ACTL20001 - 统计代写答疑辅导

标签： ACTL20001

金融代写|金融数学代写Financial Mathematics代考|MATH3090

Posted on 2022年6月21日2022年6月21日 by statistics-lab

如果你也在怎样代写金融数学Financial Mathematics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Financial Mathematics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Financial Mathematics代写方面经验极为丰富，各种代写金融数学Financial Mathematics相关的作业也就用不着说。

我们提供的金融数学Financial Mathematics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

金融代写|金融数学代写Financial Mathematics代考|MATH3090

金融代写|金融数学代写Financial Mathematics代考|Deposits and Withdrawals: Cash Flow Problems

We now look at a class of problems where the CF Worksheet is virtually required. Cash flow problems involve situations where money is deposited and/or withdrawn at various points during the time period of interest. Many of the problems we have just finished looking at involve a sequence of deposits and withdrawals over a specified period of time. Such a sequence of deposits/withdrawals is called an investment scheme. We make no assumptions as to the interest earned by these deposits and withdrawals. They may or may not earn interest and the interest rate earned may or may not be constant.

We are often interested in a number called the internal rate of return (IRR). As you may recall from our earlier work:

The internal rate of return (IR.R) is a single interest rate which represents the effective or average rate of interest earned by the investment scheme. If the IRR is used as the (constant) rate of interest for a given investment scheme the final balance will match the actual final balance.

If we assume that the account is closed by withdrawing all funds available on the date of closure the sequence would have a net present value at inception of $\$ 0$. The IRR is the (constant) interest rate which would also result in a NPV of $\$ 0 .$ The IRR is used by managers to compare the returns on various schemes. Schemes which yield higher IRR values are deemed to be more profitable than those with smaller IRR values.

We begin with situations where we know the interest rate and then proceed to problems which involve calculating the IRR. Our first examples are just slightly more complicated versions of the problems we have been solving.

金融代写|金融数学代写Financial Mathematics代考|Fixed Term Annuities-Immediate with Constant Payments

We start our analysis with an annuity-immediate which provides $n$ payments of $\$ 1$ at the end of each period. We assume further that the effective interest rate $^{2}$ is $i$ per period. For example, if the payments are made every month, we will assume that interest is compounded monthly as well. Later we will deal with cases where the payment period and interest conversion periods are not the same. For example, an annuity with monthly payments is valued using an interest rate convertible quarterly.

We compute the value of a fixed term annuity-immediate which pays $\$ 1$ each period at the time of its inception using techniques developed in Chapter 1:
Since the first payment occurs one period after inception it has a value at inception $(t=0)$ of : $1 \cdot \frac{1}{1+i}=\frac{1}{1+i}=v$. The second payment will be delayed two periods and so has a value of $v^{2}$, and so forth. The total value at inception of this annuity is given by the sum $v+v^{2}+\cdots+v^{n}$. This expression occurs so often that it has a special symbol: $a_{\text {пn, } i}$ If we are clear about the interest rate we will write it as just $a_{\text {기. }}$. Using this notation we have:
$$
a_{\text {同, } i}=v+v^{2}+v^{3}+\cdots v^{n}=\sum_{i=1}^{n} v^{i}
$$
We can simplify this expression by using the fact that it is a geometric series. We can then use Equation $1.1$ to compute its sum. Here is Equation $1.1$
$$
a+a r+a r^{2}+a r^{3}+\cdots+a r^{n-1}=a \frac{1-r^{n}}{1-r}
$$
Comparing Equation $4.1$ with Equation $1.1$ we see that $a=v, r=v$ and have:
$$
\begin{aligned}
a_{\text {䒣 }, i} &=v+v^{2}+v^{3}+\cdots v^{n} \
&=v\left(1+v+v^{2}+\cdots+v^{n-1}\right) \
&=v \frac{1-v^{n}}{1-v} \
&=v \frac{1-v^{n}}{i v} \
&=\frac{1-v^{n}}{i}
\end{aligned}
$$
Present Value $(\mathrm{PV})$ of an Annuity Immediate of $\$ 1$ per period for $\mathrm{n}$ periods
$$
a_{\bar{n}, i}=\frac{1-v^{n}}{i}
$$
To compute the value of $a_{\text {卫 , } i}$ using the TVM keys we use the $P M T$ key. Since payments cease at the end of the term of the annuity the future value is $\$ 0$. If we enter PMT $=-1$ the PV will be reported as positive (Table 4.1).

金融代写|金融数学代写Financial Mathematics代考|Fixed Term Annuities-Due

An annuity-due consists of a sequence of payments which commence immediately and are then made at the start of each subsequent period. Recall that annuities-immediate provide payments at the end of the each period. We will use $\ddot{a}{\text {ᄁ }, i}$ for the present value of an annuity-due and $\ddot{s}{\text {万n, } i}$ for the accumulated value of an annuity-due. Since an annuity-due starts one term earlier than an annuity-immediate there is a simple relationship between the present values of these two types of annuities:
Present Value of an Annuity-Due
$$
\ddot{a}{\text {出 } i}=a{\text {氞 }, i}(1+i)
$$
Accumulated Value of an Annunity-Due
$$
\vec{s}{\text {ๆ }, i}=s{\text {ๆ, }, i}(1+i)
$$
We can also compute these values directly from their definition. For the present value (PV) of an annuity-due we have:
$$
\begin{aligned}
\ddot{a}{\text {ฑ , }, i} &=1+v+v^{2}+v^{3}+\cdots v^{n-1} \ &=\frac{1-v^{n}}{1-v} \ &=\frac{1-v^{n}}{i v} \ &=\frac{1-v^{n}}{d} \end{aligned} $$ Likewise, for the accumulated value we have: $$ \begin{aligned} s{\text {畄 }, i} &=(1+i)+(1+i)^{2}+\cdots+(1+i)^{n} \
&=(1+i) \frac{(1+i)^{n}-1}{(1+i)-1} \
&=\frac{(1+i)^{n}-1}{i v} \
&=\frac{(1+i)^{n}-1}{d}
\end{aligned}
$$
Recall that $d$ is the discount rate and that $d=\frac{i}{1+i}$.

金融数学代考

金融代写|金融数学代写Financial Mathematics代考|Deposits and Withdrawals: Cash Flow Problems

我们现在来看一类实际上需要 CF 工作表的问题。现金流问题涉及在利息期间的不同时间点存入和/或提取资金的情况。我们刚刚完成的许多问题都涉及到在特定时间段内的一系列存款和取款。这样的存款/取款顺序称为投资计划。我们不对这些存款和取款所赚取的利息做任何假设。他们可能会或可能不会赚取利息，所赚取的利率可能会或可能不会是恒定的。

我们经常对一个称为内部收益率 (IRR) 的数字感兴趣。您可能还记得我们早期的工作：

内部收益率 (IR.R) 是代表投资计划所赚取的有效或平均利率的单一利率。如果 IRR 用作给定投资计划的（恒定）利率，则最终余额将与实际最终余额相匹配。

如果我们假设账户在关闭之日通过提取所有可用资金而关闭，则序列在开始时将具有净现值$0. IRR 是（恒定）利率，它也将导致 NPV$0.管理人员使用内部收益率来比较各种计划的回报。产生较高 IRR 值的计划被认为比具有较小 IRR 值的计划更有利可图。

我们从知道利率的情况开始，然后处理涉及计算 IRR 的问题。我们的第一个示例只是我们一直在解决的问题的稍微复杂的版本。

金融代写|金融数学代写Financial Mathematics代考|Fixed Term Annuities-Immediate with Constant Payments

我们从立即年金开始我们的分析，它提供n的付款$1在每个时期结束时。我们进一步假设实际利率2是一世每个时期。例如，如果每月付款，我们将假设利息也是每月复利的。稍后我们将处理付款期限和利息转换期限不同的情况。例如，按月支付的年金使用可按季度兑换的利率进行估值。

我们计算一个固定期限年金立即支付的价值$1使用第 1 章开发的技术在其开始时的每个时期：
由于第一次付款发生在开始后一个时期，因此它在开始时具有价值(吨=0)的：1⋅11+一世=11+一世=在. 第二次付款将延迟两个周期，因此价值为在2，等等。该年金开始时的总价值由总和给出在+在2+⋯+在n. 这个表达式经常出现，以至于它有一个特殊的符号：п一个星期一，一世如果我们清楚利率，我们将把它写成기一个活力。 . 使用这个符号我们有：

同一个同, 一世=在+在2+在3+⋯在n=∑一世=1n在一世
我们可以利用它是一个几何级数这一事实来简化这个表达式。然后我们可以使用方程1.1计算其总和。这是方程1.1

一个+一个r+一个r2+一个r3+⋯+一个rn−1=一个1−rn1−r
比较方程4.1用方程1.1我们看到一个=在,r=在并且有：

䒣一个䒣 ,一世=在+在2+在3+⋯在n =在(1+在+在2+⋯+在n−1) =在1−在n1−在 =在1−在n一世在 =1−在n一世
目前的价值(磷在)立即年金$1每期为n时期

一个n¯,一世=1−在n一世
计算值卫一个卫 , 一世使用我们使用的 TVM 键磷米吨钥匙。由于在年金期限结束时付款停止，因此未来价值为$0. 如果我们输入 PMT=−1PV 将被报告为正值（表 4.1）。

金融代写|金融数学代写Financial Mathematics代考|Fixed Term Annuities-Due

年金到期包括一系列付款，这些付款立即开始，然后在每个后续期间的开始时进行。回想一下，annuities-immediate 在每个期末提供付款。我们将使用ᄁ一个¨ᄁ ,一世年金到期的现值和万s¨万n, 一世为年金到期的累计值。由于年金到期比立即年金早一个学期开始，因此这两种年金的现值之间存在简单的关系：
年金到期的现值

出氞一个¨出一世=一个氞 ,一世(1+一世)
年金到期累积值

ๆๆs→其他 ,一世=s其他， ,一世(1+一世)
我们也可以直接从它们的定义中计算这些值。对于应付年金的现值 (PV)，我们有：

ฑ一个¨n , ,一世=1+在+在2+在3+⋯在n−1 =1−在n1−在 =1−在n一世在 =1−在nd同样，对于累积值，我们有：

畄s畄 ,一世=(1+一世)+(1+一世)2+⋯+(1+一世)n =(1+一世)(1+一世)n−1(1+一世)−1 =(1+一世)n−1一世在 =(1+一世)n−1d
回顾d是贴现率和d=一世1+一世.

金融代写|金融数学代写Financial Mathematics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

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

金融代写|金融数学代写Financial Mathematics代考|MATHS 1009

Posted on 2022年6月21日2022年6月21日 by statistics-lab

如果你也在怎样代写金融数学Financial Mathematics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Financial Mathematics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

金融代写|金融数学代写Financial Mathematics代考|MATHS 1009

金融代写|金融数学代写Financial Mathematics代考|Unknown Time

Problems involving an unknown value of $t$ often lead to equations which can only be solved approximately using numerical methods. The TI BA II Plus, Excel, and MAPLE all have approximation algorithms. Of the three, Excel seems to be the least accurate. While the differences are slight, for large amounts of money, Excel might lead to significant errors.

A common problem concerns replacing a sequence of payments (called an annuity – to be discussed in greater detail later) with a single payment equal to the sum of the numerical values of the other payments. The problem determine the time at which the single payment is to be made in order that the sequence of payments is equal in value to the single payment.

As our first example, suppose that payments in the amounts $p_{1}, p_{2}, p_{3}, \ldots, p_{n}$ are made at times $t_{1}, t_{2}, t_{3}, \ldots, t_{n}$. Suppose further that interest is compounded at an effective rate of $i$ per period. We are to find a time, $t$, at which a single payment of $p_{1}+p_{2}+p_{3}+\cdots+p_{n}=\sum_{k=1}^{n} p_{k}$ has the same value as this annuity. We compute our equation of value at the inception of the transaction resulting in the following equation of value. The left side of Equation $3.4$ is the value of the single payment at time $t=0$, while the right side is the value of the sequence of payments at this same time
$$
\left(\sum_{k=1}^{n} p_{k}\right) v^{t}=\sum_{k=1}^{n} p_{k} v^{t_{k}}
$$
Solving this equation for $t$ yields the following equation:
$$
t=\frac{\ln \left(\frac{\sum_{k=1}^{n} \mathbf{p}{k} v^{t{k}}}{\left(\sum_{k=1}^{n} p_{k}\right)}\right)}{\ln (v)}
$$
Example 3.5 Suppose that the rate of interest is a nominal $5.04 \%$ annually and that interest is compounded monthly. Find the time at which a single payment of $\$ 1000$ is equivalent in value to the following sequence of payments displayed in Table $3.8$.

金融代写|金融数学代写Financial Mathematics代考|Doubling Time

We are often interested in determining how long it will take a fixed deposit to increase in value by a given amount. In the case of compound interest the time required for a deposit to increase by any given factor is independent of the amount which is currently on deposit. The amount of time a given amount takes to double in value is known as the doubling time.

We begin by deriving a formula for the doubling time as a function of the rate of interest in the case that interest is compounded.

