# 金融代写|期权理论代写Mathematical Introduction to Options代考|MATH1203

## 金融代写|期权理论代写Mathematical Introduction to Options代考|THE BINOMIAL NETWORK

(i) The stock price movement over a single step of length $\delta t$ is of little use in itself. We need to construct a network of successive steps covering the entire period from now to the maturity of the option; the beginning of one such network is shown in Figure 7.2.
The procedure for using this model to price an option is as follows:
(A) Select parameters $u, d$ and $p$ which conform to equation (7.2). The most popular ways of doing this are described in the following subparagraphs.
(B) Using these values of $u$ and $d$, work out the possible values for the stock price at the final nodes at $t=T$. We could work out the stock value for each node in the tree but if the tree is European, we only need the stock values in the last column of nodes.
(C) Corresponding to each of the final nodes at time $t=T$, there will be a stock price $S_{m, T}$ where $m$ indicates the specific node in the final column of nodes.

(D) Assume the derivative depends only on the final stock price. Corresponding to the stock price at each final node, there will be a derivative payoff $f_{m, T}\left(S_T\right)$.
(E) Just as each node is associated with a stock price, each node has a derivative price. The nodal derivative prices are related to each other by the repeated use of equations (7.1). Looking at Figure $7.2$ we have
$$\begin{gathered} f_4=\mathrm{e}^{-r \delta t}\left{p f_7+(1-p) f_8\right} \ f_5=\mathrm{e}^{-r \delta t}\left{p f_8+(1-p) f_9\right} \ \vdots \ f_2=\mathrm{e}^{-r \delta t}\left{p f_4+(1-p) f_5\right} \ \vdots \end{gathered}$$
This sequence of calculations allows the present value of the option, $f_0$, to be calculated from the payoff values of the option, $f_{m, T}\left(S_T\right)$; this is commonly referred to as “rolling back through the tree”.

## 金融代写|期权理论代写Mathematical Introduction to Options代考|APPLICATIONS

(i) European Call: Jarrow-Rudd Method $\left(u=d^{-1}=\mathrm{e}^{\sigma \sqrt{d t}}\right)$ : consider the tree shown in Figure 7.4. From the specification of the option and equation (7.6), the following parameters can be calculated:
With three steps $\delta t=0.5 / 3: \quad F_{t t+\delta t}=S_t \mathrm{e}^{(r-q) \delta t}=1.01005 S_t ; \quad \mathrm{e}^{-r \delta t}=0.983$ $u=\mathrm{e}^{\sigma \sqrt{6 t}}=1.0851 ; \quad d=\mathrm{e}^{-\sigma \sqrt{6 t}}=0.9216 ; \quad p=\frac{F_{t t+\delta t}-S_t \mathrm{e}^{-\sigma \sqrt{\delta t}}}{S_t \mathrm{e}^{\sigma \sqrt{\delta t}}-S_t \mathrm{e}^{-\sigma \sqrt{8 t}}}=0.541$
Using these $u$ and $d$ factors, we can start filling in the stock prices on the tree (shown just above each node). The intermediate values of $S_t$ are not really necessary for a European option, since the option payoff only depends on the stock price at maturity; however, they are shown for ease of understanding.

The payoff values of the option are $\max \left[\left(S_T-100\right), 0\right]$ and are shown just below the final nodes. The option values at the next column of nodes to the left can be calculated as follows:
$$\begin{gathered} f(117.74,4 \text { months })=0.983{0.541 \times 27.76+(1-0.541) \times 8.51}=18.609 \ f(100.00,4 \text { months })=0.983{0.541 \times 8.51+(1-0.541) \times 0.00}=4.43 \ f(84.93,4 \text { months })=0.983{0.541 \times 0.00+(1-0.541) \times 0.00}=0.00 \end{gathered}$$

Continuing this process back to the first node (“rolling back through the tree”) finally gives a 6-month option value of 7.44. This may be compared to the Black Scholes value (equivalent to an infinite number of steps) of $7.01$. This price error is equivalent to using a volatility of $21.6 \%$ instead of $20 \%$ in the Black Scholes formula.
(ii) European Call: Cox-Ross-Rubinstein Method ( $\left.p=\frac{1}{2}\right)$ : For purposes of comparison, we reprice the same option as in the last section, using a different discretization procedure. Once again we have $\delta t=0.5 / 3$ and $\mathrm{e}^{-t 6 t}=0.983$ but now we use $p=\frac{1}{2}$ and equation (7.7), so that
$$u=\frac{F_{t t+\delta t}}{S_t}(1+\sigma \sqrt{\delta t})=1.093 ; \quad d=\frac{F_{t t+\delta t}}{S_t}(1-\sigma \sqrt{\delta t})=0.928$$
The tree is shown in Figure 7.5. This time, only the final stock prices are shown. The procedure for rolling back through the tree is identical to that in the last section, with the simplifying feature that $p=(1-p)=\frac{1}{2}$. The calculation for the top right-hand step in the diagram becomes
$$0.983 \times \frac{1}{2} \times(30.40+10.72)=20.220$$
and so on through the tree. For all intents and purposes, the final answer is identical to that of the last section (more precise numbers are $7.444$ previously and $7.438$ now).

