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Particle Filters and Bayesian Estimation · Tutorial 70 of 100

Consistency of Particle Estimates for Filtering Expectations

Learn when particle-filter estimates of bounded functions of the latent state converge, why positive normalizers and a zero-denominator convention matter, and how to interpret the result for concentration estimates.

Advanced 12 min read

What You'll Learn

  • Define the filtering distribution and its particle approximation using the same weighting and resampling steps.
  • State sufficient assumptions for consistency at any fixed observation time.
  • Prove convergence of propagated particles, weighted estimates, and multinomially resampled estimates.
  • Explain why a zero empirical normalizer needs an explicit convention.
  • Calculate concentration-state estimates and resampling variability in finite particle samples.
  • Recognize the limits of fixed-time consistency and diagnose finite-sample problems.

What Consistency Guarantees

A particle filter approximates a filtering distribution by a finite collection of simulated states and weights. The practical question is not only how to calculate an estimate, but what happens to that estimate as the particle count grows. Consistency gives a precise answer for a fixed observation time: under stated conditions, the particle estimate of the expectation of a bounded function of the state converges in probability to the filtering expectation.

This is an asymptotic statement, not a promise that a small particle set is accurate. It does not by itself control the error at a chosen particle count, or guarantee stability over an indefinitely long sequence of observations. We establish the fixed-time result for a standard bootstrap particle filter and identify which parts of the argument matter for interpreting pharmacokinetic concentration estimates.

Definition: Let \(X_k\) be the latent state at observation time \(k\), and let \(y_{1:k}\) denote the observations through time \(k\). The filtering distribution is \(\eta_k(A)=P(X_k\in A\mid y_{1:k})\). For a bounded measurable function \(\varphi\), its filtering expectation is \(\eta_k(\varphi)=\int \varphi(x)\eta_k(dx)\). In a pharmacokinetic model, \(\varphi(x)=C(x)\) can select a concentration component of the state.

Write \(M_k(x,dx')\) for the Markov transition kernel from time \(k-1\) to \(k\), and \(g_k(x')\) for the nonnegative observation likelihood at \(x'\), with the observed data held fixed. For a measurable function \(h\), write \(M_kh(x)=\int h(x')M_k(x,dx')\). The filtering recursion is

$$ \eta_k(\varphi) = \frac{\eta_{k-1}\!\left(M_k(g_k\varphi)\right)} {\eta_{k-1}\!\left(M_k g_k\right)}. $$

This formula requires a positive denominator. It first predicts the state using the transition kernel, then reweights the prediction according to the observation likelihood.

Particle Approximation and Assumptions

At time \(k-1\), suppose the particle filter has a resampled collection \(Z_{k-1}^1,\ldots,Z_{k-1}^N\). Propagate each particle independently through \(M_k\), producing \(Y_k^i\). The propagated empirical measure is \(\nu_k^N=N^{-1}\sum_{i=1}^N\delta_{Y_k^i}\). Weight the propagated particles by \(g_k(Y_k^i)\). If the total weight is positive, the weighted filtering estimate is

$$ \widehat{\eta}_k^N(\varphi) = \frac{\nu_k^N(g_k\varphi)}{\nu_k^N(g_k)}. $$

When this denominator is positive, multinomial resampling draws \(N\) particles independently from the normalized weights. These resampled particles form the collection for the next propagation step. A finite-sample convention is essential: if \(\nu_k^N(g_k)=0\), define the displayed weighted estimate to be \(0\), and set all resampled particles equal to \(Y_k^1\). This convention need not be scientifically meaningful on that event; it makes the algorithm-defined quantities well-defined. Under the assumptions below, the event becomes negligible as \(N\) grows.

Sufficient conditions for fixed-time consistency: The state space is measurable, each \(M_k\) is a Markov transition kernel, and each \(g_k\) is measurable, nonnegative, and bounded. The exact normalizer \(c_k=\eta_{k-1}(M_k g_k)\) is strictly positive at every time \(k\leq K\), for a fixed finite \(K\). The initial particles are independent with distribution \(\eta_0\). Propagation uses conditionally independent draws from \(M_k\), and resampling is multinomial when the total weight is positive, with the zero-weight convention just specified.

Boundedness is a convenient sufficient condition: it keeps the conditional variances in the proof finite. It is not the only possible route to consistency, but relaxing it requires additional moment and integrability conditions. The positive-normalizer condition ensures that the exact filtering distribution is defined at each step.

Consistency Result and Proof

Theorem: Under the sufficient conditions above, for every fixed \(k\leq K\) and every bounded measurable \(\varphi\), the weighted estimate \(\widehat{\eta}_k^N(\varphi)\) converges in probability to \(\eta_k(\varphi)\) as \(N\to\infty\). The empirical average of \(\varphi\) over the resampled particles at time \(k\) also converges in probability to \(\eta_k(\varphi)\).

We prove the result by induction. The key steps are convergence of propagated empirical averages, stability of the ratio when its limiting denominator is positive, and concentration of the resampling average around the weighted estimate.

