7 The Binomial Model
Final stock prices now take valuesS0(1+σ√
δt)Ne(r−q−12σ2)T, S0(1+σ√
δt)N−1(1−σ√
δt) e(r−q−12σ2)T, . . . , S0(1−σ√
δt)Ne(r−q−12σ2)T (iv) For completeness, we list a third discretization occasionally used:
u =exp{(r−q)δt+σ√
δt}; d =exp{(r−q)δt−σ√ δt}
Substituting in equation (7.5) and retaining only terms O[δt] gives p=12(1−12σ√ δt). The center line of the grid now has the equationScenter=S0e(r−q)t, which is the equation for the forward rate (known as theforward curve).
7.3 APPLICATIONS
Continuing this process back to the first node (“rolling back through the tree”) finally gives a 6-month option value of 7.44. This may be compared to the Black Scholes value (equivalent to an infinite number of steps) of 7.01. This price error is equivalent to using a volatility of 21.6% instead of 20% in the Black Scholes formula.
(ii)European Call: Cox–Ross–Rubinstein Method (p= 12): For purposes of comparison, we reprice the same option as in the last section, using a different discretization procedure.
Once again we haveδt =0.5/3 and e−rδt=0.983 but now we usep= 12 and equation (7.7), so that
u= Ft t+δt
St
(1 +σ√
δt)=1.093; d = Ft t+δt
St
(1 − σ√
δt)=0.928
The tree is shown in Figure 7.5. This time, only the final stock prices are shown. The procedure for rolling back through the tree is identical to that in the last section, with the simplify- ing feature that p=(1−p)= 12. The calculation for the top right-hand step in the diagram becomes
0.983×12 ×(30.40+10.72)=20.220
and so on through the tree. For all intents and purposes, the final answer is identical to that of the last section (more precise numbers are 7.444 previously and 7.438 now).
130.40 30.40 110.72
10.72 94.00 0 79.81
0 20.22
5.27
0 12.53
2.59 7.44
6 months
t=o 2 months 4 months
S = 1000
X = 100 r = 10%
q = 4%
s= 20%
t = 0.5 year
Figure 7.5 European call: Cox–Ross–Rubinstein discretization
(iii)Bushy Trees and Discrete Dividends:Suppose that instead of continuous dividends, the stock paid one fixed, discrete dividendQ. For purposes of illustration, we assume that it is paid the instant before the second nodes. The tree can be adjusted at these nodes by the shift shown in Figure 7.6.St, whatever its value, simply drops by the amount of the dividend. Unfortunately, this dislocates the entire tree as shown. The tree is said to have becomebushy.
Let us recall the original random walk on which the binomial model is based. This is described in Appendix A.1, where we see that the tree is recombining by construction since the up-stepsUand down-stepsDareadditive. In such a tree, the insertion of a constantQwould not cause a dislocation since everything to the right of this point would move down by the same amount. This would have been the case if we had constructed the tree forxi=lnSi. However,
81
7 The Binomial Model
d(uS - Q)0
u(dS - Q)0
S0
uS0
uS - Q0
dS0
Q
Q
dS - Q0
Figure 7.6 Discrete fixed dividend
we have constructed the tree forSi directly, so that the sizes of the up- and down-moves are determined by themultiplicativefactorsuandd. A discrete dividend must also be multiplicative if the tree is to remain recombining. Instead of a fixed discrete dividend we therefore use a discrete dividend whose size is proportional to the value ofStat the node in question. This is illustrated in Figure 7.7, where the dividend isku S0at the higher node where the stock price is u S0, andkd S0at the lower node. The effect of this on the following three nodes is immediately apparent: the tree recombines.
S0
uS0
dS0
uS (1 - k)0
Q = kuS0
Q = kdS0
dS (1 - k)0
duS (1 - k)0
udS (1 - k)0
Figure 7.7 Discrete proportional dividend
This proportional dividend assumption is implicit in the continuous dividend case, where each infinitesimal dividend in a periodδt isq Stδt, i.e. proportional toSt.
