Flipping a Coin

A few weeks ago, I was exposed to my first large data set. It made me want to learn some statistical theory.

Suppose that you have a coin. You flip it \(n\) times and record the number of heads. What can you say about the coin?

The Frequentist Approach

The coin has some fixed unknown probability \(\phi\) of landing heads. Consider the following experiment: flip the coin \(n\) times. Let \(H\) be the number of heads. We want to estimate bounds on \(\phi\). One way to do this is by finding functions \(f,g\) such that \[{\bf P}(\phi \in [f(H),g(H)]) \geq 0.95 \qquad \text{for all $\phi$}\] In English, this says that we are 95% sure that \(\phi\) lies between \(f(H)\) and \(g(H)\). We call \([f(H),g(H)]\) a confidence interval. The value \(0.95\) was chosen to make things more concrete. It follows from Hoeffding's Inequality that if \(X_1,\dots,X_n \sim {\rm Bernoulli}(\phi)\) are independent, then for any \(\epsilon > 0\), we have \[{\bf P}\left( \left| \frac{\sum X_i}{n} - \phi \right| > \epsilon \right) \leq 2 e^{-2n \epsilon^2}\] which implies that for any \(\epsilon > 0\), \[{\bf P} \left( \phi \in \left[ \frac{\sum X_i}{n} - \epsilon, \frac{\sum X_i}{n} + \epsilon \right] \right) \geq 1 - 2 e^{-2n\epsilon^2}.\] If we set \(\epsilon = \sqrt{log(40)/2n}\), then we get \[{\bf P} \left( \phi \in \left[ \frac{\sum X_i}{n} - \sqrt{log(40)/2n}, \frac{\sum X_i}{n} + \sqrt{log(40)/2n} \right] \right) \geq 0.95.\] Since \(H = \sum X_i\), we are 95% sure that \(\phi\) lies between \(H / n - \sqrt{log(40)/2n}\) and \(H / n + \sqrt{log(40)/2n}\). There are many other ways to calculate confidence intervals. We have chosen to use Hoeffding's Inequality because it does not place any restrictions on the sample size.

The Bayesian Approach

Write \(\Phi\) for the probability that the coin lands heads. Since we do not know the value of \(\Phi\), we consider it a random variable. We have no prior information about \(\Phi\), so our initial guess is that \(\Phi\) is uniformly distributed: the probability density function is \(f_{\Phi}(\phi) = 1\).

Suppose that we flip the coin \(n\) times and observe \(r\) heads. Call this event \(K\). We now have more information, so we should update our probability density function. We do this using Bayes law. Let \(h\) be very small. Then

\[{\bf P}(\phi \leq \Phi \leq \phi + h | K) = \frac{{\bf P}(K | \phi \leq \Phi \leq \phi + h) {\bf P}(\phi \leq \Phi \leq \phi + h)}{{\bf P}(K)}.\]

We can compute each term on the right hand side as follows: \[{\bf P}(K | \phi \leq \Phi \leq \phi + h) = \phi^r (1 - \phi)^{n-r}\] \[{\bf P}(\phi \leq \Phi \leq \phi + h) =f_{\Phi}(\phi)h = h\] \[{\bf P}(K) = \sum_i {\bf P}(K | \phi_i \leq \Phi \leq \phi_{i+1}) {\bf P}(\phi_i \leq \Phi \leq \phi_{i+1}) = \sum_i \phi_i^r (1 - \phi_i)^{n-r} \Delta \phi_i = \int_0^1 \phi^r(1 - \phi)^{n-r} d \phi\] This implies that \[f_{\Phi | K}(\phi) = \lim_{h \to 0} \frac{{\bf P}(\phi \leq \Phi \leq \phi + h | K)}{h} = \frac{\phi^r(1-\phi)^{n-r}}{\int_0^1 \phi^r(1 - \phi)^{n-r} d \phi}\] which looks like

As you can see, the new distribution is more concentrated around \(r/n\).

Discussion

As a mathematician, it is not my place to argue for either the Frequentist or Bayesian approach to this problem. As a human being, I favor the Bayesian approach. Less mental gymnastics is required to interpret pictures of density functions than is required to interpret statements like "we are 95% sure that \(\phi\) is between \(f(H)\) and \(g(H)\)". Another benefit of the Bayesian approach is that it suggests how to store unknown values in a computer: Consider them as random variables and store the density function. As you learn more information, the density function is updated.