= Solution
Integrating the positive component's <normal distribution> prior gives its <prior predictive distribution>:
$$
\boxed{Y_i\mid+,V\sim N(0,1+V).}
$$
The negative component gives $Y_i\mid-\sim N(0,1)$. Let $f_1(y)=\phi_{1+V}(y)$ and $f_0(y)=\phi_1(y)$ denote those centred normal densities. The <point-null mixture prior> assigns component odds $q/(1-q)$, so the <posterior odds> of the positive component are
$$
\boxed{O_i(y_i)=\frac q{1-q}\frac1{\sqrt{1+V}}
\exp\left\{\frac{Vy_i^2}{2(1+V)}\right\}.}
$$
Its <posterior probability> is $O_i/(1+O_i)$. Evidence depends on $y_i^2$, so large effects of either sign favour the positive component. This is the marginal distribution before observing $y_i$, not the posterior predictive law for a second observation conditional on it.
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