We have $A(t)=P_{0} a(t)=P_{0}(1+i)^{t}$. Starting with any initial time, $t_{0}$ we want to find $t$ so that $A(t)=2 A\left(t_{0}\right)$. This gives us
$$
\begin{aligned}
P_{0} a(t) &=2 P_{0} a\left(t_{0}\right) \
P_{0}(1+i)^{t} &=2 P_{0}(1+i)^{t_{0}} \
(1+i)^{t} &=2(1+i)^{t_{0}}
\end{aligned}
$$
Taking logarithms of both sides we obtain
$$
\begin{aligned}
t \ln (1+i) &=\ln (2)+t_{0} \ln (1+i) \
\left(t-t_{0}\right) &=\frac{\ln (2)}{\ln (1+i)}
\end{aligned}
$$
An easy generalization of this gives the following formula for the time required for an investment to increase by a factor of $k$ at a given rate of interest $i$ is given by
$$
\Delta t=\frac{\ln (k)}{\ln (1+i)}
$$
We can use Equation $2.17$ to obtain this same solution. We have
$$
\begin{aligned}
&A\left(t_{1}\right)=k A\left(t_{0}\right)=\frac{a\left(t_{1}\right)}{a\left(t_{0}\right)} A\left(t_{0}\right) \
&\frac{a\left(t_{1}\right)}{a\left(t_{0}\right)}=k=\frac{(1+i)^{t_{1}}}{(1+i)^{t_{0}}}=(1+i)^{t_{1}-t_{0}} \
&\ln (k)=\left(t_{1}-t_{0}\right) \ln (1+i)
\end{aligned}
$$
We then solve for $\left(t_{1}-t_{0}\right)$ as before.
Time for an Investment to Increase by a Factor of $k$
$$
t_{k}=\Delta t=\frac{\ln (k)}{\ln (1+i)}
$$
To use the TI BA II Plus, we use the TVM keys. Use $P V=1, F V=-k$ and solve for $n$. We enter PV as negative because of the TI BA II Plus convention that PV and FV must have opposite signs.

金融代写|金融数学代写Financial Mathematics代考|Finding the Rate of Interest

Problems involving an unknown rate of interest often require an approximation technique. We work several examples to give the flavor of the sorts of issues which arise. As usual the TVM keys will be our best friend!

Example 3.9 At what nominal rate of annual interest, convertible monthly, will $\$ 456$ accumulate to $\$ 500$ in three years?

Solution: We will use months as our unit of measurement to avoid an interestconversion problem.
We use Equation $2.25$ and the TI-30XS
$$
F V=P V\left(1+\frac{i^{(m)}}{m}\right)^{m t}
$$

We have $m t=36$ (months), $P V=456$, and $F V=500$
$$
\begin{gathered}
500=456\left(1+\frac{i^{(12)}}{12}\right)^{36} \
\left(1+\frac{i^{(12)}}{12}\right)=\left(\frac{500}{456}\right)^{\frac{1}{56}} \
\frac{i^{(12)}}{12}=\left(\frac{500}{456}\right)^{\frac{1}{36}}-1=.002562034
\end{gathered}
$$
Thus $i^{(12)}=12(.002562034)=0.03074440799=3.0744 \%$.

金融数学代考

金融代写|金融数学代写Financial Mathematics代考|Unknown Time

涉及未知值的问题吨通常会导致只能使用数值方法近似求解的方程。TI BA II Plus、Excel 和 MAPLE 都具有近似算法。在这三个中，Excel 似乎是最不准确的。虽然差异很小，但对于大量资金，Excel 可能会导致重大错误。

一个常见的问题涉及将一系列支付（称为年金 – 稍后将更详细地讨论）替换为等于其他支付的数值总和的单一支付。该问题确定进行单次付款的时间，以使付款顺序与单次付款的价值相等。

作为我们的第一个示例，假设支付金额为p1,p2,p3,…,pn有时制作吨1,吨2,吨3,…,吨n. 进一步假设利息以有效利率复利一世每个时期。我们要找时间，吨，其中单次支付p1+p2+p3+⋯+pn=∑ķ=1npķ与此年金具有相同的价值。我们在交易开始时计算我们的价值等式，得到以下价值等式。方程的左边3.4是当时单笔付款的价值吨=0，而右边是同时支付序列的值

(∑ķ=1npķ)在吨=∑ķ=1npķ在吨ķ
求解这个方程吨产生以下等式：
$$
t=\frac{\ln \left(\frac{\sum_{k=1}^{n} \mathbf{p} {k} v^{t {k}}}{\ left(\sum_{k=1}^{n} p_{k}\right)}\right)}{\ln (v)}
$$
示例 3.5 假设利率是名义利率5.04%每年，并且该利息每月复利。查找单次付款的时间$1000价值等同于表中显示的以下付款顺序3.8.

金融代写|金融数学代写Financial Mathematics代考|Doubling Time

我们通常对确定定期存款需要多长时间才能使价值增加给定数量感兴趣。在复利的情况下，存款增加任何给定因素所需的时间与当前存款的金额无关。给定数量的价值翻倍所需的时间称为翻倍时间。

我们首先推导在复利情况下作为利率函数的倍增时间的公式。

我们有一个(吨)=磷0一个(吨)=磷0(1+一世)吨. 从任何初始时间开始，吨0我们想找到吨以便一个(吨)=2一个(吨0). 这给了我们

磷0一个(吨)=2磷0一个(吨0) 磷0(1+一世)吨=2磷0(1+一世)吨0 (1+一世)吨=2(1+一世)吨0
两边取对数我们得到

吨ln⁡(1+一世)=ln⁡(2)+吨0ln⁡(1+一世) (吨−吨0)=ln⁡(2)ln⁡(1+一世)
对此的简单概括给出了以下公式，即投资增加 1 倍所需的时间ķ在给定的利率一世是（谁）给的

Δ吨=ln⁡(ķ)ln⁡(1+一世)
我们可以使用方程2.17以获得相同的解决方案。我们有

一个(吨1)=ķ一个(吨0)=一个(吨1)一个(吨0)一个(吨0) 一个(吨1)一个(吨0)=ķ=(1+一世)吨1(1+一世)吨0=(1+一世)吨1−吨0 ln⁡(ķ)=(吨1−吨0)ln⁡(1+一世)
然后我们解决(吨1−吨0)和以前一样。
投资增加的时间ķ

吨ķ=Δ吨=ln⁡(ķ)ln⁡(1+一世)
要使用 TI BA II Plus，我们使用 TVM 键。利用磷在=1,F在=−ķ并解决n. 由于 TI BA II Plus 约定 PV 和 FV 必须具有相反的符号，我们将 PV 输入为负数。

金融代写|金融数学代写Financial Mathematics代考|Finding the Rate of Interest

涉及未知利率的问题通常需要近似技术。我们用几个例子来说明出现的各种问题。像往常一样，TVM 键将是我们最好的朋友！

例 3.9 以多少名义年利率，每月可兑换，将$456积累到$500三年内？

解决方案：我们将使用月份作为我们的计量单位，以避免利息转换问题。
我们使用方程2.25和 TI-30XS

F在=磷在(1+一世(米)米)米吨

我们有米吨=36（月），磷在=456，和F在=500

500=456(1+一世(12)12)36 (1+一世(12)12)=(500456)156 一世(12)12=(500456)136−1=.002562034
因此一世(12)=12(.002562034)=0.03074440799=3.0744%.

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

金融代写|金融数学代写Financial Mathematics代考|Solving Problems in Interest

Posted on 2022年6月21日2022年6月21日 by statistics-lab

如果你也在怎样代写金融数学Financial Mathematics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Financial Mathematics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

金融代写|金融数学代写Financial Mathematics代考|Solving Problems in Interest

金融代写|金融数学代写Financial Mathematics代考|Measuring Time Periods: Simple Interest and Fractional Time Periods

In all of our examples so far we have simply stated a value of $t$. In most practical problems, time is expressed in terms of a date of inception and a date of termination. Thus, we might have an investment which is initiated on January 12, 2012, and terminates on September 11, 2014. In order to use our formulas we need to convert these dates into a time period. There are two methods in common use for calculating the value of $t$ corresponding to the time between two specified dates. This method uses the exact number of days and calculates $t$ based on the assumption that there are 365 days in a year.

This method assumes that each month has 30 days, resulting in a year of 360 days. If our two dates are $M_{1}, D_{1}, Y_{1}$ and $M_{2}, D_{2}, Y_{2}$ the formula for computing the number of days between two dates are $M_{1}, D_{1}, Y_{1}$ and $M_{2}, D_{2}, Y_{2}$ days for ordinary simple interest is
$$
360\left(Y_{2}-Y_{1}\right)+30\left(M_{2}-M_{1}\right)+\left(D_{2}-D_{1}\right)
$$
In both of these two cases, the easiest way to calculate the number of days between two dates is to use the DATE Worksheet on the TI BA II Plus.
We invoke this worksheet by keying in 2ND DATE (it’s above the 1). As for other worksheets, we move from entry to entry using the $\uparrow, \downarrow$. Dates are entered as mm.ddyy. Thus, March 5,2007 , would be entered as $03.0507$. It will display as $3.0507$ until you hit the ENTER key, at which point it will display as 03-05-2007.

There are four variables which are accessed by using the $\uparrow$ or $\downarrow$ keys. DT1 is the first date and is always assumed to be prior to DT2, the second date. DBD is the number of days between the two dates. The final variable is either ACT (for exact number of days) or 360 (using the ordinary simple interest method). This last variable is toggled by using 2ND SET. Here’s how to compute the number of days between March 3,2008 , and October 5,2009 , using both the ordinary simple interest rule and the exact number of days (Table $3.1$ ).
$2 N D$ Date Since there are 581 days between March 3,2014 , and October 5,2015 , using the exact method we obtain $t=\frac{581}{365}=1.592$. For the ordinary simple interest method we have $t=\frac{572}{360}=1.588$.

金融代写|金融数学代写Financial Mathematics代考|Fractional Time Periods

The equation $F V=P V(1+i)^{t}$ can be evaluated directly for any value of $t$. For example, if we want the value of an investment of $\$ 5,456$ at six and a third years earning compound annual interest of $5.6 \%$, we have (Table $3.2$ )
$$
F V=5456(1.056)^{6 \frac{1}{3}}=5456 \cdot 1.412119603=\$ 7,704.52
$$

This method gives the exact value of the accumulated amount over any time period. In some cases the interest earned over the fractional portion of a time period is computed using simple interest. Historically this practice developed due to the difficulties involved in computing the value of $(1+i)^{t}$ if $t$ is not an integer.

As an example, simple interest is sometimes used to compute the interest due at inception for a home mortgage. Home mortgages are timed to start on the first day of the month subsequent to the date of inception of the loan. If you close on a home loan on a day other than the first of the month, you are assessed simple interest on the loan amount for the number of days remaining in that month at the inception of the loan. After that, interest is computed using compound interest. A loan which closes on May 12 will be assessed interest on the remaining days in May as part of the closing costs. The loan will begin June 1 , and the first regular payment will be due at the end of June.
We can calculate the FV of an account using simple interest over the fractional period as follows. Suppose that $t=n+q$ where $n$ is a whole number and $q$ represents the fractional portion of time $(0<q<1)$ we would compute the $F V$ using simple interest for the fractional portion of time as
$$
F V=P V(1+i)^{n}(1+i q)
$$

金融代写|金融数学代写Financial Mathematics代考|Equations of Value at any Time

We will often need to compare two or more amounts of money at different points in time. If the rate of interest is not zero, this is only makes sense if we compute the values of each amount at a common point in time. This common date is called the comparison date. We can choose any date we like. An amount which occurs prior to the comparison date must be accumulated to that date while an amount which occurs after the comparison date must be discounted. It is often helpful to use a number line to place each amount at the appropriate time. This is called a time diagram. The equation which equates two different methods of computing the value of a transaction is called the equation of value. An amount which will be available $n$ periods after the comparison date is discounted ${ }^{1}$ by the factor $v^{n}$ Recall that $v=\frac{1}{1+i}$. An amount which will be available $m$ periods prior to the comparison date is accumulated by a factor of $(1+i)^{m}$.

Example 3.3 In return for the promise of a payment of $\$ 600$ at the end of eight years, a person agrees to pay $\$ 100$ now, $\$ 200$ at the end of five years, and to make one further payment at the end of ten years. What is the required final payment if the nominal rate of annual interest is $6.2 \%$ compounded semiannually?

Solution. We will measure time in six-month intervals. This gives us $i=\frac{.062}{2}=.031$ and $v=\frac{1}{1.031}$. We will use the inception date as our comparison date. We then have the following situation for each amount of money using $X$ for the (unknown) value of the final payment. See Figure 3.1. $\$ 600$ : is not available for eight years or sixteen periods. Its value at inception is thus $600 v^{16}$
$\$ 100$ : is available now. Its value is thus 100
$\$ 200$ : is not available for five years or ten periods. Its value at inception is thus $200 v^{10}$
$\$ X$ : is not available for ten years or twenty periods. Its value at inception is thus $x v^{20}$

金融数学代考

金融代写|金融数学代写Financial Mathematics代考|Measuring Time Periods: Simple Interest and Fractional Time Periods

到目前为止，在我们所有的例子中，我们都简单地陈述了一个值吨. 在大多数实际问题中，时间以开始日期和终止日期的形式表示。因此，我们可能有一项投资于 2012 年 1 月 12 日开始，并于 2014 年 9 月 11 日终止。为了使用我们的公式，我们需要将这些日期转换为时间段。有两种常用的计算价值的方法吨对应于两个指定日期之间的时间。此方法使用确切的天数并计算吨假设一年有 365 天。

此方法假设每个月有 30 天，因此一年有 360 天。如果我们的两个日期是米1,D1,是1和米2,D2,是2计算两个日期之间天数的公式是米1,D1,是1和米2,D2,是2普通单利的天数是

360(是2−是1)+30(米2−米1)+(D2−D1)
在这两种情况下，计算两个日期之间天数的最简单方法是使用 TI BA II Plus 上的日期工作表。
我们通过键入 2ND DATE（它在 1 之上）来调用此工作表。至于其他工作表，我们使用↑,↓. 日期输入为 mm.ddyy。因此， March 5,2007 将输入为03.0507. 它将显示为3.0507直到您按下 ENTER 键，此时它将显示为 03-05-2007。

有四个变量可以通过使用↑或者↓键。DT1 是第一个日期，并且始终假定在 DT2 之前，即第二个日期。DBD 是两个日期之间的天数。最后一个变量是 ACT（精确的天数）或 360（使用普通的单利法）。最后一个变量是通过使用 2ND SET 来切换的。下面是如何使用普通的单利规则和确切的天数计算 2008 年 3 月 3 日到 2009 年 10 月 5 日之间的天数（表3.1 ).
2ñD日期由于从 2014 年 3 月 3 日到 2015 年 10 月 5 日之间有 581 天，使用我们得到的确切方法吨=581365=1.592. 对于普通的单利方法，我们有吨=572360=1.588.

