(2 a+(n-1) d)}{2}
$$
示例 1.4：在最简单的情况下 $a=d=1$ 我们有卡尔弗里德里希高斯据说在六岁时证明的公式。
$$
\sum_{i=1}^n i=1+2+3+\cdots+n=\frac{n(n+1)}{2}
$$
在某些情况下，我们需要处理算术级数和几何级数的组合:
$$
A=P v+(P+Q) v^2+(P+2 Q) v^3+\cdots+(P+(n-1) Q) v^n
$$
这种情况（我们将其称为 $P-Q$ Series) 在我们有年金时出现 ${ }^3$ 从初始付款开始，然后增加 $Q$ 在每个后续期间结束时 ( $Q$ 可以是正面的也可以是负面的) 。在很多情况下 $Q$ 被添加以解释通货膨胀。为了简化这个表达式，我们首先将两边除以 $v$ ，获得:
$$
\frac{A}{v}=P+(P+Q) v+(P+2 Q) v^2+(P+3 Q) v^3+\cdots+(P+(n-1) Q) v^{n-1}
$$
从公式 $1.6$ 中减去公式 $1.5$ 得出:
$$
A\left(\frac{1}{v}-1\right)=A i=P\left(1-v^n\right)+Q\left(v+v^2+v^3+\cdots+v^n\right)-Q n v^n
$$

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

数学代写|金融数学代写Intro to Mathematics of Finance代考|MATH3090

Posted on 2022年11月4日2022年11月4日 by statistics-lab

如果你也在怎样代写金融数学Intro to Mathematics of Finance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。

我们提供的金融数学Intro to Mathematics of Finance及其相关学科的代写，服务范围广, 其中包括但不限于:

Statistical Inference 统计推断
Statistical Computing 统计计算
Advanced Probability Theory 高等概率论
Advanced Mathematical Statistics 高等数理统计学
(Generalized) Linear Models 广义线性模型
Statistical Machine Learning 统计机器学习
Longitudinal Data Analysis 纵向数据分析
Foundations of Data Science 数据科学基础

数学代写|金融数学代写Intro to Mathematics of Finance代考|The Roles of Arbitragers, Hedgers, and Speculators

Hull (2015) discusses three types of traders: ${ }^1$ arbitragers, hedgers, and speculators. Arbitragers, hedgers, and speculators, along with investors, borrowers, and entrepreneurs, all commonly use markets for financial securities and financial derivatives. Understanding the roles of arbitragers, hedgers, and speculators is helpful in understanding why derivatives have emerged and evolved into diverse forms.

An important concept before discussing these roles is short selling. The idea that a trader can take a short position in a financial derivative was introduced. It makes sense that, as a contract between two people regarding the price of an asset one of the parties to the contract will take the long side and one will take the short side. Having the long side of the contract means that one will have the risk exposure more similar to owning the asset with the other side having an exposure similar to owing the asset. Short selling is the process of obtaining the opposite exposure of owing an asset by borrowing the asset and selling it into the market at the current market price. The short seller of an asset therefore is inversely exposed to the price changes of the underlying asset: the short seller gains if the underlying stock or other asset’s market price falls and loses market value if the asset rises in value. Eventually, the short seller purchases the asset and returns it Io the entity from which it was borrowed, terminating the short sale. Ignoring dividends and the time value of money, the short seller of a stock the stuck rises by $\$ X$.

Arbitragers are traders who establish a long (or buy) position in one or more assets that they believe are relatively underpriced and one or more short (or sell) positions in assets that they believe are relatively overpriced. Arbitragers hold these positions based on their belief that the prices will revert to more proper price levels that allow the arbitrager to exit both positions at a net profit. When markets are poorly or inefficiently priced, arbitragers can earn riskless or virtually riskless profits. Successful arbitragers serve the market by correcting pricing errors thereby (and unintentionally) ensuring that less informed market participants are transacting at prices that are nearer to their proper or equilibrium values. The distinguishing feature of arbitragers is that their trades are designed to take very little risk and tend to be short term.

