Q44. Testing whether a coin is fair
[10 marks]
Suppose X\mid\theta\sim\operatorname{Binomial}(n,\theta). Compare M_0:\theta=\frac12, \qquad M_1:\theta\sim\operatorname{Beta}(a,b), with a,b>0 and equal prior model probabilities.
You may use B(a,b)=\int_0^1t^{a-1}(1-t)^{b-1}\,dt.
(a) Derive \mathbb{P}(X=x\mid M_0) and \mathbb{P}(X=x\mid M_1). [4 marks]
(b) Obtain the Bayes factor BF_{01} and the posterior probability \mathbb{P}(M_0\mid X=x). [3 marks]
(c) A student argues:
“Under a continuous Beta posterior, \mathbb{P}(\theta=1/2\mid x)=0, so Bayesian inference always rejects a fair coin.”
Explain the error. [3 marks]
Definitions. M_0 fixes \theta=1/2 and M_1 assigns a \operatorname{Beta}(a,b) prior, with a,b>0. Its density is \theta^{a-1}(1-\theta)^{b-1}/B(a,b), where B(a,b)=\int_0^1u^{a-1}(1-u)^{b-1}\,du. BF_{01} is the marginal-likelihood ratio favouring M_0. Prior model probabilities are equal; a model probability is distinct from a point probability within the continuous alternative.
(a) Marginal likelihoods. Under M_0, \theta=1/2 is fixed, so \boxed{\mathbb{P}(X=x\mid M_0)=\binom nx2^{-n}.} Under M_1, average the likelihood over the Beta prior: \begin{aligned} \mathbb{P}(X=x\mid M_1) &=\int_0^1\binom nx\theta^x(1-\theta)^{n-x} \frac{\theta^{a-1}(1-\theta)^{b-1}}{B(a,b)}\,d\theta\\ &=\frac{\binom nx}{B(a,b)} \int_0^1\theta^{a+x-1}(1-\theta)^{b+n-x-1}\,d\theta\\ &=\boxed{\binom nx\frac{B(a+x,b+n-x)}{B(a,b)}}. \end{aligned}
(b) Bayes factor and posterior model probability. The Bayes factor in favour of M_0 over M_1 is \boxed{ BF_{01} =\frac{\mathbb{P}(X=x\mid M_0)}{\mathbb{P}(X=x\mid M_1)} =2^{-n}\frac{B(a,b)}{B(a+x,b+n-x)}. } The binomial coefficients cancel in the ratio, but belong in the individual marginal likelihoods. Equal prior model probabilities give prior odds one, so posterior odds equal BF_{01}. Hence \boxed{ \mathbb{P}(M_0\mid X=x) =\frac{BF_{01}}{1+BF_{01}} =\frac{2^{-n}B(a,b)}{2^{-n}B(a,b)+B(a+x,b+n-x)}. }
(c) Error in the student’s argument. Under M_1 alone, \theta\mid x,M_1\sim\operatorname{Beta}(a+x,b+n-x), \qquad \mathbb{P}(\theta=1/2\mid x,M_1)=0, because this posterior is continuous. Every individual point has probability zero under it; that fact is not a test rejecting that particular value.
The model comparison instead assigns positive prior probability to the separate point-null model M_0. Both marginal likelihoods are positive for valid data, so \mathbb{P}(M_0\mid x)>0. In the combined model, \mathbb{P}(\theta=1/2\mid x)=\mathbb{P}(M_0\mid x), since M_0 places all its probability at 1/2 and M_1 contributes no point mass there. The student has confused a probability conditional on M_1 with the probability after allowing both models.