约翰.赫尔_期权期货和其他衍生品第八版部分习题答案汇总
期权、期货及其他衍生品（第8版）课后作业题解答（4-5章）
期权、期货及其他衍生品（第8版）课后作业题解答（4-5章）第二次作业答案4.25.2.5*e^(-y*0.5)+2.5*e^(-y*1)+……+2.5*e^(-y*4.5)+(100+2.5)*e^(-y*5)=104 求解得y=4.07%.4.26由F n=R n?n12?R n?1?n?112n12?n?112可得对应的远期利率为3.2%，3.5%，3.5%，3.45%，3.55%。
4.27a) 按3个月Libor借入，贷出2个月Libor，则套利(2/12)*0.28-(3/12)*0.01>0.b) (3/12)*r>(2/12)*0.28，即r>0.187%4.28易算出6-9月远期利率为8%，则套利策略如下：按6个月Libor利率借入100，按9个月Libor贷出100，进入6-9月远期利率合约多头借入100*e^0.025，9月到期时收益：100*e^(0.06*9/12)-100*e^(0.05*6/12)*e^(0.07*3/12)=0.264.29a) (1+0.05/2)^2=1+r, r=5.06%b) (1+0.05/2)^2=(1+r/12)^12, r=4.95%c) (1+0.05/2)^2=e*r, r=4.94%4.30a) 由(1+rm/2)^2=e^(rc)，可得相应连续复利利率为3.96%，4.45%，4.69%，4.94%。
b) 代入公式：R=2×R2?1.5×R1.5R=5.67%这是连续复利的远期利率，同样可由(1+r/2)^2=e^R，得到半年复利的远期利率5.75%。
4.31a) 0.5c*e^(-0.0396*0.5)+ 0.5c*e^(-0.0445*1)+ 0.5c*e^(-0.0469*1.5)+ (0.5c+100)*e^(-0.0494*2)=100解得：c=4.98b) 2.49*e^(-y*0.5)+ 2.49*e^(-y*1)+ 2.49*e^(-y*1.5)+(2.49+100)*e^(-y*2)=100, y=4.92%.4.32a) 100=98e^(0.5r), r=4.04%.100=95e^(r), r=5.13%.3.1e^(-0.04*0.5)+3.1e^(-0.0513)+103.1e^(-r*1.5)=101,r=5.44%.4e^(-0.04*0.5)+4e^(-0.0513)+4e^(-0.0544*1.5)+104e^(-r*2)=104, r=5.81%.b) 6-12月远期利率(5.13%*1-4.04%*0.5)/0.5=6.22%,12-18月远期利率(5.44%*1.5-5.13%*1)/0.5=6.07%,18-24月远期利率(5.81%*2-5.44%*1.5)/0.5=6.91%,c) 6月，(100+0.5c)e^(-4.04%*0.5)=100, c=4.0812月，0.5c*e^(-4.04%*0.5) +(100+0.5c)e^(-5.13%*1)=100,c=5.1818月，0.5c*e^(-4.04%*0.5) +0.5c*e^(-5.13%*1)+(100+0.5c)e^(-5.44%*1.5)=100, c=5.524月，0.5c*e^(-4.04%*0.5) +0.5c*e^(-5.13%*1)+ 0.5c*e^(-5.44%*1.5)+(100+0.5c)e^(-5.81%*2)=100, c=5.86d) 债券价格：5e^(-4.04%*0.5)+5e^(-5.13%*1)+5*e^(-5.44%*1.5)+(100+5)e^(-5.81%*2)=107.7收益率：5e^(-y*0.5)+5e^(-y*1)+5*e^(-y*1.5)+(100+5)e^(-y*2)=107.7，解得y=5.76%4.33a) D=2000?e^(?0.1?1)2000?e?0.1?1+6000?e^(?0.1?10)?1+6000?e?0.1?102000?e?0.1?1+6000?e?0.1?1010=5.95b) 可以利用公式?BB=?D?y来证明，也可以具体算出每个组合的变化百分比后比较c)B B =2000?e?0.15?1+6000?e?0.15?10?2000?e?0.10?1?6000?e?0.1 0?102000?e?0.10?1+6000?e?0.10?10=?23.8%B=5000?e?0.15?5.95?5000?e?0.10?5.950.10?5.95=?25.7%5.243个月指数期货价格：F=S*e^(r-q)t=1200*e^(0.03-0.012)*0.25=1205.46个月指数期货价格：F=S*e^(r-q)t=1200*e^(0.035-0.01)*0.5=1215.15.25F=S*e^(r-rf)t=1.4*e^(0.01-rf)*0.5=1.395，解得rf=1.72%.5.26原油属于消费商品，所以F<=(S+U)e^rt，U=3*e^-0.05，所以F<=(80+3*e^-0.05)*e^0.05=87.1.5.27a) I=1*e^(-0.08*2/12)+1*e^(-0.08*5/12)=1.95, F=(50-1.95)*e^(0.08*0.5)=50.01, 远期合约初始价值为0.b) I=1*e^(-0.08*2/12)=0.9868, F=(48-0.9868)*e^(0.08*3/12)=47.96, 远期合约价值(50.01-47.96)*e^(-0.08*3/12)=2.01.5.28组合一：借入黄金卖出，利用黄金期货锁定远期价格，则一年期F=S*(1+0.02/2)^2*e^(0.0925+0.005)=1.125*S组合二：借入现金贷款，则F=S*(1+0.11/2)^2=1.103*S所以借入黄金的利率太高。
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。
期权期货和其它衍生产品第三约翰赫尔答案
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×(29-p),后者损失为$5，800;当股票价格为（29，30），购买股票收益为$5,800$29×(p-29),购买期权损失为$5,800;当股票价格高于$30时，购买股票收益为$5,800$29×(p-29)，购买期权收益为$$5,800$29×(p-30)-5,800。
JohnHull《期货期权和衍生证券》章习题解答
