back start next[start] [1] [2] [3] [4] [5] [6] [7] [8] [9] [10] [11] [12] [13] [14] [15] [16] [17] [18] [19] [20] [21] [22] [23] [24] [25] [26] [27] [28] [29] [30] [31] [32] [33] [34] [35] [36] [37] [38] [39] [40] [41] [42] [43] [44] [45] [46] [47] [48] [49] [ 50 ] [51] [52] [53] [54] [55] [56] [57] 50 Prior to expiry the call will sell at prices, which vary directly with the stock price, above the solid line. As expiry approaches, the call price will move closer to the solid line and will reach it at expiry for the value corresponding to the stock price at that time. For example, at expiry if the stock is selling for \$60, the call is worth \$10. If the stock is selling for \$40, the call is worthless.Consider next a put which gives its owner the right to sell a share of the associated stock for \$50 at its expiry. Figure 10.3 gives a graphical illustration of the value of such a put at expiry.Figure 10.4 Value of the underlyuig stockNow consider what happens if the investor purchases a share of stock and a put. The payoff at the expiry date is the sum of Figures 10.3 and 10.4 and is illustrated in Figure 10.5. However, Figure 10.5 is exactly the same as Figure 10.2, only it is \$50 higher at all points.Figure 10.3 Value of an illustrative put at expiryPrior to expiry the put will sell at prices, which vary inversely with the stock price, above the solid line. As expiry approaches, the put price will move closer to the solid line and will reach it at expiry for the value corresponding to the stock price at that time. For example, at expiry if the stock is selling for \$40, the put is worth \$10. If the stock is selling for \$60, the put is worthless.As noted in Section 8.8, investors can either buy or sell options. Thus, if an investor thinks the security price is likely to rise, then a call should be purchased or a put should be sold. Conversely, if an investor thinks the security price is likely to fall, then a put should be purchased or a call should be sold.It is possible to show that the value of a put and a call widi the same exercise price on a given stock are related. Figure 10.4 is a graph of the results of buying a share of the stock underlying Figures 10.2 and 10.3.S 50StCKk priceFigure 10.5 Value of the combination of a stock and a put at expiry(10.39fc)= standard deviation of the force of interest on the stock or odier underlying assetInterestingly enough, the Black-Scholes formula is based on the assumption that 1 + /, follows the lognormal distribution, just as described in Sections 10.2 and 10.3. Thus, a as just defined is the standard deviation for 5j,], as defined in formula (10.18).Formula (10.38) also assumes that the underlying stock or other asset does not pay dividends prior to the expiry date. The derivation of formula (10.38) is lengthy and is given in Appendix X.Since an American call can be exercised at the expiry date and at any time prior to that date, it would seem that its value should be greater than diat for a European call. However, on a non-dividend paying stock, early exercise is never optimal. This can be seen from Figure 10.2, since the call prior to expiry sells at prices above the solid line. However, for a stock which pays dividends, early exercise may be optimal behavior in order to obtain the income from the dividend.The corresponding formula for the value of a European put is given byP = Ee-"[\-Nid)] - S[l -Nidi)] (10.40)where P is the value of the put and all other symbols are defined the same as in formula (10.38). Surprisingly, the value of an American put is greater than the value of a European put. The early exercise of a put can be optimal even when the underlying stock pays no dividends. On the one hand, a longer-term put may be worth more than a shorter-term put, since more time exists for a favorable movement in the underlying stock price. However, on the other hand, a shorter-term put may be worth more than a longer-term put due to the time value of money. The right to sell something at a favorable price in the fiiture is worth less the more distant the fiiture date. Thus, early exercise of a put may be optimal, so an American put is worth more than a European put.Empirical tests of the Black-Scholes formula suggest that the formula works reasonably well under certain conditions. It tends to develop significant errors in the following situations:1. When the exercise price E is far from the current market price S.2. For securities with volatility much above or below average, i.e. a being quite high or low.3. If the expiry date is far in the future, i.e. for high values of n.log90/100) -f- (.1 -f- .09/2)(l) 3(1)= .132d2 =log90/100) -f- (.1 - .09/2)(l).3(1)= -.168.From tables for the cumulative normal distribution we haveM-132) = .5525 (-.168) = .4333, Now substituting into formula (10.38) = (90)(.5525) - (100)(e-)(.4333) = \$10.52.10.6 A RANDOM WALK MODELIn Section 10.5 we discussed the Black-Scholes Option Pricing Model. In this section we introduce an alternative approach, the binomial model.This model is based on the premise that security prices move in a random walk. In a random walk the probabilities of price movements of a given magnitude up or down from period to period remain unchanged and, moreover, die outcome in each interval is independent of the outcomes in prior intervals. There is considerable empirical evidence that, in fact, security prices follow a random walk.We