Learning Optimal Flows for Non-Equilibrium Importance Sampling
Anonymous Author(s)
AffiliationAddress email
Abstract
Many applications in computational sciences and statistical inference require thecomputation of expectations with respect to complex high-dimensional distributionswith unknown normalization constants, as well as the estimation of these constants.
Here we develop a method to perform these calculations based on generating samplesfrom a simple base distribution, transporting them along the flow generated by avelocity field, and performing averages along these flowlines.
This non-equilibriumimportance sampling (NEIS) strategy is straightforward to implement, and can be used for calculations with arbitrary target distributions.
On the theory side wediscuss how to tailor the velocity field to the target and establish general conditionsunder which the proposed estimator is a perfect estimator, with zero-variance.
Wealso draw connections between NEIS and approaches based on mapping a basedistribution onto a target via a transport map.
On the computational side we showhow to use deep learning to represent the velocity field by a neural network andtrain it towards the zero variance optimum.
These results are illustrated numericallyon benchmark examples (with dimension up to 10), where we show that trainingthe velocity field can decrease the variance of the NEIS estimator by up to 6 ordersof magnitude compared to a vanilla estimator.
We also show that NEIS withoptimized flows achieves significant variance reduction on these examples than Neal’s annealed importance sampling (AIS).
Introduction
Given a potential function 𝑈 1 : Ω → R on the domain Ω ⊆ R 𝑑 , the main goal of this paper is toevaluate
[Equation]
The calculations of such integrals arise in many applications from several scientific fields.
For instance,
Z 1 is known as the partition function in statistical physics [22], where it is used to characterize thethermodynamic properties of a system with energy 𝑈 1 , and as the evidence in Bayesian statistics,where it is used for model selection [13].
When the dimension 𝑑 of the domain Ω is large, standard numerical quadrature methods areinapplicable to (1) and the method of choice to estimate Z 1 is Monte-Carlo sampling [23, 8].
Thisrequires expressing Z 1 as an expectation, which can be done e.g. by realizing that
[Equation]
where E 0 denotes expectation with respect to the probability density function 𝜌 0 > 0.
If 𝜌 0 is bothknown (i.e. we can evaluate it pointwise in Ω , normalization factor included) and simple to samplefrom, we can build an estimator for Z 1 by replacing the expectation at the right hand side of (2) by the empirical average of 𝑒 − 𝑈 1 / 𝜌 0 over samples drawn from 𝜌 0 .
Unfortunately, finding a density 𝜌 0 thathas the two properties above is hard: unless 𝜌 0 is well-adapted to 𝑒 − 𝑈 1 , the estimator based on (2) isterrible in general, with a standard deviation that is typically much larger than its mean or even infinite.
A similar issue arises if we want to estimate the expectation E 0 𝑓 of some function 𝑓 : Ω → R , and thetwo problems are in fact connected when 𝑓 > 0 since the second reduces to (2) for 𝑈 1 = − log ( 𝑓 𝜌 0 ) .
These difficulties have prompted the development of importance sampling strategies [15] whoseaim is to produce estimators with a reasonably low variance for Z 1 or E 0 𝑓 .
These include forexample umbrella sampling [43, 41], replica exchange (aka parallel tempering)
[15, 24], nestedsampling [37, 38], etc., which aim is to amortize the calculation of Z 1 by turning it into the calculationof several expectations of the type (2), but with better properties, that can then be recombined usingthermodynamic integration [18] or Bennett acceptance ratio method [6].
Complementary to these equilibrium techniques, non-equilibrium sampling strategies have also beenintroduced for the calculation of (1).
For example, Neal’s annealed importance sampling (AIS)
[27]based on the Jarzynski equality [16, 17, 1] uses properly weighted averages over sequences of statesevolving from samples from 𝜌 0 to calculate Z 1 , without requiring that the kernel used to generate thesestates be in detailed-balance with respect to either 𝜌 0 , or 𝜌 1 : = 𝑒 − 𝑈 1 /Z 1 , or any density interpolatingbetween these two.
Instead the weight factors are based on the probability distribution of the sequenceof states in path space.
Other non-equilibrium sampling strategies in this vein include bridge and pathsampling [14], and sequential Monte Carlo (SMC) sampling [25, 2] (as an extension of AIS).
In this paper, we analyze another non-equilibrium importance sampling (NEIS) method that hasbeen introduced in [33].
NEIS is based on generating samples from a simple base density 𝜌 0 , thenpropagating forward and backward in time along the flowlines of a velocity field, and computingaverages along these trajectories—the basic idea of the method is to use the flow induced by thisvelocity field to sweep samples from 𝜌 0 through regions in Ω that contribute most to the expectation.
As shown in [33] and recalled below, this procedure leads to consistent estimators for the calculationof Z 1 or E 0 𝑓 via a generalization of (2).
One advantage of the method, which is a rare feature amongimportance sampling strategies, is that it leads to estimators that always have lower variance thanthe vanilla estimator based on (2) [33].
The question we investigate in this paper is how low theirvariance can be made, both in theory and in practice.
Our main contributions are:
• Under mild assumptions on 𝑈 1 and 𝜌 0 , we show that if the NEIS velocity field is the gradient of apotential that satisfies a Poisson equation, the NEIS estimator for Z 1 has zero variance.• Under the same assumptions, we show that this optimal flow can be used to construct a perfecttransport map from 𝜌 0 to 𝜌 1 .
This allows us to compare NEIS with importance sampling strategiesinvolving transport maps like normalizing flows (NF) that have recently gained popularity [32, 20, 30],and highlight some potential advantages of the former over the latter.• On the practical side, we derive variational problems for the optimal velocity field in NEIS, and
show how to solve these problems by approximating the velocity by a neural network and optimizingits parameters using deep learning training strategies, similar to what is done with neural ODE [9].
• We illustrate the feasibility and usefulness of this approach by testing it on numerical examples.
Firstwe consider Gaussian mixtures in up to 10 dimensions.
In this context, we show that training thevelocity used in NEIS allows to reduce the variance of a vanilla estimator using a standard Gaussiandistribution as 𝜌 0 by up to 6 orders of magnitude.
Second we study Neal’s 10 dimensional funneldistribution [28, 2] for which the variance of the vanilla importance sampling method is infinity;training a linear dynamics with 2 parameters in NEIS can lead to an estimator with empiricalvariance less than 1.
In these examples we also show that after training NEIS leads to estimatorswith lower variance than AIS [27].
Related works.
The idea of transporting samples from 𝜌 0 to lower the variance of the vanillaestimator based on (2) is also at the core of importance sampling strategies using normalizing flows(NF)
The type of transport used in NF-based method is howeverdifferent in nature from the one used in NEIS.
With NF, one tries to construct a map that transformseach sample from 𝜌 0 into a sample from the target 𝜌 1 = 𝑒 − 𝑈 1 /Z 1 .
In contrast, NEIS uses samplesfrom 𝜌 0 as initial conditions to generate trajectories, and uses the data along these entire trajectoriesto build an estimator.
Intuitively, this means that samples likely on 𝜌 0 must become likely on 𝜌 1sometime along these trajectories rather than at a given time specified beforehand, which is easier toenforce.
NEIS bears similarities with Neal’s AIS [27], except that in NEIS the sampling is done once from𝜌 0 to generate deterministic trajectories to gather data for the estimator, whereas AIS uses randomtrajectories.
There are some methods based on AIS that optimize the transition kernel: for instance,stochastic normalizing flows (SNF) proposed in [46] incorporates NF between annealing steps; andannealed flow transport (AFT) in [2] combines NF with the sequential Monte Carlo method toprovide optimized flow transport.
These approaches require learning several maps along the annealedtransition, whereas the NEIS herein only needs to learn a single flow dynamics.
A time-discrete version of NEIS termed NEO was proposed in [42].
The current implementationof NEO iterates on a map that needs to be prescribed beforehand, but this map could perhaps beoptimized using a strategy similar to the one proposed here.
From a practical standpoint, the idea of optimizing the velocity field in NEIS using a neural networkapproximation for this field can be viewed as an application of neural ODEs [9] that uses the varianceof the NEIS estimator as the objective function to minimize.
The nature of this objective posesspecific challenges in the training procedure, which we investigate here.
Notations.
For symmetry, we will denote 𝜌 0 ( 𝑥 ) =
𝑒 − 𝑈
( 𝑥 ) with 𝑈 0 =
− log 𝜌 0 : Ω → R and
Z 0 = ∫ Ω 𝑒 − 𝑈 0 ( 𝑥 ) d 𝑥 = 1.
We denote a 𝑑 -dimensional vector filled with zeros as 0 𝑑 and the 𝑑 × 𝑑identity matrix as I 𝑑 .
⟨· , ·⟩ is the Euclidean inner product in R 𝑑 .
We assume that the domain Ω iseither an open and connected subset of R 𝑑 with smooth boundary or a 𝑑 -dimensional torus (withoutboundary).
We denote by 𝒩 ( 𝜇, Σ ) the multivariate Gaussian density with mean 𝜇 and covariancematrix Σ .
For two functions 𝑓 , 𝑔 : 𝐷 → R where 𝐷 is a domain of interest, the notation 𝑓 ≲ 𝑔 meansthat there exists a constant 𝐶 >
such that 𝑓 ( 𝑥 ) ≤
𝐶𝑔 ( 𝑥 ) for any 𝑥 ∈ 𝐷 .
Suppose T : Ω → R 𝑑 is amap and 𝜇 is a distribution, then the pushforward distribution of 𝜇 by the map T is denoted as T
# 𝜇 .
The notation |·| is the usual ℓ 2 norm for vectors and ∥·∥ is the matrix norm or functional norm.
Flow-based NEIS method
Here we recall the main ingredients of the non-equilibrium importance sampling (NEIS) methodproposed in [33].
Let b : Ω → R 𝑑 be a velocity field which we assume belongs to the vector space
[Equation]
where n is the normal vector at the boundary 𝜕 Ω .
Define the associated flow map X 𝑡 : Ω → Ω via
[Equation]
as well as the Jacobian of this map:
[Equation]
Finally, let us denote
[Equation]
for 𝑘 ∈ { 0 , 1 } , 𝑥 ∈ Ω and 𝑡 ∈ R .
NEIS is based on the following result, proven in Appendix B.1:
Proposition 2.1.
If b ∈ 𝔅 , then for any −∞ < 𝑡 − < 𝑡 + < ∞ , we have
[Equation]
[Equation]
where
[Equation]
In addition, if
[Equation]
exists for almost all 𝑥 ∼ 𝜌 0 , then
[Equation]
When b = 0 𝑑 , or 𝑡 − ↑ 0 and 𝑡 + ↓ 0, (7) reduces to (2).
The aim, however, is to choose b so that theestimator based on (7) has a better variance than the one based (2): we will show below that this canindeed be done.
For now, note that Jensen’s inequality shows that an estimator based on (10) for any bhas lower variance than the one based (2); see [33] or Proposition H.3 below for details.
Interestingly, the estimator (10) based on (9) is invariant with respect to the parametrization of theflowlines generated by the dynamics b , as shown by the following result proved in Appendix B.2:
Proposition 2.2 (An invariance property) .
Suppose b , 𝛼 b ∈ 𝔅 , where 𝛼 ∈ 𝐶 ∞ ( Ω , R ) satisfyinginf 𝑥 ∈ Ω 𝛼 ( 𝑥 ) > 0 .
Then the fields b and 𝛼 b generate the same flowlines, and A b =
A 𝛼 b where A band A 𝛼 b are the function defined in (9) using b and 𝛼 b , respectively.
Optimal NEIS
The NEIS estimator for (10) is unbiased no matter what b is. However, its performance relies on thechoice of b .
Therefore, a natural question is to find the field b that achieves the largest variancereduction.
The next result shows that an optimal b exists that leads to a zero-variance estimator:
Proposition 3.1 (Existence of zero-variance dynamics) .
Suppose Ω = [ 0 , 1 ] 𝑑 is a torus and 135 𝑈 0 ,
𝑈 1 ∈ 𝐶 ∞ ( Ω , R ) .
Let D : Ω → ( 0 , ∞) be some smooth positive function, and suppose that 136 𝑉 ∈ 𝐶 ∞ ( Ω ) solves the following Poisson’s equation on Ω
[Equation]
If the solution 𝑉 is a Morse function, then b = ∇ 𝑉 is a zero-variance velocity field: if we use it todefine (9) , we have
[Equation]
for almost all 𝑥 ∼
[Equation]
12) and as a result
[Equation]
almost all
[Equation]
This proposition is proven in Appendix E where we also make a connection between (11) and
Beckmann’s transportation problem.
We stress that the optimal b specified in Proposition 3.1 isnot unique (see Proposition E.9): however, we show below in Proposition 4.1 that, under certainconditions, all local minima of the variance (viewed as a functional of b ) are global minima.
We alsonote that the assumption that the solution is a Morse function is mostly a technicality, as discussed in Appendix E.3.
Similarly, we consider the torus in Proposition 3.1 for simplicity mainly; we expectthat the proposition will hold in general when Ω has compact closure or even when Ω = R 𝑑 , see
Appendix E for examples including that of Gaussian mixture distributions.
For illustration, the contour plot of 𝑉 and the flowlines of b = ∇ 𝑉 are shown in Figure 1 in a simpleexample in a two-dimension torus where 𝜌 0 = 1 and 𝜌 1 is a mixture density with 3 modes; their explicit expressions are given in (52); in this example, we solved (11) numerically with D = 1, see
Appendix E.5 for more details.
Some other examples where the zero-variance dynamics is explicitare discussed in Appendix F.
Connection to transport maps and normalizing flows.
The zero-variance dynamics provides atransport map T from 𝜌 0 to 𝜌 1 , as shown in:
Proposition 3.2 (Existence of a perfect generator) .
Suppose D = 1 for simplicity.
Under the sameassumption as in Proposition 3.1, let 𝑉 be the Morse function solving (11) and b = ∇ 𝑉 the associatedzero-variance dynamics.
Then there exists a continuously differentiable function 𝜘 (defined almosteverywhere on Ω ) such that
[Equation]
Furthermore, the map T ( 𝑥 ) :
= X 𝜘 ( 𝑥 ) ( 𝑥 ) is a transport map from 𝜌 0 to 𝜌 1 , i.e., T # 𝜌 0 =
𝜌 1 .
The proof is given in Appendix G.1.
This transport map T appears to have better theoretical propertiescompared with ODE-based NF sampling method [9].
For example it was shown in [10] that aone-dimensional transport map T : R → R with T ( 1 ) = − 1 and T (− 1 )
= 1 cannot be realized vianeural ODEs.
However, this is not an issue for NEIS method: we can simply choose b ( 𝑥 ) = 1 and 165 𝜘 ( 𝑥 )
=
T ( 𝑥 ) − 𝑥 , and straightforwardly verify that X 𝜘 ( 𝑥 ) ( 𝑥 )
=
T ( 𝑥 ) .
We also illustrate the validityof Proposition 3.2 via non-trivial numerical examples in Appendix G.2.
We stress however that we will not use the transport map T of Proposition 3.2 in the examples below.
Indeed, using this map would require identifying 𝜘 , which introduces an unnecessary additionalcalculation which we can avoid using the NEIS estimator. In addition, the NEIS estimator will likelyhave better properties than those based on transport maps, as we can we can think of NEIS as using atime-parameterized family of transport maps rather than a single one. In particular, the variance ofthe NEIS estimator will be small if samples likely on 𝜌 0 become likely on 𝜌 1 sometime along the
NEIS trajectories, rather than at the same fixed time for all the samples. The former seems easier tofulfill than the latter.
For example, in one-dimension, the NEIS estimator has zero variance for any bbounded away from zero, whereas building a transport map from 𝜌 0 to 𝜌 1 is already nontrivial in thatsimple case.
Variational formulations
The Poisson equation (11) admits a variational formulation:
[Equation]
If D is chosen to be a probability density function (for example D = 𝜌 0 or D = 12 ( 𝜌 1 + 𝜌 0 ) )
the twoterms in the objective in (15) are expectations which can be estimated via sampling (using e.g. directsampling for the expectation with respect to 𝜌 0 and NEIS for the one with respect to 𝜌 1 ).
This meansthat we can in principle use an MCMC estimator of (15) as empirical loss, and minimize it over all 𝑉in some parametric class.
Here however, we will follow a different strategy that allows us to directlyparametrize b instead of 𝑉 (i.e. relax the requirement that b = ∇ 𝑉 ) and simply use the variance of theestimator as objective function.
Specifically, since we quantify the performance of the estimators based on (7) and (10) by theirvariance, we can view these quantities as functionals of b that we wish to minimize.
Since theestimators are unbiased, these objectives are
[Equation]
where we defined the second moments M 𝑡 − ,𝑡 + ( b ) :
=
E 0 (cid:2) |A
𝑡 − ,𝑡 + | 2 (cid:3) and M( b ) :
= E 0 (cid:2) |A| 2 (cid:3) .
With thefinite-time objective, we know that with b = 0 𝑑 , (7) reduces to (2).
Therefore minimizing M 𝑡 − ,𝑡 + ( b )over b by gradient descent starting from b near 0 𝑑 will necessarily produce a better estimator: whilewe cannot guarantee that the variance of this optimized estimator will be zero, the experimentsconducted below indicate that it can be a several order of magnitude below that of the vanilla estimator.
For the infinite-time objective, we know that for any b , (9) leads to an estimator with a lower variancethan the one based on (2) [33].
Minimizing Var ( b ) over b using gradient descent leads to a localminimum; the next result shows that all such local minima are global:
Proposition 4.1 (Global minimum) .
Under some technical assumptions listed in Proposition H.1,if b ∗ ∈ 𝔅 is a local minimum of Var ( b ) where the functional derivative of Var ( b ) with respect to bvanishes, i.e. 𝛿 Var ( b ∗ )/ 𝛿 b =
𝑑 on Ω , then b ∗ is a global minimum and Var ( b ∗ )
= 0 .
The expression of the functional derivative 𝛿 Var ( b ∗ )/ 𝛿 b is given in Proposition D.1.
The technicalassumptions under which Proposition 4.1 hold are explained in Appendix A and the proof is given in
Appendix H.
Training towards the optimal b
Here we discuss how to use deep learning techniques to find the optimal b ; these techniques will beillustrated on numerical examples in Section 6.
Some technical details are deferred to Appendix I.
Objective.
We use the finite-time objective M 𝑡 − ,𝑡 + ( b ) in (7) with 𝑡 − ∈ [− 1 , 0 ] , 𝑡 + = 𝑡 − + 1.
Twonatural choices are 𝑡 − = 0 and 𝑡 − = − 1 / 2, which will be used below in the numerical experiments.
This leads to no loss of generality a priori since in the training scheme we put no restriction onthe magnitude that b can reach, and with large b the flow line can travel large distance even during 210 𝑡 ∈ [− 1 , 1 ] (the range of integration in 𝑠, 𝑡 in (8)).
In practice we use a time-discretized version of (8)with 2 𝑁 𝑡 discretization points, and use the standard Runge-Kutta scheme of order 4 (RK4) to integratethe ODE (4) over 𝑡 ∈ [− 1 , 1 ] using uniform time step ( Δ 𝑡 = 1 / 𝑁 𝑡 ).
Neural architecture.
We parameterize b by deep neural networks, consistent with three differentansatzes: apart from the generic one (for which we parameterize b via neural networks directly), wealso consider gradient and divergence-free forms
[Equation]
with the skew-symmetric matrix J 𝑑 : = (cid:104)
I 𝑑 / 2 − I 𝑑 / 2 0 (cid:105) (assuming that the dimension 𝑑 is an even integer).
We always use an ℓ -layer neural network with width 𝑚 for all inner layers; therefore, from now on, wesimply refer the neural network structure by a pair ( ℓ, 𝑚 ) ; see Appendix I.2 for more details.
Whenwe parametrize the potential function 𝑉 , the only difference is that the output dimension of the neuralnetwork becomes 1 instead of 𝑑 .
The activation function is chosen as the softplus function, whichhas better empirical result compared to sigmoid and has better smoothness property compared with ReLu.
The parameters inside neural networks are randomly generated for initialization.
Theoreticalresults about the gradient of M 𝑡 − ,𝑡 + ( b ) with respect to parameters are given in Appendix C andcorresponding numerical implementations are explained in Appendix I.
Direct training method.
We minimize M 𝑡 − ,𝑡 + ( b ) with respect to the parameters in the neuralnetwork using stochastic gradient descent (SGD).
In the direct training strategy we evaluate the lossempirically using mini-batches of data drawn from 𝜌 0 at every iteration step of SGD.
