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## 金融代写|金融微积分代写Financial Calculus代考|Backwards induction

In fact most of the hard work has already been done when we examined the branch model. Extending the results and intuitions of section $2.1$ to an entire binomial tree is surprisingly straightforward. The key idea is that of backwards induction – extending the construction portfolio back one tick at a time from the claim to the required starting place.

Consider, then, a general claim for our stock $S$. When we examined a single branching of our tree, we had the function $f$ dependent only on the node chosen at the end of a single tick period – here we can extend the idea of a claim to cover not only the value of $S$ at the time the claim is exercised but also the history of $S$ up until that point.
The tree structure of the stock was not entirely arbitrary – it embodies a one-to-one relationship between a node and the history of the stock’s path up to and including that node. No other history reaches that node; and trivially no other node is reached by that history. This is precisely that condition that allows us actually to associate a claim value with a particular end-node on our tree. We shall also insist on the finiteness of our tree. There must be some final tick-time at which the claim is fully determined. A condition not unreasonable in the real financial world. A general claim can be thought of as some function on the nodes at this claim time-horizon.

## 金融代写|金融微积分代写Financial Calculus代考|The two-step

We know that the expectation operator can be made to work for a single branch – here, then, we must wade through the algebra for two time-steps, three branches stuck together into a tree. If two time-steps work, then so will many.

Suppose that the interest rate over any branch is constant at rate $r$. Then there exists some set of suitable $q_j$ s such that the value of the derivative at node $j$ at tick-time $i, f(j)$, is
$$f(j)=e^{-r \delta t}\left(q_j f(2 j+1)+\left(1-q_j\right) f(2 j)\right)$$
That is the discounted expectation under $q_j$ of the time- $(i+1)$ claim values $f(2 j+1)$ and $f(2 j)$. So in our two-step tree (figure 2.4), the two forks from node 3 to nodes 6 and 7 , and from node 2 to nodes 4 and 5 , are both structurally identical to the simple one-step branch. This means that $f(3)$ comes from $f(6)$ and $f(7)$ via
$$f(3)=e^{-r \delta t}\left(q_3 f(7)+\left(1-q_3\right) f(6)\right) \text {, }$$
and similarly, $f(2)$ comes from $f(4)$ and $f(5)$, with
$$f(2)=e^{-r \delta t}\left(q_2 f(5)+\left(1-q_2\right) f(4)\right) \text {. }$$
Here $q_j$ is the probability $\left(s_j \exp (r \delta t)-s_{2 j}\right) /\left(s_{2 j+1}-s_{2 j}\right)$, so for instance
$$q_2=\frac{s_2 \exp (r \delta t)-s_4}{s_5-s_4}, \quad \text { and } \quad q_3=\frac{s_3 \exp (r \delta t)-s_6}{s_7-s_6} \text {. }$$

## 金融代写|金融微积分代写Financial Calculus代考|The two-step

$$f(j)=e^{-r \delta t}\left(q_j f(2 j+1)+\left(1-q_j\right) f(2 j)\right)$$

$$f(3)=e^{-r \delta t}\left(q_3 f(7)+\left(1-q_3\right) f(6)\right),$$

$$f(2)=e^{-r \delta t}\left(q_2 f(5)+\left(1-q_2\right) f(4)\right) .$$

$$q_2=\frac{s_2 \exp (r \delta t)-s_4}{s_5-s_4}, \quad \text { and } \quad q_3=\frac{s_3 \exp (r \delta t)-s_6}{s_7-s_6}$$

## MATLAB代写

MATLAB 是一种用于技术计算的高性能语言。它将计算、可视化和编程集成在一个易于使用的环境中，其中问题和解决方案以熟悉的数学符号表示。典型用途包括：数学和计算算法开发建模、仿真和原型制作数据分析、探索和可视化科学和工程图形应用程序开发，包括图形用户界面构建MATLAB 是一个交互式系统，其基本数据元素是一个不需要维度的数组。这使您可以解决许多技术计算问题，尤其是那些具有矩阵和向量公式的问题，而只需用 C 或 Fortran 等标量非交互式语言编写程序所需的时间的一小部分。MATLAB 名称代表矩阵实验室。MATLAB 最初的编写目的是提供对由 LINPACK 和 EISPACK 项目开发的矩阵软件的轻松访问，这两个项目共同代表了矩阵计算软件的最新技术。MATLAB 经过多年的发展，得到了许多用户的投入。在大学环境中，它是数学、工程和科学入门和高级课程的标准教学工具。在工业领域，MATLAB 是高效研究、开发和分析的首选工具。MATLAB 具有一系列称为工具箱的特定于应用程序的解决方案。对于大多数 MATLAB 用户来说非常重要，工具箱允许您学习应用专业技术。工具箱是 MATLAB 函数（M 文件）的综合集合，可扩展 MATLAB 环境以解决特定类别的问题。可用工具箱的领域包括信号处理、控制系统、神经网络、模糊逻辑、小波、仿真等。

