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Uniform convergence of the risk-set treatment fraction

Statement

For each $g \in {0, 1}$ , let $(U_{g i})_{i \geq 1}$ be IID measurable $[0, \infty)$ -valued observed times with law $P_{g}$ ; independence between the two groups is not required. Along a sequence of deterministic positive sample sizes, let $n = n_{0} + n_{1} \to \infty$ and $n_{g} / n \to ρ_{g} \in (0, 1)$ , where $ρ_{0} + ρ_{1} = 1$ . Set $s_{g} (t) : = P_{g} (U_{g 1} \geq t)$ , $y_{g} (t) = ρ_{g} s_{g} (t)$ , and $y (t) = y_{0} (t) + y_{1} (t)$ . Assume $\inf_{0 \leq t \leq τ} y (t) > 0$ . Write $Y_{g, n} (t) = \sum_{i = 1}^{n_{g}} 1 {U_{g i} \geq t}, p_{n} (t) = {\begin{cases} Y_{1, n} (t) / (Y_{0, n} (t) + Y_{1, n} (t)), & Y_{0, n} (t) + Y_{1, n} (t) > 0, \\ 0, & Y_{0, n} (t) + Y_{1, n} (t) = 0. \end{cases}$ Then, for $g = 0, 1$ , $\sup_{t \leq τ} ∣ \frac{Y_{g, n} (t)}{n} - y_{g} (t) ∣ ⟶_{P} 0,$ and $\sup_{0 \leq t \leq τ} ∣ p_{n} (t) - \overset{ˉ}{z} (t) ∣ ⟶_{P} 0, \overset{ˉ}{z} (t) : = \frac{y_{1} (t)}{y (t)} .$

Facts & Assumptions

Given: The within-group IID observed-time sequences, the deterministic allocation limits and positive pooled support in the Statement, and the groupwise uniform empirical-risk convergence.

[F1]

Uniform convergence of empirical risk-set fractions: for each group, $\sup_{t \leq τ} ∣ Y_{g, n} (t) / n_{g} - s_{g} (t) ∣ \to_{P} 0$ .

[F2]

Two-group logrank and weighted logrank scores: $Y_{g, n} (t)$ counts subjects with observed time at least $t$ ; risk at event time is counted immediately before failures at that time.

[A1]

The pooled limiting risk fraction obeys $\inf_{t \leq τ} y (t) = m > 0$ .

Proof

1.1F1givenalgebra

By [F1], for each $g$ , $\sup_{t \leq τ} ∣ \frac{Y_{g, n} (t)}{n} - y_{g} (t) ∣ \leq ∣ \frac{n_{g}}{n} - ρ_{g} ∣ \sup_{t \leq τ} \frac{Y_{g, n} (t)}{n_{g}} + ρ_{g} \sup_{t \leq τ} ∣ \frac{Y_{g, n} (t)}{n_{g}} - s_{g} (t) ∣ ⟶_{P} 0.$ The first term tends to zero deterministically, and the empirical fraction in the second term is at most one. Hence the pooled fraction $Y_{n} (t) / n$ converges uniformly in probability to $y (t)$ and the group-1 fraction $Y_{1, n} (t) / n$ converges uniformly to $y_{1} (t)$ .

2.1F2A1step 1.1algebra∎

On the event $\inf_{t \leq τ} Y_{n} (t) / n \geq m / 2$ , whose probability tends to one by [A1] and step 1.1, the denominator is positive for all $t \leq τ$ . There, $∣ \frac{Y_{1, n} (t)}{Y_{n} (t)} - \frac{y_{1} (t)}{y (t)} ∣ \leq \frac{2}{m} ∣ \frac{Y_{1, n} (t)}{n} - y_{1} (t) ∣ + \frac{2}{m} ∣ \frac{Y_{n} (t)}{n} - y (t) ∣,$ because $0 \leq y_{1} (t) / y (t) \leq 1$ . Taking suprema and using step 1.1 proves convergence to zero in probability. The convention $p_{n} = 0$ on an empty risk set is immaterial on this event.

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Sources

Lu Tian, STAT331 Unit 13: Large Sample Theory for 2-Sample Tests, PDF pp. 1-6 (standard reference, not scraped)
Michal Kulich, NMST531 Censored Data Analysis: Extended Course Notes, §5.4, PDF pp. 61-62 (standard reference, not scraped)