For simplicity,we always use 𝜌 0 ( 𝑥 ) =
( 2 𝜋 ) − 𝑑 / 2 𝑒 − 12 | 𝑥
| 2 .
Assisted training method. When local minima of 𝑈 1 are far away from the local minimum of 𝑈 0 at 230 𝑥 = 0, the direct training method by sampling data from 𝜌 0 and minimizing M 𝑡 − ,𝑡 + ( b ) fails, becausethe flowlines may not reach the importance region of 𝜌 1 due to an undesirable random initializationof b .
More specifically, if along almost all trajectories, 𝑒 − 𝑈 1 ( X 𝑠 ( 𝑥 ))
≈ 0 for 𝑠 ∈ [− 1 , 1 ] , then withlarge probability A 𝑡 − ,𝑡 + ( 𝑥 ) ≈ where 𝑥 ∼ 𝜌 0 ; as a result the empirical variance of the estimator canbe extremely small if the number of samples is small, while the true variance could be extremelylarge.
Such a mode collapse phenomenon is common in rare event simulations.
Recall that ideally, we would like to find a dynamics b such that A is approximately a constantfunction in the infinite-time case; this suggests that we are not constrained to use the base distribution𝜌 0 and minimize the functional b → Var ( b ) in (16); if fact, if b is a zero-variance dynamics, then Ais simply a constant function and E 𝑝 (cid:2) (A − ( E 𝑝 A) 2 ) (cid:3) = 0 for any distribution 𝑝 .
Motivated by this idea and to remedy the above mentioned difficulty, we use an assisted training scheme in which, atiteration 𝑖 of SGD, the loss function is
[Equation]
Here E 𝑝 𝑖 denotes expectation with respect to the probability density 𝑝 𝑖 defined as
[Equation]
where 𝜐 ∈ ( 0 , 1 ) controls the number of steps during which the training is assisted, 𝑐 ∈ ( 0 , 1 ) , 𝐿 is thetotal number of training steps, and Z : = Z 𝑡 = 1 is the time-1 map of the ODE (cid:164) Z 𝑡 ( 𝑥 ) =
− 𝜍 ∇ 𝑈 1 (cid:0) Z 𝑡 ( 𝑥 ) (cid:1)with Z 𝑡 =
( 𝑥 ) =
𝑥 and 𝜍 > 0 is a parameter.
In essence, using (17) means that, for the first 𝜐𝐿 trainingsteps, there is a probability 𝑐 𝑖 that the data 𝑥 ∼ 𝜌 0 are replaced by Z ( 𝑥 ) , so that the training methodcan better explore important regions near local minima of 𝑈 1 .
Subsequently, the assistance is turnedoff so that some subtle adjustment can be made.
If some samples from 𝜌 1 = 𝑒 − 𝑈 1 /Z 1 were availablebeforehand, we could equivalently replace Z # 𝜌 0 in (18) by the empirical distribution over thesesamples.
We emphasize that the assisted training method is only used to guide the training initiallyand the NEIS estimator for Z 1 is unbiased.
Numerical experiments
We consider three benchmark examples to illustrate the effectiveness of NEIS assisted with training.
The first two examples involve Gaussian mixtures, for which we use NEIS with 𝑡 − = 0; the thirdexample is Neal’s funnel distribution, for which we use NEIS with 𝑡 − = − 1 / 2.
In all examples,we compare the performance of NEIS with those of annealed importance sampling (AIS) [27]; thenumber of transition in AIS is denoted as 𝐾
and we refer to this method as AIS- 𝐾 below; see moredetails in Appendix I.1.
When presenting results, we have rescaled estimates so that the exact value
Z 1 = 1 for all examples.
More implementation details about training are deferred to Appendix I.2,including computational resources and costs.
Appendix I.3 includes additional figures about training.
When computing the gradient of variance with respect to parameters, we use integration-basedmethod when 𝑡 − = − 1 / 2 (for the convenience of numerical implementation) and use ODE-basedmethod when 𝑡 − =
(for higher accuracy); details about these two approximation methods aregiven in Appendix I.4 and Appendix I.5 respectively.
The codes and assets can be accessed viahttps://anonymous.4open.science/r/NEIS-4DC0/ .
An asymmetric 2 -mode Gaussian mixture in 2D. As a first illustration, we consider an asymmetric2-mode Gaussian mixture
[Equation]
(19) where e 1 = (cid:2) 10 (cid:3) , e 2 = (cid:2) 01 (cid:3) , 𝜎 1 = √ 0 .
1, 𝜆 = 5.
With this choice of parameters, the variance of thevanilla estimator based on (2) is approximately 1 .
× 10 6 .
We train over 𝐿 = 200 SGD steps usingthe loss (17) by imposing bias in the first half of the training period only (i.e. with 𝜐 = 1 / 2).
Theevolution of the variance during the second half of the training (i.e. after the bias is turned off) isshown in Figure 10 in Appendix.
As seen in the figure, when the bias is turned off (i.e., the trainingstep 100), the variance is already quite small and in most cases, the variance continues to reduce as bgets further optimized.
For the gradient form ansatz, the best training result gives a b leading to anestimator whose variance is around 0 .
Sample trajectories following the vector field b obtainedafter training are shown in Figure 2(a): as can be seen the mass at the origin is split by the flowand transported towards the two local minima of 𝑈 1 .
In Figure 2(b), we compare the NEIS schemewith the dynamics obtained from training to (a) the NEIS scheme with a naive gradient flow choiceb = −∇ 𝑈 1 ; (b) AIS with 10 or 100 transitions.
The NEIS scheme with optimized dynamics indeedperforms significantly better than other cases.
A symmetric 4 -mode Gaussian mixture in 10D. Next we consider a symmetric 4-mode Gaussianmixture in 𝑑 = 10 dimension with
[Equation]
[Figure] [Figure]
Direct GD refers to the choice b = − 5 ∇ 𝑈 1 in the scheme; NN refers to the dynamics from the training where the vector 𝜇 𝑖 =
[ 𝜆 cos ( 𝑖𝜋 2 ) 𝜆 sin ( 𝑖𝜋 2 ) 0 0 ··· 0 ] and Σ =
Diag [ 𝜎 21 𝜎 2 1 𝜎 22 𝜎 2 2 ··· 𝜎 22 ] is a diagonalmatrix.
The parameters are 𝑑 = 10, 𝜎 1 = √ 0 .
1, 𝜎 2 = √ 0 .
5 and 𝜆 = 5.
With this choice of parameters,the variance of the vanilla estimator based on (2) is [-Var /Z 
-]approximately 2 .
15 × 10 6 .
We show the training result in Figure 11.
The variance reduces to 1 ∼ 10 for the gradient formansatz after 200 SGD steps; the training results for the generic and divergence-free form are not ascompetitive as the gradient form ansatz and are therefore not presented.
In Figure 3(a), we visualize(projected) sample trajectories following the dynamics b obtained by training.
As can be seen in
Figure 3(a), the mass near the origin flows towards four local minima of 𝑈 1 , as we would intuitivelyexpect.
In Figure 3(b), we could observe that NEIS scheme with an optimized flow can achieve muchmore variance reduction compared with AIS schemes or a direct gradient descent dynamics.
A funnel distribution in 10D. We consider the following 10D funnel distribution studied in [2, 28]:for the state 𝑥
=
[ 𝑥 1 𝑥 2 ··· 𝑥 10 ] ∈ R 10 ,
[Equation]
For numerical stability, we consider the above funnel distribution restricted to a unit ball centeredat the origin with radius 25; we choose the proposal 𝜌 0 to be the standard Gaussian.
Instead of
[Figure]
heuristically parameterizing the dynamics via neural-networks, we consider the following linearansatz,
[Equation]
where 𝛼, 𝛽 are two parameters to optimize.
We choose 𝛼 = 𝛽 = 2 for initialization and use the directtraining method with 𝑡 − = − 1 / 2.
We notice that the asymmetric choice 𝑡 − = 0 can also leads intomore optimal dynamics, but its performance is not as competitive as the symmetric case 𝑡 − = − 1 / 2.
It is very likely that such a difference is due to the structure of funnel distribution: each coordinate 303 𝑥 𝑖 (1 ≤ 𝑖 ≤ 10) has mean 0 and therefore, a symmetric version can probably better weight thecontribution from both forward and backward flows.
The training result is shown in Figure 12 in
Appendix: we can observe that the error and variance are overall decreasing during the training andthe parameters 𝛼, 𝛽 tend to increase with a similar speed.
In Figure 4(b), we verify that the NEISscheme with linear dynamics after the training achieves better variance reduction compared withother methods.
Though a direct comparison with Annealed Flow Transport (AFT) [2] is difficult, the NEIS appears to show more promising results than AFT for this example; cf Figure 2(b) in [2].
Conclusion and outlook
In this work, we revisited the NEIS strategy proposed in [33] and analyzed its capabilities, bothfrom theoretical and computational standpoints.
Regarding the former, we showed that NEIS leadsto a zero-variance estimator for a velocity field b = ∇ 𝑉 with a potential 𝑉 that satisfies a certain
Poisson equation with the difference between the target and the base density as source.
Moreover,a zero-variance dynamics can be used to construct a transport map from 𝜌 0 to 𝜌 1 .
In turn, wehighlighted the connection and difference between NEIS and importance sampling strategies basedon the normalizing flows (NF).
On the computational side, we showed that the variance of the NEIS estimator can be used asobjective function to train the velocity field b .
This training procedure can be performed in practice byapproximating the velocity field by a neural network, and optimizing the parameters in this networksusing SGD, similar to what is done in the context of neural ODE but with a different objective.
Ournumerical experiments showed that this strategy is effective and can lower the variance of a vanillaestimator for Z 1 by several orders of magnitude.
While the numerical examples we used in the present paper are somewhat academic, the resultssuggest that NEIS has potential in more realistic settings.
In order to explore other applications, itwill be interesting to investigate how to best parametrize b (e.g., less parameters and non-stiff energylandscape with respect to these parameters) and how to best initiate the training procedure.
Theanswers to these questions are probably model specific and are left for future work.
Checklist
For all authors...
(a) Do the main claims made in the abstract and introduction accurately reflect the paper’s
contributions and scope?
[Yes]
(b) Did you describe the limitations of your work?
[Yes]
(c) Did you discuss any potential negative societal impacts of your work?
[N/A]
(d) Have you read the ethics review guidelines and ensured that your paper conforms to them?
[Yes]
If you are including theoretical results...
(a) Did you state the full set of assumptions of all theoretical results?
[Yes]
(b) Did you include complete proofs of all theoretical results?
[Yes]
If you ran experiments...
(a) Did you include the code, data, and instructions needed to reproduce the main
experimental results (either in the supplemental material or as a URL)?
[Yes] It can be
found in https://anonymous.4open.science/r/NEIS-4DC0/ .
(b) Did you specify all the training details (e
g., data splits, hyperparameters, how theywere chosen)?
[Yes] Details can be found in Section 5, Section 6 or Appendix I.
(c) Did you report error bars (e.g., with respect to the random seed after running experiments
multiple times)?
[Yes] We alternatively use the boxplot for multiple independentexperiments in Section 6.
(d) Did you include the total amount of compute and the type of resources used (e.g., typeof GPUs, internal cluster, or cloud provider)?
[Yes] see Appendix I.2.
If you are using existing assets (e
g., code, data, models) or curating/releasing new assets...
(a) If your work uses existing assets, did you cite the creators?
[N/A]
Our examples
do not involve special data and we only use standard Julia packages that are widely
available.
The dependence of the code will be explained in README.md in the code
https ://anonymous.4open.science/r/NEIS 4DC0/ )
(b) Did you mention the license of the assets?
[N/A]
(c) Did you include any new assets either in the supplemental material or as a URL?
[Yes]
(d) Did you discuss whether and how consent was obtained from people whose data you’re using/curating?
[N/A]
(e) Did you discuss whether the data you are using/curating contains personally identifiable
information or offensive content?
[N/A]
If you used crowdsourcing or conducted research with human subjects...
(a) Did you include the full text of instructions given to participants and screenshots, if
applicable?
[N/A]
(b) Did you describe any potential participant risks, with links to Institutional Review
Board (IRB) approvals, if applicable?
[N/A]
(c) Did you include the estimated hourly wage paid to participants and the total amount spent on participant compensation?
[N/A]
A The functional space for the infinite-time case
A.1 Assumptions
For simplicity, we make:
Assumption A.1.
(i)
The domain Ω is either • an open and connected subset of R 𝑑 with smooth boundary, • or a 𝑑 -dimensional torus (without boundary). (ii) 𝑈 0 , 𝑈 1 ∈ 𝐶 ∞ ( Ω , R ) .
(iii) Z 0 : = ∫ Ω 𝑒 − 𝑈
= 1, Z 1 < ∞ and Var ( max ) ∈ ( 0 , ∞) , where
[Equation]
[-E 𝜌 
 [ 𝑒 − 2 ( 𝑈 1 − 𝑈 0 ) ] − (Z 1 )is-]is the variance for the direct importance sampling method.
Notations
The open ball around 𝑥 with radius 𝑟 is denoted as 𝐵 𝑟 ( 𝑥 ) : =
(cid:8) 𝑦 ∈ Ω (cid:12) (cid:12) | 𝑦 − 𝑥 | < 𝑟 (cid:9) .
For any subset𝐷 ⊂ Ω , let 𝜏 + 𝐷 ( 𝑥 ) and 𝜏 − 𝐷 ( 𝑥 ) be the first hitting times to the boundary 𝜕𝐷 in the forward and backwarddirections, respectively.
More specifically,
[Equation]
For later convenience, let us denote
[Equation]
23) and thus
[Equation]
A.2 Preliminaries
Lemma A.2.
If b ∈ 𝔅 in (3) , then for any 𝑡, 𝑠 ∈ R , 𝑥 ∈ Ω and 𝑘 = 0 , 1 ,
J 𝑡 (cid:0) X
𝑠
( 𝑥 ) (cid:1) =
J 𝑡 + 𝑠 ( 𝑥 )/J
𝑠 ( 𝑥 ) ,
F ( 𝑘 ) 𝑡 (cid:0)
[Equation]
[Equation]
This lemma can be verified easily by definition and thus the proof is omitted.
A simple consequenceof (24) is the following result that we also state without proof:
Lemma A.3.
For any 𝑠 ∈ R and 𝑥 ∈ Ω ,
[Equation]
provided that these terms are well-defined.
A.3 The functional space
For the infinite-time case, we consider the following family of vector fields denoted as 𝔅 ∞ .
Definition A.4.
𝔅 ∞ is a set that contains all b ∈ 𝔅 such that Ω \ ℧ ( b ) has Lebesgue measure zero,where ℧ ( b )
⊂ Ω is the collection of points 𝑥 at which the functions
[Equation] [Equation]
are continuous on a local neighborhood of 𝑥 for 𝑘 = 0 , 1.
We use the notation ℧ ( b ) because in general, such a subset depends on the choice of b .
The mainreason behind the above definition is that we need functions in (26) to behave nicely almost everywhere.
The integrability of 𝑡 → F ( 𝑘 ) 𝑡 ( 𝑥 ) depends on the long-term behavior of the flow, which is not easy tocharacterize in general; thus we simply include the integrability into the assumption.
However, wecan indeed expect that the above conditions should hold for most interesting examples, e.g.:
Example A.5 .
If 𝑈 0 ( 𝑥 ) = | 𝑥 | 2 2 + 𝑑 2 ln ( 2 𝜋 ) and 𝑈 1 ( 𝑥 ) = | 𝑥 | 2 2 𝜎 2 (so that both 𝜌 0 and 𝜌 1 are Gaussiandensities), one optimal flow is b ( 𝑥 ) =
𝑥 (cf.
Appendix F below).
For this choice, we can directcompute that when 𝑧 ≠ 0 𝑑 , X 𝑡 ( 𝑧 ) = 𝑒 𝑡 𝑧 and hence
[Equation]
and similar expressions hold for F ( 0 ) by letting 𝜎 = 1 in above equations.
Then clearly, ℧ ( b ) =
Ω \{ 0 𝑑 } and such a dynamics belongs to 𝔅 ∞ .
However, the constant function b = 0 𝑑 ∉ 𝔅 ∞ : we caneasily verify that e.g., for this dynamics, ∫ ∞ 0 F ( 1 ) 𝑡 ( 𝑧 ) d 𝑡 = ∞ for any 𝑧 and hence ℧
( 0 𝑑 ) = ∅ ; see also Lemma A.7 below. Here are a few immediate properties of the set ℧ ( b ) :
Lemma A.6.
We have:
(i) ℧ ( b ) is an open subset of Ω .
(ii) The set ℧ ( b ) is closed under the evolution of the dynamics b ,
i
e., if 𝑥 ★ ∈ ℧ ( b ) , thenX 𝑡 ( 𝑥 ★ ) ∈ ℧ ( b ) for any 𝑡 ∈ R .
(iii)
If 𝑥 ★ ∈ ℧ ( b ) , then X 𝑡 ( 𝑥 ★ ) ∈ ℧ ( b ) for any 𝑡 ∈ R , where ℧ ( b ) is the closure of ℧ ( b ) in R 𝑑 .
(iv) 𝑥 → ∫ ∞ −∞ F ( 𝑘 ) 𝑡 ( 𝑥 )
d 𝑡 is continuous on ℧ ( b ) .
(v) ( 𝑥, 𝑡 ) → ∫ ∞ 𝑡 F ( 𝑘 ) 𝑠
( 𝑥 )
d 𝑠 is continuous on ℧ ( b ) × R .
Proof.
The part (i) follows easily from the definition of ℧ ( b ) and the part (iv) also trivially holds.
For the part (ii), notice that if 𝑦 = X 𝑡 ( 𝑥 ) , then∫ ∞
[Equation]
Since ( 𝑥, 𝑡 ) → X 𝑡 ( 𝑥 ) , ( 𝑥, 𝑡 ) → J 𝑡 ( 𝑥 ) are continuous by the smoothness assumption of b , it is clear that J 𝑡 (cid:0) X − 𝑡 ( 𝑦 ) (cid:1) is continuous on Ω .
The continuity of 𝑦 → ∫ 𝑡 0 F ( 𝑘 ) 𝑠
(cid:0) X − 𝑡 ( 𝑦 ) (cid:1)
d 𝑠 also triviallyholds on Ω .
Next, the continuity of 𝑦 → ∫ ∞ 0 F ( 𝑘 ) 𝑠
(cid:0) X − 𝑡 ( 𝑦 ) (cid:1)
d 𝑠 in a local neighborhood of X 𝑡 ( 𝑥 ★ )comes from the assumption that 𝑥 ★ ∈ ℧ ( b ) .
Thus, 𝑦 → ∫ ∞ 0 F ( 𝑘 ) 𝑠
( 𝑦 )
d 𝑠 is continuous in a localneighborhood of X 𝑡 ( 𝑥 ★ ) , if 𝑥 ★ ∈ ℧ ( b ) .
The other case for 𝑦 → ∫
−∞ F ( 𝑘 ) 𝑠 ( 𝑦 ) d 𝑠 can be similarlyverified. The part (iii) is an immediate consequence of part (ii).
Let us prove it by contradiction.
Assume thatthe conclusion in part (iii) does not hold, then there exists 𝑥 ★ ∈ ℧ ( b ) such that 𝑦 ★ : = X 𝑡 ( 𝑥 ★ ) ∉ ℧ ( b )for some 𝑡 ∈ R . By the part (ii), we know this 𝑥 ★ ∉ ℧ ( b ) and thus 𝑥 ★ ∈ 𝜕 ℧ ( b ) is in the boundary.
As℧ ( b ) is closed and 𝑦 ★ ∉ ℧ ( b ) , there must exist an 𝜖 > 0 such that for any 𝑦 ∈ R 𝑑 with | 𝑦 − 𝑦 ★ | < 𝜖 ,we have 𝑦 ∉ ℧ ( b ) .
By the smoothness assumption on b , we know there exists a 𝛿 > 0 such that for any 𝑥 ∈ 𝐵 𝛿 ( 𝑥 ★ ) , we have | X 𝑡 ( 𝑥 ) − 𝑦 ★ | < 𝜖 outside of ℧ ( b ) .
Since 𝑥 ★ is in the boundary of ℧ ( b ) ,we must be able to find an 𝑥 ∈ ℧
( b ) such that 𝑥 ∈ 𝐵 𝛿 ( 𝑥 ★ ) and by part (ii), X 𝑡 ( 𝑥 ) ∈ ℧ ( b ) .
Thus, wereach a contradiction.
