【经典】约翰赫尔期权期货其他衍生品课后习题解答完整中文版-1-20习题解答【完整版】

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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。

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

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背景介绍赫尔期权、期货及其他衍生产品是一本经典的金融学教材，已经出版了多个版本。

本文将对第10版的框架知识点进行详细介绍，并对课后习题进行解析。

框架知识点第1章期权与期权市场本章主要介绍了期权的基本概念和期权市场的基本特点。

其中包括期权的定义、期权的基本特征、期权的交易方式、期权市场的参与者和期权市场的发展趋势等内容。

第2章期权定价基础本章介绍了期权定价的基本理论。

其中包括无套利定价原理、布莱克-舒尔斯期权定价模型、期权的几何布朗运动模型和完全市场假设等内容。

此外，还介绍了期权定价模型的应用和限制。

第3章期权策略与风险管理本章介绍了期权策略的基本概念和常见的期权策略类型。

其中包括购买期权、卖出期权、期权组合策略和套利策略等内容。

此外，还介绍了期权风险管理的基本方法和相关的风险指标。

第4章期货市场与期货定价本章介绍了期货市场的基本原理和期货合约的定价方法。

其中包括期货市场的特点、期货合约的基本要素、期货定价的原理和期货定价模型等内容。

此外，还介绍了期货市场的参与者和期货交易的风险管理。

第5章期货交易策略与风险管理本章介绍了期货交易策略的基本原理和常用的期货交易策略类型。

其中包括多头策略、空头策略、套利策略和市场中性策略等内容。

此外，还介绍了期货交易的风险管理方法和基本的交易技巧。

第6章期货市场的运行与监管本章介绍了期货市场的运行机制和监管体系。

其中包括期货市场的交易流程、交易所的角色和功能、期货市场的风险管理和期货市场的监管机构等内容。

此外，还介绍了期货市场的监管规则和期货市场的发展趋势。

课后习题解析第1章期权与期权市场习题1：期权是一种金融衍生品，它的特点是什么？答：期权有两个基本特点，即灵活性和杠杆效应。

灵活性指的是期权可以灵活选择行权，可以在未来的某个时间点以特定的价格购买或者卖出标的资产。

杠杆效应指的是期权的价格相对于标的资产的价格波动比较大，可以获得倍数的投资回报。

期权期货与其他衍生产品第九版课后习题与答案Chapter(.

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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）

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。

JohnHull《期货期权和衍生证券》章习题解答

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.。

约翰.赫尔,期权期货和其他衍生品(third edition)习题答案

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。

约翰.赫尔,期权期货和其他衍生品(third edition)习题答案

8.14 执行价格为$60 的看涨期权成本为$6，相同执行价格和到期日的看跌期权成
本为$4，制表说明跨式期权损益状况。请问：股票价格在什么范围内时，
跨式期权将导致损失呢？
解：可通过同时购买看涨看跌期权构造跨式期权：max（ ST －60，0）+max(60
－ ST )－（6＋4），其损益状况为：
股价 ST
解：(a)该组合等价于一份固定收益债券多头，其损益V = C ，不随股票价格变化。（V 为组合损益，C 为期权费，下同）如图 8.2： (b)该组合等价于一份股票多头与一份固定收益债券多头，其损益V = ST + C ，与股价同向同幅度变动。（ ST 为最终股票价格，下同）如图 8.3 (c)该组合等价于一份固定收益债券多头与一份看涨期权空头，其损益为
8.18 盒式价差期权是执行价格为 X 1 和 X 2 的牛市价差期权和相同执行价格的熊市看跌价差期权的组合。所有期权的到期日相同。盒式价差期权有什么样的特征？
解：牛市价差期权由 1 份执行价格为 X 1 欧式看涨期权多头与 1 份执行价格为 X 2 的欧式看涨期权空头构成（ X 1 < X 2 ），熊市价差期权由 1 份执行价格为 X 2 的欧式看跌期权多头与 1 份执行价格为 X 1 的看跌期权空头构成，则盒式价差
8.17 运用期权如何构造出具有确定交割价格和交割日期的股票远期合约？解：假定交割价格为 K，交割日期为 T。远期合约可由买入 1 份欧式看涨期权，
同时卖空 1 份欧式看跌期权，要求两份期权有相同执行价格 K 及到期日 T。可见，该组合的损益为 ST －K，在任何情形下，其中 ST 为 T 时股票价格。假定 F 为远期合约价格，若 K=F，则远期合约价值为 0。这表明，当执行价格为 K 时，看涨期权与看跌期权价格相等。

赫尔《期权、期货及其他衍生产品》(第7版)课后习题详解(期权交易策略)

10.2　课后习题详解一、问答题1. 什么是保护性看跌期权?什么样的看涨期权头寸能等价于保护性看跌期权?What is meant by a protective put? What position in call options is equivalent to a protective put?答：保护性看跌期权是由一个看跌期权多头头寸和一个标的资产多头头寸组合而成的。

它等同于一个看涨期权多头头寸加上一定数量的现金。

这是由买卖权平价关系推出的：2. 解释熊市差价的两种构造方式。

Explain two ways in which a bear spread can be created.答：（1）熊市差价可以由两个具有相同到期日和不同执行价格的看涨期权构成：投资者出售执行价格较低的看涨期权并买入执行价格较高的看涨期权。

（2）熊市差价也可以由两个具有相同到期日和不同执行价格的看跌期权构成：投资者出售执行价格较低的看跌期权并买入执行价格较高的看跌期权。

3. 对投资者而言，什么是购买蝶式差价的良好时机?When is it appropriate for an investor to purchase a butterfly spread'?答：蝶式差价是由三个不同执行价格（K1、K2和K3）的期权头寸构成的。

当投资者认为标的股票价格很可能处于中间执行价格K2附近时，应当购买蝶式差价。

4. 一个有效期为3个月的看涨期权的执行价格分别为15美元、17.5美元及20美元，相应的期权价格分别为4美元、2美元及0.5美元。

解释如何运用这些期权构造蝶式差价。

Call options on a stock are available with strike prices of $15, $17.5, and $20 and expiration dates in three months. Their prices tire $4, $2, and $0.5 , respectively. Explain how the options can be used to create a butterfly spread. Construct a table showing how profit varies with stock price for the butterfly spread.答：投资者可以买入执行价格为15美元和20美元的看涨期权，并出售两个执行价格为17.50美元的看涨期权，从而构造出蝶式差价。

约翰.赫尔_期权期货和其他衍生品第八版部分课后思考题

思考题1.1 远期合约长头寸与短期头寸之间的区别1)长头寸是买入，短头寸是卖出2)长头寸的收益是S-K 短头寸的收益是K-S1.2 期货合约与远期合约的区别1.3 卖出一个看涨期权与买入一个看跌期权的区别1)卖出看涨期权是一种义务，买入看跌期权是一种权利2)期初现金流不同3)收益公式不同卖出看涨期权买入看跌期权靠期权费赚利润1.4 期权与期货/远期合约的区别期货/远期合约，赋予它的持有者一个义务：以某个约定的价格买入或卖出标的资产。

期权合约，赋予它的持有者一个权利：以某个约定的价格买入或卖出标的资产。

1.5对冲、投机和套利之间的区别共同点：都是通过低买高卖或者高卖低买获利，都基于对未来市场预期的判断不同点：投机风险大，看涨看跌均没有保护性套期具有保护性对冲，如果货币市场流动性没问题，风险较低2.1 什么是逐日盯市逐日盯市制度，是指结算部门在每日闭市后计算、检查保证金账户余额，通过适时发出追加保证金通知，使保证金余额维持在一定水平之上，防止负债现象发生的结算制度。

2.2 保证金制度如何可以保证投资者免受违约风险？为了保证投资者保证金账户的资金余额在任何情况下都不为负值，设置了维持保证金，若保证金账户的余额低于维持保证金，投资者就会收到保证金催付，这部分资金称为变动保证金。

