约翰.赫尔,期权期货和其他衍生品(third edition)习题答案
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。
期权期货和其他衍生品 第四章答案(双数题)
金融工程原理作业4.2 答：LIBOR 是伦敦同业银行拆出利率（London Interbank Offered Rate ）的缩写，是指银行给其他大银行提供企业资金时所收取的利率。
LIBID 是伦敦同业拆出利率（London Interbank Bid Rate ），即银行同意其他银行以LIBID 利率将资金存入自己的银行。
LIBOR 利率会略高于LIBID 利率。
4.4 解：设利率为R（a ）一年复利一次，1000（1+R ）=1100，求出R=10%（b ）一年复利两次，1000（1+R/2）2=1100，求出R=9.76%（c ）每月复利一次，共12次，1000(1+R/12)12=1100，求出 R=9.57% （d ）连续复利，1000e R =1100，求出R=9.53%4.6解：收入R K 的FRA 的价值为V 22))((12T R F K FRA e T T R R L ---=，已知，R K =9.5%，T 2=1.25，T 1=1，L=1000000，R 2=8.6%R F =121122T T T R T R --=125.15.825.16.8--⨯×100%=9%，对应的每季度复利一次的利率为R F =4（e 4/09.0﹣1）=9.102%代入得，V 25.1086.0)125.1()09102.0095.0(1000000⨯--⨯-⨯=e FRA =893.59答：该FRA 的价值为893.59美元。
4.8答：久期可以用来估算债券收益的一个微小变化对证券组合价值的影响，金融机构常常通过确保其资产平均久期等于其负债平均久期来对冲其面临的风险，但是组合对于利率较大的平行移动和非平行移动仍有风险暴露，这是久期的缺陷性。
4.10 解：已知R c =12%，m=4，则由公式得每年计息4次的利率R m =m(e Rc /m -1)= 4(e 12%/4-1)=12.182%所以每季度的利息为10000*（12.182%/4）=304.55美元4.12解：设收益率为y,年利率为8%,半年付息一次，每次付息为4元，三年期共付息6次，最后一次连本金104，可得4e y 5.0-+4e y 1.0-+4e y 5.1-+4e y 2-+4e y 5.2-+104e y 3-=104,解得，y=6.407%4.14 解：R F2=121122T T T R T R --=%41202.0203.0=--⨯ R F3=232233T T T R T R --=%1.523203.03037.0=-⨯-⨯ R F4=343344T T T R T R --=%7.5343037.04042.0=-⨯-⨯ R F5=454455T T T R T R --=%7.5454042.05045.0=-⨯-⨯ 所以，第二年、第三年、第四年、第五年的远期利率分别为4%，5.1%，5.7%和5.7%。
赫尔《期权、期货及其他衍生产品》（第7版）课后习题详解（曲率、时间与Quanto调整）
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。
期权期货和其它衍生产品第三版约翰赫尔答案
第一章1.1请解释远期多头与远期空头的区别。
答：远期多头指交易者协定将来以某一确定价格购入某种资产；远期空头指交易者协定将来以某一确定价格售出某种资产。
1.2请详细解释套期保值、投机与套利的区别。
答：套期保值指交易者采取一定的措施补偿资产的风险暴露；投机不对风险暴露进行补偿，是一种“赌博行为”；套利是采取两种或更多方式锁定利润。
1.3请解释签订购买远期价格为$50的远期合同与持有执行价格为$50的看涨期权的区别。
答：第一种情况下交易者有义务以50$购买某项资产（交易者没有选择），第二种情况下有权利以50$购买某项资产（交易者可以不执行该权利）。
1.4一位投资者出售了一个棉花期货合约，期货价格为每磅50美分，每个合约交易量为50，000磅。
请问期货合约结束时，当合约到期时棉花价格分别为（a）每磅48.20美分；（b）每磅51.30美分时，这位投资者的收益或损失为多少？答：(a)合约到期时棉花价格为每磅$0.4820时，交易者收入：（$0.5000-$0.4820）×50，000＝$900;(b)合约到期时棉花价格为每磅$0.5130时，交易者损失:($0.5130-$0.5000) ×50，000=$6501.5假设你出售了一个看跌期权，以$120执行价格出售100股IBM的股票，有效期为3个月。
IBM股票的当前价格为$121。
你是怎么考虑的？你的收益或损失如何？答：当股票价格低于$120时，该期权将不被执行。
当股票价格高于$120美元时，该期权买主执行该期权，我将损失100(st-x)。
1.6你认为某种股票的价格将要上升。
现在该股票价格为$29,3个月期的执行价格为$30的看跌期权的价格为$2.90.你有$5,800资金可以投资。
现有两种策略：直接购买股票或投资于期权，请问各自潜在的收益或损失为多少？答：股票价格低于$29时，购买股票和期权都将损失，前者损失为$5,800×(29-p),$29×(p-29),后者损失为$5，800;当股票价格为（29，30），购买股票收益为$5,800$29购买期权损失为$5,800;当股票价格高于$30时，购买股票收益为$5,800×(p-29)，$29×(p-30)-5,800。
期权 期货及其他衍生品 第 版 课后作业题解答 章
1第一次作业参考答案第1章1.26远期合约多头规定了一年后以每盎司1000美元买入黄金，到期远期合约必须执行，交易双方权利义务对等；期权合约多头规定了一年后以每盎司1000美元买入黄金的权利，到期合约可以不执行，也可以执行，交易双方权利义务不对等。
假设S T为一年以后黄金的价格，则远期合约的收益为S T-1000;期权合约的受益为S T-1100,如果S T>1000;-100,如果S T<10001.27投资人承诺在7月份以40美元的执行价格买入股票。
如果未来股票价格跌至37美元以下，则该投资人赚取的3美元期权费不足以弥补期权上的损失，从而亏损。
当未来股票价格为37-40美元时，交易对手会执行期权，此时，投资人此时同样有正收益。
如果未来股票价格高于40美元，该期权不会被对手执行，此时投资者仅赚取期权费。
1.28远期：购入三个月期限的300万欧元的欧元远期合约，并在三个月后，用到期的远期合约进行支付300万欧元。
期权：购入三个月期限的300万欧元的欧元看涨期权，如果三个月后汇率高于期权约定执行汇率，则执行该期权，反之则不执行该期权。
1.29当股票到期价格低于30美元时，两个期权合约都与不会被执行，该投资者无头寸;当股票价格高于32.5美元时,两个期权都会执行，该投资者无头寸，若股票价格在30-32.5美元之间时，该投资者买入期权会被执行，卖出的期权不会被执行，因此该投资者持有长头寸。
1.30（低买高卖）借入1000美元资金，买入黄金，同时在卖出一年期的黄金远期合约，锁定到期的价格1200美元，到期偿还本金和利息。
到期时的受益为1200-1000（1+10%）=100；收益率为100/1000=10%1.312由于期权存在杠杆效应，看张期权的风险更大，同时收益率也更高。
假设股票在S时，购入看涨期权和购入股票无差异，则100*（S-94)=2000（S-95）-9400即S=100。
如果未来股票价格高于100美元，则购入看涨期权合约则盈利更高，反之，如果未来股票价格低于100美元，则购入股票比购入期权合约盈利更多。
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。
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。
