Past exam of the mathematics course of the University of Cambridge / 2013 / ia / Paper 2 / 12F (Probability)

By the partition and the law of total probability, $\mathbb{P}(A)={\sum }_j\mathbb{P}(A\cap B_j)={\sum }_j\mathbb{P}(A\mid B_j)\mathbb{P}(B_j)$. Also $\mathbb{P}(B_i\mid A)=\mathbb{P}(A\cap B_i)/\mathbb{P}(A)$. Combining these identities proves Bayes theorem:The assumed positive probabilities make the conditional expressions well defined.

Let $B$ mean that the selected coin is biased and $F$ that it is fair. Both have prior probability $1/2$. Conditional on either coin, tosses are independent. For the observed head count, the two binomial distribution likelihoods are $\left(\genfrac{}{}{0ex}{}{n}{k}\right)p^k(1-p{)}^{n-k}$ and $\left(\genfrac{}{}{0ex}{}{n}{k}\right)2^{-n}$. Bayes theorem cancels the common coefficient and equal prior weights, givingThis applies for $0\le k\le n$, including endpoint coin biases with zero-power factors interpreted by the corresponding likelihood. The fair-coin likelihood is positive, so the denominator cannot vanish.

For a finite stopping time $n$ with target $k\ge 1$, necessarily $n\ge k$. The last toss must be a head and the preceding $n-1$ tosses must contain $k-1$ heads. The negative binomial stopping argument therefore gives likelihood $\left(\genfrac{}{}{0ex}{}{n-1}{k-1}\right)p^k(1-p{)}^{n-k}$ for the biased coin, and $\left(\genfrac{}{}{0ex}{}{n-1}{k-1}\right)2^{-n}$ for the fair coin. Applying Bayes theorem again yieldsThe two experiments have different combinatorial coefficients, but each coefficient is independent of the coin parameter. Thus posterior equality for proportional likelihoods explains their identical posterior formula. If $p=0$, a finite positive-head stopping observation has zero likelihood under the biased coin, consistently giving posterior zero. If the target is $k=0$, the experiment stops at $n=0$ without a toss and the posterior remains $1/2$; a positive stopping time would then be impossible.