如果投资者未提供变动保证金，经纪人将出售该合约来平仓。

2.3一个交易的完成，会对未平仓合约数量产生什么样的影响?若交易是开仓，数量增加，若交易是平仓，则是减少2.4一天内发生的交易数量可以超过交易结束时未平仓合约的数量吗?交易数量包括开仓数量和平仓数量，若开仓=平仓，就会使未平仓数量为02.5设计一个新的期货合约时需要考虑哪几个重要方面？选择期货合约的标的资产、合约规模、交割月份3.1对冲的本质是什么？定义：为了减低另一项投资的风险而进行的投资。

目的：选择期货头寸，从而使得自身整体的投资风险尽量呈中性。

方法：用于对冲的期货交易，与需对冲的资产交易相比，头寸相等，在将来确定的时刻，操作方向相反。

赫尔《期权、期货及其他衍生产品》(第9版)笔记和课后习题详解答案

赫尔《期权、期货及其他衍生产品》（第9版）笔记和课后习题详解答案赫尔《期权、期货及其他衍生产品》（第9版）笔记和课后习题详解完整版>精研学习?>无偿试用20％资料全国547所院校视频及题库全收集考研全套>视频资料>课后答案>往年真题>职称考试第1章引言1.1复习笔记1.2课后习题详解第2章期货市场的运作机制2.1复习笔记2.2课后习题详解第3章利用期货的对冲策略3.1复习笔记3.2课后习题详解第4章利率4.1复习笔记4.2课后习题详解第5章如何确定远期和期货价格5.1复习笔记5.2课后习题详解第6章利率期货6.1复习笔记6.2课后习题详解第7章互换7.1复习笔记7.2课后习题详解第8章证券化与2007年信用危机8.1复习笔记第9章OIS贴现、信用以及资金费用9.1复习笔记9.2课后习题详解第10章期权市场机制10.1复习笔记10.2课后习题详解第11章股票期权的性质11.1复习笔记11.2课后习题详解第12章期权交易策略12.1复习笔记12.2课后习题详解第13章二叉树13.1复习笔记13.2课后习题详解第14章维纳过程和伊藤引理14.1复习笔记14.2课后习题详解第15章布莱克-斯科尔斯-默顿模型15.1复习笔记15.2课后习题详解第16章雇员股票期权16.1复习笔记16.2课后习题详解第17章股指期权与货币期权17.1复习笔记17.2课后习题详解第18章期货期权18.1复习笔记18.2课后习题详解第19章希腊值19.1复习笔记第20章波动率微笑20.1复习笔记20.2课后习题详解第21章基本数值方法21.1复习笔记21.2课后习题详解第22章风险价值度22.1复习笔记22.2课后习题详解第23章估计波动率和相关系数23.1复习笔记23.2课后习题详解第24章信用风险24.1复习笔记24.2课后习题详解第25章信用衍生产品25.1复习笔记25.2课后习题详解第26章特种期权26.1复习笔记26.2课后习题详解第27章再谈模型和数值算法27.1复习笔记27.2课后习题详解第28章鞅与测度28.1复习笔记28.2课后习题详解第29章利率衍生产品：标准市场模型29.1复习笔记29.2课后习题详解第30章曲率、时间与Quanto调整30.1复习笔记30.2课后习题详解第31章利率衍生产品：短期利率模型31.1复习笔记31.2课后习题详解第32章HJM，LMM模型以及多种零息曲线32.1复习笔记32.2课后习题详解第33章再谈互换33.1复习笔记33.2课后习题详解第34章能源与商品衍生产品34.1复习笔记34.2课后习题详解第35章章实物期权35.1复习笔记35.2课后习题详解第36章重大金融损失与借鉴36.1复习笔记36.2课后习题详解。

赫尔《期权、期货及其他衍生产品》（第7版）课后习题详解（信用风险）

赫尔《期权、期货及其他衍⽣产品》（第7版）课后习题详解（信⽤风险）22.2　课后习题详解⼀、问答题1. 某家企业3年期的债券的收益率与类似的⽆风险债券收益率的溢差为50个基点.债券回收率为30％，估计3年内每年的平均违约密度。

The spread between the yield on a three-year corporate bond and the yield on a similar risk-free bond is 50 basis points. The recovery rate is 30%. Estimate the average default intensity per year over the three-year period.答：由⽅程（22-2）知，3年内每年的平均违约密度为：0.0050/(10.3)=0.0071或每年0.71%。

2. 在习题1中，假定同⼀家企业5年期债券的收益率与类似的⽆风险债券收益率溢差为60个基点，假定回收率也为30％，估计5年内每年的平均违约密度，由计算结果显⽰的第四年到第五年的平均违约密度为多少?Suppose that in Problem 22.1 the spread between the yield on a five-year bond issued by the same company and the yield on a similar risk-free bond is 60 basis points. Assume the same recovery rate of 30%. Estimate the average default intensity per year over the five-year period. What do your results indicate about the average default intensity in years 4 and 5?答：由等式22-2知，5年内平均违约密度为0.0060/(10-0.3)=0.0086或每年0.86%。

赫尔《期权、期货及其他衍生产品》（第7版）课后习题详解（波动率微笑）

赫尔《期权、期货及其他衍生产品》（第7版）课后习题详解（波动率微笑）18.2 课后习题详解一、问答题1．在下列情形所观察到的波动率微笑是什么形式?（a ）股票价格分布两端的尾部均没有对数正态分布肥大；（b ）股票价格分布右端的尾部比对数正态分布要肥大，右端尾部没有对数正态分布肥大。

What pattern of implied volatilities is likely to be observed whena. Both tails of the stock price distribution are less heavy than those of the lognormal distribution?b. The right tail is heavier, and the left tail is less heavier than that of alognormal distribution?答：（a ）当股票价格分布两端的尾部均没有对数正态分布肥大时，对于深度虚值或深度实值的期权而言，布莱克—斯科尔斯倾向于高估期权的价格。

这将导致类似于图18-6中的波动率微笑。

图18-6 波动率微笑（b ）相比较而言，当股票价格分布右端的尾部比对数正态分布要肥大，右端尾部没有对数正态分布肥大时，对于虚值看涨期权和实值看跌期权而言，布莱克—斯科尔斯倾向于低估期权的价格；对于虚值看跌期权和实值看涨期权而言，布莱克—斯科尔斯倾向于高估期权的价格。

这将导致隐含波动率是执行价格的增函数，即波动率微笑将向上倾斜。

2．股票的波动率微笑形式是什么?What volatility smile is observed for equities?答：观察到的股票期权的波动率微笑通常是向下倾斜的，即股票期权的隐含波动率是执行价格的减函数。

具体参见本章复习笔记。

3．标的资产价格有跳跃时会造成什么形式的波动率微笑?这种形式对于2年和3个月期限的期权中哪个更显著?What pattern of implied volatilities is likely to be caused by jumps intheunderlying asset price? Is the pattern likely to be more pronounced for a two-year option than for a three-month option?答：（1）通常标的资产价格的跳跃将使得资产价格分布的两侧比对数正态分布都要肥大。

期权期货和其他衍生品约翰赫尔第九版答案 (2)

期权期货和其他衍生品约翰赫尔第九版答案简介《期权期货和其他衍生品》是由约翰·赫尔（John C. Hull）编写的一本经典教材，是金融衍生品领域的权威参考书籍之一。

