Finite-horizon SMC error, CLTs, and degeneracy — Examples

The two child indices are independent fair draws by [F1]. They choose the same parent with probability $(1\mathrm{/}2{)}^2+(1\mathrm{/}2{)}^2=1\mathrm{/}2$. The expected number of distinct parents represented is $2\{1-(1-1\mathrm{/}2{)}^2\}=3\mathrm{/}2$, since each parent is selected at least once with probability $3\mathrm{/}4$.

If both children select one parent, identity mutation leaves two copies of that same original ancestor. Every later resampling copies represented ancestors, so it cannot restore the lost ancestor. If both lineages remain, their survival through the next call requires different parent indices. By the stipulated independence across calls and step 1.1, this has conditional probability $1\mathrm{/}2$ given the full preceding index history.

Starting from two lineages, the probability that no collapse occurs in $m$ successive resampling steps is $(1\mathrm{/}2{)}^m$ by induction on $m$ using step 2.1. Hence the probability of a single represented ancestor is $1-2^{-m}\to 1$, even though the expected offspring counts at each step remain $(1,1)$ by [F2].

The geometric proposal masses sum to one. Since $0<s<1\mathrm{/}2$, $w_0$ is finite and positive, and ${\sum }_{n\ge 0}{p}_n{w}_n=1$ by construction; hence ${\pi }_1$ is a probability law with density $w$ relative to $p$. The identity kernel gives ${\tilde{\pi }}_1=p$.

For a separate initial-proposal witness, let ${\mathrm{\Theta }}_0={\mathbb{N}}_{\ge 1}$, ${\tilde{\pi }}_0\{n\}=2^{-n}$, $c=({\sum }_{n\ge 1}{2}^{-n\mathrm{/}2}{)}^{-1}$, and ${\upsilon }_0(n)=c{2}^{n\mathrm{/}2}$. Put ${\pi }_0\{n\}=2^{-n}{\upsilon }_0(n)$ and $\varphi (n)={\mathbf{1}}_{\{n\text{ even}\}}$. Then ${\sum }_n{\tilde{\pi }}_0\{n\}{\upsilon }_0(n)=1$, so ${\pi }_0$ is a probability law and ${\upsilon }_0=d{\pi }_0\mathrm{/}d{\tilde{\pi }}_0$. Both $\varphi$ and $1$ are bounded, hence belong to ${\mathrm{\Phi }}_0$.

Choose $\delta =1$. For $n\ge 1$, $w_n\varphi (n)=w_n^{1\mathrm{/}4}$, so $\sum p_n\mathrm{\mid }{w}_n\varphi (n){\mathrm{\mid }}^3={\sum }_{n\ge 1}{p}_n{w}_n^{3\mathrm{/}4}<\mathrm{\infty }$ and $\sum p_n\mathrm{\mid }{w}_n\varphi (n){\mathrm{\mid }}^2={\sum }_{n\ge 1}{p}_n{w}_n^{1\mathrm{/}2}<\mathrm{\infty }$. Thus $k_1(w\varphi )=w\varphi \in {\mathrm{\Phi }}_0$ and $\varphi \in {\mathrm{\Phi }}_1$. Also $\sum p_n{w}_n^3<\mathrm{\infty }$ and $k_{1}w=w\in {\mathrm{\Phi }}_0$, so $1\in {\mathrm{\Phi }}_1$.

The target mean is ${\pi }_1(\varphi )={\sum }_{n\ge 1}{p}_n{w}_n^{1\mathrm{/}4}<\mathrm{\infty }$, but ${\pi }_1({\varphi }^2)={\sum }_{n\ge 1}{p}_n{w}_n^{-1\mathrm{/}2}={\sum }_{n\ge 1}{2}^{-(n+1)}{2}^{n^2\mathrm{/}2}=\mathrm{\infty },$ since the summands do not tend to zero. Therefore ${\mathrm{Var}}_{{\pi }_1}(\varphi )=\mathrm{\infty }$.