CHAPTER 13Wiener P rocesses and Itô’s LemmaPractice QuestionsProblem 13.1.What would it mean to assert that the temperature at a certain place follows a Markov process? Do you think that temperatures do, in fact, follow a Markov process?Imagine that you have to forecast the future temperature from a) the current temperature, b) the history of the temperature in the last week, and c) a knowledge ofseasonal averages and seasonal trends. If temperature followed a Markov process, the history of the temperature in the last week would be irrelevant.To answer the second part of the question you might like to consider the following scenario for the first week in May:(i) Monday to Thursday are warm days; today, Friday, is a very cold day. (ii) Monday to Friday are all very cold days.What is your forecast for the weekend? If you are more pessimistic in the case of the second scenario, temperatures do not follow a Markov process.Problem 13.2.Can a trading rule based on the past history of a stock’s price ever produce returns that are consistently above average? Discuss.The first point to make is that any trading strategy can, just because of good luck, produce above average returns. The key question is whether a trading strategy consistently outperforms the market when adjustments are made for risk. It is certainly possible that a trading strategy could do this. However, when enough investors know about the strategy and trade on the basis of the strategy, the profit will disappear.As an illustration of this, consider a phenomenon known as the small firm effect. Portfolios of stocks in small firms appear to have outperformed portfolios of stocks in large firms when appropriate adjustments are made for risk. Research was published about this in the early 1980s and mutual funds were set up to take advantage of the phenomenon. There is some evidence that this has resulted in the phenomenon disappearing.Problem 13.3.A company’s cash position, measured in millions of dollars, follows a generalized Wiener process with a drift rate of 0.5 per quarter and a variance rate of 4.0 per quarter. How high does the company’s initial cash position have to be for the company to have a less than 5% chance of a negative cash position by the end of one year?Supp ose that the company’s initial cash position is x . The probability distribution of the cash position at the end of one year is (40544)(2016)x x ϕϕ+⨯.,⨯=+.,where ()m v ϕ, is a normal probability distribution with mean m and variance v . The probability of a negative cash position at the end of one year is204x N +.⎛⎫- ⎪⎝⎭where ()N x is the cumulative probability that a standardized normal variable (with mean zero and standard deviation 1.0) is less than x . From normal distribution tables200054x N +.⎛⎫-=. ⎪⎝⎭when:20164494x +.-=-.i.e., when 45796x =.. The initial cash position must therefore be $4.58 million.Problem 13.4.Variables 1X and 2X follow generalized Wiener processes with drift rates 1μ and2μ and variances 21σ and 22σ. What process does 12X X + follow if:(a) The changes in 1X and 2X in any short interval of time are uncorrelated?(b) There is a correlation ρ between the changes in 1X and 2X in any short interval of time?