约翰.赫尔_期权期货和其他衍生品(six_edition)习题答案
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。
赫尔《期权、期货及其他衍生产品》(第7版)课后习题详解(希腊值)
17.2 课后习题详解一、问答题1．解释投资者如何对一个卖出的虚值看涨期权实施止损对冲策略。
为什么这种策略的效果并不好?Explain how a stop-loss hedging scheme can be implemented for the writer of an out-of-the money call option. Why does it provide a relatively poor hedge?答：假设期权的执行价格为10.00美元。
当期权处于实值状态时，期权的出售方将对其头寸实施完全的保护；当期权处于虚值状态时，出售方对期权头寸不采取任何对冲措施。
他试图通过以下方法实现上述策略：当期权标的资产的价格刚刚上涨至10.00美元时，买入该资产；当标的资产价格刚刚下跌至10.00美元时，卖出该资产。
该策略的问题是，它假设当资产价格从9.99美元上涨至l0.00美元时，接下来价格将会上涨至10.00美元以上。
(实际上接下来价格可能会回到9.99美元。
)类似地，它假设当资产价格从10.01美元下跌至10.00美元时，接下来价格将会下跌至10.00美元以下。
(实际上接下来价格可能会回到10.01美元。
)基于上述假设，期权出售方会在10.01美元买入而在9.99美元卖出。
然而这并不是一个好的对冲。
如果资产价格从未达到10.00美元，该交易策略的成本为零；如果资产价格多次达到10.00美元，交易策略的成本将十分高。
一个好的对冲的成本总是十分接近期权的价值。
2．一个看涨期权的Delta为0.7的含义是什么?当每个期权的Delta均为0.7时，如何使得1000份期权的短头寸组合变为Delta中性?What does it mean to assert that the Delta of a call option is 0.7? How can a short position in 1,000 options be made Delta neutral when the Delta of each答：（1）期权的Delta值（△），是指期权价格的变化与标的资产价格变化之比，衡量的是期权价格对标的资产价格变化的反应程度。
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.。
期权期货与其他衍生产品第九版课后习题与答案Chapter(.
E ( ST Se (T t var( ST S 2e2 (T t [e 2 2 2 (T t 1] Sincevar(ST E[(ST ] [ E(ST ] , it follows that E[(ST 2 ] var(ST [ E (ST ]2 so that E[( ST 2 ] S 2e2 (T t [e 2 2 (T t 1] S 2e2 (T t S 2e(2(T t In a risk-neutral world r so that ˆ [(S 2 ] S 2e(2 r 2 (T t E T Using risk-neutral valuation, the value of the derivative security at time t is ˆ [(S 2 ] e r (Tt E T S 2e(2 r 2 (T t r (T t e S 2e( r 2 (T t (b If: f S 2 e( r2 (T t 2 f S 2 (r 2 e( r (T t t 2 f 2 Se( r (T t S 2 2 f 2e( r (T t 2 S The left-hand side of the Black-Scholes–Merton differential equation is: 2 2 2 S 2 (r 2 e( r (T t 2rS 2e( r (T t2 S 2e( r (T t rS 2e( r 2 (T t rf Hence the Black-Scholes equation is satisfied. Problem 15.30. Consider an option on a non-dividend-paying stock when the stock price is $30, the exercise price is $29, the risk-free interest rate is 5% per annum, the volatility is 25% per annum, and the time to maturity is four months. a. What is the price of the option if it is a European call? b. What is the price of the option if it is an American call? c. What is the price of the option if it is a European put? d. Verify that put–call parity holds. In this case S0 30 , K 29 , r 005 , 025 and T4 12 d1 ln(30 29 (005 0252 2 4 12 04225 025 03333d2 ln(30 29 (005 0252 2 4 12 02782 025 03333 N (04225 06637 N (02782 06096 N (04225 03363 N (02782 03904 a. The European call price is 30 0663729e005412 06096 252 or $2.52. b. The American call price is the same as the European call price. It is $2.52. c. The European put price is29e005412 03904 30 03363 105 or $1.05. d. Put-call parity states that: p S c Ke rT In this case c 252 , S0 30 , K 29 , p105 and e rT 09835 and it is easy to verify that the relationship is satisfied, Problem 15.31. Assume that the stock in Problem 15.30 is due to go ex-dividend in 1.5 months. The expected dividend is 50 cents. a. What is the price of the option if it is aEuropean call? b. What is the price of the option if it is a European put? c. If the option is an American call, are there any circumstances when it will be exercised early? a. The present value of the dividend must be subtracted