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Series and Wavelet Estimation · Tutorial 459 of 989

Global Truncation Versus Spatial Thresholding

Learn how global cutoffs and spatially localized wavelet thresholds trade approximation error against estimation variance in different signal structures.

Advanced 12 min read

What You'll Learn

  • Derive the exact risk of global series truncation with the actual coefficient counts at each level.
  • Balance truncation bias and variance for coefficient energy that decays across scales.
  • Compute Gaussian hard-thresholding risk and identify the conditions for its exact formula.
  • Compare risks for a localized coefficient and for energy spread across a wavelet level.
  • Recognize why neither global truncation nor spatial thresholding uniformly dominates.

Two Ways to Control Complexity

A series estimator can reduce noise by discarding fine-scale terms. A global truncation rule makes one decision for an entire scale: keep every coefficient through level \(J\), and discard every coefficient above it. Wavelet thresholding instead makes a decision coefficient by coefficient. Because wavelets are localized in space, this allows an estimator to keep a prominent feature in one region while suppressing small coefficients elsewhere, even at the same scale.

The distinction is about how estimation complexity is controlled, not about whether one method is always preferable. When signal energy is distributed smoothly across locations and scales, a common cutoff can be effective. When a function has a small number of localized features, a global cutoff may retain many noisy coefficients just to preserve those features. But thresholding can also discard genuine small coefficients or retain noise, so its performance depends on the signal and the threshold.

Definition: In an orthonormal wavelet basis, write the target coefficients as \(\theta_{jk}\), where \(j\) indexes scale and \(k\) indexes location. In the Gaussian sequence model, the observed coefficients are \(Y_{jk}=\theta_{jk}+\tau Z_{jk}\), with independent \(Z_{jk}\sim N(0,1)\) and \(\tau>0\). Global truncation at level \(J\) estimates coefficients by \(Y_{jk}\) for \(j\leq J\) and by zero for \(j>J\). Hard thresholding at \(\lambda\geq0\) estimates each observed coefficient by \(Y_{jk}\mathbf{1}\{|Y_{jk}|>\lambda\}\).

For clarity, take levels \(j=0,1,\ldots\), and let \(N_j\) be the actual number of coefficients at level \(j\). In an ideal periodic basis, one often has \(N_j=2^j\); for other bases, including boundary-corrected bases, use the actual counts. Assume the basis is orthonormal and the target has finite squared norm, so \(\sum_{j,k}\theta_{jk}^2<\infty\). Risk below is integrated squared error, which equals squared coefficient error by orthonormality.

Risk of a Global Cutoff

A cutoff incurs variance on every retained coefficient and squared bias on every omitted coefficient. Thus its risk depends on the total energy at each scale, not on where within a scale that energy lies.

Result: In the Gaussian sequence model, the risk of global truncation at level \(J\) is \(R(J)=\tau^2\sum_{j\leq J}N_j+\sum_{j>J}\sum_k\theta_{jk}^2\).

Proof. For \(j\leq J\), the coefficient error is \(Y_{jk}-\theta_{jk}=\tau Z_{jk}\), with expected square \(\tau^2\). For \(j>J\), the estimate is zero, so the squared error is \(\theta_{jk}^2\). Parseval's identity gives the squared function error as the sum of these squared coefficient errors. Taking expectations and summing gives the displayed expression. The sums are well-defined: the retained sum is finite for finite \(J\), and the omitted sum is finite because the target is square-integrable. \(\square\)

This identity gives the basic bias–variance tradeoff. For example, suppose \(N_j\leq C_N2^j\) and the coefficient energy \(S_j=\sum_k\theta_{jk}^2\) satisfies \(S_j\leq C_\theta 2^{-2sj}\) for some \(s>0\). Then \(\sum_{j\leq J}N_j\leq 2C_N2^J\), and \(\sum_{j>J}S_j\leq C_\theta 2^{-2s(J+1)}/(1-2^{-2s})\). Consequently, for constants \(A,B\) independent of \(J\), \(R(J)\leq A\tau^2 2^J+B2^{-2sJ}\). The two terms are balanced when \(2^J\) is of order \(\tau^{-2/(2s+1)}\), giving risk of order \(\tau^{4s/(2s+1)}\). If \(\tau^2=\sigma^2/n\), this is order \(n^{-2s/(2s+1)}\), up to constants. This calculation explains why a global cutoff can work well when energy declines regularly with scale.

