Normal-Normal Posterior Mean

Problem

A normal prior with mean prior_mean and variance prior_var is updated with n observations whose sample mean is data_mean and whose known variance is data_var. The posterior mean is the precision-weighted average (prior_mean / prior_var + n * data_mean / data_var) divided by (1 / prior_var + n / data_var). Return prior_mean when n <= 0 or either variance is nonpositive.

Hint

Precisions add, and the posterior mean is the precision-weighted average of the prior mean and data mean.

Test cases

case 1

in: [0,1,1,1,1]

expected: 0.5

case 2

in: [0,1,0,1,10]

expected: 0

case 3

in: [2,4,6,1,2]

expected: 5.555555555555555

case 4

in: [5,2,10,4,0]

expected: 5

case 5

in: [1,0.5,0,2,3]

expected: 0.5714285714285714

Starter codePython

def normal_normal_posterior_mean(prior_mean, prior_var, data_mean, data_var, n):
    # Your code here
    pass

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