(a) Suppose that X 1 and X 2 equal a 1 and a 2 initially. After a time period of length T , X 1 has the probability distribution2111()a T T ϕμσ+,and 2X has a probability distribution2222()a T T ϕμσ+,From the property of sums of independent normally distributed variables, 12X X + has the probability distribution()22112212a T a T T T ϕμμσσ+++,+i.e.,22121212()()a a T T ϕμμσσ⎡⎤+++,+⎣⎦This shows that 12X X + follows a generalized Wiener process with drift rate 12μμ+and variance rate 2212σσ+.(b) In this case the change in the value of 12X X + in a short interval of time t ∆ has the probability distribution:22121212()(2)t t ϕμμσσρσσ⎡⎤+∆,++∆⎣⎦If 1μ, 2μ, 1σ, 2σ and ρ are all constant, arguments similar to those in Section 13.2 show that the change in a longer period of time T is22121212()(2)T T ϕμμσσρσσ⎡⎤+,++⎣⎦The variable,12X X +, therefore follows a generalized Wiener process with drift rate12μμ+ and variance rate 2212122σσρσσ++.Problem 13.5.Consider a variable,S , that follows the process dS dt dz μσ=+For the first three years, 2μ= and 3σ=; for the next three years, 3μ= and 4σ=. If the initial value of the variable is 5, what is the probability distribution of the value of the variable at the end of year six?The change in S during the first three years has the probability distribution (2393)(627)ϕϕ⨯,⨯=,The change during the next three years has the probability distribution (33163)(948)ϕϕ⨯,⨯=,The change during the six years is the sum of a variable with probability distribution(627)ϕ, and a variable with probability distribution (948)ϕ,. The probability distribution of the change is therefore (692748)ϕ+,+ (1575)ϕ=,Since the initial value of the variable is 5, the probability distribution of the value of the variable at the end of year six is (2075)ϕ,Problem 13.6.Suppose that G is a function of a stock price, S and time. Suppose that S σ and G σ are the volatilities of S and G . Show that when the expected return of S increases by S λσ, the growth rate of G increases by G λσ, where λ is a constant.From Itô’s lemmaG S GG S Sσσ∂=∂Also the drift of G is222212G G G S S S t S μσ∂∂∂++∂∂∂where μ is the expected return on the stock. When μ increases by S λσ, the drift of Gincreases byS GS Sλσ∂∂ orG G λσThe growth rate of G , therefore, increases by G λσ.Problem 13.7.Stock A and stock B both follow geometric Brownian motion. Changes in any short interval of time are uncorrelated with each other. Does the value of a portfolio consisting of one of stock A and one of stock B follow geometric Brownian motion? Explain your answer.Define A S , A μ and A σ as the stock price, expected return and volatility for stock A. Define B S , B μ and B σ as the stock price, expected return and volatility for stock B. Define A S ∆ and B S ∆ as the change in A S and B S in time t ∆. Since each of the two stocks follows geometric Brownian motion,A A A A A S S t S μσε∆=∆+B B B B B S S t S μσε∆=∆+where A ε and B ε are independent random samples from a normal distribution.