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。
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。
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。
JohnHull《期货、期权和衍生证券》13章习题解答
CHAPTER 13Wiener Processes 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 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 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 variancerate 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 rates1μ 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 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 lengthT , 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 rate12μμ+ and variance rate 2212σσ+.(b) In this case the change in the value of 12X X + in a short interval of timet ∆ 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 withdrift rate 12μμ+ and variance rate 2212122σσρσσ++.Problem 13.5.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 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 It ô’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 G increases 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 andvolatility 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 andB 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.The process for the stock price in equation (13.8) isS S t S μσε∆=∆+ where μ and σ are constant. Explain carefully the difference between thismodel and each 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 behaviorthan 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 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 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 It ô’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 It ô’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 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)Sϕ∆⨯.,⨯.that is,(002206164)Sϕ∆.,.(a) The expected stock price at the end of the next day is therefore 50.022 (b) The standard deviation of the stock price at the end of the next day is061540785.=.(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 cashposition 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 normalvariable [i.e., a variable with probability distribution (01)ϕ,] is less than x . From normal distribution 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 xϕ.+.,.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 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 expected instantaneous 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 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 thatIt ô’s lemma gives22dy S dt S dz μσ=+ordy y dt y dz μσ=+(b) In this case 2y S S ∂/∂=, 222y S ∂/∂=, and 0y t ∂/∂= so thatIt ô’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 thatIt ô’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 It ô’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 standard deviation T σ. In this case 050S =, 012μ=., 2T =, and 030σ=. so that the mean and standard deviation of ln T S are 2ln 50(012032)24062+.-./=. and 0320424.=., 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 . 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 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 ’s are correlated samples from standard normal distributions.。
HullOFOD9eSolutionsCh03第九版期权期货及其他衍生品课后答案汇编
Basis risk arises from the hedger’s uncertainty as to the difference between the spot price and futures price at the expiration of the hedge.