Hedgers are market participants that use financial derivatives and other instruments to offset risks that they do not wish to bear. For example, an agricultural firm may use derivatives to lock in the revenues from their output such as food products, lock down their costs of production such as energy costs, and minimize their exposure to exchange rates from their contracts to export their products. Financial derivatives allow hedgers to hedge or “lay off” those risks that they cannot control and allow them to focus on those factors over which they may have control.

Speculators are similar to arbitragers except that they typically take substantially more risk by establishing one or more positions with unhedged risks. Speculators take positions in anticipation of profits if their forecasts of future price movements are accurate. The key to successful speculators is that they gain money and are able to continue speculating only when they are able to consistently buy assets that are underpriced and sell assets that are consistently overpriced. In doing do, they can consistently earn profits while acting to balance supply and demand, drive prices toward their true values, and stabilize markets and prices.

数学代写|金融数学代写Intro to Mathematics of Finance代考|Market Prices, Risk, and Randomness

Risk is a fascinating topic that is central to financial economics. What causes risk, and what causes security prices to change? The answers to these questions and the study of financial risk in general have progressed primarily using financial math.

The terms value and price (as well as the terms valuation and pricing) are often used interchangeably in finance. This book tends to use the term value to describe how much a particular person believes an item is worth and uses the term price to describe the amount of money that people receive or pay when they exchange the item. However, the term asset pricing is used frequently in financial economics to describe very important models even when the context is more clearly described as involving asset valuation.

Since the general meaning of the terms are quite similar, this book often uses the terms price and pricing to describe values and valuation in cases where price and pricing are conventionally used.

The value of an asset tends to change through time because the preferences of people change through time and the abilities of a modern economy to meet those preferences changes through time. Agricultural prices change due to factors such as weather, energy prices change due to factors such as economic activity, and stock prices change due to factors such as predictions of future revenues and expenses. Prices respond to changes as soon as the information about those changes is revealed. The effects of a frost on orange harvests begin changing market prices of oris done.

The spot price is the price today for delivery today. The spot price of an asset such as a stock or bond reflects a consensus in the marketplace with regard to the future benefits that the asset offers. As time passes, the asset price changes because the market’s predictions of future conditions change. More generally, asset prices change because of the arrival of new information. Good news for a stock such as higher forecasts of earnings causes a positive price change, while bad news such as a disappointment regarding the sales of a new product causes a negative price change. To the extent that market participants rationally and efficiently process available information to form the current price of an asset, it follows based on the arrival of new information.
Therefore, security prices can be viewed as random variables that change through time based on the arrival of new information. Simply put, the future price of an asset can be modeled as being equal to its future expected value plus or minus price changes due to new information that becomes available to market participants.

金融数学代考

数学代写|金融数学代写Intro to Mathematics of Finance代考|The Roles of Arbitragers, Hedgers, and Speculators

Hull (2015) 讨论了三种类型的交易者：1套利者、套期保值者和投机者。套利者、套期保值者和投机者，以及投资者、借款者和企业家，通常都使用金融证券和金融衍生品市场。了解套利者、套期保值者和投机者的角色有助于理解为什么衍生品会出现并演变成多种形式。

在讨论这些角色之前，一个重要的概念是卖空。引入了交易者可以在金融衍生品中做空头寸的想法。有意义的是，作为两个人之间关于资产价格的合同，合同的一方将采取多头一方，另一方将采取空方。拥有合约的多头意味着一方的风险敞口与拥有资产更相似，而另一方的风险敞口类似于欠资产。卖空是通过借入资产并以当前市场价格将其出售到市场来获得所欠资产的相反风险敞口的过程。因此，资产的卖空者反向暴露于标的资产的价格变化：如果标的股票或其他资产的市场价格下跌，卖空者就会获利，如果资产价值上涨，卖空者就会失去市场价值。最终，卖空者购买资产并将其归还给借入资产的实体，从而终止卖空。忽略股息和货币时间价值，被困股票的卖空者上涨$X.

