I am note sure if you can omit that loop in an efficient way, but when we do not use
np.delete() but just use the indices to shrink
payoff in every iteration I already got a big speed increase on my machine:
# ... arr = np.full(N+1, d/u) arr = S*np.power(u, N) arr[1:] = arr * np.cumprod(arr[1:]) def get_payoff(s, k): return np.maximum((s - k), 0) payoff = get_payoff(arr, K) def binomial_tree(payoff,S,K,T,r,sigma,N): for i in range(0, N): payoff = coeff*(p*payoff[:-1] + (1-p)*payoff[1:]) return payoff
Note that I also changed the generation of
arr to be without a loop and changed the name of the
payload() function so that it does not interfere with the variable name.
I asked a similar question because I thought your formula was interesting and the answer from Frank Yellin also applies here (so all credits to him):
import numpy as np from scipy.stats import binom binomial = binom(p=p, n=N) pmf = binomial(np.arange(N+1)) res = coeff**n*np.sum(payoff * pmf)
In this form it is also clearer what is calculated in your loop:
the expected value of the binomial distributed random variable payoff.
solved What is wrong with this Binomial Tree Backwards Induction European Call Option Pricing Function?