Normal-Normal Posterior Variance

Problem

Under the conjugate normal model, posterior precision is the sum of the prior precision and n data precisions, so the posterior variance is 1 / (1 / prior_var + n / data_var). Return prior_var when n <= 0 or data_var is nonpositive, and 0.0 when prior_var is nonpositive.

Hint

More observations and tighter data precision shrink the posterior variance.

Test cases

case 1

in: [1,1,1]

expected: 0.5

case 2

in: [4,2,2]

expected: 0.8

case 3

in: [2,1,0]

expected: 2

case 4

in: [0.5,1,1]

expected: 0.3333333333333333

case 5

in: [0,1,1]

expected: 0

Starter codePython

def normal_normal_posterior_variance(prior_var, data_var, n):
    # Your code here
    pass

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