 Solution

Let $p_0=\mathrm{Pr}(\theta \in {\Theta }_0)$ and $p_1=1-p_0$, assuming both are positive. The prior odds are $\alpha =p_0/p_1$. Given data $x$, the posterior odds areWith conditional joint likelihood $L(\theta ;x)={\prod }_{i}f(x_i\mid \theta )$, define the hypothesis-conditional marginal densitiesThe Bayes factor for $H_0$ relative to $H_1$ is $B=m_0(x)/m_1(x)$. The normalizations $p_0,p_1$ matter: the prior distribution within each hypothesis must have total probability one. Bayes' theorem now givesConverting posterior odds to posterior probability therefore gives