We return to the call option and discretization procedure of subsection (ii), except that instead of a continuousq=4% (i.e. 2% over the 6-month period), there is a dividend of 2%×St at the second pair of nodes. The parameters are similar to those of subsection (ii) withq=0;
Ft t+δt =102.020;p=0.582;u =1.0851;d =0.9216; erδt =0.983.
The terminal values are calculated, taking into account the dividend as shown. Rolling back through the tree shown in Figure 7.8 is exactly the same as before and nothing different needs to be done at the dividend point; this was entirely handled by the adjustment in the stock price.
The initial value of the option works out to be 7.297 compared with 7.444 for the continuous dividend case. This difference gradually closes as the number of steps in the model increases.
At 25 steps it is only half as big.
The calculation was repeated using a fixed dividend of 2 paid at the same point, so that the tree did not recombine. It is not worth giving the details of the calculations, but the option
7.3 APPLICATIONS
100.00
108.51
106.34 92.16
88.51
115.38
98.00
83.24
125.20 25.20
106.34 6.34 90.32
76.71 0
0
(in 2 months) S = 1000
X = 100 r = 10%
Q = 4%
s= 20%
t = 0.5 year
Figure 7.8 European call: discrete proportional dividend
value is found to be 7.356. The difference is negligible, justifying the use of the proportional dividend model.
(iv)American Options:The European call option could of course have been priced using the Black Scholes model. Binomial trees really come into their own when pricing American options.
Consider an American put with X =110 and the remaining parameters the same as for the European call of subsection (i); the same discretization procedure is used as in that section and the results are laid out in Figure 7.9.
100.00
108.51
117.74
127.76
92.16
100.00
84.93
108.51
92.16
78.27 0
1.49
17.84
31.73 .67
8.85 10.00
23.81 25.07 4.87
16.63 17.84 10.64
S = 1000
X = 110 r = 10%
q = 4%
s= 20%
t = 0.5 year
Figure 7.9 American put: Jarrow–Rudd discretization The procedure starts the same as in subsection (i):
(A) Set up the tree and calculate the values of eachStand the terminal values of the option. This time we need to put in the intermediate stock prices for reasons which become apparent below.
(B) Calculate the terminal payoff values for the put option.
(C) Roll back through the tree calculating the intermediate option values. Starting at the top right-hand corner, we have
0.67=e−rδt(p×0+(1−p)×1.49) 83
7 The Binomial Model
(D) The next value in this column is8.85=e−rδt(p×1.49+(1−p)×17.84)
But an American put option at this point (S =100,X =110) could be exercised to give a payoff of 10.00. The value of 8.85 must therefore be replaced by 10.00. Similarly, at the bottom node in this column, the exercise value must be used.
(E) With these replacement values, the next column to the left is derived. Once again, the bottom node is calculated as
16.63=e−rδt(p×10.00+(1−p)×25.07)
This is less than the exercise value and must be replaced by 17.84, the exercise value of the American option.
(F) Finally a price of 10.64 is obtained for the option. This compares with a value of 9.29 for a similar European put.
The essence of the matter is summed up in Figure 7.10. In the next chapter we will show that a binomial tree is mathematically equivalent to a numerical solution of the Black Scholes equation. We saw in Section 6.1 that the price of an American option is only a solution of the Black Scholes equation in certain regions. Below the exercise boundary, the value of the American put is simply its intrinsic (exercise) value:
American put price=
f(St,t); solution to BS equation;Stabove exercise boundary X−St; not solution to BS equation; Stbelow exercise boundary
terminal values
Option exercised
here
t = 0
t = T
Exercise boundary
Figure 7.10 American puts
(v) While the pricing given in this section is useful for illustration, such a small number of steps would never be used for a real-life pricing. So whatisthe minimum number of steps needed to price an option in the market?
