Q21. Evidence and posterior model probabilities
[4 marks]
An analysis produces a Bayes factor of 4 in favour of model M_1 over model M_0. Before observing the data, the researcher regarded M_0 as nine times as probable as M_1. The researcher concludes:
“The Bayes factor favours M_1, so M_1 must now have the greater posterior probability.”
Assess this conclusion. Explain how the Bayes factor changes the model odds, and identify which model has the greater posterior probability.
Definitions. The Bayes factor is B_{10}=p({\boldsymbol{x}}\mid M_1)/p({\boldsymbol{x}}\mid M_0). Odds in favour of M_1 are \mathbb P(M_1)/\mathbb P(M_0), with the analogous conditional ratio for posterior odds. The two models exhaust the comparison, so their probabilities sum to one. Here B_{10}=4 and the prior odds are 1:9.
A Bayes factor updates prior model odds according to \frac{\mathbb{P}(M_1\mid{\boldsymbol{x}})}{\mathbb{P}(M_0\mid{\boldsymbol{x}})} =B_{10} \frac{\mathbb{P}(M_1)}{\mathbb{P}(M_0)}. Because M_0 was initially nine times as probable as M_1, the prior odds in favour of M_1 are \frac{\mathbb{P}(M_1)}{\mathbb{P}(M_0)}=\frac19. With B_{10}=4, the posterior odds become \boxed{ \frac{\mathbb{P}(M_1\mid{\boldsymbol{x}})}{\mathbb{P}(M_0\mid{\boldsymbol{x}})} =4\left(\frac19\right)=\frac49. } Therefore, \mathbb{P}(M_1\mid{\boldsymbol{x}})=\frac4{13}, \qquad \mathbb{P}(M_0\mid{\boldsymbol{x}})=\frac9{13}. The data favour M_1 relative to the prior because B_{10}>1, but the evidence is not strong enough to overcome the initial nine-to-one odds in favour of M_0. Thus \boxed{ \mathbb{P}(M_0\mid{\boldsymbol{x}})> \mathbb{P}(M_1\mid{\boldsymbol{x}}). } The researcher’s conclusion is therefore incorrect: a Bayes factor favouring M_1 does not necessarily make M_1 the more probable model a posteriori.