This notebook presents code and exercises from Think Bayes, second edition.
Copyright 2016 Allen B. Downey
MIT License: https://opensource.org/licenses/MIT
In [1]:
from __future__ import print_function, division
% matplotlib inline
import warnings
warnings.filterwarnings('ignore')
import math
import numpy as np
from thinkbayes2 import Pmf, Cdf, Suite, Joint
import thinkplot
In [2]:
def EvalWeibullPdf(x, lam, k):
"""Computes the Weibull PDF.
x: value
lam: parameter lambda in events per unit time
k: parameter
returns: float probability density
"""
arg = (x / lam)
return k / lam * arg**(k-1) * np.exp(-arg**k)
def EvalWeibullCdf(x, lam, k):
"""Evaluates CDF of the Weibull distribution."""
arg = (x / lam)
return 1 - np.exp(-arg**k)
def MakeWeibullPmf(lam, k, high, n=200):
"""Makes a PMF discrete approx to a Weibull distribution.
lam: parameter lambda in events per unit time
k: parameter
high: upper bound
n: number of values in the Pmf
returns: normalized Pmf
"""
xs = np.linspace(0, high, n)
ps = EvalWeibullPdf(xs, lam, k)
return Pmf(dict(zip(xs, ps)))
SciPy also provides functions to evaluate the Weibull distribution, which I'll use to check my implementation.
In [3]:
from scipy.stats import weibull_min
lam = 2
k = 1.5
x = 0.5
weibull_min.pdf(x, k, scale=lam)
Out[3]:
In [4]:
EvalWeibullPdf(x, lam, k)
Out[4]:
In [5]:
weibull_min.cdf(x, k, scale=lam)
Out[5]:
In [6]:
EvalWeibullCdf(x, lam, k)
Out[6]:
And here's what the PDF looks like, for these parameters.
In [7]:
pmf = MakeWeibullPmf(lam, k, high=10)
thinkplot.Pdf(pmf)
thinkplot.Config(xlabel='Lifetime',
ylabel='PMF')
We can use np.random.weibull to generate random values from a Weibull distribution with given parameters.
To check that it is correct, I generate a large sample and compare its CDF to the analytic CDF.
In [8]:
def SampleWeibull(lam, k, n=1):
return np.random.weibull(k, size=n) * lam
data = SampleWeibull(lam, k, 10000)
cdf = Cdf(data)
model = pmf.MakeCdf()
thinkplot.Cdfs([cdf, model])
Exercise: Write a class called LightBulb that inherits from Suite and Joint and provides a Likelihood function that takes an observed lifespan as data and a tuple, (lam, k), as a hypothesis. It should return a likelihood proportional to the probability of the observed lifespan in a Weibull distribution with the given parameters.
Test your method by creating a LightBulb object with an appropriate prior and update it with a random sample from a Weibull distribution.
Plot the posterior distributions of lam and k. As the sample size increases, does the posterior distribution converge on the values of lam and k used to generate the sample?
In [9]:
# Solution
class LightBulb(Suite, Joint):
def Likelihood(self, data, hypo):
lam, k = hypo
if lam == 0:
return 0
x = data
like = EvalWeibullPdf(x, lam, k)
return like
In [10]:
# Solution
from itertools import product
lams = np.linspace(0, 5, 101)
ks = np.linspace(0, 5, 101)
suite = LightBulb(product(lams, ks))
In [11]:
# Solution
datum = SampleWeibull(lam, k, 10)
lam = 2
k = 1.5
suite.UpdateSet(datum)
Out[11]:
In [12]:
# Solution
pmf_lam = suite.Marginal(0)
thinkplot.Pdf(pmf_lam)
pmf_lam.Mean()
Out[12]:
In [13]:
# Solution
pmf_k = suite.Marginal(1)
thinkplot.Pdf(pmf_k)
pmf_k.Mean()
Out[13]:
In [14]:
# Solution
thinkplot.Contour(suite)
Exercise: Now suppose that instead of observing a lifespan, k, you observe a lightbulb that has operated for 1 year and is still working. Write another version of LightBulb that takes data in this form and performs an update.
