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import numpy as np
import matplotlib.pyplot as plt
%matplotlib inline
import matplotlib_inline
matplotlib_inline.backend_inline.set_matplotlib_formats('svg')
import seaborn as sns
sns.set_context("paper")
sns.set_style("ticks");

Example: Inferring the probability of a coin toss from data#

We toss a coin with an unknown probability of heads \(\theta\) \(N\) times independently, and we observe the result:

\[ x_{1:N} := (x_1,\dots,x_N). \]

Assume that we have coded the result so that heads correspond to a “1” and tails to a “0.” We aim to estimate the probability of heads \(\theta\) from this dataset.

Assuming that we know nothing, we set:

\[ \theta \sim U([0,1]). \]

In terms of probability densities, this is:

\[\begin{split} p(\theta) = \begin{cases} 1,&\;\text{if}\;\theta\in [0,1].\\ 0,&\;\text{otherwise} \end{cases} \end{split}\]

Now, let’s write down the likelihood of the data. Because of the independence assumption, we have:

\[ p(x_{1:N}|\theta) = \prod_{n=1}^Np(x_n|\theta). \]

Then, each measurement is a Bernoulli with probability of success \(\theta\), i.e.,

\[ x_n | \theta \sim \operatorname{Bernoulli}(\theta). \]

In terms of probability densities, we have the likelihood:

\[\begin{split} p(x_n|\theta) = \begin{cases} \theta,&\;\text{if}\;x_n=1,\\ 1-\theta,&\;\text{otherwise}. \end{cases} \end{split}\]

Using a common mathematical trick, we can rewrite this as:

\[ p(x_n|\theta) = \theta^{x_n}(1-\theta)^{1-x_n}. \]

Work out the cases \(x_n=0\) and \(x_n=1\) to convince yourself.

Now, we can find the expression for the likelihood of the entire dataset. It is:

\[\begin{split} \begin{split} p(x_{1:N}|\theta) &= \prod_{n=1}^Np(x_n|\theta)\\ &= \prod_{n=1}^N\theta^{x_n}(1-\theta)^{1-x_n}\\ &= \theta^{\sum_{n=1}^Nx_n}(1-\theta)^{N-\sum_{n=1}^Nx_n}. \end{split} \end{split}\]

This intuitively means the probability of getting \(\sum_{n=1}^Nx_n\) heads and the rest \(N-\sum_{n=1}^Nx_n\) tails.

We can now find the posterior. It is:

\[ \text{posterior} \propto \text{likelihood}\times\text{prior}. \]

In our problem:

\[\begin{split} \begin{split} p(\theta|x_{1:N}) &\propto p(x_{1:N}|\theta)p(\theta)\\ &= \begin{cases} \theta^{\sum_{n=1}^Nx_n}(1-\theta)^{N-\sum_{n=1}^N},&\;\text{if}\;\theta\in[0,1]\\ 0,&\;\text{otherwise}. \end{cases} \end{split} \end{split}\]

And this is just the density corresponding to a Beta distribution:

\[ p(\theta|x_{1:N}) = \operatorname{Beta}\left(\theta\middle|1 + \sum_{n=1}^Nx_n, 1 + N - \sum_{n=1}^Nx_n\right). \]

Let’s try this out with some fake data. Take a fake coin which is a little bit biased:

import scipy.stats as st
theta_true = 0.8
X = st.bernoulli(theta_true)

Sample from it a number of times to generate our data = (x1, …, xN):

N = 5
data = X.rvs(size=N)
data

array([0, 1, 1, 1, 1])

Now we are ready to calculate the posterior which the Beta we have above:

alpha = 1.0 + data.sum()
beta = 1.0 + N - data.sum()
Theta_post = st.beta(alpha, beta)

And we can plot it:

fig, ax = plt.subplots()
thetas = np.linspace(0, 1, 100)
ax.plot(
    [theta_true],
    [0.0],
    'o',
    markeredgewidth=2,
    markersize=10,
    label='True value'
)
ax.plot(
    thetas,
    Theta_post.pdf(thetas),
    label=r'$p(\theta|x_{1:N})$'
)
ax.set_xlabel(r'$\theta$')
ax.set_ylabel('Probability density')
ax.set_title(f'$N={N}$')
plt.legend(loc='best', frameon=False)
sns.despine(trim=True);
../_images/0c06e9dafb44fa9ac66702954be7dac20410d28c67c3096bfbede1a245c0a9c6.svg

Questions#

  • Try \(N=0,5,10,100\) and see what happens.