advanced36 min studyLesson 714 in Mathematics

Orthogonal series estimation

Translate function estimation into noisy coefficient estimation and separate exact Parseval risk identities from approximate regression analogies.

On this page

What you’ll understand

  • Calculate the exact risk of a coefficient cutoff in a Gaussian sequence model.
  • Distinguish coefficient bias from observation noise.
  • Recognize when orthogonal-series density estimates need positivity correction.

A function can be estimated by learning its coordinates in an orthonormal basis. Keeping a few coefficients gives a simple approximation, while retaining more coefficients resolves finer features and admits more noise. Orthogonality makes this tradeoff particularly transparent under squared integrated loss. It also creates an important modeling boundary: independent Gaussian coefficient observations are exact in the Gaussian sequence experiment, but they are generally only an approximation to coefficients calculated from a finite regression sample or a density sample.

Choose the function space before choosing coefficients

Let {ϕj}j≥1\{\phi_j\}_{j\ge1} be a complete orthonormal basis of a specified real Hilbert space, such as L2[0,1]L^2[0,1] with Lebesgue measure. A square-integrable target f has coefficients θj=⟨f,ϕj⟩\theta_j=\langle f,\phi_j\rangle, and Parseval identifies its squared norm with the sum of squared coefficients. Completeness matters: orthogonality alone does not guarantee that the expansion represents every target. The measure also matters, since changing from Lebesgue measure to a predictor distribution changes the meaning of the projection and loss.

In the Gaussian sequence model, observe each coefficient plus independent centered Gaussian noise of known variance sigma squared divided by n. Here n is an information-level parameter, not the number of coefficient coordinates retained. This experiment is directly connected to Gaussian white-noise observation of a function. Moving from ordinary regression to that experiment requires a design argument or an asymptotic equivalence theorem with its own smoothness and noise conditions; a visually similar basis expansion does not establish equivalence.

Zj=θj+σnξj,ξj∼iidN(0,1),f^m=∑j=1mZjϕjZ_j=\theta_j+\frac{\sigma}{\sqrt n}\xi_j,\quad\xi_j\stackrel{\mathrm{iid}}\sim N(0,1),\qquad \widehat f_m=\sum_{j=1}^mZ_j\phi_j
The independent Gaussian coefficient model is the exact experiment in this lesson; m is a deterministic cutoff.↔ Scroll wide equations horizontally.

Parseval separates the risk without cross terms

For a fixed cutoff m, retained coordinates contribute noise variance and omitted coordinates contribute squared approximation error. Orthogonality removes cross terms in the function norm, giving an exact risk of mσ2/n+∑j>mθj2m\sigma^2/n+\sum_{j>m}\theta_j^2. The cutoff estimator is unbiased for the projected function, but generally biased for the full function. Even a perfectly calibrated confidence set for the retained coordinates can miss f if it ignores the omitted tail.

Smoothness becomes a bound on coefficient decay or on a weighted sum of coefficient squares. Such a bound controls the tail, but it does not identify every individual coefficient. Hard cutoff is only one strategy: shrinkage, thresholding and block procedures use different amounts of information about weak coordinates. If the cutoff is chosen from the same observations, the fixed-cutoff risk identity cannot simply be evaluated at the random cutoff and declared an unbiased risk formula. Selection couples the retained noise with the chosen model size.

Eθ∥f^m−f∥22=mσ2n+∑j>mθj2\mathbb E_\theta\|\widehat f_m-f\|_2^2=\frac{m\sigma^2}{n}+\sum_{j>m}\theta_j^2
This identity is exact for a deterministic cutoff under the stated coefficient-noise model. The first term is estimation variance and the second is approximation bias squared.↔ Scroll wide equations horizontally.

