Using pymc3 to fit lomax model

Question

I have a pretty simple example that doesn't seem to work. My goal is to build a Lomax model, and since PyMC3 doesn't have a Lomax distribution I use the fact that an Exponential mixed with a Gamma is a Lomax (see here):

import pymc3 as pm
from scipy.stats import lomax

# Generate artificial data with a shape and scale parameterization
data = lomax.rvs(c=2.5, scale=3, size=1000)

# if t ~ Exponential(lamda) and lamda ~ Gamma(shape, rate), then t ~ Lomax(shape, rate)
with pm.Model() as hierarchical:
    shape = pm.Uniform('shape', 0, 10)
    rate = pm.Uniform('rate', 0 , 10)
    lamda = pm.Gamma('lamda', alpha=shape, beta=rate)
    t = pm.Exponential('t', lam=lamda, observed=data)
    trace = pm.sample(1000, tune=1000)

The summary is:

>>> pm.summary(trace)
           mean        sd  mc_error   hpd_2.5  hpd_97.5   n_eff      Rhat
shape  4.259874  2.069418  0.060947  0.560821  8.281654  1121.0  1.001785
rate   6.532874  2.399463  0.068837  2.126299  9.998271  1045.0  1.000764
lamda  0.513459  0.015924  0.000472  0.483754  0.545652  1096.0  0.999662

I would expect the shape and rate estimates to be close to 2.5 and 3 respectively. I tried various non-informative priors for shape and rate, including pm.HalfFlat() and pm.Uniform(0, 100) but both resulted in worse fits. Any ideas?

score 0 · Accepted Answer · answered May 29 '18 at 19:54

Figured it out: To derive a lomax from an exponential-gamma mixture, I need to specify a lamda for each example in the dataset (lamda = pm.Gamma('lamda', alpha=shape, beta=rate, shape=len(data)). This is because the model assumes each subject in the data has its own lamda_i where lamda_i ~ Gamma(shape, rate) for every i.

Using pymc3 to fit lomax model

1 Answers1