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）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。

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。

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。

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

CH99.1 股票现价为$40。

已知在一个月后股价为$42或$38。

无风险年利率为8％（连续复利）。

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

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

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

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

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

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

即是股票期权价格变化与其标的股票价格变化的比率。

9.4某个股票现价为$50。

已知6个月后将为$45或$55。

无风险年利率为10％（连续复利）。

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

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

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。

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。

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

期权合约，赋予它的持有者一个权利：以某个约定的价格买入或卖出标的资产。

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

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

如果投资者未提供变动保证金，经纪人将出售该合约来平仓。

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

目的：选择期货头寸，从而使得自身整体的投资风险尽量呈中性。

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

赫尔《期权、期货及其他衍生产品》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。

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.。

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.。

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.。

风险管理与金融机构第四版约翰·C·赫尔附加问题(Further Problems)的答案第一章:导论1.15.假设一项投资的平均回报率为8%，标准差为14%。

另一项投资的平均回报率为12%，标准差为20%。

收益率之间的相关性为0.3。

制作一张类似于图1.2的图表，显示两项投资的其他风险回报组合。

答：第一次投资w1和第二次投资w2=1-w1的影响见下表。

可能的风险回报权衡范围如下图所示。

w1 w2μp σp0.0 1.0 12% 20%0.2 0.8 11.2% 17.05%0.4 0.6 10.4% 14.69%0.6 0.4 9.6% 13.22%0.8 0.2 8.8% 12.97%1.0 0.0 8.0% 14.00%1.16.市场预期收益率为12%，无风险收益率为7%。

市场收益率的标准差为15%。

一个投资者在有效前沿创建一个投资组合，预期回报率为10%。

另一个是在有效边界上创建一个投资组合，预期回报率为20%。

两个投资组合的回报率的标准差是多少？答：在这种情况下，有效边界如下图所示。

预期回报率为10%时，回报率的标准差为9%。

与20%的预期回报率相对应的回报率标准差为39%。

1.17.一家银行估计, 其明年利润正态分布, 平均为资产的0.8%，标准差为资产的2%。

公司需要多少股本（占资产的百分比）：（a）99%确定其在年底拥有正股本；（b）99.9%确定其在年底拥有正股本？忽略税收。

答：（一）银行可以99%确定利润将优于资产的0.8-2.33×2或-3.85%。

因此，它需要相当于资产3.85%的权益，才能99%确定它在年底的权益为正。

（二）银行可以99.9%确定利润将大于资产的0.8-3.09×2或-5.38%。

因此，它需要权益等于资产的5.38%，才能99.9%确定它在年底将拥有正权益。

1.18.投资组合经理维持了一个积极管理的投资组合，beta值为0.2。

期权期货和其它衍生产品约翰赫尔答案1.1请说明远期多头与远期空头的区别。

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

1.2请详细说明套期保值、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请说明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：第一种情形下交易者有义务以50$购买某项资产〔交易者没有选择〕，第二种情形下有权益以50$购买某项资产〔交易者能够不执行该权益〕。

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。

请问期货合约终止时，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缺失为多少？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，交易者缺失:($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。

你是如何考虑的？你的收益或缺失如何？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，我将缺失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资金能够投资。

现有两种策略：直截了当购买股票或投资于期权，请问各自潜在的收益或缺失为多少？答：股票价格低于$29时，购买股票和期权都将缺失，前者缺失为$5,800$29×(29-p),后者缺失为$5，800;当股票价格为〔29，30〕，购买股票收益为$5,800$29×(p-29),购买期权缺失为$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。

第一章绪论1、什么是期货和期权，二者区别期货合约：两个对手之间签定的一个在确定的将来时间按确定的价格购买或出售某项资产的协议期权合约：权利义务期权合约多头一方有义务在将来某一确定时刻以确定价格买进某种商品看涨期权持有者有权利选择买进某种商品成本进入期货市场无须成本（除保证金要求）期权投资者需要支付期权费才能够得到一张期权合约2、期权交易者类型套期保值者：利用衍生产品的头寸来规避该风险投机者：持有某个头寸，承担风险以获得高收益套利者：瞬态进入两个或多个市场的交易，以锁定一个无风险的收益3、在投机时，期权与期货区别杠杆效应以较小的投入来控制较大的投机头寸。

期货：潜在损失与收益都很大期权：不管市场有多么糟糕，投机者的损失不会超过所支付的期权费用。

4、其他金融衍生产品百慕大期权：在有效期的某些特定天数之内执行回顾期权：持有者能在到期日之前选择一个最好、最有利的价格作为合同价格，如果是回顾买权，那就是基础资产出现过的最低价，如果是回顾卖权，则是出现过的最高价。

当然，这种期权的期权费往往较高平均期权(亚式期权)：平均期权中平均的，可以是将合同价格就设定为资产的平均价格。

平均期权适合于那些具有经常性的收入或费用的避险者5、组合分解6、利率上限利率远期协议利率上限：金融机构将该合约出售给公司借款者，使这些借款者在借入浮动利率的贷款时，免受由于利率上升到某一特定水平之上而产生的不利影响。

如果贷款利率的确高于上限利率，那么利率上限这一衍生产品的卖者就要替借款人补足贷款利率和上限之间的利息差额远期利率协议：俩家公司规定，6个月以后开始一年期定期存款利率为7%。

如果届时不是7%，一家公司就要将由利率变化引起的未来利息现金流的变动值贴现，付给另一家第二章期货市场与远期市场的运行机制1、期货合约期货市场卖空期货合约：在交易所内达成，受一定规则约束必须执行，规定在将来某一时间和地点交收某一特定商品的一种标准化契约。

（两个对手之间签定的一个在确定的将来时间按确定的价格购买或出售某项资产的协议）期货市场：卖空：投资者委托经纪人借来别人的股票，在公开市场上将其卖掉。