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Sequential Validity and Ville's Inequality — Examples

1 · Prerequisites

2 · Summary

Examples

A finite Bernoulli likelihood ratio gives a concrete null test martingale. Its two-observation version shows an exact data-dependent first-crossing probability below Ville's bound. The counterexample separates validity under one null law from the required supermartingale condition under every law in a composite null family.

3 · Logical flowchart

4 · Definitions, theorems and proofs

None yet.

5 · Examples, counterexamples and false statements

ExampleConstruction: AI-generatedVerification: AI-generatedprecheck passjudge pass (gpt-6-luna)Open item page →

A finite-horizon likelihood-ratio test supermartingale

Example

Fix an integer $K \geq 1$ . Let $Ω = {0, 1}^{K}$ and let $P_{0}$ be uniform on $Ω$ , so the coordinate variables $X_{1}, \dots, X_{K}$ are independent Bernoulli $(1 / 2)$ . Let $Q$ be the product Bernoulli $(3 / 4)$ law and let $F_{t} = σ (X_{1}, \dots, X_{\min (t, K)})$ . The likelihood ratio of $Q$ to $P_{0}$ on $F_{t}$ is $E_{t} = \prod_{i = 1}^{\min (t, K)} ℓ (X_{i}), ℓ (1) = \frac{3}{2}, ℓ (0) = \frac{1}{2},$ with the empty product $E_{0} = 1$ . It is a nonnegative test martingale, hence a nonnegative test supermartingale, for the simple null ${P_{0}}$ .

Facts & Assumptions

Given: $K \geq 1$ , the finite sample space, laws, filtration, and process in the Example. Assume AC for the test-supermartingale convention.

[F1]

Nonnegative test supermartingales: the process must be real-valued, adapted, integrable, nonnegative, supermartingale under the null, and have initial expectation at most one.

[A1]

The Axiom of Choice: AC supplies the conditional-expectation convention used in F1.

Verification

1.1givenconstruct

For a prefix $x_{1}, \dots, x_{t}$ with $t \leq K$ , the ratio of its $Q$ -probability to its $P_{0}$ -probability is $\prod_{i = 1}^{t} (Q (X_{i} = x_{i}) / P_{0} (X_{i} = x_{i})) = \prod_{i = 1}^{t} ℓ (x_{i})$ , because every $P_{0}$ prefix has positive probability. Thus the displayed process is the finite-prefix likelihood ratio; after time $K$ it is held constant.

2.1givenstep 1.1algebra

Each $E_{t}$ is a finite nonnegative function of the first $\min (t, K)$ coordinates, hence is adapted and integrable. Conditional on any prefix before time $K$ , the next bit is fair and $E_{P_{0}} [ℓ (X_{t + 1}) ∣ F_{t}] = (3 / 2 + 1 / 2) / 2 = 1$ , so $E_{P_{0}} [E_{t + 1} ∣ F_{t}] = E_{t}$ . For $t \geq K$ the process is constant.

3.1F1A1step 1.1step 2.1∎

Since $E_{0} = 1$ , steps 1.1 and 2.1 verify every condition in F1 under the null law $P_{0}$ ; hence $E$ is a test martingale and a test supermartingale.

ExampleConstruction: AI-generatedVerification: AI-generatedprecheck passjudge pass (gpt-6-luna)Open item page →

A data-dependent threshold crossing

Example

Take two fair coin observations under the null $P_{0}$ , with the likelihood-ratio process $E_{0} = 1$ , $E_{1} = ℓ (X_{1})$ , $E_{2} = ℓ (X_{1}) ℓ (X_{2})$ and $E_{t} = E_{2}$ for $t \geq 2$ , where $ℓ (1) = 3 / 2$ and $ℓ (0) = 1 / 2$ . Set $τ = \inf {t \geq 0 : E_{t} \geq 2}$ . This is a data-dependent stopping rule. It crosses only on two heads, so $P_{0} (τ < \infty) = 1 / 4$ . Since $2 = 1 / α$ for $α = 1 / 2$ , the crossing probability is below Ville's bound $1 / 2$ .

Facts & Assumptions

Given: Two independent fair bits, their prefix filtration, the displayed likelihood-ratio process, and threshold $2$ . Assume AC for the cited test-supermartingale and optional-sampling conventions.

