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q=2 Gaussian-sequence ellipsoid detection boundary

Statement

Assume Countable Choice and the minimax detection-risk definition with $q = 2$ . For each $ε$ , suppose $Θ_{2} (r_{ε})$ is nonempty and has a finite-support fourth-power minimizer ${\tilde{η}}_{ε}$ with profile ${\tilde{η}}_{ε, j}^{2} = z_{0}^{2} σ_{j}^{2} (1 - A a_{j}^{2})_{+}$ as in the extremal-profile lemma. Put $u_{ε}^{2} = \frac{1}{2 ε^{4}} \sum_{j} {\tilde{η}}_{ε, j}^{4}, w_{ε, j} = \frac{{\tilde{η}}_{ε, j}^{2}}{\sqrt{2 \sum_{j} {\tilde{η}}_{ε, j}^{4}}}, w_{ε, 0} = \max_{j} w_{ε, j},$ and define $T_{ε} = \sum_{j} w_{ε, j} ((Y_{j} / ε)^{2} - 1), ψ_{ε, H} = 1_{{T_{ε} > H}} .$ If $u_{ε} \to 0$ , then $γ_{ε} (Θ_{2} (r_{ε})) \to 1$ and $β_{ε} (Θ_{2} (r_{ε}), α) \to 1 - α$ for every $α \in (0, 1)$ . If $u_{ε} \to \infty$ , then for every fixed $c \in (0, 1)$ the tests $ψ_{ε, c u_{ε}}$ have total error tending to zero.

For the extremal profile, define the finite sums $J_{1} = \sum_{j} σ_{j}^{4} (1 - A a_{j}^{2})_{+}, J_{2} = A \sum_{j} a_{j}^{2} σ_{j}^{4} (1 - A a_{j}^{2})_{+}, J_{0} = \sum_{j} σ_{j}^{4} (1 - A a_{j}^{2})_{+}^{2} .$ The active constraints imply $r_{ε}^{2} = A \frac{J_{1}}{J_{2}}, u_{ε}^{2} = {(\frac{r_{ε}}{ε})}^{4} \frac{J_{0}}{2 J_{1}^{2}} .$

Statement references

Ingster, Sapatinas and Suslina, Minimax signal detection in ill-posed inverse problems, Theorem 4.1 and equations (4.1)–(4.8)

Giné and Nickl, Mathematical Foundations of Infinite-Dimensional Statistical Models, §6.2.3, Theorem 6.2.11 proof, PDF pp. 506–508

Marteau and Sapatinas, A unified treatment for non-asymptotic and asymptotic approaches to minimax signal detection, §§3.1–3.2

Facts & Assumptions

Given: Countable Choice; $q = 2$ ; a nonempty class $Θ_{2} (r_{ε})$ ; a finite-support minimizing profile satisfying both active constraints; and the Gaussian sequence detection experiment.

[F1]

Minimax detection risk and separation rate: $α_{ε} (ψ) = E_{0} ψ$ , $β_{ε} (Θ_{2} (r), ψ) = \sup_{η \in Θ_{2} (r)} E_{η} (1 - ψ)$ , and $γ_{ε}$ is the infimum total error.

[F2]

The q=2 ellipsoid extremal profile minimizes the fourth-power objective: ${\tilde{η}}_{j}^{2} = z_{0}^{2} σ_{j}^{2} (1 - A a_{j}^{2})_{+}$ minimizes $\sum_{j} η_{j}^{4}$ over $Θ_{2} (r)$ and satisfies $\sum_{j} σ_{j}^{2} {\tilde{η}}_{j}^{2} = r^{2}$ , $\sum_{j} a_{j}^{2} σ_{j}^{2} {\tilde{η}}_{j}^{2} = 1$ .

[F3]

Symmetric Gaussian product-mixture testing bound: for a finite-support feasible $v$ , its symmetric sign mixture has total-variation distance at most $\frac{1}{2} \sqrt{\exp (u (v)^{2}) - 1}$ from the null; consequently it gives the displayed minimax lower bounds.

[F4]

Isonormal Gaussian white-noise inverse experiment supplies independent standard-normal coordinates with normalized density $φ (t) = (2 π)^{- 1 / 2} e^{- t^{2} / 2}$ and covariance one. We use the scalar calculation of A normal variable has finite fourth moment with this supplied law, rather than invoke its AC-based law construction: $φ' (t) = - t φ (t)$ , so integration by parts on $[- R, R]$ gives $\int t^{2} φ = [- t φ]_{- R}^{R} + \int φ$ and $\int t^{4} φ = [- t^{3} φ]_{- R}^{R} + 3 \int t^{2} φ$ . The polynomial-Gaussian boundary terms vanish as $R \to \infty$ ; the nonnegative integrals increase to their full integrals. Supplied normalization therefore gives second and fourth moments $1$ and $3$ ; symmetry gives first and third moments zero. This supplied-law calculation uses no additional choice.

[F5]

Chebyshev-Markov inequality for the integral: Chebyshev's inequality bounds a tail probability by the variance divided by the squared deviation.