金融代写|金融数学代写Financial Mathematics代考|Fractional Time Periods

方程F在=磷在(1+一世)吨可以直接评估任何值吨. 例如，如果我们想要一项投资的价值$5,456在第六年和第三年赚取复合年利息5.6%, 我们有 (表3.2 )

F在=5456(1.056)613=5456⋅1.412119603=$7,704.52

这种方法给出了任何时间段内累计金额的准确值。在某些情况下，在一个时间段的小数部分获得的利息是使用单利计算的。从历史上看，这种做法是由于计算价值所涉及的困难而发展起来的(1+一世)吨如果吨不是整数。

例如，有时使用单利来计算房屋抵押贷款开始时到期的利息。房屋抵押贷款的时间安排在贷款开始日期之后的一个月的第一天开始。如果您在当月第一天以外的一天结束房屋贷款，您将获得贷款开始时该月剩余天数的单利。之后，使用复利计算利息。5 月 12 日结束的贷款将在 5 月剩余的日子里评估利息，作为结束成本的一部分。贷款将于 6 月 1 日开始，第一笔定期付款将于 6 月底到期。
我们可以在小数期间使用单利计算账户的 FV，如下所示。假设吨=n+q在哪里n是一个整数，并且q表示时间的小数部分(0<q<1)我们将计算F在对部分时间使用单利作为

F在=磷在(1+一世)n(1+一世q)

金融代写|金融数学代写Financial Mathematics代考|Equations of Value at any Time

我们经常需要在不同的时间点比较两个或更多的金额。如果利率不为零，那么只有在我们计算每个金额在同一时间点的值时，这才有意义。这个共同的日期称为比较日期。我们可以选择任何我们喜欢的日期。发生在比较日期之前的金额必须累积到该日期，而发生在比较日期之后的金额必须贴现。使用数字线在适当的时间放置每个数量通常很有帮助。这称为时间图。将计算交易价值的两种不同方法等同起来的方程式称为价值方程式。将可用的金额n比较日期之后的期间被贴现1因数在n回顾在=11+一世. 将可用的金额米比较日期之前的期间由一个因素累加(1+一世)米.

示例 3.3 作为回报承诺支付$600八年后，一个人同意支付$100现在，$200在五年结束时，并在十年结束时再支付一次。如果名义年利率为6.2%每半年复利一次？

解决方案。我们将每隔六个月测量一次时间。这给了我们一世=.0622=.031和在=11.031. 我们将使用开始日期作为比较日期。然后，对于使用的每个金额，我们有以下情况X用于最终付款的（未知）价值。请参见图 3.1。$600: 八年或十六期不可用。因此，它在开始时的价值是600在16
$100: 现在可以了。因此其值为 100
$200: 五年或十期不可用。因此，它在开始时的价值是200在10
$X: 十年或二十年不可用。因此，它在开始时的价值是X在20

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

金融代写|金融数学代写Financial Mathematics代考|Financial Mathematics for Actuarial Science

Posted on 2022年6月21日2022年6月21日 by statistics-lab

如果你也在怎样代写金融数学Financial Mathematics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Financial Mathematics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

金融代写|金融数学代写Financial Mathematics代考|Financial Mathematics for Actuarial Science

金融代写|金融数学代写Financial Mathematics代考|Other Accumulation Functions

In theory any function $a(t)$ which is continuous, increasing, and satisfies $a(0)=1$ can serve as an accumulation function. We now consider a few problems involving non-standard accumulation functions. While these don’t show up in “real life,” they do appear on the actuarial exams.

Example 2.15 Suppose that $a(t)=.1 t^{2}+b$. The only investment is $\$ 300$ made at time $t=1$. What is the accumulated value of the investment at time $t=10 ?$

Solution: Since it is always required that $a(0)=1$ we must have $b=1$, so $a(t)=.1 t^{2}+1$. We are interested in computing $A(t)$ at $t=10$. We know that $A(t)=P_{0} a(t)$ and so need to find a value for $P_{0}$. The trick is to pretend that the $\$ 300$ is not the only investment but rather the value of the investment at time $t=1$. We then have
$$
\begin{gathered}
300=A(1)=P_{0} \cdot a(1) \
P_{0}=\frac{300}{a(1)}=\frac{300}{1.1}
\end{gathered}
$$
We then have:
$$
A(10)=P_{0} \cdot a(10)=\frac{300}{a(1)} \cdot a(10)=\frac{300}{1.1} \cdot 11=\$ 3,000
$$
The technique used in this example can be generalized to any problem involving an accumulation function. Suppose we know $A\left(t_{1}\right)$ and want to find $A\left(t_{2}\right)$. Since we do not know the value of $P_{0}$, we can’t compute $A\left(t_{2}\right)$ directly. However, we do know that $A\left(t_{1}\right)=P_{0} \cdot a\left(t_{1}\right)$ and $A\left(t_{2}\right)=P_{0} \cdot a\left(t_{1}\right)$ hence $P_{0}=\frac{A\left(t_{1}\right)}{a\left(t_{1}\right)}=\frac{A\left(t_{2}\right)}{a\left(t_{2}\right)}$. We solve this equation for $A\left(t_{2}\right)$ to obtain a very useful formula which computes the amount function at any time based on the amount at any other time.

金融代写|金融数学代写Financial Mathematics代考|Present and Future Value: Equations of Value

We now have formulas which compute the accumulated amount (also called the future value or $F V$ ) of $P_{0}$ at an interest rate of $i$ over $t$ periods. The amount $P_{0}$ is usually called the present value (PV). In many cases, we must answer the converse question: how much must be deposited at an interest rate of $i$ so that it will accumulate to a given future amount $F V$ ? As an example, we might be putting money away toward the purchase of a car or a down payment on a house.

In some cases, money is being saved toward the payment of an annual tax or other obligation. In Chapter 9 we will discuss ways in which investors seek to ensure that the funds saved will suffice to cover a set of future obligations. Banks often require mortgage holders to contribute to an account to provide funds for the payment of property taxes. These sorts of accounts are known as escrow accounts.

We begin with the calculation of the relation between the present and future value of a single deposit for one year. We want to find the amount $P V$ which is sufficient to accumulate to an amount $F V$ at the end of one year. In one period a deposit of $P V$ will accumulate to $P V(1+i)$. We want $P V(1+i)=F V$ and so $P V=F V \cdot \frac{1}{1+i}$. The term $\frac{1}{1+i}$ is called the discount factor and is referenced by the symbol $v$. Thus:
$$
\begin{aligned}
&v=\frac{1}{1+i} \
&i=\frac{1-v}{v}=\frac{1}{v}-1
\end{aligned}
$$
Over $t$ periods we have one set of formulas for simple interest and a second for compound interest. In each case, we merely solved the first equation (relating FV to PV) for the present value in order to get an equation with computes $\mathrm{PV}$ in terms of FV.
Present and Future Value: Simple Interest
$$
\begin{aligned}
&F V=P V(1+i t) \
&P V=F V \frac{1}{1+i t}
\end{aligned}
$$ Present and Future Value: Compound Interest
$$
\begin{aligned}
&F V=P V(1+i)^{t} \
&P V=F V(1+i)^{-t}=F V v^{t}
\end{aligned}
$$
Note that in each case there are four variables (PV, FV, $i$ and $t$ ). Knowing any three of the four enables us to solve for the fourth variable. As we have seen the TI BA II Plus will perform these calculations. The examples below show how this can be done.

金融代写|金融数学代写Financial Mathematics代考|Nominal and Effective Rates of Interest

We now consider cases where interest is paid either more or less often than the period used for measuring time. In many cases an annual interest rate is quoted even though interest may be compounded (or converted) more (or less) often than once per year. In order to compare such rates we must convert to a common time unit. This is typically a year, but in some problems another unit will be most useful. If the unit of measurement is one year we are computing what is called the effective annual rate of interest.

Example 2.26 Mammoth Credit offers a credit card with a monthly interest rate of $1.9 \%$. Cards $\mathrm{R}$ Us offers a card with an annual rate of $23 \%$. Which card is the better buy?

Solution: This may seem simple-multiply $1.9$ times 12 to obtain $22.8$. Hence the Mammoth card is the better deal. This is exactly what Mammoth wants us to do. In fact, Mammoth is allowed to describe their card as carrying an “APR” (Annual Percentage Rate) of $22.8 \%$. However, this simple calculation only works if interest is computed using simple interest! Since all credit card interest is computed as compound interest, we need to convert the Mammoth rate to its effective annual interest rate. To do that we need an annual rate of interest, $i$, which is equivalent to our monthly rate of $1.9 \%$. If we invest $\$ 1$ at an annual rate of interest of $i$ we will have $1+i$ after one year. On the other hand, $\$ 1$ invested at $1.9 \%$ per month for twelve months will accumulate to $1 \cdot(1.019)^{12}=1.25340$. We thus we must have $1+i=1.25340$ so that $i=.25340=25.34 \%$. The Mammoth Card has an effective annual interest rate of $25.34 \%$ making the Cards $\mathrm{R}$ Us card at $23 \%$ the better buy.

金融数学代考

金融代写|金融数学代写Financial Mathematics代考|Other Accumulation Functions

理论上任何函数一个(吨)是连续的，递增的，满足一个(0)=1可以作为累积函数。我们现在考虑一些涉及非标准累积函数的问题。虽然这些不会出现在“现实生活”中，但它们确实出现在精算考试中。

例 2.15 假设一个(吨)=.1吨2+b. 唯一的投资是$300当时制作的吨=1. 当时投资的累计价值是多少吨=10?

解决方案：因为总是要求一个(0)=1我们必须有b=1，所以一个(吨)=.1吨2+1. 我们对计算感兴趣一个(吨)在吨=10. 我们知道一个(吨)=磷0一个(吨)所以需要找到一个值磷0. 诀窍是假装$300不是唯一的投资，而是投资当时的价值吨=1. 然后我们有

300=一个(1)=磷0⋅一个(1) 磷0=300一个(1)=3001.1
然后我们有：

一个(10)=磷0⋅一个(10)=300一个(1)⋅一个(10)=3001.1⋅11=$3,000
这个例子中使用的技术可以推广到任何涉及累积函数的问题。假设我们知道一个(吨1)并想找到一个(吨2). 因为我们不知道磷0，我们无法计算一个(吨2)直接地。然而，我们确实知道一个(吨1)=磷0⋅一个(吨1)和一个(吨2)=磷0⋅一个(吨1)因此磷0=一个(吨1)一个(吨1)=一个(吨2)一个(吨2). 我们解这个方程为一个(吨2)获得一个非常有用的公式，该公式根据任何其他时间的金额计算任何时间的金额函数。

金融代写|金融数学代写Financial Mathematics代考|Present and Future Value: Equations of Value

我们现在有计算累积金额的公式（也称为未来值或F在）的磷0利率为一世超过吨期间。数量磷0通常称为现值（PV）。在许多情况下，我们必须回答相反的问题：在利率为一世这样它将累积到给定的未来数量F在? 例如，我们可能会将钱存入购买汽车或房屋的首付。

在某些情况下，省钱是为了支付年度税或其他义务。在第 9 章中，我们将讨论投资者寻求确保节省的资金足以支付一系列未来义务的方式。银行经常要求抵押贷款持有人向一个账户供款，以提供资金支付房产税。这些类型的账户被称为托管账户。

我们从计算单笔存款一年的现值和未来值之间的关系开始。我们想找到金额磷在足以累积到一定数额F在在一年结束时。在一个时期内存款磷在会累积到磷在(1+一世). 我们想要磷在(1+一世)=F在所以磷在=F在⋅11+一世. 期限11+一世称为折扣因子并由符号引用在. 因此：

在=11+一世一世=1−在在=1在−1
超过吨在此期间，我们有一套用于单利的公式和一套用于复利的公式。在每种情况下，我们只求解当前值的第一个方程（将 FV 与 PV 关联），以便得到一个计算方程磷在在 FV 方面。
现值和未来值：单利

F在=磷在(1+一世吨) 磷在=F在11+一世吨现值和未来值：复利

F在=磷在(1+一世)吨磷在=F在(1+一世)−吨=F在在吨
请注意，在每种情况下，都有四个变量（PV、FV、一世和吨）。知道这四个中的任何三个使我们能够求解第四个变量。正如我们所见，TI BA II Plus 将执行这些计算。下面的例子展示了如何做到这一点。

金融代写|金融数学代写Financial Mathematics代考|Nominal and Effective Rates of Interest

我们现在考虑支付利息的频率高于或低于用于衡量时间的期限的情况。在许多情况下，即使利息的复利（或转换）多于（或少于）每年一次，也会引用年利率。为了比较这些比率，我们必须转换成一个共同的时间单位。这通常是一年，但在某些问题中，另一个单元将是最有用的。如果计量单位是一年，我们正在计算所谓的有效年利率。

示例 2.26 Mammoth Credit 提供一张月利率为1.9%. 牌R我们提供一张年利率为23%. 买什么卡比较好？

解决方案：这可能看起来很简单——乘法1.9乘以 12 获得22.8. 因此，猛犸卡是更好的交易。这正是猛犸希望我们做的。事实上，猛犸象被允许将他们的卡描述为带有“APR”（年度百分比率）22.8%. 但是，这种简单的计算只有在使用单利计算利息时才有效！由于所有信用卡利息都是按复利计算的，因此我们需要将猛犸象利率转换为其有效年利率。为此，我们需要年利率，一世，这相当于我们的月费率1.9%. 如果我们投资$1年利率为一世我们将有1+一世一年后。另一方面，$1投资于1.9%每月十二个月将累积到1⋅(1.019)12=1.25340. 因此我们必须有1+一世=1.25340以便一世=.25340=25.34%. 猛犸卡的有效年利率为25.34%制作卡片R我们的卡在23%更好的购买。

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

金融代写|金融数学代写Financial Mathematics代考|Measuring Interest

Posted on 2022年6月21日2022年6月21日 by statistics-lab

如果你也在怎样代写金融数学Financial Mathematics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Financial Mathematics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

金融代写|金融数学代写Financial Mathematics代考|Measuring Interest

金融代写|金融数学代写Financial Mathematics代考|The Effective Rate of Interest

Definition: The effective rate of interest for a given period is the amount of money that one unit of principal invested at the start of a particular period will earn during that one period. It is assumed that interest is paid at the end of the period in question 2 .