Posted on Categories:Financial Calculus, 金融代写, 金融微积分

## avatest™帮您通过考试

avatest™的各个学科专家已帮了学生顺利通过达上千场考试。我们保证您快速准时完成各时长和类型的考试，包括in class、take home、online、proctor。写手整理各样的资源来或按照您学校的资料教您，创造模拟试题，提供所有的问题例子，以保证您在真实考试中取得的通过率是85%以上。如果您有即将到来的每周、季考、期中或期末考试，我们都能帮助您！

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## 金融代写|金融微积分代写Financial Calculus代考|Bond-only strategy

All is not lost, though. Consider a portfolio of just the cash bond. The cash bond will grow by a factor of $\exp (r \delta t)$ across the period, thus buying discount bonds to the value of $\exp (-r \delta t)[(1-p) f(2)+p f(3)]$ at the start of the period will provide a value equal to $(1-p) f(2)+p f(3)$ at the end. Why would we choose this value as the target to aim for? Because it is the expected value of the derivative at the end of the period – formally:
Expectation for a branch
Let $S$ be a binomial branch process with base value $s_1$ at time zero, downvalue $s_2$ and up-value $s_3$. Then the expectation of $S$ at tick-time 1 under the probability of an up-move $p$ is:
$$\mathbb{E}_p\left(S_1\right)=(1-p) s_2+p s_3$$
Our claim $f$ on $S$ is just as much a random variable as $S_1$ is – we can meaningfully talk of its expectation. And thus we can meaningfully aim for the expectation of the claim, via the cash bonds. This strategy of construction would at the very least be expected to break even. And the value of the starting portfolio of cash bonds might be claimed to be a good predictor of the value of the derivative at the start of the period. The price we would predict for the derivative would be the discounted expectation of its value at the end.
But of course this is just the strong law of chapter one all over again just thinly disguised as construction. And exactly as before we are missing an element of coercion. We haven’t explicitly constructed the two possible values the derivative can take: $f(2)$ and $f(3)$; we have simply aimed between them in a probabilistic sense and hoped for the best.

And we already know that this best isn’t good enough for forwards. For a stock that obeys a binomial branch process, its forward price is not suggested by the possible stock values $s_2$ and $s_3$, but enforced by the interest rate $r$ implied by the cash bond $B$ : namely $s_1 \exp (r \delta t)$. The discounted expectation of the claim doesn’t work as a pricing tool.

## 金融代写|金融微积分代写Financial Calculus代考|Stocks and bonds together

But can we do any better? Another strategy might occur to us, we have after all two instruments which we can build into a portfolio to hold for the tick-period. We tried using the guaranteed growth of the cash bond as a device for producing a particular desired value, and we chose the expected value of the derivative as our target point. But we have another instrument tied more strongly to the behaviour of both the stock and the derivative than just the cash bond. Namely the stock itself. Suppose we attempted to guarantee not an amount known in advance which we hope will stand as a reasonable predictor for the value of the derivative, but the value of the derivative itself, whatever it might be.

Consider a general portfolio $(\phi, \psi)$, namely $\phi$ of the stock $S$ (worth $\phi s_1$ ) and $\psi$ of the cash bond $B$ (worth $\psi B_0$ ). If we were to buy this portfolio at time zero, it would cost $\phi s_1+\psi B_0$.
One tick later, though, it would be worth one of two possible values:
$\phi s_3+\psi B_0 \exp (r \delta t) \quad$ after an ‘up’ move,
and $\phi s_2+\psi B_0 \exp (r \delta t) \quad$ after a ‘down’ move.
This pair of equations should intrigue us – we have two equations, two possible claim values and two free variables $\phi$ and $\psi$. We have two values $f(3)$ and $f(2)$ which we want to duplicate under the appropriate move of the stock, and we have two variables $\phi$ and $\psi$ which we can adjust. Thus the strategy can reduce to solving the following two simultaneous equations for $(\phi, \psi)$
\begin{aligned} \phi s_3+\psi B_0 \exp (r \delta t) &=f(3) \ \phi s_2+\psi B_0 \exp (r \delta t) &=f(2) \end{aligned}
Except if perversely $s_2$ and $s_3$ are identical – in which case $S$ is a bond not a stock – we have the solutions:
\begin{aligned} \phi &=\frac{f(3)-f(2)}{s_3-s_2} \ \psi &=B_0^{-1} \exp (-r \delta t)\left(f(3)-\frac{(f(3)-f(2)) s_3}{s_3-s_2}\right) \end{aligned}
What can we do with this algebraic result? If we bought this $(\phi, \psi)$ portfolio and held it, the equations guarantee that we achieve our goal – if the stock moves up, then the portfolio becomes worth $f(3)$; and if the stock moves down, the portfolio becomes worth $f(2)$. We have synthesized the derivative.

