Recursive Bayesian filtering under dominated observations

Assume AC and a finite-horizon model as in General finite-horizon state-space models by kernels. For each $1\le t\le T$, suppose the observation kernel has a supplied product-measurable density $g_t:E_t\times F_t\to [0,\mathrm{\infty }]$ with respect to a sigma-finite measure ${\nu }_t$ on $F_t$, normalized so that $G_t(x,B)={\int }_B{g}_t(x,y){\nu }_t(dy),{\int }_{F_t}{g}_t(x,y){\nu }_t(dy)=1$ for every state $x$ and measurable $B\subseteq F_t$.

The procedure takes the initial law ${\mu }_0$, transition kernels $K_t$, observation densities $g_t$ and measures ${\nu }_t$ as inputs. It defines data-indexed prediction, evidence, and filtering functions on every observation history as follows.

1. Initialize ${\pi }_0$ on the singleton empty-history space by ${\pi }_0(\ast ,A)={\mu }_0(A)$.

2. For $t=1,\dots ,T$, each history $h\in {\mathsf{F}}_{1:t-1}$ and each measurable state set $A\in {\mathcal{E}}_t$, form the predictive measure ${\widehat{\pi }}_t(h,A):={\int }_{E_{t-1}}{K}_t(x,A){\pi }_{t-1}(h,dx)\mathrm{.}$

3. For each $y\in F_t$, calculate the evidence density $z_t(h,y):={\int }_{E_t}{g}_t(x,y){\widehat{\pi }}_t(h,dx)\mathrm{.}$

4. Define the filtering measure at $(h,y)$ by ${\pi }_t(h,y,A):=\{\begin{array}{ll}{\displaystyle \frac{{\int }_A{g}_t(x,y){\widehat{\pi }}_t(h,dx)}{z_t(h,y)},}& 0<z_t(h,y)<\mathrm{\infty },\\ {\widehat{\pi }}_t(h,A),& z_t(h,y)=0\text{ or }{z}_t(h,y)=\mathrm{\infty }\mathrm{.}\end{array}$ Use ${\pi }_t$ as the input filtering function at the next time step.

The procedure returns the initialized filter ${\pi }_0={\mu }_0$ and, for every $1\le t\le T$, ${\widehat{\pi }}_t$, $z_t$, and ${\pi }_t$. It is an exact measure-valued recursion; it makes no claim that the integrals have closed forms or can be evaluated with finite computational cost. Exact state-space filtering recursion under dominated observations ↗ proves that the returned time-$t$ functions are predictive and filtering kernels for the specified finite-horizon model.