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These bounds are enough to use Bernstein’s inequality to show that ( 13 ) holds except with probability at most [MATH] Similar moment bounds hold for [MATH] and we can again use Bernstein’s inequality to show that the probability that ( 14 ) holds is at least [MATH] . Note that we can get both ( 13 ) and ( 14 ) to hold... |
Note that [MATH] under [MATH] except with probability [MATH] . With probability at least [MATH] , we have that [MATH] for all [MATH] |
Let [MATH] be a sufficiently small multiple of [MATH] Let [MATH] [MATH] be [MATH] -covers of the unit sphere in [MATH] and [MATH] be the set of multiples of [MATH] between [MATH] and [MATH] [MATH] . By a union bound, we have that for all [MATH] [MATH] and [MATH] that we have ( 13 ) and ( 14 ) with [MATH] in place of [M... |
[MATH] We have that for any unit vectors [MATH] [MATH] , there exist [MATH] and [MATH] such that [MATH] and [MATH] For any [MATH] , we have that [MATH] and the equivalent for [MATH] . We also have that [MATH] |
We have that [EQUATION] with a similar bound from below of the form [MATH] Next we show that [EQUATION] Thus, we have that ( 13 ) holds with the bound [MATH] for all unit vectors [MATH] and all [MATH] as required. A similar proof shows that ( 14 ) also holds. |
Next we show that these expectations are close under [MATH] and [MATH] Lemma C.10 For any [MATH] let [MATH] be the true distribution over [MATH] conditioned on [MATH] . Then for all [MATH] and all [MATH] and all unit vectors [MATH] , we have that |
[EQUATION] and [EQUATION] Proof. We show that the contribution of points with [MATH] to [MATH] is small and use Cauchy-Schwarz. Let [MATH] be [MATH] conditioned on the negated condition that [MATH] which happens with probability at most [MATH] . Let [MATH] and [MATH] |
and note that [EQUATION] By a similar proof, [MATH] By Cauchy-Schwarz, [MATH] and applying it again [MATH] Noting that [MATH] and [MATH] , we have that |
[EQUATION] and [EQUATION] Corollary C.11 Assuming Lemma C.9 holds, conditions 3 (iii) and (iv) of Definition C.1 hold. Proof. We need show show that these condtios hold for all [MATH] First note that [MATH] . By Lemmas C.10 |
C.9 and the triangle inequality, we have [MATH] , which is (iii). It follows using Claim A.9 that [MATH] . Again using Lemmas C.10 |
C.9 and the triangle inequality, we have [MATH] , where [MATH] which is (iv). To make the robust variance estimation via interquartile range work, we need that |
Lemma C.12 With probability at least [MATH] , for all [MATH] and [MATH] [MATH] Proof. Noting that the VC dimension of halfspaces over [MATH] is [MATH] the result follows from the VC inequality. |
Proof of Proposition C.2 By a union bound, Claim A.4 and Lemma C.3 C.4 C.9 and C.12 all hold with probability at least [MATH] . We condition on the event that they do. We have the following: |
1 (i) and (ii) follow from Claim A.4 and (iii) and (iv) follow from Lemma C.3 2 (i) and (ii) follow from Lemma C.4 and (iii) and (iv) from Lemma C.5 |
3 (i) follows from Claim A.4 and the bound on [MATH] in [MATH] (ii) follows from Lemma C.8 , which required Claim A.4 and Lemma C.4 (iii) and (iv) are given by Corollary C.11 which requires Lemma C.9 |
4 is given by Lemma C.12 This completes the proof of the proposition. Appendix D Proof of Proposition 2.4 D.1 Analysis if we Remove Samples at Any Step |
For this, we need to show that the conditions on the set [MATH] given by Definition C.1 are sufficient to guarantee that the filter steps return a set [MATH] with [MATH] |
Note that the conditions given in Definition C.1 1 and 2, satisfy the requirements for the sub-gaussian filter, such as those given in DKK 17b , but with [MATH] in place of [MATH] This ensures that if either of the steps |
and 13 return a subset [MATH] , then [MATH] For step 16 , we perform a filter similar to the one in the previous section. Note that if we replace [MATH] by [MATH] , the filter is identical to the one in the previous section. The samples in [MATH] satisfy the conditions in Definition C.1 |
(i)-(iv). Conditions (i)-(iv) exactly correspond to (i) to (iv) of Definition 2.2 Note that these conditions are sufficient for the proof of section to apply and show that if this step removes samples from [MATH] , then [MATH] . Adding [MATH] back, we obtain that [MATH] |
D.2 Analysis of Correctness if we Return [MATH] We first show that after passing steps and 13 , removing the samples in [MATH] does not affect the expectation on [MATH] too much. |
It is immediate from the definition of [MATH] that when we pass step , we have [MATH] Lemma D.1 [MATH] and [MATH] for [MATH] as in Step 5. |
Proof. If we pass the filter steps then [MATH] and [MATH] However, we can lower bound the variance of [MATH] after removing a set of size [MATH] by Corollary A.11 of DKK 17b (which uses the same technique as Lemma B.1 in this paper), as [MATH] and [MATH] . Taking the difference and scaling by [MATH] gives the lemma. |
Corollary D.2 [MATH] Proof. This follows by Cauchy-Scwartz on the expectation over [MATH] . For the [MATH] in step 5, which is the normalisation of [MATH] , we have |
[MATH] Since [MATH] , we have that [MATH] and so [MATH] Since [MATH] and [MATH] , we are done. Lemma D.3 [MATH] Proof. The case of small spectral norm in section gave that [MATH] . Translating that into the notation of this section and applying it to [MATH] with [MATH] in place of [MATH] gives that [MATH] . ∎ |
Combining Corollary D.2 and Lemma D.3 , we obtain Corollary D.4 [MATH] Proof. By the triangle inequality, [MATH] . However, recall that |
[MATH] . We thus obtain that [MATH] as long as [MATH] is sufficiently small. Appendix E Proof of Theorem 3.1 : Statistical Query Lower Bounds |
We restate Theorem 3.1 below for convenience: Theorem 3.1 No algorithm given statistical query access to [MATH] , defined as above with unknown noise and unknown variances [MATH] and [MATH] gives an output [MATH] with [MATH] on all instances unless it uses more than [MATH] calls to the |
[EQUATION] oracles for any [MATH] Recall that we use the construction of DKS17c , which intuitively says that if we have a distribution which is standard Gaussian in all except one direction then if the low degree moments match the standard Gaussian, then that direction is hard to find with an SQ algorithm. The idea is... |
The further the mean of [MATH] conditioned on [MATH] is from [MATH] , the more noise needs to be added to match the first three moments. Lemma E.2 , which is the main lemma of the lowerbound proof, shows that we can match the first three moments by adding [MATH] fraction of noise when the [MATH] has mean [MATH] in the ... |
The rest of the section formally proves Theorem 3.1 . To start, recall that [MATH] is distributed as Gaussian, restated as below: |
Lemma 3.2 Let [MATH] be the joint distribution of [MATH] where [MATH] and [MATH] where [MATH] is unknown an [MATH] . Then [MATH] where [MATH] and [MATH] |
For the simplicity of our construction, we let the variance of [MATH] to be [MATH] in all except [MATH] direction while the [MATH] direction will have smaller variacne. This is because if the corruption affects the mean in [MATH] direction by much, they also increase the variance significantly. However to match the sec... |
Lemma E.1 If we set [MATH] [MATH] where [MATH] and [MATH] for constants [MATH] , then for any [MATH] , there exists a [MATH] , such that [MATH] and [MATH] |
Proof. By the previous lemma, we have [MATH] . Given arbitrary [MATH] , there exists a [MATH] such that [MATH] We take the joint distribution of [MATH] given by the above lemma to be [MATH] . We now need to define the corrupted distribution [MATH] . We want [MATH] for any [MATH] , in [MATH] to be a distribution of the ... |
Lemma E.2 For any [MATH] [MATH] , there is a distribution [MATH] such that [MATH] agree with the first [MATH] moments of [MATH] and [MATH] |
for some distribution [MATH] and [MATH] satisfying: If [MATH] , then [MATH] and [MATH] If [MATH] , then [MATH] and [MATH] Subsection will be devoted to the proof of the above lemma. Here we show that it suffices to prove Theorem 3.1 . Similarly to the construction in DKS17c , we define [MATH] . By Lemma 3.4 of that pap... |
[EQUATION] Now we need to define a [MATH] such that [MATH] that is a contaminated version of [MATH] Lemma E.3 If we define [MATH] where |
[EQUATION] and [MATH] , then [MATH] is a distribution with [MATH] for some distribution [MATH] and under [MATH] [MATH] Proof. First, to show that [MATH] and so [MATH] are well defined, we need to show that [MATH] is finite. Indeed we have that |
[EQUATION] where we have applied Lemma E.2 and the fact that [MATH] We have that [MATH] is non-negative and integrates to [MATH] . Thus [MATH] is the joint distribution of [MATH] and [MATH] [MATH] has the distribution [MATH] with pdf [MATH] and [MATH] |
Since for any [MATH] [MATH] , where we use [MATH] to denote the pdf function of distribution [MATH] , we have that [MATH] Since [MATH] , we have furthermore that |
[MATH] We can thus write [MATH] for the distribution [MATH] with pdf [MATH] Lemma E.4 For [MATH] as in Lemma E.3 , we have [EQUATION] |
where [MATH] is the joint distribution of [MATH] and [MATH] when they are independent and [MATH] and [MATH] Proof. The chi-square divergence is expressed as: |
[EQUATION] where we have applied Lemma 3.4 of DKS17c . Recall that [MATH] and by Lemma E.2 , If [MATH] then [MATH] and if [MATH] , then [MATH] We thus have that |
[EQUATION] We thus have that [MATH] as required. Proof of Theorem 3.1 given Lemma E.2 The proof now follows that of Proposition 3.3 of DKS17c . By Lemma 3.7 of DKS17c , for any [MATH] , there is a set [MATH] of at least [MATH] unit vectors in [MATH] such that for each pair of distinct [MATH] , it holds [MATH] . Then, b... |
We thus have that the set of [MATH] for [MATH] is [MATH] -corrlated for [MATH] [MATH] . Thus applying Lemma 2.12 of DKS17c we obtain that any SQ algorithm requires at least [MATH] calls to the |
[EQUATION] oracle to find [MATH] and therfore [MATH] within better than [MATH] Note that [MATH] for any [MATH] . Hence, the total number of required queries is at least |
[MATH] This completes the proof. Appendix F Proof of Lemma E.2 We restate Lemma E.2 here for convenience. Lemma E.2 For any [MATH] [MATH] , there is a distribution [MATH] (also written as [MATH] for simplicity) such that [MATH] agree with the first [MATH] moments of [MATH] and [MATH] |
for some distribution [MATH] and [MATH] satisfying: If [MATH] , then [MATH] and [MATH] If [MATH] , then [MATH] and [MATH] We split this into a number of cases, each of which will be a mixture of three Gaussians. The parameters and weights of these Gaussians will need to be chosen to make the first three moments the sam... |
We first deal with the case when [MATH] . We will need to further split this into cases. Lemma F.1 We have the following: For [MATH] , the distribution |
[MATH] , where [MATH] are defined as [EQUATION] has first moment [MATH] , second moment [MATH] , third moment [MATH] with [MATH] |
For [MATH] , the distribution [MATH] where [MATH] are defined as [EQUATION] has first moment [MATH] , second moment [MATH] , third moment [MATH] with [MATH] |
For [MATH] , the distribution [MATH] , where [MATH] are defined as [EQUATION] has first moment [MATH] , second moment [MATH] , third moment [MATH] with [MATH] |
Proof. We verified these facts with the symbolic computation function of Mathematica. Then we consider the remaining case when [MATH] |
Lemma F.2 Given [MATH] , the distribution [MATH] has first three moments as [MATH] and that [MATH] [MATH] Proof. Let [MATH] , to simplify the problem, we require [MATH] and hence [MATH] . The first moment equation requires |
[EQUATION] The second moment condition requires [EQUATION] The third moment condition requires [EQUATION] To simplify the problem, we require |
[EQUATION] which also implies [EQUATION] by the second moment equation. We can solve for the value of [MATH] and [MATH] together using the first moment condition. The solution is the following: |
[EQUATION] The range of [MATH] is [MATH] . Assume that [MATH] , we have that [MATH] and [MATH] . Plugging the range of [MATH] into the second moment condition yields: |
[EQUATION] Solving the linear equations we get [MATH] Finally we can combine the corrupted distributions constructed in different cases into a single corrupted distribution [MATH] by the following definition. |
Definition F.3 [EQUATION] Lemma F.4 [MATH] is well-defined for any [MATH] and has first three moments as [MATH] Proof. We need to verify that in each case [MATH] lies in the correct range for [MATH] to be well-defined. When [MATH] , the solution of [MATH] is less than [MATH] . When [MATH] , the solution of [MATH] is be... |
Lemma F.5 [MATH] for some distribution [MATH] and when [MATH] [MATH] Proof. When [MATH] , we have [MATH] which yields [MATH] . When [MATH] , we have [MATH] which yields [MATH] . When [MATH] , we have [MATH] |
Finally, to complete the proof of Lemma E.2 , we will need to bound [MATH] . Notice that by Fact F.7 and Fact F.8 , we have [MATH] and [MATH] for constant [MATH] . In the case where [MATH] , it is straightforward to verify that all the means of the Gaussian distributions are [MATH] . Hence by applying Fact F.6 repeated... |
The following three technical facts regarding the chi-square distance and the Gaussian distribution can be verified easily, we omit some of the proof. Fact F.6 For distributions [MATH] and [MATH] , we have that [MATH] Proof. [EQUATION] Fact F.7 [EQUATION] Fact F.8 [EQUATION] |
