plays two roles, in most cases the same rate isapplicable and we need only input one interest rate into the model. If, however,different rates are applicable, such as would be the case with foreign currency options(the foreign currency interest rate plays one role, the domestic currency interest rateplays a different role) the model will require the input of two interest rates. This is thecase with the Garman-Kohlhagen version of the Black-Scholes Model.
The fact that interest rates play a dual role also means that the relative importanceof interest rates will vary, depending on the type of underlying instrument and thesettlement procedure. For example, interest rates have a much greater impact on thevalue of stock options than on futures options. As we raise interest rates, we increasethe forward price of stock, but leave the forward price of a futures contract unchanged.At the same time, assuming stock-type settlement, as we raise interest rates we decreasethe value of options. The option price, however, is usually very small in relation to theprice of the underlying contract.
What interest rate should a trader use when evaluating options? Most traderscannot borrow and lend at the same rate, so the correct interest rate will, in theory,depend on whether the trade will create a debit or a credit. In the former case the traderwill be interested in the borrowing rate, while in the latter case he will be interested inthe lending rate. In practice, however, the most common solution is to use the risk-freeinterest rate, i.e., the most secure rate. In the United States, the government is usuallyconsidered the most secure borrower of funds, so that the yield on a governmentsecurity with a term equivalent to the life of the option is the general benchmark. For a60-day option, use the yield on a 60-day treasury bill; for a 180-day option, use the yieldon a 180-day treasury bill.
DIVIDENDS
We did not list dividends as an input in Figure 3-4 since they are only a factor in thetheoretical evaluation of stock options, and then only if the stock is expected to pay adividend over the life of the option.
In order to accurately evaluate a stock option, a trader must know both the amountof the dividend which the stock will pay and the ex-dividend date, the date on which atrader must own the stock in order to receive the dividend. The emphasis here is onownership of the stock. A deeply in-the-money option may have many of the samecharacteristics as stock, but only ownership of the stock carries with it the right to collectthe dividend.
In the absence of other information, most traders tend to assume that a companywill continue the same dividend policy it has had in the past. If the company has beenpaying a 75¢ dividend each quarter, it will probably continue to do so. However, thisis not always a certainty. Companies sometimes increase or decrease dividends, andIntroduction to Theoretical Pricing Modets 49
occasionally omit them completely. If there is the possibility of a change in a company’sdividend policy, a trader has to consider its impact on option values. Additionally, if theex-dividend date is expected just prior to expiration, there is the danger that a delay ofseveral days will cause the ex-dividend date to fall after expiration. For purposes ofoption evaluation, this is the same as eliminating the dividend completely. In such asituation a trader ought to make a special effort to ascertain the exact ex-dividend date.
VOLATILITY
Of all the inputs required for option evaluation, volatility is the most difficult for tradersto understand. At the same time, volatility often plays the most important role in actualtrading situations. Changes in our assumptions about volatility can have a dramaticeffect on an option’s value, and the manner in which the marketplace assesses volatilitycan have an equally dramatic effect on an option’s price. For these reasons, we willdevote the next chapter to a detailed discussion of volatility.J 4%
Volatility
What is volatility and why is it so important to an option trader? The option trader, likea trader in the underlying instrument, is interested in the direction of the market. Butunlike the trader in the underlying, an option trader is also extremely sensitive to thespeed of the market. If the market for an underlying contract fails to move at a sufficientspeed, options on that contract will have less value because of the reduced likelihoodof the market going through an option’s exercise price. In a sense, volatility is a measureof the speed of the market. Markets which move slowly are low-volatility markets;markets which move quickly are high-volatility markets.
One might guess intuitively that some markets are more volatile than others.Between 1980 and 1982, the price of gold moved from $300 per ounce to $800 perounce, more than doubling its price. Yet few traders would predict that the S&P 500Index might more than double in a similar period. A commodity trader knows thatprecious metals are generally more volatile than interest rate instruments. In the sameway, a Stock trader knows that high-technology stocks tend to be more volatile thanutility stocks.
If we knew whether a market was likely to be relatively volatile, or relatively quiet,and could convey this information to a theoretical pricing model, any evaluation ofoptions on that market would be more accurate than if we simply ignored volatility.Since option models are based on mathematical formulae, we will need some methodof quantifying this volatility component so that we can feed it into the model innumerical form.
RANDOM WALKS AND NORMAL DISTRIBUTIONS
Consider for a moment the pinball maze pictured in Figure 4-1. When a ball is droppedinto the maze at the top it moves downward, pulled by gravity, through a series of nails.When the ball encounters each nail there is a 50% chance the bail will move to the right,and a 50% chance it will move left. The ball then falls down to a new level where itencounters another nail. Finally, at the bottom of the maze the ball falls into one of thetroughs.
The path the ball follows as it moves downward through the maze of nails is knownas a random walk. Once the ball enters the maze nothing can be done to artificially alterits course, nor can one predict ahead of time the path the ball will take through themaze,
5152 Chapter 4
Figure 4-1: Random Walk
 
