1、 CHAPTER 13Wiener Processes and Its 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)
2、 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 migh
3、t 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, temperatu
4、res do not follow a Markov process. Problem 13.2.Can a trading rule based on the past history of a stocks 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 k
5、ey 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.
6、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
7、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 companys 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
8、per quarter. How high does the companys 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? Suppose that the companys initial cash position is . The probability distribution of xthe cash position at the end of one year is
9、 (405)(2016)xwhere is a normal probability distribution with mean and variance . The ()mv mvprobability of a negative cash position at the end of one year is 204xNwhere is the cumulative probability that a standardized normal variable (with mean ()Nxzero and standard deviation 1.0) is less than . Fr
10、om normal distribution tables x2054Nwhen: 169xi.e., when . The initial cash position must therefore be $4.58 million. 45796xProblem 13.4.Variables and follow generalized Wiener processes with drift rates and 1X2 12and variances and . What process does follow if: 212X(a) The changes in and in any sho
11、rt interval of time are uncorrelated?12(b) There is a correlation between the changes in and in any short interval of 12Xtime?(a) Suppose that X1 and X2 equal a1 and a2 initially. After a time period of length , X1 has Tthe probability distribution 211()Tand has a probability distribution 2 22()aFro
12、m the property of sums of independent normally distributed variables, has the 12Xprobability distribution 2121aTTi.e., 21211()() This shows that follows a generalized Wiener process with drift rate and 12X 12variance rate . 2(b) In this case the change in the value of in a short interval of time has
13、 the 12Xtprobability distribution: 1212()()t t If , , , and are all constant, arguments similar to those in Section 13.2 1212show that the change in a longer period of time is T21112()() The variable, , therefore follows a generalized Wiener process with drift rate 12Xand variance rate . 122112Probl
14、em 13.5.Consider a variable, , that follows the process SdStdzFor the first three years, and ; for the next three years, and . If the 2334initial 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 during the first three
15、years has the probability distribution S(239)(627)The change during the next three years has the probability distribution (1)(48)The change during the six years is the sum of a variable with probability distribution (627)and a variable with probability distribution . The probability distribution of
16、the (9)change is therefore 62748(15)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 (207)Problem 13.6.Suppose that is a function of a stock price, and time. Suppose that and GSSGare the volatilities of and . Show that
17、when the expected return of increases by , S Sthe growth rate of increases by , where is a constant. GFrom Its lemma GSAlso the drift of is G21Stwhere is the expected return on the stock. When increases by , the drift of G Sincreases by SGor The growth rate of , therefore, increases by . GGProblem 1
18、3.7.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 , and as the stock price,
19、 expected return and volatility for stock A. SADefine , and as the stock price, expected return and volatility for stock B. Define BBand as the change in and in time . Since each of the two stocks follows ASSBtgeometric Brownian motion, AAtStBBSwhere and are independent random samples from a normal
20、distribution. AB()()ABABABStStThis cannot be written as ()()ABABABSttfor any constants and . (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) is StS
21、twhere and are constant. Explain carefully the difference between this model and each of the following: ttSttWhy is the model in equation (13.8) a more appropriate model of stock price behaviorthan any of these three alternatives? In: StStthe expected increase in the stock price and the variability
22、of the stock price are constant when both are expressed as a proportion (or as a percentage) of the stock price In: ttthe 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 stoc
23、k 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: Sttthe expected increase in the stock price is a constant proportion of the stock price while the variability is c
24、onstant in absolute terms. In: Sttthe expected increase in the stock price is constant in absolute terms while the variability of the proportional stock price change is constant. The model: StStis the most appropriate one since it is most realistic to assume that the expected percentage return and t
25、he variability of the percentage return in a short interval are constant. Problem 13.9.It has been suggested that the short-term interest rate, , follows the stochastic rprocess ()drabtcdzwhere , , and are positive constants and is a Wiener process. Describe the nature abcof this process. The drift