Worked Example: Choosing a Cutoff for Smoothly Decaying Energy

Suppose \(N_j=2^j\), \(\tau^2=0.01\), and the target's energy at level \(j\) is \(S_j=4^{-j}\) for every \(j\geq0\). Calculate the risks for cutoffs \(J=1,2,3\), including the energy beyond the cutoff.

For \(J=1\), the retained coefficient count is \(1+2=3\), so the variance is \(3(0.01)=0.03\). The omitted energy is \(\sum_{j=2}^{\infty}4^{-j}=(1/16)/(1-1/4)=1/12\). Thus \(R(1)=0.03+1/12\approx0.1133\).

For \(J=2\), the retained count is \(1+2+4=7\), giving variance \(0.07\). The omitted energy is \(\sum_{j=3}^{\infty}4^{-j}=(1/64)/(1-1/4)=1/48\). Thus \(R(2)=0.07+1/48\approx0.0908\).

For \(J=3\), the retained count is \(15\), giving variance \(0.15\), while the omitted energy is \(\sum_{j=4}^{\infty}4^{-j}=(1/256)/(1-1/4)=1/192\). Thus \(R(3)=0.15+1/192\approx0.1552\). Among these three cutoffs, \(J=2\) has the smallest risk: keeping more levels after that point adds more variance than the remaining tail energy it removes.

Spatial Thresholding and Its Risk

A wavelet coefficient is attached to both a scale and a location. Thresholding uses the observed magnitude at that particular location: a large coefficient is kept, while a small one is set to zero. It therefore does not impose the all-or-nothing scale decision made by a global cutoff. Its risk is nonlinear in the signal coefficient, however, and is not simply a fixed variance term plus a fixed tail bias.

Result: Let \(Y=\theta+\tau Z\), where \(Z\sim N(0,1)\), \(\tau>0\), and \(\lambda\geq0\). For \(\delta_\lambda(Y)=Y\mathbf{1}\{|Y|>\lambda\}\), define \(a=(\lambda-\theta)/\tau\) and \(b=(-\lambda-\theta)/\tau\). If \(\phi\) and \(\Phi\) are the standard normal density and distribution function, then \(E_\theta[(\delta_\lambda(Y)-\theta)^2]=\tau^2\{a\phi(a)+1-\Phi(a)+\Phi(b)-b\phi(b)\}+\theta^2\{\Phi(a)-\Phi(b)\}\).

Derivation. The estimator retains \(Y\) on the events \(Z>a\) and \(Z<b\). Since \(\lambda\geq0\), we have \(b\leq a\), so these two events are disjoint, and the coefficient is discarded when \(b\leq Z\leq a\). On either retained event, the estimation error is \(Y-\theta=\tau Z\); on the discarded event, it is \(-\theta\). Therefore the risk is \(\tau^2\{E[Z^2\mathbf{1}(Z>a)]+E[Z^2\mathbf{1}(Z<b)]\}+\theta^2\{\Phi(a)-\Phi(b)\}\). Integration using \(\phi'(z)=-z\phi(z)\) gives \(\int_a^\infty z^2\phi(z)\,dz=a\phi(a)+1-\Phi(a)\) and \(\int_{-\infty}^b z^2\phi(z)\,dz=\Phi(b)-b\phi(b)\), proving the formula. \(\square\)

For a finite collection of orthonormal coefficients, the total thresholding risk is the sum of these single-coefficient risks, plus the squared energy of any unobserved or deliberately omitted coefficients. This sum makes spatial distribution consequential: coefficients at the same level can have very different individual risks under thresholding, even though global truncation treats them identically.

Worked Example: A Single Localized Feature

Consider levels \(0,1,2\), with counts \(1,2,4\). Let \(\tau=0.2\), and suppose exactly one coefficient at level 2 equals \(1\), while all other coefficients are zero. Compare global cutoffs \(J=1\) and \(J=2\) with hard thresholding at \(\lambda=0.4\).