该书第九版是在第八版的基础上进行了更新和修订，以适应当前金融市场的动态变化。

本文档旨在提供《期权期货和其他衍生品第九版》的答案，帮助读者更好地理解和应用书中的知识点。

以下将按照书籍的章节顺序，逐一给出答案。

第一章期权市场的基本特征1.什么是期权？答：期权是一种金融衍生品，它赋予买方在特定时间以特定价格买入或卖出标的资产的权力，而不是义务。

可以将期权分为看涨期权和看跌期权。

2.期权的四个基本特征是什么？答：期权的四个基本特征是价格、到期日、标的资产和行权方式。

价格即期权的成交价，到期日是期权到期的日期，标的资产是期权合约要买入或卖出的资产，而行权方式则决定了期权何时可以行使。

3.什么是期权合约？答：期权合约是买卖双方约定的具体规定和条件，包括标的资产、行权价格、到期日等。

它规定了买方在合约到期前是否可以行使期权。

第二章期权定价：基础观念1.定价模型的基本原理是什么？答：期权定价模型的基本原理是假设市场是有效的，即不存在无风险套利机会。

通过建立基于风险中性概率的模型，可以计算期权的理论价值。

2.什么是风险中性概率？答：风险中性概率是指在假设市场是有效的情况下，使得在无套利条件下资产价格在期望值与当前价格之间折现的概率。

风险中性概率的使用可以将市场中的现金流折算为无风险利率下的现值。

3.什么是期权的内在价值和时间价值？答：期权的内在价值是指期权当前即时的价值，即行权价格与标的资产价格之间的差额。

时间价值是期权除去内在价值后剩余的价值，它受到时间、波动率和利率等因素的影响。

第三章期权定价模型：基础知识1.什么是布莱克斯科尔斯期权定价模型？答：布莱克斯科尔斯期权定价模型是一种用于计算欧式期权价格的数学模型。

它基于连续性投资组合原理，使用了假设市场是完全有效的和无交易成本的条件，可以通过著名的布拉克斯科尔斯公式来计算期权的价格。

赫尔《期权、期货及其他衍生产品》(第9版)笔记和课后习题详解答案

赫尔《期权、期货及其他衍生产品》（第9版）笔记和课后习题详解完整版>精研学习䋞>无偿试用20％资料全国547所院校视频及题库全收集考研全套>视频资料>课后答案>往年真题>职称考试第1章引言1.1复习笔记1.2课后习题详解第2章期货市场的运作机制2.1复习笔记2.2课后习题详解第3章利用期货的对冲策略3.1复习笔记3.2课后习题详解第4章利率4.1复习笔记4.2课后习题详解第5章如何确定远期和期货价格5.1复习笔记5.2课后习题详解第6章利率期货6.1复习笔记6.2课后习题详解第7章互换7.1复习笔记7.2课后习题详解第8章证券化与2007年信用危机8.1复习笔记8.2课后习题详解第9章OIS贴现、信用以及资金费用9.1复习笔记9.2课后习题详解第10章期权市场机制10.1复习笔记10.2课后习题详解第11章股票期权的性质11.1复习笔记11.2课后习题详解第12章期权交易策略12.1复习笔记12.2课后习题详解第13章二叉树13.1复习笔记13.2课后习题详解第14章维纳过程和伊藤引理14.1复习笔记14.2课后习题详解第15章布莱克-斯科尔斯-默顿模型15.1复习笔记15.2课后习题详解第16章雇员股票期权16.1复习笔记16.2课后习题详解第17章股指期权与货币期权17.1复习笔记17.2课后习题详解第18章期货期权18.1复习笔记18.2课后习题详解第19章希腊值19.1复习笔记19.2课后习题详解第20章波动率微笑20.1复习笔记20.2课后习题详解第21章基本数值方法21.1复习笔记21.2课后习题详解第22章风险价值度22.1复习笔记22.2课后习题详解第23章估计波动率和相关系数23.1复习笔记23.2课后习题详解第24章信用风险24.1复习笔记24.2课后习题详解第25章信用衍生产品25.1复习笔记25.2课后习题详解第26章特种期权26.1复习笔记26.2课后习题详解第27章再谈模型和数值算法27.1复习笔记27.2课后习题详解第28章鞅与测度28.1复习笔记28.2课后习题详解第29章利率衍生产品：标准市场模型29.1复习笔记29.2课后习题详解第30章曲率、时间与Quanto调整30.1复习笔记30.2课后习题详解第31章利率衍生产品：短期利率模型31.1复习笔记31.2课后习题详解第32章HJM，LMM模型以及多种零息曲线32.1复习笔记32.2课后习题详解第33章再谈互换33.1复习笔记33.2课后习题详解第34章能源与商品衍生产品34.1复习笔记34.2课后习题详解第35章章实物期权35.1复习笔记35.2课后习题详解第36章重大金融损失与借鉴36.1复习笔记36.2课后习题详解。

赫尔《期权、期货及其他衍生产品》(第7版)课后习题详解(利率衍生品标准市场模型)

28.2　课后习题详解一、问答题1. 一家企业签署了一项上限合约，合约将3个月期LIBOR利率上限定为每年10％，本金为2000万美元。

在重置日3个月的LIBOR利率为每年12％。

根据利率上限协议，收益将如何支付，付款日为何时?A company caps three-month LIBOR at 10% per annum. The principal amount is $20 million. On a reset date, three-month LIBOR is 12% per annum. What payment would this lead to under the cap? When would the payment be made?答：应支付的数量为：20000000×0.02×0.25=100000（美元），该支付应在3个月后进行。

2. 解释为什么一个互换期权可以看作是一个债券期权。

Explain why a swap option can be regarded as a type of bond option.答：互换期权是是基于利率互换的期权，它给予持有者在未来某个确定时间进入一个约定的利率互换的权利。

利率互换可以被看作是固定利率债券和浮动利率债券的交换。

因而，互换期权可以看成是固定利率债券和浮动利率债券的交换的选择权。

在互换开始时，浮动利率债券的价值等于其本金额。

这样互换期权就可以被看作是以债权的面值为执行价格、以固定利率债券为标的资产的期权。

即互换期权可以看作是一个债券期权。

3. 采用布莱克模型来对一个期限为1年，标的资产为10年期债券的欧式看跌期权定价。

假定债券当前价格为125美元，执行价格为110美元，1年期利率为每年10％，债券远期价格的波动率为每年8％，期权期限内所支付票息的贴现值为10美元。

Use Black’s model to value a one-year European put option on a 10-year bond. Assume that the current value of the bond is $125, the strike price is $110, the one-year interest rate is 10% per annum, the bond's price volatility is 8% per annum, and the present value of the coupons to be paid during the life of the option is $10.答：根据布莱克模型，F0=(125-10)e0.1×1=127.09，K=110，P（0，T）=e-0.1×1，σB=0.08和T=1.0。

JohnHull期货期权和衍生证券章习题解答(终审稿)

