Feynman–Kac targets and sequential particle filters — Examples

The vector is a probability vector, and $Np=(31\mathrm{/}10,7\mathrm{/}5,1\mathrm{/}2,0,0)=(3.1,1.4,0.5,0,0)$. By [F2], both schemes therefore have those expected counts.

In multinomial resampling each count is a sum of five independent Bernoulli indicators with success probability $p_j$, so its variance is $5p_j(1-p_j)$. The first three marginal variances are $1.178$, $1.008$, and $0.45$; the last two are zero.

Residual resampling has deterministic floors $(3,1,0,0,0)$ and fractional remainders $(0.1,0.4,0.5,0,0)$, whose sum is $R=1$. It draws one extra index with these probabilities, so the marginal count variances are $0.1(0.9)=0.09$, $0.4(0.6)=0.24$, and $0.5(0.5)=0.25$ on the first three coordinates, with variance zero on the last two.

The expected counts agree componentwise, while for these weights each of the first three residual marginal variances is smaller than its multinomial counterpart, while the last two variances are zero in both schemes. This is a calculation for this input vector, not a general variance-ordering claim.

The unnormalized path masses are ${\gamma }_1(a)=M_1(\ast ,\{a\})G_1(a)=1\mathrm{/}10$ and ${\gamma }_1(b)=M_1(\ast ,\{b\})G_1(b)=2\mathrm{/}5$. Thus $Z_1=1\mathrm{/}2$ and the normalized target masses are $(1\mathrm{/}5,4\mathrm{/}5)$.

For the bootstrap proposal, ${\rho }_0=1$ and $r_1(a)=r_1(b)=1$. The one-particle incremental weight is therefore $1\mathrm{/}5$ with probability $1\mathrm{/}2$ and $4\mathrm{/}5$ with probability $1\mathrm{/}2$, so its expectation is $1\mathrm{/}2$ but it is not constant.

For the informed proposal, $r_1(a)=(1\mathrm{/}2)\mathrm{/}(1\mathrm{/}5)=5\mathrm{/}2$ and $r_1(b)=(1\mathrm{/}2)\mathrm{/}(4\mathrm{/}5)=5\mathrm{/}8$. Multiplication by $G_1$ gives $G_1(a)r_1(a)=G_1(b)r_1(b)=1\mathrm{/}2$, so the one-particle normalizer estimate equals $Z_1$ for either sampled state.

Against counting measure, the unnormalized target density is $\tilde{p}(a)=1\mathrm{/}10$, $\tilde{p}(b)=2\mathrm{/}5$, and both proposal densities are strictly positive, so the support and finite-integrability hypotheses of [F4] hold. Setting $f=1$ in [F4] gives ${\mathbb{E}}_Q[\tilde{p}(X)\mathrm{/}q(X)]=Z_1$ after multiplying the unbiased normalized estimate by $Z_1$; the explicit bootstrap and informed weights above are exactly these ordinary importance-sampling ratios.

Let ${\mathsf{X}}_0=\{\ast \}$ and ${\mathsf{X}}_1=\{a,b\}$. Take ${\mu }_0={\nu }_0={\delta }_{\ast }$, $M_1(\ast ,\{a\})=M_1(\ast ,\{b\})=1\mathrm{/}2$, $R_1(\ast ,\{a\})=1$, $R_1(\ast ,\{b\})=0$, and $G_0=G_1=1$. By [F1], ${\gamma }_1$ is the transition law itself, $Z_1=1$, and the normalized target assigns mass $1\mathrm{/}2$ to $b$.

The support premise fails because $M_1(\ast ,\{b\})=1\mathrm{/}2$ while $R_1(\ast ,\{b\})=0$. A proposal path drawn from $Q_1={\nu }_0{R}_1$ equals $(\ast ,a)$ with probability one, so every empirical measure formed from such paths assigns mass zero to the event $\{x_1=b\}$.

If a kernel derivative $r_1$ represented $M_1(\ast ,\cdot )$ with respect to $R_1(\ast ,\cdot )$, then $M_1(\ast ,\{b\})={\int }_{\{b\}}{r}_1(\ast ,y)R_1(\ast ,dy)=0$, contradicting the value $1\mathrm{/}2$. Thus the required importance weight does not exist on this target-positive, proposal-null set, and the false claim fails.