Testing Arguments for Adaptive-Band Impossibility

Why a Confidence Band Creates a Testing Problem

A confidence band must cover an entire function, not just one number. This makes a useful impossibility argument possible: if a band is narrow and covers a function that differs noticeably from a smooth baseline, then the band cannot also contain that baseline. The band itself can therefore be converted into a test.

The challenge is to find alternatives that are separated from the baseline but statistically difficult to distinguish from it. A standard construction places small, smooth bumps at many possible locations. Each alternative differs from the baseline in supremum norm, while a mixture over the possible locations can remain close in distribution to the baseline.

The simultaneous coverage event is essential: it gives a coverage guarantee under each individual alternative, which can then be averaged over a mixture. The argument below does not require a particular way of constructing the band. It applies to any measurable band satisfying the stated coverage and width properties.

Localized Alternatives and Their Mixture

Let the baseline be \(f_0=0\). Choose a nonzero, infinitely differentiable function \(\psi\) supported on \([0,1]\), with \(\sup_u|\psi(u)|=1\). For a bandwidth \(h\), place rescaled copies of this bump on \(m\) disjoint intervals of length \(h\). For smoothness \(\beta>0\), define \(f_j(x)=c h^\beta\psi((x-t_j)/h)\), where \(t_j\) is the left endpoint of the \(j\)th interval and \(c>0\) is fixed. With disjoint supports, each \(f_j\) is a localized alternative. Under the usual Hölder norm, the factor \(h^\beta\) keeps the roughness of these alternatives bounded uniformly in \(h\), for fixed \(c\). Their distance from the baseline is \(\|f_j-f_0\|_\infty=c h^\beta\).

Write \(P_f\) for the distribution of the observations when the signal is \(f\). The likelihood ratio of \(P_{f_j}\) relative to \(P_0\) is \(L_j=\exp\{n\int_0^1 f_j(x)\,dY(x)-n\|f_j\|_2^2/2\}\). The equal-weight mixture distribution is \(\bar P=m^{-1}\sum_{j=1}^mP_{f_j}\), with likelihood ratio \(\bar L=m^{-1}\sum_j L_j\) relative to \(P_0\).

Proof. Under \(P_0\), the stochastic integrals \(\int f_j\,dY\) are jointly Gaussian with mean zero and covariance \(n^{-1}\langle f_i,f_j\rangle\), where \(\langle f_i,f_j\rangle=\int_0^1 f_i(x)f_j(x)\,dx\). The Gaussian moment-generating formula therefore gives \(E_0[L_iL_j]=\exp(n\langle f_i,f_j\rangle)\). Disjoint supports imply \(\langle f_i,f_j\rangle=0\) when \(i\ne j\), and identical rescaled bumps have equal squared norms. Since \(E_0[L_j]=1\), \[ E_0[(\bar L-1)^2] =E_0[\bar L^2]-1 =\frac{1}{m^2}\left(m e^{n\|f_1\|_2^2}+m(m-1)\right)-1 =\frac{e^{n\|f_1\|_2^2}-1}{m}. \] Finally, total variation satisfies \(\|\bar P-P_0\|_{\mathrm{TV}}=\tfrac12 E_0|\bar L-1| \le\tfrac12\sqrt{E_0[(\bar L-1)^2]}\) by Cauchy–Schwarz. This proves both claims. \(\square\)

From Indistinguishability to a Width Lower Bound

The testing step is general. Suppose a band is honest over a class containing \(f_0\) and the alternatives \(f_1,\ldots,f_m\), each of which is at least \(\delta\) away from \(f_0\) in supremum norm. Define a test \(\phi\) by \(\phi=1\) if the band does not contain \(f_0\) everywhere, and \(\phi=0\) otherwise. This test is based only on the observed data through the band.

Proof. Under \(P_0\), the event \(\phi=1\) means the band fails to cover \(f_0\). Honesty at \(f_0\) thus gives \(P_0(\phi=1)\le\alpha\). For each alternative \(f_j\), honesty gives probability at least \(1-\alpha\) to the event that the band covers \(f_j\) everywhere. If it covers \(f_j\) and \(W_n<\delta\), it cannot also contain \(f_0\) everywhere: at a point \(x\) where \(|f_j(x)-f_0(x)|\ge\delta\), an interval of width less than \(\delta\) cannot contain both values. Therefore \(P_{f_j}(\phi=1)\ge 1-\alpha-P_{f_j}(W_n\ge\delta)\). Averaging this inequality over \(j\) proves the lower bound on \(\bar P(\phi=1)\).

For any test, the difference between its rejection probabilities under two distributions is at most their total variation distance. Therefore \(\bar P(\phi=1)\le P_0(\phi=1)+\varepsilon\le\alpha+\varepsilon\). Combining this with the lower bound gives \(1-\alpha-\bar P(W_n\ge\delta)\le\alpha+\varepsilon\), or \(\bar P(W_n\ge\delta)\ge1-2\alpha-\varepsilon\), as claimed. \(\square\)

Why the Logarithmic Separation Appears

The bump calculation shows how to choose a separation that makes the mixture difficult to detect. For the \(\beta\)-smooth alternatives above, \(\|f_j\|_2^2=c^2h^{2\beta+1}\|\psi\|_2^2\). With \(m\) disjoint locations, the mixture chi-square divergence is \(\{\exp(nc^2h^{2\beta+1}\|\psi\|_2^2)-1\}/m\). If \(h\) is chosen so that \(n c^2h^{2\beta+1}\|\psi\|_2^2\) is a sufficiently small multiple of \(\log m\), the numerator grows more slowly than \(m\), and the divergence tends to zero. Because \(m\) is of order \(1/h\), this balance gives, up to constants and the usual logarithmic equivalences, \(h\asymp(\log n/n)^{1/(2\beta+1)}\) and \(\delta\asymp h^\beta=(\log n/n)^{\beta/(2\beta+1)}\).

For example, suppose the baseline belongs to a smoother Hölder class of order \(\beta_1\), and the alternatives belong to a larger class allowing order \(\beta_2<\beta_1\). A width of the smoother estimation order \(n^{-\beta_1/(2\beta_1+1)}\) is smaller than the testing separation at the rougher order, since \[ \frac{n^{-\beta_1/(2\beta_1+1)}}{(\log n/n)^{\beta_2/(2\beta_2+1)}}\longrightarrow 0. \] The theorem then explains the obstruction: if the band were narrow at the smooth baseline and honest over the rough alternatives, it would distinguish the baseline from a mixture that is asymptotically indistinguishable from it.

Common Pitfall

A common mistake is to argue that each alternative is detectable because its individual likelihood ratio has a large second moment, and then conclude that the mixture must also be detectable. This confuses a collection of separate comparisons with the single testing problem that matters. The second moment of the average likelihood ratio includes a factor \(1/m\) on the diagonal terms; with many nearly orthogonal alternatives, that factor can make the mixture close to the null even when individual alternatives have substantial information.

The opposite shortcut is also unsafe: small average width under the baseline alone does not immediately imply small width under every alternative. The valid argument first transfers the event \(W_n\ge\delta\) from the baseline to the mixture using total variation. It then combines that transfer with honesty under each alternative. Both steps are needed.