For the part (v),
[Equation]
Because ( 𝑥, 𝑡 ) → X 𝑡 ( 𝑥 ) is continuous (by the smoothness of b ∈ 𝔅 ) and 𝑧 → ∫ ∞ 0 F ( 𝑘 ) 𝑠
( 𝑧 ) d 𝑠is continuous in a local neighborhood of X 𝑡 ( 𝑥 ) ∈ ℧ ( b ) by the part (ii), it is then clear that 538 ∫ ∞ [Equation] (cid:0) X 𝑡 ( 𝑥 ) (cid:1)
d 𝑠 is continuous with respect to ( 𝑥, 𝑡 ) , and hence the conclusion follows easily.
□
A notable family of points that are excluded from ℧ ( b ) are points at which ∫ ∞
−∞ F ( 𝑘 ) 𝑡 ( 𝑥 ) d 𝑡 = ∞ .
Forinstance, these include stationary points of b and periodic orbits, which we state as:
Lemma A.7.
Given a vector field b ∈ 𝔅 , we have 𝑥 ∉ ℧
( b ) ,(i) if 𝑥 is a stationary point {+of b (i.e., b ( 𝑥 ) = 
+}
𝑑 ), or(ii)
if 𝑥 is on a periodic orbit.
Proof.
If 𝑥 is a stationary point of b , the trajectory is X 𝑡 ( 𝑥 )
=
𝑥 for all 𝑡 ∈ R .
Then it is obvious that 545 ∫
[Equation]
𝑡 = ∞ , which establishes the part (i).
For the part (ii), letus assume that the period is 𝜏 .
If 𝑐 : = ∫ 𝜏 0 ∇ · b (cid:0) X 𝑠 ( 𝑥 ) (cid:1)
d 𝑠 ≥ 0, then∫ ∞
[Equation]
where the constant 𝐶 = 𝑒 − max 0 ≤ 𝑠 ≤ 𝜏 𝑈 𝑘 (cid:0) X 𝑠 ( 𝑥 ) (cid:1)
𝑒 min 0 ≤ 𝑠 ≤ 𝜏 ∫ 𝑠 0 ∇· b (cid:0) X 𝑟 ( 𝑥 ) (cid:1) d 𝑟 .
Therefore, ∫ ∞ −∞ F ( 𝑘 ) 𝑡 ( 𝑥 ) = ∞ .
If 𝑐 : = ∫ 𝜏 0 ∇ · b (cid:0) X 𝑠 ( 𝑥 ) (cid:1)
d 𝑠 < 0, we can similarly show that ∫
−∞ F ( 0 ) 𝑡 ( 𝑥 ) d 𝑡 = ∞ , and the sameconclusion holds.
□
A.4 Perturbation of the dynamics
Next we investigate the following question: given b ∈ 𝔅 ∞ and 𝛿 b ∈ 𝐶 ∞ 𝑐 ( Ω , R 𝑑 ) , can we guaranteethat b + 𝜖𝛿 b ∈ 𝔅 ∞ for small enough 𝜖 ?
Such a conclusion trivially holds for the finite-time case;however, more underlying structures are needed for the infinite-time case, due to the fact that thelong-term behavior of the flow b can sensitively depend on the local perturbation 𝛿 b .
This question isprobably unavoidable in order to understand the mathematical structure of 𝔅 ∞ .
Notation: We shall use the notation X 𝑡 (·) to represent the flow map under b , and use X 𝜖𝑡 (·) torepresent the flow map under b 𝜖 :
= b + 𝜖𝛿 b in this section below.
Moreover, we shall use F ( 𝑘,𝜖 ) · (·) todenote the function defined in (6) for the dynamics b 𝜖 and the Jacobian of the flow J 𝜖 (·) (·) is similarlydefined.
Definition A.8 ( b -stability) .
Given b ∈ 𝔅 ∞ :(a)
For an open bounded set).
A nonempty open bounded set 𝐷 ⊂ ℧ ( b ) is said to be b -stableif:
(i)
There exists a point 𝑥 ★ ∈ 𝐷 and 𝜁 ∈ ( 0 , 1 ) such that
[Equation]
(ii)
For any 𝑥 ∈ 𝐷 , the trajectory (cid:8) X 𝑡 ( 𝑥 ) (cid:9) 𝑡 ∈ R intersects with the boundary 𝜕𝐷 at exactly two points.
(b) (For a point).
A point 𝑥 ∈ ℧ ( b ) is said to be b -stable if there exists a neighborhood 𝐵 𝜖 ( 𝑥 )such that the region 𝐵 𝜖 ( 𝑥 ) is b -stable.
Otherwise, the point 𝑥 is said to be b -unstable.
The assumption in the part (i) ensures that the trajectory 𝑡 → X 𝑡 ( 𝑥 ) will leave this region 𝐷 within afinite amount of time; see Lemma A.9 below.
The part (ii) is used to ensure that the trajectory is not(infinitely) recurrent to 𝐷 ; once the trajectory leaves 𝐷 , it will not return to 𝐷 again.
Lemma A.9.
Suppose b ∈ 𝔅 and 𝐷 ⊂ Ω is open and bounded.
If there exists a point 𝑥 ★ ∈ 𝐷 and𝜁 ∈ ( 0 , 1 ) such that
[Equation]
then for any 𝑥 ∈ 𝐷 , we have
[Equation]
Proof.
Consider the quantity ℎ 𝑡 ( 𝑥 ) :
= (cid:10) b ( 𝑥 ★ ) , X 𝑡 ( 𝑥 )− 𝑥 ★ (cid:11) for 𝑥 ∈ 𝐷 .
Then when 𝑡 ∈ (cid:0) 𝜏 − 𝐷 ( 𝑥 ) , 𝜏 + 𝐷 ( 𝑥 ) (cid:1) ,d [Equation] 𝑥 [Equation]
★
[Equation]
𝑥 ★ )
[Equation]
2 > 0 .
Therefore,
[Equation]
Then the conclusion can be immediately obtained.
□ We now state the main result of this section:
Proposition A.10.
Suppose b ∈ 𝔅 ∞ and 𝑥 ★ ∈ ℧ ( b ) is b -stable.
Then there exists an open boundedneighborhood of 𝑥 ★ , denoted as 𝐷 ⊂ ℧ ( b ) , such that for an arbitrary 𝛿 b ∈ 𝐶 ∞ 𝑐 ( 𝐷, R 𝑑 ) , there existsan 𝜖 0 > 0 and b + 𝜖𝛿 b ∈ 𝔅 ∞ for any 𝜖 ∈ ( 0 , 𝜖 0 ) .
Proof.
The main idea is that if the path (cid:8) X 𝑡 ( 𝑥 ) (cid:9) 𝑡 ∈ R passes through 𝐷 , then a small perturbationwithin a bounded time period does not affect the long-term behavior; if the path does not pass through𝐷 , then b 𝜖 = b along this path and therefore, ∫ ∞
F ( 𝑘,𝜖 ) 𝑡 ( 𝑥 )
d 𝑡 = ∫ {+∞ 
+}
F ( 𝑘 ) 𝑡 ( 𝑥 ) d 𝑡 and hence, thecontinuity is also preserved locally around 𝑥 .
Step (I): Setup and the choice of 𝐷 .
Since 𝑥 ★ ∈ ℧ ( b ) , we know | b ( 𝑥 ★ )| > 0 by Lemma A.7.
Moreover, we can find a small b -stableball 𝐵 𝜃 ( 𝑥 ★ ) by Definition A.8 with a parameter 𝜁 ∈ ( 0 , 1 ) in (27).
Then consider the followingcross-section within 𝐵 𝜃 ( 𝑥 ★ )
[Equation]
and define the streamtube 𝔗 passing through 𝑆 as
[Equation]
It is not hard to see that 𝔗 is an open subset of ℧ ( b ) by Lemma A.6.
Then let us choose 𝐷 as anarbitrary open ball around 𝑥 ★ such that [Equation]
Next let us consider an arbitrary 𝛿 b ∈ 𝐶 ∞ 𝑐 ( 𝐷, R 𝑑 ) and from now on, we shall fix 𝛿 b .
It is easy toverify that b + 𝜖𝛿 b ∈ 𝔅 for any 𝜖 ∈ R .
The non-trivial part is to check that Ω \ ℧ ( b 𝜖 ) has measurezero for sufficiently small 𝜖 and hence b 𝜖 ∈ 𝔅 ∞ .
Step (II): Choice of 𝜖 0 .
Let us pick
[Equation]
for any 𝜁 ★ ∈ ( 𝜁, 1 ) .
The main motivation is that we need (27) to hold for b 𝜖 as well.
Indeed, if 𝜖 ≤ 𝜖 0 ,
[Equation]
and for any 𝑥 ∈ 𝐵 𝜃 ( 𝑥 ★ ) ,
[Equation]
As discussed above, this property ensures that any trajectory (cid:8)
X 𝜖𝑡 ( 𝑥 ) (cid:9) 𝑡 ∈ R with 𝑥 ∈ 𝐵 𝜃 ( 𝑥 ★ ) will passthrough the boundary 𝜕𝐵 𝜃 ( 𝑥 ★ ) , and at the same time, since 𝐵 𝜃 ( 𝑥 ★ ) is b -stable, the condition (ii) in
Definition A.8 ensures that such a trajectory only intersects with the boundary
𝜕𝐵 𝜃 ( 𝑥 ★ ) at exactly 2points.
Let us denote
[Equation]
By Lemma A.9, we know
[Equation]
Step (III): Prove that Ω \ ℧ ( b 𝜖 ) has measure zero for any 𝜖 ∈ ( 0 , 𝜖 0 ) .
We will prove that for any 𝜖 ∈ ( 0 , 𝜖 0 ) ,
[Equation]
𝜖 [Equation] It could be observed that the boundary 𝜕 𝔗 has measure zero: the boundary contains flowlinesfrom a hyper-surface with dimension 𝑑 − 2, that is, 𝜕 𝔗 is a set of flowlines passing through(cid:8) 𝑦 : | 𝑦 − 𝑥 ★ | = 𝜃 / 2 , ⟨ 𝑦 − 𝑥 ★ , b ( 𝑥 ★ )⟩ = 0 (cid:9) . Hence, provided that the above equation holds, weimmediately know that
[Equation]
𝔗 𝑐 ∩ ℧ ( b ) 𝑐 (cid:1) ∪ ( 𝔗 𝑐 ∩ 𝔗 ) ⊂ ℧ ( b ) 𝑐 ∪ 𝜕 𝔗 = (cid:0) Ω \ ℧ ( b ) (cid:1) ∪ 𝜕 𝔗 has measure zero, where the superscript 𝑐 means set complement.
As a remark, from now on, weshall fix 𝜖 ∈ ( 0 , 𝜖 0 ) .
Proof of 𝔗 ⊂ ℧ ( b 𝜖 ) .
Let us pick an arbitrary 𝑥 ∈ 𝔗 and we shall prove that 𝑧 → ∫ ∞
F ( 𝑘,𝜖 ) 𝑡 ( 𝑧 ) d 𝑡 iscontinuous locally near 𝑥 .
Similarly, we can verify that 𝑧 → ∫ 0 −∞ F ( 𝑘,𝜖 ) 𝑡 ( 𝑧 ) d 𝑡 is locally continuousnear 𝑥 .
Therefore, 𝑥 ∈ ℧ ( b 𝜖 ) and thus 𝔗 ⊂ ℧ ( b 𝜖 ) .
Next we return to verify that 𝑧 → ∫ ∞ 0 F ( 𝑘,𝜖 ) 𝑡 ( 𝑧 )
d 𝑡 is continuous locally near 𝑥 .
We claim that thereexists a 𝑦 ∈ 𝑆 ∪ 𝐷 ∈ 𝐵 𝜃 / 2 ( 𝑥 ★ ) and 𝑠 ∈ R such that X 𝜖𝑠 ( 𝑦 )
= 𝑥 .
To prove this, we need to discuss twocases:
• Suppose (cid:8) X 𝜖𝑡 ( 𝑥 ) (cid:9) 𝑡 ∈ R never enters 𝐷 .
Because b 𝜖 = b on Ω \ 𝐷 , we know X 𝜖𝑡 ( 𝑥 ) =
X 𝑡 ( 𝑥 )for any 𝑡 ∈ R .
By the definition of the streamtube 𝔗 , there exists a 𝑦 ∈ 𝑆 and 𝑠 ∈ R such that𝑦 =
X − 𝑠
( 𝑥 ) =
X 𝜖 −
𝑠 ( 𝑥 ) , which immediately gives X 𝜖𝑠 ( 𝑦 ) =
𝑥 .•
Suppose (cid:8) X 𝜖𝑡 ( 𝑥 ) (cid:9) 𝑡 ∈ R enters 𝐷 at some time.
Then the above conclusion follows easily
Because b 𝜖 is smooth, for small enough 𝛿 , we can ensure that 𝐵 𝛿 ( 𝑥 ) ⊂ 𝔗 and also X 𝜖 − 𝑠
( 𝑧 ) ∈
𝐵 𝜃 / 2 ( 𝑥 ★ )for any 𝑧 ∈ 𝐵 𝛿 ( 𝑥 ) .
By (29),
[Equation]
We divide the proof of continuity of 𝑧 → ∫ ∞ 0 F ( 𝑘,𝜖 ) 𝑡 ( 𝑧 ) d 𝑡 into two cases:(a)
If 𝜏 ≤ 𝑠 , then we already know for any point 𝑧 ∈ 𝐵 𝛿 ( 𝑥 ) , we have X 𝜖𝑡 ( 𝑧 )
∉
𝐵 𝜃 ( 𝑥 ★ ) for 𝑡 ≥ 0.
Recall that b =
b 𝜖 outside of 𝐵 𝜃 ( 𝑥 ★ ) .
Hence X 𝜖𝑡 ( 𝑧 ) =
X 𝑡 ( 𝑧 ) for any 𝑡 ≥ 0, and∫ ∞
[Equation]
which is continuous on a neighbor of 𝑥 by 𝑥 ∈ ℧ ( b ) .(b)
If 𝜏 > 𝑠 , then by (30), we know X 𝜖𝑡 ( X 𝜖𝜏
−
𝑠
( 𝑧 ))
∉
𝐵 𝜃 ( 𝑥 ★ ) for any 𝑧 ∈ 𝐵 𝛿 ( 𝑥 ) and 𝑡 ≥ 0 andin particular, X 𝜖𝜏 − 𝑠 ( 𝑧 ) ∉
𝐵 𝜃 ( 𝑥 ★ ) for any 𝑧 ∈ 𝐵 𝛿 ( 𝑥 ) .
Let us rewrite∫ ∞
[Equation]
Since b 𝜖 is smooth, apparently 𝑧 → ∫ 𝜏 − 𝑠
F ( 𝑘,𝜖 ) 𝑡 ( 𝑧 ) d 𝑡 and 𝑧 → 𝐽 𝜖𝜏 − 𝑠
( 𝑧 ) are continuous onΩ .
The continuity of 𝑧 → ∫ ∞ 0 F ( 𝑘,𝜖 ) 𝑡 (cid:0) X 𝜖𝜏 − 𝑠 ( 𝑧 ) (cid:1)
d 𝑡 locally near 𝑥 can be exactly proved inthe same way as the Part (a) above for the new point X 𝜖𝜏 − 𝑠
( 𝑥 ) .
One small technical result to verify is that X 𝜖𝜏 − 𝑠 ( 𝑥 ) ∈ ℧ ( b ) in order to apply the sameargument from the Part (a).
Note that X 𝜖𝜏 − 𝑠
( 𝑥 ) =
X 𝜖𝜏 ( 𝑦 ) =
X 𝜖𝜏 − (cid:101) 𝜏 (cid:0) X 𝜖 (cid:101) 𝜏 ( 𝑦 ) (cid:1) where
[Equation]
Since X 𝜖 (cid:101) 𝜏 ( 𝑦 ) ∈
𝜕𝐵 𝜃 ( 𝑥 ★ ) , we know X 𝜖 (cid:101) 𝜏 ( 𝑦 ) ∈ ℧ ( b ) (e.g., by choosing a small enough 𝜃 ).
Outside of 𝐵 𝜃 ( 𝑥 ★ ) , we know b 𝜖 = b and thus X 𝜖𝜏 − (cid:101) 𝜏 (cid:0)
X 𝜖 (cid:101) 𝜏
( 𝑦 ) (cid:1) =
X 𝜏 − (cid:101) 𝜏 (cid:0) X 𝜖 (cid:101) 𝜏
( 𝑦 ) (cid:1) ∈ ℧ ( b ) by
Lemma A.6.
Proof of ℧ ( b )\ 𝔗 ⊂ ℧ ( b 𝜖 ) .
Let us consider an arbitrary point 𝑥 ∈ ℧ ( b )\ 𝔗 .
Note that b 𝜖 =
boutside of 𝐷 ⊂ 𝔗 .
For a local neighborhood 𝐵 𝛿 ( 𝑥 ) outside of 𝔗 , we also know for any 𝑦 ∈ 𝐵 𝛿 ( 𝑥 ) ,
X 𝜖𝑡 ( 𝑦 ) = X 𝑡 ( 𝑦 ) ∉ 𝔗 for 𝑡 ∈ R , by both the definition of 𝔗 and the construction b 𝜖 =
b outside of 𝔗 .
By the same argument as in the Part (a) above, it could be readily shown that 𝑥 ∈ ℧ ( b 𝜖 ) and thus℧ ( b )\ 𝔗 ⊂ ℧ ( b 𝜖 ) . □
B Supplementary material for Section 2B.1 Proof of Proposition 2.1
We first prove the finite-time case.
As b ∈ 𝔅 , we know that ∫ 𝑡 + 𝑡 − F ( 0 ) − 𝑠 ( 𝑥 ) d 𝑠 ≡∫ 𝑡 + 𝑡 − 𝑒 − 𝑈 0 (cid:0) X − 𝑠
( 𝑥 ) (cid:1)
J − 𝑠
( 𝑥 )
d 𝑠 < ∞ by the continuity assumption of b and Assumption A.1.
Then
[Equation]
By the change of variables 𝑥 =
X 𝑡 ( (cid:101) 𝑥 ) , we have
[Equation]
Note that as the integrand is non-negative, switching the order of time integration and space integrationis justified by Fubini–Tonelli theorem.
The proof of Proposition 2.1 for the infinite-time case is essentially the same as the finite-time case,except the followings:
• We need to replace the domain Ω in (31) by ℧
( b ) defined in Definition A.4.•
We need the continuity of 𝑥 → ∫ ∞ −∞ F ( 0 ) 𝑡
( 𝑥 ) d 𝑡 in order to use Theorem 2 in [21] to get the firstline in (31).
A generalization with discontinuity should be possible by e.g., considering piecewisecontinuous b .
However, we shall not explore this further in this work.
As a remark, the map𝑥 → X 𝑡 ( 𝑥 ) being bĳective (due to the nature of ODE flows) on ℧ ( b ) was proved in Lemma A.6(iii), when applying [21, Theorem 2].
B.2 Proof of Proposition
Consider
[Equation]
To prove the first conclusion, we need to verify that the trajectory (cid:8) X 𝑡 [Equation] 𝑥 ) [Equation]
𝑡 [Equation] R = (cid:8) Z 𝑡 ( 𝑥 ) (cid:9) 𝑡 ∈ R . From now on, let us fix 𝑥 and introduce a scalar-valued function 𝜃 by 𝜃 𝑡 : = ∫ 𝑡 0 𝛼 (cid:0)
Z 𝑠 ( 𝑥 ) (cid:1) d 𝑠
(i.e.,dd 𝑡 𝜃 𝑡 = 𝛼 (cid:0) Z 𝑡 ( 𝑥 ) (cid:1) .
By taking the time derivative, it is not hard to verify that Z 𝑡 ( 𝑥 ) =
X 𝜃 𝑡 (cid:0) 𝑥 (cid:1) asboth satisfy the same ODE.
Of course, 𝜃 also depends on 𝑥
but we shall not explicitly specify thisdependence for simplicity of notations.
Thus, the trajectory 𝑡 → Z 𝑡 ( 𝑥 ) is the same as 𝑡 → X 𝑡 ( 𝑥 )under time re-scaling specified by 𝜃 .
Next we shall prove the following lemma, which immediately leads into the second result in
Proposition 2.2.
Lemma B.1.
Suppose 𝑔 : Ω → R is a non-negative continuous function.
For any 𝑥 ∈ Ω ,
[Equation]
Proof.
Because 𝛼 is strictly positive, 𝜃 · : R → R is bĳective, and by the inverse function theorem
[Equation]
Then by the change of variables (cid:101) 𝑡 = 𝜃 𝑡 and (cid:101) 𝑠
= 𝜃 𝑠 ,∫ ∞
[Equation]
[Equation]
It is then sufficient to show that 𝜓 1 ( 𝑡 ) =
𝜓 2 ( 𝑡 ) for any 𝑡 ∈ R , where
[Equation]
It is easy to observe that 𝜓 1 ( 0 ) =
𝜓 2 ( 0 ) .
