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Localized Complexity and Stability · Tutorial 173 of 989

Star-Shaped Classes and Sub-Root Complexity

Learn why star-shapedness makes radius-localized Rademacher complexity sub-root, and how its scaling identifies the complexity–radius balance.

Advanced 10 min read

What You'll Learn

  • Define star-shaped excess-loss classes and localized Rademacher complexity.
  • Distinguish empirical localization from localization by population second moment.
  • Prove monotonicity in the localization radius.
  • Derive the square-root scaling inequality from star-shapedness.
  • Calculate localized complexities and identify why star-shapedness matters.
  • Interpret the complexity–radius balance that motivates a critical radius.

Why Star-Shapedness Matters

Localized Rademacher complexity measures how large random fluctuations can be among functions whose squared size is limited. The limit is important: functions close to a risk minimizer often have small excess risk, and restricting attention to such functions can give a sharper complexity measure than studying the entire class at once.

The key geometric condition in this tutorial is star-shapedness around zero. It says that shrinking any function in the class toward zero keeps it in the class. This lets us compare two localization radii: a function allowed at the larger radius can be shrunk until it is allowed at the smaller one. Because Rademacher averages scale linearly under this shrinking, the localized complexity inherits a square-root scaling inequality.

Definition: A function class \(\mathcal G\) is star-shaped around zero if \(0\in\mathcal G\) and, for every \(g\in\mathcal G\) and every \(a\in[0,1]\), the function \(ag\) also belongs to \(\mathcal G\). An excess-loss class has this property when its functions can be shrunk toward the zero function while remaining in the class.

For example, given a risk minimizer \(f^\star\) and a loss function \(\ell_f\), an excess-loss function has the form \(g_f(x)=\ell_f(x)-\ell_{f^\star}(x)\). The class of these functions is not automatically star-shaped: the condition requires that \(a g_f\) belong to the excess-loss class for every \(a\in[0,1]\). The functions may take negative values at individual observations even when their expectations are nonnegative.

Two Localized Complexities

Let \(X_1,\ldots,X_n\) be independent observations with common distribution \(P\), and let \(\epsilon_1,\ldots,\epsilon_n\) be independent Rademacher random variables, independent of the observations: each sign is \(+1\) or \(-1\) with probability \(1/2\). Write \(P_n g^2=n^{-1}\sum_{i=1}^n g(X_i)^2\) for the empirical second moment and \(P g^2=E[g(X)^2]\) for the population second moment.

Definition: For a fixed observed sample \(S=(X_1,\ldots,X_n)\), the empirical localized Rademacher complexity at squared-radius \(r\geq0\) is \[ \widehat\psi_S(r)=E_\epsilon\!\left[\sup_{g\in\mathcal G:\,P_n g^2\leq r}\frac{1}{n}\sum_{i=1}^n\epsilon_i g(X_i)\right]. \] The population-localized version is \[ \psi_n(r)=E_{X,\epsilon}\!\left[\sup_{g\in\mathcal G:\,P g^2\leq r}\frac{1}{n}\sum_{i=1}^n\epsilon_i g(X_i)\right]. \] The localization radius \(r\) bounds a squared size, so the corresponding function scale is \(\sqrt r\).

The two definitions localize in different ways. The empirical version restricts functions using the observed sample, while the population version uses their second moment under \(P\). In either case the random average is evaluated on the sample. For the results below, assume the functions are measurable, the suprema are measurable (for example, \(\mathcal G\) is countable), and the displayed expectations are finite. A uniformly bounded class is a convenient sufficient condition for finiteness. Since \(0\in\mathcal G\), each localized class contains zero, and each complexity is nonnegative.

Worked Example: A One-Dimensional Star-Shaped Class

Take \(n=1\), let \(X\) be constant, and let \(h(X)=2\). Let \(\mathcal G=\{ah:0\leq a\leq1\}\). Compute the population-localized complexity.

For \(g=ah\), its population second moment is \(P g^2=4a^2\). Thus the constraint \(P g^2\leq r\) permits \(0\leq a\leq \min(1,\sqrt r/2)\) when \(r\leq4\), and all \(a\in[0,1]\) when \(r\geq4\).

The Rademacher average is \(2a\epsilon_1\). For any permitted maximum \(a_{\max}\), the supremum over \(a\in[0,a_{\max}]\) is \(2a_{\max}\) if \(\epsilon_1=1\), and zero if \(\epsilon_1=-1\). Averaging over the two equally likely signs gives \(a_{\max}\). Therefore

$$ \psi_1(r)= \begin{cases} \sqrt r/2, & 0\leq r\leq4,\\ 1, & r\geq4. \end{cases} $$

For \(0<r\leq4\), the ratio \(\psi_1(r)/\sqrt r\) equals \(1/2\). For \(r\geq4\), it equals \(1/\sqrt r\), which decreases as \(r\) increases. The class’s star-shaped structure is visible directly: every allowed coefficient can be reduced toward zero.

Monotonicity and Square-Root Scaling

The first property needs no star-shapedness. Increasing \(r\) only enlarges the set over which the supremum is taken. The second property is the geometric payoff: star-shapedness lets us shrink every function admissible at a larger radius so it becomes admissible at a smaller radius.

Theorem: Suppose \(\mathcal G\) is star-shaped around zero and the complexities above are well-defined and finite. Both \(r\mapsto\widehat\psi_S(r)\) and \(r\mapsto\psi_n(r)\) are nondecreasing. Moreover, for \(0<r_1\leq r_2\), \[ \widehat\psi_S(r_1)\geq\sqrt{\frac{r_1}{r_2}}\,\widehat\psi_S(r_2), \qquad \psi_n(r_1)\geq\sqrt{\frac{r_1}{r_2}}\,\psi_n(r_2). \] Consequently, for \(r>0\), each complexity divided by \(\sqrt r\) is nonincreasing in \(r\).

Proof. First fix the sample. If \(r_1\leq r_2\), then \(\{g\in\mathcal G:P_n g^2\leq r_1\}\) is contained in \(\{g\in\mathcal G:P_n g^2\leq r_2\}\). For every fixed sign vector, the supremum over the smaller set cannot exceed the supremum over the larger set. Taking expectation over the signs proves \(\widehat\psi_S(r_1)\leq\widehat\psi_S(r_2)\). For population localization the same inclusion holds with \(P g^2\) in place of \(P_n g^2\). Taking expectation over observations and signs proves that \(\psi_n\) is nondecreasing.

Now fix \(0<r_1\leq r_2\) and set \(a=\sqrt{r_1/r_2}\), so \(a\in(0,1]\). For any \(g\in\mathcal G\) with \(P_n g^2\leq r_2\), star-shapedness gives \(ag\in\mathcal G\), and \[ P_n(ag)^2=a^2P_n g^2\leq a^2r_2=r_1. \] The Rademacher average is linear in the function: \(n^{-1}\sum_i\epsilon_i(ag)(X_i)=a n^{-1}\sum_i\epsilon_i g(X_i)\). Thus, for each sign vector, the supremum at radius \(r_1\) is at least \(a\) times the supremum at radius \(r_2\). To see this directly, the scaled versions \(ag\) of all functions admissible at \(r_2\) are among those admissible at \(r_1\), and \(a\geq0\). Taking sign expectations yields the empirical scaling inequality.

For population localization, if \(P g^2\leq r_2\), then \(P(ag)^2=a^2P g^2\leq r_1\). The same linearity argument applies to each observed sample and sign vector. Taking expectation over both proves the population scaling inequality. Dividing either inequality by \(\sqrt{r_1}\) gives that the ratio at \(r_1\) is at least the ratio at \(r_2\), as claimed. \(\square\)

Terminology: A nonnegative function \(\psi\) is called sub-root if it is nondecreasing and \(\psi(r)/\sqrt r\) is nonincreasing for \(r>0\). The theorem shows that star-shapedness makes these localized complexities sub-root. When the zero-radius complexity is defined as zero, one may also set \(\psi(0)=0\).

Worked Example: Empirical Localization on a Fixed Sample

Suppose \(n=2\), and on the observed sample a function \(h\) takes values \((1,1)\). Let \(\mathcal G=\{th:0\leq t\leq1\}\). Find the empirical localized complexity as a function of \(r\).

Here \(P_n(th)^2=t^2(1^2+1^2)/2=t^2\), so the largest allowed coefficient is \(t_{\max}=\min(1,\sqrt r)\). The Rademacher average is \(t(\epsilon_1+\epsilon_2)/2\). For the four equally likely sign pairs \((+,+),(+,-),(-,+),(-,-)\), the supremum over \(0\leq t\leq t_{\max}\) is respectively \(t_{\max},0,0,0\). Its expectation is \(t_{\max}/4\). Hence

$$ \widehat\psi_S(r)=\frac14\min(1,\sqrt r). $$

For \(0<r\leq1\), the ratio is \(1/4\); for \(r\geq1\), it is \(1/(4\sqrt r)\). It is nonincreasing. This calculation also shows why the radius is squared: the empirical second moment of \(th\) is \(t^2\), while the complexity scales as \(t\).