J o h n H u l l期货期权和衍生证券章习题解答公司内部档案编码：[OPPTR-OPPT28-OPPTL98-OPPNN08]CHAPTER 13Wiener P rocesses and It’s LemmaPractice QuestionsProblem .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 processImagine that you have to forecast the future temperature from a) the current temperature, b) the history of the temperature in thelast week, and c) a knowledge of seasonal 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 .Can a trading rule based on the past history of a stock’sprice 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 appropriateadjustments 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 .A company’s cash position, measured in millions of dollars, follows a generalized Wiener process with a drift rate of per quarter and a variance rate of per quarter. How high does thecompany’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 yearSuppose 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 normalvariable (with mean zero and standard deviation is less than x . From normal distribution tables200054x N +.⎛⎫-=. ⎪⎝⎭when:20164494x +.-=-.., when 45796x =.. The initial cash position must therefore be$ million. Problem .Variables 1X and 2X follow generalized Wiener processes withdrift rates 1μ and 2μ 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 inany 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 distributedvariables, 12X X + has the probability distribution()22112212a T a T T T ϕμμσσ+++,+.,22121212()()a a T T ϕμμσσ⎡⎤+++,+⎣⎦This shows that 12X X + follows a generalized Wiener process withdrift 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 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 processwith drift rate 12μμ+ and variance rate 2212122σσρσσ++. Problem .Consider a variable,S , that follows the processdS 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 sixThe 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 probabilitydistribution of the value of the variable at the end of year six is (2075)ϕ,Problem .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 It’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 byS λσ, the drift of G increases byS GS Sλσ∂∂ orG G λσThe growth rate of G , therefore, increases by G λσ. Problem .Stock A and stock B both follow geometric Brownian motion. Changes in any short interval of time are uncorrelated with eachother. 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 thestochastic term correspond.) Hence the value of the portfolio does not follow geometric Brownian motion. Problem .The process for the stock price in equation isS S t S μσε∆=∆+where μ and σ are constant. Explain carefully the difference between this model and each of the following:S t S S t S t S μσεμσεμσε∆=∆+∆=∆+∆=∆+Why is the model in equation a more appropriate model of stock price behaviorthan any of these three alternativesIn: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 theexpected growth rate is $5 per annum when the stock price 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 absoluteterms 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 .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 thereforecontinually 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. Thecurrent 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 .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 It’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 .Suppose that x is the yield to maturity with continuouscompounding 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 priceThe process followed by B , the bond price, is fro m It’ 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 BT 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 (Excel Spreadsheet)A stock whose price is $30 has an expected return of 9% and avolatility 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 is a random sample from a standard normal distribution.Further QuestionsProblem .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.(d)(e)With the notation in the text2()S t t Sϕμσ∆∆,∆In this case 50S =, 016μ=., 030σ=. and 1365000274t ∆=/=.. Hence(016000274009000274)50(0000440000247)S ϕϕ∆.⨯.,.⨯.=.,.and2(50000044500000247)Sϕ∆⨯.,⨯.that is,(002206164)Sϕ∆.,.(a)The expected stock price at the end of the next day is therefore (b)The standard deviation of the stock price at the end of the next0785=.(c)95% confidence limits for the stock price at the end of the next day are 500221960785and 500221960785.-.⨯..+.⨯..,4848and 5156..Note that some students may consider one trading day rather than one calendar day. Then 1252000397t ∆=/=.. The answer to (a) is then . The answer to (b) is . The answers to part (c) are and . Problem .A company’s cash position, measured in millions of dollars,follows a generalized Wiener process with a drift rate of per month and a variance rate of per month. The initial cash position is .(a)What are the probability distributions of the cash positionafter one month, six months, and one year(b)What are the probabilities of a negative cash position at theend of six months and one year(c) At what time in the future is the probability of anegative 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 standardizednormal variable [., a variable with probability distribution(01)ϕ,] is less than x . From normal distribution tables(265)00040N -.=.. Hence the probability of a negative cashposition at the end of six months is %.Similarly the probability of a negative cash position at theend of one year is(230)00107N N ⎛=-.=. ⎝or %.(c) In general the probability distribution of the cashposition 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 that220d y dx/> for this value of x. It therefore gives a minimum value for y. Hence the probability of a negative cash positionis greatest after 20 months.Problem .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 bondThe process followed by B, the bond price, is from It’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∂∂∂=;=-;=∂∂∂Hence 2202322021121()21()dB a x x s x dt sxdz x x x s s a x x dt dz x x x ⎡⎤=--+-⎢⎥⎣⎦⎡⎤=--+-⎢⎥⎣⎦The expected instantaneous rate at which capital gains are earned from the bond is therefore: 2021()s a x x x x--+ The expected interest per unit time is 1. The total expectedinstantaneous return is therefore:20211()s a x x x x --+ When expressed as a proportion of the bond price this is: 202111()s a x x x x x ⎛⎫⎛⎫--+ ⎪ ⎪⎝⎭⎝⎭20()a x x x s x =--+Problem .If S follows the geometric Brownian motion process in equation , 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 thatIt’(b) s lemma gives(c)22dy S dt S dz μσ=+ordy y dt y dz μσ=+(d) In this case 2y S S ∂/∂=, 222y S ∂/∂=, and 0y t ∂/∂= so thatIt’(e) s lemma gives(f)2222(2)2dy S S dt S dz μσσ=++or2(2)2dy y dt y dz μσσ=++(g) In this case S y S e ∂/∂=, 22S y S e ∂/∂=, and 0y t ∂/∂= so thatIt’(h) s lemma gives(i)22(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 thatIt’(e) s lemma gives (f)2()dy ry y y dt y dz μσσ=--+-or2()dy r y dt y dz μσσ=-+--Problem .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 > whenln ln 80T S >.)The variable ln T S is normally distributed with mean 20ln (2)S T μσ+-/and standard deviation σ050S =, 012μ=., 2T =, and 030σ=. so that the mean and standard deviation of ln T S are2ln 50(012032)24062+.-./=. and 00424.=., respectively. Also,ln804382=.. The probability that 80T S > is the same as theprobability that ln 4382T S >.. This is 4382406211(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 .Problem (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 . 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 equal to , , and .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 ’s are correlated samples from standard normal distributions.。