()(A B A A B B A A A B B B S S S S t S S μμσεσε∆+∆=+∆++This cannot be written as()()A B A B A B S S S S t S S μσ∆+∆=+∆++for any constants μ and σ. (Neither the drift term nor the stochastic term correspond.) Hence the value of the portfolio does not follow geometric Brownian motion.Problem 13.8.S S t S μσε∆=∆+ where μ and σ are constant. Explain carefully the difference between this model andeach of the following:S t S S t S t S μσεμσεμσε∆=∆+∆=∆+∆=∆+Why is the model in equation (13.8) a more appropriate model of stock price behavior than any of these three alternatives?In:S S t S μσε∆=∆+ the expected increase in the stock price and the variability of the stock price are constant when both are expressed as a proportion (or as a percentage) of the stock price In:S t μ∆=∆+the expected increase in the stock price and the variability of the stock price are constant in absolute terms. For example, if the expected growth rate is $5 per annum when the stockprice is $25, it is also $5 per annum when it is $100. If the standard deviation of weekly stock price movements is $1 when the price is $25, it is also $1 when the price is $100. In:S S t μ∆=∆+the expected increase in the stock price is a constant proportion of the stock price while the variability is constant in absolute terms. In:S t S μσ∆=∆+the expected increase in the stock price is constant in absolute terms while the variability of the proportional stock price change is constant. The model:S S t S μσ∆=∆+ is the most appropriate one since it is most realistic to assume that the expected percentage return and the variability of the percentage return in a short interval are constant.Problem 13.9.It has been suggested that the short-term interest rate,r , follows the stochastic process()dr a b r dt rc dz =-+where a , b , and c are positive constants and dz is a Wiener process. Describe the nature of this process.The drift rate is ()a b r -. Thus, when the interest rate is above b the drift rate is negative and, when the interest rate is below b , the drift rate is positive. The interest rate is therefore continually pulled towards the level b . The rate at which it is pulled toward this level is a . A volatility equal to c is superimposed upon the “pull” or the drift.Suppose 04a =., 01b =. and 015c =. and the current interest rate is 20% per annum. The interest rate is pulled towards the level of 10% per annum. This can be regarded as a long run average. The current drift is 4-% per annum so that the expected rate at the end of one year is about 16% per annum. (In fact it is slightly greater than this, because as the interest rate decreases, the “pull” decreases.) Superimposed upon the drift is a volatility of 15% per annum.Problem 13.10.Suppose that a stock price, S , follows geometric Brownian motion with expected return μ and volatility σ: dS S dt S dz μσ=+What is the process followed by the variable n S ? Show that n S also follows geometric Brownian motion.If ()n G S t S ,= then 0G t ∂/∂=, 1n G S nS -∂/∂=, and 222(1)n G S n n S -∂/∂=-. Using Itô’s lemma:21[(1)]2dG nG n n G dt nG dz μσσ=+-+This shows that n G S = follows geometric Brownian motion where the expected return is21(1)2n n n μσ+-and the volatility is n σ. The stock price S has an expected return of μ and the expected value of T S is 0T