Particle smoothing, backward simulation, and Rao–Blackwellized filters — Examples

The state and observation noises are independent centered Gaussians, the conditional initial covariance is $1$, the process covariance is $1$, and the observation covariance is $1>0$. Thus the model satisfies [F1] and the exact conditional Kalman interface [F2].

Conditional on $Z_1=z$, prediction gives $m_{1\mathrm{\mid }0}=0$ and $P_{1\mathrm{\mid }0}=a_z^2+1$. Hence $P_{1\mathrm{\mid }0}=1$ for $z=0$ and $P_{1\mathrm{\mid }0}=2$ for $z=1$. The innovation is zero and the innovation variances are $S_0=2$ and $S_1=3$.

By [F4], the two conditional observation densities at zero are ${\mathrm{\ell }}_0=(4\pi {)}^{-1\mathrm{/}2}$ and ${\mathrm{\ell }}_1=(6\pi {)}^{-1\mathrm{/}2}$. Multiplying by the equal prior masses and normalizing gives $P(Z_1=0\mid Y_1=0)=\sqrt{3}\mathrm{/}(\sqrt{3}+\sqrt{2})$ and $P(Z_1=1\mid Y_1=0)=\sqrt{2}\mathrm{/}(\sqrt{3}+\sqrt{2})$.

The Kalman covariance update is $P_{1\mathrm{\mid }1}=P_{1\mathrm{\mid }0}R\mathrm{/}(P_{1\mathrm{\mid }0}+R)$ here, so the two conditional posterior variances are $1\mathrm{/}2$ and $2\mathrm{/}3$; both conditional means remain zero. Mixing these exact conditional Gaussians with the mode probabilities from step 2.1 gives the displayed posterior. The filter in [F3] uses these same per-mode likelihoods and Gaussian components; only its finite particle approximation to the mode weights is random.

Substitution gives $f_1(1\mid 0)=0$ and $w_0^1=1$, so the unique backward score is zero and its normalizing denominator is zero. No division by this quantity defines a probability vector.

The exact predictive law is ${\eta }_0{M}_1={\delta }_0$, hence the terminal state $1$ has predictive probability zero. By [A1] and [F1] a conditional version on this null fiber is not determined by the model's conditional identities, and the supplied cloud at $1$ is not a possible output of the stated filter.

Because [F3] restricts the density-based backward draw to a finite positive denominator, the procedure rejects this input; a uniform choice would be an unsupported choice of conditional version, not a consequence of Bayes' formula.

Since $Y_1$ is constant, conditioning on $Y_1$ leaves the prior law of $X_0$ unchanged. The time-$0$ target for lag $1$ uses $s_0=\min (0+1,2)=1$, so [F2] and [F4] give the lagged law $\mathrm{Bernoulli}(1\mathrm{/}2)$.

For each $b\in \{0,1\}$, the event $Y_2=b$ has probability $1\mathrm{/}2$. On this event, $X_0=X_1=X_2=b$ surely. Thus the full smoothing path law is ${\delta }_{(b,b,b)}$, and its time-$0$ marginal is $\mathcal{L}(X_0\mid Y_1=0,Y_2=b)={\delta }_b$.

For $A=\{b\}$, the probabilities under the lagged law and the full smoothing marginal at time $0$ are $1\mathrm{/}2$ and $1$, respectively; for any subset $A\subseteq \{0,1\}$ their probability difference has absolute value at most $1\mathrm{/}2$. Hence ${\sup }_A\mathrm{\mid }\mathrm{Bernoulli}(1\mathrm{/}2)(A)-{\delta }_b(A)\mathrm{\mid }=1\mathrm{/}2$. This difference is between the truncated and full targets at time $0$, so particle-count growth alone cannot remove it.