Series

Regularisation — the series

12 essays on one idea, from the one that introduces it to the one that assumes the rest.
  1. 081624324048566400.250.50.751index kfilter factor fₖno regularisation: fₖ = 1truncationTikhonovthe same sum, three weightsTikhonov, relative error0.11truncation, relative error0.11no filter at all5.5·10⁸both filters are one expression with a different weightfₖ = 1 is the catastrophe

    When the answer is a choice

    A backward-stable least-squares solve of this problem returns an answer whose relative error is 5.5·10⁸. Nothing went wrong. The singular values decay exponentially with no gap anywhere in them, the data does not determine the answer, and something outside the data has to choose — which is the computation rather than a preliminary to it.

    part 1 · regularisation
  2. 081624324048566410⁻¹⁷10⁻¹⁴10⁻¹¹10⁻⁸10⁻⁵10⁻²10¹index kmagnitudethe floor: k = 32best truncation: k = 28σₖ|uₖᵀb| exact|uₖᵀb| with noisetwo different indicesthe crossing, from the data alone32the truncation that is actually best28relative error there0.11the exact coefficients never flattenthe noisy ones stop at ‖e‖/√n

    Where the answer stops being in the data

    The Picard condition finds the index where a noisy right-hand side stops carrying signal, from the data alone, with no knowledge of the answer. It lands at 32 where the truncation that actually minimises the error is 28 — and at 45 where the best is 38. It overshoots at every stop from 10% noise to 0.0001%, and it overshoots for a reason. The best truncation walks up the spectrum in a straight line, six or seven indices a decade; the crossing climbs in jumps of 11, 0, 8, 5 and 1.

    part 2 · regularisation
  3. 192123252729313335373941434500.10.20.30.40.50.60.7position jweight on the true signal at jthe blur's row, 5.89 widethe kernel, 2.82 widecomputed from A and λ alonekernel width at half height2.8components kept, Σfₖ28width × Σfₖ / n1.2deepest negative lobe-0.075no data and no truth went into this curvethe answer is the truth seen through it

    A second blur, narrower than the first

    A regularised answer is not the truth with the noise taken out. It is the truth seen through a second blur, V F Vᵀ, which depends on the operator and λ and on nothing that was measured. At the best λ for 0.1% noise its rows are 2.82 points wide against the instrument's 5.89, they dip to −0.075 on either side, and their width times the number of components kept stays between 1.10n and 1.27n across seven decades of λ. Two spikes four points apart come back as two; three apart, as one.

    part 3 · regularisation
  4. 081624324048566410⁻¹⁷10⁻¹⁴10⁻¹¹10⁻⁸10⁻⁵10⁻²10¹index kmagnitudethe floor: k = 43best truncation: k = 31σₖ|uₖᵀb| exact|uₖᵀb| with noise|uₖᵀe|, smoothedcorrelation ρ = 0.9noise energy, first 28 directions0.98tilt of the noise floor, decades1.1the crossing43the best truncation31the floor tilts, and is still a floorρ = 0.9

    Noise that spares the answer and fools the rules

    Make each noise sample remember the last one, keep its size fixed, and the best answer available gets slightly better — 0.1056 to 0.1010 — because slow noise hides in the directions where dividing by σ costs nothing. The Picard crossing still lands two dozen indices past the best truncation. What breaks is the rules. Generalised cross-validation more than doubles the best error on 14 draws of 48 instead of 3, the discrepancy principle's typical cost triples, and the two miss in opposite directions. Whitening by the covariance takes GCV back to 3.

    part 4 · regularisation
  5. 1216202428323640444810⁻¹110¹10²grid points nrelative error against the continuous signalbest truncation, 0.117best grid: n = 26best K = 28with noisenoise-freeno λ anywhere in this curvebest grid, error0.13best truncation, error0.12κ on the best grid61grid where it has doubled34every point is an unregularised solvethe grid chose the truncation

    The grid was the first filter

    A continuous deconvolution discretised on n points and solved with no regularisation at all is not unregularised. Its error against the continuous signal is least at 24, 26 and 34 points for noise of 1%, 0.1% and 0.01% per sample — beside best truncations of 24, 28 and 32 components on a 64-point grid — and within 4 to 16 per cent of their error. The grid's own filter factors sum to n exactly and fall through a half at k = n. Choosing the grid was choosing a truncation, before anybody chose a λ.

    part 5 · regularisation
  6. 122028364452606876849210010⁻¹110¹grid points nrelative error against the continuous signalbest with no λ: n = 26no λeach grid's best λTikhonov, 64 points0.1% noise, 16 drawsbest grid with no λ26its error0.1396 points with its λ0.12best λ on fine grids0.0056each point is a median over the same drawsλ belongs to the problem, not to the grid

    Where the grid hands over to λ

    An unregularised solve on a coarse grid comes within a tenth of the best Tikhonov answer on a fine one, and the pair of a grid and a λ was left unmeasured. Measured, the two do not trade. On every grid up to the best unregularised one no λ helps at all. On every grid of 40 points and more the best λ is the same to within a quarter of a decade — 3.2·10⁻² at 1% noise per sample, 10⁻³ at 0.01% — and the 96-point grid with it beats the best coarse grid by 7, 9 and 13 per cent. The grids between the two, given their own λ, land between them.

    part 6 · regularisation
  7. 10141822263034384210⁻¹10⁻⁰.⁵110⁰.⁵grid points nrelative error against the continuous signalsampled kernel, hatsintegrated, hatsintegrated, spline0.1% noise, no λsampled kernel, hats, best0.13integrated, hats, best0.12integrated, spline, best0.12truncation of 64 points0.12same nodal unknowns, same databetter at representing the noise too

