Minimax rates over smoothness classes
Match a constructive bias–variance bound with a testing lower-bound argument, while keeping experiment, dimension and loss explicit.
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The question for today
What you’ll understand
- Define minimax squared error over a fixed Sobolev ellipsoid.
- Balance cutoff variance against a smoothness-controlled tail.
- Explain why a lower bound needs statistically indistinguishable alternatives.
A bandwidth calculation suggests how quickly one estimator might improve, but it does not prove that a faster method is impossible. Minimax analysis asks for the smallest worst-case risk among all estimators over a specified function class. An upper bound requires constructing an estimator; a lower bound requires showing that the data cannot reliably distinguish enough alternatives. The answer changes with the observation experiment, smoothness, dimension and loss, so a bare convergence exponent without those choices is not a complete statistical claim.
Fix the experiment and the loss
Use the infinite Gaussian sequence experiment with independent standard normal noises. For fixed smoothness s>0 and radius L>0, take the Sobolev ellipsoid . Under squared sequence norm loss, define minimax risk as the infimum over measurable estimators of the largest expected squared error on this ellipsoid. Through an orthonormal expansion, this is a squared integrated function loss in the corresponding model.
The class is fixed before seeing the data. It limits the total high-frequency energy rather than forbidding every rough-looking sample realization. The infimum ranges over all estimators, including nonlinear ones, not just cutoffs or linear smoothers. A worst-case guarantee can be slower than the risk at a particularly simple signal. Conversely, excellent behavior at one chosen function does not prove minimax optimality. The radius and noise normalization affect constants even when the sample-size exponent stays the same.
A cutoff constructs the upper rate
For a deterministic cutoff at m, the variance is m/n. Every omitted index exceeds m, so the ellipsoid condition bounds the omitted squared energy by L squared times m to the minus twice s. Balancing these terms gives m of order n to the reciprocal of twice s plus one, with fixed radius absorbed into constants. The resulting squared-risk order is n to the minus twice s divided by twice s plus one. This is an upper bound before a lower-bound argument is supplied.
In a d-dimensional periodic function problem, the number of modes below a frequency scale grows like that scale to the power d. The corresponding isotropic Sobolev rate becomes n to the minus twice s divided by twice s plus d for squared integrated error in the analogous Gaussian experiment. This dimension effect is not removed by renaming the coefficients. Supremum loss has a different analysis and often logarithmic factors; root integrated error is the square root of the squared-risk rate. Keeping these losses separate prevents a factor-of-two error in the exponent.
Testing explains why the upper exponent is unavoidable
A lower-bound construction puts small independent sign choices into roughly m coordinates. Choose each coordinate amplitude of order one over the square root of n. Neighboring alternatives then differ by a signal comparable to the coordinate noise, keeping their Gaussian likelihoods statistically close. The ellipsoid constraint allows this many alternatives when m to the power twice s plus one is of order n. A hypercube testing argument, such as Assouad's lemma, forces an accumulated squared error of order m/n.
This is a proof strategy, not a complete derivation of sharp constants. It explains why one two-point alternative may miss the full function-estimation difficulty: many uncertain coordinates contribute simultaneously. Rate optimality also does not imply optimal constants, honest confidence bands, or adaptation to unknown s. Those are additional goals with additional restrictions. Before invoking a minimax theorem for regression or density data, check that its observation design, smoothness class, boundedness conditions and chosen loss actually match the problem at hand.
Work it through
Check a rate with an explicit upper bound
In the one-dimensional Gaussian sequence model take s=1, L=1 and n=32768. Use m=32 in the cutoff bound. Repeat with eight times as much information and compare squared and root errors.
At n=32768 and m=32, the variance bound is 32/32768=1/1024. The tail bound is 32^(−2)=1/1024. Their sum is 1/512, approximately 0.001953125.
At n=262144 choose m=64. Both variance and tail bounds are 1/4096, giving total 1/2048. The total upper bound is one-quarter of the earlier one.
The square root of that upper bound is halved. This agrees with squared-risk exponent 2/3: multiplying n by eight multiplies n^(−2/3) by one-quarter, while the root-error order changes by one-half. These are risk bounds, not realized errors for a particular sample.
The explicit cutoff calculation matches the one-dimensional s=1 minimax exponent while keeping squared loss distinct from root loss.
The living notebook
Explore the model
Interactive lab · spectral
More coordinates mean a larger estimation bill
Pause and Reset, then set both controls to mode count N=4 and coefficient decay p=1. Reset and set N=8 and p=1. At time zero, predict four additional deterministic sine coordinates with amplitudes 1/5 through 1/8. In the separate Gaussian sequence calculation, retaining eight rather than four coordinates doubles the variance term m/n. The plot itself contains no noise and neither proves the lower bound nor displays minimax risk.
- Sum at x = π/2
- 0.867
- Smallest wavelength
- 2π / 5
- Highest-mode coefficient
- 0.2
A finite sum of sin(nx − t)/nᵖ. Playback changes t; increasing the cutoff includes finer spatial scales. This is not a proof about a continuum limit or renormalization. Parameter changes pause playback and recompute the model at the current elapsed time. Reset restores both parameters and returns time to zero. Playback starts paused. Adjust either parameter, then choose Play or Step.
Look forApproximation improves with more retained modes, while estimating those modes consumes a growing statistical budget.
Pause & reason
Check your understanding
Your turn
Try it yourself
For fixed smoothness s=1, compare the normalized squared-risk rate expressions at n=4096 in dimension one and dimension two. State what the numerical comparison does and does not establish.
Take a moment to make a prediction or work it out on paper. Open a hint when you need one.
Give me a hint
Use n^(−2/(2+d)) for squared integrated loss. These expressions omit theorem-dependent constants.
Compare with the solution
For d=1 the expression is 4096^(−2/3)=1/256. For d=2 it is 4096^(−1/2)=1/64, four times larger. This compares rate expressions and demonstrates the dimensional exponent cost. It does not say the actual minimax risks equal those numbers, since radius, noise, domain, class conventions and constants remain unspecified. A pointwise or supremum-loss problem would need its own theorem.
Keep the idea
Take it with you
- A minimax statement specifies experiment, parameter class and loss.
- Cutoff variance and smoothness tails yield a constructive upper rate.
- Testing many close alternatives supplies the lower rate and exposes dimension costs.
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