1
Propagate.
Show that averages under the propagated particles converge to their exact predictive expectations.
2
Weight.
Apply propagation convergence to \(g_k\varphi\) and \(g_k\), then take their ratio using the positive limiting normalizer.
3
Resample.
Use the conditional variance of a multinomial sample average to show that resampled averages have the same limit.

At time zero, the weak law of large numbers gives \(N^{-1}\sum_i h(Z_0^i)\xrightarrow{p}\eta_0(h)\) for every bounded measurable \(h\). Now assume that at time \(k-1\), the resampled empirical average of every bounded measurable function converges to its expectation under \(\eta_{k-1}\). For bounded \(h\), conditional on the particles at \(k-1\), the propagated draws are independent and

$$ E[\nu_k^N(h)\mid Z_{k-1}^{1:N}] =\frac{1}{N}\sum_{i=1}^N M_kh(Z_{k-1}^i), \qquad \operatorname{Var}(\nu_k^N(h)\mid Z_{k-1}^{1:N}) \leq \frac{\|h\|_\infty^2}{N}. $$

Because \(M_kh\) is bounded, the induction hypothesis makes the conditional mean converge in probability to \(\eta_{k-1}(M_kh)\). To see that the random average itself has the same limit, for any \(\epsilon>0\) split the deviation into the conditional-mean deviation and the difference from that mean. Conditional Chebyshev's inequality gives

$$ P\!\left(\left|\nu_k^N(h)-\eta_{k-1}(M_kh)\right|>\epsilon\right) \leq P\!\left(\left|\frac1N\sum_i M_kh(Z_{k-1}^i)-\eta_{k-1}(M_kh)\right|>\frac{\epsilon}{2}\right) +\frac{4\|h\|_\infty^2}{N\epsilon^2}. $$

Both terms tend to zero. Thus \(\nu_k^N(h)\xrightarrow{p}\eta_{k-1}(M_kh)\). In particular, use this result with \(h=g_k\varphi\) and \(h=g_k\). The numerator and denominator of the particle ratio converge in probability to \(\eta_{k-1}(M_k(g_k\varphi))\) and \(c_k>0\), respectively.

The denominator issue cannot be omitted. Even if \(c_k>0\), a finite-particle denominator can be zero with positive probability. Since \(\nu_k^N(g_k)\xrightarrow{p}c_k>0\), however,

$$ P(\nu_k^N(g_k)=0) \leq P\!\left(|\nu_k^N(g_k)-c_k|\geq c_k\right)\longrightarrow 0. $$

On the event that the denominator is positive, the ratio converges in probability to the ratio of its limits: for example, this follows by restricting to denominators greater than \(c_k/2\), whose probability tends to one, and applying the algebra of differences of ratios. On the zero-denominator event the estimate was defined to be \(0\); because that event has probability tending to zero, this arbitrary finite value does not change convergence in probability. This proves consistency of the weighted estimate.

For resampling, conditional on the propagated particles and positive total weight, the resampled states are independent draws from the weighted empirical distribution. Their sample average of \(\varphi\) has conditional mean \(\widehat{\eta}_k^N(\varphi)\) and conditional variance at most \(\|\varphi\|_\infty^2/N\). Conditional Chebyshev's inequality shows that this average differs from the weighted estimate by a quantity converging to zero in probability. On the zero-weight event the resampled average follows the stated convention; that event has probability tending to zero. The resampled average therefore converges to \(\eta_k(\varphi)\), completing the induction. \(\square\)

Worked Concentration Examples

Worked Example: A Zero Denominator Can Occur at Finite N

Suppose the state is \(a\) or \(b\), with prior probabilities \(P(X=a)=0.4\) and \(P(X=b)=0.6\). Let the transition be the identity and let the observation likelihood be \(g(a)=1\), \(g(b)=0\). For \(\varphi(x)=\mathbf{1}\{x=a\}\), the exact normalizer is \(0.4\), and the filtering expectation is \(0.4/0.4=1\).

With \(N\) independent particles, the denominator is zero precisely when every particle is \(b\). This has probability \(0.6^N\), which is positive for every finite \(N\). Thus the unextended ratio is not defined on an event of positive probability. Under the convention that the estimate is \(0\) on that event, the estimate equals \(1\) otherwise. For any \(0<\epsilon<1\), its probability of being more than \(\epsilon\) from the exact value is \(0.6^N\), which tends to zero. The convention resolves the finite-\(N\) definition; positive exact normalizer and convergence make its effect asymptotically negligible.

