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Open frontiers in calibrated and robust learned inversion — Examples and Counterexamples
1 · Prerequisites
None. This page is self-contained.
2 · Logical flowchart
3 · Definitions, theorems and proofs
None yet.
4 · Examples, counterexamples and false statements
Exact Gaussian credible-ball coverage with a score-error budget of zero
Example
Let , , and . Given a scalar truth , observe independent , , and use the proper prior . Write , , , and , where is the standard normal cumulative distribution function. The exact posterior is and its central credible interval is . Its frequentist coverage at truth is
Facts & Assumptions
Given: The normal observation model, the proper prior, the posterior credible interval, , , and from the Example.
The likelihood is proportional to and the prior density is proportional to .
for , and is its quantile.
Verification
Multiplying the likelihood by the prior and completing the square in gives posterior precision and posterior mean . Thus the posterior variance is and the stated interval has posterior mass .
Under the fixed truth , is normal with mean and standard deviation . The event is exactly ; standardizing the two endpoints gives the displayed difference of normal cumulative probabilities.
A proper Gaussian prior gives nonuniform finite-sample coverage on an unbounded truth class
Example
Fix , , and . Observe independent under truth and use prior . Let , , and , where is the standard normal cumulative distribution function. The exact posterior credible interval is . Since , its repeated-sampling coverage at fixed tends to zero as , despite exact posterior computation and exact sampling.
Example references
Facts & Assumptions
Given: The scalar normal observation model and proper prior in the Example, with , , and .
Completing the Gaussian likelihood and prior gives posterior mean , variance , and central credible radius .
The standard normal cumulative distribution has vanishing tails: as and as .
Verification
At every fixed , lies strictly between zero and one, and is finite and independent of the observation. Under truth , has mean and fixed positive standard deviation .
Standardizing the event in the Example gives .
As , both arguments in step 1.2 tend to , so both cumulative probabilities tend to one by [A2] and their difference tends to zero. As , both arguments tend to and their difference again tends to zero. Thus the infimum coverage over is zero at each fixed .
A scalar inverse with a conditioning-dependent shift modulus
Example
Fix and . The working inverse is . Under a test operator with , the observation is for and independent . For any two truth laws on , define as the infimum of over their couplings. The risk is and it satisfies and
Example references
Facts & Assumptions
Given: , , , , independent of , and the working inverse .
The test law and source law are Borel probability measures on ; is the infimum of expected absolute distance over their couplings.
Verification
Substitution gives . Independence and make the cross term vanish, so the expected squared error is . Since and , this is at most .
For any coupling of , . Taking expectations, then the infimum over couplings, gives . Multiplication by and step 1.1 yields the claimed risk-transfer bound.
Small absolute operator error does not give a uniform inverse bound without conditioning
Example
Fix a truth bound . For every absolute operator-shift radius , take the working scalar operator , the true operator , the noiseless datum at truth , and the working inverse . Then but . Thus no recovery modulus tending to zero with the absolute operator error can hold uniformly when the baseline inverse conditioning is allowed to vanish.
Facts & Assumptions
Given: , any , working operator , true operator , truth , noiseless data , and working inverse .
In the scalar model the operator norm is absolute value, and the recovery loss is squared absolute error.
Verification
The absolute operator discrepancy is .
By step 1.1 this is an allowed perturbation. The true noiseless datum is , so . The squared reconstruction error is , independent of . Since is fixed while can be arbitrarily small, a uniform modulus depending only on absolute operator shift cannot tend to zero.
Exact iid samples estimate two-mode posterior weights
Example
Let , , and let the two-mode posterior be on . From independent exact posterior draws , define and . The one-Wasserstein error and the probability that at least one posterior mode is absent from the sample are and .
Example references
Facts & Assumptions
Given: , , , posterior , and independent draws .
is the infimum of over couplings of the two probability measures.
Verification
Put . In any coupling of and , mass at and at can be matched at zero cost; the remaining mass is and must move between the atoms, a distance . This coupling is feasible and no coupling can move less unmatched mass, so it is optimal and gives .
The count is binomial , so and . Cauchy–Schwarz gives ; step 1.1 yields the expected Wasserstein bound. The event of missing has probability , and the event of missing has probability ; for these events are disjoint, giving the displayed miss probability.
No fixed sample budget uniformly protects arbitrarily rare modes
Example
For the two-mode posterior with and , independent exact draws miss the rare mode with probability . For any fixed , this probability tends to one as . If instead with , then the probability of missing either mode is at most , so suffices to make it at most .
Example references
Facts & Assumptions
Given: , a mixture weight , independent draws from , and, for the final bound, with .
Each draw independently equals the rare atom with probability and the other atom with probability .
Verification
The event that no draw equals is the intersection of independent events, each of probability , hence has probability . Its limit as is one for every fixed finite .
The event that at least one of the two atoms is absent is contained in the union of the event “no draw” and the event “no draw.” Their probabilities are and , each at most under the displayed weight interval. The union bound gives at most . If , then , proving the sufficient budget.
A diagonal inverse has an explicit Fourier truncation error
Example
Write , where , so all sequence coordinates below have positive indices. Let , let be an integer, let satisfy , and let . Given data with , the mesh- truncated inverse is with zero output coordinates for . It satisfies the continuum error bound This separates measurement-noise amplification from the mesh truncation tail; it makes no statistical or neural-operator claim.
Example references
Facts & Assumptions
Given: , an integer , , data , and .
retains the first coordinates and sets all later coordinates to zero.
Verification
For , . The norm of the first- noise vector is at most .
The omitted truth tail satisfies . Its norm is at most .
The error vector is the sum of the retained-coordinate noise error and the omitted truth tail. The triangle inequality with steps 1.1 and 1.2 gives .
Finite samples cannot identify every Lipschitz function uniformly
Example
Let and let be arbitrary training inputs. For every deterministic learning rule that returns a prediction function from the observed pairs , there are two -Lipschitz functions agreeing with each other and taking the common value at every training input but satisfying Consequently, at the midpoint of a largest sample-free interval, the returned prediction function has absolute prediction error at least for at least one of these two indistinguishable target functions.
Example references
Facts & Assumptions
Given: , arbitrary sample sites , exact labels , and a deterministic learning rule depending only on these observed pairs.
A function is -Lipschitz when for all .
Verification
Sort the distinct sample sites together with endpoints and . The at most consecutive gaps have total length one, so one open gap has length . Let , , , and . The tent has slopes of magnitude at most one, vanishes at every sample site, and has height at . Thus are both -Lipschitz and yield identical training pairs.
Because the training pairs are identical, the deterministic rule returns the same value for both target functions. Since , one of the two errors is at least . Their uniform distance is , proving both claims.
Integrated score error can miss every Euler evaluation
Example
In the finite-dimensional Gaussian credible-ball question reproduced below, choose , , , , , and , with arbitrary positive . Take the numerical operator equal to the true operator, so . For every integer there is a jointly Borel finite score field with integrated score error exactly zero, whose Euler terminal law is for every observation. With independent terminal draws, the resulting empirical ball has frequentist coverage tending to zero at truth , although the stated uniform boundary-margin condition holds with exponent . The exact credible-ball baseline satisfies . Consequently no bound in that question can have uniformly over its original score class as . The original formulation is retained verbatim under Original question.
Example references
Original question
Fix integers , parameters , a credibility error , and a continuous schedule with . Let Observe , , with independent . Use the proper Gaussian prior , and let be its exact posterior under the stated forward operator and noise law. Let be the posterior mean and let be the unique radius for which the Euclidean ball has posterior probability . Define the exact frequentist coverage and its baseline error .
For a deterministic numerical matrix with and , let be the Gaussian posterior formed with and the same prior and noise variance. Write for the density obtained by applying the forward variance-preserving diffusion to , and set . A jointly Borel learned score field is required to satisfy For , define the reverse-time Euler chain by and, for , where the are independent vectors independent of and the observations. Draw independent copies of this chain conditionally on and denote their terminal values by . Let , let be the empirical quantile of (the smallest order statistic whose empirical cumulative mass is at least ), and put . Suppose the exact credible-ball boundary satisfies the uniform margin condition for fixed and . Determine whether a minimax-sharp nonasymptotic function exists uniformly over all such score fields and, if it does, find it as a function of , and such that with matching lower bounds that identify when . The probability includes observation noise and sampler randomness. The target concerns the additional coverage error introduced by the approximate forward operator, learned score, finite reverse discretization, and finite posterior sample; it does not assume that the exact Bayesian credible ball is already calibrated on every truth class.
