Think Bayes

Copyright 2018 Allen B. Downey

MIT License:

In [1]:
# Configure Jupyter so figures appear in the notebook
%matplotlib inline

# Configure Jupyter to display the assigned value after an assignment
%config InteractiveShell.ast_node_interactivity='last_expr_or_assign'

import numpy as np
import pandas as pd

import matplotlib.pyplot as plt

# import classes from thinkbayes2
from thinkbayes2 import Pmf, Cdf, Suite, Joint

import thinkplot

Lions and Tigers and Bears

Suppose we visit a wild animal preserve where we know that the only animals are lions and tigers and bears, but we don't know how many of each there are.

During the tour, we see 3 lions, 2 tigers and one bear. Assuming that every animal had an equal chance to appear in our sample, estimate the prevalence of each species.

What is the probability that the next animal we see is a bear?

Grid algorithm

I'll start with a grid algorithm, enumerating the space of prevalences, p1, p2, and p3, that add up to 1, and computing the likelihood of the data for each triple of prevalences.

In [2]:
class LionsTigersBears(Suite, Joint):
    def Likelihood(self, data, hypo):
        data: string 'L' , 'T', 'B'
        hypo: p1, p2, p3
        # Fill this in.

In [3]:
# Solution

class LionsTigersBears(Suite, Joint):
    def Likelihood(self, data, hypo):
        data: string 'L' , 'T', 'B'
        hypo: p1, p2, p3
        p1, p2, p3 = hypo
        if data == 'L':
            return p1
        if data == 'T':
            return p2
        if data == 'B':
            return p3

In [4]:
ps = np.linspace(0, 1, 101);

Here's a simple way to find eligible triplets, but it is inefficient, and it runs into problems with floating-point approximations.

In [5]:
from itertools import product

def enumerate_triples(ps):
    for p1, p2, p3 in product(ps, ps, ps):
        if p1+p2+p3 == 1:
            yield p1, p2, p3

As an exercise, write a better version of enumerate_triples.

In [6]:
# Solution

from itertools import product

def enumerate_triples(ps):
    for p1, p2 in product(ps, ps):
        if p1 + p2 > 1:
        p3 = 1 - p1 - p2
        yield p1, p2, p3

Now we can initialize the suite.

In [7]:
suite = LionsTigersBears(enumerate_triples(ps));

Here are functions for displaying the distributions

In [8]:
def plot_marginal_pmfs(joint):
    pmf_lion = joint.Marginal(0)
    pmf_tiger = joint.Marginal(1)
    pmf_bear = joint.Marginal(2)

    thinkplot.Pdf(pmf_lion, label='lions')
    thinkplot.Pdf(pmf_tiger, label='tigers')
    thinkplot.Pdf(pmf_bear, label='bears')

In [9]:
def plot_marginal_cdfs(joint):
    pmf_lion = joint.Marginal(0)
    pmf_tiger = joint.Marginal(1)
    pmf_bear = joint.Marginal(2)

    thinkplot.Cdf(pmf_lion.MakeCdf(), label='lions')
    thinkplot.Cdf(pmf_tiger.MakeCdf(), label='tigers')
    thinkplot.Cdf(pmf_bear.MakeCdf(), label='bears')

Here are the prior distributions

In [10]:

Now we can do the update.

In [11]:
for data in 'LLLTTB':

And here are the posteriors.

In [12]:

To get the predictive probability of a bear, we can take the mean of the posterior marginal distribution:

In [13]:


Or we can do a pseudo-update and use the total probability of the data.

In [14]:


Using the Dirichlet object

The Dirichlet distribution is the conjugate prior for this likelihood function, so we can use the Dirichlet object to do the update.

The following is a monkey patch that gives Dirichlet objects a Marginal method.

In [15]:
from thinkbayes2 import Dirichlet

def DirichletMarginal(dirichlet, i):
    return dirichlet.MarginalBeta(i).MakePmf()

Dirichlet.Marginal = DirichletMarginal

Here are the priors:

In [16]:
dirichlet = Dirichlet(3)

Here's the update.

In [17]:
dirichlet.Update((3, 2, 1))

Here are the posterior PDFs.

In [18]:

And the CDFs.

In [19]:

And we can confirm that we get the same results as the grid algorithm.