Worked Example: Filtering Mean of a Concentration State

Suppose the predicted concentration state \(C\) is \(1\) with probability \(0.7\) and \(3\) with probability \(0.3\). The observation likelihoods are \(g(1)=0.2\) and \(g(3)=0.8\). The exact normalizer is

$$ 0.7(0.2)+0.3(0.8)=0.14+0.24=0.38. $$

The filtering probabilities are \(P(C=1\mid y)=0.14/0.38\) and \(P(C=3\mid y)=0.24/0.38\). Therefore the filtering mean concentration is

$$ E[C\mid y]=\frac{1(0.14)+3(0.24)}{0.38} =\frac{0.86}{0.38}\approx 2.2632. $$

For \(N\) predicted particles, let \(A_N\) be the number whose concentration is \(1\). Then \(A_N/N\xrightarrow{p}0.7\), and the particle estimate is

$$ \widehat{\eta}^N(C) = \frac{0.2A_N+0.8(3)(N-A_N)} {0.2A_N+0.8(N-A_N)}. $$

As \(A_N/N\) tends to \(0.7\), the numerator divided by \(N\) tends to \(0.14+0.72=0.86\), and the denominator divided by \(N\) tends to \(0.38\). The ratio consequently tends to \(2.2632\), the exact filtering mean. For instance, if \(N=1000\) and \(A_N=700\), the particle ratio is \(860/380\approx2.2632\). That numerical agreement is an illustration, not a guarantee for every sample of 1000 particles.

Worked Example: Resampling Variability at a Fixed Particle Count

Suppose the normalized weights assign probability \(0.6\) to concentration \(10\) and probability \(0.4\) to concentration \(0\). The weighted filtering mean is \(0.6(10)+0.4(0)=6\). Under multinomial resampling with \(N=100\), let \(K\) be the number of resampled particles at concentration \(10\). Then \(K\sim\operatorname{Binomial}(100,0.6)\), and the resampled average concentration is \(10K/100\).

Its expectation is \(10E[K]/100=10(60)/100=6\). Since \(\operatorname{Var}(K)=100(0.6)(0.4)=24\), the variance of the resampled mean is

$$ \operatorname{Var}(10K/100)=(10/100)^2(24)=0.24, \qquad \operatorname{SD}(10K/100)=\sqrt{0.24}\approx0.490. $$

Resampling is centered on the weighted estimate, but it adds sampling variation. Increasing \(N\) reduces this conditional variance to \(100(0.6)(0.4)/N=24/N\). It does not remove other sources of particle-filter error automatically.

Interpretation, Diagnostics, and Limits

The theorem applies directly to bounded functions of a concentration state. If the state concentration is known to lie in a bounded range, \(\varphi(C)=C\) is bounded and the result says that the particle estimate of the filtered mean concentration converges in probability to the model's filtering mean. If concentration is unbounded, the theorem as stated does not cover the raw mean; a suitable extension needs assumptions controlling the tails and moments. A bounded function such as a concentration indicator \(\mathbf{1}\{C>c\}\) remains covered.

Several practical checks follow from the proof. A very small empirical total weight means the ratio can be unstable even when its limiting denominator is positive. A zero total weight makes the unextended ratio undefined and should be handled explicitly in software. A low effective sample size after weighting signals that few particles dominate; resampling may reduce duplicated low-weight particles but cannot create state diversity that was lost during propagation. These are finite-sample warnings, not contradictions of fixed-time consistency.

The result also has clear boundaries. It assumes a fixed finite number of observation times; it does not establish a uniform error bound as time grows. It addresses Monte Carlo approximation for the specified transition and likelihood, not bias from an approximate SDE transition, a misspecified observation model, or numerical error in a discretization. Exact linear-Gaussian filtering and Kalman methods can be preferable when their assumptions apply; particle filters address nonlinear or non-Gaussian cases but introduce sampling error. The earlier material on Filtering Distributions for Nonlinear State-Space Models describes the filtering target, while this result concerns convergence of its particle approximation.

Common pitfall: “The exact normalizer is positive, so the particle denominator is positive.” This is false: finite samples can miss every state with nonzero likelihood. Define the estimate on the zero-denominator event, then use convergence of the empirical denominator to a positive limit to show that the event's probability vanishes.

Check Your Understanding

Try each question, then check the solutions below it.

  1. If the exact normalizer is \(c>0\), why does the probability of a zero empirical normalizer tend to zero?
  2. For a bounded \(h\), what controls the difference between a propagated empirical average and its conditional mean?
  3. In the concentration example with prior probabilities \(0.7,0.3\) and likelihoods \(0.2,0.8\), calculate the filtering probability of concentration \(3\).
  4. Does fixed-time consistency prove that a particle filter will remain accurate over an arbitrarily long sequence of observations? Explain.

Solutions. (1) A zero empirical normalizer differs from \(c\) by at least \(c\), so its probability is at most \(P(|\nu_k^N(g_k)-c|\geq c)\), which tends to zero by convergence in probability. (2) Conditional independence gives conditional variance at most \(\|h\|_\infty^2/N\); conditional Chebyshev's inequality makes the difference vanish in probability. (3) The probability is \(0.3(0.8)/[0.7(0.2)+0.3(0.8)]=0.24/0.38\approx0.6316\). (4) No. The result holds for each fixed finite time horizon; long-time stability requires additional assumptions and analysis.

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