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Exact Bernoulli crossings in a sequential likelihood-ratio test

Example

Assume AC. Consider IID observations $X_{i} \in {0, 1}$ . Under $H_{0}$ , let $P_{0} (X_{i} = 1) = 1 / 4$ ; under $H_{1}$ , let $P_{1} (X_{i} = 1) = 3 / 4$ . Use the procedure Sequential likelihood-ratio test with $α = β = 1 / 4$ . A type-I error means deciding $H_{1}$ when $P_{0}$ governs the observations; a type-II error means deciding $H_{0}$ when $P_{1}$ governs them. The terminal type-I and type-II error probabilities are each $1 / 10$ , and the test stops almost surely under either law.

Facts & Assumptions

Given: A common measurable space $(Ω, F)$ carrying the ${0, 1}$ -valued observations, the Bernoulli laws and thresholds in the Example, AC, and the sequential likelihood-ratio procedure. For applying the procedure, use $F_{0} = {\emptyset, Ω}$ and $F_{t} = σ (X_{1}, \dots, X_{t})$ for $t \geq 1$ . Compute both terminal error probabilities and verify almost-sure stopping.

[A1]

The Axiom of Choice: AC states that every family of nonempty sets has a choice function.

[F1]

Identical distribution and IID families: IID observations are mutually independent and share a common law under each hypothesis.

[F2]

Sequential likelihood-ratio test: the procedure requires mutually equivalent restrictions on every $F_{t}$ , including $t = 0$ ; the upper boundary is $1 / α$ and decides $H_{1}$ , while the lower boundary is $β$ and decides $H_{0}$ .

[F3]

Finite-sample likelihood ratio for iid observations: for IID observations, $L_{t} = \prod_{i = 1}^{t} d Q_{1} / d Q_{0} (X_{i})$ is the finite-time likelihood-ratio density.

[F4]

Terminal error bounds for a sequential likelihood-ratio test: the two terminal error probabilities are bounded by $α$ and $β$ .

[F5]

Iid drift conditions imply almost-sure test termination: finite absolute log-likelihood means and strict negative/positive drifts imply almost-sure termination under $P_{0}$ and $P_{1}$ .

Verification

1.1givenconstruct

The one-observation likelihood ratio is $λ (1) = (3 / 4) / (1 / 4) = 3$ and $λ (0) = (1 / 4) / (3 / 4) = 1 / 3$ . Both laws give positive mass to both outcomes, so they are equivalent.

2.1F1F2F3step 1.1givenalgebra

For $t \geq 1$ , set $G_{t} = \prod_{i = 1}^{t} λ (X_{i})$ and choose the algorithm's representatives to be $L_{t} = G_{t}$ , with $L_{0} = 1$ . By [F3], $G_{t}$ is a version of the finite-time density. It is finite and strictly positive because each factor is either $3$ or $1 / 3$ , so for every $A \in F_{t}$ , $P_{1} (A) = \int_{A} G_{t} d P_{0}$ implies $P_{1} (A) = 0$ exactly when $P_{0} (A) = 0$ . At $t = 0$ , both restrictions to the trivial sigma-algebra assign zero to $\emptyset$ and one to $Ω$ , so they are equivalent and $L_{0} = 1$ is their density. The algorithm's finite-time equivalence hypotheses therefore hold for every $t \geq 0$ . Also, each nonempty atom of $F_{t}$ is a fiber for a binary word and has positive $P_{0}$ -probability by [F1]; consequently any other $F_{t}$ -measurable density representative agrees with $G_{t}$ everywhere, since it agrees $P_{0}$ -almost surely.

3.1A1givenF5step 1.1step 2.1algebra

The log increment is $Z (1) = \log 3$ and $Z (0) = - \log 3$ . Hence $E_{Q_{0}} ∣ Z ∣ = E_{Q_{1}} ∣ Z ∣ = \log 3 < \infty$ , $E_{Q_{0}} Z = - (1 / 2) \log 3 < 0$ , and $E_{Q_{1}} Z = (1 / 2) \log 3 > 0$ . The hypotheses of [F5] hold by steps 1.1 and 2.1, so its general termination theorem gives almost-sure stopping under both laws.

3.2step 1.1step 2.1algebra

Put $S_{t} = \sum_{i = 1}^{t} X_{i}$ . By step 2.1 and the values in step 1.1, $L_{t} = \prod_{i = 1}^{t} λ (X_{i}) = 3^{S_{t}} (1 / 3)^{t - S_{t}} = 3^{2 S_{t} - t}$ .

4.1F2step 3.2algebra

The boundaries are $4$ and $1 / 4$ . Starting a two-observation block at $L = 1$ , the first observation gives $L = 3$ or $1 / 3$ , both strictly between the boundaries. A repeated outcome in the second observation gives $L = 9$ for $11$ or $L = 1 / 9$ for $00$ , crossing the upper or lower boundary; a mixed pair multiplies by $1$ and returns to $L = 1$ . Induction over successive disjoint pairs shows that the procedure stops at the first pair with equal outcomes.

5.1F1step 4.1algebra

Under either hypothesis, a pair is mixed with probability $2 (1 / 4) (3 / 4) = 3 / 8$ . The disjoint pairs are independent by [F1], so the probability that the first $k$ pairs are all mixed is $(3 / 8)^{k} \to 0$ . Thus the pair calculation also verifies almost-sure stopping.

5.2F1step 4.1algebra

Under $P_{0}$ , each disjoint pair is $11$ with probability $1 / 16$ , $00$ with probability $9 / 16$ , and mixed with probability $3 / 8$ . The event of a type-I error is that the first pair that is not mixed is $11$ , so $P_{0} (type-I error) = \sum_{k = 1}^{\infty} (3 / 8)^{k - 1} (1 / 16) = (1 / 16) / (1 - 3 / 8) = 1 / 10$ . Under $P_{1}$ , the probability of $00$ is $1 / 16$ and of a mixed pair is again $3 / 8$ , so the event of a type-II error gives the same geometric sum and probability $1 / 10$ .

6.1A1F4step 2.1step 5.2algebra∎

Since $1 / 10 \leq 1 / 4 = α = β$ , these exact terminal errors satisfy both bounds in [F4]; the hypotheses for that theorem's procedure hold by step 2.1.

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