Problem 3.6. Suppose that the standard deviation of quarterly changes in the prices of a commodity is $0.65, the standard deviation of quarterly changes in a futures price on the commodity is $0.81, and the coefficient of correlation between the two changes is 0.8. What is the optimal hedge ratio for a three-month contract? What does it mean?
Problem 3.4. Under what circumstances does a minimum-variance hedge portfolio lead to no hedging at all?
A minimum variance hedge leads to no hedging when the coefficient of correlation between the futures price changes and changes in the price of thect hedge is one that completely eliminates the hedger’s risk. A perfect hedge does not always lead to a better outcome than an imperfect hedge. It just leads to a more certain outcome. Consider a company that hedges its exposure to the price of an asset. Suppose the asset’s price movements prove to be favorable to the company. A perfect hedge totally neutralizes the company’s gain from these favorable price movements. An imperfect hedge, which only partially neutralizes the gains, might well give a better outcome.
JohnHull《期货、期权和衍生证券》13章习题解答
CHAPTER 13Wiener P rocesses and It?’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 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 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 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 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 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 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 ? 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. (d)(e) 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 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∂∂∂=;=-;=∂∂∂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 It?’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 It?’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 It?’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 It?’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 ?. 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 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 ?’s are correlated samples from standard normal distributions.。
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.。
赫尔《期权期货及其他衍生产品》第3章(第八版)
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为什么要对冲股权组合？
• 如果想暂时离开市场一会儿, 那么对冲可以避免出售资产和购买组合 的成本。
• 假设你投资组合的β均值为1.0。你认为你的投资组合挑选较为完美, 在任何市场情形下都能战胜市场。对冲可以确保你的收益为无风险收 益与投资组合超额收益（相对于市场组合）之和。
期权、期货及其他衍生产品（第八版） Copyright © John C.Hull 2012
• 如果在对冲上有损失、在标的资产上有收益, 如何解释这种情形较为困难。
期权、期货及其他衍生产品（第八版） Copyright © John C.Hull 2012
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基差风险
• 基差=被对冲风险的即期价格-用于对冲的期货合约的价格 • 对冲被平仓时基差的不确定, 导致了基差风险的产生。
期权、期货及其他衍生产品（第八版） Copyright © • 选择与对冲的到期日最近, 但仍长于对冲到期日的交割月份； • 当被对冲的资产与期货的标的资产不吻合时, 选择期货价格与被对冲
资产的价格相关性最高的期货合约。这就是交叉对冲。
期权、期货及其他衍生产品（第八版） Copyright © John C.Hull 2012
8
最小方差对冲比率的计算(p41)
2
拥护对冲的观点
• 公司应该集中精力发展自身的主要业务, 采取措施将由利率、汇率和 其他市场变量所引起的风险降至最低。
期权、期货及其他衍生产品（第八版） Copyright © John C.Hull 2012
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反对对冲的观点
• 股东通常持有充分分散的投资组合, 他们可以 自己对冲面临的风险；
• 当竞争对手选择不对冲风险时, 这会导致对冲 风险的上升；
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HullOFOD9eSolutionsCh03第九版期权期货及其他衍生品课后答案
CHAPTER 3Hedging Strategies Using FuturesPractice QuestionsProblem 3.1.Under what circumstances are (a) a short hedge and (b) a long hedge appropriate?A short hedge is appropriate when a company owns an asset and expects to sell that asset in the future. It can also be used when the company does not currently own the asset but expects to do so at some time in the future. A long hedge is appropriate when a company knows it will have to purchase an asset in the future. It can also be used to offset the risk from an existing short position.Problem 3.2.Explain what is meant by basis risk when futures contracts are used for hedging.Basis risk arises from the hedger’s uncertainty as to the difference between the spot price and futures price at the expiration of the hedge.Problem 3.3.Explain what is meant by a perfect hedge. Does a perfect hedge always lead to a better outcome than an imperfect hedge? Explain your answer.A perfect hedge is one that completely eliminates the hedger’s risk. A p erfect hedge does not always lead to a better outcome than an imperfect hedge. It just leads to a more certain outcome. Consider a company that hedges its exposure to the price of an asset. Suppose the asset’s price movements prove to be favorable to the company. A perfect hedge totally neutralizes the company’s gain from these favorable price movements. An imperfect hedge, which only partially neutralizes the gains, might well give a better outcome.Problem 3.4.Under what circumstances does a minimum-variance hedge portfolio lead to no hedging at all?A minimum variance hedge leads to no hedging when the coefficient of correlation between the futures price changes and changes in the price of the asset being hedged is zero.Problem 3.5.Give three reasons why the treasurer of a company might not hedge the company’s exposure to a particular risk.