For \(J=1\), the variance from the three retained zero coefficients is \(3(0.04)=0.12\). The level-2 signal coefficient is omitted, contributing squared bias \(1\). Hence \(R(1)=1.12\). For \(J=2\), all seven coefficients are retained, so the risk is \(7(0.04)=0.28\).

For thresholding, the signal coefficient has \(a=(0.4-1)/0.2=-3\) and \(b=(-0.4-1)/0.2=-7\). The exact risk formula gives approximately \(0.040764\). For each zero coefficient, \(a=2\) and \(b=-2\), so its risk is \(0.04\{2\phi(2)+1-\Phi(2)+\Phi(-2)+2\phi(2)\}\approx0.010458\). There are six zero coefficients, contributing approximately \(0.06275\). Total thresholding risk is therefore about \(0.040764+0.06275=0.10351\), smaller than either global-cutoff risk in this example. It keeps the isolated feature without paying variance for every coefficient at its scale.

Worked Example: The Same Energy Spread Across a Scale

Keep the same seven coefficients, noise level \(\tau=0.2\), and threshold \(\lambda=0.4\), but now let four level-2 coefficients each equal \(0.5\), with all other coefficients zero. The total level-2 signal energy is still \(4(0.5^2)=1\), the same as for one coefficient equal to \(1\).

For a coefficient equal to \(0.5\), \(a=(0.4-0.5)/0.2=-0.5\) and \(b=(-0.4-0.5)/0.2=-4.5\). Substitution in the risk formula gives approximately \(0.09775\) per signal coefficient. The four such coefficients contribute approximately \(4(0.09775)=0.3910\).

The three zero coefficients at levels 0 and 1 contribute \(3(0.010458)\approx0.03137\). Thus total thresholding risk is approximately \(0.4224\), greater than the \(0.10351\) in the localized example. Yet the global risks at \(J=1\) and \(J=2\) are unchanged: they depend on the level's total energy, not its spatial distribution. At \(J=1\) the risk is \(1.12\); at \(J=2\) it is \(0.28\). The calculation illustrates that thresholding benefits from sufficiently prominent isolated coefficients, but a threshold that works well for one pattern need not work as well when the same energy is spread among smaller coefficients.

What the Comparison Establishes

Global truncation and spatial thresholding respond to different features of the coefficient array. Truncation responds to scale: if a fine level is retained, all its coefficients are included, whether large or negligible. Thresholding responds to individual observed magnitudes: coefficients at one scale may be treated differently. Accordingly, truncation risk is governed by the variance cost of all retained coordinates and the signal energy in omitted levels; thresholding risk accumulates coordinate-specific errors.

This is why an isolated local feature can be expensive for a global cutoff: retaining it may require retaining many other coefficients at the same and coarser scales. Conversely, if a signal's coefficients are numerous but individually moderate, a threshold can remove genuine signal along with noise. A favorable calculation for one localized signal is not a uniform dominance result. The best choice depends on the coefficient structure and on how the cutoff or threshold is selected.

Common pitfall: It is tempting to write the truncation variance as \(\tau^2\sum_{j\leq J}2^j\) and call it exact in every wavelet basis. That expression assumes exactly \(2^j\) coefficients at level \(j\). With boundary adjustments or another basis, the exact variance is \(\tau^2\sum_{j\leq J}N_j\), using the actual counts. Likewise, the exact hard-threshold risk formula above assumes \(\lambda\geq0\); without that condition, the two retained-tail events need not be disjoint.
Key takeaway: Global truncation is governed by energy across scales; spatial wavelet thresholding can respond to where energy occurs within a scale. Compare both approximation and estimation error, and do not infer that either procedure must dominate for every coefficient pattern.

Check Your Understanding

Use the definitions and risk calculations above to answer the following questions.

  1. Why must the exact variance term for a global cutoff use the actual number \(N_j\) of coefficients at each retained level?
  2. If \(S_j\leq C_\theta2^{-2sj}\) and \(N_j\) is of order \(2^j\), which two terms are balanced to choose a cutoff?
  3. In the localized-feature example, what variance and bias contributions produce the risk at \(J=1\)?
  4. Why can two signals with the same energy at a scale have the same global-truncation risk but different thresholding risks?
  5. What role does the condition \(\lambda\geq0\) play in the exact hard-threshold risk derivation?
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