套利者是在他们认为相对低估的一种或多种资产中建立多头（或买入）头寸并在他们认为相对高估的资产中建立一个或多种空头（或卖出）头寸的交易者。套利者持有这些头寸是因为他们相信价格将恢复到更合适的价格水平，从而允许套利者以净利润退出两个头寸。当市场定价不佳或效率低下时，套利者可以获得无风险或几乎无风险的利润。成功的套利者通过纠正定价错误来为市场服务，从而（并且无意中）确保信息较少的市场参与者以更接近其适当或均衡价值的价格进行交易。

对冲者是使用金融衍生品和其他工具来抵消他们不想承担的风险的市场参与者。例如，一家农业公司可以使用衍生品来锁定其产出（如食品）的收入，锁定其生产成本（如能源成本），并尽量减少其产品出口合同中的汇率风险。金融衍生品允许套期保值者对冲或“放弃”他们无法控制的风险，并允许他们专注于他们可能控制的那些因素。

投机者与套利者相似，只是他们通常通过建立一个或多个未对冲风险的头寸来承担更大的风险。如果投机者对未来价格走势的预测准确，他们就会在预期利润的情况下建仓。成功的投机者的关键是他们赚钱，并且只有当他们能够持续购买被低估的资产并出售一直被高估的资产时，他们才能继续投机。在这样做的过程中，他们可以在平衡供需、推动价格接近其真实价值、稳定市场和价格的同时持续赚取利润。

数学代写|金融数学代写Intro to Mathematics of Finance代考|Market Prices, Risk, and Randomness

风险是一个引人入胜的话题，是金融经济学的核心。是什么导致了风险，又是什么导致了证券价格的变化？这些问题的答案和一般金融风险的研究主要使用金融数学取得了进展。

术语价值和价格（以及术语估值和定价）在金融中经常互换使用。这本书倾向于使用价值这个词来描述一个特定的人认为一件物品的价值，并使用价格这个词来描述人们在交换物品时收到或支付的金额。然而，资产定价一词在金融经济学中经常用于描述非常重要的模型，即使上下文被更清楚地描述为涉及资产估值。

由于这些术语的一般含义非常相似，因此本书经常使用术语价格和定价来描述通常使用价格和定价的情况下的价值和估值。

资产的价值往往会随着时间而变化，因为人们的偏好会随着时间而变化，而现代经济满足这些偏好的能力也会随着时间而变化。农产品价格因天气等因素而变化，能源价格因经济活动等因素而变化，股票价格因未来收入和支出预测等因素而变化。一旦有关这些变化的信息被披露，价格就会对变化做出反应。霜冻对橙子收成的影响开始改变 oris done 的市场价格。

现货价格是今天交割的价格。股票或债券等资产的现货价格反映了市场对资产提供的未来收益的共识。随着时间的推移，资产价格会发生变化，因为市场对未来状况的预测会发生变化。更一般地说，资产价格会随着新信息的到来而发生变化。对股票的好消息（例如更高的收益预测）会导致价格上涨，而坏消息（例如对新产品销售的失望）会导致价格下跌。只要市场参与者合理有效地处理可用信息以形成资产的当前价格，它就会基于新信息的到来而遵循。
因此，证券价格可以被视为随机变量，随着新信息的到来而随时间变化。简而言之，资产的未来价格可以建模为等于其未来预期价值加上或减去市场参与者可获得的新信息导致的价格变化。

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

数学代写|金融数学代写Intro to Mathematics of Finance代考|ACTL20001

Posted on 2022年11月4日2022年11月4日 by statistics-lab

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我们提供的金融数学Intro to Mathematics of Finance及其相关学科的代写，服务范围广, 其中包括但不限于:

Statistical Inference 统计推断
Statistical Computing 统计计算
Advanced Probability Theory 高等概率论
Advanced Mathematical Statistics 高等数理统计学
(Generalized) Linear Models 广义线性模型
Statistical Machine Learning 统计机器学习
Longitudinal Data Analysis 纵向数据分析
Foundations of Data Science 数据科学基础