If the period is one year, the effective rate of interest is also referred to as the effective annual rate of interest. This is sometimes called the APR (annual percentage rate) when expressed as a percentage. An annual effective rate of .05 would correspond to an APR of $5 \%$. As we shall see, the APR reported by various consumer credit companies is not always the same as the effective annual rate. As a result, we will use the term effective annual interest rate (or effective monthly interest rate or …) except during our discussion of the Truth in Lending Laws at the end of Chapter $6 .$

We will use the symbol $i$ for the effective rate of interest. Again, unless stated otherwise, the period is assumed to be one year. In terms of $a(t)$ we have $i=a(1)-a(0)=a(1)-1$ so we have:
Effective Rate of Interest, $i_{1}$ in period 1
$$
i_{1}=a(1)-1
$$
We can express $i$ in terms of $A(t)$ as well. Since $A(t)=P_{0} a(t)$ we have:
$$
\begin{aligned}
i_{1} &=\frac{a(1)-a(0)}{a(0)} \
&=\frac{\frac{A(1)}{P_{0}}-\frac{A(0)}{P_{0}}}{\frac{A(0)}{P_{0}}} \
&=\frac{A(1)-A(0)}{A(0)}
\end{aligned}
$$
Example 2.4: A deposit of $\$ 550$ earns $\$ 45$ in interest at the end of one year.
What is the effective annual interest rate?
Solution: We use Equation $2.7$
$$
i_{1}=\frac{A(1)-A(0)}{A(0)}=\frac{595-550}{550}=.08182
$$
This would typically be expressed as a percentage: $8.182 \%$.

金融代写|金融数学代写Financial Mathematics代考|Simple Interest

The first method of computing interest we will consider assumes that $a(t)$ is a linear function of $t$. This method is known as simple interest. It was in common use prior to the advent of calculators and computers since it is very easy to compute. It is still used in a few cases, most notably in the case of fractional time periods.

In the case of simple interest we know that $a(t)$ is a linear function and so its graph is a line. We are given two points on this line so it is easy to obtain an expression for the value of $a(t)$ at any time $t$. The slope of our line is given by
$$
\frac{(1+i)-1}{1-0}=i
$$
Since the line contains the point $(0,1)$ it’s equation is given by:
Accumulation Function for Simple Interest
$$
a(t)=1+i t
$$
In the case of simple interest, the accumulation function is a linear function with slope $i$. Figure $2.1$ is a plot of the accumulation function for the case that $i=.05$ :The amount function is just as easy to compute:
Amount Function for Simple Interest
$$
A(t)=F V=P V \cdot(1+i t)
$$
This is a line with y-intercept $=P V$ and slope $=P V_{i}$.

金融代写|金融数学代写Financial Mathematics代考|Compound Interest

Simple interest computes the interest in each period based solely on the amount of the initial deposit. If there are no withdrawals during the life of the transaction, the amount on deposit increases over time while the amount of interest paid at the end of each period remains constant. As we just saw this results in a declining effective rate of interest when we use simple interest. If $\$ 400$ is deposited at $5 \%$ simple interest the amount on deposit after three years is $\$ 460$. However (using simple interest) the interest paid at the end of year 4 is still only $\$ 20$ (5\% of $\$ 400)$. If interest was paid based on the amount on deposit at the end of year 3 , the interest at the end of year 4 would be $(.05) \cdot \$ 460=\$ 23$

When the interest paid at the end of a given period is based on the accumulated value of the principal at the start of that period rather than on the amount of the original deposit we obtain what is known as compound interest.

To find the formula for the accumulation function in the case of compound interest, we compute the earned interest at the end of each period, add this to the previous balance, and use that number as the principal in computing the interest for the next period. See Table 2.1.

This leads us to believe that the following formula holds for the accumulation function in the case of compound interest
Accumulation Function: Compound Interest
$$
a(n)=(1+i)^{n}
$$
You can prove Equation $2.13$ for whole numbers using mathematical induction. We have calculated this result using only integral values for the time on deposit. If interest is only paid at integral multiples of the period, $a(n)$ is a step function:
$$
a(n)= \begin{cases}1 & 0 \leq n<1 \ 1+i & 1 \leq n<2 \ (1+i)^{2} & 2 \leq n<3 \ (1+i)^{3} & 3 \leq n<4 \ \text { etc.. } & \end{cases}
$$
If we assume that interest can be collected at any time in the life of the investment, it’s natural to extend this step function to include real $t \geq 0$. We thus obtain the function $a(t)=(1+i)^{t}$ defined for all real numbers $t$ which is the continuous, differentiable extension of Equation $2.3$
Accmulation Function for Compound Interest
$$
a(t)=(1+i)^{t}
$$

金融数学代考

金融代写|金融数学代写Financial Mathematics代考|The Effective Rate of Interest

定义：给定时期的有效利率是在特定时期开始时投资的一单位本金在该时期内将赚取的金额。假设利息在问题 2 的期末支付。

如果期限为一年，则实际利率也称为实际年利率。当以百分比表示时，这有时称为 APR（年度百分比率）。0.05 的年有效利率相当于 APR5%. 正如我们将看到的，各个消费信贷公司报告的 APR 并不总是与实际年利率相同。因此，我们将使用术语有效年利率（或有效月利率或……），除非我们在本章末尾讨论贷款法的真相6.

我们将使用符号一世为有效利率。同样，除非另有说明，否则该期限假定为一年。按照一个(吨)我们有一世=一个(1)−一个(0)=一个(1)−1所以我们有：
有效利率，一世1在第 1 期

一世1=一个(1)−1
我们可以表达一世按照一个(吨)也是。自从一个(吨)=磷0一个(吨)我们有：

一世1=一个(1)−一个(0)一个(0) =一个(1)磷0−一个(0)磷0一个(0)磷0 =一个(1)−一个(0)一个(0)
例 2.4：存款$550赚$45一年结束时的利息。
什么是有效年利率？
解决方案：我们使用方程式2.7

一世1=一个(1)−一个(0)一个(0)=595−550550=.08182
这通常表示为百分比：8.182%.

金融代写|金融数学代写Financial Mathematics代考|Simple Interest

我们将考虑的第一种计算兴趣的方法假设一个(吨)是一个线性函数吨. 这种方法被称为单利。它在计算器和计算机出现之前就已经普遍使用，因为它很容易计算。它仍然在少数情况下使用，尤其是在分数时间段的情况下。

在简单兴趣的情况下，我们知道一个(吨)是一个线性函数，所以它的图形是一条线。我们在这条线上有两个点，因此很容易获得值的表达式一个(吨)随时吨. 我们线的斜率由下式给出

(1+一世)−11−0=一世
由于该线包含该点(0,1)它的方程由下式给出：
单利的累积函数

一个(吨)=1+一世吨
在简单的情况下，累积函数是具有斜率的线性函数一世. 数字2.1是累积函数的图一世=.05：金额函数同样容易计算：
Amount Function for Simple Interest

一个(吨)=F在=磷在⋅(1+一世吨)
这是一条带有 y 截距的线=磷在和坡度=磷在一世.

金融代写|金融数学代写Financial Mathematics代考|Compound Interest

单利仅根据初始存款金额计算每个期间的利息。如果在交易期间没有提款，存款金额会随着时间的推移而增加，而在每个期间结束时支付的利息金额保持不变。正如我们刚刚看到的，当我们使用单利时，这会导致有效利率下降。如果$400存放在5%单利三年后的存款金额为$460. 但是（使用单利）在第 4 年末支付的利息仍然只有$20（5\% 的$400). 如果按第 3 年末的存款金额支付利息，则第 4 年末的利息为(.05)⋅$460=$23

当在给定期间结束时支付的利息基于该期间开始时本金的累积值而不是原始存款金额时，我们获得了所谓的复利。

为了在复利的情况下找到累积函数的公式，我们计算每个期末赚取的利息，将其添加到上一个余额中，并使用该数字作为计算下一期利息的本金。见表 2.1。

这使我们相信，在复利累积函数的情况下，以下公式适用于
累积函数：复利

一个(n)=(1+一世)n
你可以证明方程2.13使用数学归纳法计算整数。我们仅使用存款时间的整数值计算了此结果。如果利息只支付期间的整数倍，一个(n)是阶跃函数：

一个(n)={10≤n<1 1+一世1≤n<2 (1+一世)22≤n<3 (1+一世)33≤n<4 ETC..
如果我们假设可以在投资生命周期的任何时间收取利息，那么很自然地扩展这个阶梯函数以包括真实的吨≥0. 因此我们得到函数一个(吨)=(1+一世)吨为所有实数定义吨这是方程的连续、可微扩展2.3
复利累积函数

一个(吨)=(1+一世)吨

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

金融代写|金融数学代写Financial Mathematics代考|Newton’s Method

Posted on 2022年6月21日2022年6月21日 by statistics-lab

如果你也在怎样代写金融数学Financial Mathematics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Financial Mathematics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

金融代写|金融数学代写Financial Mathematics代考|Newton’s Method

金融代写|金融数学代写Financial Mathematics代考|Also Called the Newton-Raphson 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.

金融代写|金融数学代写Financial Mathematics代考|Approximations Using Taylor Series

If $f(x)$ is a function which has derivatives of all orders $\left(f^{\prime}, f^{\prime \prime}, f^{\prime \mu^{\prime}}\right.$ 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 $1.3$ ).

金融代写|金融数学代写Financial Mathematics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 $2 \mathrm{ND}$ function for this button is $e^{x}$.
BASIC IDENTITIES
$$
\begin{aligned}
\ln (a b) &=\ln (a)+\ln (b) \
\ln \left(a^{r}\right) &=r \ln (a) \
\ln \left(\frac{a}{b}\right) &=\ln (a)-\ln (b) \
\frac{d \ln (x)}{d x} &=\frac{1}{x}, \quad \frac{d \ln (1+i)}{d i}=\frac{1}{1+i} \
\ln \left(e^{x}\right) &=e^{\ln (x)}=x \
\frac{d e^{x}}{d x} &=e^{x} \
\int e^{u} d u &=e^{u}+c \
\int \frac{1}{u} d u &=\ln (|u|)+c
\end{aligned}
$$
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.

Note: Don’t round intermediate values in your calculations. It is appropriate to round the final answer. For example $\$ 145.8802$ would be reported as $\$ 145.88$.
The TI BA II Plus does not have an $n^{\text {th }}$ root button, so you need to use the $y^{x}$ button with $x=\frac{1}{3}$. If you don’t know the decimal value of $\frac{1}{x}$ use the $\frac{1}{x}$ key. Here is how the calculation looks on the TI BA II Plus for $1.08^{\frac{1}{7}}, 1.08$, $y^{x}, 7, \frac{1}{x}=$ Result: $1.011055 .$

金融数学代考

金融代写|金融数学代写Financial Mathematics代考|Also Called the Newton-Raphson 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。

金融代写|金融数学代写Financial Mathematics代考|Approximations Using Taylor Series

如果F(X)是一个具有所有阶导数的函数(F′,F′′,F′μ′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(一个)+F′(一个)(X−一个)2+F(2)(一个)2!(X−一个)2+⋯+F(n)(一个)n!(X−一个)n+⋯
在上面的表达式中F(n)(一个)指的是nth 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(一个)通过使用前几个术语1.12.

例 1.6：使用公式的前四项1.12和一个=0近似罪⁡(X)
解决方案：

罪⁡(X)=罪⁡(0)+罪′⁡(0)(X−0)+罪′′⁡(0)2!(X−0)2+⋯+罪′′′⁡(0)3!(X−0)3 =0+因⁡(0)(X−0)+−罪⁡(0)2!(X−0)2+−因⁡(0)3!(X−0)2 =X−X36
正如下面的草图所示，这个近似值对于X接近于 0（图1.3 ).

金融代写|金融数学代写Financial Mathematics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和，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- 这2ñD这个按钮的功能是和X.
基本身份

ln⁡(一个b)=ln⁡(一个)+ln⁡(b) ln⁡(一个r)=rln⁡(一个) ln⁡(一个b)=ln⁡(一个)−ln⁡(b) dln⁡(X)dX=1X,dln⁡(1+一世)d一世=11+一世 ln⁡(和X)=和ln⁡(X)=X d和XdX=和X ∫和在d在=和在+C ∫1在d在=ln⁡(|在|)+C
例 1.9：求解(1.05)n=2.
解决方案：我们采取ln双方获得n⋅ln⁡(1.05)=ln⁡(2). 因此，n=ln⁡(2)ln⁡(1.05)≈14.21.
例 1.10：求解一世:

(1+一世)3=1+3⋅(.05)=1.15.
解决方案：我们采取ln双方获得3ln⁡(1+一世)=ln⁡(1.15). 这给了我们ln⁡(1+一世)=ln⁡(1.15)3=0.04658731412. 因此，1+一世=和0.04658731412=1.047689553。因此一世=.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+一世)=1.153=(1.15)13= 1.047689553.