(a) If the company’s competitors are not hedging, the treasurer might feel that the company will experience less risk if it does not hedge. (See Table 3.1.) (b) The shareholders might not want the company to hedge because the risks are hedged within their portfolios. (c) If there is a loss on the hedge and a gain from the company’s exposure to the underlying asset, the treasurer might feel that he or she will have difficulty justifying the hedging to other executives within the organization.Problem 3.6.Suppose that the standard deviation of quarterly changes in the prices of a commodity is $0.65, the standard deviation of quarterly changes in a futures price on the commodity is $0.81, and the coefficient of correlation between the two changes is 0.8. What is the optimal hedge ratio for a three-month contract? What does it mean?The optimal hedge ratio is 065080642081..⨯=.. This means that the s ize of the futures position should be 64.2% of the size of the company’s exposure in a three-month hedge.Problem 3.7.A company has a $20 million portfolio with a beta of 1.2. It would like to use futures contracts on a stock index to hedge its risk. The index futures is currently standing at 1080, and each contract is for delivery of $250 times the index. What is the hedge that minimizes risk? What should the company do if it wants to reduce the beta of the portfolio to 0.6?The formula for the number of contracts that should be shorted gives 20000000128891080250,,.⨯=.⨯ Rounding to the nearest whole number, 89 contracts should be shorted. To reduce the beta to 0.6, half of this position, or a short position in 44 contracts, is required.Problem 3.8.In the corn futures contract, the following delivery months are available: March, May, July, September, and December. State the contract that should be used for hedging when the expiration of the hedge is in a) June, b) July, and c) JanuaryA good rule of thumb is to choose a futures contract that has a delivery month as close aspossible to, but later than, the month containing the expiration of the hedge. The contracts that should be used are therefore(a) July(b)September(c)MarchProblem 3.9.Does a perfect hedge always succeed in locking in the current spot price of an asset for a future transaction? Explain your answer.No. Consider, for example, the use of a forward contract to hedge a known cash inflow in a foreign currency. The forward contract locks in the forward exchange rate — which is in general different from the spot exchange rate.Problem 3.10.Explain why a short hedger’s position improves when the basis strengthens unexpectedly and worsens when the basis weakens unexpectedly.The basis is the amount by which the spot price exceeds the futures price. A short hedger is long the asset and short futures contracts. The value of his or her position therefore improves as the basis increases. Similarly, it worsens as the basis decreases.Problem 3.11.Imagine you are the treasurer of a Japanese company exporting electronic equipment to the United States. Discuss how you would design a foreign exchange hedging strategy and the arguments you would use to sell the strategy to your fellow executives.The simple answer to this question is that the treasurer should1.Estimate the company’s future cash flows in Japanese yen and U.S. dollars2.Enter into forward and futures contracts to lock in the exchange rate for the U.S. dollarcash flows.However, this is not the whole story. As the gold jewelry example in Table 3.1 shows, the company should examine whether the magnitudes of the foreign cash flows depend on the exchange rate. For example, will the company be able to raise the price of its product in U.S. dollars if the yen appreciates? If the company can do so, its foreign exchange exposure may be quite low. The key estimates required are those showing the overall effect on the company’s profitability of changes in the exchange rate at various times in the future. Once these estimates have been produced the company can choose between using futures and options to hedge its risk. The results of the analysis should be presented carefully to other executives. It should be explained that a hedge does not ensure that profits will be higher. It means that profit will be more certain. When futures/forwards