Leverage in Simple Regression

Problem

Return the leverage of observation i, h_i = 1/n + (x_i - mean(x))^2 / sum((x - mean(x))^2). Return 0.0 when n < 2, i is out of range, or x has zero variance.

Hint

Leverage measures how far x_i sits from the center of the predictor.

Test cases

case 1

in: [[1,2,3],0]

expected: 0.8333333333333333

case 2

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

expected: 0.3333333333333333

case 3

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

expected: 0.8333333333333333

case 4

in: [[5,5,5],0]

expected: 0

case 5

in: [[1],0]

expected: 0

Starter codePython

def leverage(x, i):
    # Your code here
    pass

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