 
If enough balls are dropped into the maze, we might begin to get a distribution ofballs similar to that in Figure 4-2. Most of the balls tend to congregate near the centerof the maze, with a decreasing number of balls ending up in troughs further away fromthe center. The distribution which results from dropping many balls into our maze isreferred to as a normal, or bell-shaped, distribution.
If we were to drop an infinite number of balls into the maze we might approximatethe distribution with a normal, or bell-shaped, curve such as the one overlaid on thedistribution in Figure 4-2. Such a curve is symmetrical (if we flipped it from right to leftit would look the same), it has its peak in the center, and its tails always flare down andaway from the center.
Normal distribution curves are used to describe the likely outcomes of randomevents. For example, the curve in Figure 4-2 might also represent the results of flippingacoin 15 times. Each outcome, or trough, would represent the number of heads whichoccurred after each 15 flips. An outcome in trough zero would represent zero heads and15 tails; an outcome in trough 15 would represent 15 heads and zero tails. Of course,we would be surprised to flip a coin 15 times and get all heads or all tails. Assuming theVolatility 53
Figure 4-2: Normal Distribution
 
 
coin is perfectly balanced; some outcome in between, perhaps 8 heads and 7 tails, or 9heads and 6 tails, seems more likely.
Suppose we change our maze slightly by closing off a row of nails so that each timea ball encounters a nail and goes either left or right, it must drop down two levels beforeit encounters another nail. If we drop enough balls into the maze we may end up witha distribution represented by the curve in Figure 4-3. Since the sideways movement ofthe balls is restricted, the curve will have a higher peak and narrower tails than the curvein Figure 4-2. In spite of its altered shape, this curve still represents a normal distribution,although one with slightly different characteristics.
Finally, we might block off some of the spaces between nails so that each time aball drops down a level it must move two nails left or right before it can drop down to4 new level. Again, if we drop enough balls into the maze we may get a distributionWhich resembles the curve in Figure 4-4. This curve, while still a normal distributionCurve, will have a much lower peak and its tails will spread out much more quickly thanthe curves in either Figure 4-2 or 4-3.54 Chapter 4
Figure 4-3: Low Volatility Distribution
[lig itl
[lig (ETI
   
 
 
Suppose we now think of the ball’s sideways movement as the up and down pricemovement of an underlying contract, and the ball’s downward movement as thepassage of time. If we assume that each day the underlying contract can move up ordown $1, the price distribution after 15 days might be represented by the curve in Figure4-2. If we assume the price can move up or down $1 every two days, the pricedistribution after 15 days might be represented by the curve in Figure 4-3. And if weassume that each day the price can move up or down $2, the price distribution mightbe represented by the curve in Figure 4-4.
With the underlying contract presently at $100 and 15 days to expiration, howmight we evaluate a $105 call? One way is to assume that prices follow a random walkthrough time and that one of the curves in Figures 4-2, 4-3, or 4-4 represents the likelydistribution after 15 days. The comparative value of the $105 call under these threescenarios is shown in Figure 4-5. If we assume a distribution similar to Figure 4-3, wecan see that the underlying contract has very little chance of reaching $105. Conse-quently, the value of the $105 call will be low. If we assume a distribution similar toFigure 4-2, there is an increased probability of the underlying reaching $105, and thiswill increase the value of the $105 call. Finally, if we assume a distribution similar toVolatility 55
Figure 4-4: High Volatility Distribution
at .
. Se ,(ee —9@ - —.
ee
LeDee ®,
_ a e
 
   
 
 
 
 
 