## 金融代写|金融微积分代写Financial Calculus代考|Bond-only strategy

$$\mathbb{E}_p\left(S_1\right)=(1-p) s_2+p s_3$$

## 金融代写|金融微积分代写Financial Calculus代考|Stocks and bonds together

$\phi s_3+\psi B_0 \exp (r \delta t)$ 在“向上”移动之后，

$$\phi s_3+\psi B_0 \exp (r \delta t)=f(3) \phi s_2+\psi B_0 \exp (r \delta t) \quad=f(2)$$

$$\phi=\frac{f(3)-f(2)}{s_3-s_2} \psi \quad=B_0^{-1} \exp (-r \delta t)\left(f(3)-\frac{(f(3)-f(2)) s_3}{s_3-s_2}\right)$$

## MATLAB代写

MATLAB 是一种用于技术计算的高性能语言。它将计算、可视化和编程集成在一个易于使用的环境中，其中问题和解决方案以熟悉的数学符号表示。典型用途包括：数学和计算算法开发建模、仿真和原型制作数据分析、探索和可视化科学和工程图形应用程序开发，包括图形用户界面构建MATLAB 是一个交互式系统，其基本数据元素是一个不需要维度的数组。这使您可以解决许多技术计算问题，尤其是那些具有矩阵和向量公式的问题，而只需用 C 或 Fortran 等标量非交互式语言编写程序所需的时间的一小部分。MATLAB 名称代表矩阵实验室。MATLAB 最初的编写目的是提供对由 LINPACK 和 EISPACK 项目开发的矩阵软件的轻松访问，这两个项目共同代表了矩阵计算软件的最新技术。MATLAB 经过多年的发展，得到了许多用户的投入。在大学环境中，它是数学、工程和科学入门和高级课程的标准教学工具。在工业领域，MATLAB 是高效研究、开发和分析的首选工具。MATLAB 具有一系列称为工具箱的特定于应用程序的解决方案。对于大多数 MATLAB 用户来说非常重要，工具箱允许您学习应用专业技术。工具箱是 MATLAB 函数（M 文件）的综合集合，可扩展 MATLAB 环境以解决特定类别的问题。可用工具箱的领域包括信号处理、控制系统、神经网络、模糊逻辑、小波、仿真等。

Posted on Categories:Financial Calculus, 金融代写, 金融微积分

## avatest™帮您通过考试

avatest™的各个学科专家已帮了学生顺利通过达上千场考试。我们保证您快速准时完成各时长和类型的考试，包括in class、take home、online、proctor。写手整理各样的资源来或按照您学校的资料教您，创造模拟试题，提供所有的问题例子，以保证您在真实考试中取得的通过率是85%以上。如果您有即将到来的每周、季考、期中或期末考试，我们都能帮助您！

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## 金融代写|金融微积分代写Financial Calculus代考|Expectation pricing

Consider playing the following game – someone tosses a coin and pays you one dollar for heads and nothing for tails. What price should you pay for this prize? If the coin is fair, then heads and tails are equally likely – about half the time you should win the dollar and the rest of the time you should receive nothing. Over enough plays, then, you expect to make about fifty cents a go. So paying more than fifty cents seems extravagant and less than fifty cents looks extravagant for the person offering the game. Fifty cents, then, seems about right.

Kolmogorov 强数定律

## MATLAB代写

MATLAB 是一种用于技术计算的高性能语言。它将计算、可视化和编程集成在一个易于使用的环境中，其中问题和解决方案以熟悉的数学符号表示。典型用途包括：数学和计算算法开发建模、仿真和原型制作数据分析、探索和可视化科学和工程图形应用程序开发，包括图形用户界面构建MATLAB 是一个交互式系统，其基本数据元素是一个不需要维度的数组。这使您可以解决许多技术计算问题，尤其是那些具有矩阵和向量公式的问题，而只需用 C 或 Fortran 等标量非交互式语言编写程序所需的时间的一小部分。MATLAB 名称代表矩阵实验室。MATLAB 最初的编写目的是提供对由 LINPACK 和 EISPACK 项目开发的矩阵软件的轻松访问，这两个项目共同代表了矩阵计算软件的最新技术。MATLAB 经过多年的发展，得到了许多用户的投入。在大学环境中，它是数学、工程和科学入门和高级课程的标准教学工具。在工业领域，MATLAB 是高效研究、开发和分析的首选工具。MATLAB 具有一系列称为工具箱的特定于应用程序的解决方案。对于大多数 MATLAB 用户来说非常重要，工具箱允许您学习应用专业技术。工具箱是 MATLAB 函数（M 文件）的综合集合，可扩展 MATLAB 环境以解决特定类别的问题。可用工具箱的领域包括信号处理、控制系统、神经网络、模糊逻辑、小波、仿真等。