Bayes formula for a finite mixture with continuous observation

For j=0 the integral of $g_0$ is one by [F2]. For j=1, the translation $t=y-1$ has derivative one; [F6] with continuous outer phi, then [F8] and [F7] on increasing compact intervals, gives $\int g_1=\int \varphi =1$. The same calculation on $(-\mathrm{\infty },z]$ gives ${\int }_{-\mathrm{\infty }}^z{g}_j={\int }_{-\mathrm{\infty }}^{z-j}\varphi$, the CDF of the translated standard law [F3]. Therefore [F9] identifies it as $N(j,1)$. Both functions are positive measurable and finite. By [F5] they define probability likelihoods; measurability in the discrete parameter j is automatic because the parameter set is finite. The joint density $(j,y)\mapsto g_j(y)$ is measurable since each of its two sections is, and the two slices are measurable. The declared AC [F4] covers all analytic choice assumptions.

Take prior $\pi =p{\delta }_0+(1-p){\delta }_1$ on the two-point space and observation Lebesgue measure, which is sigma-finite. In [F1] the marginal density is $m(y)=p{g}_0(y)+(1-p)g_1(y)>0$ and finite for every real y. Its total mass is $p+(1-p)=1$ by step 1.1. The posterior mass of {1} is therefore the first displayed ratio, and that of {0} is $1-w(y)$; these sum to one. Dividing numerator and denominator by $g_0(y)>0$ and computing $g_1(y)\mathrm{/}{g}_0(y)=\exp ((y^2-(y-1{)}^2)\mathrm{/}2)=e^{y-1\mathrm{/}2}$ gives the second ratio. At $y=1\mathrm{/}2$ the exponential is one, so $w(1\mathrm{/}2)=1-p$. For p=1/2, for example, $w(0)=1\mathrm{/}(1+e^{1\mathrm{/}2})$ and $w(1)=e^{1\mathrm{/}2}\mathrm{/}(1+e^{1\mathrm{/}2})$. Finally for every Borel B, ${\int }_{B}w(y)m(y)dy=(1-p){\int }_B{g}_1(y)dy=P(J=1,Y\in B)$, directly verifying the posterior event calculation.