Renewal Expected Count

Problem

Expected number of renewals by time horizon for a renewal process whose interarrival distribution is interarrival_pmf, where interarrival_pmf[i] is the probability that a lifetime equals i + 1. Sum P(S_k <= horizon) over k >= 1 using repeated convolution of the pmf. Return 0.0 for an empty distribution or horizon <= 0.

Hint

The renewal function is the sum of the convolution powers evaluated up to the horizon.

Test cases

case 1

in: [[1],5]

expected: 5

case 2

in: [[0.5,0.5],2]

expected: 1.25

case 3

in: [[0.2,0.8],3]

expected: 1.36

case 4

in: [[],5]

expected: 0

Starter codePython

def renewal_expected_count(interarrival_pmf, horizon):
    # Your code here
    pass

4 test cases · run your code against them in the editor