Facts & Assumptions
Given: The scalar parameter choices in the Example and the original score-error, Euler and empirical-ball definitions reproduced below it.
Write and . A standard Gaussian has density , mean zero, variance one and tail probability tending to zero at infinity.
Let and let solve for a standard Gaussian . Continuity and strict positivity of give existence and uniqueness of .
Verification
Given: The scalar model and [A1]–[A2].
Completing the square in gives posterior , so its exact credible interval is . Under truth , and the exact coverage is . This is a continuous strictly positive function of , and is at most one. By compactness of it has a positive minimum . Therefore .
Put and let . Away from this finite set of times, use the exact noised-posterior score. At each set . This finite-valued jointly Borel field has zero integrated score error: its discrepancy from the exact score is supported on the finite, Lebesgue-null set in time. The noised posterior has mean and variance : the integrating-factor solution of the linear forward equation is , whose independent Gaussian integral has variance (use dyadic Brownian-increment sums and pass their Gaussian characteristic functions to the mean-square limit). Its exact score is therefore affine in , so the inner Gaussian error integral is finite at every time. The zero-error conclusion holds for every observation and hence uniformly over all truths.
The variable has a density bounded by uniformly in . The event is contained in two intervals of total length at most . Thus its probability is at most . The required margin holds, for example with , and .
At every Euler step the deterministic update becomes . Thus each new state is , independently of its input and of , and the terminal law is . For , write the independent terminal draws as . If , then the empirical center differs from one by at most and every empirical distance to that center is at most . Hence the empirical radius has that same upper bound.
The Gaussian density gives for : integrate on and use . The union bound then gives . On the complementary event the center is at least and the radius at most , so for all sufficiently large the empirical interval excludes truth zero. Its coverage therefore tends to zero, and its deviation from tends to . Since both error norms are zero and , a uniformly vanishing would contradict this strict gap.
Scope of the resolution
This resolves the requested uniformly vanishing-error guarantee negatively under the original score hypothesis. Time-integrated error does not control pointwise values at a deterministic solver grid. A constant worst-case coverage bound remains possible; no impossibility of all numerical functions is asserted. No Bayesian calibration claim is imported from a conformal theorem.
Worst-law ellipsoid recovery has no training-size rate
Example
For the robust minimax value in the original question reproduced below, fix , , , and . Put There is , depending only on , such that These bounds hold for every training size and truth-law shift radius. For fixed positive , along any sequence of the remaining parameters the risk tends to zero if and only if both and . In particular, increasing with fixed positive noise does not force the worst-law risk to zero. The original question, including both distinct shift classes, is retained verbatim under Original question.
Example references
Original question
Fix an integer , parameters , a shift radius , and an operator radius . Let , with norm , and define the compact truth class Let have its product Borel sigma-algebra. A sequence observation with forward operator is , where the are independent variables; this defines a probability law on even though the white-noise sequence need not lie in . Write for the Borel probability laws on a Borel set . For a Borel probability law on , the training data are with independent and . At test time, for another Borel probability law on and the unknown forward operator is , where is bounded linear and . Write for the bounded linear operators from to itself. The learner receives and the test observation , but not .
For probability laws on , define their Wasserstein distance by where is the set of couplings with those marginals. A learner is a Borel map . For fixed training data its test risk is where is an independent standard Gaussian sequence. Define the robust minimax value Determine matching upper and lower bounds for as a function of , and identify necessary and sufficient joint regimes in which it tends to zero. In particular, the bound must distinguish source-to-test truth-law shift from perturbation of the conditional measurement law induced by ; no access to is available at deployment.
Facts & Assumptions
Given: All data spaces, training laws, test laws, learners and shift classes in the original question. Let denote the jth coordinate vector. Let be the standard normal distribution function and put .
A scalar observation with uniform on and independent standard Gaussian has minimum sign-testing error for : the two Gaussian densities cross at zero, and choosing the larger density pointwise minimizes the error integral. Their ratio is .
For vectors in a real Hilbert space, . This identity follows by expansion of the squares.
Verification
Given: The original robust minimax model and [A1]–[A2].
The zero estimator gives risk at most for every allowed law and operator. If , put and . Use the Borel finite-coordinate estimator for and zero otherwise; it ignores the training data. Under any test truth its retained error is and its omitted error is . These components are orthogonal, and the centered noise is independent of the truth. Hence its expected squared error is at most . Because and , the last two terms are at most . If , and the zero estimator already gives the asserted upper bound. Taking the better of the zero and cutoff rules proves the displayed upper bound uniformly over both shift classes.
For the noise lower bound first suppose . Set and . These laws have Wasserstein distance zero and are admissible at every . The training sample is independent of the fresh test sign, and all other test coordinates are independent pure noise. Given those ancillary variables, testing the fresh sign reduces to [A1] with . Thus any prediction of its first coordinate has squared error at least whenever its induced sign test is wrong, and its integrated risk is at least . At a zero predicted coordinate, choose either sign; the squared error is still .
Put . If , define , and take the same source and test law . Compare the two allowed operators and , whose norms are at most . In the first world, the test truth has observation mean ; in the second, truth has the same mean because . The joint law of the full training sample and this conditional test observation is identical in the two cases, including every noisy coordinate. Calling that common law , the corresponding two full test risks satisfy and . Therefore [A2] gives . This holds for every Borel learner and all training sizes. If , the asserted zero operator lower bound is immediate.
If , use from step 1.1 and . On the coordinate block , put and let be the uniform law on the vectors having independent coordinates on this block and zero elsewhere. Each such vector lies in the ellipsoid because . Again choose . For every block coordinate, its test sign is independent of the whole training sample and of all other test coordinates. Its signal amplitude is , by the definition of and . Applying the density-crossing argument in [A1], conditional on all ancillary data, gives sign error at least . Summing the coordinate squared errors gives risk at least . Finite averaging over the sign prior and integration over the independent ancillary observations suffice; no choice of an unknown prior by the learner is being assumed.
Steps 1.2 and 2.1 give with , and step 1.3 gives . The maximum of these bounds is at least , proving the lower bound with that explicit positive . For fixed , the upper bound tends to zero when , and the lower bound forces both limits whenever the risk tends to zero. Both lower constructions have , so they remain admissible even at zero truth-law shift; the upper construction ignores the training law entirely. This explains why and do not enter the matching-order rate for this worst-law formulation.
Scope of the resolution
This gives matching-order upper and lower bounds for the exact original worst-law problem, including simultaneous unknown relative-operator shift and truth-law Wasserstein shift. It neither identifies the exact minimax constant nor says that finite-sample distribution learning is unimportant for a fixed restricted prior family. The source paper's conditional-law DRO bound is adjacent context; the two shift mechanisms and finite sign-prior proofs above are local and do not substitute its finite-dimensional hypotheses.
Ideal-marginal score error allows unbounded finite-Euler Wasserstein error
Example
For the separated-mixture question reproduced below, there are admissible fixed one-dimensional parameters with , fixed positive score-error allowance , and every fixed integer , for which the supremum of the displayed expected fidelity loss over the original finite-valued Borel score class is , for every finite . Each individual field in the construction has a finite-first-moment Euler chain and obeys the required score bound at every time. Thus the original ideal-marginal hypothesis supplies no finite uniform bound for the requested loss, however large the fixed budget is. The original question is retained verbatim under Original question.
Example references
Original question
Fix dimension , component count , noise scale , diffusion horizon , and a continuous schedule . Assume the parameter choices below admit at least one component configuration and mixture weight vector. Consider a posterior density on of the form Here denotes the normal density with mean and covariance matrix . The component means satisfy and for , and every eigenvalue of each positive-definite lies in for fixed . Assume and . For , let be the density of under where is a standard -dimensional Brownian motion, and let be its exact score. Let be a supplied jointly Borel, finite-valued score field satisfying the integrated error bound Consider fields whose Euler chain has finite first moment. For , use the reverse-time Euler convention and with independent of and of one another. Generate independent terminal values and let ; one score evaluation per reverse step gives online budget . Define disjoint mode regions , target masses , and empirical masses . For probability measures on , let be the infimum of over all couplings with the specified marginals. Determine whether the integrated score-error assumption alone yields finite minimax-sharp nonasymptotic upper and lower bounds, uniformly over the stated score-field class, for as a function of . In particular, characterize the compute regimes that guarantee simultaneous control of distributional error and missed-mode probability uniformly over this family. This asks about score-based reverse diffusion with finite computation; it does not assume that the posterior is unimodal or that a good posterior mean certifies its mode weights.
