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Finite-support risk-set proportion limits

Statement

For every positive integer sample size $n$ , let $(T_{i}, C_{i})$ be IID subject pairs with $T_{i}$ independent of $C_{i}$ , and let the event-time support be $0 \leq a_{1} < \dots < a_{m}$ . Fix $j$ and suppose $P (T \geq a_{j}) > 0$ and $P (C \geq a_{j}) > 0$ . Put $q_{j} = P (T = a_{j} ∣ T \geq a_{j})$ , $r_{j} = P (U \geq a_{j})$ , $Y_{n} (a_{j}) = \sum_{i = 1}^{n} 1 {U_{i} \geq a_{j}}$ , and $d N_{n} (a_{j}) = \sum_{i = 1}^{n} 1 {U_{i} = a_{j}, δ_{i} = 1}$ . Then $Y_{n} (a_{j}) / n \to_{P} r_{j} = P (T \geq a_{j}) P (C \geq a_{j}) > 0$ , $d N_{n} (a_{j}) / n \to_{P} P (T = a_{j}) P (C \geq a_{j})$ , and the ratio $d N_{n} (a_{j}) / Y_{n} (a_{j})$ , set to zero when its denominator is zero, converges in probability to $q_{j}$ .

Facts & Assumptions

Given: The right-censoring convention, IID subject pairs, within-subject independence of $T$ and $C$ , and positive risk probability at $a_{j}$ .

[F1]

Under right censoring, $U_{i} = \min (T_{i}, C_{i})$ and $δ_{i} = 1 {T_{i} \leq C_{i}}$ (Event-time observation, censoring, and truncation schemes); the at-risk process is $Y_{i} (u) = 1 {U_{i} \geq u}$ (Nelson–Aalen, Kaplan–Meier, and Greenwood estimators).

[F2]

IID means a common law and mutual independence across the entire subject family. (Identical distribution and IID families)

[A1]

Within each subject, $T_{i}$ and $C_{i}$ are independent, and $P (T \geq a_{j}) P (C \geq a_{j}) > 0$ .

Proof

1.1A1F1given

Let $D_{i j} = 1 {U_{i} = a_{j}, δ_{i} = 1}$ . By [F1], $Y_{i} (a_{j}) = 1 {T_{i} \geq a_{j}, C_{i} \geq a_{j}}$ and $D_{i j} = 1 {T_{i} = a_{j}, C_{i} \geq a_{j}}$ . Independence in [A1] therefore gives $E [Y_{i} (a_{j})] = r_{j} = P (T \geq a_{j}) P (C \geq a_{j})$ and $E [D_{i j}] = P (T = a_{j}) P (C \geq a_{j})$ .

2.1F2step 1.1algebra

The two indicators are bounded IID sequences by [F2]. For either sequence, the variance of its sample mean is at most $1 / (4 n)$ , so Chebyshev's inequality gives convergence in probability to its expectation. Since there are finitely many support points, these convergences hold jointly over $j = 1, \dots, m$ .

3.1A1step 2.1algebra∎

Since $r_{j} > 0$ , with probability tending to one $Y_{n} (a_{j}) / n > r_{j} / 2$ . On that event the ratio is the quotient of the two convergent proportions from step 2.1, and its limit is $\frac{P (T = a_{j}) P (C \geq a_{j})}{P (T \geq a_{j}) P (C \geq a_{j})} = P (T = a_{j} ∣ T \geq a_{j}) = q_{j}$ . The zero-denominator event has probability tending to zero by the same risk-proportion convergence, so the stated zero convention does not change the limit.

Depends on

Used by

Finite-support consistency of Nelson–Aalen and Kaplan–Meier Theorem

Dependency tree · two levels

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Sources

Qiqing Yu, Survival Analysis (Lecture Notes), §4.2.2, PDF pp. 35–36 (standard reference, not scraped)