Path storage in the particle filter

Abstract

This article considers the problem of storing the paths generated by a particle filter and more generally by a sequential Monte Carlo algorithm. It provides a theoretical result bounding the expected memory cost by T+CNlogN where T is the time horizon, N is the number of particles and C is a constant, as well as an efficient algorithm to realise this. The theoretical result and the algorithm are illustrated with numerical experiments.

Appendix: Proof of Lemma 1

Let $N\in\mathbb{N}$ and ε∈(0,1), define (u _k ) _k≥0 as in the statement of the lemma and define g _N,ε as in Eq. (6). We are interested in ∑ _k≥0(u _k −1). Note first that g _N,ε is contracting and is such that g _N,ε (1)=1, so that u _k goes to 1 using Banach fixed-point theorem. The contraction coefficient of g _N,ε can be bounded by

$$\begin{aligned} \sup_{x}\bigl\lvert g_{N,\varepsilon}'(x)\bigr\rvert & \leq g_{N,\varepsilon }'(1)=1-\frac{\varepsilon^{2}}{2N}<1, \end{aligned}$$

however this contraction coefficient depends on N and a direct use of it yields a bound on ∑ _k≥0(u _k −1) that is not in NlogN.

Note also that even though u _k goes to 1, we can focus on the partial sum $\sum_{k=0}^{\sigma_{2}}(u_{k}-1)$ where σ ₂=inf{k:u _k ≤2}, because $\sum_{k=\sigma_{2}}^{\infty}(u_{k}-1)$ is essentially bounded by N. Indeed note that for 1≤u≤2 we have (ε ³/6N ²)u(u−1)(u−2)≤0 so that

$$\begin{aligned} u_{k}-1 & \leq u_{k-1}-1-\frac{\varepsilon^{2}}{2N}u_{k-1}(u_{k-1}-1) \\ &\leq (u_{k-1}-1) \biggl(1-\frac{\varepsilon^{2}}{2N}\biggr), \end{aligned}$$

hence $\sum_{k=\sigma_{2}}^{\infty}(u_{k}-1)\leq(2N/ \varepsilon^{2})$. Therefore we can focus on bounding $\sum_{k=0}^{\sigma_{2}}(u_{k}-1)$ by NlogN. Let us split this sum into partial sums, where the first partial sum is over indices k such that N/2≤u _k ≤N, the second is over indices k such that N/4≤u _k ≤N/2, etc. More formally, we introduce $(k_{j})_{j=0}^{J}$ such that k ₀=0, k ₁=inf{k:u _k ≤N/2},…,k _j =inf{k:u _k ≤N/2^j}, up to k _J =inf{k:u _k ≤N/2^J} where J is such that N/2^J≤2, or equivalently logN/log2−1≤J. For instance we take J=⌊logN/log2⌋. Thus we have split $\sum_{k=0}^{\sigma_{2}}(u_{k}-1)$ into J partial sums of the form $\sum_{k=k_{j}}^{k_{j+1}-1}(u_{k}- 1)$ and we are now going to bound each of these partial sum by the same quantity C(ε)N for some C(ε) that depends only on ε.

To do so, we consider the time needed by (u _k ) _k≥0 to decrease from a value N/m _j to a value N/m _j+1, with m _j+1>m _j ; we will later take m _j =2^j and m _j+1=2^j+1. Note that for any m we have

$$\begin{aligned} &{g_{N,\varepsilon} \biggl(\frac{N}{m} \biggr) }\\ &{\quad = \frac{N}{m} \biggl(1-\frac{1}{m} \biggl[\frac{\varepsilon^{2}}{2}- \frac {m\varepsilon^{2}}{2N}-\frac{\varepsilon^{3}}{6m}+\frac{\varepsilon ^{3}}{2N}-\frac{m\varepsilon^{3}}{3N^{2}} \biggr] \biggr).} \end{aligned}$$