from the stock price. This gives a new stock price of: 30 05e0125005 295031 and ln(295031 29 (005 0252 2 03333 d1 03068 025 03333 ln(29503129 (005 0252 2 03333 d2 01625 025 03333 N (d1 06205 N (d2 05645 The price of the option is therefore 295031 06205 29e005412 05645 221 or $2.21. b. Because N (d103795 N (d2 04355 the value of the option when it is a European put is29e005412 04355 295031 03795 122or $1.22. c. If t1 denotes the time when the dividend is paid: K (1 e r (T t1 29(1 e00502083 03005 This is less than the dividend. Hence the option should be exercised immediately before the ex-dividend date for a sufficiently high value of the stock price. Problem 15.32. Consider an American call option when the stock price is $18, the exercise price is $20, the time to maturity is six months, the volatility is 30% per annum, and the risk-free interest rate is 10% per annum. Two equal dividends are expected during the life of the option, with ex-dividend dates at the end of two months and five months. Assume the dividends are 40 cents. Use Black’s approximation and the DerivaGem software to value the option. Suppose now that the dividend is D on each ex-dividend date. Use the results in the Appendix to determine how high D can be without the American option being exercised early. We first value the option assuming that it is not exercised early, we set the time to maturity equal to 0.5. There is a dividend of 0.4 in 2 months and 5 months. Other parameters are S0 18 , K 20 , r 10% , 30% . DerivaGem gives the price as 0.7947. We next value the option assuming that it is exercised at the five-month point just before the final dividend. DerivaGem gives the price as 0.7668. The price given by Black’s approximation is therefore 0.7947. (DerivaGem also shows that the American option price calculated using the binomial model with 100 time steps is 0.8243. It is never optimal to exercise theoption immediately before the first ex-dividend date when D1 K[1 e r (t2 t1 ] where D1 is the size of the first dividend, and t1 and t2 are the times of the first and second dividend respectively. Hence we must have: D1 20[1 e(01025 ] that is, D1 0494 It is never optimal to exercise the option immediately before the second ex-dividend date when: D2 K (1 e r (T t2 where D2 is the size of the second dividend. Hence we must have: D2 20(1 e0100833 that is, D2 0166 It follows that the dividend can be as high as 16.6 cents per share without the American option being worth more than the corresponding European option.。
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】
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
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。
期权期货和其它衍生产品约翰赫尔答案
资料范本本资料为word版本，可以直接编辑和打印，感谢您的下载期权期货和其它衍生产品约翰赫尔答案地点：__________________时间：__________________说明：本资料适用于约定双方经过谈判，协商而共同承认，共同遵守的责任与义务，仅供参考，文档可直接下载或修改，不需要的部分可直接删除，使用时请详细阅读内容第一章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)。
赫尔《期权期货及其他衍生产品》第1章(第八版)讲述
期权、期货及其他衍生产品（第八版） Copyright © John C. Hull 2012
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对冲基金( 见业界事例1-2，p8)
• 对冲基金受到的约束与共同基金不同，一般不对外公 布持有的证券组合。
• 共同基金必须
– 披露投资策略 – 在任意时刻允许份额赎回 – 杠杆率受到限制 – 不能持有空头头寸
20
2. 石油:另外一种套利机会?
假定:
- 石油的即期价格为95美元 - 1年期原油期货的标价为80美元 - 1年期的美元利率为 5% - 原油的储存成本为每年2%
是否存在套利机会?