will introduce the binomial model for option pricing, which is based on a random walk, with a simple illustration. A stock is currendy selling for \$70. We assume that in one year the price must either be \$55 or \$95. We seek to find the value of a European call which expires at the end of one year with an exercise price of \$75. We will assume that the one-year rate of interest is 10%.If the stock falls to \$55, the call will be worthless. If the stock rises to \$95, the call will be worth \$20 at expiry.Now compare these outcomes with those that would result from buying one share of stock and borrowing \$50. If the stock falls to \$55, we can sell it and use the proceeds to pay off the loan for 50(1.1) = \$55 leaving a zero balance.Example 10.9 Use the Black-Scholes formula to find the value of a European call expiring in one year with an exercise price of \$100 for a stock currently selling for \$90. The standard deviation for the continuous rate of return for this stock is .3, and the risk-free force of interest is 10%.Applying formulas (10.39a) and (10.39b) we haveFigure 10.7 Illustrative open binomial latticeIn practice, the period until expiry is fixed. Thus, in applying the binomial method as we increase n we apply the method more times over shorter intervals., For example, if the period until expiry is one year, we could choose n = 1,2, 4, 12 , 52, 365 depending on whether we were considering prices changesj annually, semiannually, quarterly, monthly, weekly, or daily.Although formula (10.41) is general in nature, it does not give us any guidance concerning a reasonable choice of k. It can be shown that an excelle choice of is given by = e"/ - 1 (lO.-iwhere a is defined the same as in the Black-Scholes formula and h is the of the time interval between successive steps. Under formula (10.42) securitie with higher variances will have higher values of it, and conversely, certainly is reasonable.The derivation of formula (10.42) is based on the lognormal distributic exactly as assumed in the derivation of the Black-Scholes formula. In fact, can be shown that the limit as -»0 of the binomial method under theC= e-i[(.5466)*(163.988 - 100) -f-(.5466)3(.4534)( 121.486 - 100)] = \$10.93.Thus, die binomial mediod applied on a quarterly basis produces a value for die call of \$10.93, in comparison to die value of \$10.52 given by die Black-Scholes formula. If die binomial method is applied more times over shorter periods, the answer will approach I given by the Black-Scholes formula. This result is illustrated in the exercises.[0.7 SCENARIO TESTINGfe! In practical applications for many of the situations described in Chapters 9 10 an approach called scenario testing has become quite popular. This chnique is, in essence, a type of sophisticated simulation which requires nsive computer capability to implement, since the volume of calculations is tically quite high. In this approach, each "scenario" refers to a different ern of future interest rates, usually called an interest rate path. For example, consider an analysis involving die matching of assets and bilities for a financial institution. It is necessary to estimate fiiture cash flows from the assets and fiiture cash outflows from the liabilities. These cash inflows and outflows may involve inherent uncertainty as to ~unt, probability of payment, and timing. It is important to recognize in then applying formula (10.41) we determine the value oi p as above for the one-)eriod case. All the other values are known and the formula can be evaluated, fhe formula may be tedious to calculate by hand, but can easily be evaluated ly computer.The random walk sequence as illustrated in Figure 10.6 is often called a lattice. Figure 10.6 is an example of a closed lattice, which will have n + I possible outcomes after n periods. These are represented by the n + \ terms in the binomial expansion in formula (10.41).A more complex model is the open lattice, which is illustrated in Figure 10.7. An open lattice will result when the upside and downside factors are not reciprocals of each other as above. An open lattice is more complex and will have 2" endpoints rather than n + 1 endpoints. However, it can still readily be evaluated by computer.conditions gives the same answer as the Black-Scholes formula. These results are stated without proof. Readers interested in pursuing the derivation of these results are referred to papers by J. C. Cox, S. Ross, and M. Rubinstein (1979) and R. P. Clancy (1985) listed in the bibliography.Example 10.10 Rework Example 10.9 using the binomial method applied in four steps, i.e. on a quarterly basis. From formula (10.42) we haveA: = -1 = e-25 , 16183We now need to compute p on a basis which eliminates risk-free arbitrage. Our upside value is 90(1 + k) = 104.565 and our downside value is 90(1 -t- ky = 77.464. An equation of value to determine p is104.565p -I- 77.464(1 -p) = 90e-25() = 92.278which can be solved to give p = .5466.We have -t- 1 = 5 possible outcomes:90(1 + k)* = 163.98890(1 + kf = 121.48690 = 90.00090(1 -t- k)- = 66.67490(1-I-*)-= 49.394 We now apply formula (10.41) to obtain[start] [1] [2] [3] [4] [5] [6] [7] [8] [9] [10] [11] [12] [13] [14] [15] [16] [17] [18] [19] [20] [21] [22] [23] [24] [25] [26] [27] [28] [29] [30] [31] [32] [33] [34] [35] [36] [37] [38] [39] [40] [41] [42] [43] [44] [45] [46] [47] [48] [49] [ 50 ] [51] [52] [53] [54] [55] [56] [57]