Figure 4-4, there is a very real likelihood that the $105 call could finish in-the-money.Asa result, the value of the option will increase dramatically.
If we assume only that the price movement of an underlying contract follows arandom walk, and nothing about the likely direction of movement, the curves in Figures4-2, 4-3, and 4-4 might represent possible price distributions in a moderate-volatility,low-volatility, and high-volatility market, respectively. In a low-volatility market, priceMovement is severely restricted, and consequently options will command relatively lowPremiums. In a high-volatility market the chances for extreme price movement is greatlyincreased, and options will command high premiums.
Since the different price distributions in Figure 4-5 are symmetrical, it may seemthat increased volatility should have no effect on an option’s value. After all, increasedvolatility may increase the likelihood of large upward movement, but this should beoffset by the equally greater likelinood of large downward movement. Here, however,there is an important distinction between an option position and an underlying position.Unlike an underlying contract, an option’s potential loss is limited. No matter how farthe market drops, a call option can only go to zero. In our example, whether the marketfinishes at $80 or $104 at expiration, the $105 call will be worthless. However, if webuy the underlying contract at $100, there is a tremendous difference between the56 Chapter 4
Figure 4-5: Price Distribution at Expiration
 
present priceof underlying
            
low volatilitydistribution —~
moderate volatilitydistribution ~.
exerciseprice
 
 
 