HullOFOD9eSolutionsCh01第九版期权期货及其他衍生品课后答案

CHAPTER 1IntroductionPractice QuestionsProblem 1.1.What is the difference between a long forward position and a short forward position?When a trader enters into a long forward contract, she is agreeing to buy the underlying asset for a certain price at a certain time in the future. When a trader enters into a short forward contract, she is agreeing to sell the underlying asset for a certain price at a certain time in the future.Problem 1.2.Explain carefully the difference between hedging, speculation, and arbitrage.A trader is hedging when she has an exposure to the price of an asset and takes a position in a derivative to offset the exposure. In a speculation the trader has no exposure to offset. She is betting on the future movements in the price of the asset. Arbitrage involves taking a position in two or more different markets to lock in a profit.Problem 1.3.What is the difference between entering into a long forward contract when the forward price is $50 and taking a long position in a call option with a strike price of $50?In the first case the trader is obligated to buy the asset for $50. (The trader does not have a choice.) In the second case the trader has an option to buy the asset for $50. (The trader does not have to exercise the option.)Problem 1.4.Explain carefully the difference between selling a call option and buying a put option.Selling a call option involves giving someone else the right to buy an asset from you. It gives you a payoff ofmax(0)min(0)T T S K K S --,=-, Buying a put option involves buying an option from someone else. It gives a payoff of max(0)T K S -,In both cases the potential payoff is T K S -. When you write a call option, the payoff is negative or zero. (This is because the counterparty chooses whether to exercise.) When you buy a put option, the payoff is zero or positive. (This is because you choose whether to exercise.)Problem 1.5.An investor enters into a short forward contract to sell 100,000 British pounds for US dollars at an exchange rate of 1.5000 US dollars per pound. How much does the investor gain or lose if the exchange rate at the end of the contract is (a) 1.4900 and (b) 1.5200?(a)The investor is obligated to sell pounds for 1.5000 when they are worth 1.4900. Thegain is (1.5000−1.4900) ×100,000 = $1,000.(b)The investor is obligated to sell pounds for 1.5000 when they are worth 1.5200. Theloss is (1.5200−1.5000)×100,000 = $2,000Problem 1.6.A trader enters into a short cotton futures contract when the futures price is 50 cents per pound. The contract is for the delivery of 50,000 pounds. How much does the trader gain or lose if the cotton price at the end of the contract is (a) 48.20 cents per pound; (b) 51.30 cents per pound?(a)The trader sells for 50 cents per pound something that is worth 48.20 cents per pound.Gain ($05000$04820)50000$900=.-.⨯,=.(b)The trader sells for 50 cents per pound something that is worth 51.30 cents per pound.Loss ($05130$05000)50000$650=.-.⨯,=.Problem 1.7.Suppose that you write a put contract with a strike price of $40 and an expiration date in three months. The current stock price is $41 and the contract is on 100 shares. What have you committed yourself to? How much could you gain or lose?You have sold a put option. You have agreed to buy 100 shares for $40 per share if the party on the other side of the contract chooses to exercise the right to sell for this price. The option will be exercised only when the price of stock is below $40. Suppose, for example, that the option is exercised when the price is $30. You have to buy at $40 shares that are worth $30; you lose $10 per share, or $1,000 in total. If the option is exercised when the price is $20, you lose $20 per share, or $2,000 in total. The worst that can happen is that the price of the stock declines to almost zero during the three-month period. This highly unlikely event would cost you $4,000. In return for the possible future losses, you receive the price of the option from the purchaser.Problem 1.8.What is the difference between the over-the-counter market and the exchange-traded market? What are the bid and offer quotes of a market maker in the over-the-counter market?The over-the-counter market is a telephone- and computer-linked network of financial institutions, fund managers, and corporate treasurers where two participants can enter into any mutually acceptable contract. An exchange-traded market is a market organized by an exchange where the contracts that can be traded have been defined by the exchange. When a market maker quotes a bid and an offer, the bid is the price at which the market maker is prepared to buy and the offer is the price at which the market maker is prepared to sell.Problem 1.9.You would like to speculate on a rise in the price of a certain stock. The current stock price is $29, and a three-month call with a strike of $30 costs $2.90. You have $5,800 to invest.Identify two alternative strategies, one involving an investment in the stock and the other involving investment in the option. What are the potential gains and losses from each?One strategy would be to buy 200 shares. Another would be to buy 2,000 options. If the share price does well the second strategy will give rise to greater gains. For example, if the share price goes up to $40 you gain [2000($40$30)]$5800$14200,⨯--,=,from the second strategy and only 200($40$29)$2200⨯-=,from the first strategy. However, if the share price does badly, the second strategy gives greater losses. For example, if the share price goes down to $25, the first strategy leads to a loss of 200($29$25)$800⨯-=,whereas the second strategy leads to a loss of the whole $5,800 investment. This example shows that options contain built in leverage.Problem 1.10.Suppose you own 5,000 shares that are worth $25 each. How can put options be used to provide you with insurance against a decline in the value of your holding over the next four months?You could buy 50 put option contracts (each on 100 shares) with a strike price of $25 and an expiration date in four months. If at the end of four months the stock price proves to be less than $25, you can exercise the options and sell the shares for $25 each.Problem 1.11.When first issued, a stock provides funds for a company. Is the same true of anexchange-traded stock option? Discuss.An exchange-traded stock option provides no funds for the company. It is a security sold by one investor to another. The company is not involved. By contrast, a stock when it is first issued is sold by the company to investors and does provide funds for the company.Problem 1.12.Explain why a futures contract can be used for either speculation or hedging.If an investor has an exposure to the price of an asset, he or she can hedge with futures contracts. If the investor will gain when the price decreases and lose when the price increases, a long futures position will hedge the risk. If the investor will lose when the price decreases and gain when the price increases, a short futures position will hedge the risk. Thus either a long or a short futures position can be entered into for hedging purposes.If the investor has no exposure to the price of the underlying asset, entering into a futures contract is speculation. If the investor takes a long position, he or she gains when the asset’s price increases and loses when it decreases. If the investor takes a short position, he or she loses when the asset’s price increases and gains when it decreases.Problem 1.13.Suppose that a March call option to buy a share for $50 costs $2.50 and is held until March. Under what circumstances will the holder of the option make a profit? Under what circumstances will the option be exercised? Draw a diagram showing how the profit on a long position in the option depends on the stock price at the maturity of the option.The holder of the option will gain if the price of the stock is above $52.50 in March. (This ignores the time value of money.) The option will be exercised if the price of the stock isabove $50.00 in March. The profit as a function of the stock price is shown in Figure S1.1.Figure S1.1:Profit from long position in Problem 1.13Problem 1.14.Suppose that a June put option to sell a share for $60 costs $4 and is held until June. Under what circumstances will the seller of the option (i.e., the party with a short position) make a profit? Under what circumstances will the option be exercised? Draw a diagram showing how the profit from a short position in the option depends on the stock price at the maturity of the option.The seller of the option will lose money if the price of the stock is below $56.00 in June. (This ignores the time value of money.) The option will be exercised if the price of the stock is below $60.00 in June. The profit as a function of the stock price is shown in Figure S1.2.Figure S1.2:Profit from short position in Problem 1.14Problem 1.15.It is May and a trader writes a September call option with a strike price of $20. The stock price is $18, and the option price is $2. Describe the investor’s cash flo ws if the option is held until September and the stock price is $25 at this time.The trader has an inflow of $2 in May and an outflow of $5 in September. The $2 is the cash received from the sale of the option. The $5 is the result of the option being exercised. The investor has to buy the stock for $25 in September and sell it to the purchaser of the optionfor $20.Problem 1.16.A trader writes a December put option with a strike price of $30. The price of the option is $4. Under what circumstances does the trader make a gain?The trader makes a gain if the price of the stock is above $26 at the time of exercise. (This ignores the time value of money.)Problem 1.17.A company knows that it is due to receive a certain amount of a foreign currency in four months. What type of option contract is appropriate for hedging?A long position in a four-month put option can provide insurance against the exchange rate falling below the strike price. It ensures that the foreign currency can be sold for at least the strike price.Problem 1.18.A US company expects to have to pay 1 million Canadian dollars in six months. Explain how the exchange rate risk can be hedged using (a) a forward contract and (b) an option.The company could enter into a long forward contract to buy 1 million Canadian dollars in six months. This would have the effect of locking in an exchange rate equal to the current forward exchange rate. Alternatively the company could buy a call option giving it the right (but not the obligation) to purchase 1 million Canadian dollars at a certain exchange rate in six months. This would provide insurance against a strong Canadian dollar in six months while still allowing the company to benefit from a weak Canadian dollar at that time. Problem 1.19.A trader enters into a short forward contract on 100 million yen. The forward exchange rate is $0.0090 per yen. How much does the trader gain or lose if the exchange rate at the end of the contract is (a) $0.0084 per yen; (b) $0.0101 per yen?a)The trader sells 100 million yen for $0.0090 per yen when the exchange rate is $0.0084⨯.millions of dollars or $60,000.per yen. The gain is 10000006b)The trader sells 100 million yen for $0.0090 per yen when the exchange rate is $0.0101⨯.millions of dollars or $110,000.per yen. The loss is 10000011Problem 1.20.The CME Group offers a futures contract on long-term Treasury bonds. Characterize the investors likely to use this contract.Most investors will use the contract because they want to do one of the following: a) Hedge an exposure to long-term interest rates.b) Speculate on the future direction of long-term interest rates.c) Arbitrage between the spot and futures markets for Treasury bonds.This contract is discussed in Chapter 6.Problem 1.21.“Options and futures are zero -sum games.” What do you think is meant by this statement?The statement means that the gain (loss) to the party with the short position is equal to the loss (gain) to the party with the long position. In aggregate, the net gain to all parties is zero.Problem 1.22.Describe the profit from the following portfolio: a long forward contract on an asset and a long European put option on the asset with the same maturity as the forward contract and a strike price that is equal to the forward price of the asset at the time the portfolio is set up.The terminal value of the long forward contract is:0T S F -where T S is the price of the asset at maturity and 0F is the delivery price, which is the same as the forward price of the asset at the time the portfolio is set up). The terminal value of the put option is:0max (0)T F S -,The terminal value of the portfolio is therefore00max (0)T T S F F S -+-,0max (0]T S F =,-This is the same as the terminal value of a European call option with the same maturity as the forward contract and a strike price equal to 0F . This result is illustrated in the Figure S1.3. The profit equals the terminal value of the call option less the amount paid for the put option. (It does not cost anything to enter into the forward contract.Figure S1.3: Profit from portfolio in Problem 1.22Problem 1.23.In the 1980s, Bankers Trust developed index currency option notes (ICONs). These are bonds in which the amount received by the holder at maturity varies with a foreign exchange rate. One example was its trade with the Long Term Credit Bank of Japan. The ICON specified that if the yen –U.S. dollar exchange rate,T S , is greater than 169 yen per dollar at maturity(in 1995), the holder of the bond receives $1,000. If it is less than 169 yen per dollar, the amount received by the holder of the bond is 1691000max 010001T S ⎡⎤⎛⎫,-,,-⎢⎥ ⎪⎝⎭⎣⎦ When the exchange rate is below 84.5, nothing is received by the holder at maturity. Show that this ICON is a combination of a regular bond and two options.Suppose that the yen exchange rate (yen per dollar) at maturity of the ICON is T S . The payofffrom the ICON is1000if 169169100010001if 8451690if 845T T T T S S S S ,>⎛⎫,-,-.≤≤ ⎪⎝⎭<.When 845169T S .