Facts & Assumptions
Given: Choose , with and . Choose , , , , , , , , and . The extraneous positive parameter called in the original formulation may be arbitrary; it is not used by its displayed sampler.
The integrating factor in the scalar forward equation gives . The stochastic integral is the mean-square limit of deterministic dyadic sums of independent Gaussian increments; these sums are centered Gaussian, with variances tending to , and are independent of . The limit is therefore Gaussian with that variance and independent of , as follows by taking limits of its Gaussian characteristic functions. At time , the noised target is the symmetric Gaussian mixture with means and common variance . Its exact score is a posterior-weighted average of and , so and is continuous. These formulas follow directly by differentiating the sum of the two Gaussian densities.
A convolution with a nondegenerate Gaussian has a strictly positive continuous density: continuity follows from the bounded Gaussian derivative, and positivity from its positive kernel. Gaussian densities have finite first moments and exponentially decaying tails.
Verification
Given: The fixed parameters and Gaussian identities in [A1]–[A2].
Use the exact score at all grid times before the last two updates. All those states have finite first moment because each exact-score drift has linear growth by [A1]. Let be the state before the penultimate update, corresponding to ideal time . Its law has a strictly positive continuous density : for it is exactly , and otherwise the last preceding nondegenerate Euler noise gives [A2]. Since is also positive and continuous, the density ratio has positive minimum and finite maximum on .
For every sufficiently small , choose the centered interval with . Such an interval exists by continuity and positivity of that density. Then . Put and add to the score at time . Its squared ideal-marginal error is exactly . Set and . After this update the state is .
Let , and let . On the state lies in , so . The ideal density is a fixed Gaussian mixture. Since the interval width is constant and its center tends to infinity, its Gaussian density formula gives constants , independent of small , such that . For example, once , every point of the interval is at distance at least from either component mean; use the largest fixed variance in its Gaussian exponent and the constant interval length.
At the final evaluation time add to the exact score, where . This second squared ideal-marginal error is also . At all other times use the exact score. These modifications at two distinct times define a jointly Borel finite-valued field with the original supremum-in-time score bound. Every individual field has finite first moments at all Euler steps: it is an affine-growth exact score plus two bounded, finite-amplitude additions.
The first moment of is bounded uniformly as : the added term has expectation at most , while and are fixed and finite. The terminal state is . By [A1] the absolute expectation of the first two terms is uniformly bounded; . Therefore , because and the exponential bound in step 3.1 dominates every power of .
Every coupling of two finite-first-moment probabilities satisfies by the integral triangle inequality. Thus because the target mean is zero. Taking expectations gives , which tends to infinity by step 5.1 for every finite . The other two terms in the requested fidelity loss are nonnegative. Its supremum is consequently infinite despite fixed positive score-error allowance and fixed admissible target parameters.
Scope of the resolution
The negative answer concerns every positive error allowance and every finite J>=2. It does not assert the same obstruction at J=1 or at exactly zero error. The mechanism is mismatch between ideal marginal mass and actual chain mass, not merely missing a mode by iid sampling. No numerical observation or source-specific failure experiment is used as the proof.
Coordinate-free Fourier inverse maps have an odd-mesh risk floor
Example
Consider the complete original Fourier architecture and diagonal inverse question reproduced below. For every odd integer , every , and every admissible bounded parameter class with a measurable empirical minimizer, its risk satisfies The same bound holds for every measurable learner whose realized predictor belongs to the stated coordinate-free Fourier class, whether or not it minimizes empirical risk. On the other hand, the unrestricted risk tends to zero as for fixed , independently of . Thus no growth of can make this original architecture attain the vanishing unrestricted risk along an odd-mesh subsequence. This is a partial resolution: no complete joint bounded-class rate, or corresponding fixed positive even-mesh floor, is asserted. The original question is retained under Original question with explicit positive coordinate indices and empty-class conventions.
Example references
Original question
Let carry normalized Lebesgue measure and fix the real orthonormal Fourier basis Use this basis to identify the real Hilbert space with . Fix integers and parameters , and define The inverse map is . For , let retain the first Fourier coefficients and let append zero coordinates.