are used both the downside and upside are eliminated. With options a premium is paid to eliminate only the downside.Problem 3.12.Suppose that in Example 3.2 of Section 3.3 the company decides to use a hedge ratio of 0.8. How does the decision affect the way in which the hedge is implemented and the result?If the hedge ratio is 0.8, the company takes a long position in 16 December oil futures contracts on June 8 when the futures price is $88.00. It closes out its position on November 10. The spot price and futures price at this time are $90.00 and $89.10. The gain on the futures position is(89.10 − 88.00) × 16,000 = 17,600The effective cost of the oil is therefore20,000 × 90 – 17,600 = 1,782, 400or $89.12 per barrel. (This compares with $88.90 per barrel when the company is fully hedged.)Problem 3.13.“If the minimum -variance hedge ratio is calculated as 1.0, the hedge must be perfect." Is this statement true? Explain your answer.The statement is not true. The minimum variance hedge ratio is S Fσρσ It is 1.0 when 05=.ρ and 2S F =σσ. Since 10<.ρ the hedge is clearly not perfect.Problem 3.14.“If there is no basis risk, the minimum variance hedge ratio is always 1.0." Is this statement true? Explain your answer.The statement is true. Using the notation in the text, if the hedge ratio is 1.0, the hedger locks in a price of 12F b +. Since both 1F and 2b are known this has a variance of zero and must be the best hedge.Problem 3.15“For an asset where futures prices for contracts on the asset are usually less than spot prices, long hedges are likely to be particularly attractive." Explain this statement.A company that knows it will purchase a commodity in the future is able to lock in a price close to the futures price. This is likely to be particularly attractive when the futures price is less than the spot price.Problem 3.16.The standard deviation of monthly changes in the spot price of live cattle is (in cents per pound)1.2. The standard deviation of monthly changes in the futures price of live cattle for the closestcontract is 1.4. The correlation between the futures price changes and the spot price changes is 0.7. It is now October 15. A beef producer is committed to purchasing 200,000 pounds of live cattle on November 15. The producer wants to use the December live-cattle futures contracts to hedge its risk. Each contract is for the delivery of 40,000 pounds of cattle. What strategy should the beef producer follow?The optimal hedge ratio is 12070614..⨯=.. The beef producer requires a long position in 20000006120000⨯.=, lbs of cattle. The beef producer should therefore take a long position in 3 December contracts closing out the position on November 15.Problem 3.17.A corn farmer argues “I do not use futures co ntracts for hedging. My real risk is not the price of corn. It is that my whole crop gets wiped out by the weather.”Discuss this viewpoint. Should the farmer estimate his or her expected production of corn and hedge to try to lock in a price for expected production?If weather creates a significant uncertainty about the volume of corn that will be harvested, the farmer should not enter into short forward contracts to hedge the price risk on his or her expected production. The reason is as follows. Suppose that the weather is bad and the farmer’sproduction is lower than expected. Other farmers are likely to have been affected similarly. Corn production overall will be low and as a consequence the price of corn will be relatively high. The farmer’s problems arising from the bad harvest will be made worse by losses on the short futures position. This problem emphasizes the importance of looking at the big picture when hedging. The farmer is correct to question whether hedging price risk while ignoring other risks is a good strategy.Problem 3.18.On July 1, an investor holds 50,000 shares of a certain stock. The market price is $30 per share. The investor is interested in hedging against movements in the market over the next month and decides to use the September Mini S&P 500 futures contract. The index is currently 1,500 and one contract is for delivery of $50 times the index. The beta of the stock is 1.3. What strategy should the investor follow? Under what circumstances will it be profitable?A short position in 50000301326501500,⨯.