约翰.赫尔,期权期货和其他衍生品(third edition)习题答案

12.1 一个证券组合当前价值为＄1000万，β值为1.0，S&P100目前位于250，解释一个执行价格为240。

标的物为S&P100的看跌期权如何为该组合进行保险？当S&P100跌到480，这个组合的期望价值是10 ×（480/500）=＄9.6million.买看跌期权10，000，000/500=20，000可以防止这个组合下跌到＄9.6million下的损失。

因此总共需要200份合约12.2 “一旦我们知道了支付连续红利股票的期权的定价方法，我们便知道了股票指数期权、货币期权和期货期权的定价”。

请解释这句话。

一个股票指数类似一个连续支付红利的股票12.3 请说明日圆看涨期权与日圆期货看涨期权的不同之处一个日元的看涨期权给了持有者在未来某个时刻以确定的价格购买日圆的权利，一个日圆远期看涨期权给予持有者在未来时刻远期价格超过特定范围按原先价格购买日圆的权利。

如果远期齐权行使，持有者将获得一个日圆远期和约的多头。

12.4请说明货币期权是如何进行套期保值的？12.5 计算3个月期，处于平价状态的欧式看涨股票指数期权的价值。

指数为250。

无风险年利率为10%，指数年波动率为18%，指数的年红利收益率为3%。

一个日元的看涨期权给了持有者在未来某个时刻以确定的价格购买日圆的权利，一个日圆远期看涨期权给予持有者在未来时刻远期价格超过特定范围按原先价格购买日圆的权利。

如果远期齐权行使，持有者将获得一个日圆远期和约的多头。

12.6 有一美式看涨期货期权，期货合约和期权合约同时到期。

在任何情况下期货期权比相应的标的物资产的美式期权更值钱？当远期价格大于即期价格时，美式远期期权在远期和约到期前的价值大于相对应的美式期权/12.7 计算5个月有效期的欧式看跌期货期权的价值。

期货价格为＄19，执行价格为＄20，无风险年利率为12%。

期货价格的年波动率为20%。

本题中12.8 假设交易所构造了一个股票指数。

金融工程原理作业4.2 答：LIBOR 是伦敦同业银行拆出利率（London Interbank Offered Rate ）的缩写，是指银行给其他大银行提供企业资金时所收取的利率。

LIBID 是伦敦同业拆出利率（London Interbank Bid Rate ），即银行同意其他银行以LIBID 利率将资金存入自己的银行。

LIBOR 利率会略高于LIBID 利率。

4.4 解：设利率为R（a ）一年复利一次，1000（1+R ）=1100，求出R=10%（b ）一年复利两次，1000（1+R/2）2=1100，求出R=9.76%（c ）每月复利一次，共12次，1000(1+R/12)12=1100，求出 R=9.57% （d ）连续复利，1000e R =1100，求出R=9.53%4.6解：收入R K 的FRA 的价值为V 22))((12T R F K FRA e T T R R L ---=，已知，R K =9.5%，T 2=1.25，T 1=1，L=1000000，R 2=8.6%R F =121122T T T R T R --=125.15.825.16.8--⨯×100%=9%，对应的每季度复利一次的利率为R F =4（e 4/09.0﹣1）=9.102%代入得，V 25.1086.0)125.1()09102.0095.0(1000000⨯--⨯-⨯=e FRA =893.59答：该FRA 的价值为893.59美元。

4.8答：久期可以用来估算债券收益的一个微小变化对证券组合价值的影响，金融机构常常通过确保其资产平均久期等于其负债平均久期来对冲其面临的风险，但是组合对于利率较大的平行移动和非平行移动仍有风险暴露，这是久期的缺陷性。

4.10 解：已知R c =12%，m=4，则由公式得每年计息4次的利率R m =m(e Rc /m -1)= 4(e 12%/4-1)=12.182%所以每季度的利息为10000*（12.182%/4）=304.55美元4.12解：设收益率为y,年利率为8%,半年付息一次，每次付息为4元，三年期共付息6次，最后一次连本金104，可得4e y 5.0-+4e y 1.0-+4e y 5.1-+4e y 2-+4e y 5.2-+104e y 3-=104,解得，y=6.407%4.14 解：R F2=121122T T T R T R --=%41202.0203.0=--⨯ R F3=232233T T T R T R --=%1.523203.03037.0=-⨯-⨯ R F4=343344T T T R T R --=%7.5343037.04042.0=-⨯-⨯ R F5=454455T T T R T R --=%7.5454042.05045.0=-⨯-⨯ 所以，第二年、第三年、第四年、第五年的远期利率分别为4%，5.1%，5.7%和5.7%。

