无循环迭代二项式更新 [英] Iterative Binomial Update without Loop

问题描述

这可以不循环吗?

import numpy as np
n = 10
x = np.random.random(n+1)
a, b = 0.45, 0.55

for i in range(n):
    x = a*x[:-1] + b*x[1:]

我在二项式期权定价模型有关，但我不太理解这个话题.诚实的.我只是对公式和x的迭代更新/收缩感到着迷，想知道是否可以不循环而完成.但是我无法解决这个问题，我不确定这是否可能.

I came across this setup in another question. There it was a covered by a little obscure nomenclature. I guess it is related to Binomial options pricing model but don't quite understand the topic to be honest. I just was intrigued by the formula and this iterative update / shrinking of x and wondered if it can be done without a loop. But I can not wrap my head around it and I am not sure if this is even possible.

让我认为可能有用的是这种变化

What makes me think that it might work is that this vatiaton

n = 10
a, b = 0.301201, 0.59692
x0 = 123
x = x0
for i in range(n):
    x = a*x + b*x
# ~42

实际上只是x0*(a + b)**n

print(np.allclose(x, x0*(a + b)**n))
# True

推荐答案

您正在计算:

sum( a ** (n - i) * b ** i * x[i] * choose(n, i) for 0 <= i <= n)

[这是伪代码，而不是Python.]我不确定将其转换为Numpy的最佳方法.

[That's meant to be pseudocode, not Python.] I'm not sure of the best way to convert that into Numpy.

choose(n, i)是n!/ (i! (n-i)!)，而不是numpy选择功能.

choose(n, i) is n!/ (i! (n-i)!), not the numpy choose function.

使用@mathfux的评论，一个人可以做

Using @mathfux's comment, one can do

import numpy as np
from scipy.stats import binom

binomial = binom(p=p, n=n)
pmf = binomial(np.arange(n+1))
res = np.sum(x * pmf)

所以

res = x.copy()
for i in range(n):
    res = p*res[1:] + (p-1)*res[:-1]

只是二项式分布随机变量x的期望值.

is just the expected value of a binomial distributed random variable x.

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