I was inspired by a recent paper by Boubert & Everall 2020 to consider the inference of the probability of a biased coin.

The problem is the following: given a table of $N$ coin flip experiments where you have, for each experiment, the number of times it was flipped and the number of times it showed heads, we want to infer the probability that this coin shows head.

Let’s call

• $i = 1,\ldots,N$: index of the experiments
• $n_i$: the number of times the coin was flipped in the $i$-th experiment
• $k_i$: the number of times it showed head in the $i$-th experiment
• $\theta$: the probability that this coin shows head (what we want to know).

The probability that the coin shows head $k$ times is described by a Binomial distribution, $B(n_i,\theta)$:

$$k_i \sim B(n_i,\theta)$$

For the prior of $\theta$, let’s assume uniform between 0 and 1.

The model is quite simple in stan.

# import libraries
import os, pickle
import pystan
import numpy as np
from scipy import stats
%matplotlib inline
import matplotlib.pyplot as plt
import seaborn as sns
plt.style.use('smoh')
plt.rc('figure', facecolor='w')

cache_pickle = 'model.pkl'
if not os.path.exists(cache_pickle):
model = pystan.StanModel(model_code='''
data {
// number of coins
int<lower=0> N;
// number of flips
int n[N];
int k[N];
}
parameters {
real<lower=0,upper=1> theta;
}
model {
for (i in 1:N) {
k[i] ~ binomial(n[i], theta);
}
}
''')
with open(cache_pickle, 'wb') as f:
pickle.dump(model, f)
else:

reading model from disk


We can simulate data with scipy.stats.binom. If we have no truncation on the data, different experiments do not matter. If the coin was flipped 20 times in a hundred experiment, it is the same as one experiment of flipping it 2000 times.

theta = 0.63
N = 100
n_i = np.ones(N, dtype=int)*20
k_i = stats.binom.rvs(n=n_i, p=theta)
data = dict(N=N, n=n_i, k=k_i)

# combine all experiment to one
data_one = dict(N=1, n=[n_i.sum()], k=[k_i.sum()])

ropt = model.optimizing(data=data)
ropt_one = model.optimizing(data=data_one)
print('theta from {} experiments of {} flips = {:.3f}'.format(N, n_i,ropt['theta']))
print('theta from {} experiments of {} flips = {:.3f}'.format(1, n_i.sum(),ropt['theta']))

theta from 100 experiments of 20 flips = 0.638
theta from 1 experiments of 2000 flips = 0.638


In fact, the uniform distribution is a special case of Beta distribution $\mathrm{Beta}(\alpha,,\beta)$ when $\alpha=\beta=1$. Because Beta distributions are conjugate priors of Binomial distribution, the posterior distribution is also a Beta distribution with $\mathrm{Beta}(\alpha+k,,\beta+n-k)$ where $n=\Sigma n_i$ and $k=\Sigma k_i$. The posterior mean is $\frac{\alpha+k}{\alpha+\beta+n}$, which approaches $\frac{k}{n}$ as $n \to \infty$, as it should intuitively.

We can check that stan posterior sample matches this expectation.

r1 = model.sampling(data=data)
r2 = model.sampling(data=data_one)

fig, ax = plt.subplots(figsize=(4,3))
sns.kdeplot(r1['theta'], label='$(N, n_i)=(100,20)$',ax=ax)
sns.kdeplot(r2['theta'], label='$(N, n_i)=(1,2000)$', ax=ax)

x = np.linspace(*ax.get_xlim(), 64)
pdfx= stats.beta(1+k_i.sum(), 1+n_i.sum()-k_i.sum()).pdf(x)
ax.plot(x, pdfx, 'k-', label=r'$\mathrm{Beta}(\alpha+k,\,\beta+n-k)$');

ax.axvline(theta, lw=1)
ax.legend(loc='upper right', fontsize=12);
ax.set(ylim=(0,70), xlabel=r'$\theta$', ylabel='pdf'); Now, what if it turns out that the experiment is not reported at all when the number of times it showed head is less than $k_t$ times? As explained in Boubert & Everall 2020, we need to adjust the likelihood function by dividing it by $$\int_{k_t}^n p(k |n,\theta)$$ (so that it will integrate to 1 from $k_t$ not $1$).

Stan supports truncated data for 1D distributions so the modification is simple:

cache_pickle = 'modelth.pkl'
if not os.path.exists(cache_pickle):
modelth = pystan.StanModel(model_code='''
data {
// number of coins
int<lower=0> N;
// number of flips
int n[N];
int k[N];
// truncation threshold
int kthresh;
}
parameters {
real<lower=0,upper=1> theta;
}
model {
for (i in 1:N) {
k[i] ~ binomial(n[i], theta) T[kthresh,];
}
}
''')
with open(cache_pickle, 'wb') as f:
pickle.dump(modelth, f)
else:

reading model from disk


Now, because a record of an experiment might be missing, it is no longer valid to treat the sum of experiments as one ‘master’ experiment.