Define

$$\beta(N,m,\varepsilon)=\frac{\varepsilon^{2}}{2}-\frac{m\varepsilon ^{2}}{2N}-\frac{\varepsilon^{3}}{6m}+ \frac{\varepsilon^{3}}{2N}-\frac {m\varepsilon^{3}}{3N^{2}} $$

and note that for any N≥6 and m≤N/2 we have

$$\underline{\beta}(\varepsilon):=\frac{\varepsilon^{2}}{4}\leq\beta (N,m,\varepsilon), $$

which is clear upon noticing that β(N,m,ε) as a function of m on [1,N/2] is concave and thus reaches its minimum in 1 or N/2 (and this minimum is greater than ε ²/4, provided N≥6). For any x≥N/m _j+1 we can check that

$$g_{N,\varepsilon}(x)\leq\frac{g_{N,\varepsilon}(N/m_{j+1})}{N/m_{j+1}}\times x $$

by noticing that g _N,ε is concave and that g _N,ε (x)≤x for x∈[0,N]. Hence for k≥0 such that u _k−1≥N/m _j+1, we have

$$u_{k}\leq \biggl(1-\frac{1}{m_{j+1}}\underline{\beta}(\varepsilon) \biggr)u_{k-1}. $$

Now suppose that for some k _j ≥0 we have $u_{k_{j}}\leq N/m_{j}$. Then let us find K such that $u_{k_{j}+K}\leq N/m_{j+1}$. It is sufficient to find K such that

$$\begin{aligned} &{ \biggl(1-\frac{1}{m_{j+1}}\underline{\beta}(\varepsilon) \biggr)^{K}\frac {N}{m_{j}}\leq\frac{N}{m_{j+1}}} \\ &{\quad \Leftrightarrow\quad K\geq\log \frac{m_{j+1}}{m_{j}} \biggl(-\log \biggl(1-\frac{1}{m_{j+1}}\underline{\beta}( \varepsilon) \biggr) \biggr)^{-1}.} \end{aligned}$$

Finally by using

$$\forall x\in(0,1)\quad\frac{1}{x}-1\leq\frac{1}{-\log(1-x)}\leq \frac{1}{x} $$

we conclude that K defined as

$$K= \biggl\lceil \biggl(\log\frac{m_{j+1}}{m_{j}} \biggr)\frac {m_{j+1}}{\underline{\beta}(\varepsilon)} \biggr\rceil $$

guarantees the inequality $u_{k_{j}+K}\leq N/m_{j+1}$. In other words (u _k ) _k≥0 needs less than K steps to decrease from N/m _j to N/m _j+1. Summing the terms between k _j and k _j +K, we obtain

$$\begin{aligned} \sum_{k=k_{j}}^{k_{j}+K}u_{k} & \leq K \frac{N}{m_{j}} \leq \biggl[ \biggl(\log \frac{m_{j+1}}{m_{j}} \biggr)\frac {m_{j+1}}{\underline{\beta}(\varepsilon)}+1 \biggr]\frac{N}{m_{j}}. \end{aligned}$$

Taking m _j =2^j and m _j+1=2^j+1, we have k _j+1≤k _j +K and thus obtain

$$\begin{aligned} \sum_{k=k_{j}}^{k_{j+1}}u_{k}\leq\sum _{k=k_{j}}^{k_{j}+K}u_{k} & \leq \biggl[ (\log2 )\frac{2}{\underline{\beta}(\varepsilon )}+\frac{1}{2^{j}} \biggr]N = C(\varepsilon)N \end{aligned}$$

with C(ε) independent of N. We have thus bounded the full sum by

$$\begin{aligned} \sum_{k\geq0}(u_{k}-1) & \leq\sum _{k=0}^{\sigma_{2}}(u_{k}-1)+\sum _{k\geq\sigma_{2}}(u_{k}-1) \\ & \leq \biggl\lceil \frac{\log N}{\log2} \biggr\rceil C( \varepsilon)N+\frac {2N}{\varepsilon^{2}} \leq D(\varepsilon) N \log N \end{aligned}$$

for some D(ε) independent of N.