赫尔《期权、期货及其他衍⽣产品》（第7版）课后习题详解（曲率、时间与Quanto调整）29.2　课后习题详解⼀、问答题1. 解释你如何去对⼀个在5年后付出100R 的衍⽣产品定价，其中R 是在4年后所观察到的1年期利率(按年复利)。

当⽀付时间在第4年时，会有什么区别?当⽀付时间在第6年时，会有什么区别?Explain how you would value a derivative that pays off 100R in five years where R is the one-year interest rate (annually compounded) observed in four years. What difference would it make if the payoff were in four years? What difference would it make if tile payoff were in six years?答：衍⽣产品的价值是，其中P(0，t)是⼀个t 期零息债券的价格，为期限在和之间的远期利率，以年复利计息。

当⽀付时间在第4年时，价值为，其中c 为由教材中⽅程（29-2）得到的曲率调整。

曲率调整公式为：其中，是远期利率在时间和之间的波动率。

表达式100(R4,5 + c)为在⼀个远期风险中性的世界中，⼀个4年后到期的零息债券的预期收益。

如果在6年后进⾏⽀付，由教材中的⽅程（29-4）得到其价值为：其中，ρ为(4,5)和(4,6)远期利率之间的相关系数。

作为估计，假定，近似计算其指数函数，得到衍⽣产品的价值为：。

2. 解释在下⾯情况下，有没有必要做出任何曲率或时间调整?(a)要对⼀种期权定价，期权每个季度⽀付⼀次，数量等于5年的互换利率超出3个⽉LIBOR利率的部分(假如超出的话)，本⾦为100美元，收益发⽣在利率被观察到后的90天。

(b)要对⼀种差价期权定价，期权每季度⽀付⼀次，数量等于3个⽉的LIBOR利率减去3个⽉的短期国库券利率，收益发⽣在利率被观察后的90天。

第一章1.1请解释远期多头与远期空头的区别。

答：远期多头指交易者协定将来以某一确定价格购入某种资产；远期空头指交易者协定将来以某一确定价格售出某种资产。

1.2请详细解释套期保值、投机与套利的区别。

答：套期保值指交易者采取一定的措施补偿资产的风险暴露；投机不对风险暴露进行补偿，是一种“赌博行为”；套利是采取两种或更多方式锁定利润。

1.3请解释签订购买远期价格为$50的远期合同与持有执行价格为$50的看涨期权的区别。

答：第一种情况下交易者有义务以50$购买某项资产（交易者没有选择），第二种情况下有权利以50$购买某项资产（交易者可以不执行该权利）。

1.4一位投资者出售了一个棉花期货合约，期货价格为每磅50美分，每个合约交易量为50，000磅。

请问期货合约结束时，当合约到期时棉花价格分别为（a）每磅48.20美分；（b）每磅51.30美分时，这位投资者的收益或损失为多少？答：(a)合约到期时棉花价格为每磅$0.4820时，交易者收入：（$0.5000-$0.4820）×50，000＝$900;(b)合约到期时棉花价格为每磅$0.5130时，交易者损失:($0.5130-$0.5000) ×50，000=$6501.5假设你出售了一个看跌期权，以$120执行价格出售100股IBM的股票，有效期为3个月。

IBM股票的当前价格为$121。

你是怎么考虑的？你的收益或损失如何？答：当股票价格低于$120时，该期权将不被执行。

当股票价格高于$120美元时，该期权买主执行该期权，我将损失100(st-x)。

1.6你认为某种股票的价格将要上升。

现在该股票价格为$29,3个月期的执行价格为$30的看跌期权的价格为$2.90.你有$5,800资金可以投资。

现有两种策略：直接购买股票或投资于期权，请问各自潜在的收益或损失为多少？答：股票价格低于$29时，购买股票和期权都将损失，前者损失为$5,800×(29-p),$29×(p-29),后者损失为$5，800;当股票价格为（29，30），购买股票收益为$5,800$29购买期权损失为$5,800;当股票价格高于$30时，购买股票收益为$5,800×(p-29)，$29×(p-30)-5,800。

