Bayes factorization of the smoothing backward kernel

Fix $t<T$ and condition on $Y_{1:t}=y_{1:t}$. The conditional pair law of $(X_t,X_{t+1})$ is the ordered kernel integral ${\eta }_t(dx)M_{t+1}(x,d{x}^{\mathrm{\prime }})$ by [F3] and [F5]. In the density case this pair law has density $f_{t+1}(x^{\mathrm{\prime }}\mid x)$ with respect to ${\eta }_t\times {\lambda }_{t+1}$, and its second marginal has density $q_{t+1}(x^{\mathrm{\prime }})$. Applying [F4] gives the displayed $B_t$ on $D_{t+1}$ and the supplied filler ${\eta }_t$ off it. The conditional-density result proves that $B_t$ is an everywhere probability kernel and the conditional law of $X_t$ given $(X_{t+1},y_{1:t})$ on its predictive support.

In the full path factorization [F5], after fixing $X_{t+1}=x^{\mathrm{\prime }}$ and the observations through time $T$, every factor after time $t+1$ depends only on $x^{\mathrm{\prime }}$ and the future states. Those factors cancel from the conditional distribution of $X_t$; the observation factor at time $t+1$ is also constant in $X_t$. The remaining factors are exactly the conditional pair law in step 1.1, so the conditional distribution of $X_t$ given $(X_{t+1:T},Y_{1:T})$ equals $B_t(X_{t+1},\cdot )$ almost surely. With a general transition kernel the same cancellation leaves the supplied regular conditional kernel, without evaluating a density or particle-point singleton.

Apply the conditional chain rule successively from $t=T-1$ down to $0$. Step 2.1 identifies each conditional factor with $B_t(X_{t+1},\cdot )$, while the terminal factor is the filtering law ${\eta }_T$. Integrating first over rectangles in the finite product state space gives the displayed product kernel; [F6] makes each iterated kernel measurable, and [F7] extends the conditional identities to all Borel path events. Thus the product is a version of the finite-horizon smoothing kernel. The argument concerns histories under their actual observation law and makes no pointwise uniqueness claim on null histories.