26、rate is . Thus, when the interest rate is above the drift rate is ()abrbnegative and, when the interest rate is below , the drift rate is positive. The interest rate is btherefore continually pulled towards the level . The rate at which it is pulled toward this level is . A volatility equal to is su
27、perimposed upon the “pull” or the drift. acSuppose , and and the current interest rate is 20% per annum. 04105The interest rate is pulled towards the level of 10% per annum. This can be regarded as a long run average. The current drift is % per annum so that the expected rate at the end of one year
28、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, , follows geometric Brownian motion with expected Sreturn and vola
29、tility : dtSdzWhat is the process followed by the variable ? Show that also follows geometric nnBrownian motion. If then , , and . Using ()nGSt0Gt1nS22(1)nGSIts lemma: 2()dndtzThis shows that follows geometric Brownian motion where the expected return is n 21()and the volatility is . The stock price
30、 has an expected return of and the expected Svalue of is . The expected value of is TS0Te nT212()0TeProblem 13.11.Suppose that is the yield to maturity with continuous compounding on a zero-xcoupon bond that pays off $1 at time . Assume that follows the process Tx0()dxadtszwhere , , and are positive
31、 constants and is a Wiener process. What is the process a0xsfollowed by the bond price? The process followed by , the bond price, is from Its lemma: B201()Bdaxsxdtsxzt Since: ()xTtethe required partial derivatives are ()()22()2)(xTttxTtBtetBtxHence: 2201()()()dBaxTtxsTtBdsxTtdz Problem 13.12 (Excel
32、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
33、 the random sample change.The process is tStS20.09.Where 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 s
34、tock 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 i
35、n the text 2()St:In this case , , and . Hence 50S16031365074t(27492)5)S:and 2(045047)S:that is, (61)(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 is 0615478(c) 95% confidence limits for the stock p
36、rice at the end of the next day are 5021960785and021978i.e., 48and516Note that some students may consider one trading day rather than one calendar day. Then . The answer to (a) is then 50.032. The answer to (b) is 0.945. The 1250397tanswers to part (c) are 48.18 and 51.88. Problem 13.14.A companys c
37、ash 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? (
38、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: (2016)(2106)9()(3)(b) The chance of a random sample from being negative is (
39、2605)9Nwhere is the cumulative probability that a standardized normal variable i.e., a ()Nxvariable with probability distribution is less than . From normal distribution (01)xtables . Hence the probability of a negative cash position at the end 26504of six months is 0.40%. Similarly the probability
40、of a negative cash position at the end of one year is 32(30)17196Nor 1.07%. (c) In general the probability distribution of the cash position at the end of months is x(2016)xThe probability of the cash position being negative is maximized when: 016xis minimized. Define 11223122320504()xyxdxxThis is z
41、ero when and it is easy to verify that for this value of 020dyx. It therefore gives a minimum value for . Hence the probability of a negative cash x yposition is greatest after 20 months. Problem 13.15.Suppose that is the yield on a perpetual government bond that pays interest at the xrate of $1 per
42、 annum. Assume that is expressed with continuous compounding, that interest xis paid continuously on the bond, and that follows the process 0()dxdtszwhere , , and are positive constants and is a Wiener process. What is the process a0xsfollowed by the bond price? What is the expected instantaneous re
43、turn (including interest and capital gains) to the holder of the bond? The process followed by , the bond price, is from Its lemma: B201()BBdaxsxdtsxzt In this case xso that: 2310BBt xHence 203221()1daxsdtsztzxThe expected instantaneous rate at which capital gains are earned from the bond is therefo
44、re: 20()saThe expected interest per unit time is 1. The total expected instantaneous return is therefore: 201()xxWhen expressed as a proportion of the bond price this is: 201()sa20()axsProblem 13.16.If follows the geometric Brownian motion process in equation (13.6), what is the Sprocess followed by
45、 (a) y = 2S, (b) y=S2 , (c) y=eS, and (d) y=er(T-t)/S. In each case express the coefficients of dt and dz in terms of rather than S. y(a) In this case , , and so that Its lemma gives S200yt2dStdzor y(b) In this case , , and so that Its lemma gives 2y2yt22()dSdSzor yt(c) In this case , , and so that
46、Its lemma gives Sye2Se 0y2()Sddtezor ln(llnyy(d) In this case , , and ()2rTtySeS2()32rTtyeSso that Its lemma gives ()rTtyt)dydtzor 2(ryProblem 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 g
47、reater than 80 in two years? (Hint when .) 80TSln80TSThe variable is normally distributed with mean and standard 20ln()STdeviation . In this case , , , and so that the mean and 0512T3standard deviation of are and , lnTSl(3)46042respectively. Also, . The probability that is the same as the probability 8432 8Tthat . This is ln432TS 80211(075)4NNwhere is the probability that a normally distributed variable with mean zero and ()Nxstandard deviation 1 is less than . From the tables at the back of the bo