In [15]:
# Solution
class LightBulb2(Suite, Joint):
def Likelihood(self, data, hypo):
lam, k = hypo
if lam == 0:
return 0
x = data
like = 1 - EvalWeibullCdf(x, lam, k)
return like
In [16]:
# Solution
from itertools import product
lams = np.linspace(0, 10, 101)
ks = np.linspace(0, 10, 101)
suite = LightBulb2(product(lams, ks))
In [17]:
# Solution
suite.Update(1)
Out[17]:
In [18]:
# Solution
pmf_lam = suite.Marginal(0)
thinkplot.Pdf(pmf_lam)
pmf_lam.Mean()
Out[18]:
In [19]:
# Solution
pmf_k = suite.Marginal(1)
thinkplot.Pdf(pmf_k)
pmf_k.Mean()
Out[19]:
Exercise: Now let's put it all together. Suppose you have 15 lightbulbs installed at different times over a 10 year period. When you observe them, some have died and some are still working. Write a version of LightBulb that takes data in the form of a (flag, x) tuple, where:
flag is eq, it means that x is the actual lifespan of a bulb that has died.flag is gt, it means that x is the current age of a bulb that is still working, so it is a lower bound on the lifespan.To help you test, I will generate some fake data.
First, I'll generate a Pandas DataFrame with random start times and lifespans. The columns are:
start: time when the bulb was installed
lifespan: lifespan of the bulb in years
end: time when bulb died or will die
age_t: age of the bulb at t=10
In [20]:
import pandas as pd
lam = 2
k = 1.5
n = 15
t_end = 10
starts = np.random.uniform(0, t_end, n)
lifespans = SampleWeibull(lam, k, n)
df = pd.DataFrame({'start': starts, 'lifespan': lifespans})
df['end'] = df.start + df.lifespan
df['age_t'] = t_end - df.start
df.head()
Out[20]:
Now I'll process the DataFrame to generate data in the form we want for the update.
In [21]:
data = []
for i, row in df.iterrows():
if row.end < t_end:
data.append(('eq', row.lifespan))
else:
data.append(('gt', row.age_t))
for pair in data:
print(pair)
In [22]:
# Solution
class LightBulb3(Suite, Joint):
def Likelihood(self, data, hypo):
lam, k = hypo
if lam == 0:
return 0
flag, x = data
if flag == 'eq':
like = EvalWeibullPdf(x, lam, k)
elif flag == 'gt':
like = 1 - EvalWeibullCdf(x, lam, k)
else:
raise ValueError('Invalid data')
return like
In [23]:
# Solution
from itertools import product
lams = np.linspace(0, 10, 101)
ks = np.linspace(0, 10, 101)
suite = LightBulb3(product(lams, ks))
In [24]:
# Solution
suite.UpdateSet(data)
Out[24]:
In [25]:
# Solution
pmf_lam = suite.Marginal(0)
thinkplot.Pdf(pmf_lam)
pmf_lam.Mean()
Out[25]:
In [26]:
# Solution
pmf_k = suite.Marginal(1)
thinkplot.Pdf(pmf_k)
pmf_k.Mean()
Out[26]:
Exercise: Suppose you install a light bulb and then you don't check on it for a year, but when you come back, you find that it has burned out. Extend LightBulb to handle this kind of data, too.
In [27]:
# Solution
class LightBulb4(Suite, Joint):
def Likelihood(self, data, hypo):
lam, k = hypo
if lam == 0:
return 0
flag, x = data
if flag == 'eq':
like = EvalWeibullPdf(x, lam, k)
elif flag == 'gt':
like = 1 - EvalWeibullCdf(x, lam, k)
elif flag == 'lt':
like = EvalWeibullCdf(x, lam, k)
else:
raise ValueError('Invalid data')
return like
Exercise: Suppose we know that, for a particular kind of lightbulb in a particular location, the distribution of lifespans is well modeled by a Weibull distribution with lam=2 and k=1.5. If we install n=100 lightbulbs and come back one year later, what is the distribution of c, the number of lightbulbs that have burned out?
In [28]:
# Solution
# The probability that any given bulb has burned out comes from the CDF of the distribution
p = EvalWeibullCdf(1, lam, k)
p
Out[28]:
In [29]:
# Solution
# The number of bulbs that have burned out is distributed Binom(n, p)
n = 100
from thinkbayes2 import MakeBinomialPmf
pmf_c = MakeBinomialPmf(n, p)
thinkplot.Pdf(pmf_c)
Exercise: Now suppose that lam and k are not known precisely, but we have a LightBulb object that represents the joint posterior distribution of the parameters after seeing some data. Compute the posterior predictive distribution for c, the number of bulbs burned out after one year.
In [30]:
# Solution
n = 100
t_return = 1
metapmf = Pmf()
for (lam, k), prob in suite.Items():
p = EvalWeibullCdf(t_return, lam, k)
pmf = MakeBinomialPmf(n, p)
metapmf[pmf] = prob
In [31]:
# Solution
from thinkbayes2 import MakeMixture
mix = MakeMixture(metapmf)
thinkplot.Pdf(mix)
In [ ]: