Finite-horizon SMC empirical laws converge for integrable tests

At time $0$, [F2] and the finite particle law [F7] give IID proposal particles with law ${\tilde{\pi }}_0$. For $h\in L^1({\tilde{\pi }}_0)$, [F4] gives ${\eta }_0^N(h)\to {\tilde{\pi }}_0(h)$. If $\varphi \in L^1({\pi }_0)$, then ${\upsilon }_0\varphi$ is integrable under ${\tilde{\pi }}_0$ and its integral is ${\pi }_0(\varphi )$ by [F1]; also ${\tilde{\pi }}_0({\upsilon }_0)=1$. Applying [F4] to numerator and denominator and dividing on the event that the denominator is positive gives ${\omega }_0^N(\varphi )\to {\pi }_0(\varphi )$. That event has probability tending to one because the denominator average converges to $1$.

Fix $t\ge 1$ and suppose the three convergences hold at $t-1$. First let $h$ be bounded. By [F2] and [F7], conditional on the selected parents, the $X_t^i$ are independent with laws $k_t({\check{X}}_{t-1}^i,\cdot )$; measurability of $k_{t}h$ follows from [F8]. Thus ${\eta }_t^N(h)-{\check{\eta }}_{t-1}^N(k_{t}h)$ has conditional mean zero and conditional variance at most $\mathrm{\parallel }h{\mathrm{\parallel }}_{\mathrm{\infty }}^2\mathrm{/}N$. By [F5] this difference tends to zero in probability. Since $k_{t}h$ is bounded and ${\pi }_{t-1}(k_{t}h)={\tilde{\pi }}_t(h)$ by [F1], the induction hypothesis proves ${\eta }_t^N(h)\to {\tilde{\pi }}_t(h)$.

The selected-law base at time zero follows as well. For bounded $\varphi$, [F3] makes the conditional selected mean equal to ${\omega }_0^N(\varphi )$. Under multinomial selection its conditional variance is at most $\mathrm{\parallel }\varphi {\mathrm{\parallel }}_{\mathrm{\infty }}^2\mathrm{/}N$; under residual selection the deterministic floor counts have no variance and the at most $N$ remaining categorical draws give the same bound. By [F5] and step 1.1, ${\check{\eta }}_0^N(\varphi )\to {\pi }_0(\varphi )$ in probability. For integrable $\varphi$, truncate at $M$. The conditional expected selected average of $\mathrm{\mid }\varphi \mathrm{\mid }{\mathbf{1}}_{\{\mathrm{\mid }\varphi \mathrm{\mid }>M\}}$ is its weighted empirical average by [F3], which converges to the target tail integral by step 1.1. Conditional Markov [F6], using $P(\text{tail average}>ϵ)\le E[\min (1,{\omega }_0^N(\text{tail})\mathrm{/}ϵ)]$ and bounded continuity of $x\mapsto \min (1,x\mathrm{/}ϵ)$, makes the limiting tail probability vanish as $M\to \mathrm{\infty }$. Combining with the bounded truncations proves the selected convergence for all ${\pi }_0$-integrable tests.

For general $h\in L^1({\tilde{\pi }}_t)$, let $h_M=h{\mathbf{1}}_{\{\mathrm{\mid }h\mathrm{\mid }\le M\}}$. The conditional mean of ${\eta }_t^N(\mathrm{\mid }h-h_M\mathrm{\mid })$ is $Y_N={\check{\eta }}_{t-1}^N(k_t\mathrm{\mid }h-h_M\mathrm{\mid })$, measurable by [F8]. The function inside this last empirical average is integrable under ${\pi }_{t-1}$ and its integral is $a_M={\tilde{\pi }}_t(\mathrm{\mid }h-h_M\mathrm{\mid })$. The induction hypothesis gives $Y_N\to a_M$ in probability. Conditional Markov [F6] bounds $P({\eta }_t^N(\mathrm{\mid }h-h_M\mathrm{\mid })>ϵ)$ by $E[\min (1,Y_N\mathrm{/}ϵ)]$. For fixed $M$, bounded continuity of $y\mapsto \min (1,y\mathrm{/}ϵ)$ and convergence in probability to the constant $a_M$ imply that this expectation tends to $\min (1,a_M\mathrm{/}ϵ)$: for any $\eta >0$, split on $\mathrm{\mid }{Y}_N-a_M\mathrm{\mid }\le \eta$ and its complement, use continuity on the first event and convergence in probability on the second. Since $a_M\to 0$ by integrability, the limiting tail probability vanishes. Applying the same truncation to the target integral and combining with step 1.2 proves the proposal empirical convergence for every integrable $h$.

Apply step 2.2 to $h={\upsilon }_t$ and $h={\upsilon }_t\varphi$, where $\varphi \in L^1({\pi }_t)$. Their limiting integrals are $1$ and ${\pi }_t(\varphi )$ by [F1]. The ratio consequently converges to ${\pi }_t(\varphi )$ whenever its denominator is positive; the denominator-zero event has probability tending to zero because its normalized sum converges to $1$. The uniform fallback in [F2] on that event cannot change convergence in probability.

For bounded $\varphi$, the conditional expectation of ${\check{\eta }}_t^N(\varphi )$ is ${\omega }_t^N(\varphi )$ by [F3]. In multinomial resampling the conditional variance is at most $\mathrm{\parallel }\varphi {\mathrm{\parallel }}_{\mathrm{\infty }}^2\mathrm{/}N$. In residual resampling the deterministic floor counts contribute no variance, and the remaining $R\le N$ categorical draws give conditional variance at most $\mathrm{\parallel }\varphi {\mathrm{\parallel }}_{\mathrm{\infty }}^2\mathrm{/}N$ as well. Chebyshev [F5] and step 3.1 prove ${\check{\eta }}_t^N(\varphi )\to {\pi }_t(\varphi )$ for bounded $\varphi$.

For $\varphi \in L^1({\pi }_t)$, truncate to ${\varphi }_M=\varphi {\mathbf{1}}_{\{\mathrm{\mid }\varphi \mathrm{\mid }\le M\}}$. By [F3], the conditional expected selected average of $\mathrm{\mid }\varphi -{\varphi }_M\mathrm{\mid }$ is exactly ${\omega }_t^N(\mathrm{\mid }\varphi -{\varphi }_M\mathrm{\mid })$. Step 3.1 applied to this integrable tail and conditional Markov [F6] make the limiting probability of a tail deviation at most $\min (1,{\pi }_t(\mathrm{\mid }\varphi -{\varphi }_M\mathrm{\mid })\mathrm{/}ϵ)$, which tends to zero as $M\to \mathrm{\infty }$. Combining with step 4.1 proves the selected empirical convergence for all integrable $\varphi$.

Steps 1.2, 2.2, 3.1, 4.1, and 5.1 establish the induction step from $t-1$ to $t$. Steps 1.1 and 2.1 give the base case, so the assertions hold for every fixed finite time. No conclusion is uniform in $t$.