## 金融代写|期权理论代写Mathematical Introduction to Options代考|MATH424

2022年10月14日

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

In the first section of this chapter we considered a simple one-step model with two possible outcomes. Then in the following section we turned our attention to a more general, continuous model, but we still only considered a single short step $\delta S_t$ over a period $\delta t$. These models not only gave insights into a general approach for solving previously intractable problems (risk neutrality); they also yielded the fundamental differential equation governing all options. We now extend the analysis from one to two steps and in the process we derive the central result which underlies the whole of the modern options industry.
(i) Beginning of First Step: We buy an option and hedge it with delta units of the underlying stock. We start with zero wealth so any cash surplus or deficit is borrowed or deposited with a bank. We have already seen from equation (4.11) that our position may be represented by
$$f_{S_t t}-\Delta_{S_t t} S_t+B_{S_t t}=0$$
Consider two concrete examples

• A call option valued at 10 when the stock price is 100 which has a delta of 0.5. The delta of the call is positive so the hedge is to short stock. Putting numbers into the last equation gives
$$10-0.5 \times 100+B_{S_t t}=0 \quad \text { or } \quad B_{S_t t}=+40$$
Shorting the stock means borrowing stock and selling it. This process generates 50 of cash but the option cost us 10 ; the net of the two is a cash surplus of 40 which we place on deposit.
• A put option worth 10 when the stock price is 100; delta is -0.5. The delta of a put is negative, so the hedge is to buy stock. Our equation now becomes
$$10+0.5 \times 100+B_{S_t t}=0 \quad \text { or } \quad B_{S_t t}=-60$$
This time we buy the option for 10 but also need to spend 50 on the stock hedge. Our total outlay is 60 which needs to be borrowed.

## 金融代写|期权理论代写Mathematical Introduction to Options代考|EXAMPLES OF DYNAMIC HEDGING

(i) The theory developed in the last section called for rebalancing of the hedge at infinitesimally small time intervals, but this is obviously not possible in practice. The example we consider is a 1-year call option for which we rebalance the hedge once a month; in real life, we would rebalance the hedge more often. The columns of Table $4.1$ are as follows.
(A) $S_t$ : Assuming the stock price starts at 100 , we have generated a scenario of stock prices after 1 month, 2 months, …, 12 months. These values are calculated from equation (3.7), making the risk-neutral substitution $m=(r-q)-\frac{1}{2} \sigma^2$. In this particular example, we have taken $r=6 \%, q=3 \%, \sigma=25 \%$ so that
$$S_{\text {month } i+1}=S_{\text {month } i} \exp \left{3 \% \times \frac{1}{12}-\frac{1}{2} \times(25 \%)^2+25 \% \times \sqrt{\frac{1}{12} z_{i+1}}\right}$$
where $z_{i+1}$ is a random variable drawn from a standard normal population. Such variables are easy to generate in a spreadsheet using formulas discussed in Chapter 10. An infinite number of paths can be generated in this column simply by pressing the button which allocates the new set of random numbers for Tables $4.1$ and $4.2$. We have simply chosen a couple of paths which are good illustrations of the present subject.
(B) $\Delta_{S_t t}$ : The deltas shown in the third column are calculated from the Black Scholes model and correspond to the stock prices of column 2 and the time left to maturity.

The last three columns correspond to the portfolio $f_{S_t t}-\Delta_{S_t t} S_t+B_{S_t t}=0$, which as we have seen should have value zero at every point in time. The first line of this part of the table is constructed as follows.
(C) $f_{S_0 0}:$ On day 1, when the stock price is $100.00$, we buy an option for its fair value of $11.01$. This fair value is obtained from the Black Scholes model.
(D) $\Delta_{S_0 0} \times S_0$ : We have already calculated the delta, and this is the number of shares that is shorted to hedge the option. The cash we receive as a result of this short is shown in this column.
(E) $B_{S_0 0}$ : The amount of cash available for depositing in the cash account is the difference of the last two items.
The remainder of the last three columns is filled in as follows.
(F) $\Delta_{S_t t} \times S_t$ : Each month, observe the new share price and calculate an appropriate delta (columns 2 and 3 ).
(G) The change in the cash account is the sum of three items:

• Interest on the cash surplus received for the previous month;
• Dividends on the stock borrowed in the previous month;
• Stock bought or sold to readjust the hedge.

# 期权理论代写

## 金融代写|期权理论代写期权数学介绍代考|DYNAMIC HEDGING

.

$$f_{S_t t}-\Delta_{S_t t} S_t+B_{S_t t}=0$$

• 当股价为100时，价值为10的看涨期权，其δ值为0.5。看涨期权的delta是正的，所以对冲是做空股票。把数字代入最后一个等式得到
$$10-0.5 \times 100+B_{S_t t}=0 \quad \text { or } \quad B_{S_t t}=+40$$
做空股票意味着借入股票并卖出。这个过程产生了50美元的现金，但期权花费了我们10美元;两者加起来的现金盈余是40，这是我们存入的。股价为100时价值10的看跌期权;等于-0.5。看跌期权的δ是负的，所以对冲是买股票。我们的等式现在变成了
$$10+0.5 \times 100+B_{S_t t}=0 \quad \text { or } \quad B_{S_t t}=-60$$
这一次我们以10的价格购买期权，但也需要花费50在股票对冲上。我们的总费用是60，需要借用。

## 金融代写|期权理论代写期权数学介绍代考|动态套期保值示例

.

(A) $S_t$:假设股价从100开始，我们生成了一个1个月，2个月，…，12个月后的股价情景。这些值由式(3.7)计算，使风险中性替代$m=(r-q)-\frac{1}{2} \sigma^2$。在这个特殊的例子中，我们取$r=6 \%, q=3 \%, \sigma=25 \%$，因此
$$S_{\text {month } i+1}=S_{\text {month } i} \exp \left{3 \% \times \frac{1}{12}-\frac{1}{2} \times(25 \%)^2+25 \% \times \sqrt{\frac{1}{12} z_{i+1}}\right}$$
，其中$z_{i+1}$是从标准正态总体中提取的随机变量。这些变量很容易在电子表格中使用第10章中讨论的公式生成。只需按下为表$4.1$和$4.2$分配新随机数的按钮，就可以在该列中生成无限多的路径。
(B) $\Delta_{S_t t}$:第三列所示的δ是根据Black Scholes模型计算出来的，对应于第2列的股票价格和剩余的到期时间

(C) $f_{S_0 0}:$第一天，当股票价格为$100.00$时，我们以其公允价值$11.01$购买期权。
(D) $\Delta_{S_0 0} \times S_0$:我们已经计算了delta，这是为对冲期权而做空的股票数量。
(E) $B_{S_0 0}$:可存入现金账户的现金金额是后两项的差额。
(F) $\Delta_{S_t t} \times S_t$:每个月，观察新股票的价格并计算一个适当的增量(列2和3)。
(G)现金账户的变化是三个项目的总和: 上月收到的现金盈余的利息;上月借入的股票的股息;买卖股票以调整套期

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