While the answer to this depends on the specific option being priced, solutions are typically distributed as shown in Figure 7.11. The principle features are as follows:
(A) The solid appearance of the left-hand graph comes about because the answers obtained change more sharply in going from an odd number of steps to an even number than they do between successive odd or even numbers of steps. When the option price is plotted
7.3 APPLICATIONS
against the number of binomial steps, the result therefore zig-zags between the envelopes made up of odd and even numbers of steps.
(B) The reason for this is intuitively apparent from Figure A2.2 in the Appendix: asnincreases from 5 to 6, the way in which the binomial distribution is “fitted” to the normal distribution changes radically. Forn=5, the binomial distribution has two equal maximum probability values while forn =6, there is only a single maximum probability; yet whenngoes from 6 to 8, the only change is that a couple of extra bars are squeezed in, giving a slightly better approximation to the normal curve. One would therefore expect smooth transitions for the sequencesn= . . .5,7,9, . . .andn = . . .6,8,10, . . .but jumps when going from odd to even to odd.
(C) In most circumstances, the answer obtained for n steps is improved on by taking the average of the answers ofnsteps and (n+1) steps.
(D) Even when the average of successive steps is taken, the value oscillates, with decreasing amplitude, around the analytical answer. However it is clear that beyond about 50 steps, the answer is close enough for most commercial purposes.
Number of steps 6.05
6.10 6.15 6.20 6.25 6.30 6.35
6 50 100 150 200 250 300
6.05 6.10 6.15 6.20 6.25 6.30 6.35
6 50 100 150 200 250 300
PRICES OF OPTION AVERAGE OF SUCCESSIVE PRICES
Figure 7.11 European call option priced with varying number of binomial steps:S=100;X=110;
r=10%;q=4%;σ =20%;t=1 year
(vi)Greeks: There are two possible approaches to calculating these, depending on the circum- stances. Imagine a structured product salesman working on the price of a complex OTC option for a client. He might typically be doing his pricing with a 100-step binomial model pro- grammed into a spreadsheet. After tinkering around for a while he establishes a price, and as a final step he works out the Greek parameters. The easiest way to do this is by numerical differentiation.
His Greeks might then look as follows, puttingδS=S/1000,δt=T/1000:
= ∂f(S0,t)
∂S0
= f(1.001×S0, T)− f(0.999×S0, T) 2S0
×1000 = ∂2f(S0,t)
∂S02 = f(1.001×S0,T) + f(0.999×S0,T) − 2 f(S0,T)
S02 ×10002
= ∂f(S0,t)
∂t = f(S0,0.999×T) − f(S0,T)
T ×1000
This lazy but practical way of finding the Greeks works fine if only one option is being valued.
The tree has to be calculated three times, but so what?
85
7 The Binomial Model
C
A D
E
B
t= -2dt
t=0 Figure 7.12 Binomial Greeks
Take instead the case of binomial models which are used to evaluate books containing hundreds of different options. Tripling the number of calculations in order to calculate the Greeks as above may be unacceptably time consuming. An alternative approach is illustrated in Figure 7.12. Suppose the solid part of the tree is the first couple of steps in the calculation of an option price. While leaving the number of steps between now and maturity unchanged, we can add another two steps backward in time; this is the dotted part of the tree. With this small addition, the Greeks can be calculated from a single tree as follows:
A= fD− fE
SD−SE A=
fD− fA
SD−SA
− fA− fE
SA−SE
1
2(SD−SE) A= fB− fC
4 dt
(vii) Finally, the reader should consider just how powerful a tool the binomial model really is: a few examples should illustrate how we have extended the range of structures that can be priced.
These should now be quite within the reader’s ability to model:
(A) The strike price could be made a function of time, e.g. an American call with the strike accreting at a constant rate.
(B) The option need not be either European or American but could be Bermudan: e.g. a 5-year option, exercisable only in the first six months of each year.
(C) The payoff may be a non-linear function of the stock price: e.g. an option of the form f =
0 ifST <X (S−X)2 ifST >X