注意：不要在计算中四舍五入中间值。对最终答案进行四舍五入是合适的。例如$145.8802将被报告为$145.88.
TI BA II Plus 没有nth 根按钮，所以你需要使用是X按钮X=13. 如果你不知道十进制值1X使用1X钥匙。以下是 TI BA II Plus 上的计算结果1.0817,1.08, 是X,7,1X=结果：1.011055.

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

金融代写|金融数学代写Financial Mathematics代考|Overview and Mathematical Prerequisites

Posted on 2022年6月21日2022年6月21日 by statistics-lab

如果你也在怎样代写金融数学Financial Mathematics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Financial Mathematics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

金融代写|金融数学代写Financial Mathematics代考|Overview and Mathematical Prerequisites

金融代写|金融数学代写Financial Mathematics代考|Calculators and Computers

Because the calculations are often tedious if done on a standard calculator, you should purchase a financial calculator, especially if you intend to take the SOA/CAS common exam on Financial Mathematics (FM). Here is a list of the acceptable calculators as provided by the SOA website:

Calculators – For all exams (except EA exams): only the following Texas Instrument calculator models may be used:

BA-35 TI-30Xa
BA II Plus* TI-30XIIS*

BA II Plus Professional Edition* TI-30XIIB*
TI-30XS MultiView* TI-30XB MultiView*
Upon entrance to the exam room, candidates must show the supervisor that the memory has been cleared. For the BA II Plus and the BA II Plus Professional Edition, clearing will reset the calculator to the factory default settings.

You may bring 2 calculators into the exam room. They should be these two:
TI BA II Plus Professional Edition TI-30XS MultiView
Keystroke instruction will be provided within this text for the TI BA II Plus and TI BA II Plus Professional as well as the TI-30XS MultiView. Prior to using your TI BA II Plus you should go to the Appendix: Basic Setup and change some of the default settings. Most of the examples in the text will be worked out using the TI BA II Plus, and the keystrokes will be presented as part of the solution.

Please be aware that not all of the problems on the FM exam can be solved using only the financial functions on the calculator. You will need to know the formulas and how to use them! Indeed, some of the questions asked require that you recognize various forms of the important formulas.

金融代写|金融数学代写Financial Mathematics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
$$
\begin{aligned}
&a=\text { first term } \
&v=\text { common multiple } \
&n=\text { number of terms. }
\end{aligned}
$$

金融代写|金融数学代写Financial Mathematics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.”

金融数学代考

金融代写|金融数学代写Financial Mathematics代考|Calculators and Computers

由于使用标准计算器进行计算通常很乏味，因此您应该购买金融计算器，特别是如果您打算参加 SOA/CAS 金融数学 (FM) 普通考试。以下是 SOA 网站提供的可接受的计算器列表：

计算器 – 对于所有考试（EA 考试除外）：只能使用以下德州仪器计算器模型：

BA-35 TI-30Xa
BA II Plus* TI-30XIIS*

BA II Plus 专业版* TI-30XIIB*
TI-30XS 多视图* TI-30XB 多视图*
进入考场后，考生必须向考官出示记忆已被清除的情况。对于 BA II Plus 和 BA II Plus 专业版，清除会将计算器重置为出厂默认设置。

您可以将 2 个计算器带入考场。它们应该是这两个：
TI BA II Plus 专业版 TI-30XS MultiView
按键说明将在本文中提供，用于 TI BA II Plus 和 TI BA II Plus Professional 以及 TI-30XS MultiView。在使用 TI BA II Plus 之前，您应该转到附录：基本设置并更改一些默认设置。文本中的大多数示例将使用 TI BA II Plus 进行编写，并且按键将作为解决方案的一部分呈现。

请注意，并非仅使用计算器上的财务功能即可解决 FM 考试中的所有问题。您将需要知道公式以及如何使用它们！事实上，有些问题要求你认识各种形式的重要公式。

金融代写|金融数学代写Financial Mathematics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）。如果我们计算未来某个时间支付的累计值，则结果称为年金的未来值（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：

∑一世=0n−1一个在一世=一个+一个在+一个在2+⋯+一个在n−1=一个1−在n1−在
这里一个是初始项，并且在是公倍数1.
如果|在|<1然后林n→∞在n=0我们还可以计算无限系列支付的总和（称为永续年金）：

∑一世=0∞一个在一世=林n→∞一个1−在n1−在=一个1−在

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一世=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

一个= 第一学期在= 公倍数 n= 项数。

金融代写|金融数学代写Financial Mathematics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。

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

金融代写|金融数学代写Financial Mathematics代考|ACTL20001

Posted on 2022年6月7日2022年6月7日 by statistics-lab

如果你也在怎样代写金融数学Financial Mathematics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Financial Mathematics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

金融代写|金融数学代写Financial Mathematics代考|ACTL20001

金融代写|金融数学代写Financial Mathematics代考|The Cox, Ross and Rubinstein model

We will now illustrate the different concepts introduced above using a specific case of financial market. This is a discretized version of the Black and Scholes model.
The market is considered to be made up of a risk-free asset $S_{n}^{0}=(1+r)^{n}$ and a single risky asset $S^{1}$ with the dynamic
$$
S_{0}^{1}=1, \quad S_{n+1}^{1}=S_{n}^{1} T_{n+1}, n \geq 0,
$$
where $\left(T_{n}\right){1 \leq n \leq N}$ is a sequence of random variable taking only two values $1+d$ and $1+u$ with $-1{0}={\emptyset, \Omega}, \mathcal{F}{n}=\sigma\left(T{1}, \ldots, T_{n}\right), 1 \leq n \leq N .
$$
In particular, $\mathcal{F}{N}=\sigma\left(T{1}, \ldots, T_{N}\right)=\mathcal{P}(\Omega)$ is the set of subsets of $\Omega$.
We will now characterize viable markets in this model. In order to do this, we start by studying risk-neutral probabilities.

PROPOSITION 5.2.-The discounted prices $\left(\widetilde{S}{n}^{1}\right){0 \leq n \leq N}$ are a martingale under a probability $\mathbb{P}^{}$ if and only if, for any $0 \leq n \leq N-1$, we have $$ \mathbb{E}^{}\left[T_{n+1} \mid \mathcal{F}{n}\right]=1+r, $$ with $\mathbb{E}^{}$ denoting the expectation for the probability $\mathbb{P}^{}$.
PROOF.- Let us proceed through double implication.
Let us assume that the realized price is a martingale. For any $0 \leq n \leq N-1$, we then have
$$
\begin{aligned}
& \mathbb{E}^{}\left[\widetilde{S}{n+1}^{1} \mid \mathcal{F}{n}\right]=\widetilde{S}{n}^{1} \
\Longrightarrow & \mathbb{E}^{}\left[\frac{S_{n+1}^{1}}{(1+r)^{n+1}} \mid \mathcal{F}{n}\right]=\frac{S{n}^{1}}{(1+r)^{n}} \
\Longrightarrow & \mathbb{E}^{*}\left[S_{n}^{1} T_{n+1} \mid \mathcal{F}{n}\right]=S{n}^{1}(1+r)
\end{aligned}
$$

$$
\begin{aligned}
&\Longrightarrow S_{n}^{1} \mathbb{E}^{}\left[T_{n+1} \mid \mathcal{F}{n}\right]=S{n}^{1}(1+r) \
&\Longrightarrow \mathbb{E}^{}\left[T_{n+1} \mid \mathcal{F}{n}\right]=(1+r) \end{aligned} $$ since $S{n}^{1}$ is the $\mathcal{F}{n}$-measurable. Conversely, if for any $0 \leq n \leq N-1$, we have $\mathbb{E}^{}\left[T{n+1} \mid \mathcal{F}{n}\right]=1+r$; therefore $$ \begin{aligned} & \mathbb{E}^{}\left[T{n+1} \mid \mathcal{F}{n}\right]=(1+r) \ \Longrightarrow & S{n}^{1} \mathbb{E}^{}\left[T_{n+1} \mid \mathcal{F}{n}\right]=S{n}^{1}(1+r) \
\Longrightarrow & \mathbb{E}^{}\left[\frac{S_{n+1}^{1}}{(1+r)^{n+1}} \mid \mathcal{F}{n}\right]=\frac{S{n}^{1}}{(1+r)^{n}} \
\Longrightarrow & \mathbb{E}^{*}\left[\widetilde{S}{n+1}^{1} \mid \mathcal{F}{n}\right]=\widetilde{S}{n}^{1} \end{aligned} $$ therefore $\left(\widetilde{S}{n}^{1}\right)$ is indeed a martingale, as the measurability and integrability conditions are satisfied.

金融代写|金融数学代写Financial Mathematics代考|Portfolio optimization

We now study a portfolio optimization problem in the Cox, Ross and Rubinstein model.

Let $V_{0}$ be the wealth of an investor at the time 0 . The investor can invest their money either in a risky asset or in a risk-free asset, following an admissible strategy. We use $\phi_{n}^{0}$ and $\phi_{n}^{1}$ to denote the number of shares in the risk-free asset and the number of shares in the risky asset, respectively, held between the time $n-1$ and $n$. Let $\pi_{n}$ be the proportion of the wealth invested in the risky asset between the instants $n-1$ and $n$, that is,
$$
\pi_{n}=\frac{\phi_{n}^{1} S_{n-1}^{1}}{V_{n-1}}
$$
1) Express $\phi_{n}^{0}$ and $\phi_{n}^{1}$ as the functions of $\pi_{n}, S_{n-1}^{0}, S_{n-1}^{1}$ and $V_{n-1}$ for any $n$.
2) Derive from this that for any $n$, the wealth at the time $n$, after the evolution of the prices and before the redistribution of the portfolio has the value:
$$
V_{n}=\left(\pi_{n} T_{n}+\left(1-\pi_{n}\right)(1+r)\right) V_{n-1}
$$
3) On the same graph and for the same random sampled trajectory, represent the evolution of the risk-free asset and the evolution of the wealth for the following two strategies:

a) The proportion of the wealth invested in the risky asset is fixed over time, at $1 / 4$.
b) The proportion of the wealth invested in the risky asset only takes the values 0 and 1 . It takes the value 1 when the price of a risky asset strictly exceeds that of the risk-free asset, and takes the value 0 when the risky asset is small than or equal to the risk-free asset, while remaining predictable.

We will take the following parameters: initial wealth $V_{0}=1$, the risky asset evolves as per the Cox, Ross and Rubinstein model with parameters $d=-2 \%$, $u=10 \%$ and $q=0.52$, interest rate $r=4 \%$ and duration of investment: $N=100$ periods.

Now consider that the investor’s utility function is logarithmic and we wish to find the strategy $\pi^{}=\left(\pi_{n}^{}\right){1 \leq n \leq N}$ which maximizes the expectation of the utility of the wealth at maturity $N$ : $$ \sup {\pi \text { admissible }} \mathbb{E}\left[\log V_{N}(\pi)\right]
$$
We will accept that the optimal strategy is constant over time and maximize the expression
$$
q \log \left(\pi^{}(u-r)+1+r\right)+(1-q) \log \left(\pi^{}(d-r)+1+r\right)
$$
We wish to compare the performances of strategies 1 and 2 , given above, and that of the optimal strategy. We will use the same parameters as for the previous question.
4) Write a function optimal $(u, d, r, q)$, which calculates the value of $\pi^{*}$ the optimal proportion to invest in the risky asset. We can restrict ourselves to five decimals. What do we find for our parameters?
5) On the same graph, trace a trajectory of the logarithm of the wealth at each instant for the optimal strategy, and for strategies 1 and 2 . Does the optimal strategy always give the same result? Why?
6) On the same graph, trace the expectation of the logarithm of the wealth at each instant, for each of the three strategies. We will calculate the expectation using the Monte Carlo method. Which is the best strategy? Discuss.

金融代写|金融数学代写Financial Mathematics代考|Portfolio optimization with withdrawal

In this section, the investor is allowed to withdraw a proportion $c_{n}$ of their wealth at the instant $n$, after updating the asset prices, but before the redistribution of their portfolio for the next investment period. Therefore, they only reinvest the

non-withdrawn part. The corresponding investment-withdrawal strategy $\left(\pi_{n}, c_{n}\right)$ is no longer self-financed, but it must remain predictable and the wealth after the withdrawal must be positive or zero at each instant. Therefore, it can thus be shown that the new wealth at the time $n$ after the evolution of the prices and when the value of the withdrawal is
$$
V_{n}(\pi, c)=\prod_{i=1}^{n}\left(1-c_{i}\right)\left(\pi_{i} T_{i}+\left(1-\pi_{i}\right)(1+r)\right),
$$
such that the value of the wealth withdrawn at the instant $n$ is $R_{n}(\pi, c)$ with
$$
R_{n}(\pi, c)= \begin{cases}\frac{c_{n} V_{n}(\pi, c)}{1-c_{n}} & \text { if } c_{n} \neq 1, \ V_{n-1}(\pi, c)\left(\pi_{n} T_{n}+\left(1-\pi_{n}\right)(1+r)\right) & \text { if not. }\end{cases}
$$
1) Graphically represent the evolution of the wealth for the investment strategy with the following withdrawal policy:
a) The proportion of the wealth invested in the risky asset is fixed over time, at $1 / 4$,
b) The proportion of the wealth withdrawn at each instant is equal to $1 \%$ over the interval $[1,80], 5 \%$ over the interval $] 80,90], 10 \%$ over the interval $] 90,95], 25 \%$ over the interval $] 95,100[$ and upon maturity, all the remaining wealth is withdrawn.
We will take the following parameters: the initial wealth $V_{0}=1$, the risky asset evolves as per the Cox, Ross and Rubinstein model with parameters $d=-2 \%$, $u=10 \%$ and $q=0.52$, interest rate $r=4 \%$, duration of investment: $N=100$ periods.

