Iterative Binomial Update Without Loop

Can this be done without a loop? 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:] I came across this setup in a

Solution 1:

You are calculating:

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

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

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

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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)

So

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

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