S e μ. The expected value of n T S is212[(1)]0n n n T n S eμσ+-Problem 13.11.Suppose that x is the yield to maturity with continuous compounding on a zero-coupon bond that pays off $1 at time T . Assume that x follows the process0()dx a x x dt sx dz =-+where a , 0x , and s are positive constants and dz is a Wiener process. What is the process followed by the bond price?The process followed by B , the bond price, is from Itô’s lemma:222021()2B B B B dB a x x s x dt sxdz x t x x ⎡⎤⎢⎥⎢⎥⎢⎥⎣⎦∂∂∂∂=-+++∂∂∂∂Since: ()x T t B e --=the required partial derivatives are()()22()22()()()()x T t x T t x T t Bxe xB t BT t e T t B x B T t e T t B x------∂==∂∂=--=--∂∂=-=-∂ Hence:22201()()()()2dB a x x T t x s x T t Bdt sx T t Bdz⎡⎤⎢⎥⎢⎥⎢⎥⎣⎦=---++---Problem 13.12 (Excel Spreadsheet)A stock whose price is $30 has an expected return of 9% and a volatility of 20%. In Excel simulate the stock price path over 5 years using monthly time steps and random samples from a normal distribution. Chart the simulated stock price path. By hitting F9 observe how the path changes as the random sample change.The process ist S t S S ∆⨯ε⨯⨯+∆⨯⨯=∆20.009.0Where ∆t is the length of the time step (=1/12) and ε is a random sample from a standard normal distribution.Further QuestionsProblem 13.13.Suppose that a stock price has an expected return of 16% per annum and a volatility of 30% per annum. When the stock price at the end of a certain day is $50, calculate the following:(a) The expected stock price at the end of the next day.(b) The standard deviation of the stock price at the end of the next day. (c) The 95% confidence limits for the stock price at the end of the next day.With the notation in the text2()St t S ϕμσ∆∆,∆In this case 50S =, 016μ=., 030σ=. and 1365000274t ∆=/=.. Hence(016000274009000274)50(0000440000247)Sϕϕ∆.⨯.,.⨯.=.,.and2(50000044500000247)S ϕ∆⨯.,⨯.that is, (002206164)S ϕ∆.,.(a)(b) The standard deviation of the stock price at the end of the next day is 0785=. (c) 95% confidence limits for the stock price at the end of the next day are 500221960785and 500221960785.-.⨯..+.⨯. i.e.,4848and 5156..Note that some students may consider one trading day rather than one calendar day. Then 1252000397t ∆=/=.. The answer to (a) is then 50.032. The answer to (b) is 0.945. The answers to part (c) are 48.18 and 51.88.Problem 13.14.A company’s cash position, measured in millions of dollars, follows a generalized Wiener process with a drift rate of 0.1 per month and a variance rate of 0.16 per month. The initial cash position is 2.0.(a) What are the probability distributions of the cash position after one month, six months, and one year?(b) What are the probabilities of a negative cash position at the end of six months and one year?(c) At what time in the future is the probability of a negative cash position greatest?(a) The probability distributions are:(2001016)(21016)ϕϕ.+.,.=.,.(20060166)(26096)ϕϕ.+.,.⨯=.,.(201201612)(32196)ϕϕ.+.,.⨯=.,.