[A1]

The Axiom of Countable Choice ( ${A C}_{ω}$ ): Countable Choice is the assumption of the upstream Gaussian sequence model.

Proof

1.1A1F1F2F4algebra

The profile has finite support by the assumption on $S$ . Its weights satisfy $2 \sum_{j} w_{ε, j}^{2} = 1$ , so $\sum_{j} w_{ε, j}^{2} = 1 / 2$ . Under a parameter $η$ , write $μ_{j} = η_{j} / ε$ ; then $Y_{j} / ε = ξ_{j} + μ_{j}$ . Independence and [F4] give $m_{η} : = E_{η} T_{ε} = \sum_{j} w_{ε, j} μ_{j}^{2}, {Var}_{η} (T_{ε}) = 1 + 4 \sum_{j} w_{ε, j}^{2} μ_{j}^{2} \leq 1 + 4 w_{ε, 0} m_{η} .$ In particular, under the null the statistic has mean zero and variance one.

1.2F1F3given

Suppose $u_{ε} \to 0$ . Apply [F3] to $v = {\tilde{η}}_{ε}$ . The resulting total-variation bound tends to zero, so every test has total error at least $1 - o (1)$ and every level- $α$ test has maximal type-II error at least $1 - α - o (1)$ . The constant test $ψ \equiv 0$ gives total error $1$ , and the constant randomized test $ψ \equiv α$ gives level $α$ and maximal type-II error $1 - α$ . These upper bounds prove both limits.

2.1A1F1F2step 1.1algebra

For any $η \in Θ_{2} (r_{ε})$ , put $x_{j} = η_{j}^{2}$ and ${\tilde{x}}_{j} = {\tilde{η}}_{j}^{2}$ . Since $1 - A a_{j}^{2} \leq 0$ off the profile support, $\sum_{j} {\tilde{x}}_{j} x_{j} = z_{0}^{2} \sum_{j \in S} σ_{j}^{2} (1 - A a_{j}^{2}) x_{j} \geq z_{0}^{2} (r_{ε}^{2} - A) = \sum_{j} {\tilde{x}}_{j}^{2},$ where the inequality follows from $\sum_{j} σ_{j}^{2} x_{j} \geq r_{ε}^{2}$ and $\sum_{j} a_{j}^{2} σ_{j}^{2} x_{j} \leq 1$ . Thus $m_{η} \geq \frac{\sum_{j} {\tilde{η}}_{j}^{4}}{ε^{2} \sqrt{2 \sum_{j} {\tilde{η}}_{j}^{4}}} = u_{ε} .$

3.1F1F5step 1.1step 2.1algebra

Suppose $u_{ε} \to \infty$ and fix $c \in (0, 1)$ . Applying [F5] under the null gives $α_{ε} (ψ_{ε, c u_{ε}}) \leq (c u_{ε})^{- 2} \to 0$ . For any alternative, step 2.1 and step 1.1 give $P_{η} (T_{ε} \leq c u_{ε}) \leq \frac{1 + 4 w_{ε, 0} m_{η}}{(m_{η} - c u_{ε})^{2}} \leq \frac{1}{(1 - c)^{2} u_{ε}^{2}} + \frac{4 w_{ε, 0}}{(1 - c)^{2} u_{ε}} .$ The final bound uses that $m / (m - c u)^{2}$ decreases for $m \geq u$ and $w_{ε, 0} \leq 1 / \sqrt{2}$ , which follows from $\sum_{j} w_{ε, j}^{2} = 1 / 2$ . It tends to zero uniformly over $Θ_{2} (r_{ε})$ , proving consistency.

4.1F2givenalgebra∎

Substituting the profile into the active constraints gives $r_{ε}^{2} = z_{0}^{2} J_{1}$ and $1 = z_{0}^{2} J_{2} / A$ . Hence $r_{ε}^{2} = A J_{1} / J_{2}$ . Also $\sum_{j} {\tilde{η}}_{j}^{4} = z_{0}^{4} J_{0}$ , and eliminating $z_{0}^{2} = r_{ε}^{2} / J_{1}$ gives $u_{ε}^{2} = (r_{ε} / ε)^{4} J_{0} / (2 J_{1}^{2})$ .

Depends on

Used by

Sharp Gaussian asymptotics under a vanishing largest weight Corollary
Mildly ill-posed analytic q=2 detection rate Example
Mildly ill-posed Sobolev q=2 detection rate Example
Severely ill-posed analytic q=2 detection rate Example
Severely ill-posed Sobolev q=2 detection rate Example

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Sources

Yuri I. Ingster, Theofanis Sapatinas and Irina A. Suslina, Minimax signal detection in ill-posed inverse problems, §2 and §4.1, PDF pp. 4–7 (standard reference, not scraped)
Evarist Giné and Richard Nickl, Mathematical Foundations of Infinite-Dimensional Statistical Models, §6.2.3, PDF pp. 506–508 (standard reference, not scraped)
Clément Marteau and Theofanis Sapatinas, A unified treatment for non-asymptotic and asymptotic approaches to minimax signal detection, §§3.1–3.2 (standard reference, not scraped)