⨯=⨯,contracts is required. It will be profitable if the stock outperforms the market in the sense that its return is greater than that predicted by the capital asset pricing model.Problem 3.19.Suppose that in Table 3.5 the company decides to use a hedge ratio of 1.5. How does the decision affect the way the hedge is implemented and the result?If the company uses a hedge ratio of 1.5 in Table 3.5 it would at each stage short 150 contracts. The gain from the futures contracts would be.1=⨯50.2$5570.1per barrel and the company would be $0.85 per barrel better off than with a hedge ratio of 1.Problem 3.20.A futures contract is used for hedging. Explain why the daily settlement of the contract can give rise to cash flow problems.Suppose that you enter into a short futures contract to hedge the sale of an asset in six months. If the price of the asset rises sharply during the six months, the futures price will also rise and you may get margin calls. The margin calls will lead to cash outflows. Eventually the cash outflows will be offset by the extra amount you get when you sell the asset, but there is a mismatch in the timing of the cash outflows and inflows. Your cash outflows occur earlier than your cash inflows.A similar situation could arise if you used a long position in a futures contract to hedge the purchase of an asset at a future time and the asset’s price fe ll sharply. An extreme example of what we are talking about here is provided by Metallgesellschaft (see Business Snapshot 3.2).Problem 3.21.An airline executive has argued: “There is no point in our using oil futures. There is just as much chance that the price of oil in the future will be less than the futures price as there is that it will be greater than this price.” Discuss the executive’s viewpoint.It may well be true that there is just as much chance that the price of oil in the future will be above the futures price as that it will be below the futures price. This means that the use of a futures contract for speculation would be like betting on whether a coin comes up heads or tails. But it might make sense for the airline to use futures for hedging rather than speculation. The futures contract then has the effect of reducing risks. It can be argued that an airline should not expose its shareholders to risks associated with the future price of oil when there are contracts available to hedge the risks.Problem 3.22.Suppose the one-year gold lease rate is 1.5% and the one-year risk-free rate is 5.0%. Both rates are compounded annually. Use the discussion in Business Snapshot 3.1 to calculate the maximum one-year forward price Goldman Sachs should quote for gold when the spot price is $1,200.Goldman Sachs can borrow 1 ounce of gold and sell it for $1200. It invests the $1,200 at 5% so that it becomes $1,260 at the end of the year. It must pay the lease rate of 1.5% on $1,200. This is $18 and leaves it with $1,242. It follows that if it agrees to buy the gold for less than $1,242 in one year it will make a profit.Problem 3.23.The expected return on the S&P 500 is 12% and the risk-free rate is 5%. What is the expected return on the investment with a beta of (a) 0.2, (b) 0.5, and (c) 1.4?a)00502(012005)0064.+.⨯.-.=. or 6.4%b)00505(012005)0085.+.⨯.-.=. or 8.5%c)00514(012005)0148.+.⨯.-.=. or 14.8%Further QuestionsProblem 3.24.It is now June. A company knows that it will sell 5,000 barrels of crude oil in September.It uses the October CME Group futures contract to hedge the price it will receive. Each contract is on 1,000 barrels of ‘‘light sweet crude.’’ What position should it take? What price risks is it still exposed to after taking the position?It should short five contracts. It has basis risk. It is exposed to the difference between the October futures price and the spot price of light sweet crude at the time it closes out its position in September. It is also possibly exposed to the difference between the spot price of light sweet crude and the spot price of the type of oil it is selling.Problem 3.25.Sixty futures contracts are used to hedge an exposure to the price of silver. Each futures contract is on 5,000 ounces of silver. At the time the hedge is closed out, the basis is $0.20per ounce. What is the effect of the bas is on the hedger’s financial position if (a) the traderis hedging the purchase of silver and (b) the trader is hedging the sale of silver?The excess of the spot over the futures at the time the hedge is closed out is $0.20 per ounce. If the trader is hedging the purchase of silver, the price paid is the futures price plus the basis. Thetrader therefore loses 60×5,000×$0.20=$60,000. If the trader is hedging the sales of silver, the price received is the futures price plus the basis. The trader therefore gains $60,000.Problem 3.26.A trader owns 55,000 units of a particular asset and decides to hedge the value of her position with futures contracts on another related asset. Each futures contract is on 5,000 units. The spot price of the asset that is owned is $28 and the standard deviation of the change in this price over the life of the hedge is estimated to be $0.43. The futures price of the related asset is $27 and the standard deviation of the change in this over the life of the hedge is $0.40. The coefficient of correlation between the spot price change and futures price change is 0.95.