Other sampling experiments change coefficient uncertainty

For iid density observations, the average of a basis function evaluated at the observations is an unbiased estimate of its density coefficient. Its variance is the variance of that basis function under the unknown density divided by sample size, and distinct estimated coefficients need not be independent. A finite series can integrate correctly yet become negative. Taking a positive part and renormalizing produces a density when its positive integral is finite and nonzero, but changes both coefficients and the exact risk calculation.

For regression with irregular predictors, the empirical basis Gram matrix can differ substantially from the population identity. Weighted least squares or a basis orthogonal with respect to the design may be useful, but the design still determines what is well observed. Statistical regularization and deterministic Fourier convergence are related through approximation error, yet one does not replace the other. State the observation experiment, coefficient covariance and loss before importing a coefficient-risk formula into a function-fitting algorithm.

A small coefficient is cheaper to omit

In the exact Gaussian sequence model let n=100, sigma=1, and coefficients theta1=1, theta2=0.4, theta3=0.05, with all later coefficients zero. Compare deterministic cutoffs m=1,2,3.

  1. At m=1 the retained variance is 0.01 and the omitted squared coefficients sum to 0.16+0.0025. The risk is 0.1725.

  2. At m=2 the variance is 0.02 and the remaining squared bias is 0.0025, giving risk 0.0225. At m=3 the bias is zero but the variance is 0.03.

  3. The best of these three cutoffs is two. Including the third coordinate spends noise variance 0.01 to remove signal energy only 0.0025, increasing expected squared error despite reducing deterministic approximation error.

The result

Resolving every nonzero feature is not automatically optimal for statistical squared error; its energy must be compared with its observation noise.

Explore the model

Interactive lab

Interactive lab · spectral

Inspect the coordinates a cutoff retains

Pause and Reset, then set both controls to mode count N=3 and coefficient decay p=2. The finite sine series retains amplitudes 1,1/4,1/9. Reset and set N=6 and p=2; predict added amplitudes 1/16,1/25,1/36. Keep time zero for a spatial comparison. These coefficients are deterministic; the instrument contains no Gaussian observation noise and does not evaluate statistical risk.

Inspect the coordinates a cutoff retainsA 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. Current readings: Sum at x = π/2: 0.867; Smallest wavelength: 2π / 5; Highest-mode coefficient: 0.2. All parameters are adjustable with the labeled controls below.01.573.144.716.28−2.56−1.2801.282.56x · radiansFinite sine sum
Finite sumHighest included mode
Sum at x = π/2
0.867
Smallest wavelength
2π / 5
Highest-mode coefficient
0.2
t = 0 s

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 forIncreasing a cutoff reduces deterministic truncation error, while a statistical estimator also pays for noisy added coordinates.

Check your understanding

For a deterministic cutoff under the exact independent Gaussian sequence model, what is its squared L² risk?

Try it yourself

Take theta_j=j^(−2), sigma=1 and n=256. Use an integral bound on the omitted tail to bound the risk at m=4. Minimize the corresponding continuous upper bound in m.

Take a moment to make a prediction or work it out on paper. Open a hint when you need one.

Give me a hint

For decreasing x^(−4), the tail sum after m is at most the integral from m to infinity. Differentiate m/n+1/(3m³).

Compare with the solution

The tail is at most 1/(3m³), so at m=4 the risk is at most 4/256+1/192=1/48, approximately 0.0208333. The derivative of the upper bound is 1/n−1/m⁴, which vanishes at m=n^(1/4)=4. Balancing these two bounds gives order n^(−3/4) for this coefficient-tail example; it is an upper-bound calculation for this sequence, not a universal minimax claim for every smoothness class.

Take it with you

  • The basis, measure and observation experiment determine the coefficient problem.
  • A fixed cutoff has an exact variance-plus-tail risk in Gaussian sequence observations.
  • Data-dependent selection and density positivity corrections change that exact calculation.

Read further

Original sources and deeper explanations to follow at your own pace.

Optional progress, saved in this browser when available. Revisit any lesson at any time.

Lesson not marked complete.