(b) The chance of a random sample from (26096)ϕ.,. being negative is(265)N N ⎛=-. ⎝where ()N x is the cumulative probability that a standardized normal variable [i.e., avariable with probability distribution (01)ϕ,] is less than x . From normaldistribution tables (265)00040N -.=.. Hence the probability of a negative cash position at the end of six months is 0.40%.Similarly the probability of a negative cash position at the end of one year is(230)00107N N ⎛=-.=. ⎝or 1.07%.(c) In general the probability distribution of the cash position at the end of x months is(2001016)x x ϕ.+.,.The probability of the cash position being negative is maximized when:is minimized. Define11223122325025250125(250125)y x xdy x xdxx x----==+.=-.+.=-.+.This is zero when 20x=and it is easy to verify that 220d y dx/>for this value of x. It therefore gives a minimum value for y. Hence the probability of a negative cash position is greatest after 20 months.Problem 13.15.Suppose that x is the yield on a perpetual government bond that pays interest at the rate of $1 per annum. Assume that x is expressed with continuous compounding, that interest is paid continuously on the bond, and that x follows the process()dx a x x dt sx dz=-+where a,x, and s are positive constants and dz is a Wiener process. What is the process followed by the bond price? What is the expected instantaneous return (including interest and capital gains) to the holder of the bond?The process followed by B, the bond price, is from Itô’s lemma:222021()2B B B BdB a x x s x dt sxdzx t x x⎡⎤⎢⎥⎢⎥⎢⎥⎣⎦∂∂∂∂=-+++∂∂∂∂In this case1Bx=so that:222312B B Bt x x x x∂∂∂=;=-;=∂∂∂Hence2202322021121()21()dB a x x s x dt sxdzx x xs sa x x dt dzx x x⎡⎤=--+-⎢⎥⎣⎦⎡⎤=--+-⎢⎥⎣⎦The expected instantaneous rate at which capital gains are earned from the bond is therefore:2021()sa x xx x--+The expected interest per unit time is 1. The total expected instantaneous return is therefore:20211()sa x xx x--+When expressed as a proportion of the bond price this is:202111()sa x xx x x⎛⎫⎛⎫--+ ⎪⎪⎝⎭⎝⎭20()ax x x s x=--+Problem 13.16.If S follows the geometric Brownian motion process in equation (13.6), what is the process followed by (a) y = 2S, (b) y=S 2 , (c) y=e S , and (d) y=e r(T-t)/S. In each case express the coefficients of dt and dz in terms of y rather than S.(a) In this case 2y S ∂/∂=, 220y S ∂/∂=, and 0y t ∂/∂= so that Itô’s lemma gives 22dy S dt S dz μσ=+or dy y dt y dz μσ=+(b) In this case 2y S S ∂/∂=, 222y S ∂/∂=, and 0y t ∂/∂= so that Itô’s lemma gives2222(2)2dy S S dt S dz μσσ=++ or2(2)2dy y dt y dz μσσ=++ (c) In this case S y S e ∂/∂=, 22S y S e ∂/∂=, and 0y t ∂/∂= so that Itô’s lemma gives22(2)S S S dy Se S e dt Se dz μσσ=+/+ or22[ln (ln )2]ln dy y y y y dt y y dz μσσ=+/+(d) In this case ()2r T t y S e S y S -∂/∂=-/=-/, 22()3222r T t y S e S y S -∂/∂=/=/, and()r T t y t re S ry -∂/∂=-/=- so that Itô’s lemma gives2()dy ry y y dt y dz μσσ=--+- or2()dy r y dt y dz μσσ=-+--Problem 13.17.A stock price is currently 50. Its expected return and volatility are 12% and 30%,respectively. What is the probability that the stock price will be greater than 80 in two years? (Hint 80T S > when ln ln 80T S >.)The variable ln T S is normally distributed with mean 20ln (2)S T μσ+-/ and standarddeviation σ050S =, 012μ=., 2T =, and 030σ=. so that the meanand standard deviation of ln T S are 2ln 50(012032)24062+.