We now consider, once again, that the investor’s utility function is logarithmic and we want to find the investment strategy with the withdrawal $\left(\pi_{n}, c_{n}\right)$ that maximizes the expectation of the cumulative sum of the withdrawal utility up to the date of maturity $N$ :
$$
\max {\left(\pi{n}, c_{n}\right)} \mathbb{E}\left[\sum_{n=1}^{N} \log \left(R_{n}(\pi, c)\right)\right] .
$$
We will admit that the investment strategy with optimal withdrawal is given by
$-\pi_{n}=\pi^{}$ for any $1 \leq n \leq N$, $-c_{n}=\frac{1}{N+1-n}$ for any $1 \leq n \leq N$, with the same $\pi^{}$ as in the earlier practical exercise.

金融数学代考

金融代写|金融数学代写Financial Mathematics代考|The Cox, Ross and Rubinstein model

我们现在将使用一个特定的金融市场案例来说明上面介绍的不同概念。这是 Black 和 Scholes 模型的离散版本。
市场被认为是由无风险资产组成的小号n0=(1+r)n和单一的风险资产小号1与动态

小号01=1,小号n+11=小号n1吨n+1,n≥0,
在哪里(吨n)1≤n≤ñ是一个随机变量序列，只取两个值1+d和1+在和−10=∅,Ω,Fn=σ(吨1,…,吨n),1≤n≤ñ.我np一个r吨一世C在l一个r,\mathcal{F}{N}=\sigma\left(T{1}, \ldots, T_{N}\right)=\mathcal{P}(\Omega)一世s吨H和s和吨○Fs在bs和吨s○F\欧米茄$。
我们现在将在这个模型中描述可行的市场。为了做到这一点，我们首先研究风险中性概率。

提案 5.2.-折扣价(小号~n1)0≤n≤ñ是概率下的鞅磷当且仅当，对于任何0≤n≤ñ−1，我们有

和[吨n+1∣Fn]=1+r,和和表示概率的期望磷.
证明——让我们继续进行双重暗示。
让我们假设实现的价格是一个鞅。对于任何0≤n≤ñ−1，然后我们有

和[小号~n+11∣Fn]=小号~n1 ⟹和[小号n+11(1+r)n+1∣Fn]=小号n1(1+r)n ⟹和∗[小号n1吨n+1∣Fn]=小号n1(1+r)

⟹小号n1和[吨n+1∣Fn]=小号n1(1+r) ⟹和[吨n+1∣Fn]=(1+r)自从小号n1是个Fn- 可测量的。相反，如果对于任何0≤n≤ñ−1，我们有和[吨n+1∣Fn]=1+r; 所以

和[吨n+1∣Fn]=(1+r) ⟹小号n1和[吨n+1∣Fn]=小号n1(1+r) ⟹和[小号n+11(1+r)n+1∣Fn]=小号n1(1+r)n ⟹和∗[小号~n+11∣Fn]=小号~n1所以(小号~n1)确实是鞅，因为满足可测性和可积性条件。

金融代写|金融数学代写Financial Mathematics代考|Portfolio optimization

我们现在研究 Cox、Ross 和 Rubinstein 模型中的投资组合优化问题。

让在0成为投资者当时的财富 0 。投资者可以按照可接受的策略将资金投资于风险资产或无风险资产。我们用φn0和φn1分别表示在这段时间内持有的无风险资产的股份数量和风险资产的股份数量n−1和n. 让圆周率n是瞬间之间投资于风险资产的财富比例n−1和n，那是，

圆周率n=φn1小号n−11在n−1
1) 快递φn0和φn1作为函数圆周率n,小号n−10,小号n−11和在n−1对于任何n.
2）由此得出对于任何n，当时的财富n，在价格演变之后和投资组合重新分配之前具有以下值：

在n=(圆周率n吨n+(1−圆周率n)(1+r))在n−1
3) 在同一张图上，对于同一个随机抽样轨迹，分别代表以下两种策略的无风险资产的演变和财富的演变：

a) 投资于风险资产的财富比例随着时间的推移是固定的，在1/4.
b) 投资于风险资产的财富比例仅取值 0 和 1 。当风险资产的价格严格超过无风险资产的价格时取值为 1，当风险资产小于或等于无风险资产的价格时取值为 0，同时保持可预测性。

我们将采用以下参数：初始财富在0=1，风险资产按照 Cox、Ross 和 Rubinstein 模型的参数演化d=−2%, 在=10%和q=0.52，利率r=4%投资期限：ñ=100期间。

现在考虑投资者的效用函数是对数的，我们希望找到策略圆周率=(圆周率n)1≤n≤ñ使到期时财富效用的期望最大化ñ :

支持圆周率可接受的和[日志⁡在ñ(圆周率)]
我们将接受最优策略随着时间的推移是恒定的，并使表达式最大化

q日志⁡(圆周率(在−r)+1+r)+(1−q)日志⁡(圆周率(d−r)+1+r)
我们希望比较上面给出的策略 1 和 2 的性能以及最优策略的性能。我们将使用与上一个问题相同的参数。
4）写一个函数优化(在,d,r,q)，它计算的值圆周率∗投资于风险资产的最佳比例。我们可以将自己限制在小数点后五位。我们发现我们的参数是什么？
5) 在同一张图上，为最优策略以及策略 1 和 2 在每个时刻追踪财富对数的轨迹。最优策略总是给出相同的结果吗？为什么？
6) 在同一张图上，针对三种策略中的每一种，追踪每个时刻财富对数的期望值。我们将使用蒙特卡罗方法计算期望值。哪个是最好的策略？讨论。

金融代写|金融数学代写Financial Mathematics代考|Portfolio optimization with withdrawal

在本节中，允许投资者提取一部分Cn他们此刻的财富n，在更新资产价格之后，但在下一个投资期重新分配其投资组合之前。因此，他们只会再投资

非撤回部分。对应的投资退出策略(圆周率n,Cn)不再自筹资金，但它必须保持可预测性，并且提取后的财富必须在每一刻都是正数或零。因此，由此可以证明当时的新财富n在价格演变之后以及提款的价值为

在n(圆周率,C)=∏一世=1n(1−C一世)(圆周率一世吨一世+(1−圆周率一世)(1+r)),
使得在瞬间提取的财富价值n是Rn(圆周率,C)和

Rn(圆周率,C)={Cn在n(圆周率,C)1−Cn 如果 Cn≠1, 在n−1(圆周率,C)(圆周率n吨n+(1−圆周率n)(1+r)) 如果不。
1) 以图形方式表示具有以下退出政策的投资策略的财富演变：
a) 投资于风险资产的财富比例随着时间的推移是固定的，在1/4,
b) 每一刻提取的财富比例等于1%在区间内[1,80],5%在区间内]80,90],10%在区间内]90,95],25%在区间内]95,100[到期后，所有剩余的财富都将被提取。
我们将采用以下参数：初始财富在0=1，风险资产按照 Cox、Ross 和 Rubinstein 模型的参数演化d=−2%, 在=10%和q=0.52，利率r=4%，投资期限：ñ=100期间。

我们现在再次考虑，投资者的效用函数是对数的，我们希望找到退出的投资策略(圆周率n,Cn)使截至到期日的提款效用累积总和的期望最大化ñ :

最大限度(圆周率n,Cn)和[∑n=1ñ日志⁡(Rn(圆周率,C))].
我们承认最优退出的投资策略由下式给出
−圆周率n=圆周率对于任何1≤n≤ñ, −Cn=1ñ+1−n对于任何1≤n≤ñ, 同圆周率和前面的实际练习一样。

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

金融代写|金融数学代写Financial Mathematics代考|STAT2032

Posted on 2022年6月7日2022年6月7日 by statistics-lab

如果你也在怎样代写金融数学Financial Mathematics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Financial Mathematics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

金融代写|金融数学代写Financial Mathematics代考|STAT2032

金融代写|金融数学代写Financial Mathematics代考|Financial assets

Let us consider a financial market comprising of a risk-free asset $\left(S_{n}^{0}\right){0 \leq n \leq N}$ and $d \geq 1$ risky assets $\left(S{n}^{0}\right){0 \leq n \leq N}$ for $1 \leq i \leq d$. The price of the risky assets is represented by stochastic processes. The risk-free asset dynamic is $$ \begin{aligned} &S{0}^{0}=1 \
&S_{n}^{0}=(1+r)^{n}, n \geq 1
\end{aligned}
$$
where $r$ is the interest rate on the market. The risky assets have a random dynamic that is unspecified for the moment. As we need to compare the prices of the assets on different dates, we may sometimes wish to ignore the effect of the depreciation of currency, which is considered to be linked to the risk-free rate of interest.

DEFINITION 5.1.-The discounted prices of assets are the prices divided by the current value of the risk-free asset
$$
\widetilde{S}{n}^{i}=\frac{S{n}^{i}}{S_{n}^{0}}=\frac{S_{n}^{i}}{(1+r)^{n}}, \quad n \geq 0
$$
In mathematical terms, discounting prices is equivalent to considering that the risk-free interest rate is zero.

金融代写|金融数学代写Financial Mathematics代考|Investment strategies

Let us now consider an investor who wishes to invest money in the market. In order to define their investment strategy, they must know, at each instant, the number of shares invested in each asset.

DEFINITION 5.2.-An investment strategy is a process $\Phi=\left(\Phi_{n}\right){1 \leq n \leq N}$ that corresponds to the quantities $\Phi{n}=\left(\phi_{n}^{0}, \phi_{n}^{1}, \ldots, \phi_{n}^{d}\right) \in \mathbb{R}^{d+1}$ of each asset held by an investor between the instant $n-1$ and the instant $n$. It is a predictable process: for any $1 \leq n \leq N$, and $0 \leq i \leq d, \phi_{n}^{i}$ is the $\mathcal{F}_{n-1}$ measurable.

An investment strategy must be predictable, since decisions on how to distribute the portfolio between the instants $n-1$ and $n$ can be based only on the information available up to the instant $n-1$. In other words, there is no insider trading and the investor has no information on the future of the market.

Given the current prices of assets and an investment strategy, it is possible to calculate the value of the investor’s portfolio at every instant.

DEFINITION 5.3.- The value, at an instant $n$, of a portfolio that applies the investment strategy $\Phi$ is
$$
V_{n}(\Phi)=\sum_{i=1}^{d} \phi_{n}^{i} S_{n}^{i}=<\Phi_{n}, S_{n}>
$$
where $$ is the scalar product of $x$ and $y$, and $S_{n}=\left(S_{n}^{0}, S_{n}^{1}, \ldots, S_{n}^{d}\right)$.
The above notation may be ambiguous. Indeed, using our notations, $V_{n}(\Phi)$ represents the wealth at time $n$, just after the asset prices have been updated and before the portfolio is redistributed for the next period. We may also wish to consider the wealth at the instant $n$ after the redistribution of the portfolio. In fact, in most cases that we will consider, these two quantities are equal.

DEFINITION 5.4.-An investment strategy $\Phi$ is said to be self-financed if for any $1 \leq$ $n \leq N-1$, we have
$$
\sum_{i=1}^{d} \phi_{n}^{i} S_{n}^{i}=\sum_{i=1}^{d} \phi_{n+1}^{i} S_{n}^{i},
$$
or $\left\langle\Phi_{n}, S_{n}\right\rangle=\left\langle\Phi_{n+1}, S_{n}\right\rangle$, using condensed notation.
A self-financed strategy is, therefore, a strategy where at each step, the entire wealth is reinvested without withdrawal or an exogenous infusion of money.

We will now describe self-financing strategies through a specific decomposition of the realized wealth.

PROPOSITION 5.1.-An investment strategy $\Phi$ is self-financed if and only if, for any $1 \leq n \leq N-1$, we have
$$
\widetilde{V}{n}(\Phi)=\widetilde{V}{0}(\Phi)+\sum_{k=1}^{n}<\Phi_{k},\left(\widetilde{S}{k}-\widetilde{S}{k-1}\right)>.
$$
PROOF.- We proceed using double implication.
Let us first assume that $\Phi$ is self-financed. We then have, by definition, $\left\langle\Phi_{n}, S_{n}>\right.$ $=\left\langle\Phi_{n+1}, S_{n}\right\rangle$. Upon dividing by $S_{n}^{0}$, we obtain $\left\langle\Phi_{n}, S_{n}\right\rangle=\left\langle\Phi_{n+1}, \widetilde{S}{n}\right\rangle$. Therefore, using the self-financing relation, we obtain $$ \begin{aligned} \tilde{V}{n}(\Phi) &=\left\langle\Phi_{n}, \tilde{S}{n}>\right.\ &=\left\langle\Phi{0}, \widetilde{S}{0}>+\sum{k=1}^{n}<\Phi_{k}, \widetilde{S}{k}>-<\Phi{k-1}, \tilde{S}_{k-1}>\right.
\end{aligned}
$$

金融代写|金融数学代写Financial Mathematics代考|Arbitrage

We will now establish a link between viable financial markets and martingales. The term viable is used here to signify the impossibility of a definite increase in wealth with an initial investment of zero. Let us formalize this.