The weighted covariance $V_1(\varphi )$ is finite: the identity mutation has zero conditional variance, $w(\varphi -{\pi }_1\varphi )\in L^2(p)$ by step 2.1 and the finite mean in step 2.2, and the initial multinomial covariance at time zero is finite because ${\upsilon }_0=1$ and $g=w(\varphi -{\pi }_1\varphi )\in L^2(p)$. Yet the post-multinomial formula in [F1] adds the infinite target variance from step 2.2. Thus the source class and $1\in {\mathrm{\Phi }}_1$ do not suffice for the claimed finite post-resampling covariance.

Let $q={\pi }_0(\varphi )$. Positive proposal and target mass occur on both parities, so $0<q<1$. For every even $n$, ${\tilde{\pi }}_0\{n\}{\upsilon }_0(n{)}^2(\varphi (n)-q{)}^2=c^2(1-q{)}^2>0$, so the sum over even $n$ diverges. Thus ${\tilde{\pi }}_0({\upsilon }_0^2(\varphi -q{)}^2)=\mathrm{\infty }$ and [F2] gives $V_0(\varphi )=\mathrm{\infty }$, despite $\varphi ,1\in {\mathrm{\Phi }}_0$. This independently refutes the claimed initial weighted-estimator covariance finiteness.

For $N$ IID proposal particles, let $M_N$ be the number at $A$ and $p_N=M_N\mathrm{/}N$. The weighted empirical estimate is $S_N=\frac{2M_N}{2M_N+(2\mathrm{/}3)(N-M_N)}=\frac{3p_N}{1+2p_N}\mathrm{.}$ Since $M_N$ is binomial with parameters $(N,1\mathrm{/}4)$, [F3] gives $\sqrt{N}(p_N-1\mathrm{/}4)\Rightarrow N(0,3\mathrm{/}16)$. The identity $\sqrt{N}(S_N-1\mathrm{/}2)=2\sqrt{N}(p_N-1\mathrm{/}4)\mathrm{/}(1+2p_N)$ and [F4] yield $Z_N:=\sqrt{N}(S_N-1\mathrm{/}2)\Rightarrow Z\sim N(0,1\mathrm{/}3)$. Moreover, $Z_N>0$ exactly when $p_N>1\mathrm{/}4$.

The limiting-weight candidate in [F1] gives $r(2)=0$ and $r(2\mathrm{/}3)=2\mathrm{/}3$. Because $\varphi (B)=0$, both ${\tilde{\pi }}_0(r({\upsilon }_0){\varphi }^2)$ and ${\tilde{\pi }}_0(r({\upsilon }_0)\varphi )$ are zero, so $R_0^{\mathrm{c}\mathrm{a}\mathrm{n}\mathrm{d}}(\varphi )=0$. If used as a universal residual CLT variance, it predicts a post-resampling Gaussian limit with variance $V_0(\varphi )={\tilde{\pi }}_0[{\upsilon }_0^2(\varphi -1\mathrm{/}2{)}^2]=1\mathrm{/}3$.

Put ${\bar{w}}_N=2\mathrm{/}3+(4\mathrm{/}3)p_N$ and $u_A=2\mathrm{/}{\bar{w}}_N$, $u_B=(2\mathrm{/}3)\mathrm{/}{\bar{w}}_N$. On $1\mathrm{/}4<p_N<1\mathrm{/}3$, the normalized residual sizes are $u_A\in (1,2)$ and $u_B\in (0,1)$; hence each $A$ particle receives one deterministic child, each $B$ particle receives none, and $R=N-M_N$ children are sampled from the remainders. If $X_N$ is the number of residual children at $A$, then conditionally on the proposal sample, $X_N\sim \mathrm{B}\mathrm{i}\mathrm{n}(R,a_N)$ with $a_N=M_N(u_A-1)\mathrm{/}R$. As $p_N\downarrow 1\mathrm{/}4$, $R\mathrm{/}N\to 3\mathrm{/}4$ and $a_N\to 1\mathrm{/}3$.

On that event, the post-resampling mean is $(M_N+X_N)\mathrm{/}N$ and its conditional mean is $S_N$. A Taylor expansion of the binomial characteristic function, uniform for $0\le a_N\le 1$ and $R\le N$, gives $\mathbb{E}\text{\hspace{0.17em}⁣}[\exp \text{\hspace{0.17em}⁣}\left(iu\frac{X_N-R{a}_N}{\sqrt{N}}\right)|M_N]=\exp \text{\hspace{0.17em}⁣}(-\frac{u^2}{2}\frac{R}{N}{a}_N(1-a_N)+O(N^{-1\mathrm{/}2}))\mathrm{.}$ Thus, on the positive branch $p_N>1\mathrm{/}4$, the extra resampling fluctuation has limiting variance $(3\mathrm{/}4)(1\mathrm{/}3)(2\mathrm{/}3)=1\mathrm{/}6$.

On the negative branch with $p_N<1\mathrm{/}4$ and $p_N\to 1\mathrm{/}4$, one has $u_A\in (2,3)$ and $u_B\in (0,1)$. The floor for each $A$ particle is two, and its total fractional remainder is $M_N(u_A-2)=O_P(\sqrt{N})$, because $u_A-2=2(1\mathrm{/}{\bar{w}}_N-1)=O_P(\mathrm{\mid }{p}_N-1\mathrm{/}4\mathrm{\mid })$. Conditional variance of the residual $A$ count is at most this total remainder. Consequently the variance of $\sqrt{N}$ times the post-resampling proportion about its weighted conditional mean is $O_P(N^{-1\mathrm{/}2})$, so the extra resampling fluctuation tends to zero in probability on the negative branch. The events $p_N\mathrm{\notin }(1\mathrm{/}5,1\mathrm{/}3)$ have probability tending to zero by [F3].

Put $U_N=\sqrt{N}({\check{\eta }}_0^N(\varphi )-S_N)$, the post-resampling fluctuation centered at the weighted mean, and condition on the proposal count $M_N$. For any real $v,s$, steps 2.1, 2.2 and the negligible exceptional event show that $\mathbb{E}[e^{is{U}_N}\mid M_N]$ converges in probability to $e^{-s^2\mathrm{/}12}$ when $Z_N>0$ and to $1$ when $Z_N<0$. Since $Z_N\Rightarrow Z$ and $P(Z=0)=0$, [F5] gives $\mathbb{E}[e^{iv{Z}_N+is{U}_N}]⟶\mathbb{E}\text{\hspace{0.17em}⁣}\left[e^{ivZ}({\mathbf{1}}_{\{Z>0\}}{e}^{-s^2\mathrm{/}12}+{\mathbf{1}}_{\{Z<0\}})\right]\mathrm{.}$ This is the joint characteristic function of $(Z,{\mathbf{1}}_{\{Z>0\}}W)$ for an independent $W\sim N(0,1\mathrm{/}6)$, so the post-resampling limit is $Z+{\mathbf{1}}_{\{Z>0\}}W$. Its mean is zero and its third moment is $3\mathbb{E}[Z{\mathbf{1}}_{\{Z>0\}}]\mathbb{E}[W^2]>0$, so it is not Gaussian. This contradicts the variance-$1\mathrm{/}3$ Gaussian predicted in step 1.2.

For $N=3$, the event that the proposal sample contains one $A$, one $C$, and one $B$ has probability $3!(1\mathrm{/}4)(1\mathrm{/}4)(1\mathrm{/}2)=3\mathrm{/}16>0$. On this event the total weight is four and the normalized weights are $(1\mathrm{/}2,1\mathrm{/}2,0)$.