Let us consider the time derivative of 𝜓 {+
+}
[Equation]
[Equation]
It is clear that 𝜓 2 ( 𝑡 ) satisfies the same ODE and thus 𝜓 1 = 𝜓 2 .
□
B.3 Remarks on the discrete-time analogy of (7)
Let us briefly explain how (7) connects to the method in [42] by time discretization.
Suppose𝑁 − = 𝑡 − / ℎ and 𝑁 + = 𝑡 + / ℎ , where ℎ ≪ 1 is the time step size; for simplicity of notation, let us assumethat 𝑁 − and 𝑁 + are simply integers.
By discretizing the time integration for the finite-time NEISscheme in (7),
[Equation]
, where we denote T (·) =
X − ℎ (·)
[Equation]
and J T − 𝑗 ( 𝑥 ) : =
[Equation]
is the Jacobian for the map T − 𝑗 .
Then by choosing 𝑁 − = − 𝐾 and 𝑁 + = 0, we have
[Equation]
which are Eqs.
(8) and (10) in arXiv v1 of [42].
C The first-order perturbation of the variance for the finite-time scheme
Here we study how the variance (or equivalently, the second moment) of the estimator depends on b ,since the performance of the finite-time NEIS scheme (7) largely depends on this choice.
Morespecifically, in the following proposition, we study how the second moment changes under a smallperturbation of b .
The expression (32) below will be useful for training optimal dynamics in Section 5(see also Appendix I for details).
Proposition C.1.
Suppose b ∈ 𝔅 and for any perturbation 𝛿 b ∈ 𝐶 ∞ 𝑐
( Ω , R 𝑑 ) , denote b 𝜖 :
= b + 𝜖𝛿 b .
Then,d
[Equation]
where for 𝑘 ∈ { 0 , 1 } , we define
[Equation]
33) and Y 𝑡 ( 𝑥 ) is the solution of the following ODE:
[Equation]
The expression of the functional derivative 𝛿 M 𝑡 − ,𝑡 + ( b ) 𝛿 b (not presented in this work) for the finite-timecase can be derived in the same way as Lemma D.8 below for the infinite-time case.
However, itappears that the expression of 𝛿 M 𝑡 − ,𝑡 + ( b ) 𝛿 b is too complicated to provide useful analytical results.
Proof.
Let us perturb b by 𝜖𝛿 b where 𝜖 ≪ 1.
Let us consider
[Equation]
with a fixed initial condition X 0 ( 𝑥 ) =
X 𝜖
( 𝑥 ) =
𝑥 .
For a small 𝜖 , we can expect that X 𝑡 ( 𝑥 )
≈
X 𝜖𝑡 ( 𝑥 )and also we know X 0 𝑡 ( 𝑥 ) ≡ X 𝑡 ( 𝑥 ) .
Define F ( 𝑘,𝜖 ) 𝑡 ( 𝑥 ) for the dynamics b 𝜖 in the same way as in (6),namely,
[Equation]
By these notations,
[Equation]
Then we take the derivative of M 𝑡 − ,𝑡 + ( b 𝜖 ) with respect to 𝜖 :
[Equation]
  
[Equation]
Next, we need to compute dd 𝜖 F ( 𝑘,𝜖 ) 𝑡 ( 𝑥 ) .
Let us first consider the perturbation to the trajectory.
Let
Y 𝜖𝑡 ( 𝑥 ) :
= dd 𝜖 (cid:0) X
𝜖𝑡
( 𝑥 ) (cid:1) and then
[Equation]
When 𝜖 = 0, Y 0 𝑡 ( 𝑥 ) is the solution to
[Equation]
Now we are ready to explicitly write down dd 𝜖 F ( [Equation] ) 𝑡 ( 𝑥 ) .
It is straightforward to derive that
[Equation]
When we let 𝜖 =
[Equation]
we have
[Equation]
[-Y 
-]
[Equation]
Finally, we arrive at the conclusion by combining previous results and dropping the superscript in
Y 0 𝑡 ( 𝑥 ) for simplicity.
□
D The first-order perturbation of the variance for the infinite-time scheme
The goal of this section is to derive the functional derivative of M( b ) with respect to b , denoted as𝛿 M( b ) 𝛿 b , defined as follows: for any 𝛿 b ∈ 𝐶 ∞ 𝑐 ( Ω , R 𝑑 ) such that b + 𝜖𝛿 b ∈ 𝔅 ∞ for small enough 𝜖 , wehave
[Equation]
Since M and Var only differ by a constant (which is independent of b ), it is apparent that 𝛿 Var ( b ) 𝛿 b ≡𝛿 M( b ) 𝛿 b .
Proposition D.1.
The functional derivative 𝛿 M( b ) 𝛿 b : Ω → R 𝑑 has the following form [Equation] ( b )
[Equation]
Remark D.2 .
The proof of the last formula is slightly formal: for instance, conditions on b to ensurethe existence of ∇A are not discussed.
Recall the expression of B from (23).
The proof of Proposition D.1 is given in Section D.2: it relieson a few explicit formulas, that we state first.
D.1 Some explicit formulas
We need some explicit formulas for Y 𝑡 ( 𝑥 ) (34) and G ( 𝑘 ) 𝑡 ( 𝑥 ) (33).
We notice that G ( 𝑘 ) depends on
Y · (·) and 𝛿 b linearly, and Y 𝑡 ( 𝑥 ) also depends on 𝛿 b linearly.
Therefore, the first step is to rewrite theexpression of Y 𝑡 ( 𝑥 ) more explicitly in terms of 𝛿 b .
Lemma D.3.
Suppose the dynamics b ∈ 𝔅 and 𝛿 b ∈ 𝐶 ∞ 𝑐
( Ω , R 𝑑 ) .
Then we have
[Equation]
The kernel C 𝑡,𝑠 ( 𝑥 ) ∈ R 𝑑 × 𝑑 has the following form
[Equation]
where exp T ← is the chronological time-ordered operator exponential.
Proof.
By plugging the ansatz (37) into (34), we immediately know that C 𝑠,𝑠 ( 𝑥 ) =
I 𝑑 for all 𝑠 ∈ Rand
[Equation]
This linear ODE has an explicit solution as in (38), by introducing time-ordered operator exponential. □
Next, we shall rewrite G ( 𝑘 ) 𝑡 ( 𝑥 ) .
Lemma D.4.
We can rewrite G ( 𝑘 ) 𝑡 ( 𝑥 ) as follows
[Equation]
where V ( 𝑘 ) 𝑡,𝑠 ( 𝑥 ) is defined as
[Equation]
𝑟.
Proof.
Recall the expression of G ( 𝑘 ) 𝑡 ( 𝑥 ) from (33).
After plugging (37), we have
G
[Equation]
𝑘
[Equation] 𝑡 [Equation] 𝑡 (cid:10)
[Equation]
[Equation]
[Equation]
) (cid:0) X 𝑠 ( 𝑥
[Equation]
(cid:1)
[Equation]
Y 𝑠 ( 𝑥 ) (cid:11) d 𝑠 + ∫ 𝑡 (∇ ·
[Equation]
) (cid:0) X 𝑠 ( 𝑥 ) (cid:1) d 𝑠 (cid:170)(cid:174)(cid:174)(cid:174) (cid:172) (37) =
F ( 𝑘 ) 𝑡 ( 𝑥 ) (cid:169)(cid:173)(cid:173)(cid:173)
(cid:171) − ∫ 𝑡 (cid:68) C 𝑡,𝑠 ( 𝑥 ) 𝑇 ∇ 𝑈 𝑘 (cid:0) X 𝑡 ( 𝑥 ) (cid:1) , 𝛿 b (cid:0)
X 𝑠 ( 𝑥 ) (cid:1)(cid:69) d 𝑠 + ∫ 𝑡 (∇ · 𝛿 b ) (cid:0) X 𝑠 ( 𝑥 ) (cid:1)
d 𝑠 + ∫ 𝑡 ∫ 𝑠 (cid:68) C 𝑠,𝑟 ( 𝑥 ) 𝑇 ∇(∇ · b ) (cid:0) X 𝑠 ( 𝑥 ) (cid:1) , 𝛿 b (cid:0) X 𝑟 ( 𝑥 ) (cid:1)(cid:69) d 𝑟 d 𝑠 (cid:170)(cid:174)(cid:174)(cid:174) (cid:172) =
F ( 𝑘 ) 𝑡 ( 𝑥 ) (cid:169)(cid:173)(cid:173)(cid:173)
(cid:171) − ∫ 𝑡 (cid:68) C 𝑡,𝑠
[Equation]
□
Then we present a few properties, which will be useful when computing the functional derivative𝛿 M( b ) 𝛿 b .
The following lemma shows how C 𝑡, − 𝑠 (·) and V ( 𝑘 ) 𝑡, − 𝑠 (·) change under the dynamical evolution
X 𝑠 (·) .
Lemma D.5.
When 𝑡 ≤ − 𝑠 ≤ 0 or 𝑡 ≥ − 𝑠 ≥ 0 ,
[Equation]
Proof.
We first consider the term C 𝑡, − 𝑠 (cid:0) X 𝑠 ( 𝑥 ) (cid:1) .
When 𝑡 ≥ − 𝑠 ≥ 0,
[Equation]
[-C (cid:101) 𝑡 , (cid:101) C 
-]
The case for 𝑡 ≤ − 𝑠 ≤ 0 can be similarly verified.
Recall from (41) that
[Equation]
where to get the third line, we use the above formula (42) about C 𝑡, − 𝑠 (cid:0) X 𝑠 ( 𝑥 ) (cid:1) .
□
The following lemma connects C 𝑡, 0 and V ( 𝑘 ) 𝑡, 0 with gradients.
Lemma D.6.
For any 𝑥 ∈ Ω and 𝑡 ∈ R , we have
[Equation]
As a consequence, det (cid:0) C 𝑡, 0 ( 𝑥 ) (cid:1) =
J 𝑡 ( 𝑥 ) .
Proof.
We fix an index 1 ≤ 𝑗 ≤ 𝑑 and consider the dynamics dd 𝑡 (cid:102) X 𝜖𝑡 ( 𝑥 ) = b (cid:0)(cid:102) X 𝜖𝑡 ( 𝑥 ) (cid:1) , (cid:102)
{+X 𝜖 
+}
( 𝑥 ) =
𝑥 + 𝜖 e 𝑗where e 𝑗 is a vector with the 𝑗 th element to be one, and all other elements to be zero. Clearly,(cid:102)
X 0 𝑡 ( 𝑥 ) =
X 𝑡 ( 𝑥 ) .
Let (cid:101)
Y 𝜖𝑡 ( 𝑥 ) :
= dd 𝜖 (cid:102)
X 𝜖𝑡 ( 𝑥 ) .
Then
[Equation]
Moreover, when 𝜖 = 0, we have
[Equation]
whose solution is simply 𝑗 th column of C 𝑡, 0 ( 𝑥 ) .
Besides, the 𝑗 th column of ∇ 𝑥 X 𝑡 ( 𝑥 ) is given by
[Equation]
By combining above results, we easily know that ∇ 𝑥 X 𝑡 ( 𝑥 ) =
C 𝑡, 0 ( 𝑥 ) .
Next, for 𝑘 ∈ { 0 , 1 } and any 𝑥 ∈ Ω ,
[Equation]
□
D.2 Proof of Proposition D.1
We list without proof the following result for the infinite-time case, which can be derived in the sameway as Proposition C.1.
Lemma D.7.
Let b 𝜖 :
= b + 𝜖𝛿 b .
Suppose that for small enough 𝜖 , we have b 𝜖 ∈ 𝔅 ∞ .
Then
[Equation]
where G ( 𝑘 ) 𝑡 ( 𝑥 ) is defined in (33) for 𝑘 = 0 , 1 .
Lemma D.8.
The functional derivative of the second moment for the infinite-time case is
[Equation]
) where
[Equation]
𝑠 [Equation] F
[Equation]
[Equation]
if 𝑠 [Equation] G ∞
[Equation]   [Equation]
𝑠
[Equation] [Equation]
if 𝑠
[Equation]
[Equation]
When 𝑠 = 0, S ∞ 0 (·) , G ∞ 0 (·) are not specified above, because they will not affect the functionalderivative 𝛿 M( b ) 𝛿 b by changing values at a single point.
Proof.
In Lemma D.7, we need to simplify the term ∫ ∞
−∞ G ( 𝑘 ) 𝑡 ( 𝑥 ) d 𝑡 .
By plugging the formula of
G ( 𝑘 ) from (40), we have
[Equation]
By plugging the last equation into Lemma D.7 and with straightforward simplification
[Equation]
we can verifythat
[Equation]
[Table]
[Equation]
The integration by parts in the last line is valid because 𝛿 b vanishes at the boundary 𝜕 Ω .
By comparingthe last equation with (35), we can immediately obtain (45) after straightforward simplifications.
□
Then we need to simplify S ∞− 𝑠 (cid:0) X 𝑠 ( 𝑥 ) (cid:1) and G ∞− 𝑠 (cid:0) X 𝑠 ( 𝑥 ) (cid:1) .
Lemma D.9.
S ∞− 𝑠 (cid:0) X 𝑠 ( 𝑥 ) (cid:1) and G ∞− 𝑠 (cid:0)
X 𝑠 ( 𝑥 ) (cid:1) has the following form
[Equation]
𝑠 ( 𝑥 ) (cid:1)
[Equation]
( 𝑥 ) (cid:1)
B( 𝑥 ) ∫−∞ F ( 0 ) 𝑡 ( 𝑥 )
d 𝑡, if 𝑠 > 0 ,
A( 𝑥 )
B( 𝑥 ) ∫ ∞
F ( 1 ) 𝑡 ( 𝑥 ) d 𝑡 − (cid:0) A( 𝑥 ) (cid:1)
B( 𝑥 ) ∫ ∞
F ( 0 ) 𝑡 ( 𝑥 ) d 𝑡, if 𝑠 < 0 .
Proof.
When 𝑠 >
[Equation]
[Equation]
which is clearly independent of 𝑠 from this expression.
Similarly
[Equation]
when 𝑠 < 0,
[Equation]
which is again independent of 𝑠 .
We can similarly simplify G ∞− 𝑠 (cid:0) X 𝑠 ( 𝑥 ) (cid:1) .
□
By plugging the formula in Lemma D.9 into (45),∫ ∞
[Equation]
Besides,∫
[Equation]
By combining previous results [Equation]
[Equation]
d 𝑡 ∫ ∞ [Equation] [Equation] 𝑥
[Equation] 𝑡 [Equation] F ( 0 )
[Equation] 𝑥 [Equation]
−∞
[Equation] ( 1 ) 𝑡 ( 𝑥 [Equation]
𝑡 [Equation] =
[Equation] [Equation] [Equation] [Equation]
[Equation] 𝑡
[Equation] [Equation] ( 0 ) 𝑡 ( 𝑥 )
[Equation]
𝑡 [Equation]
[Equation] [Equation] ( 1 ) 𝑡 (
[Equation]
d 𝑡 [Equation]
∞ F ( 0 ) 𝑡 ( 𝑥 )
d 𝑡 ∫−∞ F ( 1 ) 𝑡 ( 𝑥 ) d 𝑡 (cid:17)
[Equation]
[Equation]
𝑡
[Equation]
−∞ F (
[Equation]
𝑡 ( 𝑥 ) d 𝑡 − ∫−∞ ∇F ( 0 ) 𝑡 ( 𝑥 ) d 𝑡 ∫ ∞
F ( 1 ) 𝑡 ( 𝑥 ) d 𝑡 (cid:17) .
It can be directly verify that
[Equation]
By plugging this into the expression of the functional derivative and dividing both sides by AB ,
B
[Equation]
[Equation]
[Equation]
F
[Equation]
We keep the terms involving ∫ ∇F ( 1 ) 𝑡 ( 𝑥 ) d 𝑡 untouched and we only try to simplify terms
involving∫ ∇F ( 0 ) 𝑡 ( 𝑥 ) d 𝑡 . The coefficient for ∫ ∞ 0 ∇F ( 0 ) 𝑡 ( 𝑥 ) d 𝑡 is [Equation]
[Table]
[Equation] [Table]
[Equation] −∞ F ( 0 ) 𝑡 ( 𝑥 ) d 𝑡 ∫−∞ F ( 1 ) 𝑡 ( 𝑥 ) d 𝑡 (cid:170)(cid:174)(cid:174)(cid:174)(cid:174) (cid:172) = 1 B( 𝑥 ) (cid:16) ∫−∞ F ( 0 ) 𝑡 ( 𝑥 ) d 𝑡 ∫ ∞
F ( 1 ) 𝑡 ( 𝑥 ) d 𝑡 − ∫ ∞
F ( 0 ) 𝑡 ( 𝑥 )
d 𝑡 ∫−∞ F ( 1 ) 𝑡 ( 𝑥 ) d 𝑡 (cid:17) .
Similarly, the coefficient for ∫ {+
+}
−∞ ∇F ( 0 ) 𝑡 ( 𝑥 )
d 𝑡 is
B( 𝑥 ) (cid:169)(cid:173)(cid:173)(cid:173)(cid:173) (cid:171) ∫ ∞
−∞ F ( 1 ) 𝑡 ( 𝑥 ) d 𝑡 ∫ ∞ F ( 0 ) 𝑡 ( 𝑥 )
d 𝑡 + 2 ∫−∞ F ( 0 ) 𝑡 ( 𝑥 ) d 𝑡 ∫ ∞
F ( 1 ) 𝑡 ( 𝑥 ) d 𝑡 − 2 ∫ ∞
F ( 0 ) 𝑡 ( 𝑥 ) d 𝑡 ∫−∞ F ( 1 ) 𝑡 ( 𝑥 ) d 𝑡 − ∫ ∞ −∞ F ( 0 ) 𝑡 ( 𝑥 ) d 𝑡 ∫ ∞
F ( 1 ) 𝑡 ( 𝑥 )
d 𝑡 (cid:170)(cid:174)(cid:174)(cid:174)(cid:174) (cid:172) = 1 B( 𝑥 ) (cid:16) ∫−∞ F ( 0 ) 𝑡 ( 𝑥 ) d 𝑡 ∫ ∞ F ( 1 ) 𝑡 ( 𝑥 ) d 𝑡 − ∫ [-
-]∞
F ( 0 ) 𝑡 ( 𝑥 )
d 𝑡 ∫−∞ F ( 1 ) 𝑡 ( 𝑥 ) d 𝑡 (cid:17)
Hence,
B( 𝑥 ) 2 A( 𝑥 ) × 𝛿 M( b )
𝛿 b ( 𝑥 ) =
(cid:16) ∫−∞ F ( 0 ) 𝑡 ( 𝑥 )
[Equation]
After straightforward simplification, we obtain (36).
□
E Proof of Proposition 3.1 and discussion about its assumptions andimplications
In this section, we prove Proposition 3.1, and discuss its assumptions (in particular the Morse functioncondition) as well as some of it implications (including in settings that go beyond the ones in theproposition).
In particular, we solve the Poisson equation (11) when 𝜌 0 is the standard Gaussiandensity in R 𝑑 and 𝜌 1 is the density of a Gaussian mixture distribution.
E.1 Proof of Proposition 3.1 when D = 1
We proceed in three steps:
Step 1: We shall first establish the limiting behavior of the dynamics.
More specifically, the trajectory 𝑡 → X 𝑡 ( 𝑥 ) will converge to a local maximum of 𝑉 in the forwarddirection and converge to a local minimum in the backward direction, except at a set of points withmeasure zero.
Lemma E.1.
For any 𝑥 ∈ Ω , we have lim | 𝑡 |→∞ | b ( X 𝑡 ( 𝑥 ))
| = 0 .
Proof.
We only need to show one direction 𝑡 → ∞ and the other case follows similarly.
If this does nothold, then there exists 𝜖 > 0 and a monotone increasing sequence { 𝑡 𝑘 } ∞ 𝑘 = 1 [-is a zero-variance dynamics iﬀ (cid:101) 𝜌 
-]such that (cid:12)(cid:12) b (cid:0) X 𝑡 𝑘 ( 𝑥 ) (cid:1)(cid:12)(cid:12) ≥ 𝜖and lim 𝑘 →∞ 𝑡 𝑘 = ∞ .
