Thresholding in the Gaussian Sequence Model

From Wavelet Coefficients to a Gaussian Sequence

Wavelet methods turn a function-estimation problem into estimation of many coefficients. In a sparse representation, a function may have only a few substantial coefficients, while most are zero or small. Noise can make those small coefficients look nonzero. Coordinatewise thresholding addresses this by deciding, coefficient by coefficient, whether to retain an observation and, if so, how much to shrink it.

The Gaussian sequence model isolates this decision. It is also directly connected to a Gaussian white noise model: for an orthonormal wavelet basis $\{\psi_{jk}\}$, the coefficient of a signal $f$ is $\theta_{jk}=\langle f,\psi_{jk}\rangle$. If the observation is $dY(x)=f(x)\,dx+\varepsilon\,dW(x)$, then the projected observations satisfy $Y_{jk}=\theta_{jk}+\varepsilon Z_{jk}$, where the $Z_{jk}$ are independent standard normal random variables. In a finite-dimensional analysis, we retain a finite index set $I$ of such coefficients.

Definition: The finite Gaussian sequence model observes $Y_i=\theta_i+\sigma Z_i$, for $i\in I$, where the unknown coefficients are $\theta_i$, the noise scale $\sigma>0$ is known, and the $Z_i$ are independent $N(0,1)$ random variables. The estimation target is the vector $\theta=(\theta_i:i\in I)$.

Under squared coefficient loss, an estimator $\hat\theta=(\hat\theta_i:i\in I)$ is assessed by $\sum_{i\in I}(\hat\theta_i-\theta_i)^2$. For an orthonormal basis and a reconstruction using this same finite index set, this is also the squared $L^2$ error of the reconstructed coefficient contribution. In the white noise model, the noise standard deviation $\sigma$ is $\varepsilon$. In regression, transformed coefficients may only approximately follow this model, or may have correlated errors unless the design and transformation justify independence.

Two Coordinatewise Thresholding Rules

A threshold should be expressed relative to the noise scale. Write it as $t=\lambda\sigma$, where $\lambda>0$ is dimensionless. A large observed coefficient is less plausibly explained by noise alone than a small one, but a threshold is a decision rule, not a test that reveals with certainty whether a coefficient is truly nonzero.

Definition: Hard thresholding keeps an observation if its magnitude exceeds $t$, and otherwise sets it to zero: $$ \hat\theta_i^{\,H}=Y_i\,\mathbf{1}\{|Y_i|>t\}. $$ Soft thresholding sets observations of magnitude at most $t$ to zero and moves the others toward zero by $t$: $$ \hat\theta_i^{\,S}=\operatorname{sign}(Y_i)(|Y_i|-t)_+, $$ where $a_+=\max(a,0)$ and $\operatorname{sign}(0)=0$.

Both rules can produce exact zeros, which is useful when the coefficient pattern is sparse. Their behavior differs for retained observations: hard thresholding leaves them unchanged, whereas soft thresholding subtracts $t$ from their magnitude. Thus hard thresholding has no additional shrinkage bias for a retained observation, but it jumps at the threshold. Soft thresholding changes continuously as the observation crosses the threshold, but it shrinks every retained value.

Worked Example: A Noise-Only Coefficient

Suppose $\theta=0$, $\sigma=1$, and $\lambda=2$, so $Y=Z\sim N(0,1)$ and $t=2$. Hard thresholding returns a nonzero estimate precisely when $|Z|>2$. By symmetry, $$ P(\hat\theta^{\,H}\ne 0)=P(|Z|>2)=2\Phi(-2)\approx 2(0.02275)=0.0455. $$ So about 4.55% of noise-only observations survive this threshold.

The expected squared error under the zero signal is the estimator's variance in this example. Hard thresholding has error $Z^2$ on $|Z|>2$ and zero otherwise. Soft thresholding has error $(|Z|-2)^2$ on that same event and zero otherwise. The two rules therefore keep the same observations, but soft thresholding gives smaller estimates for those it keeps.

What Sparsity Changes

A coefficient vector is sparse when relatively few coordinates are substantial and many are zero or close to zero. It is dense when many coordinates carry signal, even if each individual coefficient is modest. These descriptions are about the pattern of the unknown coefficients, not about how many observed values happen to exceed a threshold.