-./=. and 00424.=., respectively. Also, ln804382=.. The probability that 80T S > is the same as the probability that ln 4382T S >.. This is4382406211(0754)0424N N .-.⎛⎫-=-. ⎪.⎝⎭where ()N x is the probability that a normally distributed variable with mean zero and standard deviation 1 is less than x . From the tables at the back of the book (0754)0775N .=. so that the required probability is 0.225.Problem 13.18 (See Excel Worksheet)Stock A, whose price is $30, has an expected return of 11% and a volatility of 25%. Stock B, whose price is $40, has an expected return of 15% and a volatility of 30%. The processes driving the returns are correlated with correlation parameter ρ. In Excel, simulate the two stock price paths over three months using daily time steps and random samples from normal distributions. Chart the results and by hitting F9 observe how the paths change as the random samples change. Consider values of ρ equal to 0.50, 0.75, and 0.95.The processes aret S t S S A A A A ∆⨯ε⨯⨯+∆⨯⨯=∆25.011.0t S t S S B B B B ∆⨯ε⨯⨯+∆⨯⨯=∆30.015.0Where ∆t is the length of the time step (=1/252) and the ε’s are correlated samples from standard normal distributions.。
约翰.赫尔,期权期货和其他衍生品(third edition)习题答案
CH99.1 股票现价为$40。
已知在一个月后股价为$42或$38。
无风险年利率为8％（连续复利）。
执行价格为$39的1个月期欧式看涨期权的价值为多少？ 解：考虑一资产组合：卖空1份看涨期权；买入Δ份股票。
若股价为$42，组合价值则为42Δ－3；若股价为$38，组合价值则为38Δ 当42Δ－3＝38Δ，即Δ＝0.75时，组合价值在任何情况下均为$28.5，其现值为：，0.08*0.0833328.528.31e −=即：－f ＋40Δ＝28.31 其中f 为看涨期权价格。
所以，f ＝40×0.75－28.31＝$1.69另解：（计算风险中性概率p ） 42p －38（1－p ）＝，p ＝0.56690.08*0.0833340e期权价值是其期望收益以无风险利率贴现的现值，即： f ＝（3×0.5669＋0×0.4331）＝$1.690.08*0.08333e−9.2 用单步二叉树图说明无套利和风险中性估值方法如何为欧式期权估值。
解：在无套利方法中，我们通过期权及股票建立无风险资产组合，使组合收益率等价于无风险利率，从而对期权估值。
在风险中性估值方法中，我们选取二叉树概率，以使股票的期望收益率等价于无风险利率，而后通过计算期权的期望收益并以无风险利率贴现得到期权价值。
9.3什么是股票期权的Delta ？解：股票期权的Delta 是度量期权价格对股价的小幅度变化的敏感度。
即是股票期权价格变化与其标的股票价格变化的比率。
9.4某个股票现价为$50。
已知6个月后将为$45或$55。
无风险年利率为10％（连续复利）。
执行价格为$50，6个月后到期的欧式看跌期权的价值为多少？ 解：考虑如下资产组合，卖1份看跌期权，买Δ份股票。
若股价上升为$55，则组合价值为55Δ；若股价下降为$45，则组合价值为：45Δ－5 当55Δ＝45Δ－5，即Δ＝－0.50时，6个月后组合价值在两种情况下将相等，均为$－27.5，其现值为：，即：0.10*0.5027.5$26.16e −−=− －P ＋50Δ＝－26.16所以，P ＝－50×0.5＋26.16＝$1.16 另解：求风险中性概率p0.10*0.505545(1)50p p e+−= 所以，p ＝0.7564看跌期权的价值P ＝0.10*0.50(0*0.75645*0.2436)$1.16e −+=9.5 某个股票现价为$100。
约翰.赫尔,期权期货和其他衍生品(third edition)习题答案
8.14 执行价格为$60 的看涨期权成本为$6，相同执行价格和到期日的看跌期权成
本为$4，制表说明跨式期权损益状况。请问：股票价格在什么范围内时，
跨式期权将导致损失呢？
解：可通过同时购买看涨看跌期权构造跨式期权：max（ ST －60，0）+max(60
－ ST )－（6＋4），其损益状况为：
股价 ST
解：(a)该组合等价于一份固定收益债券多头，其损益V = C ，不随股票价格变化。 （V 为组合损益，C 为期权费，下同）如图 8.2： (b)该组合等价于一份股票多头与一份固定收益债券多头，其损益V = ST + C ， 与股价同向同幅度变动。（ ST 为最终股票价格，下同）如图 8.3 (c)该组合等价于一份固定收益债券多头与一份看涨期权空头，其损益为
8.18 盒式价差期权是执行价格为 X 1 和 X 2 的牛市价差期权和相同执行价格的熊 市看跌价差期权的组合。所有期权的到期日相同。盒式价差期权有什么样的 特征？
解：牛市价差期权由 1 份执行价格为 X 1 欧式看涨期权多头与 1 份执行价格为 X 2 的欧式看涨期权空头构成（ X 1 < X 2 ），熊市价差期权由 1 份执行价格为 X 2 的 欧式看跌期权多头与 1 份执行价格为 X 1 的看跌期权空头构成，则盒式价差
8.17 运用期权如何构造出具有确定交割价格和交割日期的股票远期合约？ 解：假定交割价格为 K，交割日期为 T。远期合约可由买入 1 份欧式看涨期权，
