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An exact logrank randomization p-value with unequal assignment weights

Statement

Fix a follow-up horizon $τ = 3$ and three subjects. For every allowed assignment $z \in {e_{1}, e_{2}, e_{3}}$ , where $γ (e_{k})_{i} = 1 {i = k}$ is the binary group map that treats subject $k$ , prescribe the full records $(U_{i} (z), δ_{i} (z)) = (i, 1)$ for $i = 1, 2, 3$ . Thus the subjects have uncensored event times $1, 2, 3$ under every assignment, satisfying the full-record sharp null. Exactly one subject is treated. The allowed assignments $e_{1}, e_{2}, e_{3}$ treat subject 1, 2, or 3, respectively, with probabilities $1 / 2, 1 / 3, 1 / 6$ . Predeclare unit weight and the two-sided absolute unweighted logrank score before observing the assignment. Enumerate all scores and inclusive assignment-tail p-values and verify superuniformity under this unequal assignment law.

Facts & Assumptions

Given: The full-record schedule $(U_{i} (z), δ_{i} (z)) = (i, 1)$ for every $i \in {1, 2, 3}$ and every allowed assignment $z \in {e_{1}, e_{2}, e_{3}}$ , the binary group map $γ (e_{k})_{i} = 1 {i = k}$ , horizon $τ = 3$ , the declared assignment law, and the predeclared unit-weight absolute logrank ordering in Two-group logrank and weighted logrank scores.

[F1]

Two-group logrank and weighted logrank scores: at each event time, using the binary group map $γ (z)$ , the score contribution is $O_{j} - d_{j} Y_{1 j} / Y_{j}$ , the treated event count minus the total event count times the treated fraction of the current risk set. Here $d_{j} = 1$ at every event time.

[F2]

The logrank ordering gives a valid Fisher randomization p-value: under a full-record sharp null, the inclusive assignment-tail p-value uses the actual assignment weights and is superuniform.

[F3]

Finite randomized treatment-assignment models: assignment weights are nonnegative and sum to one on the finite allowed assignment set.

Proof

1.1F1F2givenalgebra

The prescribed full record for each subject is identical under all three allowed assignments, so the full-record sharp-null hypothesis in [F2] holds. Under the map $γ (e_{1})$ , the event at time 1 contributes $1 - 1 / 3 = 2 / 3$ and the later risk sets contain no treated subject, so $U (γ (e_{1})) = 2 / 3$ . Under $γ (e_{2})$ , time 1 contributes $- 1 / 3$ , and time 2 contributes $1 - 1 / 2 = 1 / 2$ , giving $U (γ (e_{2})) = 1 / 6$ . Under $γ (e_{3})$ , time 1 contributes $- 1 / 3$ , time 2 contributes $- 1 / 2$ , and time 3 contributes zero, giving $U (γ (e_{3})) = - 5 / 6$ . These are the three distinct scores; their absolute values are $2 / 3, 1 / 6, 5 / 6$ .

2.1F2F3step 1.1algebra

The inclusive absolute-tail probability at $e_{1}$ is $1 / 2 + 1 / 6 = 2 / 3$ , at $e_{2}$ it is $1$ , and at $e_{3}$ it is $1 / 6$ , because the absolute scores are distinct and the assignment weights are $1 / 2, 1 / 3, 1 / 6$ . Therefore the p-value takes $2 / 3, 1, 1 / 6$ with respective probabilities $1 / 2, 1 / 3, 1 / 6$ .

3.1F2F3step 2.1algebra∎

Its distribution function is $0$ for $0 \leq u < 1 / 6$ , $1 / 6$ for $1 / 6 \leq u < 2 / 3$ , $2 / 3$ for $2 / 3 \leq u < 1$ , and $1$ at $u = 1$ . On each interval this is at most $u$ , including the endpoints. Thus the exact finite assignment calculation is superuniform under the declared unequal-weight law, as in [F2].

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