Binomial trees are perfect. Well, the Cox-Ross-Rubinstein One-Level Model is perfect to introduce the concept of a risk-neutral measure, which is (and was also not for me) not easy to understand for beginners. When there are no dividends, constant interest rates and constant volatilities, then N-level binomial trees recombine.
Recombining N-level trees need little storage and are easy to implement, even for instruments with early exercise rights. If done properly (from the algorithmic point of view), they are quite efficient.
A readable piece od Mathematica code for a European call would be
BinomialEuropeanCall[S_, K_, r_, sigma_, T_, n_] :=
Module[{dt, a, up, down, P, Q, BinomTree, value, level},
dt = T/n;
a = Exp[r*dt];
up = Exp[sigma*Sqrt[dt]]*a; down = Exp[-sigma*Sqrt[dt]]*a;
P = (a - down)/(up - down); Q = 1 - P;
P = P*Exp[-r*dt]; Q = Q*Exp[-r*dt];
BinomTree = Table[Max[S*down^node*up^(n - node) - K, 0], {node, 0, n}];
Do[BinomTree =
Table[{P, Q}.{BinomTree[[node]], BinomTree[[node + 1]]},
{node, 1, level}]; {level, n, 1, -1}];
value = BinomTree[[1]];
Clear[BinomTree];
value]
Nevertheless, there must be some disadvantages of binomial trees, otherwise my blog colleague Michael Aichinger would not have any reason to tell you about, say, Finite Elements and other really clever methods.
Even or odd?
One disadvantage of naïve implementations of binomial trees is that the option value depends quite heavily (and oscillatory) on the number of levels used as the following figure exhibits:
For vanilla instruments, these oscillations could be repaired by calculating the mean of two neighbouring values. But when discontinuous conditions are of importance (like in knock-out options), things get even worse. The following figure shows the value of an up-and out call option calculated by binomial trees and its binomial delta.
Recombining N-level trees need little storage and are easy to implement, even for instruments with early exercise rights. If done properly (from the algorithmic point of view), they are quite efficient.
A readable piece od Mathematica code for a European call would be
BinomialEuropeanCall[S_, K_, r_, sigma_, T_, n_] :=
Module[{dt, a, up, down, P, Q, BinomTree, value, level},
dt = T/n;
a = Exp[r*dt];
up = Exp[sigma*Sqrt[dt]]*a; down = Exp[-sigma*Sqrt[dt]]*a;
P = (a - down)/(up - down); Q = 1 - P;
P = P*Exp[-r*dt]; Q = Q*Exp[-r*dt];
BinomTree = Table[Max[S*down^node*up^(n - node) - K, 0], {node, 0, n}];
Do[BinomTree =
Table[{P, Q}.{BinomTree[[node]], BinomTree[[node + 1]]},
{node, 1, level}]; {level, n, 1, -1}];
value = BinomTree[[1]];
Clear[BinomTree];
value]
Nevertheless, there must be some disadvantages of binomial trees, otherwise my blog colleague Michael Aichinger would not have any reason to tell you about, say, Finite Elements and other really clever methods.
Even or odd?
One disadvantage of naïve implementations of binomial trees is that the option value depends quite heavily (and oscillatory) on the number of levels used as the following figure exhibits:
For vanilla instruments, these oscillations could be repaired by calculating the mean of two neighbouring values. But when discontinuous conditions are of importance (like in knock-out options), things get even worse. The following figure shows the value of an up-and out call option calculated by binomial trees and its binomial delta.
The true (Black-Scholes) delta should become negative above 115 or so. But the calculated binomial delta is positive in most of the points and negative only at the sharp peaks downwards. So, if one delta-hedges, he/she might end up with a situation of increased instead of decreased market risk. Well done, trees!I will return to trinomial trees in my next blog entry.
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