 
high volatilitydistribution ‘NSo .Os NTsee Sere ——
market finishing at $80 or $104. With an underlying contract all outcomes areimportant; with an option, only those outcomes which result in the option finishingin-the-money are important. In Figure 45, we are only concerned with price outcomesto the right of the exercise price. Everything else is zero.
This leads to an important distinction between evaluation of an underlying contractand evaluation of an option. If we assume that prices are distributed along a normaldistribution curve, the value of an underlying contract depends on where the peak of thecurve is located, while the value of an option depends on how fast the curve spreads out.
MEAN AND STANDARD DEVIATION
Suppose we want to use the concept of price movement based on normal distributioncurves in a theoretical pricing model. To do this we need a method of describing thecharacteristics of the curve to the model. Since the model is based on mathematics, weneed to describe the curve in numerical! terms so that we can feed the numbers into themodel.
Fortunately, a normal distribution curve can be fully described with two numbers,the mean and the standard deviation. If we know that a distribution is normal, and wealso know these two numbers, then we know all the characteristics of the distribution.
Graphically, we can interpret the mean as the location of the peak of the curve, andthe standard deviation as a measure of how fast the curve spreads out. Curves whichVolatility 57
spread out very quickly, such as Figure 4-4, have a high standard deviation; curveswhich spread out very slowly, such as Figure 4-3, have a low standard deviation.
While the mean is nothing more than the average outcome, and therefore a familiarconcept for many traders, the standard deviation may not be quite so familiar. Indeed,one need not know how either of these numbers are calculated in order to successfullytrade options. (For those who are interested, a more detailed discussion appears inAppendix B.) What is important to an option trader is the interpretation of thesenumbers, in particular what a mean and standard deviation suggest in terms of likelyprice movement.
Let’s go back to Figure 4-2 and consider the troughs numbered 0 to 15 at thebottom. We suggested that these numbers might represent the number of headsresulting from 15 flips of a coin. Alternatively, they might also represent the number oftimes a ball went right at each nail as it dropped down through the maze. The firsttrough is assigned zero since any ball which ends there must have gone left at everynail, The last trough is assigned 15 since any ball which ends there must have gone rightat every nail.
Suppose we are told that the mean and standard deviation in Figure 4-2 are 7.50and 3.00, respectively. What does this tell us about the distribution? (The actual meanand standard deviation of 7.51 and 2.99 are calculated in Appendix B. Here, forsimplicity, we will round to 7.50 and 3.00.} The mean tells us the average outcome. Ifwe add up all the outcomes and divide by the number of occurrences, the result will De7.50. In terms of the troughs, the average result will fall half-wey between troughs 7and 8. (Of course this is not an actual possibility. However, we noted in Chapter 3 thatthe average outcome does not have to be an actual possibility for any one outcome.)
The standard deviation not only describes how fast the distribution spreads out; italso tells us something about the likelihood of a ball ending up in a specific trough orgroup of troughs. In particular, the standard deviation tells us the probability of a ballending up in a trough which is a specified distance from the mean. For example, weMay want to know the likelihood of a ball falling down through the maze and endingup in a trough lower than 5 or higher than 10. We can answer the question by askinghow many standard deviations the ball must move away from the mean, and thendetermine the probability associated with that number of standard deviations.
The exact probability associated with any specific number of standard deviationscan be found in mathematical tables in most books on statistics. Alternatively, suchprobabilities can be closely approximated using an appropriate formula (see AppendixB). For option traders the following approximations will be useful:
+1 standard deviation takes in approximately 68.3% (about 24) of all occurrences
+2 standard deviations takes in approximately 95.4% (about !9%o) of all occur-rences
+3 standard deviations takes in approximately 99.7% (about 369470) of all occur-rences58 Chapter 4
Note that each number of standard deviations is preceded by a plus or minus sign.Because normal distributions are symmetrical, the likelihood of up movement and downmovement is identical.
Now let’s try to answer our question about the likelihood of getting a ball in atrough lower than 5 or higher than 10. We can designate the divider between troughs7 and 8 as the mean of 7!/. If the standard deviation is 3, what troughs are within onestandard deviation of the mean? One standard deviation from the mean is 714 +3 =4 to 104%. Again interpreting 14 as the divider between troughs, we can see thattroughs 5 through 10 fall within one standard deviation of the mean. We know that onestandard deviation takes in about 2/3 of all occurrences, so we can conclude that outof every three balls we drop into the maze, two should end up in troughs 5 through 10.What is left over, one out of every three balls, will end up in one of the remainingtroughs, 0-4 and 11-15. Hence, the answer to our original question about the likelihoodof getting a ball in a trough lower than 5 or higher than 10 is about 1 chance in 3, orabout 33%. (The exact answer is 100% — 68.3%, or 31.7%.} This is shown in Figure 4-6.
Let’s try another calculation, but this time we can think of the problem as a wager.Suppose someone offers us 30 to 1 odds that we can’t drop a ball into the maze and getit specifically in troughs 14 or 15. Is this bet worth making? One characteristic ofstandard deviations is that they are additive. In our example, if one standard deviationis 3, then two standard deviations are 6, Two standard deviations from the mean istherefore 712 + 6 = 1% to 13%. We can see in Figure 4-6 that troughs 14 and 15 lieoutside two standard deviations. Since the probability of getting a result within twostandard deviations is approximately 19 out of 20, the probability of getting a resultbeyond two standard deviations is 1 chance in 20. Therefore 30 to 1 odds may seemvery favorable. Recall, however, that beyond two standard deviations also includestroughs 0 and 1. Since normal distributions are symmetrical, the chances of getting aball specifically in troughs 14 or 15 must be half of 1 chance in 20, or about 1 chancein 40. At 30 to 1 odds the bet must be a bad one since the odds do not sufficientlycompensate us for the risk involved.
In Chapter 3 we said that one logical approach to option evaluation is to assign aprobability to an infinite number of possible price outcomes for an underlying contract.Then, if we multiply each possible price outcome by its associated probability we canuse the results to calculate an option’s theoretical value. The problem is in dealing withan infinite number of price outcomes and probabilities, since an infinite number ofanything is not easy to work with. Fortunately, the characteristics of normal distribu-tions have been so closely studied that formulas have been developed which facilitatethe computation of both the probabilities associated with every point along a normaldistribution curve, as well as the area under various portions of the curve. If we assumethat prices of an underlying instrument are normally distributed, these formulas repre-sent a unique set of tools with which we can solve for an option’s theoretical value. Thisis one of the reasons Black and Scholes adopted the normal distribution assumption aspart of their model.Volatility 59
eeFigure 4-6+1 standard deviation = 68.3% (2/3)
Mean = 7.50
+2 standard deviations = 95.4% (19/20)Standard Deviation = 3.00 -
£3 standard deviations = 99.7% (369/370)
 