1第一次作业参考答案第1章1.26远期合约多头规定了一年后以每盎司1000美元买入黄金，到期远期合约必须执行，交易双方权利义务对等；期权合约多头规定了一年后以每盎司1000美元买入黄金的权利，到期合约可以不执行，也可以执行，交易双方权利义务不对等。

假设S T为一年以后黄金的价格，则远期合约的收益为S T-1000;期权合约的受益为S T-1100,如果S T>1000;-100,如果S T<10001.27投资人承诺在7月份以40美元的执行价格买入股票。

如果未来股票价格跌至37美元以下，则该投资人赚取的3美元期权费不足以弥补期权上的损失，从而亏损。

当未来股票价格为37-40美元时，交易对手会执行期权，此时，投资人此时同样有正收益。

如果未来股票价格高于40美元，该期权不会被对手执行，此时投资者仅赚取期权费。

1.28远期：购入三个月期限的300万欧元的欧元远期合约，并在三个月后，用到期的远期合约进行支付300万欧元。

期权：购入三个月期限的300万欧元的欧元看涨期权，如果三个月后汇率高于期权约定执行汇率，则执行该期权，反之则不执行该期权。

1.29当股票到期价格低于30美元时，两个期权合约都与不会被执行，该投资者无头寸;当股票价格高于32.5美元时,两个期权都会执行，该投资者无头寸，若股票价格在30-32.5美元之间时，该投资者买入期权会被执行，卖出的期权不会被执行，因此该投资者持有长头寸。