What do we expect for our inference of $\theta$ if we use the truncated data yet use the old (wrong) model that does not account for missing data? The data will be biased towards high $k_i$ and this will bias $\theta$ high as well.

Let’s simulate this truncated data as well and see if this is the case.

theta = 0.63
N = 100
n_i = np.ones(N, dtype=int)*10
k_i = stats.binom.rvs(n=n_i, p=theta)
k_t = 5

# repeat until i have N experiments replacing truncated
while (k_i<k_t).sum()>0:
data = dict(N=N, n=n_i, k=k_i,)

replacing 20
replacing 2

r1 = model.sampling(data=data)
data['kthresh'] = k_t
r2 = modelth.sampling(data=data)

fig, ax = plt.subplots(figsize=(4,3))
sns.kdeplot(r1['theta'], label='old model',ax=ax)
sns.kdeplot(r2['theta'], label='truncated data model', ax=ax)

ax.axvline(theta, lw=1)
ax.legend(loc='upper left', fontsize=12);
ax.set(ylim=(0,40), xlabel=r'$\theta$', ylabel='pdf'); Indeed, the old model that does not account for missing data infers a higher $\theta$ than we should whereas the new model recovers the correct value.

This will be more prounounced as we repeat more experiments and the posterior distribution gets sharper.

rs_old = []
rs_new = []
for N in [100, 1000]:
n_i = np.ones(N, dtype=int)*10
k_i = stats.binom.rvs(n=n_i, p=theta)
k_t = 5

# repeat until i have N experiments replacing truncated
while (k_i<k_t).sum()>0:
data_tmp = dict(N=N, n=n_i, k=k_i,)

%time r1tmp = model.sampling(data=data_tmp)
data_tmp['kthresh'] = k_t
%time r2tmp = modelth.sampling(data=data_tmp)
rs_old.append(r1tmp)
rs_new.append(r2tmp)

CPU times: user 91.2 ms, sys: 447 ms, total: 538 ms
Wall time: 715 ms
CPU times: user 102 ms, sys: 425 ms, total: 526 ms
Wall time: 1.67 s
CPU times: user 95.7 ms, sys: 440 ms, total: 536 ms
Wall time: 1.17 s
CPU times: user 134 ms, sys: 466 ms, total: 600 ms
Wall time: 9.09 s

fig, ax = plt.subplots(figsize=(4,3))
for i, (rr1, rr2) in enumerate(zip(rs_old, rs_new)):
sns.kdeplot(rr1['theta'], c='C0', lw=1 if i>0 else 2)
sns.kdeplot(rr2['theta'], ax=ax, c='C1',lw=1 if i>0 else 2)

ax.axvline(theta, lw=1)
ax.set(xlabel=r'$\theta$', ylabel='pdf'); The correct model converges to the true value while the wrong model will bias $\theta$ higher.

Note that if $\theta n_i$ (the expectation value of $k_i$ for $i$-th experiment) is much larger than the threshold $k_t$, than the threshold is not going to filter out much data and it would not matter if we use the corret model that accounts for the missing data or not.

What is acutally happening here?

The posterior distribution is

$$p(\theta| {n_i,k_i}) \propto \prod _{i=1}^{N} p (k_i |n_i,,\theta) p(\theta)$$ where $$p (k_i| n_i, \theta) = \frac{1}{1 - p(k_i \lt 5 | n_i, \theta)} {n \choose k} \theta^k \theta^{n-k}$$

Assuming uniform prior, $p(\theta) = 1$ between 0 and 1, let’s have a look at what the normalization factor that applies the data truncation does.

fig, (ax1,ax2) = plt.subplots(1,2, figsize=(8,3))
thetas = np.linspace(0.01,0.99,201)
pdfs=[]
n_flip = 10

for k in np.arange(k_t, 11):
ax1.plot(thetas, stats.binom.pmf(k, n_flip, thetas), '-',lw=1,
c=plt.cm.plasma((k-k_t)/(n_flip-k_t)));

I = 1-stats.binom.cdf(k_t-1 , n_flip, thetas)
ax1.plot(thetas, I, c='k', label=r'$1 - p(k_i < 5\,|\,n_i,\,\theta)$')
ax1.legend(loc='upper left', fontsize=12)
ax1.set(ylim=(None, 1.2))
for k in np.arange(k_t, 11):
ax2.plot(thetas, stats.binom.pmf(k, 10, thetas)/I, '-', lw=2,
c=plt.cm.plasma((k-k_t)/(n_flip-k_t)), label='k={}'.format(k));

ax2.legend(loc=(0.2,0.6), fontsize=12, ncol=2)
ax2.set(ylim=(None, 1.6))
for cax in (ax1, ax2):
cax.set(xlabel=r'$\theta$') For each possible $k_i \ge k_t = 5$ upto $n_i=10$, the plot above shows the usual Binomial probability mass function (pmf) on the left. The normalization factor is the reciprocal of the black line, which shows $1 - p(k_i \lt 5 | n_i, \theta)$. After the correction is applied, each pmf on the left of the same color is changed to that on the right.

