# BDA3 Chapter 2 Exercise 2

Here’s my solution to exercise 2, chapter 2, of Gelman’s Bayesian Data Analysis (BDA), 3rd edition. There are solutions to some of the exercises on the book’s webpage.

We are given the following information about the two coins.

\begin{align} p(C_1) &= 0.5 & p(H \mid C_1) &= \dbern(H \mid 0.6) \\ p(C_2) &= 0.5 & p(H \mid C_2) &= \dbern(H \mid 0.4) \end{align}

The posterior probability of each coin given two tails is:

\begin{align} p(C_1 \mid TT ) &\propto p(TT \mid C_1) \cdot p(C_1) \\ &= \left(\frac{2}{5}\right)^2 \frac{1}{2} \\ &= \frac{2}{25} \end{align} \begin{align} p(C_2 \mid TT ) &\propto p(TT \mid C_2) \cdot p(C_2) \\ &= \left(\frac{3}{5}\right)^2 \frac{1}{2} \\ &= \frac{9}{50} \end{align}

Both of the previous probabilities are normalised by the same constant. Since $$p(C_1 \mid TT) + p(C_2 \mid TT) = 1$$, the normalising constant is $$\frac{2}{25} + \frac{9}{50} = \frac{13}{50}$$. Thus

$p(C_1 \mid TT) = \frac{4}{13} \qquad \text{and} \qquad p(C_2 \mid TT) = \frac{9}{13}.$

Let $$y$$ be the number of additional spins until the next head. Conditional on a coin, $$y$$ is geometrically distributed. So the expected number of spins before the next head is:

\begin{align} \mathbb E(y \mid TT) &= \frac{4}{13}\mathbb E(y \mid C_1) + \frac{9}{13}\mathbb E(y \mid C_1) \\ &= \frac{4}{13}\frac{5}{3} + \frac{9}{13}\frac{5}{2} \\ &= \frac{20}{39} + \frac{45}{26} \\ &= \frac{175}{78}, \end{align}

which is 2.24359.