For a sparse pattern, thresholding can remove many noise-dominated coordinates while retaining a few strong ones. For a dense pattern of weak coefficients, the same rule may set many genuine signals to zero. Soft thresholding adds shrinkage to retained coefficients, which can help when retained observations contain noise but can also suppress real signal. There is no coordinatewise rule that can infer the entire pattern perfectly from noisy observations.

Worked Example: A Sparse Signal With One Strong Coefficient

Take four coefficients with $\theta/\sigma=(4,0,0,0)$, threshold $t=\sigma$, and noise realization $Z=(0,-0.4,0.6,1.2)$. The observed vector is $Y/\sigma=(4,-0.4,0.6,1.2)$. The signal is sparse: only its first coefficient is nonzero.

Hard thresholding retains observations with magnitude greater than $\sigma$, giving $\hat\theta^{\,H}/\sigma=(4,0,0,1.2)$. The errors divided by $\sigma$ are $(0,0,0,1.2)$, so the total squared error is $1.44\sigma^2$.

Soft thresholding gives $\hat\theta^{\,S}/\sigma=(3,0,0,0.2)$. Its errors divided by $\sigma$ are $(-1,0,0,0.2)$, so its total squared error is $(1+0.04)\sigma^2=1.04\sigma^2$. In this realization the soft rule improves on hard thresholding by shrinking the retained noise-only fourth coordinate, though it also shrinks the true first coefficient. The unthresholded estimate $Y$ has squared error $(0^2+(-0.4)^2+0.6^2+1.2^2)\sigma^2=1.96\sigma^2$.

Worked Example: A Dense Pattern of Weak Coefficients

Now let $\theta/\sigma=(0.8,0.8,0.8,0.8)$, use $t=\sigma$, and suppose $Z=(0,0.3,-0.1,0.5)$. The observed vector is $Y/\sigma=(0.8,1.1,0.7,1.3)$. All four true coefficients are nonzero, but each is smaller than the threshold.

Hard thresholding gives $\hat\theta^{\,H}/\sigma=(0,1.1,0,1.3)$. Its errors are $(-0.8,0.3,-0.8,0.5)$ in units of $\sigma$, and its total squared error is $(0.64+0.09+0.64+0.25)\sigma^2=1.62\sigma^2$.

Soft thresholding gives $\hat\theta^{\,S}/\sigma=(0,0.1,0,0.3)$. The errors are $(-0.8,-0.7,-0.8,-0.5)$, yielding squared error $(0.64+0.49+0.64+0.25)\sigma^2=2.02\sigma^2$. The unthresholded vector has squared error $(0^2+0.3^2+(-0.1)^2+0.5^2)\sigma^2=0.35\sigma^2$. This particular realization illustrates the cost of thresholding a dense weak signal; it does not show that unthresholded estimation is generally preferable.

Risk Separates Across Coordinates

The coordinatewise structure makes total squared-error risk easy to express. This identity does not say which threshold is best; it says that, for coordinatewise estimates, the total risk is the sum of the risks for individual coefficients.

Theorem: In the finite Gaussian sequence model, let $\hat\theta_i=g_i(Y_i)$ be measurable coordinatewise estimators with finite second moments. Under squared coefficient loss, $$ E_\theta\!\left[\sum_{i\in I}(\hat\theta_i-\theta_i)^2\right] = \sum_{i\in I}E_\theta[(g_i(Y_i)-\theta_i)^2]. $$

Expand the loss.
The squared coefficient loss is a finite sum of nonnegative terms, one for each coordinate.

Take expectations.
Each term has finite expectation by the assumed second moments. Linearity of expectation for a finite sum gives the sum of the coordinatewise expectations.

E_\theta\!\left[\sum_{i\in I}(\hat\theta_i-\theta_i)^2\right] = \sum_{i\in I}E_\theta[(\hat\theta_i-\theta_i)^2]. \quad \square

Independence is part of the stated Gaussian sequence model, but this finite-sum identity itself does not require independence. Independence matters for other distributional questions; it is not needed to exchange a finite sum and expectation.

Exact Null Risk: A Comparison at One Signal Value

A simple exact calculation makes the different shrinkage behavior concrete. Write $\phi$ and $\Phi$ for the standard normal density and distribution function, and $Q(\lambda)=\Phi(-\lambda)$. At a zero coefficient, the risk of either rule is its expected squared estimate. The next result compares those risks at that particular signal value only.

