Thompson Linear Posterior

Problem

Update a scalar Bayesian linear regression posterior after observing xs and ys.

post_precision = prior_precision + sum(x^2) / obs_var and post_mean = (prior_precision * prior_mean + sum(x * y) / obs_var) / post_precision. Return [post_mean, post_precision]; with no data return the prior.

Hint

Precision adds; the posterior mean is the precision-weighted average of prior and data.

Test cases

case 1

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

expected: [0.9333333333333333,15]

case 2

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

expected: [0.6666666666666666,1.5]

case 3

in: [2,0.5,[],[],0.5]

expected: [2,0.5]

case 4

in: [2,0.5,[1,-1,2],[3,-3,6],0.5]

expected: [2.96,12.5]

Starter codePython

def thompson_linear_posterior(prior_mean, prior_precision, xs, ys, obs_var):
    # Your code here
    pass

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