Having written about pricing American-style options on a binomial tree in q, I thought it would be instructive to do the same in Python and NumPy. Here is the code:

import functools as ft
import numpy as np
def BPTree(n, S, u, d):
r = [np.array([S])]
for i in range(n):
r.append(np.concatenate((r[-1][:1]*u, r[-1]*d)))
return r
def GBM(R, P, S, T, r, b, v, n):
t = float(T)/n
u = np.exp(v * np.sqrt(t))
d = 1./u
p = (np.exp(b * t) - d)/(u - d)
ptree = BPTree(n, S, u, d)[::-1]
R_ = ft.partial(R, np.exp(-r*t), p)
return reduce(R_, map(P, ptree))[0]
def American(D, p, a, b): return np.maximum(b, D*(a[:-1]*p + a[1:]*(1-p)))
def VP(S, K): return np.maximum(K - S, 0)
ABM = ft.partial(GBM, American)

There is a minor deviation from the q code: we are allowing d to be specified in BPTree. But otherwise, they are doing the same thing. Performance (as measured in ipython) isn’t too far-off either:

In [1]: from binomial import *
In [2]: %timeit ABM(ft.partial(VP,K=102.0), 100.0, 1.0, 0.08, 0.08, 0.2, 1000)
10 loops, best of 3: 38.4 ms per loop
In [3]: ABM(ft.partial(VP,K=102.0), 100.0, 1.0, 0.08, 0.08, 0.2, 1000)
Out[3]: 6.2215001602514555

Note the similarity between the q and Python code. The similarity is a result of using NumPy and functools which enabled Python to perform array-oriented computation and partial function application. We did use a loop in BPTree as Python/NumPy does not seem to have the same “scan” operation as q. I suppose we could have created a numpy.ufunc to use accumulate()… but the loop felt cleaner and more Pythonic.

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14 April, 2014 at 5:24 am

Hi there! I am trying to understand your code but there are a few issues I don’t understand. I’ve posted a question on StackOverflow about it and it would be fantastic if you can explain it http://stackoverflow.com/questions/23048564/american-option-pricing-binomial-model-in-python . Thank you!