Considerd
[Equation]
𝑥
[Equation]
∇ 𝑉
[Equation]
X 𝑡
[Equation]
𝑥
[Equation]
≥ − 2 𝐶 (cid:12)(cid:12)
[Equation]
X 𝑡 ( 𝑥 ) (cid:1)(cid:12)(cid:12) 2 , where 𝐶 : = sup 𝑥 ∈ Ω (cid:13)(cid:13) ∇ 2 𝑉 ( 𝑥 ) (cid:13)(cid:13) < ∞ .
[-= 𝑒 − 𝑈 1 ( 𝑥 ) + 𝜖𝑥 1 𝑒 − 𝑥 41 − 𝑥 42 , where 𝜖 : = 12 𝜋 (cid:0) 2 sup 𝑦 ∈ R | 𝑦 | 𝑒 − 𝑦 
-]
Therefore, (cid:12)(cid:12) b (cid:0) X 𝑡 ( 𝑥 ) (cid:1)(cid:12)(cid:12) ≥ (cid:12)(cid:12) b (cid:0) X 𝑠 ( 𝑥 ) (cid:1)(cid:12)(cid:12) 𝑒 − 𝐶 ( 𝑡 − 𝑠 ) , for all 𝑡 ≥ 𝑠 ≥ 0.
[-F.1 Proof of Proposition 3.1 when D = 
-]
Without loss of generality, we can ensure that 𝑡 𝑘 − 𝑡 𝑘 − 1 ≥ 1.
Hence,
[Equation]
which will diverge to infinity as 𝑡 → ∞ .
This contradicts with the boundedness of 𝑉 and thus theassumption does not hold.
As a remark, the notation 𝜒 𝐴 () is an indicator function for the set 𝐴 .
□
Lemma E.2.
Suppose 𝑥 is not on the stable or unstable manifold of a saddle point.
Then the trajectory{ X 𝑡 ( 𝑥 )} 𝑡 ∈ R must converge to a local maximum of 𝑉 in the forward direction and a local minimum of 807 𝑉 in the backward direction.
Proof.
Since the torus Ω
=
[ 0 , 1 ] 𝑑 is bounded, the trajectory { X 𝑡 ( 𝑥 )} 𝑡 ≥ 0 must be bounded and thereexists an increasing sequence { 𝑡 𝑘 } ∞ 𝑘 = 1 such that (cid:8) X 𝑡 𝑘 ( 𝑥 ) (cid:9) ∞
𝑘 = 1 is convergent by Bolzano-Weierstrasstheorem and let us denote the limit as 𝑥 ★ .
By Lemma E.3, we know b ( 𝑥 ★ ) =
𝑑 and thus 𝑥 ★ is acritical point.
By the assumption, 𝑥 ★ is not a saddle point nor a local minimum, that is, 𝑥 ★ must bea local maximum of 𝑉 .
By the assumption that the critical point 𝑥 ★ has a non-degenerate Hessian,after the trajectory enters its basin of attraction (containing an open ball around 𝑥 ★ ), the trajectory{ X 𝑡 ( 𝑥 )} 𝑡 ∈ R will eventually converge to 𝑥 ★ .
The backward direction can be proved in a similarway.
□
Lemma E.3.
Under the same assumption as in Lemma E.2, we know ∫ ∞
F ( 𝑘 ) 𝑡 ( 𝑥 ) d 𝑡 < ∞ and∫
−∞ F ( 𝑘 ) 𝑡 ( 𝑥 ) d 𝑡 < ∞ .
In particular, lim | 𝑡 |→∞ F ( 𝑘 ) 𝑡 ( 𝑥 ) =
and lim | 𝑡 |→∞ J 𝑡 ( 𝑥 ) = 0 .
Proof.
Without loss of generality, we only consider the forward branch.
Since 𝑉 is assumed to bea Morse function, the Hessian ∇ 2 𝑉 ( 𝑥 ★ ) < 0 is non-degenerate and 0 > tr (cid:0) ∇ 2 𝑉 ( 𝑥 ★ ) (cid:1) =
Δ 𝑉 ( 𝑥 ★ ) .
Therefore,
[Equation]
Since 𝜌 𝑘 = 𝑒 − 𝑈 𝑘 /Z 𝑘 are bounded, we know∫ ∞
[Equation]
𝑡 [Equation] where 𝐶 : = sup 𝑥 ∈ Ω max { 𝑒 − 𝑈 0 ( 𝑥 ) , 𝑒 − 𝑈 1 ( 𝑥 ) } < ∞ herein.
From (46), there exists 𝛽 > 0 and 𝜏 > 0such that ∇ · b ( X 𝑠 ( 𝑥 )) ≤ − 𝛽 for all 𝑠 ≥ 𝜏 , then if 𝑡 ≥ 𝜏 ,∫ 𝑡
[Equation]
( 𝑡 − 𝜏 ) , and therefore,
[Equation]
In particular, when 𝑡 ≥ 𝜏 ,
[Equation]
which converges to zero exponentially fast as 𝑡 → ∞ .
The same conclusion holds for F ( 𝑘 ) 𝑡 ( 𝑥 ) ≡𝑒 − 𝑈 𝑘 (cid:0) X 𝑡 ( 𝑥 ) (cid:1) J 𝑡 ( 𝑥 ) when 𝑡 → ∞ .
□
Step 2: We verify that b = ∇ 𝑉 is a zero-variance dynamics.
We need to show that ∫ ∞ −∞ ( 𝜌 1 − 𝜌 0 ) (cid:0) X 𝑡 ( 𝑥 ) (cid:1) J 𝑡 ( 𝑥 ) d 𝑡 = 0 almost everywhere on Ω .
Let us consider∫
[Equation]
The last line comes from Lemma E.3.
Step 3: We prove that b = ∇ 𝑉 ∈ 𝔅 ∞ in the sense of Definition A.4, i.e., such a gradient ascenddynamics is a valid one for the infinite-time NEIS scheme.
If we can find two open subsets 𝐷 1 , 𝐷 2 such that 𝑥 → ∫ ∞
F ( 𝑘 ) 𝑡 ( 𝑥 )
d 𝑡 is continuous on 𝐷 1 ,𝑥 → ∫ 0 −∞ F ( 𝑘 ) 𝑡 ( 𝑥 ) d 𝑡 is continuous on 𝐷 2 , and both Ω \ 𝐷 1 and Ω \ 𝐷 2 have Lebesgue measure zero,then clearly ℧ ( b ) ⊃
𝐷 1 ∩ 𝐷 2 and b ∈ 𝔅 ∞ .
Due to the symmetric role of forward and backward branches of trajectories, it is then sufficient toprove the following lemma.
Lemma E.4.
There exists an open subset 𝐷 ⊂ Ω such that 𝑥 → ∫ ∞ 0 F ( 𝑘 ) 𝑡 ( 𝑥 ) is continuous and Ω \ 𝐷has measure zero.
Proof.
Let us denote the local maxima of 𝑉 as 𝒳 1 , 𝒳 2 , · · · , 𝒳 𝑟 .
The index 𝑟 < ∞ because 𝑉 is a
Morse function and Ω is compact.
Since ∇ 2 𝑉 ( 𝒳 𝑖 ) < 0 for 1 ≤ 𝑖 ≤ 𝑟 , there exists a local neighborhood𝐵 𝛿 𝑖 ( 𝒳 𝑖 ) such that lim 𝑡 →∞ X 𝑡 ( 𝑧 ) =
𝒳 𝑖 if 𝑧 ∈ 𝐵 𝛿
𝑖
( 𝒳 𝑖 ) .
Hence, it is not hard to characterize the basinof attraction of 𝒳 𝑖
[Equation]
which is open.
Then define an open subset 𝐷 : = ∪ 𝑟𝑖 = 1 𝑂 𝑖 .
By Lemma E.2, we know Ω \ 𝐷 has Lebesguemeasure zero, since there is only a finite number of critical points and the stable manifold of saddlepoints has measure zero.
Next, we still need to verify 𝑧 → ∫ ∞ 0 F ( 𝑘 ) 𝑡 ( 𝑧 )
d 𝑡 is continuous at an arbitrarypoint 𝑥 ∈ 𝐷 .
Since 𝐷 = ∪ 𝑟𝑖 = 1 𝑂 𝑖 and 𝑂 𝑖 are open and disjoint, it is sufficient to verify this conclusionfor 𝑥 ∈ 𝑂 𝑖 for an arbitrary index 𝑖 .
By the smoothness of 𝑉 , there exists a local neighborhood 𝑂 𝛿 : =
(cid:8) 𝑧 ∈ 𝑂 𝑖 : 𝑉 ( 𝒳 𝑖 ) − 𝛿 < 𝑉 ( 𝑧 ) ≤𝑉 ( 𝒳 𝑖 ) (cid:9) such that ( ∇· b )( 𝑧 ) ( ∇· b )( 𝒳 𝑖 ) ∈
( 12 , 32 ) for every 𝑧 ∈ 𝑂 𝛿 .
Let 𝑀 =
sup { max { 𝑒 − 𝑈 1 ( 𝑥 ) , 𝑒 − 𝑈 0 ( 𝑥 ) } : 𝑥 ∈ Ω } .
We know 𝑀 < ∞ because Ω is compact and 𝑈 0 ,𝑈 1 are smooth.
Define 𝜏 : = inf { 𝑡 ≥ 1 : X 𝑡 ( 𝑥 ) ∈ 𝑂 𝛿 } .
Due to the smoothness of b , for any integer 𝑗 ≥ 2, we can choose a small neighborhood 𝐵 𝛿 𝑗 ( 𝑥 ) suchthat X 𝑡 ( 𝑧 ) ∈ 𝑂 𝛿 for all 𝑡 ≥ 𝑗𝜏 and for all 𝑧 ∈ 𝐵 𝛿 𝑗 ( 𝑥 ) (note that 𝑂 𝛿 is automatically a trapping regionof b by construction).
Then
[Equation]
[Equation]
For an arbitrary 𝜖 > 0, by Lemma E.3, we can pick 𝑗 large enough such that J 𝑗𝜏 ( 𝑥 ) < −( ∇· b )( 𝒳 𝑖 )
𝑀 𝜖 6 .
Next we can accordingly pick 𝛿 𝑗 small enough such that J 𝑗𝜏 ( 𝑧 ) − J 𝑗𝜏 ( 𝑥 ) < −( ∇· b )( 𝒳 𝑖 )
𝑀 𝜖 6 for all𝑧 ∈
𝐵 𝛿 𝑗 ( 𝑥 ) .
In this way, we can ensure that
[Equation]
Furthermore, we can choose 𝛿 𝑗 (possibly even smaller) so that
[Equation]
The continuity of 𝑧 → ∫ 𝑗𝜏
F ( 𝑘 ) 𝑡 ( 𝑧 ) d 𝑡 can be easily established due to the differentiability of𝑧 → F ( 𝑘 ) 𝑡 ( 𝑧 ) .
By combining previous results, for each 𝜖 > 0, we can find a 𝛿 𝑗 such that for any𝑧 ∈
𝐵 𝛿 𝑗 ( 𝑥 ) ,
[Equation]
∞
[Equation]
This proves the continuity of 𝑧 → ∫ ∞ 0 F ( 𝑘 ) 𝑡 ( 𝑧 ) d 𝑡 at the point 𝑥 .
□
E.2 A remark about the general case
We shall prove that b = D∇
𝑉 is a zero-variance dynamics where 𝑉 solves (11).
Notice that J 𝑡 ( 𝑥 ) = 𝑒 ∫ 𝑠 0 ∇·(D 𝑉 ) (cid:0) X 𝑠 ( 𝑥 ) (cid:1)
d 𝑠 and∫ ∞[-−∞ (cid:0) 𝜌 [Equation] ( X 𝑡 ( 𝑥 )) − 𝜌 
 ( X 𝑡 ( 𝑥 )) (cid:1)-]
[Equation]
As long as J 𝑡 ( 𝑥 ) vanishes when | 𝑡 | → ∞ , then such a dynamics b =
D∇
𝑉 is indeed a zero-variancedynamics.
Since D is strictly positive, it does not change the concavity of the local extreme points: forinstance, if 𝑥 ★ is a local minimum of 𝑉 (with ∇ 𝑉 ( 𝑥 ★ )
=
𝑑 and ∇ 2 𝑉 ( 𝑥 ★ ) > 0), then ∇ b ( 𝑥 ★ ) =
∇ 𝑉 ( 𝑥 ★ )D( 𝑥 ★ ) 𝑇
+ D( 𝑥 ★ )∇ 2 𝑉 ( 𝑥 ★ )
=
D( 𝑥 ★ )∇ 2 𝑉 ( 𝑥 ★ ) > 0 is still a source for the dynamics b = D∇ 𝑉 .
Therefore, the same argument as in the case D = 1 applies herein to establish the validity that
J 𝑡 ( 𝑥 ) → 0 as | 𝑡 | → ∞ .
E.3 A remark about the existence of Morse function
In Poisson’s equation (11), a Morse function 𝑉 does not always exist for an arbitrary smooth densityfunction 𝜌 1 , e.g., when 𝜌 1 = 𝜌 0 , D = 1, we know 𝑉 = 0 is the solution of (11) but 𝑉 = 0 is not a
Morse function.
However, since Morse functions are dense in 𝐶 ∞ ( Ω , R ) [4], we can always find a
Morse function such that the dynamics b
= ∇ 𝑉 behaves almost like a zero-variance dynamics, whichis summarized in the next proposition.
Proposition E.5.
Suppose Ω = [ 0 , 1 ] 𝑑 is a torus and 𝑈 0 , 𝑈 1 ∈ 𝐶 ∞ ( Ω , R ) .
Without loss of generality,assume Z
= Z 1 = 1 .
For any 𝜖 ∈ ( 0 , 1 ) , there exists a Morse function 𝑉 such that the dynamicsb = ∇ 𝑉 provides an estimator 1 − 𝜖 ≤ A( 𝑥 ) ≤ 1 + 𝜖 for almost all 𝑥 ∈ Ω in NEIS method.
Consequently, the variance Var ( b ) ≤ 3 𝜖 .
Proof.
Denote 𝜃 : = inf { 𝜌 0 ( 𝑥 ) : 𝑥 ∈ Ω } ≡ 𝑒 − sup
{ 𝑈 0 ( 𝑥 ) : 𝑥 ∈ Ω } > 0 since 𝑈 0 is smooth and Ω is compact.
Since both 𝜌 𝑘 = 𝑒 − 𝑈 𝑘 are smooth for 𝑘 = 0 , 1, one could approximate 𝜌 1 − 𝜌 0 by trigonometricpolynomials 𝑇 𝑁 ( 𝑥 ) =
(cid:205) | 𝜇 | ∞ ≤ 𝑁,𝜇 ≠ 0 𝑑 𝑎 𝜇 𝑒 𝑖 2 𝜋 ⟨ 𝜇,𝑥 ⟩ such that
[Equation]
where 𝑎 𝜇 ∈ C are Fourier coefficients, 𝜇 ∈ Z 𝑑 and 𝑁 ∈ N ; see e.g., [7, Theorem 16].
LetΨ 𝑁 ( 𝑥 ) =
(cid:205) | 𝜇 | ∞ ≤ 𝑁,𝜇 ≠ 0 𝑑 𝑎 𝜇 − 4 𝜋 2 | 𝜇 | 2 𝑒 𝑖 2 𝜋 ⟨ 𝜇,𝑥 ⟩ ∈ 𝐶 ∞ ( Ω , R ) .
It is clear that ΔΨ 𝑁 = 𝑇 𝑁 .
As Morsefunctions are dense, we can find a Morse function 𝑉 such that ∥ 𝑉 − Ψ 𝑁 ∥ 𝐶 2 ( Ω ) < 𝜖 𝜃 2 [4, Proposition1.2.4], and in particular, ∥ Δ 𝑉 − ΔΨ 𝑁
∥ 𝐶 0 ( Ω ) < 𝜖 𝜃 2 .
Therefore, [Equation]
By Proposition 3.1, we know that b = ∇ 𝑉 is a zero-variance dynamics for (cid:101) 𝜌 1 : =
𝜌 0 + Δ 𝑉 .
The
Proposition 3.1 is proved under the assumption that densities are positive smooth functions forconvenience and it is straightforward to verify that it also holds if 𝜌 1 is an arbitrary smooth functionin Proposition 3.1.
In particular, using the same argument in Appendix E.1 Step 2, we have for almostall 𝑥 ∼ 𝜌 0 ,
[Equation]
Hence,
[Equation]
Since we assumed Z 1 = 1,
[Equation]
□
E.4 Solution of Poisson’s equation (11) for Gaussian mixtures
Lemma E.6.
One solution of the Poisson’s equation Δ 𝑉 =
𝐶𝑒 − | 𝑥 − 𝜇 | 2 2 𝜎 2 with 𝑑 ≥ 2 on Ω = R 𝑑 is𝑉 ( 𝑥 ) =
𝑓 (| 𝑥 − 𝜇 |) , where the function 𝑓 : R + → R has the derivative
[Equation]
where Γ (cid:0) 𝑎, 𝑥 (cid:1) : = ∫ 𝑥 0 𝑡 𝑎 − 1 𝑒 − 𝑡 d 𝑡 is the lower incomplete gamma function.
Proof.
Without loss of generality, let 𝜇 =
𝑑 .
Then a natural radial solution is given as 𝑉 ( 𝑥 ) =
𝑓 (| 𝑥 |)for some scalar-valued function 𝑓 .
The above Poisson’s equation becomes
[Equation]
By some straightforward computation,
[Equation]
(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32) (cid:123)(cid:122) (cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32) (cid:125) ≡
[Equation] (cid:0) 𝑑 [Equation] 2 , | 𝑟 | 2 2 𝜎
[Equation]
□ Proposition E.7. Suppose 𝑉 solves the following Poisson’s equation on Ω = R 𝑑Δ 𝑉 = 𝜌 1 − 𝜌where
[Equation]
and 𝜇 𝑖 ≠ 𝜇 𝑗 if 𝑖 ≠ 𝑗 .
Then one solution for the gradient flow dynamics b = ∇ 𝑉 is given as
[Equation]
Proof.
We just need to apply the last lemma and the formula ∇ 𝑉 ( 𝑥 ) =
𝑓 ′ (| 𝑥 |) | 𝑥 | 𝑥 if 𝑉 ( 𝑥 ) =
𝑓 (| 𝑥 |) .
□
Proposition E.8.
Suppose b is given in (51) .
Then
[Equation]
otherwise .
The limiting dynamics 𝑥 → lim (cid:0) max 𝑛𝑖 = 1 𝜎 𝑖 (cid:1) → 0 b ( 𝑥 ) is continuous on the region R 𝑑 \{ 𝜇 𝑖 } 𝑛𝑖 = 1 .
Proof.
If 𝑥 = 𝜇 𝑗 , thenb ( 𝑥 ) =
2 𝜋 𝑑 /(cid:18)
[Equation]
When 𝜎 𝑖 → 0 for all 𝑖 ,we know Γ (cid:0) 𝑑 / 2 , | 𝜇 𝑗 − 𝜇 𝑖 |
2 𝜎 2 𝑖 (cid:1) →
Γ ( 𝑑 / 2 ) when 𝑖 ≠ 𝑗
and hence we have the aboveresult.
Similarly, we can obtain the expression when 𝑥 ≠ 𝜇 𝑗 for any 𝑗 .
□
E.5 Example: Poisson’s equation yields a zero-variance dynamics
The example for Figure 1 is
[Equation]
The periodic boundary condition in Proposition 3.1 helps to ease the technicalities in proving that thegradient dynamics b
= ∇ 𝑉 from solving the Poisson’s equation (11) is a zero-variance dynamics, byremoving the effect from the boundary 𝜕 Ω .
The same conclusion, however, should hold if 𝑉 solvesthe Poisson’s equation with Neumann boundary condition:
[Equation]
We consider the same model (52) and Ω = ( 0 , 1 ) 2 .
The potential 𝑉 and flowlines of b = ∇ 𝑉 arevisualized in Figure 5 and we can numerically verify that Z
= A( 𝑥 ) for almost all 𝑥 ∈ Ω .
E.6 Non-uniqueness of zero-variance dynamics
Recall from Proposition 2.2 that there are certain degrees of freedom to choose the dynamics: for agiven b 1 , if we choose b 2 = 𝛼 b 1 where 𝛼 ∈ 𝐶 ∞ ( Ω , R ) is strictly positive, then this function 𝛼 can beabsorbed into the time re-scaling and it does not affect the variance of the sampling scheme.
However,even if we remove this parameterization redundancy, zero-variance dynamics may still not be unique,e.g., due to the geometric rotational symmetry.
Proposition E.9 (Non-uniqueness) .
For given 𝜌 0 and 𝜌 1 , there might exist more than one zero-variancedynamics (let us say b 1 , b 2 ) but there is no scalar-valued function 𝛼 such that b 2 = 𝛼 b 1 .