Result: If $\theta=0$, $Y=\sigma Z$ with $Z\sim N(0,1)$, and $t=\lambda\sigma$, then $$ E_0[(\hat\theta^{\,H})^2] =2\sigma^2\{\lambda\phi(\lambda)+Q(\lambda)\}, $$ and $$ E_0[(\hat\theta^{\,S})^2] =2\sigma^2\{(1+\lambda^2)Q(\lambda)-\lambda\phi(\lambda)\}. $$

Derivation. Under the zero signal, hard thresholding has squared estimate $\sigma^2 Z^2\mathbf{1}\{|Z|>\lambda\}$. By symmetry its expectation is $2\sigma^2\int_\lambda^\infty z^2\phi(z)\,dz$. Since $\phi'(z)=-z\phi(z)$, integration by parts gives $\int_\lambda^\infty z^2\phi(z)\,dz=\lambda\phi(\lambda)+Q(\lambda)$. This proves the hard-threshold formula.

For soft thresholding the squared estimate is $\sigma^2(|Z|-\lambda)^2\mathbf{1}\{|Z|>\lambda\}$, so its expectation is $2\sigma^2\int_\lambda^\infty(z-\lambda)^2\phi(z)\,dz$. Expanding the square, using $\int_\lambda^\infty z^2\phi(z)\,dz=\lambda\phi(\lambda)+Q(\lambda)$ and $\int_\lambda^\infty z\phi(z)\,dz=\phi(\lambda)$, gives $$ 2\sigma^2\{\lambda\phi(\lambda)+Q(\lambda)-2\lambda\phi(\lambda)+\lambda^2Q(\lambda)\} = 2\sigma^2\{(1+\lambda^2)Q(\lambda)-\lambda\phi(\lambda)\}. \quad \square $$

Worked Example: Comparing Null Risks at One Noise Standard Deviation

Set $\lambda=1$. The standard normal values are $\phi(1)\approx0.24197$ and $Q(1)\approx0.15866$. The hard-threshold risk at $\theta=0$ is $$ 2\sigma^2\{0.24197+0.15866\}\approx0.80126\sigma^2. $$ The soft-threshold risk is $$ 2\sigma^2\{2(0.15866)-0.24197\} \approx0.15070\sigma^2. $$ At a zero coefficient, soft thresholding has lower risk here because it reduces the magnitude of every retained noise observation. This comparison is not a general ranking over nonzero signals: soft thresholding also introduces shrinkage for genuine coefficients.

Practical Interpretation and a Common Pitfall

Thresholding matters because it provides a direct way to exploit coefficient sparsity without estimating every coefficient by an unrestricted noisy value. In a wavelet reconstruction, setting coefficients to zero can remove noise across many locations and scales; retaining strong coefficients can preserve prominent structure. For a dense weak pattern, however, a threshold that is effective at removing noise-only coordinates can also erase real signal. The examples and null-risk calculation show why conclusions depend on the coefficient pattern and on the loss being considered.

Common mistake: treating thresholding as an automatic sparse-signal detector. The mistaken argument is: “If $|Y_i|\leq t$, then $\theta_i=0$; if $|Y_i|>t$, then $\theta_i\ne0$.” This does not follow from the model. A zero coefficient can produce a large noise observation, and a nonzero coefficient can produce a small observation. Thresholding is an estimation rule whose performance is assessed over repeated noise realizations, not a guarantee about the status of an individual coefficient.

A second practical caution is to check the noise model before applying these calculations. The independent, equal-variance Gaussian sequence representation follows directly from orthonormal projection in the stated white noise model. A regression transformation need not yield independent coefficient errors with a common variance. Applying the same threshold without checking the transformed noise scale and dependence can change the rule's behavior.

Key takeaway: Hard thresholding deletes small observations and leaves retained values unchanged; soft thresholding deletes the same small observations and shrinks retained values. Their consequences depend on noise scale, signal strength, and whether coefficients are sparse or dense.

Check Your Understanding

Use the model and definitions above to answer these questions.

In the Gaussian sequence model, what assumptions make wavelet coefficient observations independent with a common noise variance?
For an observation with magnitude $1.4t$, compare the magnitudes of its hard- and soft-thresholded estimates.
Why can a dense pattern of weak nonzero coefficients be poorly represented by a threshold that removes many noise-only coefficients?
Which part of the finite-sum risk identity requires independence of the coordinates, and why?
Why does the lower null risk of soft thresholding at a chosen threshold not establish that soft thresholding has lower risk for every $\theta$?