(a) What is the minimum variance hedge ratio?(b) Should the hedger take a long or short futures position?(c) What is the optimal number of futures contracts with no tailing of the hedge?(d) What is the optimal number of futures contracts with tailing of the hedge?(a) The minimum variance hedge ratio is 0.95×0.43/0.40=1.02125.(b) The hedger should take a short position.(c) The optimal number of contracts with no tailing is 1.02125×55,000/5,000=11.23 (or 11 when rounded to the nearest whole number)(d) The optimal number of contracts with tailing is 1.012125×(55,000×28)/(5,000×27)=11.65 (or 12 when rounded to the nearest whole number).Problem 3.27.A company wishes to hedge its exposure to a new fuel whose price changes have a 0.6correlation with gasoline futures price changes. The company will lose $1 million for each 1 cent increase in the price per gallon of the new fuel over the next three months. The new fuel's price change has a standard deviation that is 50% greater than price changes in gasoline futures prices. If gasoline futures are used to hedge the exposure what should the hedge ratio be? What is the company's exposure measured in gallons of the new fuel? What position measured in gallons should the company take in gasoline futures? How many gasoline futures contracts should be traded? Each contract is on 42,000 gallons.Equation (3.1) shows that the hedge ratio should be 0.6 × 1.5 = 0.9. The company has anexposure to the price of 100 million gallons of the new fuel. It should therefore take a position of 90 million gallons in gasoline futures. Each futures contract is on 42,000 gallons. The number of contracts required is therefore9.2142000,42000,000,90 or, rounding to the nearest whole number, 2143.Problem 3.28.A portfolio manager has maintained an actively managed portfolio with a beta of 0.2. During the last year the risk-free rate was 5% and equities performed very badly providing a return of−30%. The portfolio manage produced a return of −10% and claims that in the circumstances it was good. Discuss this claim.When the expected return on the market is −30% the expected return on a portfolio with a beta of 0.2 is0.05 + 0.2 × (−0.30 − 0.05) = −0.02or –2%. The actual return of –10% is worse than the expected return. The portfolio manager done 8% worse than a simple strategy of forming a portfolio that is 20% invested in an equity index and 80% invested in risk-free investments. (The manager has achieved an alpha of –8%!)Problem 3.29. (Excel file)The following table gives data on monthly changes in the spot price and the futures price for a certain commodity. Use the data to calculate a minimum variance hedge ratio.Denote i x and i y by the i -th observation on the change in the futures price and the change in the spot price respectively.096130i i x y =.=.∑∑222447423594i i x y =.=.∑∑2352i i x y =.∑ An estimate of F σ is05116=. An estimate of S σ is04933=.An estimate of ρ is0981=.The minimum variance hedge ratio is049330981094605116SF.=.⨯=..σρσProblem 3.30.It is July 16. A company has a portfolio of stocks worth $100 million. The beta of the portfolio is 1.2. The company would like to use the December futures contract on a stock index to change beta of the portfolio to 0.5 during the period July 16 to November 16. The index is currently1,000, and each contract is on $250 times the index.a)What position should the company take?b)Suppose that the company changes its mind and decides to increase the beta of theportfolio from 1.2 to 1.5. What position in futures contracts should it take?a)The company should short(1205)1000000001000250.-.⨯,,⨯or 280 contracts.b)The company should take a long position in(1512)1000000001000250.-.⨯,,⨯or 120 contracts.Problem 3.31. (Excel file)A fund manager has a portfolio worth $50 million with a beta of 0.87. The manager is concerned about the performance of the market over the next two months and plans to use three-month futures contracts on the S&P 500 to hedge the risk. The current level of the index is 1250, one contract is on 250 times the index, the risk-free rate is 6% per annum, and the dividend yield on the index is 3% per annum. The current 3 month futures price is 1259.a)What position should the fund manager take to eliminate all exposure to the market overthe next two months?b)Calculate the effect of your strategy on the fund manager’s returns if the level of themarket in two months is 1,000, 1,100, 1,200, 1,300, and 1,400. Assume that the one-month futures price is 0.25% higher than the index level at this time.a) The number of contracts the fund manager should short is 50000000087138201259250,,.