From: Bayesian Models for Astrophysical Data, Cambridge Univ. Press

(c) 2017,  Joseph M. Hilbe, Rafael S. de Souza and Emille E. O. Ishida 

 

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Code 10.6 Multivariate Gaussian mixed model in Python, using Stan, for accessing the relationship between luminosity, period, and color in early-type contact binaries

===================================================================================

import numpy as np
import pandas as pd
import pystan 
import statsmodels.api as sm

# Data
path_to_data = 'https://raw.githubusercontent.com/astrobayes/BMAD/master/data/Section_10p3/PLC.csv'

# read data
data_frame = dict(pd.read_csv(path_to_data))

# prepare data for Stan
data = {}
data['x1'] = np.array(data_frame['logP'])
data['x2'] = np.array(data_frame['V_I'])
data['y'] = np.array(data_frame['M_V'])
data['nobs'] = len(data['x1'])
data['type'] = np.array([1 if item == data_frame['type'][0] else
                                      for item in data_frame['type']])
data['M'] = 3
data['K'] = data['M'] - 1

# Fit
stan_code="""
data{
    int<lower=0> nobs;                # number of data points
    int<lower=1> M;                    # number of linear predicor coefficients
    int<lower=1> K;                     # number of  distinct populations
    vector[nobs] x1;                      # obs log period
    vector[nobs] x2;                      # obs color V-I
    vector[nobs] y;                        # obs luminosity
    int type[nobs];                         # system type (near/genuine contact)
}
parameters{
    matrix[M,K] beta;                    # linear predictor coefficients
    real<lower=0> sigma[K];        # scatter around linear predictor
    real mu0;
    real sigma0;
}
model{
    vector[nobs] mu;                        # linear predictor

    for (i in 1:nobs) {
        if (type[i] == type[1])
            mu[i] = beta[1,2] + beta[2,2] * x1[i] + beta[3,2] * x2[i];
        else mu[i] = beta[1,1] + beta[2,1] * x1[i] + beta[3,1] * x2[i];
    }

    # priors and likelihood
    mu0 ~ normal(0, 100);
    sigma0 ~ gamma(0.001, 0.001);

    for (i in 1:K) {
        sigma[i] ~ gamma(0.001, 0.001);
        for (j in 1:M) beta[j,i] ~ normal(mu0,sigma0);
    }

    for (i in 1:nobs){
        if (type[i] == type[1]) y[i] ~ normal(mu[i], sigma[2]);
        else y[i] ~ normal(mu[i], sigma[1]);
    }
}
"""

# Run mcmc
fit = pystan.stan(model_code=stan_code, data=data, iter=5000, chains=3,
                            warmup=2500, thin=1, n_jobs=3)

# Output
nlines = 13                                          # number of lines in screen output

output = str(fit).split('\n')


for item in output[:nlines]:
    print(item) 

===================================================================================

Output on screen:

Inference for Stan model: anon_model_fa17b80801723fc7926798026c342239.
3 chains, each with iter=5000; warmup=2500; thin=1; 
post-warmup draws per chain=2500, total post-warmup draws=7500.

                    mean      se_mean         sd      2.5%       25%        50%        75%       97.5%      n_eff        Rhat
beta[0,0]      -1.01          3.9e-3      0.26     -1.53      -1.17      -1.01        -0.83           -0.5       4541          1.0
beta[1,0]      -3.31             0.01      0.94     -5.19      -3.92        -3.3        -2.69         -1.45       6476          1.0
beta[2,0]       7.26             0.02      1.23       4.85       6.45       7.28          8.06          9.66       5041          1.0
beta[0,1]      -0.41         2.2e-3      0.15      -0.72      -0.51      -0.41        -0.32         -0.11       4684          1.0
beta[1,1]      -3.19         7.2e-3      0.57      -4.31      -3.57      -3.19        -2.79         -2.09       6266          1.0
beta[2,1]       8.48            0.01      0.79       6.89        7.97       8.48         8.99         10.06       4834          1.0
sigma[0]       0.62          1.2e-3     0.09       0.47        0.55       0.61         0.67           0.82       5800          1.0
sigma[1]       0.42          6.6e-4     0.05       0.34        0.39       0.42        0.46            0.54       6344          1.0