Proposal mixing ratios alone do not bound SMC asymptotic variance over time

For every pair of preceding states, both $g(x\mid x^{\mathrm{\prime }})\mathrm{/}g(x\mid x^{\mathrm{\prime }\mathrm{\prime }})$ and $q_t(x\mid x^{\mathrm{\prime }})\mathrm{/}{q}_t(x\mid x^{\mathrm{\prime }\mathrm{\prime }})$ equal $1$. The likelihood satisfies $0<1\le f(\ast \mid x)\le 1$, and $\mathrm{\Delta }\varphi =1$. The filtering target at every $t\ge 1$ has last-state law ${\pi }_t(X_t=1)=1\mathrm{/}2$, while the proposal last-state law is $q_t$. Its normalized incremental Radon–Nikodym weight is ${\upsilon }_t(1)=1\mathrm{/}(2{ϵ}_t)$ and ${\upsilon }_t(0)=1\mathrm{/}(2(1-{ϵ}_t))$.

For every preceding path, the forward image of the centered test is $E_t(\varphi -1\mathrm{/}2)={\sum }_{x=0}^1{q}_t(x\mid x_{t-1}){\upsilon }_t(x)({\mathbf{1}}_{\{x=1\}}-1\mathrm{/}2)={\sum }_{x=0}^{1}g(x\mid x_{t-1})({\mathbf{1}}_{\{x=1\}}-1\mathrm{/}2)=0.$ Consequently $E_{k+1:t}(\varphi -{\pi }_t\varphi )=0$ for every $k<t$, so all earlier-time terms in [F2] vanish.

The remaining $k=t$ term in [F2] is $V_t(\varphi )={ϵ}_t{\left(\frac{1}{2{ϵ}_t}\right)}^2\frac{1}{4}+(1-{ϵ}_t){\left(\frac{1}{2(1-{ϵ}_t)}\right)}^2\frac{1}{4}=\frac{1}{16{ϵ}_t}+\frac{1}{16(1-{ϵ}_t)}\mathrm{.}$ Since ${ϵ}_t=2^{-t-2}\downarrow 0$, $V_t(\varphi )\to \mathrm{\infty }$ despite all the stated ratio, likelihood, and oscillation bounds. Thus those conditions alone do not imply the claimed time-uniform variance bound.