temperatures do, in fact, follow a Markov process?
11.2. Can a trading rule based on the past history of a stock’s price ever produce returns that areconsistently above average? Discuss.
11.3. A company’s cash position (in millions of dollars) follows a generalized Wiener process with adrift rate of 0.5 per quarter and a variance rate of 4.0 per quarter. How high does the company’sinitial cash position have to be for the company to have a less than 5% chance of a negative cashposition by the end of one year?
11.4. Variables X, and X, follow generalized Wiener processes with drift rates jz, and jz. andvariances a; and o3. What process docs X; + X> follow if:
a. The changes in X, and X, in any short interval of time are uncorrelated?b. There is a correlation p between the changes in X, and X, in any short interval of time?
11.5. Consider a variable S that follows the processdS = jpdt+oadz
For the first three years, j: = 2 and o — 3; for the next three years, 4. = 3 and o = 4. If the initialvalue of the variable is 5, what is the probability distribution of the value of the variable at theend of year 6?
11.6. Suppose that G is a function of a stock price, S, and time. Suppose that os and og are thevolatilities of S and G. Show that when the expected return of S increases by Ags, the growth rateof G increases by Aog, where A is a constant.230
CHAPTER 11
 
 
 
11.7. Stock A and stock B both follow geometric Brownian motion. Changes in any short interval oftime are uncorrelated with each other. Does the value of a portfolio consisting of one of stock Aand one of stock B follow geometric Brownian motion? Explain your answer.
11.8. The process for the stock price in equation (11.8) is
bS = wS St + aSeV stwhere yz and o are constant. Expiain carefuliy the difference between this modei and each of thefollowing:
5S = pdt + aeJSt
5S = pS bt +oeV St
6S = dt + oSeV/StWhy is the model in equation (11.8) a more appropriate model of stock price behavior than anyof these three alternatives?
11.9. It has been suggested that the short-term interest rate, r, follows the stochastic process
dr = a(b —r)dt+redzwhere a, b, and ¢ are positive constants and dz is a Wiener process. Describe the nature of thisprocess.11.10. Suppose that a stock price, S, follows geometric Brownian motion with expected return yp andvolatility o:dS = pS dt+oaSdzWhat is the process followed by the variable 5"? Show that S” also follows geometric Brownianmotion.11.11. Suppose that x is the yield to maturity with continuous compounding on a zero-coupon bondthat pays off $1 at time 7. Assume that x follows the processdx = a(xy — x) dt + sxdzwhere a, xp, and s are positive constants and dz is a Wiener process. What is the process followedby the bond price?ASSIGNMENT QUESTIONS11.12. Suppose that a stock price has an expected return of 16% per annum and a volatility of 30% per
11.13.
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 dayb. The standard deviation of the stock price at the end of the next dayc. The 95% confidence limits for the stock price at the end of the next day
A company’s cash position (in millions of dollars) follows a generalized Wiener process with adrift 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?Model of the Behavior of Stock Prices 231
 
11.14.
11.15.
11.16.
Suppose that x is the yield on a perpetual government bond that pays interest at the rate of $1 perannum. Assume that x is expressed with continuous compounding, that interest is paidcontinuously on the bond, and that x follows the process
dx = a(x — x) dt + sxdzwhere a, xj, 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 capitalgains) to the holder of the bond?
If S follows the geometric Brownian motion process in equation (11.12), what is the processfollowed by:
a. y= 28?b. y= 8"?c. y=e?
d. y= e959In each case express the coefficients of dt and dz in terms of y rather than S.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: S; > 80when In S; > In 80.)232 CHAPTER 11
APPENDIX 11A
Derivation of It6’s Lemma
In this appendix, we show how Ité’s lemma can be regarded as a natural extension of other,simpler results. Consider a continuous and differentiable function G of a variable x. If 8x is a smallchange in x and 8G is the resulting small change in G, a well-known result from ordinary calculusis
dG5G bx (11A.1)
In other words, 5G is approximately equal to the rate of change of G with respect to x multipliedby dx. The error involves terms of order 6x’. If more precision is required, a Taylor series expansionof 6G can be used: :
4. ,@G,, dG, ,
For a continuous and differentiable function G of two variables x and y, the result analogous toequation (11A.1) isaG aGa — bx+— 6 .6G ax x ay y (11A.2)
and the Taylor series expansion of 8G is
25G = 29 gy 4 18 524
by + aG GC.»ax ay 2 axe axa
bxdy+4 by eee H1A3y Yt aT y ( )
 