1.30（低买高卖）借入1000美元资金，买入黄金，同时在卖出一年期的黄金远期合约，锁定到期的价格1200美元，到期偿还本金和利息。

到期时的受益为1200-1000（1+10%）=100；收益率为100/1000=10%1.312由于期权存在杠杆效应，看张期权的风险更大，同时收益率也更高。

假设股票在S时，购入看涨期权和购入股票无差异，则100*（S-94)=2000（S-95）-9400即S=100。

如果未来股票价格高于100美元，则购入看涨期权合约则盈利更高，反之，如果未来股票价格低于100美元，则购入股票比购入期权合约盈利更多。

CHAPTER 13Wiener P rocesses and Itô’s LemmaPractice QuestionsProblem 13.1.What would it mean to assert that the temperature at a certain place follows a Markov process? Do you think that temperatures do, in fact, follow a Markov process?Imagine that you have to forecast the future temperature from a) the current temperature, b) the history of the temperature in the last week, and c) a knowledge ofseasonal averages and seasonal trends. If temperature followed a Markov process, the history of the temperature in the last week would be irrelevant.To answer the second part of the question you might like to consider the following scenario for the first week in May:(i) Monday to Thursday are warm days; today, Friday, is a very cold day. (ii) Monday to Friday are all very cold days.What is your forecast for the weekend? If you are more pessimistic in the case of the second scenario, temperatures do not follow a Markov process.Problem 13.2.Can a trading rule based on the past history of a stock’s price ever produce returns that are consistently above average? Discuss.The first point to make is that any trading strategy can, just because of good luck, produce above average returns. The key question is whether a trading strategy consistently outperforms the market when adjustments are made for risk. It is certainly possible that a trading strategy could do this. However, when enough investors know about the strategy and trade on the basis of the strategy, the profit will disappear.As an illustration of this, consider a phenomenon known as the small firm effect. Portfolios of stocks in small firms appear to have outperformed portfolios of stocks in large firms when appropriate adjustments are made for risk. Research was published about this in the early 1980s and mutual funds were set up to take advantage of the phenomenon. There is some evidence that this has resulted in the phenomenon disappearing.Problem 13.3.A company’s cash position, measured in millions of dollars, follows a generalized Wiener process with a drift rate of 0.5 per quarter and a variance rate of 4.0 per quarter. How high does the company’s initial cash position have to be for the company to have a less than 5% chance of a negative cash position by the end of one year?Supp ose that the company’s initial cash position is x . The probability distribution of the cash position at the end of one year is (40544)(2016)x x ϕϕ+⨯.,⨯=+.,where ()m v ϕ, is a normal probability distribution with mean m and variance v . The probability of a negative cash position at the end of one year is204x N +.⎛⎫- ⎪⎝⎭where ()N x is the cumulative probability that a standardized normal variable (with mean zero and standard deviation 1.0) is less than x . From normal distribution tables200054x N +.⎛⎫-=. ⎪⎝⎭when:20164494x +.-=-.i.e., when 45796x =.. The initial cash position must therefore be $4.58 million.Problem 13.4.Variables 1X and 2X follow generalized Wiener processes with drift rates 1μ and2μ and variances 21σ and 22σ. What process does 12X X + follow if:(a) The changes in 1X and 2X in any short interval of time are uncorrelated?(b) There is a correlation ρ between the changes in 1X and 2X in any short interval of time?(a) Suppose that X 1 and X 2 equal a 1 and a 2 initially. After a time period of length T , X 1 has the probability distribution2111()a T T ϕμσ+,and 2X has a probability distribution2222()a T T ϕμσ+,From the property of sums of independent normally distributed variables, 12X X + has the probability distribution()22112212a T a T T T ϕμμσσ+++,+i.e.,22121212()()a a T T ϕμμσσ⎡⎤+++,+⎣⎦This shows that 12X X + follows a generalized Wiener process with drift rate 12μμ+and variance rate 2212σσ+.(b) In this case the change in the value of 12X X + in a short interval of time t ∆ has the probability distribution:22121212()(2)t t ϕμμσσρσσ⎡⎤+∆,++∆⎣⎦If 1μ, 2μ, 1σ, 2σ and ρ are all constant, arguments similar to those in Section 13.2 show that the change in a longer period of time T is22121212()(2)T T ϕμμσσρσσ⎡⎤+,++⎣⎦The variable,12X X +, therefore follows a generalized Wiener process with drift rate12μμ+ and variance rate 2212122σσρσσ++.Problem 13.5.Consider a variable,S , that follows the process dS dt dz μσ=+For the first three years, 2μ= and 3σ=; for the next three years, 3μ= and 4σ=. If the initial value of the variable is 5, what is the probability distribution of the value of the variable at the end of year six?The change in S during the first three years has the probability distribution (2393)(627)ϕϕ⨯,⨯=,The change during the next three years has the probability distribution (33163)(948)ϕϕ⨯,⨯=,The change during the six years is the sum of a variable with probability distribution(627)ϕ, and a variable with probability distribution (948)ϕ,. The probability distribution of the change is therefore (692748)ϕ+,+ (1575)ϕ=,Since the initial value of the variable is 5, the probability distribution of the value of the variable at the end of year six is (2075)ϕ,Problem 13.6.Suppose that G is a function of a stock price, S and time. Suppose that S σ and G σ are the volatilities of S and G . Show that when the expected return of S increases by S λσ, the growth rate of G increases by G λσ, where λ is a constant.From Itô’s lemmaG S GG S Sσσ∂=∂Also the drift of G is222212G G G S S S t S μσ∂∂∂++∂∂∂where μ is the expected return on the stock. When μ increases by S λσ, the drift of Gincreases byS GS Sλσ∂∂ orG G λσThe growth rate of G , therefore, increases by G λσ.Problem 13.7.Stock A and stock B both follow geometric Brownian motion. Changes in any short interval of time are uncorrelated with each other. Does the value of a portfolio consisting of one of stock A and one of stock B follow geometric Brownian motion? Explain your answer.Define A S , A μ and A σ as the stock price, expected return and volatility for stock A. Define B S , B μ and B σ as the stock price, expected return and volatility for stock B. Define A S ∆ and B S ∆ as the change in A S and B S in time t ∆. Since each of the two stocks follows geometric Brownian motion,A A A A A S S t S μσε∆=∆+B B B B B S S t S μσε∆=∆+where A ε and B ε are independent random samples from a normal distribution.