 Solution

Put ${\kappa }_i=n/(n+q_i)$. The posterior mean of the Gaussian practical-null mixture isNear zero, $B_{01}(0)>1$, so the narrow component has high posterior probability and ${\kappa }_0$ is small when $n_0\gg n$. This pulls the Bayesian model averaging posterior toward zero. For the given ratios, writing $z=\sqrt{n}y$ givesThus the order statement $y=O(n^{-1/2})$ alone does not guarantee strong pull toward zero: $z=3$ already has the opposite model preference.

As $|y|$ grows, $V_1>V_0$ makes $B_{01}(y)\to 0$, and the mixture model approaches $N({\kappa }_{1}y,1/(n+q_1))$. Its center is exactly ${\kappa }_{1}y$, and approximately $y$ only for a sufficiently diffuse wide prior, meaning $q_1\ll n$. Here ${\kappa }_1=100/101$, so the relative displacement is about one percent. Large observations alone do not remove this finite-prior shrinkage.