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Survival jumps multiply by one minus the atomic hazard

Statement

Assume Countable Choice. Let $T \geq 0$ have CDF $F_{T} (t) = P (T \leq t)$ and survival $S_{T} (t) = P (T > t)$ . For any $t \geq 0$ with $S_{T} (t^{-}) = P (T \geq t) > 0$ , define the atomic hazard $q_{T} (t) : = \frac{P (T = t)}{P (T \geq t)} .$ Then $S_{T} (t) = S_{T} (t^{-}) (1 - q_{T} (t)) .$

In particular, if $m \geq 1$ and the law is supported on a finite set of distinct times $0 \leq t_{1} < \dots < t_{m}$ and each listed point has positive mass, put $q_{j} : = q_{T} (t_{j})$ . Then $S_{T} (t_{j}) = \prod_{r = 1}^{j} (1 - q_{r}), 1 \leq j \leq m .$ For laws with both continuous and atomic parts, the pointwise jump identity still holds at every atom; this finite-support product is not a formula for the continuous decrease between atoms.

Facts & Assumptions

Given: Countable Choice, a nonnegative event time $T$ , and the stated finite support condition when the product formula is used.

[F1]

$F_{T} (t) = P (T \leq t)$ and $S_{T} (t) = P (T > t) = 1 - F_{T} (t)$ for $t \geq 0$ ; $S_{T} (t^{-}) : = P (T \geq t)$ . Cumulative distribution function of a real random variable, Parametric survival and reliability models

[F2]

Under Countable Choice, the law of $T$ satisfies $P (T = t) = F_{T} (t) - F_{T} (t^{-})$ . Probability laws correspond to distribution functions, Interval formulas and atoms for a Lebesgue-Stieltjes measure

[A1]

Countable Choice is used by [F2] for the CDF/measure correspondence and atom formula. The Axiom of Countable Choice ( ${A C}_{ω}$ )

Proof

technique · direct

1.1A1F1F2algebra

By [F2], $P (T = t) = F_{T} (t) - F_{T} (t^{-})$ . Also $P (T \geq t) = P (T > t) + P (T = t)$ , so [F1] gives $S_{T} (t^{-}) = 1 - F_{T} (t^{-}), P (T = t) = S_{T} (t^{-}) - S_{T} (t) .$

2.1step 1.1givenalgebra

Divide the identity in step 1.1 by the positive quantity $S_{T} (t^{-})$ . Using the definition of $q_{T} (t)$ gives $q_{T} (t) = 1 - \frac{S_{T} (t)}{S_{T} (t^{-})},$ which is equivalent to $S_{T} (t) = S_{T} (t^{-}) (1 - q_{T} (t))$ .

3.1step 2.1givenalgebra

In the finite-support case, $S_{T} (t_{1}^{-}) = 1$ , since $T \geq t_{1}$ . For $j > 1$ , no probability mass lies in $(t_{j - 1}, t_{j})$ , so $S_{T} (t_{j}^{-}) = P (T \geq t_{j}) = P (T > t_{j - 1}) = S_{T} (t_{j - 1})$ . Applying step 2.1 successively yields $S_{T} (t_{j}) = S_{T} (t_{j - 1}) (1 - q_{j}) = \prod_{r = 1}^{j} (1 - q_{r}),$ where the initial factor is $S_{T} (t_{1}^{-}) = 1$ .

4.1step 2.1step 3.1∎

Step 2.1 proves the jump identity for every atom with positive pre-jump survival, and step 3.1 proves the finite atomic product formula.

Depends on

Used by

Exponential and Weibull hazards with right-censored exposure Example

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Sources

Qiqing Yu, Survival Analysis (Lecture Notes), §2.1, PDF p. 10 (standard reference, not scraped)
NIST/SEMATECH Engineering Statistics Handbook, Chapter 8, §8.1.2.3, PDF p. 16 (standard reference, not scraped)