()(A B A A B B A A A B B B S S S S t S S μμσεσε∆+∆=+∆++This cannot be written as()()A B A B A B S S S S t S S μσ∆+∆=+∆++for any constants μ and σ. (Neither the drift term nor the stochastic term correspond.) Hence the value of the portfolio does not follow geometric Brownian motion.Problem 13.8.S S t S μσε∆=∆+ where μ and σ are constant. Explain carefully the difference between this model andeach of the following:S t S S t S t S μσεμσεμσε∆=∆+∆=∆+∆=∆+Why is the model in equation (13.8) a more appropriate model of stock price behavior than any of these three alternatives?In:S S t S μσε∆=∆+ the expected increase in the stock price and the variability of the stock price are constant when both are expressed as a proportion (or as a percentage) of the stock price In:S t μ∆=∆+the expected increase in the stock price and the variability of the stock price are constant in absolute terms. For example, if the expected growth rate is $5 per annum when the stockprice is $25, it is also $5 per annum when it is $100. If the standard deviation of weekly stock price movements is $1 when the price is $25, it is also $1 when the price is $100. In:S S t μ∆=∆+the expected increase in the stock price is a constant proportion of the stock price while the variability is constant in absolute terms. In:S t S μσ∆=∆+the expected increase in the stock price is constant in absolute terms while the variability of the proportional stock price change is constant. The model:S S t S μσ∆=∆+ is the most appropriate one since it is most realistic to assume that the expected percentage return and the variability of the percentage return in a short interval are constant.Problem 13.9.It has been suggested that the short-term interest rate,r , follows the stochastic process()dr a b r dt rc dz =-+where a , b , and c are positive constants and dz is a Wiener process. Describe the nature of this process.The drift rate is ()a b r -. Thus, when the interest rate is above b the drift rate is negative and, when the interest rate is below b , the drift rate is positive. The interest rate is therefore continually pulled towards the level b . The rate at which it is pulled toward this level is a . A volatility equal to c is superimposed upon the “pull” or the drift.Suppose 04a =., 01b =. and 015c =. and the current interest rate is 20% per annum. The interest rate is pulled towards the level of 10% per annum. This can be regarded as a long run average. The current drift is 4-% per annum so that the expected rate at the end of one year is about 16% per annum. (In fact it is slightly greater than this, because as the interest rate decreases, the “pull” decreases.) Superimposed upon the drift is a volatility of 15% per annum.Problem 13.10.Suppose that a stock price, S , follows geometric Brownian motion with expected return μ and volatility σ: dS S dt S dz μσ=+What is the process followed by the variable n S ? Show that n S also follows geometric Brownian motion.If ()n G S t S ,= then 0G t ∂/∂=, 1n G S nS -∂/∂=, and 222(1)n G S n n S -∂/∂=-. Using Itô’s lemma:21[(1)]2dG nG n n G dt nG dz μσσ=+-+This shows that n G S = follows geometric Brownian motion where the expected return is21(1)2n n n μσ+-and the volatility is n σ. The stock price S has an expected return of μ and the expected value of T S is 0T S e μ. The expected value of n T S is212[(1)]0n n n T n S eμσ+-Problem 13.11.Suppose that x is the yield to maturity with continuous compounding on a zero-coupon bond that pays off $1 at time T . Assume that x follows the process0()dx a x x dt sx dz =-+where a , 0x , and s are positive constants and dz is a Wiener process. What is the process followed by the bond price?The process followed by B , the bond price, is from Itô’s lemma:222021()2B B B B dB a x x s x dt sxdz x t x x ⎡⎤⎢⎥⎢⎥⎢⎥⎣⎦∂∂∂∂=-+++∂∂∂∂Since: ()x T t B e --=the required partial derivatives are()()22()22()()()()x T t x T t x T t Bxe xB t BT t e T t B x B T t e T t B x------∂==∂∂=--=--∂∂=-=-∂ Hence:22201()()()()2dB a x x T t x s x T t Bdt sx T t Bdz⎡⎤⎢⎥⎢⎥⎢⎥⎣⎦=---++---Problem 13.12 (Excel Spreadsheet)A stock whose price is $30 has an expected return of 9% and a volatility of 20%. In Excel simulate the stock price path over 5 years using monthly time steps and random samples from a normal distribution. Chart the simulated stock price path. By hitting F9 observe how the path changes as the random sample change.The process ist S t S S ∆⨯ε⨯⨯+∆⨯⨯=∆20.009.0Where ∆t is the length of the time step (=1/12) and ε is a random sample from a standard normal distribution.Further QuestionsProblem 13.13.Suppose that a stock price has an expected return of 16% per annum and a volatility of 30% per annum. When the stock price at the end of a certain day is $50, calculate the following:(a) The expected stock price at the end of the next day.(b) The standard deviation of the stock price at the end of the next day. (c) The 95% confidence limits for the stock price at the end of the next day.With the notation in the text2()St t S ϕμσ∆∆,∆In this case 50S =, 016μ=., 030σ=. and 1365000274t ∆=/=.. Hence(016000274009000274)50(0000440000247)Sϕϕ∆.