Fix a globally Lipschitz nonpolynomial function , called the activation. For , form the scalar input field . A pointwise affine lift , with and , creates channels. For fields with positive integer channel counts , apply an integer number of layers indexed by : Here acts coordinatewise, and are pointwise channel weights, and is a Fourier multiplier meaning , where is computed componentwise, with matrices satisfying and for for some integer cutoff . A pointwise affine output map gives a real scalar field, and the network output is . Let be the union of these maps over architectures whose depth, channel counts, Fourier cutoff, and number of real scalar parameters are each at most the integer , with every trainable parameter bounded in absolute value by . A parameter array used at mesh size is the same array at every other mesh size. For fixed , there are finitely many allowed architecture indices; equip their disjoint union of bounded real parameter boxes with its Borel structure.
Write for the Borel probability laws on a Borel set . For a Borel probability law on , draw independent training truths and observations with all noises independent, and write . A selector is Borel into the finite disjoint union of architecture indices and their bounded parameter boxes, and its realized map minimizes the empirical loss. For each with , assume there is at least one such selector that, from , returns a minimizer of over ; let be the set of all such selectors, and set it to when . At test time, draw an independent and , and define Use the convention , so an empty architecture class has infinite ERM risk. Also define by the same infimum over all Borel predictors from to , with . Determine matching minimax upper and lower rates for jointly in . In particular, characterize the joint growth regimes of and that separate statistical estimation, neural-operator approximation, and continuum truncation error, and determine when the bounded shared-parameter Fourier architecture attains the unrestricted rate .
Facts & Assumptions
Given: The original real Fourier basis , , and all higher complete cosine/sine pairs; the architecture has only constant pointwise channel weights and biases, Fourier multipliers and a scalar pointwise Lipschitz activation. Its parameters may depend on all training data but are fixed during test evaluation.
Translation is an isometry of , sends to for , and preserves each full two-dimensional cosine/sine frequency pair. The first-m-coordinate subspace is translation-invariant when is odd.
A vector with distribution is invariant under every orthogonal matrix: its density is a constant times , and orthogonal changes of variables preserve norm and volume.
by completing the square.
Verification
Given: The original architecture, training and test model, and [A1]–[A3].
Each Fourier multiplier commutes with translation because the kth translated Fourier coefficient is multiplied by the scalar , which commutes with its channel matrix. Every constant channel matrix, constant bias and coordinatewise scalar activation also commutes with translation. By induction through the lift, all layers and output map, the realized scalar output field is translation-equivariant. The layer fields are in : a globally Lipschitz activation has linear growth, the pointwise weights are finite and the finite Fourier multipliers are bounded. For odd , projection onto the first m coefficients commutes with translation by [A1]. Hence the zero-extended output map satisfies , where is the orthogonal translation matrix on the retained coefficients. This holds for every individual parameter array, also after data-dependent selection.
Put , choose and , and let . These truths lie in the ellipsoid because their weighted squared norms are at most . Their forward coefficient vectors are respectively and , related by . Conditional on any training sample, the test truth is still an independent equal-probability draw from these two values. Couple the corresponding test noises as and ; by [A2] both have the required law.
To see unrestricted vanishing without training, put . Output for and zero otherwise. Its risk for every truth is at most . It is a Borel predictor and is allowed in the unrestricted class. This proves the contrast with the positive odd-mesh architecture floor.
For fixed training data, let be its selected predictor. By step 1.1, . Translating the first output and truth preserves their norm. Consequently the conditional average of the two squared risks is , which is at least by [A3]. This lower bound is independent of the training sample. Integrating over it, taking the worst law and then the infimum over selectors proves the stated floor.
The mismatch arises because the diagonal inverse rescales the first cosine and sine by different factors and , while the coordinate-free architecture must respect translation between them. It is independent of finite training or parameter-size limitations. The proof deliberately uses odd so both the input noise law and retained output subspace are translation-invariant; it supplies no identical assertion for even , whose last frequency pair is incomplete.
Scope of the resolution
The odd-mesh consistency subquestion is answered negatively, and the unrestricted vanishing comparison is proved. The full requested joint N,m,S,B minimax rate is not resolved here; in particular even meshes break the complete-pair symmetry used by the lower bound, and different activations/parameter budgets may have distinct approximation behavior. This is not the EIT question: the generalized EIT architecture has coordinate/chart lifting, so this coordinate-free translation obstruction is not transferred to it.
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