⨯=.⨯ Rounding to the nearest whole number, 138 contracts should be shorted.b) The following table shows that the impact of the strategy. To illustrate the calculations in the table consider the first column. If the index in two months is 1,000, the futures price is 1000×1.0025. The gain on the short futures position is therefore(1259100250)2501388849250$-.⨯⨯=,,The return on the index is 3212⨯/=0.5% in the form of dividend and250125020%-/=- in the form of capital gains. The total return on the index istherefore 195%-.. The risk-free rate is 1% per two months. The return is therefore205%-. in excess of the risk-free rate. From the capital asset pricing model we expect the return on the portfolio to be 0872*******%%.⨯-.=-. in excess of the risk-free rate. The portfolio return is therefore 16835%-.. The loss on the portfolio is01683550000000.⨯,, or $8,417,500. When this is combined with the gain on the futures the total gain is $431,750.Problem 3.32.It is now October 2014. A company anticipates that it will purchase 1 million pounds of copper in each of February 2015, August 2015, February 2016, and August 2016. The company has decided to use the futures contracts traded in the COMEX division of the CME Group to hedge its risk. One contract is for the delivery of 25,000 pounds of copper. The initial margin is $2,000per contract a nd the maintenance margin is $1,500 per contract. The company’s policy is to hedge 80% of its exposure. Contracts with maturities up to 13 months into the future are considered to have sufficient liquidity to meet the company’s needs. Devise a hedging stra tegy for the company. (Do not make the “tailing” adjustment described in Section 3.4.)Assume the market prices (in cents per pound) today and at future dates are as follows. What is the impact of the strategy you propose on the price the company pays for copper? What is the initial margin requirement in October 2014? Is the company subject to any margin calls?To hedge the February 2015 purchase the company should take a long position in March 2015 contracts for the delivery of 800,000 pounds of copper. The total number of contracts required is 8000002500032,/,=. Similarly a long position in 32 September 2015 contracts is required to hedge the August 2015 purchase. For the February 2016 purchase the company could take a long position in 32 September 2015 contracts and roll them into March 2016 contracts during August 2015. (As an alternative, the company could hedge the February 2016 purchase by taking a long position in 32 March 2015 contracts and rolling them into March 2016 contracts.) For the August 2016 purchase the company could take a long position in 32 September 2015 and roll them into September 2016 contracts during August 2015.The strategy is therefore as followsOct. 2014: Enter into long position in 96 Sept. 2015 contractsEnter into a long position in 32 Mar. 2015 contractsFeb 2015: Close out 32 Mar. 2015 contractsAug 2015: Close out 96 Sept. 2015 contractsEnter into long position in 32 Mar. 2016 contractsEnter into long position in 32 Sept. 2016 contractsFeb 2016: Close out 32 Mar. 2016 contractsAug 2016: Close out 32 Sept. 2016 contractsWith the market prices shown the company pays.+.⨯.-.=.3690008(3723036910)37156for copper in February, 2015. It pays.+.⨯.-.=.3650008(3728036480)37140for copper in August 2015. As far as the February 2016 purchase is concerned, it loses 3728036480800.-.=. on the .-.=. on the September 2015 futures and gains 37670364301240 February 2016 futures. The net price paid is therefore.+.⨯.-.⨯.=.377000880008124037348.-.=. on theAs far as the August 2016 purchase is concerned, it loses 3728036480800.-.=. on the September 2016 futures. The September 2015 futures and gains 38820364202400net price paid is therefore388000880008240037520.+.⨯.-.⨯.=.The hedging strategy succeeds in keeping the price paid in the range 371.40 to 375.20.In October 2014 the initial margin requirement on the 128 contracts is 1282000$⨯, or $256,000. There is a margin call when the futures price drops by more than 2 cents. This happens to the March 2015 contract between October 2014 and February 2015, to the September 2015 contract between October 2014 and February 2015, and to the September 2015 contract between February 2015 and August 2015. (Under the plan above the March 2016 contract is not held between February 2015 and August 2015, but if it were there would be a margin call during this period.)。