Wild Bootstrap Standard Error

Problem

Return the wild bootstrap standard error of the mean: center the data, multiply each centered value by a random sign from the seeded Lehmer RNG (an even draw means +1, an odd draw means -1), and take the population standard deviation of the resulting bootstrap means. Return 0.0 for empty data or n_boot <= 0.

Hint

Random signs preserve heteroscedasticity without permuting values.

Test cases

case 1

in: [[1,2,3,4,5],200,7]

expected: 0.6105603983227211

case 2

in: [[5,5,5],100,1]

expected: 0

case 3

in: [[1,2],100,3]

expected: 0.3054095610815091

case 4

in: [[],10,1]

expected: 0

case 5

in: [[10,20,30],50,42]

expected: 4.349201714746692

Starter codePython

def wild_bootstrap_se(data, n_boot, seed):
    # Your code here
    pass

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