# Source: arxiv 1806.00237 # Title: Accounting for model errors in iterative ensemble smoothers # Sections: all # Downloaded: 2026-03-03T05:16:52.432875+00:00 |
Accounting for model errors in iterative ensemble smoothers Abstract In the strong-constraint formulation of the history-matching problem, we assume that all the model errors relate to a selection of uncertain model input parameters. One does not account for additional model errors that could result from, e.g., exclude... |
Introduction It is standard to assume the model to be perfect when we use ensemble smoothers for solving inverse problems. This paper addresses the problem of consistently including the additional effect of stochastic model errors in different ensemble smoothers. In particular, we consider methods such as Ensemble Smoo... |
There is a vast literature on solving the data-assimilation problem in the presence of model errors (see, e.g., the reviews by Carrassi and Vannitsem 2016 Harlim 2017 . We traditionally characterize the data-assimilation problem as being either a weak-constraint or a strong-constraint problem, dependent on whether we i... |
In Eknes and Evensen 1997 the variational formulation was solved for a weak-constraint parameter and state estimation problem using the representer method by Bennett 1992 , and in Evensen 2003 different methods including ensemble methods were used to solve a weak-constraint state- and parameter-estimation problem. A co... |
At this point, we should mention that we do not distinguish between model errors and model bias. In fact, with time-correlated stochastic model errors, and if the correlation becomes perfect (equal to one), then the model errors become equivalent to a constant bias. Fortunately, the procedure outlined in this paper can... |
In a recent paper by Sakov et al. 2018 , the Iterative EnKF was reformulated to allow for additive model errors. However, for the history-matching problem, we need to account for more general representations of the error term since the solution of the reservoir simulator depends nonlinearly on the errors in the rate da... |
We will start by defining the standard strong-constraint history-matching problem, and after that, we move on to the general weak-constraint formulation. We then formally derive the ES, ESMDA, and IES, in the presence of model errors. The different smoother methods are used in a simple example to demonstrate the consis... |
Standard history-matching problem The strong-constraint formulation given by Evensen 2018 is attractive because it simplifies the inverse problem and efficient ensemble smoothers can be defined. A first fundamental assumption is that we have a perfect deterministic forward model where the prediction [MATH] only depends... |
[EQUATION] In a reservoir history-matching problem, the model operator is the reservoir simulation model, which predicts the observed production of oil, water, and gas, from the reservoir. Thus, given the true parametrization of [MATH] , the true prediction of [MATH] is precisely determined by the model in Eq. ( ). Als... |
[EQUATION] From evaluating the model operator [MATH] , given a realization of the uncertain model parameters [MATH] , we uniquely determine a realization of predicted measurements [MATH] (corresponding to the real measurements [MATH] ). Here [MATH] is the number of parameters and [MATH] the number of measurements. We w... |
In history matching, it is common to define a prior distribution for the uncertain parameters since we usually will have more degrees of freedom in the parameters, than we have independent information in the measurements. Bayes’ theorem gives the joint posterior pdf for [MATH] and [MATH] as |
[EQUATION] In the case of no model errors, the transition density [MATH] becomes the Dirac delta function, and we can write [EQUATION] |
We are interested in the marginal pdf for [MATH] , which we obtain by integrating Eq. ( ) over [MATH] , giving [EQUATION] When introducing the normal priors |
[EQUATION] we can write Eq. ( ) as [EQUATION] Note that the posterior pdf in Eq. ( ) is non-Gaussian due to the nonlinear model [MATH] Maximizing [MATH] is equivalent to minimizing the cost function |
[EQUATION] Most methods for history matching apply the assumptions of a perfect model and Gaussian priors, and they attempt to solve either one of Eqs. ( ) or ( ). |