In [20]:


Exercise: Implement this model using MCMC. You might want to start with this example.

In [21]:
import warnings
warnings.simplefilter('ignore', FutureWarning)

import pymc3 as pm

Here's the data.

In [22]:
observed = [0,0,0,1,1,2]
k = len(Pmf(observed))
a = np.ones(k)

array([1., 1., 1.])

Here's the MCMC model:

In [23]:
# Solution

model = pm.Model()

with model:
    ps = pm.Dirichlet('ps', a, shape=a.shape)
    xs = pm.Categorical('xs', ps, observed=observed, shape=1)

$$ \begin{array}{rcl} ps &\sim & \text{Dirichlet}(\mathit{a}=array)\\\text{xs} &\sim & \text{Categorical}(\mathit{p}=f(f(\text{ps.T},~f(f(\text{ps}))))) \end{array} $$

In [24]:
# Solution

with model:
    start = pm.find_MAP()
    step = pm.Metropolis()
    trace = pm.sample(1000, start=start, step=step, tune=1000)

logp = -5.8985, ||grad|| = 1.118: 100%|██████████| 7/7 [00:00<00:00, 1525.99it/s]
Multiprocess sampling (4 chains in 4 jobs)
Metropolis: [ps]
Sampling 4 chains: 100%|██████████| 8000/8000 [00:01<00:00, 6107.20draws/s]
The number of effective samples is smaller than 25% for some parameters.

Check the traceplot

In [25]:

And let's see the results.

In [26]:
def plot_trace_cdfs(trace):
    rows = trace['ps'].transpose()

    cdf_lion = Cdf(rows[0])
    cdf_tiger = Cdf(rows[1])
    cdf_bear = Cdf(rows[2])

    thinkplot.Cdf(cdf_lion, label='lions')
    thinkplot.Cdf(cdf_tiger, label='tigers')
    thinkplot.Cdf(cdf_bear, label='bears')

In [27]:

And compare them to what we got with Dirichlet:

In [28]:

Using a Multinomial distribution

Here's another solution that uses a Multinomial distribution instead of a Categorical. In this case, we represent the observed data using just the counts, [3, 2, 1], rather than a specific sequence of observations [0,0,0,1,1,2].

I suspect that this is a better option; because it uses a less specific representation of the data (without losing any information), I would expect the probability space to be easier to search.

This solution is based on this excellent notebook from Will Koehrsen.

In [29]:
animals = ['lions', 'tigers', 'bears']
c = np.array([3, 2, 1])
a = np.array([1, 1, 1])

array([1, 1, 1])

In [30]:
warnings.simplefilter('ignore', UserWarning)

with pm.Model() as model:
    # Probabilities for each species
    ps = pm.Dirichlet('ps', a=a, shape=3)
    # Observed data is a multinomial distribution with 6 trials
    xs = pm.Multinomial('xs', n=6, p=ps, shape=3, observed=c)

In [31]:

$$ \begin{array}{rcl} ps &\sim & \text{Dirichlet}(\mathit{a}=array)\\xs &\sim & \text{Multinomial}(\mathit{n}=6, \mathit{p}=f(\text{ps},~f(f(\text{ps})))) \end{array} $$

In [32]:
with model:
    # Sample from the posterior
    trace = pm.sample(draws=1000, tune=1000)

Auto-assigning NUTS sampler...
Initializing NUTS using jitter+adapt_diag...
Multiprocess sampling (4 chains in 4 jobs)
NUTS: [ps]
Sampling 4 chains: 100%|██████████| 8000/8000 [00:02<00:00, 3157.96draws/s]

In [33]:

In [34]:

The results look good. We can use summary to get the posterior means, and other summary stats.

In [35]:
summary = pm.summary(trace)
summary.index = animals

mean sd mc_error hpd_2.5 hpd_97.5 n_eff Rhat
lions 0.446750 0.155029 0.002386 0.163030 0.744781 3833.026759 0.999522
tigers 0.330735 0.150189 0.002113 0.060451 0.616149 3769.939544 0.999898
bears 0.222515 0.132449 0.002385 0.009386 0.469737 3754.170642 1.000137

We can also use plot_posterior to get a better view of the results.

In [36]:
ax = pm.plot_posterior(trace, varnames = ['ps']);

for i, a in enumerate(animals):

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