≤≤ the payoff can be written 1690002000TS ,,-The payoff from an ICON is the payoff from:(a) A regular bond(b) A short position in call options to buy 169,000 yen with an exercise price of 1/169 (c) A long position in call options to buy 169,000 yen with an exercise price of 1/84.5 This is demonstrated by the following table, which shows the terminal value of the various components of the positionProblem 1.24.On July 1, 2011, a company enters into a forward contract to buy 10 million Japanese yen on January 1, 2012. On September 1, 2011, it enters into a forward contract to sell 10 million Japanese yen on January 1, 2012. Describe the payoff from this strategy.Suppose that the forward price for the contract entered into on July 1, 2011 is 1F and thatthe forward price for the contract entered into on September 1, 2011 is 2F with both 1F and 2F being measured as dollars per yen. If the value of one Japanese yen (measured in USdollars) is T S on January 1, 2012, then the value of the first contract (in millions of dollars)at that time is110()T S F -while the value of the second contract at that time is:210()T F S -The total payoff from the two contracts is therefore122110()10()10()T T S F F S F F -+-=-Thus if the forward price for delivery on January 1, 2012 increased between July 1, 2011 and September 1, 2011 the company will make a profit. (Note that the yen/USD exchange rate is usually expressed as the number of yen per USD not as the number of USD per yen)Problem 1.25.Suppose that USD-sterling spot and forward exchange rates are as follows :What opportunities are open to an arbitrageur in the following situations?(a) A 180-day European call option to buy £1 for $1.52 costs 2 cents.(b) A 90-day European put option to sell £1 for $1.59 costs 2 cents.Note that there is a typo in the problem in the book. 1.42 and 1.49 should be 1.52 and 1.59 in the last two lines of the problem s(a) The arbitrageur buys a 180-day call option and takes a short position in a 180-day forward contract. If T S is the terminal spot rate, the profit from the call option is 02.0)0,52.1max(--T SThe profit from the short forward contract isT S -5518.1The profit from the strategy is thereforeT T S S -+--5518.102.0)0,52.1max(orT T S S -+-5318.1)0,52.1max(This is1.5318−S T when S T <1.520.0118 when S T >1.52This shows that the profit is always positive. The time value of money has been ignored in these calculations. However, when it is taken into account the strategy is still likely to be profitable in all circumstances. (We would require an extremely high interest rate for $0.0118 interest to be required on an outlay of $0.02 over a 180-day period.)(b) The trader buys 90-day put options and takes a long position in a 90 day forwardcontract. If T S is the terminal spot rate, the profit from the put option is02.0)0,59.1max(--T SThe profit from the long forward contract isS T −1.5556The profit from this strategy is therefore5556.102.0)0,59.1max(-+--T T S Sor5756.1)0,59.1max(-+-T T S SThis isS T −1.5756 when S T >1.590.0144 when S T <1.59The profit is therefore always positive. Again, the time value of money has been ignored but is unlikely to affect the overall profitability of the strategy. (We would require interest rates to be extremely high for $0.0144 interest to be required on an outlay of $0.02 over a 90-day period.)Problem 1.26.A trader buys a call option with a strike price of $30 for $3. Does the trader ever exercise the option and lose money on the trade. Explain.If the stock price is between $30 and $33 at option maturity the trader will exercise the option, but lose money on the trade. Consider the situation where the stock price is $31. If the trader exercises, she loses $2 on the trade. If she does not exercise she loses $3 on the trade. It is clearly better to exercise than not exercise.Problem 1.27.A trader sells a put option with a strike price of $40 for $5. What is the trader's maximum gain and maximum loss? How does your answer change if it is a call option?The trader’s maximum gain from the put option is $5. The maximum loss is $35,corresponding to the situation where the option is exercised and the price of the underlying asset is zero. If the option were a call, the trader’s maxim um gain would still be $5, but there would be no bound to the loss as there is in theory no limit to how high the asset price could rise.Problem 1.28.``Buying a put option on a stock when the stock is owned is a form of insurance.'' Explain this statement.If the stock price declines below the strike price of the put option, the stock can be sold for the strike price.Further QuestionsProblem 1.29.On May 8, 2013, as indicated in Table 1.2, the spot offer price of Google stock is $871.37 and the offer price of a call option with a strike price of $880 and a maturity date ofSeptember is $41.60. A trader is considering two alternatives: buy 100 shares of the stock and buy 100 September call options. For each alternative, what is (a) the upfront cost, (b)the total gain if the stock price in September is $950, and (c) the total loss if the stockprice in September is $800. Assume that the option is not exercised before September andif stock is purchased it is sold in September.a)The upfront cost for the stock alternative is $87,137. The upfront cost for the optionalternative is $4,160.b)The gain from the stock alternative is $95,000−$87,137=$7,863. The total gain fromthe option alternative is ($950-$880)×100−$4,160=$2,840.c)The loss from the stock alternative is $87,137−$80,000=$7,137. The loss from theoption alternative is $4,160.Problem 1.30.What is arbitrage? Explain the arbitrage opportunity when the price of a dually listed mining company stock is $50 (USD) on the New York Stock Exchange and $52 (CAD) on the Toronto Stock Exchange. Assume that the exchange rate is such that 1 USD equals 1.01 CAD. Explain what is likely to happen to prices as traders take advantage of this opportunity. Arbitrage involves carrying out two or more different trades to lock in a profit. In this case, traders can buy shares on the NYSE and sell them on the TSX to lock in a USD profit of52/1.01−50=1.485 per share. As they do this the NYSE price will rise and the TSX price will fall so that the arbitrage opportunity disappearsProblem 1.31 (Excel file)Trader A enters into a forward contract to buy an asset for $1000 in one year. Trader B buys a call option to buy the asset for $1000 in one year. The cost of the option is $100. What is the difference between the positions of the traders? Show the profit as a function of the price of the asset in one year for the two traders.Trader A makes a profit of S T 1000 and Trader B makes a profit of max (S T 1000, 0) –100 where S T is the price of the asset in one year. Trader A does better if S T is above $900 as indicated in Figure S1.4.Figure S1.4: Profit to Trader A and Trader B in Problem 1.31Problem 1.32.In March, a US investor instructs a broker to sell one July put option contract on a stock. The stock price is $42 and the strike price is $40. The option price is $3. Explain what the investor has agreed to. Under what circumstances will the trade prove to be profitable? What are the risks?The investor has agreed to buy 100 shares of the stock for $40 in July (or earlier) if the party on the other side of the transaction chooses to sell. The trade will prove profitable if the option is not exercised or if the stock price is above $37 at the time of exercise. The risk to the investor is that the stock price plunges to a low level. For example, if the stock price drops to $1 by July , the investor loses $3,600. This is because the put options are exercised and $40 is paid for 100 shares when the value per share is $1. This leads to a loss of $3,900 which is only a little offset by the premium of $300 received for the options.Problem 1.33.A US company knows it will have to pay 3 million euros in three months. The current exchange rate is 1.3500 dollars per euro. Discuss how forward and options contracts can be used by the company to hedge its exposure.The company could enter into a forward contract obligating it to buy 3 million euros in three months for a fixed price (the forward price). The forward price will be close to but not exactly the same as the current spot price of 1.3500. An alternative would be to buy a call option giving the company the right but not the obligation to buy 3 million euros for a particular exchange rate (the strike price) in three months. The use of a forward contract locks in, at no cost, the exchange rate that will apply in three months. The use of a call option provides, at a cost, insurance against the exchange rate being higher than the strike price. Problem 1.34. (Excel file)A stock price is $29. An investor buys one call option contract on the stock with a strike price of $30 and sells a call option contract on the stock with a strike price of $32.50. The market prices of the options are $2.75 and $1.50, respectively. The options have the same maturity date. Describe the investor's position.This is known as a bull spread (see Chapter 12). The profit is shown in Figure S1.5.Figure S1.5: Profit in Problem 1.34Problem 1.35.The price of gold is currently $1,400 per ounce. The forward price for delivery in one year is $1,500. An arbitrageur can borrow money at 4% per annum. What should the arbitrageur do? Assume that the cost of storing gold is zero and that gold provides no income.The arbitrageur should borrow money to buy a certain number of ounces of gold today and short forward contracts on the same number of ounces of gold for delivery in one year. This means that gold is purchased for $1,400 per ounce and sold for $1,500 per ounce. Interest on the borrowed funds will be 0.04×$1400 or $56 per ounce. A profit of $44 per ounce will therefore be made.Problem 1.36.The current price of a stock is $94, and three-month call options with a strike price of $95 currently sell for $4.70. An investor who feels that the price of the stock will increase istrying to decide between buying 100 shares and buying 2,000 call options (20 contracts). Both strategies involve an investment of $9,400. What advice would you give? How high does the stock price have to rise for the option strategy to be more profitable?The investment in call options entails higher risks but can lead to higher returns. If the stock price stays at $94, an investor who buys call options loses $9,400 whereas an investor who buys shares neither gains nor loses anything. If the stock price rises to $120, the investor who buys call options gains⨯--=,$2000(12095)940040600An investor who buys shares gains$⨯-=,100(12094)2600The strategies are equally profitable if the stock price rises to a level, S, where⨯-=--100(94)2000(95)9400S SorS=100The option strategy is therefore more profitable if the stock price rises above $100.Problem 1.37.On May 8, 2013, an investor owns 100 Google shares. As indicated in Table 1.3, the share price is about $871 and a December put option with a strike price $820 costs $37.50. The investor is comparing two alternatives to limit downside risk. The first involves buying one December put option contract with a strike price of $820. The second involves instructing a broker to sell the 100 shares as so on as Google’s price reaches $820. Discuss the advantages and disadvantages of the two strategies.The second alternative involves what is known as a stop or stop-loss order. It costs nothing and ensures that $82,000, or close to $82,000, is realized for the holding in the event the stock price ever falls to $820. The put option costs $3,750 and guarantees that the holding can be sold for $8,200 any time up to December. If the stock price falls marginally below $820 and then rises the option will not be exercised, but the stop-loss order will lead to the holding being liquidated. There are some circumstances where the put option alternative leads to a better outcome and some circumstances where the stop-loss order leads to a better outcome.If the stock price ends up below $820, the stop-loss order alternative leads to a better outcome because the cost of the option is avoided. If the stock price falls to $800 in November and then rises to $850 by December, the put option alternative leads to a betteroutcome. The investor is paying $3,750 for the chance to benefit from this second type of outcome.Problem 1.38.A bond issued by Standard Oil some time ago worked as follows. The holder received no interest. At the bond’s maturity the company promised to pay $1,000 plus an additionalamount based on the price of oil at that time. The additional amount was equal to the product of 170 and the excess (if any) of the price of a barrel of oil at maturity over $25. The maximum additional amount paid was $2,550 (which corresponds to a price of $40 per barrel). Show that the bond is a combination of a regular bond, a long position in call options on oil with a strike price of $25, and a short position in call options on oil with a strike price of $40.Suppose T S is the price of oil at the bond’s maturity. In addition to $1000 the Standard Oilbond pays:$250$40$25170(25)$402550T T T T S S S S <:>>:->:,This is the payoff from 170 call options on oil with a strike price of 25 less the payoff from 170 call options on oil with a strike price of 40. The bond is therefore equivalent to a regular bond plus a long position in 170 call options on oil with a strike price of $25 plus a short position in 170 call options on oil with a strike price of $40. The investor has what is termed a bull spread on oil. This is discussed in Chapter 12.Problem 1.39.Suppose that in the situation of Table 1.1 a cor porate treasurer said: “I will have £1 million to sell in six months. If the exchange rate is less than 1.52, I want you to give me 1.52. If it is greater than 1.58 I will accept 1.58. If the exchange rate is between 1.52 and 1.58, I will sell the sterling for the exchange rate.” How could you use options to satisfy the treasurer?You sell the treasurer a put option on GBP with a strike price of 1.52 and buy from the treasurer a call option on GBP with a strike price of 1.58. Both options are on one million pounds and have a maturity of six months. This is known as a range forward contract and is discussed in Chapter 17.Problem 1.40.Describe how foreign currency options can be used for hedging in the situation considered in Section 1.7 so that (a) ImportCo is guaranteed that its exchange rate will be less than 1.5700, and (b) ExportCo is guaranteed that its exchange rate will be at least 1.5300. Use DerivaGem to calculate the cost of setting up the hedge in each case assuming that the exchange rate volatility is 12%, interest rates in the United States are 5% and interest rates in Britain are 5.7%. Assume that the current exchange rate is the average of the bid and offer in Table 1.1.ImportCo should buy three-month call options on $10 million with a strike price of 1.5700. ExportCo should buy three-month put options on $10 million with a strike price of 1.5300. In this case the spot foreign exchange rate is 1.5543 (the average of the bid and offer quotes in。