Proof.
We construct an example to prove the non-uniqueness: let 𝑑 = 2, 𝑈 0 ( 𝑥 ) =
| 𝑥 | 2 / 2 + ln ( 2 𝜋 ) and 932 𝑈 1 be given by
[Equation]
We can easily verify that | 𝑥 | 𝑒 − 𝑥 4 < 34 𝑒 − 𝑥 2 / 2 for any 𝑥 ∈ R .
Then we know 716 𝑒 − 𝑈 0 ( 𝑥 ) < 𝑒 − 𝑈 1 ( 𝑥 ) <2516 𝑒 − 𝑈 0 ( 𝑥 ) .
Therefore, 𝑈 1 is clearly well-defined and we can easily know Z 1 = 1.
For the dynamicsb ( 𝑥 ) = (cid:2) 𝑣 1 𝑣 2 (cid:3) with 𝑣 2 1 + 𝑣 2 2 > 0, we have J 𝑡 ( 𝑥 )
=
1 for any 𝑥 ∈ Ω , 𝑡 ∈ R , and∫ ∞
[Equation]
When either 𝑣 1 = 0 or 𝑣 2 = 0, we can easily verify that
[Equation]
Therefore, the variance is zero for two dynamics with orthogonal direction b = (cid:2) 10 (cid:3) and b = (cid:2) 01 (cid:3) .
It isclear that there is no scalar-valued function 𝛼 such that b 2 = 𝛼 b 1 , and thus the non-uniqueness isestablished.
□
E.7 Connection to the Beckmann’s problem.
The Poisson’s equation (11) with D = 1 is the Euler-Lagrange equation associated with
[Equation]
when 𝑝 = 2.
The variational problem in (54) is known as Beckman’s problem of continuoustransportation [5]; when 𝑝 = 1, it is also related to optimal transport in 𝑊 1 Wasserstein distance[35, 36].
F Explicitly solvable zero-variance dynamics
In this section, we provide some examples with explicitly solvable zero-variance dynamics. Throughoutthis section, we consider Ω = R 𝑑 .
[Table]
𝑥 )
=see
Section F
[Table]
[Equation]
1 ( 𝑥 − 𝜛 )/b
( 𝑥 ) =
Λ 𝑥
+ 𝑣 with Λ = ln (cid:0) Σ
[Equation]
1 / 2 (cid:1) , 𝑣 = − (cid:0) I 𝑑 − Σ 1 / 2 (cid:1)
[Equation]
ln (cid:0) Σ − 1 / 2 (cid:1) 𝜛. see
Section F. general 𝑑 𝜌 0 and 𝜌 1 has the same marginal distribution on the orthogonal subspace of { 𝑐𝑣 : 𝑐 ∈ R } b ( 𝑥 ) =
𝑣 see
Section F.
F.1 Some general properties
Given a b ∈ 𝔅 and the distribution 𝜌 0 , we study the family of 𝑈 1 such that b is a zero-variancedynamics.
Given an ODE flow map X 𝜏 (·) based on the dynamics b , let us introduce
[Equation]
[Equation]
Then 𝜌 ∝ 𝑒 − 𝑈 𝜏 is the push-forward distribution of the flow map X 𝜏 (·) , i.e., 𝜌 = (cid:0) X 𝜏 (·) (cid:1)
# 𝜌 0 .
Thefamily of distributions that can be written as a linear combination of such pushforward distributionscan be characterized by
[Equation]
We have
Proposition F.1.
For every 𝑈 1 ∈ 𝔉 , the variance Var ( b ) (if well-defined) is exactly zero, i.e., if thedistribution 𝜌 1 ∝ 𝑒 − 𝑈 1 is a linear combination of push-forward distributions by the ODE flows maps,then Z 1 can be estimated with zero-variance by the infinite-time NEIS scheme.
This proposition is proven in Section F.5 below.
In words it says that, if we can learn a perfectneural ODE such that 𝜌 1 = X 𝜏 (·) # 𝜌 0 for some 𝜏 , then such a dynamics is also the optimal one (i.e.,zero-variance dynamics) for the infinite-time NEIS scheme.
Conversely, if 𝑈 ∉ 𝔉 , then is it stillpossible that Var ( b ) = 0?
The answer is positive:
Proposition F.2.
The exists a dynamics b ∈ 𝔅 and a 𝜌 1 ∝ 𝑒 − 𝑈 1 such that Z 1 can be computed withzero-variance but 𝜌 1 does not need to be a linear combination of (cid:8) X 𝜏 (·) # 𝜌 0 (cid:9) 𝜏 ∈ R (namely, 𝑈 1 ∉ 𝔉 ).
This proposition is proven in Section F.6 below.
F.2 Flows for the 1D case
Let us consider b = 1.
Then we can compute Z 1 via the infinite-time NEIS scheme with zero-variancefor arbitrary potentials 𝑈 0 and 𝑈 1 .
This could be verified via direct computation: J 𝑡 ( 𝑥 ) = 1 and
X 𝑡 ( 𝑥 ) =
𝑥 + 𝑡 for any 𝑡, 𝑥 ∈ R , and thus for an arbitrary 𝑥 ,
[Equation]
Therefore, the variance is exactly zero.
Another perspective to understand this comes from Proposition F.1.
For b = 1, we know 𝑒 − 𝑈 𝜏 ( 𝑥 ) =
𝑒 − 𝑈 0 ( 𝑥 − 𝜏 ) in (55).
By Proposition F.1, if the potential 𝑈 1 can be expressed as follows
[Equation]
then Z 1 can be computed with zero-variance, where 𝑓 is a tempered distribution and ∗ means theconvolution.
Then it is sufficient to show the existence of such a 𝑓 for a generic 𝑈 1 .
For a given potential 𝑈 1 , we can solve the above equation for 𝑓 using Fourier transform; morespecifically,
[Equation]
where ℱ (− 1 ) are (inverse)
Fourier transform.
F.3 Linear flows for Gaussian distributions
Proposition F.3.
Suppose 𝑈 0 ( 𝑥 ) =
| 𝑥 | 2 / 2 + 𝑑 2 ln ( 2 𝜋 ) and 𝑈 1 ( 𝑥 ) =
( 𝑥 − 𝜛 ) 𝑇 Σ − 1 ( 𝑥 − 𝜛 )/ 2 , where thecovariance matrix Σ is non-degenerate.
A zero-variance linear dynamics is b ( 𝑥 ) =
Λ 𝑥 + 𝑣 with
[Equation]
Before presenting detailed proofs, let us make a few remarks about zero-variance dynamics in (56):• If we further
let Σ =
( 1 − 𝜖 )
I 𝑑 where 𝜖 ≪ 1 is an asymptotic parameter,
then the above choice(56) can be approximated as follows:
[Equation]
The leading order dynamics dd 𝑡 X 𝑡 ( 𝑥 )
≈ − 𝜛 is consistent with the parallel velocity case below in
Section F.4.
• For the above Gaussian case, the dynamics (56) can be regarded as the gradient flow dynamics of
the
following quadratic potential
[Equation]
[Table]
Indeed, it is not hard to guess that the optimal dynamics might have a linear form in orderto transport Gaussians.
It is natural to guess that b = −∇ 𝑈 1 or b = −∇( 𝑈 1 − 𝑈 0 ) might bezero-variance dynamics, but it could be verified that neither of them are zero-variance dynamics.
Proof of Proposition F.3.
Suppose we consider a family of linear dynamics
[Equation]
Then X 𝑡 ( 𝑥 ) =
𝑒 Λ 𝑡 𝑥 + Λ − 1 (cid:0) 𝑒 Λ 𝑡 − I 𝑑 (cid:1) 𝑣 and (∇ · b )( 𝑥 ) =
tr ( Λ ) .
Therefore, 𝑈 − 𝜏 defined in (55) has thefollowing form for any 𝜏 ∈ R ,
[Equation]
By Proposition F.1, the variance is zero if
[Equation]
where 𝐶 is some constant.
This condition can be simplified as
[Equation]
Thus, we just need to ensure
[Equation]
by matching the order of 𝑥 and, more specifically, we can choose Λ and 𝑣 as in (56).
□
F.4 Flows with parallel velocity
As we have mentioned, in the 1D case, the choice b = 1 gives a zero-variance estimator for theinfinite-time NEIS scheme.
A straightforward generalization is to consider the following parallelvelocity case
[Equation]
where 𝑣 ∈ R 𝑑 and 𝛼 is a positive scalar-valued function.
Due to Proposition 2.2, it suffices to considerb ( 𝑥 ) =
𝑣 .
As we can always rotate the coordinate without affecting partition functions, without lossof generality, let us assume 𝑣 = e 1 for simplicity, where e 1 is a vector with the first element to be 1and zeros otherwise.
For an arbitrary initial proposal 𝑥 , only the first coordinate 𝑥 1 is changing under
[Figure]
the dynamics b = e 1 .
The estimator essentially works like the 1D case above:
[Equation]
where (cid:101) 𝜌
𝑘 is the marginal distribution of 𝜌 𝑘 for the subspace R 𝑑 − 1 by tracing out the first coordinate.
{+Therefore, the dynamics b = e 1 is a zero-variance dynamics iff (cid:101) 𝜌 
+}
= (cid:101) 𝜌 1 .
Proposition F.4.
Suppose b ( 𝑥 ) =
𝛼 ( 𝑥 ) 𝑣 where 𝛼 : R 𝑑 → R with inf 𝑥 ∈ Ω 𝛼 ( 𝑥 ) > 0 and 𝑣 ∈ R 𝑑 .
Sucha dynamics b gives a zero-variance estimator iff 𝜌 0 and 𝜌 1 have the same marginal distribution onthe orthogonal space of (cid:8) 𝑐𝑣 : 𝑐 ∈ R (cid:9) .
F.5 Proof of Proposition F.1
Lemma F.5.
Fix the potential 𝑈 0 and a dynamics b ∈ 𝔅 .
If ∫ ∞−∞ 𝑒 − 𝑈𝑘
[Equation]
J
[Equation]
d 𝑡
[Equation]
𝐶 𝑘 is aconstant function for 𝑘 ∈ { 2 , 3 } , then any mixture of 𝑈 2 and 𝑈 3 , given below, also ensures that
[Equation]
J 𝑡
[Equation]
𝑡 ∫ ∞ −∞ 𝑒 − 𝑈
[Equation]
X 𝑡 ( 𝑥 ) (cid:1) J 𝑡 ( 𝑥 )
d 𝑡 is a constant function, as long as 𝑈 is a valid potential function:
[Equation]
Proof.
We can easily observe that
[Equation]
for some constants 𝐶 2 = 𝜔 2 ∫ Ω 𝑒 − 𝑈 2 and 𝐶 3 = 𝜔 3 ∫ Ω 𝑒 − 𝑈 3 .
□
[Figure]
Lemma F.6.
For the distribution 𝜌 1 ∝ 𝑒 − 𝑈 𝜏 + 𝐶 , the partition function Z 1 can be computed withzero-variance.
[Figure]
Proof.
We only need to verify the case 𝐶 = 0.
Then
[Figure]
[Equation]
This means b is a zero-variance dynamics for the distribution 𝜌 1 .
□
Proof of Proposition F.1.
Combine the above two lemmas.
□
F.6 Proof of Proposition F.
Let us consider a 2D example with 𝑈 0 ( 𝑥 ) =
| 𝑥 | 2 2 + ln ( 2 𝜋 ) and b = (cid:2) 10 (cid:3) .
Then J 𝑡 ( 𝑥 ) = 1, and𝑈 𝜏 ( 𝑥 ) =
( 𝑥 1 − 𝜏 ) 2 + 𝑥
+ ln ( 2 𝜋 ) where 𝑥 = (cid:2)
𝑥 1 𝑥 2 (cid:3) .
Let us consider
[Equation]
where 𝜖 : = 12 𝜋 (cid:0) 2 sup 𝑦 ∈ R | 𝑦 | 𝑒 − 𝑦 4 +( 𝑦 − 1 ) 2 / 2 sup 𝑦 ∈ R 𝑒 − 𝑦 4 + 𝑦 2 / 2 (cid:1) − 1 > 0.
It could be straightforwardlyverified that 𝑈 1 is a well-defined potential and
[Equation]
Therefore, b is a zero-variance dynamics for 𝜌 0 and 𝜌 1 .
However, 𝑈 1 ∉ 𝔉 because ∫ ∞
−∞ 𝑓 ( 𝜏 ) 𝑒 − 𝑈 𝜏
[Equation]
𝑥 ) d 𝜏must be a separable function, whereas 𝑈 1 is not.
G Proof of Proposition 3.2 and more discussion
G.1 Proof of Proposition 3.
We proceed in three steps:
Step 1: Let us first assume the existence of 𝜘 ∈ 𝐶 1 ( 𝐷, R ) where 𝐷 is an open subset of Ω and Ω \ 𝐷has measure zero.
We shall verify that T # 𝜌
=
𝜌 1 almost everywhere.
From (14), let us replace 𝑥 by X 𝜖 ( 𝑥 ) ,0 (24) =
J 𝜖 ( 𝑥 ) (cid:16) ∫
[Equation]
By straightforward simplification,
[Equation]
By taking the derivative with respect to 𝜖 at 𝜖 = 0,
[Equation]
To show that T # 𝜌 0 = 𝜌 1 , we need to verify that 𝜌 0 ( 𝑥 ) =
𝜌 1 ( T ( 𝑥 ))
J T ( 𝑥 ) where J T ( 𝑥 ) = 1038 (cid:12)(cid:12) det [Equation] 𝑥 [Equation]
𝜘
[Equation]
[Equation] 𝑥 )
[Equation]
Therefore, it remains to prove that
[Equation]
By the direct computation,
[Equation]
By the matrix determinant lemma,
[Equation]
where we used (38) and Lemma D.6 to get the second line.
The last equation verifies (57) andtherefore, T # 𝜌 0 =
𝜌 1 almost everywhere.
Step 2: We shall explain 𝐷 and establish the existence of such 𝜘 ∈ 𝐶 1 ( 𝐷, R ) .
Suppose we denote the local minima of 𝑉 as 𝒳 1 , 𝒳 2 , · · · , 𝒳 𝑎 , and local maxima of 𝑉 as 𝒴 1 , 𝒴 2 , · · · , 𝒴 𝑏 ,where 𝑎, 𝑏 ∈ N .
Then in the proof of Proposition 3.1, we have already mentioned that
[Equation]
some [Equation] is an open subset of ℧ ( b ) and Ω \ 𝐷 has measure zero, due to the assumption that 𝑉 is a Morsefunction.
Consider the following function 𝐿 ( 𝑥, 𝑡 ) : 𝐷 × R → R , defined as
[Equation]
We can observe that
{+• lim 𝑡 →−∞ 𝐿 ( 𝑥, 𝑡 ) = ∫ 
+}
−∞ 𝜌 0 ( X 𝑠 ( 𝑥 )) J 𝑠 ( 𝑥 ) d 𝑠 > 0. • lim 𝑡 →∞ 𝐿 ( 𝑥, 𝑡 ) = ∫−∞
{+𝜌 
+}
( X 𝑠 ( 𝑥 )) J 𝑠 ( 𝑥 ) d 𝑠 − ∫ ∞ [Equation] J 𝑠
[Equation]
d 𝑠 [Equation] − ∫ ∞ [Equation]
𝜌
[Equation] X [Equation] where the second equality comes from the fact that b = ∇ 𝑉 is a zero-variance dynamics; see e.g.,
• With fixed 𝑥 , the function 𝑡 → 𝐿 ( 𝑥, 𝑡 ) is continuously differentiable and is strictly monotonically
decreasing.
These imply that for each 𝑥 ∈ 𝐷 , there exists a unique 𝜘 ( 𝑥 ) such that 𝐿 (cid:0) 𝑥, 𝜘 ( 𝑥 ) (cid:1) = 0 by the intermediatevalue theorem.
Therefore, 𝜘 is well-defined via (14).
Next, we need to prove that such a function 𝜘 ∈ 𝐶 1 ( 𝐷, R ) , which can be immediately obtained by theimplicit function theorem [34], provided that we can prove 𝐿 ∈ 𝐶 1 ( 𝐷 × R , R ) .
Due to the smoothness assumption on 𝜌 0 and 𝜌 1 , it is clear that 𝜕 𝑡 𝐿 ( 𝑥, 𝑡 ) = − 𝜌 1 ( X 𝑡 ( 𝑥 ))
J 𝑡 ( 𝑥 ) < 0 exists and is continuous.
[Figure]
Therefore, the task becomes to prove that ∇ 𝑥 𝐿 ( 𝑥, 𝑡 ) exists and is continuous.
Since it is clear that
[Figure]
𝜌 1 ( X 𝑠 [Equation] 𝑥 ))
J 𝑠 ( 𝑥 ) d
𝑠 is continuously differentiable with respect to 𝑥 , it is then sufficient to prove
[Figure]
that
[Figure]
[Equation]
is continuously differentiable for 𝑘 ∈ { 0 , 1 } .
[Figure]
Step 3: Prove that 𝐺 𝑘 ∈ 𝐶 1 ( 𝐷, R ) .
[Figure]
In Section E.1, we have proved that 𝐺 𝑘 is continuous; see Lemma E.4 in particular.
Next, we first
[Figure]
verify that 𝐺 𝑘 is differentiable and then verify that ∇ 𝐺 𝑘 is also continuous.
[Figure]
Part 1: 𝐺 𝑘 is differentiable.
[Figure]
We want to verify that 𝐺 𝑘 is differentiable and in particular
[Figure]
[Equation]
Let us consider an arbitrary 𝑥 ∈ 𝐷 .
Without loss of generality, suppose its limit in the backward branch [Figure] is 𝒳 1 = lim 𝑡 →−∞ X 𝑡 ( 𝑥 ) .
Let us focus on a local neighborhood 𝐵 𝛿 ( 𝑥 ) such that lim 𝑡 →−∞ X 𝑡 ( 𝑧 ) = 𝒳
[Figure] for all 𝑧 ∈ 𝐵 𝛿 ( 𝑥 ) .
Such a small 𝛿 exists because 𝒳 1 is a strict local minimum and b = ∇ 𝑉 is smooth [Figure] from the assumption that 𝑉 is a Morse function.
In order to verify that 𝐺 𝑘 is differentiable (i.e., (58)), [Figure] by the Leibniz rule (see e.g., [19, Thm. 6.28]), it is sufficient to prove that there exists an integrable
[Figure] function 𝑄 : (−∞ , 0 ] → R such that [Figure]
[Equation]
By direct computation,
[Equation]
Since the domain Ω is assumed to be a torus
[Equation]
we know 𝜌 𝑘 [Equation] ∇ 𝑈 𝑘 and ∇( ∇ · b ) are uniformly boundedon the domain Ω [Equation]
Therefore,
[Equation] [Equation]
Recall from (5) and (38)
that
[Equation]
Recall that X 𝑠
( 𝑧 )
→ 𝒳 1 as 𝑠 → −∞ ; the value ∇ · b ( 𝒳 1 ) and the hessian matrix ∇ b ( 𝒳 1 ) are bothstrictly positive, which imply that J 𝑠 ( 𝑥 ) and (cid:13)(cid:13) C 𝑠, 0 ( 𝑧 ) (cid:13)(cid:13) are both decaying as 𝑠 → −∞ .
Let us denote 𝜐 > 0 as the smallest eigenvalue of ∇ b ( 𝒳 1 ) .
For a given 𝜖 with 0 < 𝜖 < min (cid:8) ∇ ·b ( 𝒳 1 ) , 𝜐 (cid:9) , let us define
[Equation]
which is an open neighborhood of 𝒳 1 .
We can find a subset of E , denoted as E trap , such that E trap isa trapping region of the dynamics b for the backward branch.
Thus we can find a negative time 𝜏(which possibly depends on 𝜖 ) such that X 𝑠 ( 𝑥 ) ∈
E trap for all 𝑠 ≤ 𝜏 .
If we choose a 𝛿 small enough,then we can even ensure that
[Equation]
due to the smoothness of b and the construction that E 1 is a trapping region.
Therefore, when 𝑠 ≤ 𝜏 ,
J 𝑠 ( 𝑧 ) decays to zero exponentially fast as 𝑠 → −∞ with a rate at least ∇ · b ( 𝒳 1 ) − 𝜖 ; similarly, thematrix norm (cid:13)(cid:13)
C 𝑠, 0 ( 𝑧 ) (cid:13)(cid:13) also decays to zero exponentially fast as 𝑠 → −∞ with a rate at least 𝜐 − 𝜖 .