In essense, the correction compensates for the fact that the observation of $k_i$ heads may be statistical fluctuation above the mean at lower $\theta$. We see that if $\theta$ is large enough to put typical number of $k_i$ above the threshold, the correction disappears, i.e., it goes to 1, and it is most dramatic for low $\theta$ at $k_i$ near the threshold.

Since we know the shape of the pdf for each $i$, we can multiply all pdf to get the posterior (without normalization).

pdfs=[]
for n_flip, k in zip(data['n'], data['k']):
I = 1-stats.binom.cdf(k_t-1, n_flip, thetas)
pdfs.append(stats.binom.pmf(k, n_flip, thetas)/I)

a=np.log(np.array(pdfs)).sum(axis=0)
imax = a.argmax()
# np.exp(-9)=0.0001234...; ignore ranges where pdf falls below this time the peak value
ii = a-a[imax]>-9
a-=a.max()   # arbitrary scaling...
a=np.exp(a)
a[~ii]=0

fig, ax =plt.subplots(figsize=(4,3))
plt.plot(thetas, a*20, label=r'$\propto \prod_i \, p(k_i,\,|\,n_i,\,\theta)$')
sns.kdeplot(r2['theta'], label='from stan')
plt.axvline(theta)
plt.legend(fontsize=12);
plt.xlabel(r'$\theta$'); The two match as they should.

If there is an absolute threhold $k>k_t$, this will affect experiments where you expect $k$ to be near $k_t$ whereas large-$n_i$ experiments will not be affected.

Let’s carry out a comparison to explore this: collect 1000 experiments in total, but

• in case 1: throw 10 times in 100 experiments
• in case 2: throw 100 times in 10 experiments

and set the threshold at $k_t = 5$ with $\theta=0.2$ (Deliberately chosen to see the effect of data truncation).

Although the total number of flips is the same, there is much higher chance that all experiments in case 1 will not be reported as we would typically expect $n_i \theta = 2$ heads with some spread.

def simulate(theta, N, n, k_t):
'''
Throw n flips each time.
Threshold at k>=k_t.
Fill until I have N records.
'''
n_i = np.ones(N, dtype=int)*n
k_i = stats.binom.rvs(n=n_i, p=theta)

# repeat until i have N experiments replacing truncated
while (k_i<k_t).sum()>0:
print("theta={:.3f} N={:5d} n={:5d} k_t={:3d}: replaced {}".format(
return dict(N=N, n=n_i, k=k_i,kthresh=k_t)

theta=0.2
data1 = simulate(theta, 100, 10, 5)
data2 = simulate(theta, 10, 100, 5)
data3 = simulate(theta, 500, 10, 5)

theta=0.200 N=  100 n=   10 k_t=  5: replaced 3228
theta=0.200 N=   10 n=  100 k_t=  5: replaced 0
theta=0.200 N=  500 n=   10 k_t=  5: replaced 13591


Indeed, to collect 100, there were >2500 experiments that were truncated because $k_i$ did not make the threshlold.

result = {}
result['$(N,\,n_i)$ = (100, 10)'] = modelth.sampling(data=data1)
result['$(N,\,n_i)$ = (10, 100)'] = modelth.sampling(data=data2)
result['$(N,\,n_i)$ = (500, 10)'] = modelth.sampling(data=data3)

fig, ax = plt.subplots(figsize=(5,4))
for k, v in result.items():
sns.kdeplot(v['theta'], label=k, ax=ax)
ax.axvline(theta, c='k', lw=1);
ax.legend(loc='upper right', fontsize=12);
ax.set(xlabel=r'$\theta$'); As expected, in this case it’s better to have $(N, n_i)=(10, 100)$ than $(N, n_i)=(100, 10)$ although the total number of flips is the same. However, even in the case where each experiment is at the risk of being discarded due to data truncation, if we have enough data and we account for the data truncation, we can still recover more $\theta$ more precisely: we start to recover $\theta$ with comparable confidence interval when 10-flip experiment records are collected 5 times more ($N=500$).

This cross-over between large-$N$-small-$n_i$ and small-$N$-large-$n_i$ depends on values of $\theta$ and $k_t$.

Tags: Statistics