赫尔《期权、期货及其他衍生产品》(第7版)课后习题详解(股票期权的性质)

9.2　课后习题详解一、问答题1. 列出影响期权价格的6个因素。

List the six factors affecting stock option prices.答：影响股票期权价格的六个因素是：当前股票价格、执行价格、无风险利率、波动率、期权期限和股息。

具体可参见本章复习笔记。

2. 一个无股息股票的看涨期权的期限为4个月，执行价格为25美元，股票的当前价格为28美元，无风险利率为每年8％，期权的下限为多少?What is a lower bound for the price of a four-month call option on a non-dividend-paying stock when the stock price is $28, the strike price is $25, and the risk-free interest rate is 8% per annum?答：根据无股息股票的看涨期权价格下限的公式：S0－Ke-rT。

，K=25，r=8%，T=0.3333，则：其中，S0=283. 一个无股息股票的看跌期权的期限为1个月，执行价格为15美元，当前股票价格为12美元，无风险利率为每年6％时，期权的下限为多少?What is a lower bound for the price of a one-month European put option on a non-dividend-paying stock when the stock price is $12, the strike price is $15, and the risk-free interest rate is 6% per annum?答：根据无股息股票的看跌期权价格下限的公式：Ke-rT－S0。

K=15，r=6%，T=0.08333，则：其中，S0=12，4. 列举两个原因来说明为什么无股息股票的美式看涨期权不应被提前行使。

赫尔《期权、期货及其他衍生产品》(第7版)课后习题详解(利率)

4.2　课后习题详解一、问答题1. 一个银行的利率报价为每年14％，每季度复利一次。

在以下不同的复利机制下对应的利率是多少?（a）连续复利；（b）一年复利一次。

A bank quotes you an interest rate of 14% per annumwith quarterlycompounding.What is the equivalent rate with (a) continuous compounding and (b) annual compounding?答：(a)等价的连续复利利率为：，即每年13.76％。

(b)按年计复利的利率为：，即每年14.75％。

2. LIBOR与LIBID的含义是什么?哪一个更高?What is meant by LIBOR and LIBID. Which is higher?答：LIBOR是伦敦同业银行拆出利率，它是一家银行提供给其他银行资金所要求的利率。

LIBID是伦敦同业银行拆入利率，它是一家银行愿意接受的从其他银行借款的利率。

一般情况下，LIBOR比LIBID高。

3. 6个月期与一年期的零息利率均为10％。

一个剩余期限还有18个月，券息利率为8％（刚刚付过半年一次的利息）的债券，收益率为10.4％的债券价格为多少?18个月的零息利率为多少?这里的所有利率均为每半年复利一次利率。

The six-month and one-year zero rates are both 10% per annum. For a bond that lasts 18 months and pays a coupon of 8% per annum (with a coupon payment having just been made), the yield is 10.4% per annum. What is the bond’s price? What is the 18-month zero rate? All rates are quoted withsemiannual compounding.答：考虑票面价值为100美元的债券。

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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×(29-p),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×(p-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 $29×(p-29)，购买期权收益为$$5,800$29×(p-30)-5,800。