⨯.,.⨯.=.,.and2(50000044500000247)S ϕ∆⨯.,⨯.that is, (002206164)S ϕ∆.,.(a)(b) The standard deviation of the stock price at the end of the next day is 0785=. (c) 95% confidence limits for the stock price at the end of the next day are 500221960785and 500221960785.-.⨯..+.⨯. i.e.,4848and 5156..Note that some students may consider one trading day rather than one calendar day. Then 1252000397t ∆=/=.. The answer to (a) is then 50.032. The answer to (b) is 0.945. The answers to part (c) are 48.18 and 51.88.Problem 13.14.A company’s cash position, measured in millions of dollars, follows a generalized Wiener process with a drift rate of 0.1 per month and a variance rate of 0.16 per month. The initial cash position is 2.0.(a) What are the probability distributions of the cash position after one month, six months, and one year?(b) What are the probabilities of a negative cash position at the end of six months and one year?(c) At what time in the future is the probability of a negative cash position greatest?(a) The probability distributions are:(2001016)(21016)ϕϕ.+.,.=.,.(20060166)(26096)ϕϕ.+.,.⨯=.,.(201201612)(32196)ϕϕ.+.,.⨯=.,.(b) The chance of a random sample from (26096)ϕ.,. being negative is(265)N N ⎛=-. ⎝where ()N x is the cumulative probability that a standardized normal variable [i.e., avariable with probability distribution (01)ϕ,] is less than x . From normaldistribution tables (265)00040N -.=.. Hence the probability of a negative cash position at the end of six months is 0.40%.Similarly the probability of a negative cash position at the end of one year is(230)00107N N ⎛=-.=. ⎝or 1.07%.(c) In general the probability distribution of the cash position at the end of x months is(2001016)x x ϕ.+.,.The probability of the cash position being negative is maximized when:is minimized. Define11223122325025250125(250125)y x xdy x xdxx x----==+.=-.+.=-.+.This is zero when 20x=and it is easy to verify that 220d y dx/>for this value of x. It therefore gives a minimum value for y. Hence the probability of a negative cash position is greatest after 20 months.Problem 13.15.Suppose that x is the yield on a perpetual government bond that pays interest at the rate of $1 per annum. Assume that x is expressed with continuous compounding, that interest is paid continuously on the bond, and that x follows the process()dx a x x dt sx dz=-+where a,x, and s are positive constants and dz is a Wiener process. What is the process followed by the bond price? What is the expected instantaneous return (including interest and capital gains) to the holder of the bond?The process followed by B, the bond price, is from Itô’s lemma:222021()2B B B BdB a x x s x dt sxdzx t x x⎡⎤⎢⎥⎢⎥⎢⎥⎣⎦∂∂∂∂=-+++∂∂∂∂In this case1Bx=so that:222312B B Bt x x x x∂∂∂=;=-;=∂∂∂Hence2202322021121()21()dB a x x s x dt sxdzx x xs sa x x dt dzx x x⎡⎤=--+-⎢⎥⎣⎦⎡⎤=--+-⎢⎥⎣⎦The expected instantaneous rate at which capital gains are earned from the bond is therefore:2021()sa x xx x--+The expected interest per unit time is 1. The total expected instantaneous return is therefore:20211()sa x xx x--+When expressed as a proportion of the bond price this is:202111()sa x xx x x⎛⎫⎛⎫--+ ⎪⎪⎝⎭⎝⎭20()ax x x s x=--+Problem 13.16.If S follows the geometric Brownian motion process in equation (13.6), what is the process followed by (a) y = 2S, (b) y=S 2 , (c) y=e S , and (d) y=e r(T-t)/S. In each case express the coefficients of dt and dz in terms of y rather than S.(a) In this case 2y S ∂/∂=, 220y S ∂/∂=, and 0y t ∂/∂= so that Itô’s lemma gives 22dy S dt S dz μσ=+or dy y dt y dz μσ=+(b) In this case 2y S S ∂/∂=, 222y S ∂/∂=, and 0y t ∂/∂= so that Itô’s lemma gives2222(2)2dy S S dt S dz μσσ=++ or2(2)2dy y dt y dz μσσ=++ (c) In this case S y S e ∂/∂=, 22S y S e ∂/∂=, and 0y t ∂/∂= so that Itô’s lemma gives22(2)S S S dy Se S e dt Se dz μσσ=+/+ or22[ln (ln )2]ln dy y y y y dt y y dz μσσ=+/+(d) In this case ()2r T t y S e S y S -∂/∂=-/=-/, 22()3222r T t y S e S y S -∂/∂=/=/, and()r T t y t re S ry -∂/∂=-/=- so that Itô’s lemma gives2()dy ry y y dt y dz μσσ=--+- or2()dy r y dt y dz μσσ=-+--Problem 13.17.A stock price is currently 50. Its expected return and volatility are 12% and 30%,respectively. What is the probability that the stock price will be greater than 80 in two years? (Hint 80T S > when ln ln 80T S >.)The variable ln T S is normally distributed with mean 20ln (2)S T μσ+-/ and standarddeviation σ050S =, 012μ=., 2T =, and 030σ=. so that the meanand standard deviation of ln T S are 2ln 50(012032)24062+.-./=. and 00424.=., respectively. Also, ln804382=.. The probability that 80T S > is the same as the probability that ln 4382T S >.. This is4382406211(0754)0424N N .-.⎛⎫-=-. ⎪.⎝⎭where ()N x is the probability that a normally distributed variable with mean zero and standard deviation 1 is less than x . From the tables at the back of the book (0754)0775N .=. so that the required probability is 0.225.Problem 13.18 (See Excel Worksheet)Stock A, whose price is $30, has an expected return of 11% and a volatility of 25%. Stock B, whose price is $40, has an expected return of 15% and a volatility of 30%. The processes driving the returns are correlated with correlation parameter ρ. In Excel, simulate the two stock price paths over three months using daily time steps and random samples from normal distributions. Chart the results and by hitting F9 observe how the paths change as the random samples change. Consider values of ρ equal to 0.50, 0.75, and 0.95.The processes aret S t S S A A A A ∆⨯ε⨯⨯+∆⨯⨯=∆25.011.0t S t S S B B B B ∆⨯ε⨯⨯+∆⨯⨯=∆30.015.0Where ∆t is the length of the time step (=1/252) and the ε’s are correlated samples from standard normal distributions.。

