BIC Difference

Problem

Return BIC2 - BIC1, where BIC = -2*log_likelihood + k*ln(n). Return 0.0 when n <= 0. Negative values favor the second model.

Hint

The penalty k*ln(n) grows with the sample size.

case 1

in: [-100,2,-110,5,100]

expected: 33.8155105579643

case 2

in: [-50,1,-50,1,10]

expected: 0

case 3

in: [-10,3,-12,2,50]

expected: 0.0879769945718536

case 4

in: [-200,10,-180,10,1000]

expected: -40

case 5

in: [-5,0,-6,0,5]

expected: 2

Starter codePython

import math
def bic_difference(ll1, k1, ll2, k2, n):
    # Your code here
    pass

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