DEFINITION 5.6.-An investment strategy is said to be an arbitrage strategy or arbitrage opportunity if it is an admissible strategy, with an initial value of zero and a final, non-zero value:
$-V_{0}(\Phi)=0$ $-V_{n}(\Phi) \geq 0$ for any $0 \leq n \leq N$ $-$ there exist $n$ and $\omega \in \Omega$ such that $V_{n}(\omega)>0$
Let us note that the last condition would simply translate to $\mathbb{P}\left(V_{n}>0\right)>0$ if the universe $\Omega$ was not finite.

DEFINITION 5.7.-A financial market is said to be viable if there are no arbitrage opportunities.

We will now see that the absence of an arbitrage opportunity translates into the fact that discounted risky assets are martingales, except for a suitably changed probability.
DEFINITION 5.8. – A probability $\mathbb{P}^{*}$ on $(\Omega, \mathcal{F})$ is a risk-neutral probability if, under this probability, all discounted risky assets are martingales.

THEOREM 5.1.- A financial market is viable if and only if there exists a risk-neutral probability $\mathbb{P}^{*}$ equivalent ${ }^{l}$ to $\mathbb{P}$.

PROOF. – It is assumed that $\mathbb{P}^{}$ exists and that there exists an admissible strategy $\Phi$ such that $V_{0}(\Phi)=0$. Under $\mathbb{P}^{},\left(\widetilde{S}{\mathrm{r}}\right)$ is a martingale and since $\Phi$ is self-financed, we have $$ \tilde{V}{n}(\Phi)=\tilde{V}{0}(\Phi)+\sum{k=1}^{n}<\Phi_{k},\left(\widetilde{S}{k}-\widetilde{S}{k-1}\right)>
$$
It can thus be seen that $\left(V_{n}(\Phi)\right)$ is a martingale transform, since $\Phi$ is predictable; therefore, it is a martingale under $\mathbb{P}^{*}$. In particular, we have
$$
\mathbb{E}\left[\tilde{V}{n}(\Phi)\right]=\mathbb{E}\left[\tilde{V}{0}(\Phi)\right]=0
$$
for any $n$. In addition, since $\widetilde{V}{n}(\Phi) \geq 0$ for any $n$, because $\Phi$ is admissible, we necessarily have that $\widetilde{V}{n}(\Phi(\omega)=0$ for all $\omega \in \Omega$. Thus, no arbitrage opportunity exists.

金融数学代考

金融代写|金融数学代写Financial Mathematics代考|Financial assets

让我们考虑一个由无风险资产组成的金融市场(小号n0)0≤n≤ñ和d≥1风险资产(小号n0)0≤n≤ñ为了1≤一世≤d. 风险资产的价格由随机过程表示。无风险资产动态是

小号00=1 小号n0=(1+r)n,n≥1
在哪里r是市场上的利率。风险资产具有随机动态，目前未指定。由于我们需要比较不同日期的资产价格，我们有时可能希望忽略货币贬值的影响，这被认为与无风险利率有关。

定义 5.1.-资产的折现价格是价格除以无风险资产的现值

小号~n一世=小号n一世小号n0=小号n一世(1+r)n,n≥0
在数学上，贴现价格等同于考虑无风险利率为零。

金融代写|金融数学代写Financial Mathematics代考|Investment strategies

现在让我们考虑一位希望在市场上投资的投资者。为了定义他们的投资策略，他们必须在每一刻都知道投资于每种资产的股票数量。

定义 5.2.-投资策略是一个过程披=(披n)1≤n≤ñ对应于数量披n=(φn0,φn1,…,φnd)∈Rd+1投资者持有的每项资产在瞬间之间n−1和瞬间n. 这是一个可预测的过程：对于任何1≤n≤ñ，和0≤一世≤d,φn一世是个Fn−1可衡量的。

投资策略必须是可预测的，因为决定如何在瞬间分配投资组合n−1和n只能基于当前可用的信息n−1. 换句话说，没有内幕交易，投资者也没有关于市场未来的信息。

给定当前的资产价格和投资策略，可以计算出投资者每时每刻的投资组合价值。

定义 5.3.- 瞬间的价值n, 应用投资策略的投资组合披是

在n(披)=∑一世=1dφn一世小号n一世=<披n,小号n>
其中 $$ 是的标量积X和是，和小号n=(小号n0,小号n1,…,小号nd).
上面的符号可能是模棱两可的。事实上，使用我们的符号，在n(披)代表当时的财富n，就在资产价格更新之后和投资组合重新分配到下一个时期之前。我们也不妨考虑一下当下的财富n投资组合重新分配后。事实上，在我们将考虑的大多数情况下，这两个量是相等的。

定义 5.4.-投资策略披据说是自筹资金，如果有的话1≤ n≤ñ−1，我们有

∑一世=1dφn一世小号n一世=∑一世=1dφn+1一世小号n一世,
或者⟨披n,小号n⟩=⟨披n+1,小号n⟩，使用压缩符号。
因此，自筹资金策略是这样一种策略，在该策略中，每一步都将全部财富进行再投资，而无需撤出或外来资金注入。

我们现在将通过对已实现财富的具体分解来描述自筹资金策略。

命题 5.1.-一种投资策略披当且仅当，对于任何1≤n≤ñ−1，我们有

在~n(披)=在~0(披)+∑ķ=1n<披ķ,(小号~ķ−小号~ķ−1)>.
证明-我们继续使用双重暗示。
我们首先假设披是自筹资金。然后，根据定义，我们有，⟨披n,小号n> =⟨披n+1,小号n⟩. 除以小号n0，我们获得⟨披n,小号n⟩=⟨披n+1,小号~n⟩. 因此，利用自筹资金关系，我们得到

在~n(披)=⟨披n,小号~n> =⟨披0,小号~0>+∑ķ=1n<披ķ,小号~ķ>−<披ķ−1,小号~ķ−1>

金融代写|金融数学代写Financial Mathematics代考|Arbitrage

我们现在将在可行的金融市场和鞅之间建立联系。这里使用的术语“可行”表示在初始投资为零的情况下不可能明确增加财富。让我们将其形式化。

定义 5.6.-如果投资策略是可接受的策略，其初始值为 0，最终值为非零值，则该投资策略被称为套利策略或套利机会：
−在0(披)=0 −在n(披)≥0对于任何0≤n≤ñ −存在n和ω∈Ω这样在n(ω)>0
让我们注意，最后一个条件将简单地转换为磷(在n>0)>0如果宇宙Ω不是有限的。

定义 5.7.-如果没有套利机会，金融市场被认为是可行的。

我们现在将看到，没有套利机会会转化为这样一个事实，即贴现的风险资产是鞅，除了适当改变的概率。
定义 5.8。– 一个概率磷∗上(Ω,F)是风险中性概率，如果在此概率下，所有贴现的风险资产都是鞅。

定理 5.1.- 当且仅当存在风险中性概率时，金融市场才是可行的磷∗相等的l至磷.

证明。– 假设磷存在并且存在可接受的策略披这样在0(披)=0. 在下面磷,(小号~r)是鞅，因为披是自筹资金，我们有

在~n(披)=在~0(披)+∑ķ=1n<披ķ,(小号~ķ−小号~ķ−1)>
由此可以看出(在n(披))是鞅变换，因为披是可预测的；因此，它是一个鞅磷∗. 特别是，我们有

和[在~n(披)]=和[在~0(披)]=0
对于任何n. 此外，由于在~n(披)≥0对于任何n，因为披是可以接受的，我们必然有在~n(披(ω)=0对所有人ω∈Ω. 因此，不存在套利机会。

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

金融代写|金融数学代写Financial Mathematics代考|MATH3090

Posted on 2022年6月7日2022年6月7日 by statistics-lab

如果你也在怎样代写金融数学Financial Mathematics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Financial Mathematics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

金融代写|金融数学代写Financial Mathematics代考|The Doob decomposition

It is possible to obtain a martingale starting from any process.
THEOREM $4.2$ (Doob decomposition theorem).-Let $X=\left(X_{n}\right){n \in \mathbb{N}}$ be a stochastic process that is adapted to the filtration $\left(\mathcal{F}{n}\right){n \in \mathbb{N}}$ and integrable. It can then be uniquely decomposed in the form $$ X{n}=X_{0}+M_{n}+A_{n}
$$
with $M_{0}=A_{0}=0, M=\left(M_{n}\right){n \in \mathbb{N}}$ is a martingale, and $A=\left(A{n}\right)_{n \in \mathbb{N}}$ is a predictable process, which is called the compensator of $X$.

PROOF.- Existence We write $A_{0}=0$,
$$
A_{n+1}=A_{n}+\mathbb{E}\left[X_{n+1}-X_{n} \mid \mathcal{F}{n}\right]=\sum{k=1}^{n} \mathbb{E}\left[X_{k+1}-X_{k} \mid \mathcal{F}{k}\right], $$ $M{0}=0$ and $M_{n}=X_{n}-A_{n}$ for $n \geq 1 .$
We then directly have that $\left(A_{n}\right)$ is predictable and $\left(M_{n}\right)$ is adapted. Since the $X_{n}$ are integrable, $A_{n}$ and $M_{n}$ are also adaptable. Furthermore
$$
\begin{aligned}
\mathbb{E}\left[M_{n+1} \mid \mathcal{F}{n}\right] &=\mathbb{E}\left[X{n+1}-A_{n+1} \mid \mathcal{F}{n}\right] \ &=\mathbb{E}\left[X{n+1} \mid \mathcal{F}{n}\right]-A{n+1} \
&=\mathbb{E}\left[X_{n+1} \mid \mathcal{F}{n}\right]-A{n}-\mathbb{E}\left[X_{n+1}-X_{n} \mid \mathcal{F}{n}\right] \ &=\mathbb{E}\left[X{n+1} \mid \mathcal{F}{n}\right]-A{n}-\mathbb{E}\left[X_{n+1} \mid \mathcal{F}{n}\right]+X{n} \
&=X_{n}-A_{n} \
&=M_{n}
\end{aligned}
$$
thus $M=\left(M_{n}\right){n \in \mathbb{N}}$ is a martingale. Unicity Let us assume that there are two such decompositions: $$ X{n}=X_{0}+M_{n}+A_{n}=X_{0}+M_{n}^{\prime}-A_{n}^{\prime} .
$$
Then, $A_{0}-A_{0}^{\prime}=0$ and since the processes are predictable
$$
\begin{aligned}
A_{n+1}-A_{n+1}^{\prime} &=\mathbb{E}\left[A_{n+1}-A_{n+1}^{\prime} \mid \mathcal{F}{n}\right] \ &=\mathbb{E}\left[M{n+1}-M_{n+1}^{\prime} \mid \mathcal{F}{n}\right] \ &=M{n}-M_{n}^{\prime} \
&=A_{n}-A_{n}^{\prime}
\end{aligned}
$$
because we have martingales. Therefore, $A_{n}=A_{n}^{\prime}$ for any $n$, and consequently, $M_{n}=M_{n}^{\prime}$ for any $n$. Thus, we do have unicity.
COROLLARY 4.1.- In the Doob decomposition of $X=\left(X_{n}\right){n \in \mathbb{N}}$, $-\left(A{n}\right)$ is increasing if and only if $X=\left(X_{n}\right){n \in \mathbb{N}}$ is a submartingale; $-\left(A{n}\right)$ is decreasing if and only if $X=\left(X_{n}\right){n \in \mathbb{N}}$ is a supermartingale; $-\left(A{n}\right)$ is null if and only if $X=\left(X_{n}\right){n \in \mathbb{N}}$ is a martingale. ProOF.- The final point is a consequence of the first two. By definition, we have $A{n+1}=A_{n}+\mathbb{E}\left[X_{n+1}-X_{n} \mid \mathcal{F}{n}\right]$ and $\mathbb{E}\left[X{n+1}-X_{n} \mid \mathcal{F}_{n}\right]$ is positive for a submartingale, negative for a supermartingale.

金融代写|金融数学代写Financial Mathematics代考|Stopping time

We now introduce the concept of stopping time. Informally, a stopping time corresponds to a random date, thus a date that is unknown in advance, but such that at any instant it is possible to say whether or not the date has passed.