CH99.1 股票现价为$40。

已知在一个月后股价为$42或$38。

无风险年利率为8％（连续复利）。

执行价格为$39的1个月期欧式看涨期权的价值为多少？ 解：考虑一资产组合：卖空1份看涨期权；买入Δ份股票。

若股价为$42，组合价值则为42Δ－3；若股价为$38，组合价值则为38Δ 当42Δ－3＝38Δ，即Δ＝0.75时，组合价值在任何情况下均为$28.5，其现值为：，0.08*0.0833328.528.31e −=即：－f ＋40Δ＝28.31 其中f 为看涨期权价格。

所以，f ＝40×0.75－28.31＝$1.69另解：（计算风险中性概率p ） 42p －38（1－p ）＝，p ＝0.56690.08*0.0833340e期权价值是其期望收益以无风险利率贴现的现值，即： f ＝（3×0.5669＋0×0.4331）＝$1.690.08*0.08333e−9.2 用单步二叉树图说明无套利和风险中性估值方法如何为欧式期权估值。

解：在无套利方法中，我们通过期权及股票建立无风险资产组合，使组合收益率等价于无风险利率，从而对期权估值。

在风险中性估值方法中，我们选取二叉树概率，以使股票的期望收益率等价于无风险利率，而后通过计算期权的期望收益并以无风险利率贴现得到期权价值。

9.3什么是股票期权的Delta ？解：股票期权的Delta 是度量期权价格对股价的小幅度变化的敏感度。

即是股票期权价格变化与其标的股票价格变化的比率。

9.4某个股票现价为$50。

已知6个月后将为$45或$55。

无风险年利率为10％（连续复利）。

执行价格为$50，6个月后到期的欧式看跌期权的价值为多少？ 解：考虑如下资产组合，卖1份看跌期权，买Δ份股票。

若股价上升为$55，则组合价值为55Δ；若股价下降为$45，则组合价值为：45Δ－5 当55Δ＝45Δ－5，即Δ＝－0.50时，6个月后组合价值在两种情况下将相等，均为$－27.5，其现值为：，即：0.10*0.5027.5$26.16e −−=− －P ＋50Δ＝－26.16所以，P ＝－50×0.5＋26.16＝$1.16 另解：求风险中性概率p0.10*0.505545(1)50p p e+−= 所以，p ＝0.7564看跌期权的价值P ＝0.10*0.50(0*0.75645*0.2436)$1.16e −+=9.5 某个股票现价为$100。

