Sickness Absence Durations: Kaplan-Meier, the Exponential Assumption and Informative Censoring
An insurer studies the duration in months from the start of a sickness absence to return to work for 12 income protection claimants. The recorded durations are 1, 2, 2, 4, 5*, 6, 7, 8*, 9, 10*, 12 and 15*, where an asterisk marks a censored observation: the claimant was still absent, or had left the insurer's book, at the end of follow-up. Two of the censored claimants left the book after taking overseas employment; the other two were still absent when the study closed. No claimant was lost to follow-up before month 5. The claims manager fits an exponential distribution by maximum likelihood, obtaining a mean duration of 10.1 months and a median of 7.0 months, and proposes to use the exponential median as the central estimate for reserving. You are asked to produce a non-parametric estimate of the survival function, to check the constant-hazard assumption, and to advise on whether the censoring can be assumed non-informative.
1. Construct the Kaplan-Meier estimate of the survival function and give the estimated median duration. Compare $\hat S(12)$ with the exponential model's value.
Read the worked answer
Ordered event times are 1, 2, 2, 4, 6, 7, 9, 12 (eight events). Risk sets account for the censoring at 5, 8, 10 and 15.
| $t$ | at risk $n$ | events $d$ | $1-d/n$ | $\hat S(t)$ |
|---|---|---|---|---|
| 1 | 12 | 1 | 11/12 | 0.9167 |
| 2 | 11 | 2 | 9/11 | 0.7500 |
| 4 | 9 | 1 | 8/9 | 0.6667 |
| 6 | 7 | 1 | 6/7 | 0.5714 |
| 7 | 6 | 1 | 5/6 | 0.4762 |
| 9 | 4 | 1 | 3/4 | 0.3571 |
| 12 | 2 | 1 | 1/2 | 0.1786 |
The censoring at month 5 removes a claimant from the risk set before the event at month 6 (8 remain, then 7), and similarly at 8 (5 then 4) and at 10 (3 then 2). After month 12 only one claimant, censored at 15, remains, so the curve is not estimated beyond that point.
The median is the smallest $t$ with $\hat S(t)\le0.5$: $\hat S(6)=0.571>0.5$ and $\hat S(7)=0.476\le0.5$, so the KM median is 7 months - coincidentally the same as the exponential median.
The exponential model with mean 10.1 gives $\hat S(12)=e^{-12/10.1}=0.305$, against KM $0.179$. The models agree at the median but diverge badly in the tail, which is precisely where the reserve sits.
2. Test the constant-hazard assumption using the observed data. What would you fit instead, and what would you check?
Read the worked answer
Compare empirical hazard rates over intervals, using events divided by total exposure within the interval.
- Months 0-4: four events (at 1, 2, 2, 4). Exposure is $\sum\min(T_i,4)=41$ claimant-months, so $\hat h=4/41=0.0976$ per month.
- Months 9-15: two events (at 9 and 12). The risk set entering month 9 is the claimant with the event at 9, the censored claimant at 10, the claimant with the event at 12 and the censored claimant at 15; exposure over [9,15] is $0+1+3+6=10$, so $\hat h=2/10=0.200$ per month.
The hazard roughly doubles, which is consistent with recovery becoming more likely as absences lengthen (or with the long-term absentees being a selected frailer group), and inconsistent with the constant $\hat\lambda=8/81=0.0988$ implied by the exponential fit.
I would fit a Weibull model, $h(t)=\lambda\gamma t^{\gamma-1}$, with $\gamma>1$, or a Gompertz if the rise is exponential in time, and check by plotting $\log\hat H(t)$ from the Nelson-Aalen estimator against $\log t$ - a straight line supports Weibull. If claimant covariates (age, occupation, waiting period) are available, a Cox model with a non-parametric baseline is preferable to a parametric form chosen by eye.
3. Advise on whether censoring can be treated as non-informative here, quantify the risk, and state what you would do.
Read the worked answer
Kaplan-Meier requires that, conditional on covariates, the censoring time be independent of the event time. The two claimants who left for overseas employment are the concern. The most plausible reason for emigration is that they had recovered and found work abroad; if so, their censoring is associated with a return to work rather than independent of it. Those claimants then leave the risk set without being counted as events, so the hazard in later months is underestimated and $\hat S$ is too high - the reserve is understated. The two administratvely censored claimants still absent at study closure are benign provided the closure date is unrelated to recovery.
Quantify the uncertainty before drawing conclusions: by Greenwood's formula, $\operatorname{Var}(\hat S(7))=\hat S(7)^2\sum d_i/[n_i(n_i-d_i)]=0.2268\times0.0988=0.0224$, so SE 0.150 and an approximate 95% interval for $S(7)$ of roughly 0.18 to 0.77. With eight events the estimate is very imprecise; two misclassified censored observations could move it materially.
Actions: trace the overseas claimants and record whether they returned to work and when; if the withdrawal is genuinely informative, model recovery and withdrawal as competing risks and report the cumulative incidence of recovery, $\int \hat S(u),d\hat\Lambda_{\text{recovery}}(u)$, rather than $1-\hat S$; and present a sensitivity analysis in which the withdrawals are treated as events.