The scaled normalized weights are $(N{p}_A,N{p}_C,N{p}_B)=(3\mathrm{/}2,3\mathrm{/}2,0)$, so their floor counts are $(1,1,0)$ and their fractional remainders are $(1\mathrm{/}2,1\mathrm{/}2,0)$. By [F2] one residual child is drawn, with equal probability from $A$ and $C$; hence the offspring-count vector is random although the limiting proposal remainder integral is zero. This contradicts the Statement.

The proposal-to-target ratios are $w(A)=\pi (A)\mathrm{/}q(A)=99\mathrm{/}25$ and $w(B)=\pi (B)\mathrm{/}q(B)=1\mathrm{/}75$, and $q(A)w(A)+q(B)w(B)=1$. The event of exactly one $A$ in four IID proposal draws has probability $\left(\genfrac{}{}{0px}{}{4}{1}\right)(1\mathrm{/}4)(3\mathrm{/}4{)}^3=27\mathrm{/}64>0$.

The four unnormalized weights sum to $99\mathrm{/}25+3\mathrm{/}75=4$, so their normalized values are $99\mathrm{/}100,1\mathrm{/}300,1\mathrm{/}300,1\mathrm{/}300$. By [F2], ${\mathrm{ESS}}_w={((99\mathrm{/}100{)}^2+3(1\mathrm{/}300{)}^2)}^{-1}=\frac{7500}{7351}\approx \mathrm{1.0203.}$ The first particle carries $99\mathrm{\%}$ of the observed weight, which explains the low ESS; the calculation concerns weight concentration only.

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.

Every transition row is $(1\mathrm{/}2,1\mathrm{/}2)$, so the target and proposal transitions coincide and the likelihood contributes no correction. By [F1], ${\upsilon }_s=1$, hence $B=1$, and ${\pi }_s$ is uniform for every $s\ge 0$. Since the row does not depend on its predecessor, [F2] gives $\delta (K)=0$; its density rows are identical, so [F3] also applies with $C=1$.

On the closed last-state subspace the forward operator is $E_s{h}_s=K{h}_s$, which maps a last-state test to its uniform average. The true transition and proposal density versions relative to counting measure are all $1\mathrm{/}2$, so [F4] applies with $C=1$, $A=1$, and $\rho =0$. No contraction of arbitrary path-history functions is asserted. The test has $\mathrm{\Delta }\varphi =1$ and is centered by $1\mathrm{/}2$. Every earlier backward image is zero; the final $k=t$ term in the variance expansion is ${\mathrm{Var}}_{{\pi }_t}(\varphi )=1\mathrm{/}4$. Hence $V_t(\varphi )=1\mathrm{/}4$ for every $t$.

Applying [F5] with $A=B=1$, $\rho =0$ and $\mathrm{\Delta }\varphi =1$ gives $V_t(\varphi )\le 1$, uniformly in $t$; the exact variance $1\mathrm{/}4$ from step 1.2 satisfies the bound. This verifies a time-uniform asymptotic-variance result under explicit mixing and weight conditions.

The identity kernel leaves the uniform law invariant and has incremental weight one. Thus [F1] gives $E_{k+1:t}(\varphi -{\pi }_t\varphi )=\varphi -1\mathrm{/}2$ for every $k\le t$. Since $(\varphi -1\mathrm{/}2{)}^2=1\mathrm{/}4$ everywhere, each of the $t+1$ summands in [F2] equals $1\mathrm{/}4$. Therefore $V_t(\varphi )=(t+1)\mathrm{/}4$.

By [F3], at every fixed $t$ the weighted estimate has a valid CLT with variance $(t+1)\mathrm{/}4$, and the post-resampling estimate has variance $(t+2)\mathrm{/}4$. Both variance sequences are unbounded as $t$ ranges over the integers. The fixed-horizon theorem therefore supplies no uniform-in-time variance bound or finite-particle accuracy guarantee.