赫尔《期权、期货及其他衍⽣产品》复习笔记及课后习题详解（利率期货）【圣才出品】第6章利率期货6.1 复习笔记1．天数计算和报价惯例天数计算常表⽰为X/Y，计算两个⽇期间获得的利息时，X定义了两个⽇期间天数计算的⽅式，Y定义了参照期内总天数计算的⽅式。

两个⽇期间获得的利息为：（两个⽇期之间的天数/参考期限的总天数）×参考期限内所得利息在美国常⽤的三种天数计算惯例为：①实际天数/实际天数；②30/360；③实际天数/360。

（1）美国短期债券的报价货币市场的产品报价采⽤贴现率⽅式，该贴现率对应于所得利息作为最终⾯值的百分⽐⽽不是最初所付出价格的百分⽐。

⼀般来讲，美国短期国债的现⾦价格与报价的关系式为：P＝360（100－Y）/n其中，P为报价，Y为现⾦价格，n为短期债券期限内以⽇历天数所计算的剩余天数。

（2）美国长期国债美国长期国债是以美元和美元的1/32为单位报出的。

所报价格是相对于⾯值100美元的债券。

报价被交易员称为纯净价，它与现⾦价有所不同，交易员将现⾦价称为带息价格。

⼀般来讲，有以下关系式：现⾦价格＝报价（即纯净价）＋从上⼀个付息⽇以来的累计利息2．美国国债期货（1）报价超级国债和超级国债期货合约的报价与长期国债本⾝在即期市场的报价⽅式相同。

（2）转换因⼦当交割某⼀特定债券时，⼀个名为转换因⼦的参数定义了空头⽅的债券交割价格。

债券的报价等于转换因⼦与最新成交期货价格的乘积。

将累计利息考虑在内，对应于交割100美元⾯值的债券收⼊的现⾦价格为：最新的期货成交价格×转换因⼦＋累计利息（3）最便宜可交割债券在交割⽉份的任意时刻，许多债券可以⽤于长期国债期货合约的交割，这些可交割债券有各式各样的券息率及期限。

空头⽅可以从这些债券中选出最便宜的可交割债券⽤于交割。

因为空头⽅收到的现⾦量为：最新成交价格×转换因⼦＋累计利息买⼊债券费⽤为：债券报价＋累计利息因此最便宜交割债券是使得：债券报价－期货的最新报价×转换因⼦达到最⼩的债券。

赫尔期权、期货及其他衍生产品第10版框架知识点及课后习题解析一、简介本文为赫尔期权、期货及其他衍生产品第10版的框架知识点及课后习题解析文档。

旨在帮助读者全面理解并掌握本教材中涉及的重要知识点，并通过课后习题的解析巩固所学内容。

二、框架知识点1. 期权•期权基本概念•期权合约要素•期权价格与影响因素•期权交易与结算•期权交易策略与风险管理2. 期货•期货市场与交易所•期货合约与交易机制1•期货价格与影响因素•期货交易策略与风险管理3. 其他衍生产品•互换（Swap）市场与交易•互换合约与交易机制•互换价格与影响因素•互换交易策略与风险管理4. 衍生产品交易策略•覆写（Covered Call）策略•裸写（Naked Writing）策略•敞口套利（Arbitrage）策略•期权组合策略•期货交易策略5. 风险管理•期权风险管理•期货风险管理•风险度量与序列风险三、课后习题解析1. 期权习题解析•习题1：某期权的看涨期权合约标的资产价格为$50，执行价为$55，期权价格为$3，到期日为3个月。

计算看涨期权合约是否处于实值、虚值或平值状态。

•习题2：假设某期权的看涨期权合约标的资产价格为$100，执行价为$90，到期日为6个月。

如果该看涨期权合约的价格为$8，在到期日当天标的资产价格为$110，计算期权的收益情况。

2. 期货习题解析•习题3：某期货合约的标的资产价格为$80，合约单位为100股，头寸为多头。

当期货价格上涨10%时，计算头寸的盈亏情况。

•习题4：某期货合约的标的资产价格为$100，合约单位为100股，头寸为空头。

当期货价格下跌5%时，计算头寸的盈亏情况。

3. 其他衍生产品习题解析•习题5：某互换合约的固定利率为5%，标的资产为未来6个月的黄金价格。

当前6个月黄金期货价格为$1500/盎司。

计算该互换合约的价格。

•习题6：某互换合约的固定利率为4%，标的资产为未来9个月的原油价格。

当前9个月原油期货价格为$70/桶。