Therefore, on the region (−∞ , 0 ] × 𝐵 𝛿 ( 𝑥 ) ,
[Equation]
[Equation]
By plugging the above estimates into (61), we know that
[Equation]
[Equation] The function 𝑒 (cid:0) ∇· b ( 𝒳 1 )− 𝜖 (cid:1) 𝑠 is integrable on (−∞ , 0 ] and this serves as the function 𝑄 that we need in(59) with some multiplicative constant.
Part 2: ∇ 𝐺 𝑘 is continuous.
Let us denote 𝐻 ( 𝑧, 𝑠 ) :
= ∇ 𝑧 (cid:0) 𝜌 𝑘 ( X 𝑠 ( 𝑧 ))
J 𝑠 ( 𝑧 ) (cid:1) .
From (60), we can observe that ∇ 𝑧 𝐻 ( 𝑧, 𝑠 ) iscontinuous with respect to 𝑧 with
[Equation]
Lemma G
[Equation]
For a given smooth vector field 𝑊 ∈ 𝐶 ∞ ( Ω , R 𝑑 ) [Equation] there exists a constant 𝐶 such that forany 𝑧 ∈ 𝐵 𝛿 ( 𝑥 ) and 𝑠 ∈ (−∞ , 0 ] , we have
[Equation]
By this lemma and the estimates in (64),
[Equation]
which readily leads into the continuity of ∇ 𝐺 𝑘 based on (58)
[Equation]
Part 3: Proof of Lemma G.
By direct computation
[Equation]
Since 𝑊 is assumed to be smooth on the torus Ω , we know 𝑊 and ∇ 𝑊 are uniformly bounded.
Besides, previously, we know that (cid:13)(cid:13) C 𝑠, 0 ( 𝑧 ) (cid:13)(cid:13) ≲ 𝑒 (cid:0)
𝜐 − 𝜖 (cid:1) 𝑠 .
Therefore, we only need to prove that for any 1 ≤ ℓ ≤ 𝑑 ,
[Equation]
From (39), we have
[Equation]
Since (cid:13)(cid:13) C 𝑠, 0 ( 𝑧 ) (cid:13)(cid:13) ≲ 𝑒 ( 𝜐 − 𝜖 ) 𝑠 from (63), the source term ∥ S ( 𝑠, 𝑧 )∥ also decays exponentially fast withrate 2 ( 𝜐 − 𝜖 ) as 𝑠 → −∞ , namely,
[Equation]
By rewriting (66) in the integral form and by (38), we have
[Equation]
To prove Lemma G.1, we only need to consider the case 𝑠 ≪ 0.
Suppose we consider 𝑠 ≤ 𝜏 only;recall the role of 𝜏 in (62).
Then we could obtain that when 𝑠 ≤ 𝑟 ≤ 𝜏
[Equation]
where exp T → is the antichronological time-ordered operator exponential.
We can separate the aboveintegral form using 𝜏 and obtain the following estimates: when 𝑠 ≤ 𝜏 ,
[Equation]
[Table]
This verifies (65) for any 𝑧 ∈ 𝐵 𝛿 ( 𝑥 ) and 𝑠 ≤ 𝜏 .
When 𝑠 ∈ [− 𝜏, 0 ] , we can simply choose the prefactorlarge enough so that (65) holds, as 𝜏 is a finite value.
Hence, Lemma G.1 is verified.
G.2 Examples
We elaborate on Proposition 3.2 using concrete examples.
For these examples, we estimate 𝜘 from(14) either analytically or numerically, and then we validate Proposition 3.2 by comparing 𝜌 1 and theempirical distribution of X 𝜘 ( 𝑥 ) ( 𝑥 ) where 𝑥 ∼ 𝜌 0 .
G.2.1 Gaussian examples in 1D
For the case 𝑈 0 ( 𝑥 ) =
| 𝑥 | 2 2 + 12 ln ( 2 𝜋 ) and 𝑈 1 ( 𝑥 ) =
| 𝑥 − 𝜔 | 2 2 𝜎 2 with 𝜎 < 1, from Table 1, we already knowthat a zero-variance dynamics is b ( 𝑥 )
=
𝑥 − 𝜔 1 − 𝜎 .
Then X 𝑡 ( 𝑥 ) =
𝑒 𝑡 𝑥 − ( 𝑒 𝑡 − 1 ) 𝜔 1 − 𝜎 for any 𝑡, 𝑥 ∈ Rand J 𝑡 ( 𝑥 ) =
𝑒 𝑡 for any 𝑡 ∈ R .
By the direct computation
[Equation]
By solving 𝜘 in (14) using the last equation, we have
[Equation]
which is independent of the state 𝑥 .
G.2.2 Three-mode mixture on a 2D torus
We consider the model (52) on the torus Ω
=
[ 0 , 1 ] 2 and recall that the zero-variance dynamicshas been shown in Figure 1.
Then the contour plot of 𝜘 is visualized in Figure 7.
Moreover, in thesame figure, the empirical distributions of X 𝜘 ( 𝑥 ) ( 𝑥 ) and contour plots of 𝜌 1 are provided, whichnumerically verifies Proposition 3.2.
G.2.3 An example on ( 0 , 1 ) 2 with Neumann
condition
We consider the model
[Figure]
[Equation]
on the domain Ω = ( 0 , 1 ) 2 and the potential 𝑉 automatically solves the Poisson’s equation with
Neumann boundary condition (see (53)) by the above construction.
Apart from verifying thatb ( 𝑥 ) = ∇ 𝑉 ( 𝑥 ) is a zero-variance dynamics numerically, we can also observe that Proposition 3.2holds in this case; see Figure 8. [Figure] [Figure]
H Proof of Proposition 4.
Below is the full version of Proposition 4.1.
Proposition H.1 (Local minimum) .
Assume that(i)
(Local minimum).
b ∈ 𝔅 ∞ is a (non-trivial) local minimum of Var, namely, Var ( b ) <
Var ( max ) and if there is a perturbation 𝛿 b ∈ 𝐶 ∞ 𝑐 ( Ω , R 𝑑 ) such that b + 𝜖𝛿 b ∈ 𝔅 ∞ forsufficiently small 𝜖 , then Var ( b + 𝜖𝛿 b ) ≥
Var ( b ) .(ii)
(Continuity assumption).
The functional derivative 𝛿 Var ( b ) 𝛿 b is continuous and 𝛿 Var ( b )
𝛿 b = [-
 𝑑-]0 𝑑on Ω .
In particular, ∇A exists and is continuous on Ω .(iii)
(Technical assumptions).
℧ ( b ) =
Ω (see Definition A.4) and the set of b -unstable points (see
Definition A.8) has Lebesgue measure zero.
Moreover, the set (cid:8) 𝑥 ∈ Ω : ∇( 𝑈 1 − 𝑈 0 )( 𝑥 ) =
𝑑 (cid:9)has Lebesgue measure zero.
Then b is a zero-variance dynamics for the infinite-time NEIS scheme, i.e., Var ( b ) = 0 .
Recall the formula of 𝛿 Var ( b ) 𝛿 b from (36):𝛿 Var ( b )
[Equation]
Here is a sketch of the main idea behind the proof.
If ∇A =
𝑑 on the domain Ω , then A is a constantfunction.
Hence, b provides a zero-variance estimator for the infinite-time NEIS scheme and it must be a global minimum of the functional Var as well.
If the other term ∫ ∞ 0 F ( 0 ) 𝑡
[Equation]
[Equation]
( 𝑥 ) d 𝑡 = 0 locally on an open subset, then it could be shown that this isequivalent to ⟨∇( 𝑈 1 − 𝑈 0 )( 𝑥 ) , b ( 𝑥 )⟩ =
(see Lemma H.4 and Lemma H.5)
Thus, such a dynamics bshould not be optimal, because b is perpendicular to the gradient of potential difference and such a bdoes not explore the local structure (cf.
Lemma H.3).
This intuition leads into the following idea: if bis a local minimum of Var, then we should only have ∇A =
𝑑 almost everywhere on Ω ; otherwise,we should be able to perturb b so that the dynamics b can better explore the landscapes of 𝑈 0 and
𝑈 1and the variance can be further reduced; see Proposition H.6.
Remark H.2 .
As we work on the domain Ω only, we shall consider the topological space for thedomain Ω instead of R 𝑑 .
H A characterization of the global maximum
Proposition H.3.
If the dynamics b ∈ 𝔅 ∞ , then
Var ( b ) ≤ Var ( max ) .
The equality can be achieved iff
[Equation]
Proof.
We only need to prove that M( b ) ≤ E 𝜌 0 [ 𝑒 − 2 ( 𝑈 1 − 𝑈 0 ) ] and the equality can be achieved iff (71)holds 𝜌 0 -almost surely.
By Jensen’s inequality,
[Equation]
By taking the expectation E 𝑥 ∼ 𝜌 0 (cid:2) · (cid:3) for both sides and by (10) for new potentials (cid:101) 𝑈 1 = 2 𝑈 1 − 𝑈 0 and(cid:101) 𝑈
= 𝑈 0 , we immediately have the inequality:
[Equation]
Next in order to achieve the equality, we need the equality to hold in the above Jensen’s inequality:𝑡 → 𝑒 −
[Equation]
𝑈 1 − 𝑈 0 (cid:1) [Equation] X 𝑡 ( 𝑥 ) (cid:1) is a constant function .
By taking derivative with respect to 𝑡 , we immediately know that
[Equation]
-almost surely .
By the continuity assumption on 𝑈 1 ,𝑈 0 and b , we obtain (71).
Conversely, when (71) holds,
[Equation]
Hence, 𝑈 1 (cid:0) X 𝑡 ( 𝑥 ) (cid:1) − 𝑈 0 (cid:0) X 𝑡 ( 𝑥 ) (cid:1) =
𝑈 1 ( 𝑥 ) − 𝑈 0 ( 𝑥 ) for any 𝑡 ∈ R .
Then
[Equation]
Hence, the equality in (70) holds under the condition (71).
□
H.2 Some observations about the functional derivative
We need some simplified understanding of the condition ∫ ∞ 0 F ( 0 ) 𝑡
[Equation]
[Equation]
( 𝑥 ) d 𝑡 arising from 𝛿 Var ( b ) 𝛿 b =
𝑑 , which are presented in the followingtwo lemmas.
[Equation]
[Equation]
( 𝑥 ) d 𝑡 is equivalent to
[Equation]
Proof.
[Equation] [Equation]
both
[Equation]
[Equation]
□ [Figure] The following result provides a simplified characterization of the equality (72). [Figure]
Lemma H.5. [Figure] [Equation] Suppose the condition in (72) holds for any 𝑥 in an open set 𝐷 .
Then
[Equation]
(ii) Conversely, if (73) holds for 𝐷
= Ω , then b is a global maximum of M (as well as Var).
Proof.
The part (ii) immediately follows from Lemma H.3.
Next we prove the part (i).
From previousresults, the condition is that
[Equation]
After replacing 𝑥 by X 𝑠 ( 𝑥 ) ∈ 𝐷 in the above equation and by (24), we have
[Equation]
where 𝜏 + 𝐷 ( 𝑥 ) and 𝜏 − 𝐷 ( 𝑥 ) defined in (22) are hitting times for the forward and backward branches to theboundary of 𝐷 .
Note that the right hand side of the last equation depends on 𝑠 , whereas the left handside does not.
Let us take the derivative with respect to 𝑠 and with straightforward simplifications, weobtain
[Equation]
By (6), we know exp
[Equation] 𝑠 ( 𝑥 ) (cid:1)(cid:17) =
A( 𝑥 ) .
Again, the left hand side depends on 𝑠whereas the right hand side does not.
So let us take the derivative with respect to 𝑠 again and wehave ∇( 𝑈 1 − 𝑈 0 ) (cid:0) X 𝑠 ( 𝑥 ) (cid:1) · b (cid:0) X 𝑠 ( 𝑥 ) (cid:1)
=
for any 𝑥 ∈ 𝐷 and 𝑠 ∈ (cid:0) 𝜏 − 𝐷 ( 𝑥 ) , 𝜏 + 𝐷 ( 𝑥 ) (cid:1) .
Then (73) followsimmediately.
□
Lemma H.5 leads into the following intuition: if there is a certain open subset 𝐷 in which∇( 𝑈 1 − 𝑈 0 ) · b = 0, then such a dynamics b should not be a local minimum of Var, because such a bcannot explored the landscape structure of 𝑈 1 on 𝐷 .
This intuition is more rigorously formulated inthe following proposition.
Proposition H.6.
Consider a dynamics b ∈ 𝔅 ∞ .
Suppose 𝐷 ⊂ ℧ ( b ) is nonempty and open.
Let𝐾 :
=
Ω \ 𝐷 .
Assume that
(i) For any 𝑥 ∈ 𝐷 , we have b ( 𝑥 ) · ∇( 𝑈 1 − 𝑈 0 )
( 𝑥 ) = 0 .
(ii) For any 𝑡 ∈ R and 𝑥 ∈ 𝐷 , we have X 𝑡 ( 𝑥 ) ∉ 𝐾 ◦ , i.e., trajectories from 𝐷 are confined inside𝐷 .(iii)
There exists a b -stable point 𝑥 ★ ∈ 𝐷 such that ∇( 𝑈 1 − 𝑈 0 )( 𝑥 ★ ) ≠ 0 𝑑 .
Then such a dynamics b must not be a local minimum of the second moment M (as well as thevariance).
Proof.
Step (I): The first goal is to find a smooth function 𝛿 b ∈ 𝐶 ∞ 𝑐 ( 𝐷, R 𝑑 ) and 𝜖 0 > 0 such that𝜌 0 ( 𝐸 ) > 0 where 𝐸 : =
(cid:8) 𝑥 ∈ 𝐷 (cid:12)(cid:12) (cid:10) 𝛿 b ( 𝑥 ) , ∇( 𝑈 1 − 𝑈 0 ) ( 𝑥 ) (cid:11) ≠ 0 (cid:9) ⊂ supp ( 𝛿 b ) ; (74a) b + 𝜖𝛿 b ∈ 𝔅 ∞ , ∀ 𝜖 ∈ ( 0 , 𝜖 0 ) ; (74b) dist (cid:0) 𝐸, 𝜕𝐷 (cid:1) > 0 . (74c)
By the assumption (iii) of this proposition and Proposition A.10, we know there is an open ball𝐵 𝜆 ( 𝑥 ★ ) ⊂
𝐷 such that for any 𝛿 b ∈ 𝐶 ∞ 𝑐 (cid:0) 𝐵 𝜆 ( 𝑥 ★ ) , R 𝑑 (cid:1) , b + 𝜖𝛿 b ∈ 𝔅 ∞ for small enough 𝜖 and thus(74b) is satisfied.
It is clear that we can easily choose 𝜆 small enough so that (74c) holds.
Next, the task is to find a smooth function 𝛿 b supported on 𝐵 𝜆 ( 𝑥 ★ ) such that (74a) holds.
Since∇( 𝑈 1 − 𝑈 0 )( 𝑥 ★ ) ≠ 0 𝑑 and 𝑈 0 ,𝑈 1 are assumed to be smooth, we can choose 𝜆 small enough such thatthere exists a 𝜉 < 1 and
[Equation]
By the triangle inequality,
[Equation]
It is well-known that
[Equation]
is a smooth function compactly supported on 𝐵 1 ( 0 ) .
Then let us consider
[Equation]
𝜆
[Equation]
It is clear that 𝛿 b 1 is compactly supported on 𝐵 𝜆 1 ( 𝑥 ★ ) ⊂ 𝐵 𝜆 ( 𝑥 ★ ) and for any 𝑥 ∈ 𝐵 𝜆 1 ( 𝑥 ★ ) , we have⟨ 𝛿 b 1 ( 𝑥 ) , ∇( 𝑈 1 − 𝑈 0 )( 𝑥 )⟩ >
so that (74a) clearly holds.
Next, we still need to further smooth out𝛿 b 1 (see e.g., Appendix C of [12]) by introducing
[Equation]
where 𝜑 𝜀 ( 𝑥 ) :
= 1 𝜀 𝑑 𝜑 ( 𝑥 / 𝜀 ) .
By choosing 0 < 𝜀 < 𝜆 − 𝜆 1 , we can ensure that the smooth function 𝛿 b 2is compactly supported on 𝐵 𝜆 ( 𝑥 ★ ) .
It is also not hard to show that (74a) still holds for 𝛿 b 2 : for any 1226 𝑥 ∈ 𝐵 𝜆 1 ( 𝑥 ★ ) , [Equation]
[Equation]
In summary, 𝛿 b 2 constructed above satisfies all requirements.
Step (II): We prove that b is not a local minimum by showing that M( b + 𝜖𝛿 b ) < M( b ) for any 1229 𝜖 ∈ ( 0 , 𝜖 0 ) , where 𝛿 b satisfies all conditions in the Step (I).
By the construction of 𝛿 b , we know b 𝜖 :
=
b + 𝜖𝛿 b does not change the velocity field at 𝜕𝐷 , andtherefore, X 𝜖𝑡 ( 𝑥 ) ∉ 𝐾 ◦ for any 𝑥 ∈ 𝐷 still holds for the dynamics
b 𝜖 i.e., trajectories from 𝐷 will notenter 𝐾 ◦ .
As an immediately consequence, trajectories 𝑡 → X 𝜖𝑡 ( 𝑥 ) with 𝑥 ∈ 𝐾 ◦ will not enter 𝐷 (as
ODE trajectories are reversible).
The slightly technical part is to consider trajectories 𝑡 → X 𝜖𝑡 ( 𝑥 )with 𝑥 ∈ 𝜕𝐷 ≡ 𝜕𝐾 , where the trajectory 𝑡 → X 𝜖𝑡 ( 𝑥 ) evolves under the dynamics b 𝜖 .
Let us considertwo disjoint sets:
[Equation]
By such a construction, we can observe that for any trajectory 𝑡 → X 𝜖𝑡 ( 𝑥 ) with 𝑥 ∈ (cid:101) 𝐾 (or 𝑥 ∈ (cid:101) 𝐷 ), itmust remain inside (cid:101) 𝐾 (or (cid:101) 𝐷 ).
In other words, the flows within (cid:101) 𝐷 and (cid:101) 𝐾 are completely separatedfrom each other.
Moreover, because 𝛿 b is only supported on 𝐸 ⊂ 𝐷 which is completely inside 𝐷 by(74c), we know the above definitions of these two sets (cid:101) 𝐷 and (cid:101)
𝐾 are independent of 𝜖 ∈ [ 0 , 𝜖 0 ) .
Let us use A to denote the function defined in (23) corresponding to the dynamics b and use A 𝜖to denote the one corresponding to the perturbed dynamics b 𝜖 .
Recall the assumption (i) thatb ( 𝑥 ) · ∇( 𝑈 1 − 𝑈 0 )
( 𝑥 ) for all 𝑥 ∈ 𝐷 .
By the fact that 𝜕𝐷 = 𝜕𝐾 , and by the continuity of b , ∇ 𝑈 0 , and∇ 𝑈 1 , we know
[Equation]
For any trajectory 𝑡 → X 𝑡 ( 𝑥 ) with 𝑥 ∈ (cid:101) 𝐷 , we can easily show that 𝑒 −( 𝑈 1 − 𝑈 0 ) (cid:0) X 𝑡 ( 𝑥 ) (cid:1) =
𝑒 −( 𝑈 1 − 𝑈 0 )
( 𝑥 )for all 𝑡 ∈ R , and thus we have A( 𝑥 ) =
𝑒 −( 𝑈 1 − 𝑈 0 )
( 𝑥 ) for any 𝑥 ∈ (cid:101) 𝐷 (the same calculation, in fact, hasbeen performed in the proof of Lemma H.3).
Hence,
[Equation]
where 𝜒 𝐴 (·) is an indicator function for a set 𝐴 .
Next we consider the trajectory 𝑡 → X 𝜖𝑡 ( 𝑥 ) with 𝑥 ∈ (cid:101) 𝐾 .
Since such a trajectory never enters (cid:101) 𝐷 ⊃ 𝐷and b 𝜖 = b on Ω \ 𝐷 , we know X 𝜖𝑡 ( 𝑥 ) =
X 𝑡 ( 𝑥 ) for any 𝑡 ∈ R and 𝑥 ∈ (cid:101) 𝐾 , and thus A =
A 𝜖 on (cid:101) 𝐾for any 𝜖 ∈ [ 0 , 𝜖 0 ) .
Hence, E 𝜌 0 (cid:104) 𝜒 (cid:101) 𝐾 (·) (cid:0) A(·) (cid:1) 2 (cid:105) =
E 𝜌 0 (cid:104) 𝜒 (cid:101) 𝐾 (·) (cid:0)
A 𝜖 (·) (cid:1) 2 (cid:105) .