In the limit as 5x and dy tend to zero, equation (11A.3) becomes
aG=—— dx+—d HA4dG == + oy? (11A.4)
We now extend equation (11A.4) to cover functions of variables following Ité processes.Suppose that a variable x follows the It6 process in equation (11.4), that is,
dx = a(x, tht + b(x, dz (11.5)
and that G is some function of x and of time t. By analogy with equation (11A.3), we can write
aG aG 18G.,, aG \PG 5= 5 8. —— dx dt+ 43— tee 11A.66G 5 bx +a 8+ 755 n+ aa OF +agp ort ( )Equation (11A.5) can be discretized to8x = a(x, t) dt + B(x, Nev Etor, if arguments are dropped,8x = adt + bev St (11A.7)
This equation reveals an important difference between the situation in equation (11A.6) and thesituation in equation (11A.3). When limiting arguments were used to move from equation (11A.3)Model of the Behavior of Stock Prices 233
to equation (11A.4), terms in 4x” were ignored because they were second-order terms. Fromequation (11A.7), we have8x” = b’e? 8t + terms of higher order in 8t (11A.8)
This shows that the term involving 3x? in equation (1 1A.6) has a component that is of order ét andcannot be ignored.The variance of a standardized normal distribution is 1.0. This means that
B(?) — [EP =
where E denotes the expected value. Because E(e) = 0, it follows that E(e?) = 1. The expectedvalue of € dt is therefore dt. It can be shown that the variance of ¢ st is of order 5¢? and that, as aresult of this, we can treat € 6t as nonstochastic and equal to its expected value of 8¢ as 6t tends tozero. It follows from equation (11A.8) that dx” becomes nonstochastic and equal to b” dt as dt tendsto zero. Taking limits as 6x and 6¢ tend to zero in equation (11A.6) and using this last result, weobtain
aG +26 = G
_ a _ 2dG == dx + dt + 5 FB at (114.9)
This is It6’s lemma. If we substitute for dx from equation (11A.5), then equation (11A.9) becomes
iG aG |G, aGd —Ga (ote ths waa e bdz 
CHAPTER 1 2
THE BLACK-SCHOLES MODEL
In the early 1970s, Fischer Black, Myron Scholes, and Robert Merton made a major breakthroughin the pricing of stock options.! This involved the development of what has become known as theBlack-Scholes model. The model has had a huge influence on the way that traders price and hedgeoptions. It has also been pivotal to the growth and success of financial engineering in the 1980sand 1990s. In 1997, the importance of the model was recognized when Robert Merton and MyronScholes were awarded the Nobel prize for economics. Sadly, Fischer Black died in 1995; otherwisehe also would undoubtedly have been one of the recipients of this prize.
This chapter shows how the Black-Scholes model for valuing European call and put options ona non-dividend-paying stock is derived. We explain how volatility can be either estimated fromhistorical data or implied from option prices using the model. We show how the risk-neutralvaluation argument introduced in Chapter 10 can be used. We also show how the Black-Scholesmodel can be extended to deal with European call and put options on dividend-paying stocks and
sresent some results on the sricine of American ca tH ostions on dividend paving stockspresent some resuits On the pricing Of American cau Options On Givicena-paying stocKs.
12.1 LOGNORMAL PROPERTY OF STOCK PRICES
 