 
A
2 “1 +1 +2standard standard mean standard standarddeviations deviation 7.50 deviation deviations
1.50 4.50 10.50 13.50
 60 Chapter 4
UNDERLYING PRICE AS THE MEAN OF A DISTRIBUTION
Now that we have decided to describe prices in terms of a normal distribution, how dowe feed this distribution into a theoretical pricing model? Since all normal distributionscan be described by a mean and the standard deviation, in some way we must feed thesetwo numbers into our pricing model.
When we enter the present price of an underlying instrument we are actuallyentering the mean of a normal distribution curve. An important assumption in theBlack-Scholes Model is that, in the long run, a trade in the underlying instrument willjust break even. It will neither make money nor lose money. Given this assumption, themean of the normal distribution curve assumed in the model must be the price at whicha trade in the underlying instrument, either a purchase ora sale, would just break even.What is that price? The answer depends on the type of underlying instrument.
Suppose a trader purchases a futures contract at 100 and holds the position for threemonths. Where does the price of the futures contract have to be at the end of threemonths for the trader to break even? Since futures contracts entail no carrying costs,nor do they pay dividends, the break even price three months from now is exactly theoriginal trade price of 100.
Now suppose that a trader purchases a $100 stock and holds it for three months.Where does the stock price have to be at the end of the holding period for the trader tobreak even? Since the purchase of stock requires immediate payment, the break evenprice will have to include the cost of carrying a $100 debit for three months. If interestrates are 8% annually the carrying cost on $100 for three months is 42 x 8% x $100= $2. Therefore, the stock price must be $102 at the end of three months for the tradeto break even. If the stock will pay a dividend of $1 during the holding period, then thestock price need only be $101 for the trade to break even.
Note that this is exactly how we calculated the forward price of a contract inChapter 3, And indeed this is exactly the type of calculation built into the various formsof the Black-Scholes Model. When we enter an underlying price into the Black-Scholesmodel, based on the type of underlying instrument, interest rates, and dividends, themodel calculates the forward price of the underlying instrument at expiration and makesthis price the mean of a normal distribution curve.
VOLATILITY AS A STANDARD DEVIATION
In addition to the mean, we also need a standard deviation to fully describe a normaldistribution curve. This is entered in the form ofa volatility. With some slight modifica-tions, which we will discuss shortly, we can define the volatility number associated withan underlying instrument as a one standard deviation price change, in percent, at theend of a one-year period.
For example, suppose that an underlying futures contract is currently trading at 100and has a volatility of 20%. Since this represents a one standard deviation price change,one year from now we expect the same futures contract to be trading between 80 and120 (100 + 20%) approximately 68% of the time, between 60 and 140 (100 + (2 x 20%))Volatility 61
approximately 95% of the time, and between 40 and 160 (100 + (3 x 20%}) approxi-mately 99.7% of the time.
If the underlying contract is a stock currently trading at $100, then the 20%volatility will have to be based on the forward price of the stock at the end of one year.If interest rates are 8% and the stock pays no dividends, the one-year forward price willbe $108. Now a one standard deviation price change is 20% x $108 = $21.60. So oneyear from now we would expect the same stock to be trading between $86.40 and$129.60 ($108 + $21.60] approximately 68% of the time, between $64.80 and$151.30 ($108 + (2 x $21.60)) approximately 95% of the time, and between $43.20and $172.90 ($108 + (3 x $21.60)} approximately 99.7% of the time.
Suppose we come back at the end of one year and find our futures contract, whichwe thought had a volatility of 20%, trading at 35. Does this mean our volatility of 20%was wrong? A price change of more than three standard deviations may be unlikely, butone shouldn’t confuse unlikely and impossible. Flipping a perfectly balanced coin 15times may result in 15 heads, even though the odds against it are more than 32,000 to1. If 20% was the right volatility, the odds of the futures price going from 100 to 35 oneyear later is more than 1,500 to 1. But one chance in 1,500 is not impossible, andperhaps this was the one time in 1,500 when the price would indeed end up at 35. Ofcourse, it is also possible that we had the wrong volatility. But we wouldn’t know thiswithout looking at price changes of the futures contract over many years so that we havea representative price distribution.
LOGNORMAL DISTRIBUTIONS
