Session 2: data, seeds and reproducibility#

Zoo models: Boltzmann wealth exchange, Bouchaud-Mezard

Goals#

  • Collect model-level data with a DataCollector and turn it into a pandas DataFrame.

  • Understand why every draw must come from model.rng or model.random, and check that the same seed gives the same run.

  • Tell apart a transient from a stationary state, and average only the latter.

Before the session: the collecting data tutorial and Checking a model.

Exercises#

2.1 Run Bouchaud-Mezard twice with rng=7 and once with rng=8. Show that the first two Gini series are identical and the third is not.

Solution 
from modelly.zoo.bouchaud_mezard.model import BouchaudMezard, BouchaudMezardScenario


def gini_series(seed):
    model = BouchaudMezard(BouchaudMezardScenario(rng=seed, steps=300))
    model.run()
    return model.datacollector.get_model_vars_dataframe()["gini"]


print(gini_series(7).equals(gini_series(7)), gini_series(7).equals(gini_series(8)))

2.2 Replace self.rng.standard_normal(...) in a copy of the model by numpy.random.standard_normal(...). Run modelly lint on the copy. What does it report, and why does it matter?

Solution

MDL001 random draw outside the model's generators. The global numpy generator is not seeded by the scenario, so two runs with the same seed differ, and runs in worker processes may even share a stream.

2.3 Plot the Gini of the Boltzmann exchange over time. How many steps does it take to settle? Estimate its stationary mean from the second half only, and compare with model.stationary_gini().

Solution 
from modelly.zoo.kinetic_wealth.model import KineticScenario, KineticWealth

model = KineticWealth(KineticScenario(rng=3, steps=1000))
model.run()
series = model.datacollector.get_model_vars_dataframe().iloc[:, 0]
series.plot()
print(series.iloc[500:].mean(), model.stationary_gini())

The Gini rises for a few hundred steps, then fluctuates around the exact value.

2.4 Save the Bouchaud-Mezard model halfway with model.save(path), load it with BouchaudMezard.load(path) and run both to the end. Are the final wealth vectors equal?

Solution

Yes, element for element: a checkpoint holds the state of both generators, so the resumed run draws exactly the same numbers. See Checkpoints.