The model of stock price behavior used by Black, Scholes, and Merton is the model we developedin Chapter 11. It assumes that percentage changes in the stock price in a short period of time arenormally distributed. Define:
yw: Expected return on stock
a: Volatility of the stock price
The mean of the percentage change in time 5t is 4 6t and the standard deviation of this percentagechange is o/5t, so that
* ~ b( dt, o Vt) (12.1)
where dS is the change in the stock price S in time dt, and @(m, s) denotes a normal distributionwith mean m and standard deviation s.
| See F. Black and M. Scholes, “The Pricing of Options and Corporate Liabilities,” Jourrial of Political Economy, 81(May/June 1973), 637-659; R. C. Merton, “Theory of Rational Option Pricing,” Bed! Journal of Economics andManagement Science, 4 (Spring 1973), 141-83.
234The Black-Scholes Model 235
As shown in Section 11.7, the model implies that
2In Sp — In Sy ~ of (u -S)r. ovT |
From this it follows that
c ry 2 4OT 2Ing al(e x)? ovT | (12.2)ando*In Sp ~ afin So + ( ~ 5) T, ovT] (12.3)
where S7 is the stock price at a future time T and Sp is the stock price at time zero.Equation (12.3) shows that In S; is normally distributed. This means that S$; has a lognormaldistribution.
Example12.1 Consider a stock with an initial price of $40, an expected return of 16% per annum,and a volatility of 20% per annum. From equation (12.3), the probability distribution of the stockprice, S;, in six months’ time is given by
In Sy ~ ofln 40 + (0.16 — 0.27/2) x 0.5, 0.2V0.5]
In Sp ~ 6(3.759, 0.141)
There is a 95% probability that a normally distributed variable has a value within 1.96 standarddeviations of its mean. Hence, with 95% confidence,3.759 — 1.96 x 0.141 < InS; < 3.759 + 1.96 x 0.141
This can be written
g3:759-1.96x0.141 3.759+1,96x0.141
<Sp<e
or32.55 < Sp < 56.56
Thus, there is a 95% probability that the stock price in six months will lie between 32.55 and 56.56.
A variable that has a lognormal distribution can take any value between zero and infinity.Figure [2.1 illustrates the shape of a lognormal distribution. Unlike the normal distribution, it
/\
Figure 12.1 Lognorma! distribution236 CHAPTER 12
is skewed so that the mean, median, and mode are all different. From equation (12.3) and the
properties of the lognormal distribution, it can be shown that the expected value, E(S;), of Sr is: 2
given by
E(Sr) = Sye“” (12.4)
This fits in with the definition of 4 as the expected rate of return. The variance, var(S;), of S; canbe shown to be given by
var(Sp) = She"? (e"? — 1) (12.5)Exemple 12.2 Consider a stock where the current price is $20, the expected return is 20% perannum, and the volatility is 40% per annum. The expected stock price in one year, E(Sy), and the
variance of the stock price in one year, var(S;), are given by
E(Sp) = 20e°?*! = 24.43
var(Sp) = 4000210" «1 _ 1) = 103.54
The standard deviation of the stock price in one year is 4103.54, or 10.18.
12.2 THE DISTRIBUTION OF THE RATE OF RETURN
 
