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Diagonal scalar bounds miss off-diagonal matrix size

Example

Assume AC and fix $δ \in (0, 1)$ . For each real $r > 0$ , let $A_{t} = A^{(r)} = (\begin{array}{cc} 0 & r \\ r & 0 \end{array})$ for every $t \geq 1$ . Define the Euclidean operator norm by $∥ A ∥_{o p} = \sup_{∥ x ∥_{2} = 1} ∥ A x ∥_{2}$ , where $∥ x ∥_{2} = \sqrt{x_{1}^{2} + x_{2}^{2}}$ . Both diagonal entry streams are identically zero; with scalar bounds $a = b = 0$ and the constant predictable forecasts ${\hat{X}}_{t} = 0$ , the scalar theorem returns $[0, 0]$ for both coordinates at every time, independently of $r$ . Yet $∥ A^{(r)} ∥_{o p} = r$ , which is unbounded across this family. An operator-norm confidence sequence at level $1 - δ$ consists of matrix estimates ${\hat{A}}_{t}$ and almost-surely finite nonnegative radii $R_{t}$ satisfying $P (∥ A_{t} - {\hat{A}}_{t} ∥_{o p} \leq R_{t} for every t) \geq 1 - δ$ . Any rule using only the two diagonal interval streams (and, if randomized, auxiliary randomness whose law is independent of $r$ ) has the same time-one output law for every $r$ . Such a rule cannot have the stated coverage uniformly over this family. A matrix-valued conclusion needs separate assumptions and an argument for the matrix process.

Example references

The scalar theorem's domain and interval guarantee

Facts & Assumptions

Given: Assume AC and fix $δ \in (0, 1)$ . The deterministic symmetric matrix family and any estimation rule restricted to its diagonal interval streams, with auxiliary randomness independent of $r$ , are as in the Example.

[F1]

Empirical-Bernstein confidence sequence under adaptive sampling: with bounded observations and predictable forecasts, the theorem gives simultaneous coverage for the running scalar conditional-mean target.

[F2]

Empirical-Bernstein confidence-sequence interval update: when $a = b$ , the update returns $[a, a]$ at every time.

[F3]

Continuity from above when one set has finite measure: for decreasing measurable sets under a finite measure, their measures converge to the measure of their intersection.

[A1]

The Axiom of Choice: AC is explicitly assumed by the scalar theorem.

Verification

1.1givenF1F2A1

For every $r > 0$ , each diagonal entry of $A^{(r)}$ is the constant zero. With scalar bounds $a = b = 0$ and forecast ${\hat{X}}_{t} = 0$ , each coordinate process satisfies the theorem's hypotheses; [F2] gives the returned interval $[0, 0]$ at every time, independently of $r$ .

1.2givenalgebra

For every vector $x \in R^{2}$ , $(A^{(r)})^{T} A^{(r)} = r^{2} I$ gives $∥ A^{(r)} x ∥_{2}^{2} = r^{2} ∥ x ∥_{2}^{2}$ , hence $∥ A^{(r)} ∥_{o p} = r$ .

2.1givenF3step 1.2algebra∎

Let $({\hat{A}}_{1}, R_{1})$ have the common time-one output law from the Example, with $R_{1}$ finite almost surely. For each integer $n \geq 1$ , the reverse triangle inequality gives $P (∥ A^{(n)} - {\hat{A}}_{1} ∥_{o p} \leq R_{1}) \leq P (∥ {\hat{A}}_{1} ∥_{o p} + R_{1} \geq n)$ . The events on the right decrease to the null event where a finite random variable is infinite; [F3] makes their probabilities tend to zero. Since $1 - δ > 0$ , some $n$ has time-one coverage probability below $1 - δ$ , and the all-times event is contained in the time-one event. Thus no such rule gives uniform coverage over the family.

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