A random variable has a linear Gaussian conditional probability distribution on its set of parents if
where are generally known as weights.
Suppose we have some data where each tuple is an observation of and . Then we can estimate the parameters using maximum likelihood relatively easily.
The detailed mathematical explanation of how this is done can be found in Section 17.2.4 of Koller and Friedman. Here we just provide which works by using the sufficient statistics outlined in that method (and hence is quite fast!).
from typing import Any, Sequence import numpy as np import pandas as pd def mle(self, node: Any, parents: Sequence[Any], data: pd.DataFrame): ''' Find maximum likelihood estimate for `mean`, `variance` and `weights` given the `data` and the dependencies on the `parents`. Parameters ---------- data : pandas.DataFrame A DataFrame with one row per observation and one column per variable. Returns ------- A two-tuple where the first argument is the vector of weights and the second is the variance. Notes ----- Maximum likelihood estimation of parameters is computed using the *sufficient statistics* approach described in section 17.2.4 of _. References ---------- ..  D. Koller and N. Friedman, Probabilistic graphical models: principles and techniques. Cambridge, MA: MIT Press, 2009. ''' M = len(data) k = len(parents) x_sum = data[node].sum() u_sums = data[list(parents)].sum().to_numpy() xu_sums = [(data[node] * data[p]).sum() for p in parents] uu_sums = [[(data[ui] * data[uj]).sum() for uj in parents] for ui in parents] # solve A*beta = b A = np.block([[np.reshape([M], (1, 1)), np.reshape(u_sums, (1, k))], [np.reshape(u_sums, (k, 1)), np.reshape(uu_sums, (k, k))]]) b = [x_sum] + xu_sums beta = np.linalg.solve(A, b) # extract parameters mean, weights = beta, beta[1:] x_var = data[node].var() cov_d = data[list(parents)].cov() var = x_var - sum([ sum([ weights[i] * weights[j] * cov_d[pi][pj] for j, pj in enumerate(parents) ]) for i, pi in enumerate(parents) ]) return beta, var