## Binomial and Trinomial Lattices

A popular method for finding the value of a derivative security is the binomial lattice method of Section 12.6 The method is straightfor ward and leads to reasonably accurate results, even if the time divisions are crude (say, 10 or so time periods over the remaining time interval) However, it is also possible to use other tree and lattice structures. For example, a good choice is to use a trinomial lattice, as shown in Figure 1.3 5. For a given number of time periods, the trinomial lattice has more nodes than a binomial lattice and hence can produce a better approximation to the continuous solution

At first it might seem that a trinomial lattice cannot replace a binomial lattice because it is impossible to replicate three possible outcomes using only two securities: the stock and the risk-free asset. This is correct; replication is not possible. Hence the trinomial lattice cannot be used as a basis for options theory However, once the theory is deduced by other methods (such as the Biack-Scholes method), we can seek alternative ways to implement it A trinomial lattice is a convenient structure for implementing the risk-neutral pricing formula.

To set up a suitable trinomial lattice refer to Figure 1.3 6, which shows one piece of the lattice. There are three paths leaving a node, with probabilities ppi, and p?,. The three resulting nodes represent multiplication of the stock value by u, 1, and d, respectively, where we set d = 1/íí, so that an up followed by a down is equal to 1

To assign the parameters of the trinomial lattice we can arbitrarily select a value for u Then if the mean value for one step is to be 1 + ¡i At and the variance is to be

FIGURE 13.5 Trinomial lattice. A trinomial lattice can give a more accurate representation than a binomial lattice for the same number of steps

u FIGURE 13.6 One piece of a trinomial lattice, in this lattice we must have c/ = 1/u so that the nodes recombine after two steps d a1 At, we select the probabilities to satisfy

Pi + P2 + Pi = 1 ltp\ -f P2 + dps = 1 + (JL At u2P\ + P2 + d2pi> = a2 At 4- (i -f- [J. At)2

(The last line represents E(.,v2) = var(.v) + E(.v)2, where .v is the random factor by which the stock price is multiplied in one period ) This is just a system of three linear equations to be solved for the three probabilities Once these probabilities are found, we have a good approximation to the underlying stock dynamics. (Note that we are implicitly using the dynamics of (11.19).)

To use this lattice for pricing, we must instead use the risk-neutral probabilities <7i, <72. <73, These are found by solving the same set of equations (1.3. 23), but with the mean value changed from fi At to / At Once the risk-neutral probabilities are found, the lattice can be solved backward, just as in the binomial procedure.

Example 13.7 (The 5-month call) Let us find the price of the 5-month call option of Example 12,3 using a trinomial lattice, just to compare the results We have S(0) = $62, K ~ $60, / = 10%, and a = 20%, The time to expiration is 5 months = ,416667, To set up the lattice we must select a value of it and solve the equations (13.23) for the probabilities (when p is set to /) in the equations. The choice of u requires a bit of experimentation, since for some values the resulting risk-neutral probabilities may not be positive. For example, using a = 1.06 leads to q\ — .57, — —.03, and <73 — .46 instead we use it = 1.1031277 and qx = .20947, <72 = .64896, and q2 = 14156 This leads to theTattice shown in Figure 13 7 Note that the value of the option obtained is $5.83, which is slightly closer to the Black-Scholes result of $5.80 than is the price of $5 85 determined by a binomial lattice.3

The lattice of Figure 13,7 has the stock value listed above each node and the option value listed below each node. The final option values are just max(0, 5' — K) The option values at other nodes are found by discounted risk-neutral pricing For

3In this example we assumed monthly compounding, while the Biack-Scholes formula implicitly assumes continuous compounding We can also use the equivalent continuous compounding rate in the example, and the result differs by only one-tenth of a cent from $5 8.3

101 28

101 28

example, the value at the top node after 4 months is (1 + .10/12)"'(cj\ x 41.28 + q2 x 31 81 + f/3 x 2.3 2.3) = .32 31 If in this calculation the stock values 101.28, 91 81, and 8.3.2.3 were used instead of the option values, the result would be the stock value of 91 81, but of course it is not necessary to use this backward procedure for the stock prices

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