The lognormal property of stock prices can be used to provide information on the probabilitydistribution of the continuously compounded rate of return earned on a stock between times zeroand 7. Define the continuously compounded rate of return per annum realized between times zeroand T as y. It follows that
Sr = Spe”?so thatlL, Sp= In— 12.6n TF in So ( )It follows from equation (12.2) that2oc 6~ - =>, Se 12.77 alu 7 *) (12.7)
Thus, the continuously compounded rate of return per annum is normally distributed with meanpp—o°/2 and standard deviation o//T?
? For a discussion of the properties of the lognormal distribution, see J, Aitchison and J. A. C. Brown, The LognormalDistribution, Cambridge University Press, Cambridge, 1966.
3 As T increases, the standard deviation of n declines, To understand the reason for this, consider two cases: T = |and T = 20. We are more certain about the average return per year over 20 years than we are about the return in anyone year.The Black-Scholes Model 237
Example12.3 Consider a stock with an expected return of 17% per annum and a volatility of 20%per annum. The probability distribution for the actual rate of return (continuously compounded)realized over three years is normal with mean
0.270.17 — >= 0.15or 15% per annum and standard deviation02 89.1155Ware
or 11.55% per annum. Because there is a 95% chance that a normally distributed variable will liewithin 1.96 standard deviations of its mean, we can be 95% confident that the actual return realizedover three years will be between —7.6% and +37.6% per annum.
12.3. THE EXPECTED RETURN
 
 
The expected return, 1, required by investors from a stock depends on the riskiness of the stock.The higher the risk, the higher the expected return. It also depends on the level of interest rates inthe economy. The higher the level of interest rates, the higher the expected return required on anygiven stock. Fortunately, we do not have to concern ourselves with the determinants of w in anydetail. It turns out that the value of a stock option, when expressed in terms of the value of theunderlying stock, does not depend on p at all. Nevertheless, there is one aspect of the expectedreturn from a stock that frequently causes confusion and is worth explaining.
Equation (12.1) shows that pdt is the expected percentage change in the stock price in a veryshort period of time 6t. This means that yz is the expected return in a very short period of time 6t. Itis natural to assume that y is also the expected continuously compounded return on the stock overa relatively long period of time. However, this is not the case. The continuously compoundedreturn realized over 7 years is1 Sy—Inand equation (12.7) shows that the expected value of this is 4 — 07 /2.
The reason for the distinction between the yz in equation (12.1) and the p — 07/2 in equation (12.7)is subtle but important. We start with equation (12.4):
Taking logarithms, we getIn[E(Sp)] = In(Sp) + 7
It is now tempting to setInfE(S;)] = E{in(S7)]
so that E[In(S;)] — In(Sp) = «7, or EfIn(S;/So)| = wT. However, we cannot do this because In is anonlinear function. In fact In[E(S;)] > Efin(S7)], so that Efln(S;/So)] < wT. This is consistent withequation (12.7).
Suppose we consider a very large number of very short periads of time of length 5t. Define 5, asthe stock price at the end of the ith interval and 6S; as S;,, — S;. Under the assumptions we are238 CHAPTER 12
 
making for stock price behavior, the average of the returns on the stock in each interval is close
to yz. In other words, j is close to the arithmetic mean of the 5S,/S;. However, the expected return
over the whole period covered by the data, expressed with a compounding period of 6t, is close to2 4
p—o°/2, not p.
Example 12.4 Suppose that the following is a sequence of returns per annum on a stock,measured using annual compounding:15%, 20%, 30%, —20%, 25%
The arithmetic mean of the returns, calculated by taking the sum of the returns and dividing by S,is 14%. However, an investor would actually earn less than 14% per annum by leaving the moneyinvested in the stock for five years. The dollar value of $100 at the end of the five years would be
100 x 1.15 x 1.20 x 1.30 x 0.80 x 1.25 = $179.40By contrast, a 14% return with annual compounding would give100 x 1.14° = $192.54The actual average return earned by the investor, with annual compounding, is
(1.7940)! — 1 = 0,124or 12.4% per annum.
The arguments in this section show that the term expected return is ambiguons. It can refer eitherto p or to p—o*/2. Unless otherwise stated, it will be used to refer to p. throughout this book.
12.4 VOLATILITY
 