Is it reasonable to assume that the prices of an underlying instrument are normallydistributed? Beyond the question of the exact distribution of prices in the real world,the normal distribution assumption has one serious flaw. A normal distribution curve issymmetrical. Under a normal distribution assumption, for every possible upward movein the price of an underlying instrument there must be the possibility of a downwardmove of equal magnitude. If we allow for the possibility of a $50 instrument rising $75to $125, we must also allow for the possibility of the instrument dropping $75 toa priceof -$25. Since it is impossible for traditional stocks and commodities to take on negativePrices, the normal distribution assumption is clearly flawed. What can we do about this?
Up to now we have defined volatility in terms of the percent changes in the priceof an underlying instrument. In this sense an interest rate and volatility are similar inthat they both represent a rate of return. The primary difference between interest andVolatility is that interest generally accrues at a positive rate, while volatility represents acombination of positive and negative rates of return. If one invests money at a fixedinterest rate, the value of the principal will always grow. But if one invests in anUnderlying instrument with a volatility other than zero, the instrument may go eitherup or down in price. Volatility, which is a standard deviation, says nothing about thedirection of movement.
Since volatility represents a rate of return, an important consideration is the mannerin which the rate of return is calculated. For example, suppose we were to invest $1,00062 Chapter 4
for one year at an annual interest rate of 12%. How much would we have at the end ofone year? The answer depends on how the 12% interest on our investment is paid out,
Value after
Rate of Payment One Year Total Yield12% once a year $1,120.00 12%
6% twice a year $1,123.60 12.36%3% every three months $1,125.51 12.55%1% every month $1,126.83 12.68%12%/52 every week $1,127.34 12.73%12%/365 every day $1,127.47 12.75%12% compounded continuously $1,127.50 12.75%
AS interest is paid more often, even though it is paid at the same rate of 12% peryear, the total yield on the investment increases. The yield is greatest when interest ispaid continuously. In this case it is just as if interest were paid at every possible momentin time.
Although less common, we can do the same type of calculation using a negativeinterest rate. For example, suppose we were to lose 12 percent annually on our $1,000investment (interest rate = -12%). How much would we have at the end of a year? Theanswer depends on the frequency at which our losses accrue.
Value after
Rate of Loss , One Year Total Yield—12% once a year $880.00 -12%—6% twice a year $883.60 -11.64%-3% every three months $885.29 —11.47%-1% every month $886.38 -11.36%-12%/52 every week $886.80 -11.32%-12%/365 every day $886.90 -11.31%-12% compounded continuously $886.92 11.31%
In the case of a negative interest rate, as lasses are compounded more frequently,even though at the same rate of -12% per year, the loss becomes smaller, and so doesthe negative yield.
In the same way that interest can be compounded at different intervals, volatilitycan also be compounded at different intervals. For purposes of theoretical pricing ofoptions, volatility is assumed to compound continuously, just as if the price changes inthe underlying instrument, either up or down, were taking place continuously but at anannual rate corresponding to the volatility number associated with the underlyinginstrument.
What would happen if at every moment in time the price of an underlying couldgo up or down a given percent, and that these up and down movements were normallydistributed? When price changes are assumed to be normally distributed, the continu-Volatility 63
ous compounding of these price changes will cause the prices at maturity to belognormally distributed. Such a distribution is skewed toward the upside becauseupside prices resulting from a positive rate of return will be greater, in absolute terms,than downside prices resulting from a negative rate of return (Figure 4-7), In our interestrate example, a continuously compounded rate of return of +12% yields a profit of$127.50 after one year, while a continuously compounded rate of return of -12% yieldsa loss of only $113.08. If the 12% were a volatility, then a one standard deviationupward price change at the end of one year would be +$127.50, while a one standarddeviation downward price change would be -$113.08. Even though the rate of returnwas a constant 12%, the continuous compounding of the 12% yielded different upwardand downward moves.
The Black-Scholes Model is a continuous time model. It assumes that the volatilityof an underlying instrument is constant over the life of the option, but that this volatilityis continuously compounded. These two assumptions mean that the possible prices ofthe underlying instrument at expiration of the option are lognormally distributed. It alsoexplains why options with higher exercise prices carry more value than options withlower exercise prices, where both exercise prices appear to be an identical amount awayfrom the price of the underlying instrument. For example, suppose a certain underlyingcontract is trading at exactly 100. If there are no interest considerations and we assume
Figure 4-7: Lognormal Distribution
lognormaldistribution
normaldistribution
 64 Chapter 4
a normal distribution of possible prices, then the 110 call and the 90 put, both being10% outof-the-money, ought to have identical theoretical values. But under thelognormal assumption in the Black-Scholes