By the same argument as in
Lemma H.3 (by treating (cid:101) 𝐷 as the domain),
[Equation]
where the equality is achieved only if ⟨ b 𝜖 , ∇( 𝑈 1 − 𝑈 0 )⟩ = 0 on (cid:101) 𝐷 .
Note that on (cid:101) 𝐷 ,
[Equation]
[Equation] However, due to the fact that ⟨ 𝛿 b , ∇( 𝑈 1 − 𝑈 0 )⟩ ≠ 0 for some open subset of 𝐷 with strictly positive𝜌 0 -measure (as constructed in the Step (I)), the equality M( b 𝜖 ) =
M( b ) cannot be achieved and thus
[Equation]
Since we can find a local perturbation 𝛿 b such that M( b 𝜖 ) <
M( b ) for any 𝜖 ∈ ( 0 , 𝜖 0 ) , then b mustnot be a local minimum of M .
□
H.4 Proof of Proposition [-H.
-]H.
By Proposition D.1, we know [-either∇A( 𝑥 ) = 
 𝑑 , or ∫ ∞-]either∇A
[Equation]
𝑥
[Equation]
or
[Equation]
Define
[Equation]
which is a closed subset of Ω by the continuity assumption.
Hence, 𝐷 : = Ω \ 𝐾 is open and by Lemma H.5 (i), we know b · ∇( 𝑈 1 − 𝑈 0 ) = 0 on 𝐷 .
Here are a fewcases to discuss.
Case (I) :
𝐾 ◦ = ∅ .
If 𝐾 ◦ = ∅ , then we claim that b must be a global maximum and this contradicts with the assumptionthat Var ( b ) <
Var ( max ) .
It is not hard to observe that 𝐾 ◦ = ∅ implies that 𝐷
= Ω .
By continuity, weknow b · ∇( 𝑈 1 − 𝑈 0 ) = 0 on 𝐷 = Ω .
Then by Lemma H.5 part (ii), b must be a global maximum.
Case (II) : 𝐷 = ∅ .
If 𝐷 = ∅ (i.e., 𝐾 = Ω ), then it is clear that A is a constant function, and thus the variance Var ( b ) = 0.
This means b is a zero-variance dynamics.
Case (III) :
𝐾 ◦ ≠ ∅ and 𝐷 ≠ ∅
In order to use Proposition H.6, we need to deal with the case that some trajectories 𝑡 → X 𝑡 ( 𝑥 ) for 1272 𝑥 ∈ 𝐷 might enter 𝐾 ◦ .
Let us introduce
[Equation]
(75) which essentially contains all points in 𝐷 whose trajectories enter 𝐾 ◦ at some time.
We can easilyshow that 𝑆 is open: because b is assumed to be smooth, the trajectories are continuous under a smallperturbation.
Then the new disjoint sub-regions to consider are (cid:101) 𝐾 : =
𝐾 ◦ ∪ 𝑆 and (cid:101) 𝐷 : = Ω \ (cid:101) 𝐾 .
Wecollect some facts for clarity:
• (cid:101) 𝐾 ◦
=
𝐾 ◦ ∪ 𝑆 ≠ ∅ ; • (cid:101) 𝐷 ⊂ 𝐷 , which immediately implies that b · ∇( 𝑈 1 − 𝑈 0 ) = 0 on (cid:101) 𝐷 .
• If 𝑥 ∈ (cid:101) 𝐷 , then X 𝑡 ( 𝑥 ) must not enter (cid:101) 𝐾 ◦ .
Indeed, if not, then we have either 𝑡 → X 𝑡 ( 𝑥 ) entering
𝐾 ◦ (which contradicts with 𝑥 ∉ 𝑆 ), or X 𝑡 ( 𝑥 ) entering 𝑆 (which still means X 𝑡 ( 𝑥 ) will enter 𝐾 ◦
due to the reversibility of deterministic trajectories).
We need to discuss two cases:
(a) Firstly, let us consider (cid:101) 𝐷 = ∅ , i.e., (cid:101) 𝐾 = Ω and thus (cid:101) 𝐾 ◦ =
Ω by the assumption that Ω is open
in the topology of the space R 𝑑 .
The connectivity granted by the definition of 𝑆 in (75)
implies that we can divide (cid:101) 𝐾 ◦ into a countable number of sub-regions on which the function
A is a constant.
More specifically, for 𝑥 ∈ Ω , define
[Equation]
Obviously, 𝑥 ∈ 𝑅 𝑥 .
By the invariance of A under the dynamical flow (see (25)) and the
definition of 𝑆 (75), we have ∪ 𝑥 ∈ 𝐾 ◦ 𝑅 𝑥 ⊃ (cid:101) 𝐾 ◦
= Ω , which implies that ∪ 𝑥 ∈
𝐾 ◦ 𝑅 𝑥
= Ω .
• Suppose 𝑥, 𝑦 ∈ 𝐾 ◦ and 𝑅 𝑥 ∩ 𝑅 𝑦 ≠ ∅ , then there exists a 𝑧 ∈ 𝑅 𝑥 ∩ 𝑅 𝑦 such that 𝑧
connects to both 𝑥 and 𝑦 via a continuous path and thus A( 𝑥 ) =
A( 𝑧 ) =
A( 𝑦 ) .
It is
then clear that 𝑅 𝑥 = 𝑅 𝑦 via treating 𝑧 as a bridge.
Therefore, ∪ 𝑥 ∈ 𝐾 ◦ 𝑅 𝑥 can be simplified
as ∪ 𝑥 ∈ 𝐸 𝑅 𝑥 where { 𝑅 𝑥 } 𝑥 ∈ 𝐸 are disjoint and 𝐸 ⊂ 𝐾 ◦ .•
Next, we can show that 𝐸 is countable.
Since 𝐾 ◦ is open and ∇A( 𝑥 ) =
𝑑 on 𝐾 , wecan easily verify that for each 𝑥 ∈ 𝐾 ◦ , there exists a local neighbor 𝐵 𝛿 ( 𝑥 ) ⊂
𝑅 𝑥 and dueto the fact that Q 𝑑 is dense and countable, 𝐸 is at most countable.
To summarize, we have [Equation]
where { 𝑅 𝑥 } 𝑥 ∈ 𝐸 are disjoint and 𝐸 is countable.
As A is a constant function on 𝑅 𝑥 by definition of 𝑅 (·) , A is a step function on Ω .
By thecontinuity of A from the assumption, and also Ω is a connected open domain, we readilyknow A must be a constant function on Ω instead, and such b must be a zero-variancedynamics. b
) Next, let us consider the case (cid:101) 𝐷 ≠ ∅ .
By the assumption that ∇( 𝑈 1 − 𝑈 0 ) =
𝑑 only on a setwith Lebesgue measure zero, we know there must exist 𝑦 ∈ (cid:101) 𝐷 such that ∇
( 𝑈 1 − 𝑈 0 )( 𝑦 ) ≠ 0 𝑑 .
By the assumption that b -unstable points has Lebesgue measure zero, there must exist ab -stable
point
𝑥 ★ (around 𝑦 ) such that ∇( 𝑈 1 − 𝑈 1 )( 𝑥 ★ ) ≠ 0 𝑑 .
Then Proposition H.6 tells usthat b must not be a local minimum, which contradicts with the assumption.
I Supplementary material for numerical experiments
I.1 Details about AIS.
For AIS method, we use equally-spaced temperature distribution, i.e., 𝜋 𝑘 ∝ 𝜌 1 − 𝛽 𝑘𝜌 𝛽 𝑘 1 , where𝛽 𝑘 = 𝑘 / 𝐾 for 0 ≤ 𝑘 ≤ 𝐾 ; for each transition step, we use Metropolis-adjusted Langevin algorithmwith time step 0 .
1 to generate the chain.
I.2 More implementation details
Neural network architecture..
We use the following ℓ -layer neural network [11, 31] to parameterizethe dynamics b : Ω → R 𝑑 during training:
[Equation]
where ℓ is the layer depth (one output layer and ℓ − 1 hidden layers), 𝑓 𝑗 (·)
=
𝜂 ( 𝑊 𝑗 (·) + 𝑏 𝑗 ) for𝑗
= 1 , 2 , · · · , ℓ − 1, 𝜂 is the activation function, 𝑊 𝑗 ∈ R 𝑛 𝑗 × R 𝑛 𝑗 + 1 , 𝑏 𝑗 ∈ R 𝑛 𝑗 + 1 for 𝑗 = 1 , 2 , · · · , ℓ .
When we choose 𝑛 𝑗 = 𝑚 for all 2 ≤ 𝑗 ≤ ℓ , we refer such a neural network as ( ℓ, 𝑚 ) -architecturewhich is mentioned in Section 5.
More specifically, let us take ℓ = 2 as an example: an ( ℓ, 𝑚 ) =
( 2 , 𝑚 ) architecture for a generic ansatzfor b : Ω → R 𝑑 refers to the following parameterization
[Equation]
where weights 𝑊 1 ∈ R 𝑑 × R 𝑚 , 𝑊 2 ∈ R 𝑚 × R 𝑑 and bias vectors 𝑏 1 ∈ R 𝑚 , 𝑏 2 ∈ R 𝑑 .
For the gradient-form ansatz and divergence-free form ansatz, as we essentially need to parameterizea potential 𝑉 : Ω → R , the ( ℓ, 𝑚 ) -architecture for 𝑉 refers to the following choice when ℓ = 2, [Equation]
where 𝑊 2 ∈ R 𝑚 × R , 𝑊 1 ∈ R 𝑑 × R 𝑚 , 𝑏 1 ∈ R 𝑚 .
The bias vector in the output layer is chosen as zero(i.e.
, 𝑏 2 = 0) because 𝑏 2 is a redundant parameter after taking the gradient.
The b in gradient formand divergence-free form using such an architecture for 𝑉 refers tob =
[Equation] 𝑉, (gradient form) ,
[Equation]
where J 𝑑 : = (cid:104)
I 𝑑 / 2 − I 𝑑 / 2 0 (cid:105) (assuming that the dimension 𝑑 is an even integer).
Initialization and trial repetition. When we parameterize the flow via neural networks, weightsand biases in the neural network are randomly generated.
Therefore in the below, we consider twoindependent trials (associated with two random initialization of b ) for the same neural networkarchitecture characterized by a pair ( ℓ, 𝑚 ) .
Optimization algorithm..
We use the SGD algorithm to optimize parameters for b .
The Armĳoline search algorithm (see e.g. [3, 44]) is used to find the learning rate.
The way to approximate theloss function M 𝑡 − ,𝑡 + ( b ) ≡ E 0 (cid:2) |A
𝑡 − ,𝑡 + | 2 (cid:3) (see Section 4) and in particular its gradient with respect toparameters in b will be explained in Appendix I.4 and I.5.
Computational resources and costs..
The training is conducted with four AMD Opteron [-6272 (2.
-]6272(2.1 GHz) (64 cores): it took around 1.5 hours on average to train the asymmetric 2-mode Gaussianmixture ( 𝑑 = 2) and it took around 5 hours on average to train the the symmetric 4-mode Gaussianmixture ( 𝑑 = 10) for independent trials; in the funnel distribution ( 𝑑 = 10), it took around 20 mins. [Table]
[Table] [Table]
[Figure]
I.4 An integration-based forward propagation method to compute the estimator and itsgradient
To estimate A 𝑡 − ,𝑡 + ( 𝑥 ) in (8) with 𝑡 + = 𝑡 − + 1 and 𝑡 − ∈ [− 1 , 0 ] , the most straightforward approach is tocompute F ( 𝑘 ) 𝑡 ( 𝑥 ) at time grid points and then employ some integration scheme like the trapezoidalquadrature method.
More specifically, we first discretize the time interval [− 1 , 1 ] by 2 𝑁 𝑡 + 1 equally-spaced points𝑡 𝑚 :
= 1 𝑁 𝑡 𝑚 where − 𝑁 𝑡 ≤ 𝑚 ≤ 𝑁 𝑡 .
Then we use classical ODE integration schemes (we use RK4) topropagate the dynamics X 𝑡 ( 𝑥 ) both forward and backward in time to estimate the following quantities:
[Equation]
Then
[Equation]
where
[Equation]
is the trapezoidal quadrature scheme.
Then similarly, we can use the trapezoidal quadrature schemeto estimate ∫ 𝑡 𝑗 − 𝑡 − 𝑡 𝑗 − 𝑡 + F ( 0 ) 𝑠
( 𝑥 ) d 𝑠 given the values F ( 0 ) 𝑡 𝑚 ( 𝑥 ) for − 𝑁 𝑡 ≤ 𝑚 ≤ 𝑁 𝑡 and the time 𝑡 𝑗 with 𝑡 𝑗 ∈ [ 𝑡 − , 𝑡 + ] .
After we have approximated values for
F ( 1 ) 𝑡𝑗 ( 𝑥 )
[Equation]
𝑠 , the trapezoidal quadrature scheme is utilized again to approximate A 𝑡 − ,𝑡 + ( 𝑥 ) .
The computational cost is mostly dominated bythe ODE integration, e.g., propagating X 𝑡 ( 𝑥 ) or evaluating ∇ · b ( X 𝑠 ( 𝑥 )) in general.
Therefore, thisstraightforward integration-based method has linear computational cost with respect to 𝑁 𝑡 and is thusexpected to be optimal.
Using the same principle, we can estimate the gradient of the second moment with respect to parameters the the [Figure]
I.5 An ODE-based forward propagation method to compute the estimator and its gradient for
[Figure] 𝑡 − = 0 , 𝑡 + = [Figure]
To simplify notations, let us introduce [Figure] [Equation] 𝑥 ) d 𝑠, and thus, A 𝑡 − ,𝑡 + ( 𝑥 ) (8) = ∫ 𝑡 + 𝑡 −
[Equation]
which implicitly depends on b .
Let us denote 𝛼 𝑡 ( 𝑥 ) : = ∫ 𝑡 𝑡 − F ( 1 )
𝑠 ( 𝑥 ) ℬ 𝑠 ( 𝑥 )
d 𝑠 .
[Figure]
I.5.1 ODE dynamics to compute the estimator
Then the estimator A 0 , 1 is simply 𝛼 1 ( 𝑥 ) .
To compute A 0 , 1 , we simply need to run the following
[Figure]
ODE:
[Figure]

[Equation]

[Equation]
where X lag 𝑡 ( 𝑥 ) :
= X 𝑡 − 1 ( 𝑥 ) and J lag 𝑡 ( 𝑥 ) :
=
J 𝑡 − 1 ( 𝑥 ) .
Moreover, in order to obtain the initial condition,we shall run the dynamics backward in time, i.e., we shall simulate the following ODE on [ 0 , 1 ] ,
[Equation]
where the superscript 𝑅 means the reversed process.
As a remark, the auxiliary backward ODE (78)has dimension 𝑑 + 2 and the ODE (77) has dimension 2 𝑑 + 4.
compute
Denote 𝜗 as a vector containing parameters in b and let 𝑁 𝑝 be the number of parameters.
Inwhat follows, we also vectorize all quantities involving 𝜗 , e.g., for each parameter 𝜗 𝑗 , there is acorresponding Y ( 𝑗 ) 𝑡 ( 𝑥 ) in (34) and we simply use the notation Y 𝑡 ( 𝑥 ) to be a matrix whose 𝑗 th columnis Y 𝑗𝑡 ( 𝑥 ) to save notations; the same convention applies to other quantities.
Next, we shall similarly re-write the expression ∇ 𝜗 A 0 , 1 (implicitly given in ∇ 𝜗 M 0 , 1 ( b ) (32)) interms of outputs from an ODE.
Using the same idea, let us introduce
[Equation]
and then we can rewrite quantities involved inside ∇ 𝜗 M 0 , 1 ( b ) as follows:the evolution of X 𝑡 ( 𝑥 ) , X lag 𝑡 ( 𝑥 ) , J 𝑡 ( 𝑥 ) , J lag 𝑡 ( 𝑥 )
[Equation]
ℬ 𝑡 ( 𝑥 ) , 𝛼 𝑡 ( 𝑥 ) in (77) ,
[Equation]
(cid:171) 𝑒 − 𝑈
[Equation]
(cid:0) X 𝑡 ( 𝑥 ) (cid:1)
[Equation]
( 𝑥 ) (cid:0) − ∇
[Equation]
(cid:0) X [Equation] ( 𝑥 ) (cid:1)
[-𝑇 Y 𝑅𝑡 ( 𝑥 ) , H 𝑅 0 ( 𝑥 ) = 
 𝑁 𝑝 , d d 𝑡 L 𝑅𝑡 ( 𝑥 ) = −∇ 𝜗 (∇ 𝑥 · b ) (cid:0)-][Equation]
Y 𝑡 ( 𝑥 )
[-L 𝑅 0 ( 𝑥 ) = 
 𝑁 𝑝 , d d 𝑡 Y 𝑅𝑡 ( 𝑥 ) = −∇ b (cid:0) X-][Equation]
H 𝑡 ( 𝑥 ) [Equation]
L 𝑡 ( 𝑥 ) (cid:1) [Equation] 𝑥 ) [Equation]
(cid:0)
[Equation]
[Equation]
𝑡
[Equation]
𝑥
[Equation]
𝑡 [Equation] 𝑥 ) (cid:0) [Equation]
[Equation]
(cid:170)(cid:174)(cid:174)(cid:174)(cid:174)(cid:174)(cid:174) (cid:172) ,
D 0 ( 𝑥 )
[Equation]
0 𝑁 𝑝 , d d 𝑡 g
[Equation]
𝑥
[Equation]
𝑥 [Equation] 𝑥 ) [-= 
-][Equation]
(cid:170)(cid:174)(cid:174)(cid:174)(cid:174)(cid:174)(cid:174)
[Equation]
) , d d 𝑡 H 𝑡 ( 𝑥 )
[Equation]
(cid:16) ∇( ∇ · b ) (cid:0) X 𝑡 ( 𝑥 ) (cid:1)(cid:17)
𝑇 Y 𝑡 ( 𝑥 ) ,
H 0 (
[Equation]
where the initial conditions come from solving an auxiliary backward equation on [ [Equation] [Equation] given asfollows:the evolution of X 𝑅 𝑡 ( 𝑥 ) , J 𝑅𝑡 ( 𝑥 ) , ℬ 𝑅 𝑡 ( 𝑥 ) in (78)
[Equation]
d
[Equation]
L 𝑅
[Equation]
d
[Equation] Y 𝑅 [Equation] As a remark, we combine the auxiliary states Y 𝑡 ( 𝑥 ) for all parameters together, so that all three terms
Y 𝑡 ( 𝑥 ) , Y lag 𝑡 ( 𝑥 ) , Y 𝑅𝑡 ( 𝑥 ) have the dimension 𝑑 × 𝑁 𝑝 .
Below is a table that summarizes the dimensionof all components.
[Table]
In the end, the outputs we need are
A 0 , 1 ( 𝑥 ) [Equation]
𝛼 [Equation] ( 𝑥 ) [Equation]
A
[Equation] [Equation]
D
[Equation]
( 𝑥 [Equation] so that ∇ 𝜗 M 0 , 1 ( b ) = 2 E 𝜌 0 (cid:2) 𝛼 1 ( 𝑥 ) D 1 ( 𝑥 ) (cid:3) .
[-I.
-]I.6 A discussion on the backward propagation for differentiation
In the formulas (32) and (33), a crucial step is to evaluate
[Equation]
[Equation]
for time 𝑡 ∈ [− 𝑇,𝑇 ] where 𝑇 = 𝑡 + − 𝑡 − .
For any fixed 𝑡 , this value can be efficiently measured by theadjoint equation with backward propagation proposed in [9].
For instance, let us consider 𝑡 > 0 and
[Equation]
Moreover, it can be easily verified by the definition (38) that for 𝑠 ∈ [ 0 , 𝑡 ] ,
[Equation]
) (cid:1) .
The above two equations are Eqs.
(4) & (5) in [9].
This can eliminate the need to simulate Y 𝑡 ( 𝑥 ) ,whose memory cost scales like O( 𝑑𝑁 𝑝 ) .
However, in the finite-time NEIS scheme, we need toestimate Φ ( 𝑡, 𝑥 ) not only for a fixed 𝑡 but also for 𝑡 ∈ [− 1 , 1 ] .
Therefore, the computational costscales like O( 𝑁 2 𝑡 ) where 𝑁 𝑡 is the number of time-discretization (or say the depth of the flow map inthe normalizing flow context).
As a comparison, the forward propagation method uses more memorybut could be computationally cheaper as it only needs to visit the whole trajectory for O( 1 ) times (seethe table below).
[Table]