    A better discretisation is a weaker filter

    A coarse grid's error has two sources — how well the discrete operator approximates the integral, and how well the grid's function represents the answer — and the grid essay could not separate them. Changed one at a time they separate: integrating the kernel against the hat functions takes a fifth off the 12-point error, reading the answer as a cubic spline takes 15 per cent more, and both roughly double the condition number on every grid. At 0.1% noise the spline discretisation's unregularised solve on 26 points reaches the best truncation of a 64-point grid to 0.3%. At 1% it is worse than the crude grid.

    part 7 · regularisation
  8. 1220283644526068768492100101214161820grid points nrelative error against the continuous signal, per centsampled kernel, hatsintegrated, hatsintegrated, splinespline, no λ1% noise per sample, median of sixteen drawssampled kernel, hats: 16 points0.18integrated, hats: 16 points0.16integrated, spline: 16 points0.16sampled kernel, hats: 96 points0.14integrated, hats: 96 points0.14integrated, spline: 96 points0.14spline, no λ, its best grid0.15dotted: the 96-point answerevery point is a grid with its own λ

    The grid on which the discretisation stops mattering

    Without regularisation, integrating the blur's kernel against cubic splines beat sampling it at 0.1% noise and lost to it at 1%. Give each discretisation its own best λ on every grid and the difference shrinks to nothing where grids are fine — 0.17, 0.28 and 0.20 per cent apart on 96 points at the three noise levels, with every discretisation choosing the same λ — and stays at 17 to 18 per cent on 16 points. The choice between them is a choice of how coarse a grid can be: at 0.1% noise the integrated discretisations reach the fine-grid answer on 26 points and the sampled one needs 40.

    part 8 · regularisation
  9. on 96 pointsno breakpoint0.12two jumps0.05doubled knots0.13one box0.013the box on fewer points16 points0.0526 points0.02248 points0.00710203040506070809010010⁻²10⁻¹.⁵10⁻¹grid pointsrelative error, best λno breakpointtwo jumpsdoubled knotsone boxevery reading spends the same n unknownsand is given its own best λ on each grid

    A corner the penalty can afford

    Every smooth reading of the deconvolution's grid needed about forty points and then stopped improving, and the step was the suspect. Give the step one coefficient of its own and forty-eight points reach an error of 0.0070 at 0.1% noise, against 0.118 for the best smooth reading on ninety-six — the step was most of the error. But the same step given two coefficients recovers half as well, and given a doubled node at each edge it recovers worse than no breakpoint at all, while representing the signal to 0.07%. What decides is what the penalty is charged for the corner, and whether the data can say where it is.

    part 9 · regularisation
  10. worst draw on any grid, % over the oraclediscrepancy13GCV4·10⁶L-curve291median on 96 points, % overdiscrepancy3.4GCV0.18L-curve282030405060708090100110¹grid pointserror ÷ the oracle's, same drawdiscrepancy, mediandiscrepancy, worstGCV, medianGCV, worstL-curve, mediansolid: the median draw; dashed: the worst of sixteenthe axis stops at twenty times the oracle

    The data count their dimensions, not the step's

    Every grid in the deconvolution essays was chosen with the answer in hand, and so was every λ. From the data alone, the discrepancy principle's worst draw is within 16 per cent of the oracle on every grid from 16 points to 96; generalised cross-validation is better on the median draw and, on grids of thirty points and more, has draws thousands of times worse. And the data can say how many dimensions they carry — about 20, 25 and 29 at three noise levels, one number once the grid exceeds it — but not how many more the step needs: the grid that count chooses is 14 to 19 per cent worse than forty points at the lower two.

    part 10 · regularisation
  11. worst draw on any grid, ÷ the oracleGCV4·10⁴rightmost minimum1.6discrepancy1.1draws more than twice the oracleGCV10rightmost minimum02030405060708090100110¹10²10³10⁴grid pointsworst error ÷ the oracle'sGCV, worst drawrightmost minimumdiscrepancythe worst of sixteen draws on each gridthe axis stops at ten thousand times the oracle

    The minimum on the right

    Generalised cross-validation's worst draws on a fine grid were all one mistake: a second dip in its function at λ near zero, deeper than the real minimum. The proposed repair was a residual threshold, one number, refusing any λ whose residual falls too far below the real minimum's. Measured over 528 draws, a threshold of one half still lets two hundredfold misses through; only the extreme value, which is no threshold at all but the rule 'take the rightmost local minimum', removes all eleven. It costs nothing on the coarse grids where the dip is the right answer, and on the collection's own problem over five thousand draws it turns 243 tenfold misses into 17.

    part 11 · regularisation
  12. square systemfloor minima21interior dips28four samples per unknownfloor minima0interior dips4051015202530samples per unknowndraws of 24011.251.524minimum at the floorinterior dipGCV over 10× the oracleover 100×horizontal axis doubles at each gridlinethe floor goes at once and the dip does not

    More samples take the floor and leave the dip

    GCV's catastrophic misses on fine grids were blamed on squareness: on an n × n system the residual and n − t both reach zero as λ does, and their ratio can dip there. With more samples than unknowns neither reaches zero, and the prediction was that the dip would be gone by construction. Half of it is. The minimum at the floor of the scale, 21 draws in 240 on square systems, is gone at every ratio. The interior dip is not — 28, 20, 13, 8 and 4 draws at one to four samples per unknown — and at four per unknown one draw still misses the oracle by 877 times.

    part 12 · regularisation

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