[F1]

Nonnegative test supermartingales: a test supermartingale is nonnegative, adapted, integrable, supermartingale under the null, and has initial expectation at most one.

[F2]

Ville's inequality for nonnegative test supermartingales: for $α \in (0, 1)$ , the probability that a nonnegative test supermartingale ever reaches $1 / α$ is at most $α$ .

[F3]

Discrete stopping time: a discrete-time map is a stopping time when each event ${τ \leq n}$ belongs to $F_{n}$ .

[A1]

The Axiom of Choice: AC is inherited by the conditional-expectation clauses used in F1 and F2.

Verification

1.1givenF1A1algebra

The one-step likelihood factor has null conditional mean $(3 / 2 + 1 / 2) / 2 = 1$ after either possible prefix. Thus $E_{0} = 1$ , $E_{1}$ and $E_{2}$ are nonnegative, adapted, integrable, and satisfy the supermartingale equality; the process is constant after time two. It is a test supermartingale under $P_{0}$ by F1.

2.1givenF2F3choosealgebra∎

For every $n \geq 0$ , ${τ \leq n} = ⋃_{s = 0}^{n} {E_{s} \geq 2}$ . This is empty for $n < 2$ and equals ${X_{1} = X_{2} = 1} \in F_{2} \subseteq F_{n}$ for $n \geq 2$ , so $τ$ is a stopping time by F3. The values $(E_{1}, E_{2})$ on $(X_{1}, X_{2}) = (1, 1), (1, 0), (0, 1), (0, 0)$ are respectively $(3 / 2, 9 / 4), (3 / 2, 3 / 4), (1 / 2, 3 / 4), (1 / 2, 1 / 4)$ . Thus the crossing event ${τ < \infty}$ is exactly ${X_{1} = X_{2} = 1}$ and has probability $1 / 4$ . Since $2 = 1 / α$ for $α = 1 / 2$ , F2 gives the general upper bound $1 / 2$ , and the exact probability is below it.

CounterexampleConstruction: AI-generatedVerification: AI-generatedprecheck passjudge pass (gpt-6-luna)Open item page →

Validity under one null law does not give a composite-null test supermartingale

Statement refuted

The claim “a nonnegative process that is a supermartingale under one member of a composite null family is a test supermartingale for the whole family” is false.

Facts & Assumptions

Given: The two Bernoulli laws, filtration, process, and composite null family in the Counterexample. Assume AC for the test-supermartingale convention.

[F1]

Nonnegative test supermartingales: a test supermartingale must satisfy the supermartingale and initial-expectation conditions under every null law.

[A1]

The Axiom of Choice: AC is inherited by the conditional-expectation convention in F1.

Counterexample

Let $Ω = {0, 1}$ , let $P_{1 / 4} (X = 1) = 1 / 4$ and $P_{3 / 4} (X = 1) = 3 / 4$ , and take the composite null family $P_{0} = {P_{1 / 4}, P_{3 / 4}}$ . Let $F_{0} = {\emptyset, Ω}$ and $F_{t} = 2^{Ω}$ for $t \geq 1$ . Set $M_{0} = 1$ and $M_{t} = 2 X$ for $t \geq 1$ .

1.1givenF1A1algebra

Under $P_{1 / 4}$ , $F_{0}$ is trivial, so $E_{P_{1 / 4}} [M_{1} ∣ F_{0}] = E_{P_{1 / 4}} [2 X] = 1 / 2 \leq M_{0}$ . At all later times $M_{t}$ is constant in time. The process is nonnegative, adapted, integrable, and starts at one, so it is a test supermartingale for the singleton null ${P_{1 / 4}}$ .

1.2givenalgebra

Under $P_{3 / 4}$ , $E_{P_{3 / 4}} [M_{1} ∣ F_{0}] = E_{P_{3 / 4}} [2 X] = 3 / 2 > M_{0} = 1$ , so the supermartingale inequality fails already at time zero.

2.1F1step 1.2∎

By F1, a test supermartingale for $P_{0}$ must satisfy the supermartingale condition under both laws. Step 1.2 shows failure under $P_{3 / 4}$ , so the proposed process is not a test supermartingale for the composite null.