How the Properties Point to a Critical Radius

A localized complexity bound is often compared with the squared radius itself. The radius describes the size permitted for functions; the complexity describes the scale of their random fluctuations. A critical radius is a scale at which these quantities balance, commonly expressed through an equation such as \(\psi(r)=r\), or through the point beyond which \(\psi(r)\leq r\). This tutorial establishes the shape properties that make such a comparison useful; determining a fixed point and using it in a risk bound require additional arguments.

Sub-root behavior gives a simple way to read the comparison. The complexity is nondecreasing, but it grows no faster than the square-root scaling inequality permits: shrinking the radius by a factor \(c\) shrinks the complexity by at least a factor \(\sqrt c\), for \(0<c\leq1\). Thus the ratio \(\psi(r)/\sqrt r\) decreases as the localization region expands. This is a structural statement, not a numerical rate by itself; calculating an actual critical radius still requires a bound or formula for \(\psi\).

Worked Example: Why Star-Shapedness Cannot Be Dropped

Consider a fixed sample of size \(n=2\). Describe functions by their two observed values, and let \(\mathcal G\) consist of the zero vector, \(u=(1,0)\), and \(v=(1,1)\). This class is not star-shaped, since, for example, \(\tfrac12v\notin\mathcal G\). Compute its empirical localized complexity at \(r=1/2\) and \(r=1\).

The empirical squared sizes are \(P_nu^2=1/2\) and \(P_nv^2=1\). At radius \(1/2\), only zero and \(u\) are available. The average for \(u\) is \(\epsilon_1/2\); taking the supremum with zero gives \(1/2\) when \(\epsilon_1=1\) and zero otherwise. Therefore \(\widehat\psi_S(1/2)=1/4\).

At radius \(1\), both nonzero vectors are available. For sign pairs \((+,+),(+,-),(-,+),(-,-)\), the supremum of the averages over zero, \(u\), and \(v\) is \(1,1/2,0,0\), respectively. Their mean is \(3/8\), so \(\widehat\psi_S(1)=3/8\). The ratios are

$$ \frac{\widehat\psi_S(1/2)}{\sqrt{1/2}}=\frac{1}{2\sqrt2} \quad\text{and}\quad \frac{\widehat\psi_S(1)}{\sqrt1}=\frac38. $$

Since \(1/(2\sqrt2)<3/8\), the ratio increases between these radii, contrary to the sub-root property. The class is still nested under increasing localization, so monotonicity holds; it is the square-root scaling conclusion that fails. Without star-shapedness, shrinking \(v\) to a smaller-radius function need not produce a member of the class.

Common Pitfall

  • Confusing monotonicity with sub-root scaling. Nested localized sets always make the supremum nondecreasing with radius. That fact alone does not show that \(\psi(r)/\sqrt r\) decreases. The latter conclusion uses star-shapedness and the linear scaling of the Rademacher average.
  • Assuming an excess-loss class is star-shaped automatically. The definition of excess loss does not guarantee that \(ag_f\) is another excess loss. Verify the closure property for the class under consideration rather than inferring it from the word “excess.”
  • Treating \(r\) as an unsquared function size. The constraint is on \(P g^2\) or \(P_n g^2\). Scaling a function by \(a\) scales this radius quantity by \(a^2\), which is why the complexity comparison involves \(\sqrt{r_1/r_2}\).
Key takeaway: Localization gives monotonicity because the admissible sets are nested. Star-shapedness adds the decisive scaling relation, making radius-localized Rademacher complexity sub-root and providing the shape needed when comparing complexity with radius.

Check Your Understanding

Use the definitions and scaling argument to answer the following questions.

  1. State the condition that makes a function class star-shaped around zero.
  2. Why does increasing the localization radius make either localized complexity nondecreasing, even if the class is not star-shaped?
  3. For \(0<r_1\leq r_2\), what scaling factor maps a function localized at \(r_2\) into one localized at \(r_1\)?
  4. In the fixed-sample example with values \((1,1)\), why does the complexity scale as \(\sqrt r\) for \(0\leq r\leq1\)?
  5. Which property fails in the three-function example: monotonicity, square-root scaling, or both? Explain why.
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