The volatility of a stock, o, is a measure of our uncertainty about the returns provided by thestock. Stocks typically have a volatility between 20% and 50%.
From equation (12.7), the volatility of a stock price can be defined as the standard deviation ofthe return provided by the stock in one year when the return is expressed using continuouscompounding.
When T is small, equation (12.1) shows that o/T is approximately equal to the standarddeviation of the percentage change in the stock price in time 7. Suppose that o = 0.3, or30% per annum, and the current stock price is $50. The standard deviation of the percentagechange in the stock price in one week is approximately
f13 = = 4.16%0 x 35 16%
A one-standard-deviation move in the stock price in one week is therefore 50 x 0.0416, or $2.08.Equation (12.1) shows that our uncertainty about a future stock price, as measured by itsstandard deviation, increases—at least approximately—with the square root of how far ahead we
* If we define the gross return as one plus the regular return, the gross return over the whole period covered by thedata is the geometric average of the gross returns in each time interyal of length ét—not the arithmetic average. Thegeometric average is less than the arithmetic average.The Black-Scholes Model 239
are looking. For example, the standard deviation of the stock price in four weeks is approximatelytwice the standard deviation in one week.Estimating Volatility from Historical Data
To estimate the volatility of a stock price empirically, the stock price is usually observed at fixedintervals of time (e.g., every day, week, or month).Define:
n+1: Number of observations§;: Stock price at end of ith (( = 0, 1,...,1) intervalt: Length of time interval in years
w=mn(g")fori=1,2,...,7.
The usual estimate, s, of the standard deviation of the u,’s is given by
and let
 
 
or
   
     
1
1 Cn hn 2pollu" ay (i, u)
where a is the mean of the u;’s.From equation (12.2), the standard deviation of the u;’s is o./t. The variable s is therefore anestimate of o./t. It follows that o itself can be estimated as 6, where
SsVtThe standard error of this estimate can be shown to be approximately 6//2n.
Choosing an appropriate value for n is not easy. More data generally lead to more accuracy,but o does change over time and data that are too old may not be relevant for predicting the future.A compromise that seems to work reasonably well is to use closing prices from daily data over themost recent 90 to 180 days. An often-used rule of thumb is to set n equal to the number of days towhich the volatility is to be applicd. Thus, if the volatility estimate is to be used to valuc a two-yearoption, then daily data for the last two years are used. More sophisticated approaches to estimatingvolatility involving GARCH models are discussed in Chapter 17.
An important issue is whether time should be measured in calendar days or trading days whenvolatility parameters are being estimated and used. Later in this chapter, we show that empiricalresearch indicates that trading days should be used. In other words, days when the exchange isclosed should be ignored for the purposes of the volatility calculation.
o=
Example12.5 Table 12.1 shows a possible sequence of stock prices during 21 consecutive tradingdays. In this caseva = 0.09531 and » u? = 0.00326240 CHAPTER 12
Table 12.1 Computation of volatility
 
Closing stock Price relative Daily returnDay price ($) Si/Si-t uj = In(S;/S;-1)
0 20.00
1 20.10 1.00500 0.004992 19.90 0.99005 —0.010003 20.00 1.00503 0.005014 20.50 1.02500 0.024695 20.25 0.98780 —0.012276 20.90 1.03210 0.031597 20.90 1.00000 0.000008 20.90 1.00000 0.000009 20.75 0.99282 —0.0072010 20.75 1.00000 0.0000011 21,00 1.01205 0.0119812 21.10 1.00476 0.0047513 20.90 0.99052 —0.0095214 20.90 1.00000 0.0000015 21.25 1.01675 0.0166116 21.40 1.00706 0.0070317 21.40 1.00000 0.0000018 21.25 0.99299 —0.0070319 21.75 1.02353 0.02326
20 22.00 1.01149 0.01143
and the estimate of the standard deviation of the daily return is
0.00326 _ 0.09531"19 380
 
