I'm given two models:
$M_1: X \sim N(x|\theta_0,1)$
$M_2: X \sim N(x|\theta,1) \land \pi(\theta|M_2) = N(\theta|\theta_0,1)$
Assume that, $\pi(M_1) = \pi(M_2) = 1/2$.
I need to compute the Bayes factor and study its asymptotic behaviour.
$B_{21}(x) = \frac{m(x|M_1)}{m(x|M_2)}$
Clearly, $m(x|M_1) = \frac{1}{(\sqrt{2\pi})^n} e^{-n(\overline{x}-\theta_0)^2/2} e^{-nS^2/2}$
$$m(x|M_2) = \int_\mathbb{R} f(x|\theta,M_2) \pi(\theta|M_2) d \theta = \int_\mathbb{R} N(x|\theta,1) N(\theta|\theta_0,1) d\theta = \frac{e^{-nS^2/2}}{(\sqrt{2\pi})^{n+1}}\int_\mathbb{R} e^{-n(\overline{x} - \theta)^2/2}e^{-(\theta-\theta_0)^2/2}d\theta$$
But how do can I integrate this term?
A simpler approach may be related with this question.
Assuming the formula in link
The Bayes factor assuming the formula in the link would be $B_{21}(x) = \frac{1}{\sqrt{n+1}}exp\Big(\frac{n^2(\overline{x}-\theta_0)^2}{2(n+1)}\Big)$ and it remains to study how does this distribution behave asymptotically for the different assumed models.
Under the assmuption that $M_1$ holds one has that $\sqrt{n} (\overline{x} - \theta_0) \to N(0,1/I(\theta_0))$ by assymptotic normality of the m.l.e. Therefore, the square converges to a $\chi_1^2$ and the bayes factor goes to zero. For the other assumption one uses the central limit theorem in the Lindeberg's-Levy formulation and finds that the bayes factor goes to $\infty$.
This gives the consistency of the test.