数学代写|金融数学代写Intro to Mathematics of Finance代考|The Role of Financial Assets

Financial assets are the complement to real assets. Financial assets are contractual or indirect claims. Where a real asset directly provides consumption, a financial asset is typically a claim to cash flows and is therefore an indirect claim on consumption. Chapter 2 introduces two major types of financial assets: bonds (as well as other fixed income securities) and stocks.
Bonds and stocks are financial assets. There is one other major and important type of financial asset: financial derivatives. Financial derivatives are financial contracts involving two parties: a buyer or long position, and a seller or short position. Each contract represents a zero-sum game wherein the buyer’s gain is the seller’s loss and vice versa. Finally, the derivative’s cash flows (payoffs) depend on the uncertain price of the contract’s underlying asset at a specified future date.

Accordingly, financial derivatives differ from traditional stocks and bonds in key ways. Financial derivatives are contracts between two (and in a few cases more) parties. The payoff(s) between the parties is derived from (hence the name derivative) or determined by the value of a specified asset that underlies the derivative. For example, an investor who holds bonds of an airline company may enter into a financial derivative with an investment bank that requires the investor to make a series of fixed payments to the bank in return for a promise by the bank that if the airline company defaults on the bonds the bank will make a large cash payment to the company to offset the losses due to the default. In this case, the investor is using a financial derivative to buy financial protection against default from an investment bank.

Financial securities and other financial assets are part of the financial system of a modern economy. Figure $1.1$ illustrates the role of the financial system as a conduit between people and the real assets that meet their needs and desires for consumption.

Financial securities, financial markets, financial institutions, corporations, and even governments are simply concepts in our minds that help organize and structure a society by determining who has the rights to the benefits generated by real assets. Corporations, governments, and other institutions do not produce or consume goods – people do. People are more efficient at producing goods and services, and benefiting from those good and services, when they organize themselves using concepts such as corporations, unions, governments, financial institutions, and so forth. The hallmark of societies with highly successful economies is that they have well-developed financial systems and institutions.

数学代写|金融数学代写Intro to Mathematics of Finance代考|The Roles of Financial Mathematics and Financial Derivatives

Financial mathematics has provided powerful tools for the development, understanding, and use of the wide spectrum of innovative financial products that have exploded in availability since the 1970s. These financial products have tremendously expanded our abilities to exchange, manage, control, and understand economic risk. Economic risk is intangible and tends to be difficult to understand. Yet it is clear that those economies that develop the greatest skills and tools for dealing with economic risk are best able to harness the incredible power of economic trade and growth to meet the needs and wants of a large society. Financial mathematics lies at the heart of that past success and our ability to create future opportunities.
For example, a large operating firm such as an airline company faces a number of huge economic uncertainties regarding their revenues and expenses. What effects will fuel costs, labor costs, and financing costs have on their profitability and, ultimately, the firm’s ability to continue to provide transportation services? What factors will determine the revenues for forthcoming quarters and years? Will exchange rates and airplane prices change in directions that will prevent the airline company from purchasing new equipment to replace aging aircraft or to open new routes?

Financial derivatives can help the airline company control for risks external to the firm such as changing energy costs, interest rates, and exchange rates. By offsetting or hedging the effects of these otherwise uncontrollable external variables, the company can focus their attention on dealing with those matters over which they have direct control: operating their firm with efficiency, safety, and high-quality service.

Financial derivatives have also played roles in creating or exacerbating financial crises at the international, individual investor, and firm levels. Clearly, derivatives are powerful tools that when used improperly can be as damaging as they are beneficial when used properly.
The key to effective management of financial risk is effective valuation. In finance, valuation of assets in general and financial derivatives in particular involves valuing prospective cash flows based on the timing of those cash flows and their risk. A good definition of finance is that it is the economics of time and risk. This chapter begins with highly simplified examples that ignore the effects of the timing of cash flows and the risk of cash flows on current values. Then the analyses will be expanded to include the time value of money (using forward contracts as an example) and the potential effects of risk on asset values (using options as an example).