 
= 0.01216
or 1.216%. Assuming that there are 252 trading days per year, t= 1/252 and the data give anestimate for the volatility per annum of 0.01216./252 = 0.193, or 19.3%. The standard error of this
estimate is0.193
2x 20
= 0.031
 
or 3.1% per annum.
This analysis assumes that the stock pays no dividends, but it can be adapted to accommodatedividend-paying stocks. The return, u;, during a time interval that includes an ex-dividend day isgiven by
§;+D
4; = In ——
Sitwhere D is the amount of the dividend. The return in other time intervals is still
Si
i—t
 
u; tnThe Black-Scholes Model , 241
However, as tax factors play a part in determining returns around an ex-dividend date, it isprobably best to discard altogether data for intervals that include an ex-dividend date.
12.5 CONCEPTS UNDERLYING THE BLACK-SCHOLES-MERTONDIFFERENTIAL EQUATION
 
 
The Black-Scholes—Merton differential equation is an equation that must be satisfied by the priceof any derivative dependent on a non-dividend-paying stock. The equation is derived in the nextsection. Here we consider the nature of the arguments we will use.
The arguments are similar to the no-arbitrage arguments we used to value stock options inChapter 10 for the situation where stock price movements are binomial. They involve setting up atiskless portfolio consisting of a position in the derivative and a position in the stock. In absence ofarbitrage opportunities, the return from the portfolio must be the risk-free interest rate, r. Thisleads to the Black-Scholes—Merton differential equation.
The reason a riskless portfolio can be set up is that the stock price and the derivative price areboth affected by the same underlying source of uncertainty: stock price movements. In any shortperiod of time, the price of the derivative is perfectly correlated with the price of the underlyingstock. When an appropriate portfolio of the stock and the derivative is established, the gain or lossfrom the stock position always offsets the gain or loss from the derivative position so that theoverall value of the portfolio at the end of the short period of time is known with certainty.
Suppose, for example, that at a particular point in time the relationship between a small changein the stock price, 8S, and the resultant smail change in the price of a European call option, 8c, isgiven by
bc = 0.488
This means that the slope of the line representing the relationship between c and S is 0.4, asindicated in Figure 12.2. The riskless portfolio would consist of:
1. A long position in 0.4 share2. A short position in one call option
There is one important difference between the Black-Scholes~Merton analysis and our analysis
Call,price, c J
Slope = 0.4
Stock price,
>
 
Figure 12.2 Relationship between ¢ and S242 CHAPTER 12
using a binomial model in Chapter 10. In the former, the position in the stock and the derivative isriskless for only a very short period of time. (In theory, it remains riskless only for an instantaneouslyshort period of time.) To remain riskless, it must be adjusted, or rebalanced, frequently.’ Forexample, the relationship between dc and 5S in our example might change from 3c = 0.485 today todc = 0.588 in two weeks. This would mean that, in order to maintain the riskless position, an extra0.1 share would have to be purchased for each call option sold. It is nevertheless true that the returnfrom the riskless portfolio in any very short period of time must be the risk-free interest rate. This isthe key element in the Black~Scholes analysis and leads to their pricing formulas.
AssumptionsThe assumptions we use to derive the Black-Scholes—Merton differential equation are as follows:
. The stock price follows the process developed in Chapter 11 with 4 and o constant.. The short selling of securities with full use of proceeds is permitted.
. There are no transactions costs or taxes. All securities are perfectly divisible.
. There are no dividends during the life of the derivative.
. There are no riskless arbitrage opportunities.
. Security trading is continuous.
a Dw why
. The risk-free rate of interest, r, is constant and the same for all maturities.
As we discuss in later chapters, some of these assumptions can be relaxed. For example, o and rcan be a known function of t. We can even allow interest rates to be stochastic providing that thestock price distribution at maturity of the option is still lognormal.
of MPeRNIATI THE PIAMY eErmrunire FD£.U VOENRIVAITIV Mo DLAUNT-VUIIVLLCOTNWICN
DIFFEREN NTIAL EQUATIO
ai
 
 
The stock price process we are assuming is the one we developed in Section 11.3:dS = pSdt+oSdz (12.8)
Suppose that f is the price of a call option or other derivative contingent on S. The variable fmust be some function of S and t. Hence, from equation (11.14),
— (8