For this strong-constraint problem, Evensen 2018 explained how Eqs. ( ) or ( ) can be approximately solved using the ES, ESMDA, and IES. We can interpret these methods to approximately sample the posterior pdf in Eq. ( ), and we can easily derive them as an ensemble of minimizing solutions of the cost function in Eq. (... |
[EQUATION] This approach relates to the papers on Ensemble Randomized Likelihood (EnRML) Kitanidis 1995 Oliver et al. 1996 Note that the minimizing solutions will not precisely sample the posterior non-Gaussian distribution. |
ES solution We obtain the ES solution by first sampling the prior parameters [MATH] and the perturbed measurements [MATH] from [EQUATION] |
We then compute the model predictions [MATH] from [EQUATION] We define the covariance matrix between two vectors [MATH] and [MATH] as the expectation |
[EQUATION] while in the ensemble methods we introduce the sample mean [EQUATION] and the sample covariance between two arbitrary vectors [MATH] and [MATH] as |
[EQUATION] Evensen 2018 derived the ES update equations for the parameters [MATH] and the predictions [MATH] as [EQUATION] but see also the alternative derivation in Secs. 4.3 4.5 below regarding the consistency of using the sample covariance [MATH] in Eq. ( 17 ). To compute the ES solution, we start from an initial en... |
by evaluating the model in Eq. ( 13 ) for each realization [MATH] . Then we compute the prior ensemble covariance between the parameters and the predicted measurements |
[MATH] and the ensemble covariance of the predicted measurements [MATH] and use them in the ES update in Eq. ( 17 ). Finally, we can recompute the model prediction using the updated parameters [MATH] in Eq. ( 18 ). |
As an alternative to solving the model in Eq. ( 18 ) for [MATH] , it is possible to compute the updated prediction directly from an update equation |
[EQUATION] and in the case with a linear model, the result would be identical to solving Eq. ( 18 ). However, due to the nonlinearity of the deterministic model, the Eqs. ( 18 ) and 19 ) will give different results for [MATH] . The Eq. ( 19 ) is just the standard ES update [MATH] given the prior forecast ensemble for [... |
and the perturbed measurements [MATH] of [MATH] . We will later show that, for nonlinear models, there may be a benefit of computing [MATH] indirectly through integration of the model in Eq. ( 18 ), initialized with [MATH] , rather than using the direct update in Eq. ( 19 ). Also, the indirect update in Eq. ( 18 ) allo... |
General weak constraint problem Lets now look at a formulation where we assume that the model depends nonlinearly on the model errors [MATH] as well as the parameters [MATH] , i.e., |
[EQUATION] The model operator can be somewhat general, including a recursion or time steps, and the model error is a vector of noise components that could represent, e.g., the time-correlated noise in rate data used to force a reservoir simulation model over a specific period. Thus, there is a significant difference be... |
We assume that we have prior pdfs for the model errors and the parameters [MATH] and [MATH] . It is common, but not necessary, to assume that the model errors and parameters are independent so we can write the joint prior pdf as [MATH] The transition density for the model evolution is |
[EQUATION] and we obtain the joint pdf by multiplying the prior with the transition density, [EQUATION] The likelihood function for the measurements given the prediction [MATH] is [MATH] thus, the posterior conditional pdf becomes |
[EQUATION] Since [MATH] is given by the model in Eq. ( 20 ) as soon as we know [MATH] and [MATH] , we can compute the marginal density |
[EQUATION] 4.1 Gaussian priors We now assume Gaussian priors [EQUATION] and [EQUATION] where [MATH] would normally be zero. The likelihood for the measurements becomes |
[EQUATION] and we write the marginal posterior as [EQUATION] Maximizing the posterior pdf in Eq. ( 28 ) is equivalent to minimizing the cost function |
[EQUATION] As in the strong constraint case we sample the priors and define a cost function for each sample realization, and we obtain the weak constraint analog to the strong-constraint cost function in Eq. ( 10 ), i.e., |
[EQUATION] 4.2 Stationarity condition To develop a consistent set of equations for the different methods, it is simpler to rewrite the cost function for an augmented variable [MATH] . We can then define the covariance |
[EQUATION] where we allow for correlations between [MATH] and [MATH] since this correlation becomes important in the iterative methods, and we can rewrite the cost function in Eq. ( 30 ) as |
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