同时卖空 1 份欧式看跌期权，要求两份期权有相同执行价格 K 及到期日 T。 可见，该组合的损益为 ST －K，在任何情形下，其中 ST 为 T 时股票价格。 假定 F 为远期合约价格，若 K=F，则远期合约价值为 0。这表明，当执行价 格为 K 时，看涨期权与看跌期权价格相等。
赫尔《期权、期货及其他衍生产品》教材精讲(期货市场的运作机制)【圣才出品】
第2章期货市场的运作机制第一部分本章要点1．期货市场的具体运作机制。
将讨论合约条款的约定、保证金账户的运作、交易所的组织结构、市场监管规则、期货报价方式及期货的财会和税务处理等内容。
我们还将讨论，并解释造成这两类合约所实现的收益不同的原因。
2．期货与远期合约的不同之处。
第二部分重难点导学2.1 背景知识假定现在是3月5日，一位纽约的投资者指示他的经纪人买入5000蒲式耳玉米，资产交割时间在7月份，经纪人会马上将这一指令通知芝加哥交易所。
同时，假定另一位在堪萨斯州的投资者指示她的经纪人卖出5000蒲式耳玉米，资产交割时间也是7月份，她的经纪人也会马上将客户的指令通知芝加哥交易所。
这时双方会同意某个交易价格，从而交易成交。
几个概念：期货的长头寸（long futures position）；期货的短头寸（short futures position）。
期货价格（futures price）；期货合约的平仓。
2.2 期货合约的规定·资产品种·合约的规模·交割地点·交割时间有时交割资产的备选方案也会被说明。
2.2.1 资产商品期货标的资产的质量可能有很大差别。
金融期货合约中的标的资产定义通常明确。
2.2.2 合约的规模合约的规模（contract size）：每一合约中交割资产的数量。
2.2.3 交割的安排交割地点必须由交易所指定，这对那些运输费用昂贵的商品尤其重要。
当选用交割地点时，期货的短头寸方所收入的价格会随着交割地点的不同而得以调整。
期货交割地点离商品生产地越远，交割价格越高。
2.2.4 交割月份期货合约通常以其交割月份来命名。
交易所必须指定交割月份内的准确交割时间。
对于许多期货而言，交割时间区间为整个月。
2.2.5 报价交易所定义报价的方式各不相同，例如，在美国，原油期货报价以美元和美分计量，而中长期国债期货报价是以1美元以及1/32美元来报价的。
2.2.6 价格和头寸的限额跌停（limit down）：价格下跌的金额等于每日价格限额。
赫尔《期权、期货及其他衍生产品》教材精讲(期权市场机制)【圣才出品】
第9章期权市场机制第一部分本章要点期权市场组织机构、市场专用语、市场的交易过程等。
第二部分重难点导学9.1 期权的类型看涨期权（call option）给期权持有者在将来一定时刻以一定的价格买入某种资产的权利；看跌期权（put option）给期权持有者在将来一定时刻以一定价格卖出某资产的权利。
美式期权可在到期之日前的任何时刻行使；欧式期权只能在到期日才能行使。
9.1.1 看涨期权例9-1：考虑某投资者买入当前价格为98美元，执行价格为100美元的100股股票的欧式看涨期权，期权的到期日为4个月，购买一股股票的期权价格为5美元。
图9-1 买入一股股票的看涨期权收益曲线9.1.2 看跌期权例9-2：考虑一个购买了执行价格为70美元的100股股票的欧式看跌期权。
假定当前股票价格为65美元，期权的到期日为3个月，卖出一股股票的期权价格为7美元。
图9-2 买入看跌期权的收益曲线9.2 期权头寸一方为取得期权的长头寸方（即买入期权），另一方为取得期权的短头寸方[即卖出期权或对期权进行承约（written the option）]。
期权交易共有4种头寸形式：·看涨期权长头寸；·看跌期权长头寸；·看涨期权短头寸；·看跌期权短头寸。
图9-3及图9-4显示了期权承约人的盈亏与股票价格之间的关系。
图9-3 卖出看涨期权的收益图（期权价格=5美元，执行价格=100美元）图9-4 卖出看跌期权的收益图（期权价格=7美元，执行价格=70美元）K——执行价格，S T——标的资产的到期价格，在S T>K时，行使期权；在S T≤K时，将不行使期权。
欧式看涨期权的长头寸的收益为max（S T-K，0）欧式看涨期权的短头寸的收益为-max（S T-K，0）=min（K-S T，O）欧式看跌期权的长头寸的收益为max（K-S T，0）欧式看跌期权的短头寸的收益为-max（K-S T，0）=min（S T-K，0）图9-5展示了期权的收益图形。
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计算题
1.1一个投资者进入了一个远期合约的短头寸，在该合约中投资者能够以1.4000的汇率（美
元/英镑）卖出1000 000英镑。
当远期合约到期时的汇率为以下数值时⑴ 1.3900，⑵
1.4200，投资者的损益分别为多少？
（1.4-1.39）*1000000=1000
（1.4-1.42）*1000000=-2000
1.2 当前黄金市价为每盎司$1,000,一个一年期的远期合约的执行价格为$1,200,一个套利者
能够以每年10%的利息借入资金，套利者应该如何做才能达到套利目的？
假设黄金存储费为0，同时黄金不会带来任何利息收入
借入1000美元买入黄金，一年后以1200美元卖出
（1200-1000）- （1000*10%）= 100
2.1 一家公司进入一个合约的短头寸，在合约中公司以每莆式耳450美分卖出5000蒲式耳
小麦。
初始保证金为3000美元，维持保证金为2000美元，价格如何变化会导致保证金的催收？在什么情况下公司可以从保证金账户提取1500美元？
如果合约亏损1000美元，则会导致保证金催收，假设价格为S时收到了保证金催收，则（S1-4.5）*5000=1000 S1=470美分，即如果未来价格高于470美分会收到保证呢过金催收通知
假设价格为S2时可以从保证金账户提取1500美金，则（4.5-S2）*5000=1500 S2=420美分即价格跌至420美分时，可以从保证金账户中提取1500美元
3.1 假定今天是7月16日，一家公司持有价值1亿美元的股票组合，此股票组合的beta系
数是1.2，这家公司希望采用CME在12月份到期SP500股指期货在7月16日至11月16日之间将beta系数由1.2变为0.5，SP500股指当前价值为1000，而每一份期货合约面值为250美元与股指的乘积。
1）公司应做什么样的交易2）Beta由1.2变为1.5，公司应该持有什么样的头寸
1)该公司需要空头（1.2-0.5）*100000000/1000*250=280份合约
2)该公司需要多头（1.5-1.2）*100000000/1000*250=120份合约
3.2 P52 3.29
4.1
P71 4.30
4.2 P71 4.32
5.1 9.1
10.1
10.2
假设你是一家杠杆比例很高的公司的经理及唯一所有者。
所有的债务在1年后到期。
如果那时公司的价值高于债务的面值，你就可以偿还债务。
如果公司的价值小于债务的面值，你就必须宣布破产，让债务人拥有公司。
a) 将公司的价值作为期权的标的物，描述你的头寸状况。
b)按照以公司价值为标的物的期权的形式，描述债务人的头寸状况。
c)你应当如何做来提高你头寸的价值？
11.1
11.2
6.1
7.1。