期权、期货及其他衍生产品（第八版） Copyright © John C. Hull 2012
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期权
• 看涨期权：其持有者有权在将来某一特定时间 以某一确定价格（ 执行价格）买入某种资产。
卖出价 1.4411 1.4413 1.4415 1.4422
期权、期货及其他衍生产品（第八版） Copyright © John C. Hull 2012
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远期价格
• 合约的远期价格是今天约定的合约支付价 格（使合约价值为零的支付价格）；
• 对不同期限的远期合约而言，远期价格也 不同（如表1-1所示） 。
期权、期货及其他衍生产品（第八版） Copyright © John C. Hull 2012
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期权、期货及其他衍生产品（第八版） Copyright © John C. Hull 2012
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表1-2 谷歌股票看涨期权在2010年6月15日的价格
(P6)
执行 2010年7月
价格
买入价
2010年7月 卖出价
2010年9月 买入价
赫尔《期权、期货及其他衍生产品》复习笔记及课后习题详解（凸性、时间与Quanto调整）【圣才出品】
赫尔《期权、期货及其他衍生产品》复习笔记及课后习题详解（凸性、时间与Quanto调整）【圣才出品】第30章凸性、时间与Quanto调整30.1 复习笔记1．凸性调整考虑对这样一种产品定价，其收益依赖于在收益发生时间点所观察到的债券收益率。
通常一个变量的远期值是通过一个在时间T收益为S T－K的远期合约来计算的，它是对应于使合约价值为0的价格K。
一般来讲，远期债券收益率是远期债券价格所隐含的利率。
假定B T是在时间T的一个债券价格，y T为其收益率。
B T与y T之间（债券定价）的关系式为：B T＝G（y T）定义B F为时间T到期的合约在时间0的远期债券价格，y F为时间0的远期债券收益率。
由定义得出：B F＝G（y F）函数G为非线性函数。
这意味着，当将来债券价格的期望值等于远期债券价格时（于是我们在一个对于时间T到期的零息债券为风险中性世界里），将来的债券期望收益率并不等于远期债券收益率。
这一点可通过图30-1来说明。
假定只有三种可能的债券价格B1、B2和B3，假如债券价格的间隔是相同的，即B2－B1＝B3－B2。
债券的远期价格是债券的期望值B2。
由债券价格，可以计算出三个具有相同可能性的收益率：y1、y2和y3。
这些收益率之间的间隔并不相同。
变量y2为远期债券的收益率，这是因为它对应于远期债券价格。
债券收益率的期望值为y1、y2和y3的平均值，显然该平均值大于y2。
图30-1 在时间T 时债券价格与债券收益率的关系对于一个收益依赖于时间T 的债券收益率的衍生产品，可以通过以下过程来定价：（a ）在对于时间T 到期的零息债券为远期风险中性的世界里计算收益的期望值；（b ）以当前期限为T 的无风险利率进行贴现。
在所考虑的世界里，债券价格期望值等于远期价格。
因此，需要计算当债券价格期望值等于远期价格时债券收益率的期望值。
债券收益率的期望值可以由以下近似式表示()()()2212F T T F F y F G y E y y y T G y ''=-'σ （30-1）式中G ′和G ″表示函数G 的一阶和二阶偏导数，E T 表示在一个对于计价单位P （t ，T ）为远期风险中性世界里的期望值，σy 为远期收益率的波动率。
赫尔《期权、期货及其他衍生产品》重大金融损失与借鉴)【圣才出品】
第35章重大金融损失与借鉴衍生产品市场的规模达数万亿美元。
不管从哪个角度来看，这个行业都是一个十分成功的行业，它确实满足了很多人的需求。
前美联储主席艾伦·格林斯潘在2003年5月曾说过以下一席话：采用多种多样的衍生产品，以及采用更为完善的方法来检测和管理风险已经成为大金融媒介机构强化其负载能力的主要动力。
35.1 定义风险额度风险额度：所有的公司（金融以及非金融公司）都必须对自身所能承担的风险有一个清晰、明确的定义。
公司应该制定管理程序来保证额度的贯彻执行。
整体的风险额度应该由董事会建立，然后这些额度应被转化为负责特定风险管理人员的特定额度。
每天的风险报告应该是对将来市场变化所带来盈亏的一个预测，风险报告所预测的数值要与实际损失进行比较以便保证报告所用定价工具的准确性。
35.1.1 认真对待风险额度在盈利时对于违反交易额度的惩罚与在亏损时对于违犯交易额度的惩罚要一视同仁，否则的话交易员在交易损失之后会加大自身的交易量来取得盈利，并希望人们忘记自己的交易损失。
典型的实例：奥兰治县的西特仑在1991～1993年曾给这个城市带来了巨大的盈利，城市以他的交易收入作为运转资金，人们这时因为盈利而忽略西特仑所承担的风险，不幸的是1994年西特仑的损失远超过了他若干年的盈利。
35.1.2 不要认为你会猜透市场有的交易员对于市场预测可能比其他人更为优秀，但是他不可能永远正确。
交易员能在所有预测中有60%正确就已经相当不错了，一个交易员出类拔萃的战绩（像20世纪90年代初的西特仑）很可能是由于运气，而并非交易技巧。
假定某金融机构雇用了16名交易员，其中一名交易员在过去一年中的每一个季度都盈利，这位交易员是否应拿到更多的奖金呢？他的交易额度是否应该增加呢？第一个问题的答案无疑是肯定的，但第二个问题的答案应该是否定的。
在四个季度中均盈利的概率为0.54，即1/16，这意味着即使完全出于随机，在过去一年里每16个交易员中至少有一位每个季度“都会盈利”。
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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。