金融数学代考

数学代写|金融数学代写Intro to Mathematics of Finance代考|The Role of Financial Assets

金融资产是实物资产的补充。金融资产是合同或间接债权。在实物资产直接提供消费的情况下，金融资产通常是对现金流的债权，因此是对消费的间接债权。第 2 章介绍了两种主要类型的金融资产：债券（以及其他固定收益证券）和股票。
债券和股票是金融资产。还有另一种主要且重要的金融资产类型：金融衍生品。金融衍生品是涉及两方的金融合约：买方或多头头寸，以及卖方或空头头寸。每份合约都代表一个零和游戏，其中买方的收益是卖方的损失，反之亦然。最后，衍生品的现金流（收益）取决于合约标的资产在特定未来日期的不确定价格。

因此，金融衍生品在关键方面不同于传统的股票和债券。金融衍生品是两方（在少数情况下甚至更多）之间的合同。各方之间的收益来自（因此称为衍生品）或由衍生品所依据的特定资产的价值确定。例如，持有航空公司债券的投资者可能与投资银行签订金融衍生品，要求投资者向银行支付一系列固定款项，以换取银行承诺如果航空公司违约对于债券，银行将向公司支付大笔现金，以弥补违约造成的损失。在这种情况下，投资者使用金融衍生品从投资银行购买金融保护以防止违约。

金融证券和其他金融资产是现代经济金融体系的一部分。数字1.1说明了金融系统作为人们与满足其消费需求和愿望的实物资产之间的管道的作用。

金融证券、金融市场、金融机构、公司甚至政府只是我们头脑中的概念，它们通过确定谁有权获得实物资产产生的利益来帮助组织和构建一个社会。公司、政府和其他机构不生产或消费商品——人们会。当人们使用诸如公司、工会、政府、金融机构等概念组织起来时，人们在生产商品和服务并从这些商品和服务中受益时会更有效率。经济高度成功的社会的标志是它们拥有发达的金融体系和机构。

数学代写|金融数学代写Intro to Mathematics of Finance代考|The Roles of Financial Mathematics and Financial Derivatives

金融数学为开发、理解和使用自 1970 年代以来广泛使用的创新金融产品提供了强大的工具。这些金融产品极大地扩展了我们交换、管理、控制和理解经济风险的能力。经济风险是无形的，往往难以理解。然而很明显，那些发展出最强大的技能和工具来应对经济风险的经济体最有能力利用经济贸易和增长的强大力量来满足大社会的需求。金融数学是过去成功的核心，也是我们创造未来机会的能力的核心。
例如，航空公司等大型运营公司在收入和支出方面面临许多巨大的经济不确定性。燃料成本、劳动力成本和融资成本对其盈利能力以及最终对公司继续提供运输服务的能力有何影响？哪些因素将决定未来季度和年度的收入？汇率和飞机价格的变化是否会阻止航空公司购买新设备来更换老化的飞机或开辟新航线？

金融衍生品可以帮助航空公司控制公司外部的风险，例如不断变化的能源成本、利率和汇率。通过抵消或对冲这些无法控制的外部变量的影响，公司可以将注意力集中在处理他们可以直接控制的事项上：以高效、安全和高质量的服务运营公司。

金融衍生品也在国际、个人投资者和公司层面造成或加剧金融危机方面发挥了作用。显然，衍生工具是强大的工具，如果使用不当，它们的破坏性可能与正确使用时的益处一样大。
有效管理财务风险的关键是有效估值。在金融领域，一般资产和金融衍生品的估值涉及根据这些现金流的时间及其风险对预期现金流进行估值。金融的一个很好的定义是它是时间和风险的经济学。本章从高度简化的示例开始，这些示例忽略了现金流时间的影响和现金流对当前价值的风险。然后将分析扩展到包括货币时间价值（以远期合约为例）和风险对资产价值的潜在影响（以期权为例）。

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