DEFINITION 4.2.- Let $T: \Omega \longrightarrow \mathbb{N} \cup{+\infty}$ be a random variable. It is said that $T$
EXAMPLE 4.7.- Let $T$ be a constant positive random variable. Then, $T$ is an $n$, we have
$$
(T \leq n)=\left{\begin{array}{l}
\emptyset \text { if } T>n \
\Omega \text { if } T \leq n
\end{array}\right.
$$
As any $\sigma$-algebra contains the empty set and $\Omega$, we have $(T \leq n) \in \mathcal{F}{n}$. EXAMPLE 4.8.-Let $\left(X{n}\right){n \in \mathbb{N}}$ be a sequence of real random variables. Let $A$ be a Borel set in $\mathbb{R}$ and consider the random variable $$ T=\inf \left{k \in \mathbb{N} ; X{k} \in A\right},
$$

using the convention $\inf \emptyset=+\infty$. Thus, $T$ is a stopping time with respect to the natural filtration of $\left(X_{n}\right){n \in \mathbb{N}}$, called the hitting time of the set $A$. Indeed, for any $n \in \mathbb{N}$, we have $$ (T \leq n)=\left(X{0} \in A\right) \cup \ldots \cup\left(X_{n-1} \in A\right) \cup\left(X_{n} \in A\right) \in \mathcal{F}{n} $$ because $\left(\mathcal{F}{n}\right){n \in \mathbb{N}}$ is the natural filtration of $\left(X{n}\right){n \in \mathbb{N}}$. Proposition 4.3.-A random variable $T: \Omega \longrightarrow \mathbb{N} \cup{+\infty}$ is an PROOF.- Let us assume that $T$ is an $\left(\mathcal{F}{n}\right){n \in \mathbb{N}^{-s t o p p i n g}}$ time. Then, for any $n \in \mathbb{N}^{*}$, we have: $(T \leq n) \in \mathcal{F}{\mathrm{n}}$ and $(T \leq n-1) \in \mathcal{F}{\mathrm{n}-1} \subset \mathcal{F}{\mathrm{n}}$, and consequently,
$$
(T=n)=(T \leq n) \cap(T \leq n-1)^{c} \in \mathcal{F}{n}, $$ because a $\sigma$-algebra is closed under finite intersection and complements. It is now assumed that for any $k \in \mathbb{N},(T=k) \in \mathcal{F}{k}$. Therefore, for any fixed $n$, we have
$$
(T \leq n)=\bigcup_{k=0}^{n}(T=k) \in \bigcup_{k=0}^{n} \mathcal{F}{k}=\mathcal{F}{n},
$$
because a filtration is an increasing sequence of $\sigma$-algebras. Consequently, $T$ is indeed an $\left(\mathcal{F}{n}\right){n \in \mathbb{N}}$-stopping time
DEFINITION 4.3.- Consider an $\left(\mathcal{F}{n}\right){n \in \mathbb{N}^{-s t o p p}}$-sime, T. The set
$$
\mathcal{F}{T}=\left{A \in \mathcal{F} ; \forall n \in \mathbb{N}, A \cap(T \leq n) \in \mathcal{F}{n}\right}
$$
is called the $\sigma$-algebra of events observable up until time $T$

金融代写|金融数学代写Financial Mathematics代考|Stopped martingales

Let us now see how to combine the concept of a martingale and that of stopping time.

Proposition 4.5.-Let $\left(Z_{n}\right){n \in \mathbb{N}}$ be an $\left(\mathcal{F}{n}\right){n \in \mathbb{N}}$-martingale and let $T$ be an $$ Z{n}^{T}=Z_{T \wedge n}=Z_{n} \mathbb{1}{{T>n}}+Z{T} \mathbb{1}{{T \leq n}} ; n \in \mathbb{N} . $$ Then, $\left(Z{n}^{T}\right){n \in \mathbb{N}}$ is an $\left(\mathcal{F}{n}\right){n \in \mathbb{N}}-$ martingale. PROOF.- We have $$ Z{T \wedge n}=Z_{0}+\sum_{l=1}^{T \wedge n}\left(Z_{l}-Z_{l-1}\right)=Z_{0}+\sum_{l \geq 1} \mathbb{1}{{l \leq T \wedge \wedge}}\left(Z{l}-Z_{l-1}\right) .
$$
Since $\mathbb{1}{{l \leq T \wedge n}}=\mathbb{1}{{l \leq T}} \mathbb{1}{{l \leq n}}$ and $\mathbb{1}{{l \leq n}}=1$ if and only if $l=1, \ldots ., n$, and 0 if $l>n$, it follows that
$$
Z_{T \wedge n}=Z_{0}+\sum_{l=1}^{n} \mathbb{1}{{l \leq T}}\left(Z{l}-Z_{l-1}\right) .
$$
On the contrary, for $k \leq n$, we have:
$$
\begin{aligned}
&\mathbb{E}\left[Z_{T \wedge n} \mid \mathcal{F}{k}\right] \ &=\mathbb{E}\left[Z{0} \mid \mathcal{F}{k}\right]+\sum{l=1}^{n} \mathbb{E}\left[\mathbb{1}{{l \leq T}}\left(Z{l}-Z_{l-1}\right) \mid \mathcal{F}{k}\right] \ &=\mathbb{E}\left[Z{0} \mid \mathcal{F}{k}\right]+\sum{l=1}^{k} \mathbb{E}\left[\mathbb{1}{{l \leq T}}\left(Z{l}-Z_{l-1}\right) \mid \mathcal{F}{k}\right] \ &=\sum{l=k+1}^{n} \mathbb{E}\left[\mathbb{1}{{l \leq T}}\left(Z{l}-Z_{l-1}\right) \mid \mathcal{F}_{k}\right]
\end{aligned}
$$

For the first two terms, because the random variables $Z_{0}$ and $\mathbb{1}{{l \leq T}}\left(Z{l}-Z_{l-1}\right)$ are $\mathcal{F}{k}$-measurable, for any $l \leq k$, we have $$ \mathbb{E}\left[Z{0} \mid \mathcal{F}{k}\right]=Z{0} \text { et } \sum_{l=1}^{k} \mathbb{E}\left[\mathbb{1}{{l \leq T}}\left(Z{l}-Z_{l-1}\right) \mid \mathcal{F}{k}\right]=\sum{l=1}^{k} \mathbb{1}{{l \leq T}}\left(Z{l}-Z_{l-1}\right)
$$
For the third term, it can be written,
$$
\begin{aligned}
\sum_{l=k+1}^{n} \mathbb{E}\left[\mathbb{1}{{l \leq T}}\left(Z{l}-Z_{l-1}\right) \mid \mathcal{F}{k}\right] &=\sum{l=k+1}^{n} \mathbb{E}\left[\mathbb{E}\left[\mathbb{1}{{l \leq T}}\left(Z{l}-Z_{l-1}\right) \mid \mathcal{F}{l-1}\right] \mid \mathcal{F}{k}\right] \
&=\sum_{l=k+1}^{n} \mathbb{E}\left[\mathbb{1}{{l \leq T}} \mathbb{E}\left[\left(Z{l}-Z_{l-1}\right) \mid \mathcal{F}{l-1}\right] \mid \mathcal{F}{k}\right] \
&=0,
\end{aligned}
$$
because, as $\left(Z_{n}\right)$ is an $\left(\mathcal{F}{n}\right){n \in \mathbb{N}}$-martingale, we have $\mathbb{E}\left[\left(Z_{l}-Z_{l-1}\right) \mid \mathcal{F}{l-1}\right]=0$, for any $l \geq 1$. It follows from this that $$ \mathbb{E}\left[Z{T \wedge n} \mid \mathcal{F}{k}\right]=Z{0}+\sum_{l=1}^{k} \mathbb{1}{{l \leq T}}\left(Z{l}-Z_{l-1}\right)
$$
and consequently, for any $k \leq n$, we have
$$
\mathbb{E}\left[Z_{T \wedge n} \mid \mathcal{F}{k}\right]=Z{T \wedge k} .
$$

金融数学代考

金融代写|金融数学代写Financial Mathematics代考|The Doob decomposition

可以从任何过程开始获得鞅。
定理4.2(Doob 分解定理).-让X=(Xn)n∈ñ是一个适应过滤的随机过程(Fn)n∈ñ且可积。然后它可以被唯一地分解为形式

Xn=X0+米n+一个n
和米0=一个0=0,米=(米n)n∈ñ是鞅，并且一个=(一个n)n∈ñ是一个可预测的过程，称为补偿器X.

证明-存在我们写一个0=0,

一个n+1=一个n+和[Xn+1−Xn∣Fn]=∑ķ=1n和[Xķ+1−Xķ∣Fķ],米0=0和米n=Xn−一个n为了n≥1.
然后我们直接有(一个n)是可预测的并且(米n)被改编。由于Xn是可积的，一个n和米n也具有适应性。此外

和[米n+1∣Fn]=和[Xn+1−一个n+1∣Fn] =和[Xn+1∣Fn]−一个n+1 =和[Xn+1∣Fn]−一个n−和[Xn+1−Xn∣Fn] =和[Xn+1∣Fn]−一个n−和[Xn+1∣Fn]+Xn =Xn−一个n =米n
因此米=(米n)n∈ñ是鞅。单一性让我们假设有两种这样的分解：

Xn=X0+米n+一个n=X0+米n′−一个n′.
然后，一个0−一个0′=0并且由于过程是可预测的

一个n+1−一个n+1′=和[一个n+1−一个n+1′∣Fn] =和[米n+1−米n+1′∣Fn] =米n−米n′ =一个n−一个n′
因为我们有鞅。所以，一个n=一个n′对于任何n，因此，米n=米n′对于任何n. 因此，我们确实具有唯一性。
推论 4.1.- 在 Doob 分解中X=(Xn)n∈ñ, −(一个n)增加当且仅当X=(Xn)n∈ñ是亚鞅；−(一个n)当且仅当X=(Xn)n∈ñ是超鞅；−(一个n)为空当且仅当X=(Xn)n∈ñ是鞅。证明。- 最后一点是前两个的结果。根据定义，我们有一个n+1=一个n+和[Xn+1−Xn∣Fn]和和[Xn+1−Xn∣Fn]对下鞅为正，对上鞅为负。

金融代写|金融数学代写Financial Mathematics代考|Stopping time

我们现在介绍停止时间的概念。非正式地，停止时间对应于随机日期，因此是一个预先未知的日期，但是在任何时刻都可以说该日期是否已经过去。

定义 4.2.- 让吨:Ω⟶ñ∪+∞是一个随机变量。据说吨
例 4.7.- 让吨是一个恒定的正随机变量。然后，吨是一个n, 我们有
$$
(T \leq n)=\left{

∅ 如果吨>n Ω 如果吨≤n\正确的。

一个s一个n是$σ$−一个lG和br一个C○n吨一个一世ns吨H和和米p吨是s和吨一个nd$Ω$,在和H一个在和$(吨≤n)∈Fn$.和X一个米磷大号和4.8.−大号和吨$(Xn)n∈ñ$b和一个s和q在和nC和○Fr和一个lr一个nd○米在一个r一世一个bl和s.大号和吨$一个$b和一个乙○r和ls和吨一世n$R$一个ndC○ns一世d和r吨H和r一个nd○米在一个r一世一个bl和T=\inf \left{k \in \mathbb{N} ; X{k} \in A\right},
$$

使用约定信息∅=+∞. 因此，吨是关于自然过滤的停止时间(Xn)n∈ñ，称为该组的击球时间一个. 的确，对于任何n∈ñ，我们有

(吨≤n)=(X0∈一个)∪…∪(Xn−1∈一个)∪(Xn∈一个)∈Fn因为(Fn)n∈ñ是自然过滤(Xn)n∈ñ. 命题 4.3.-一个随机变量吨:Ω⟶ñ∪+∞是一个证明。-让我们假设吨是一个(Fn)n∈ñ−s吨○pp一世nG时间。那么，对于任何n∈ñ∗，我们有：(吨≤n)∈Fn和(吨≤n−1)∈Fn−1⊂Fn，因此，

(吨=n)=(吨≤n)∩(吨≤n−1)C∈Fn,因为一个σ-代数在有限交集和补集下是封闭的。现在假设对于任何ķ∈ñ,(吨=ķ)∈Fķ. 因此，对于任何固定n，我们有

(吨≤n)=⋃ķ=0n(吨=ķ)∈⋃ķ=0nFķ=Fn,
因为过滤是一个递增的序列σ-代数。最后，吨确实是一个(Fn)n∈ñ- 停止时间
定义 4.3.- 考虑一个(Fn)n∈ñ−s吨○pp-sime, T. 集合

\mathcal{F}{T}=\left{A \in \mathcal{F} ; \forall n \in \mathbb{N}, A \cap(T \leq n) \in \mathcal{F}{n}\right}\mathcal{F}{T}=\left{A \in \mathcal{F} ; \forall n \in \mathbb{N}, A \cap(T \leq n) \in \mathcal{F}{n}\right}
被称为σ- 直到时间可观察到的事件的代数吨

金融代写|金融数学代写Financial Mathematics代考|Stopped martingales

现在让我们看看如何将鞅的概念和停止时间的概念结合起来。

命题 4.5.-让(从n)n∈ñ豆(Fn)n∈ñ-鞅并让吨豆

从n吨=从吨∧n=从n1吨>n+从吨1吨≤n;n∈ñ.然后，(从n吨)n∈ñ是一个(Fn)n∈ñ−鞅。证明-我们有

从吨∧n=从0+∑l=1吨∧n(从l−从l−1)=从0+∑l≥11l≤吨∧∧(从l−从l−1).
自从1l≤吨∧n=1l≤吨1l≤n和1l≤n=1当且仅当l=1,….,n, 如果是 0l>n，它遵循

从吨∧n=从0+∑l=1n1l≤吨(从l−从l−1).
相反，对于ķ≤n，我们有：

和[从吨∧n∣Fķ] =和[从0∣Fķ]+∑l=1n和[1l≤吨(从l−从l−1)∣Fķ] =和[从0∣Fķ]+∑l=1ķ和[1l≤吨(从l−从l−1)∣Fķ] =∑l=ķ+1n和[1l≤吨(从l−从l−1)∣Fķ]

对于前两项，因为随机变量从0和1l≤吨(从l−从l−1)是Fķ- 可测量的，适用于任何l≤ķ，我们有

和[从0∣Fķ]=从0 和 ∑l=1ķ和[1l≤吨(从l−从l−1)∣Fķ]=∑l=1ķ1l≤吨(从l−从l−1)
对于第三项，可以写成，

∑l=ķ+1n和[1l≤吨(从l−从l−1)∣Fķ]=∑l=ķ+1n和[和[1l≤吨(从l−从l−1)∣Fl−1]∣Fķ] =∑l=ķ+1n和[1l≤吨和[(从l−从l−1)∣Fl−1]∣Fķ] =0,
因为，作为(从n)是一个(Fn)n∈ñ-马丁格尔，我们有和[(从l−从l−1)∣Fl−1]=0, 对于任何l≥1. 由此可知

和[从吨∧n∣Fķ]=从0+∑l=1ķ1l≤吨(从l−从l−1)
因此，对于任何ķ≤n，我们有

和[从吨∧n∣Fķ]=从吨∧ķ.

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