1.7假设你拥有5,000股每股价值$25的股票，如何运用看跌期权来确保你的股票价值在未来的四个月中不会受到股价下跌的影响。

答：通过购买5，000份价格为$25，期限为4个月的看跌期权来保值。

1.8一种股票在首次发行时会为公司提供资金。

请说明CBOE股票期权是否有同样的作用。

答：股票期权不为公司提供资金，它只是交易者之间相互买卖的一种证券，公司并不参与交易。

1.9请解释为什么远期合同既可用来投机又可用来套期保值？答：如果投资者预期价格将会上涨，可以通过远期多头来降低风险暴露，反之，预期价格下跌，通过远期空头化解风险。

如果投资者资产无潜在的风险暴露，远期合约交易就成为投机行为。

1.10假设一个执行价格为$50的欧式看涨期权价值$2.50,并持有到期。

在何种情况下期权的持有者会有盈利？在何种情况下，期权会被执行？请画图说明期权的多头方的收益是如何随期权到期日的股价的变化而变化的。

答：由欧式看涨期权多头的损益计算公式：max(,0)T S X −-2.5=-52.5,该欧式看涨期权的持有者在标的资产的市场价格大于$52.5时，会有盈利；当标的资产的市场价格高于$50时，期权就会被执行。

图形如下：T S 损益T S 0 52.51.11假设一欧式看跌期权执行价格为$60，价值为$4.00并持有到期。

在何种情况下，期权持有者（即空头方）会有盈利？在何种情况下，期权会被执行？请画图说明期权的空头方的收益是如何随期权到期日的股价的变化而变化的。

答：由欧式看跌期权多头的损益计算公式：max(,0)T X S −-4=56-,该欧式看跌期权的持有者在标的资产的市场价格低于$56时，会有盈利；当标的资产的市场价格低于$60时，期权就会被执行。

图形如下：T S 损益T S 0 561.12一位投资者出售了一个欧式9月份到期的看涨期权，执行价格为$20。

现在是5月，股票价格为18,期权价格为$20,现在是5月，股票价格为$18,期权价格为$2如果期权持有到期，并且到期时的股票价格为$25,请描述投资者的现金流状况。

答：由欧式看涨期权空头的损益计算公式：max(,0)T X S −+2=20-25+2=-3，投资者到期时将损失$3。

1.13一位投资者出售了一个欧式12月份到期的看跌期权，执行价格为$30,期权价值为$4。

在什么情况下，投资者会有盈利？答：当市场价格高于$20时，该看跌期权不被执行，投资者盈利为$4,当市场价格为（30，34）时，投资者盈利为4-(30-)。

T S1.14请说明在1.4节中描述的标准石油公司的债券是一种普通债券、一个执行价格为$25的基于油价的看涨期权的多头和一个执行价格为$40的基于油价的看涨期权的空头的组合。

max(25,0)T S −+,若市场价格高于$25,低于$40，则投资者损失为$25；若市场价格高于或等于$40,投资者收入为-25+40-+=+15;因此，该组合为一种普通债券，一个执行价格为$25的看涨期权多头和一个执行价格为$40的看涨期权空头。

min(40,0)T S −+T S T S T S T S T S 1.15一家公司将在4个月后收到一笔外币。

选用哪种期权合约，可以进行套期保值？答：通过购入四月期的看跌期权进行套期保值。

1.16黄金的现价为每盎司$500。

一年后交割的远期价格为每盎司$700。

一位套期保值者可以10％的年利率借到钱。

套利者应当如何操作才能获利？假设储存黄金费用不计。

答：套利者以10%的年利率借入货币，购买黄金现货，卖出黄金远期，一年后交割收益为700-（1＋10％）1.17芝加哥交易所提供标的物为长期国债的期货合约。

请描述什么样的投资者会使用这种合约。

答：投资者预期长期利率下降的套期保值者；长期利率的投机者以及在现货和期货市场套利者，可购买该期货合约。

1.18一种股票的现价为$94,执行价格为$95的3个月期的看涨期权价格为$4.70。

一位投资者预计股票价格将要上升，正在犹豫是购买100股股票，还是购买20份看涨期权（每份合约为100股）。

两种策略都须投资$9,400。

你会给他什么建议？股票价格上升到多少时，购买期权会盈利更大？答：购买股票盈利更大些。

由：max(,0)T S X C −−=20max(95,0)9,400T S −− 因此，当股票价格高于$94009520+=$565时，期权会盈利更大。

1.19“期权和期货是零合游戏”你是怎样理解这句话的？答：这句话是说期权和期货的一方损失程度等于另一方的盈利程度，总的收入为零。

1.20请描述下述组合的损益：同时签订一项资产的远期多头合约和有同样到期日的基于该项资产的欧式看跌期权的多头，并且在构造该组合时远期价格等于看跌期权的执行价格。

答：-+max(-,0),当>时，收入为-,当<时，收入为0。

T S X X T S T S X T S X T S X 1.21说明在1.4节中描述的ICON 是由一种普通股票和两种期权组合而成。

答：假设ICON 中外汇汇率为，则ICON 的收益为T S 1000，若<XT S1000-a(-X),当T S 1000T X S X a+>> 0，当1000T X S a+< 因此，ICON 的收益来自：（a）普通债券（b）执行价格为X 的欧式空头看涨期权（c）执行价格为1000X a+的欧式多头看涨期权如下图所示：普通债券空头看涨期权多头看涨期权总收益T S < 1000 0 0 1000X 1000X a+>> 1000 -a(-) 0 1000-a(-) T S X T S X T S X T S >+X 1000a 1000 -a(-) -a(--T S X T S X 1000a) 0 1.22说明在1.4节中描述的范围远期合约可由两种期权组合而成。

如何构造价值为零的范围远期合约？答:假设用范围远期合约去购买一单位的外汇，为汇率，则T S （a ）若<，支付T S 1X 1X （b ）若>,支付T S 1X 2X （c ）若1X ≤T S ≤2X ，支付即期利率范围远期合约可以看作由一个执行价格为的空头看跌期权和一个执行价格为的多头看涨期权组成。

1X 2X 如下表所示：外汇成本看跌期权价值看涨期权价值净成本 T S < －－（－） 0 － 1X T S 1X T S 1X 1X << － 0 0 － T S 2X T S T S 2X < － 0 －－T S T S T S 2X 2X 由于范围远期合约看跌期权与看涨期权头寸在建立初相等，因此构建范围远期合约不需要成本。

1.23某公司在1996年7月1日签订了一份远期合约，在1997年1月1日，购买1000万日元。

1996年9月1日，又签订了在1997年1月1日出售1000万日元的远期合约。

请描述这项策略的损益。

答：第一份远期合约的收益为-T S 1F ,第二份远期合约的收益为-T S 2F ,因此总收益为2F -1F 。

1.24假设英镑兑美元的即期和远期汇率如表1.1所示。

在下列情况中，投资者会有何获利机会？（A）一个180天的欧式看涨期权执行价格为1英镑兑1.5700美元，成本2美分。

（B）一个90天的欧式看跌期权执行价格为1英镑兑1.6400美元，成本2美分。

答：交易者通过卖出（A），90天后买入（B）来套利。

则（A）合约的损失为+0.02＝0.0118,(B)合约盈利为min(,0)T X S −max(,0)T X S −-0.02=0.0144,净收益为0.0026。

1.25请解释下面这句话：“一个远期合约的多头等价于一个欧式看涨期权的多头和一个欧式看跌期权的空头。

”答：由欧式看涨期权和看跌期权的损益公式得，一个欧式看涨期权的多头和一个欧式看跌期权的空头组合的损益为：max(,0)T S X −+ min(,0)T X S −,当>时，总收入为-+-=0;当<时，总收入亦为0。

与远期合约多头相一致。

T S X T S X X T S T S X 第二章2.1请说明未平仓合约数与交易量的区别。

答：未平仓合约数既可以指某一特定时间里多头合约总数，也可以指空头合约总数，而交易量是指在某一特定时间里交易的总和约数。

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