# 期权理论代考

## 金融代写|期权理论代写期权数学介绍代考|THE BINOMIAL NETWORK

. THE . THE

(A)选择符合式(7.2)的参数$u, d$和$p$。
(B)使用$u$和$d$的这些值，在$t=T$的最终节点上计算出股票价格的可能值。我们可以计算出树中每个节点的股票价值，但如果树是欧洲的，我们只需要最后一列节点中的股票价值。
(C)与时间$t=T$的每个最终节点对应，将有一个股票价格$S_{m, T}$，其中$m$表示节点最后一列中的特定节点。

(D)假设衍生品只取决于最终股价。对应于最后每个节点的股价，会有一个衍生品收益$f_{m, T}\left(S_T\right)$ .
(E)正如每个节点都与股价相关联，每个节点都有一个衍生品价格。节点导数价格通过重复使用式(7.1)相互关联。看看图$7.2$，我们有
$$\begin{gathered} f_4=\mathrm{e}^{-r \delta t}\left{p f_7+(1-p) f_8\right} \ f_5=\mathrm{e}^{-r \delta t}\left{p f_8+(1-p) f_9\right} \ \vdots \ f_2=\mathrm{e}^{-r \delta t}\left{p f_4+(1-p) f_5\right} \ \vdots \end{gathered}$$

## 金融代写|期权理论代写选项数学介绍代考|APPLICATIONS

. application . application

(i) European Call: Jarrow-Rudd Method $\left(u=d^{-1}=\mathrm{e}^{\sigma \sqrt{d t}}\right)$:考虑图7.4所示的树。根据期权和方程(7.6)的规范，可以计算出以下参数:
$\delta t=0.5 / 3: \quad F_{t t+\delta t}=S_t \mathrm{e}^{(r-q) \delta t}=1.01005 S_t ; \quad \mathrm{e}^{-r \delta t}=0.983$$u=\mathrm{e}^{\sigma \sqrt{6 t}}=1.0851 ; \quad d=\mathrm{e}^{-\sigma \sqrt{6 t}}=0.9216 ; \quad p=\frac{F_{t t+\delta t}-S_t \mathrm{e}^{-\sigma \sqrt{\delta t}}}{S_t \mathrm{e}^{\sigma \sqrt{\delta t}}-S_t \mathrm{e}^{-\sigma \sqrt{8 t}}}=0.541 使用这些u和d因子，我们可以开始在树(显示在每个节点的上方)上填充股票价格。对于欧洲期权来说，S_t的中间值并不是必需的，因为期权的支付只取决于到期时的股价;然而，它们是为了便于理解而显示出来的 期权的收益值是\max \left[\left(S_T-100\right), 0\right]，显示在最后的节点下面。左边下一列节点的选项值可以计算如下:$$ \begin{gathered} f(117.74,4 \text { months })=0.983{0.541 \times 27.76+(1-0.541) \times 8.51}=18.609 \ f(100.00,4 \text { months })=0.983{0.541 \times 8.51+(1-0.541) \times 0.00}=4.43 \ f(84.93,4 \text { months })=0.983{0.541 \times 0.00+(1-0.541) \times 0.00}=0.00 \end{gathered} $$继续这个过程回到第一个节点(“通过树回滚”)，最终得到6个月的选项值7.44。这可以与7.01的布莱克·斯科尔斯值(相当于无限步数)相比较。这个价格误差相当于在Black Scholes公式中使用波动率21.6 \%而不是20 \%。 (ii) European Call: Cox-Ross-Rubinstein方法(\left.p=\frac{1}{2}\right):为了进行比较，我们使用了不同的离散化程序，对上一节中的相同期权重新定价。我们仍然有\delta t=0.5 / 3和\mathrm{e}^{-t 6 t}=0.983，但是现在我们使用p=\frac{1}{2}和等式(7.7)，因此$$ u=\frac{F_{t t+\delta t}}{S_t}(1+\sigma \sqrt{\delta t})=1.093 ; \quad d=\frac{F_{t t+\delta t}}{S_t}(1-\sigma \sqrt{\delta t})=0.928 $$树如图7.5所示。这一次，只显示最终股票价格。回滚树的过程与上一节中的过程相同，其简化特性是p=(1-p)=\frac{1}{2}。图中右上角步骤的计算变成$$ 0.983 \times \frac{1}{2} \times(30.40+10.72)=20.220$$，以此类推。无论如何，最终的答案都与上一节的答案相同(更精确的数字以前是$7.444$，现在是$7.438\$)

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