Why Estimate the Variance of Cumulative Hazard?
A Nelson-Aalen cumulative hazard estimate is built by adding estimated hazard increments across event times. The estimate describes the accumulated hazard, but it is also subject to sampling variation: a different sample of individuals could produce different failure counts and risk-set sizes. A variance estimate quantifies that uncertainty on the cumulative-hazard scale and can be used to assess the precision of the estimate.
The calculation is local at each observed failure time. It uses the number of failures at that time and the number of individuals at risk immediately before it. A key point is that tied failures share one risk set: the failure count is squared nowhere in the contribution, and the tied failures are not processed one at a time with successively smaller denominators.
Here, “at risk” means that the individual has not failed and has not been censored before the event time in question. A person censored before \(t_j\) is not included in \(Y_j\); a person censored after \(t_j\) is included in the risk set just before \(t_j\). Censoring times do not themselves add variance contributions because the cumulative hazard estimate has increments at event times.
The formula is an estimate of the sampling variance of \(\widehat H(t)\), not the variance of individual lifetimes or of the event count by itself. It is commonly used when censoring is non-informative for the event process and the observed risk sets are appropriate for estimating the hazard. Like other large-sample variance estimates, it is an approximation to sampling uncertainty, not an exact finite-sample guarantee.
Why the Event-Time Contributions Have This Form
The Nelson-Aalen increment at event time \(t_j\) is \(d_j/Y_j\). In the counting-process formulation, the variation in that increment is approximately the variation in the event count, scaled by \(1/Y_j\) on each side of a variance calculation. The estimated event-count variation at that time is represented by \(d_j\). Thus the estimated variance contribution is \(d_j(1/Y_j)^2=d_j/Y_j^2\). Contributions are added over event times because the cumulative estimate is a sum of increments and, under the usual counting-process assumptions, the increment variations accumulate.
This gives a practical rule: at every failure time, take the number of failures, divide by the square of the risk-set size immediately before that time, and then sum the results. The estimate of the cumulative hazard itself uses \(d_j/Y_j\), which is a different quantity. Keeping those two denominators distinct prevents a frequent calculation error.
When \(d_j>1\), all failures tied at \(t_j\) are counted together in the numerator, while the denominator remains the square of the risk set just before \(t_j\). For example, if \(Y_j=12\) and \(d_j=3\), the variance contribution is \(3/12^2=3/144\). It is not calculated as three separate sequential contributions with denominators \(12^2\), \(11^2\), and \(10^2\).
Worked Example: Summing Contributions Across Three Event Times
In a hypothetical follow-up study, the observed event times have the following risk-set sizes and failure counts. Find the Nelson-Aalen cumulative hazard estimate and its variance estimate through the third event time, and then calculate the estimated standard error.
At the first event time, \(Y_1=20\) and \(d_1=2\). At the second, \(Y_2=15\) and \(d_2=3\). At the third, \(Y_3=10\) and \(d_3=1\). Assume independent, non-informative censoring for the event process.
Method. Calculate the hazard increment \(d_j/Y_j\) and variance contribution \(d_j/Y_j^2\) at each time, then sum the columns separately.
Cumulative hazard estimate. The three increments are \(2/20=0.10\), \(3/15=0.20\), and \(1/10=0.10\). Therefore, \(\widehat H(t)=0.10+0.20+0.10=0.40\).
Variance contributions. They are \(2/20^2=2/400=0.005\), \(3/15^2=3/225=1/75=0.013333\ldots\), and \(1/10^2=1/100=0.01\). Their sum is \(1/200+1/75+1/100\). With denominator 600, this is \(3/600+8/600+6/600=17/600=0.028333\ldots\).
Estimated standard error. Use the unrounded variance sum: \(\sqrt{17/600}\approx0.168325\). The estimated standard error is approximately 0.1683, to four decimal places.
Check and trap avoided. Each variance term uses the failure count divided by the squared risk set; the middle term is \(3/225\), not \(3/15\). The variance is positive and the standard error squared returns \(17/600\), as required.
Handling Tied Failures
Ties occur when two or more individuals have the same recorded failure time. For this variance calculation, group the tied failures into one event-time record. The value \(Y_j\) is measured immediately before that time, so it includes all individuals still at risk just before the tied failures occur. Then use the full tied count \(d_j\) in the numerator.
This convention is consistent with the Nelson-Aalen increment \(d_j/Y_j\) at a tied time. It also makes clear why tied failures are not treated as if the first failure occurred just before the second: the data record one event time, not an observed ordering within that time.
Worked Example: A Tied Event Time and a Later Failure
A hypothetical cohort has 12 individuals at risk immediately before time 4, when 3 failures occur together. After that time, 1 individual is censored before time 7. At time 7, 2 failures occur among the 8 individuals then at risk. Calculate the cumulative hazard estimate and variance estimate through time 7, and find the estimated standard error.
Time 4. The hazard increment is \(3/12=1/4=0.25\). Its variance contribution is \(3/12^2=3/144=1/48=0.0208333\ldots\).
Risk set at time 7. After the 3 failures at time 4, 9 individuals remain. The stated censoring removes 1 before time 7, leaving \(Y_2=8\). The later failure count is \(d_2=2\).
Time 7. The hazard increment is \(2/8=1/4=0.25\). Its variance contribution is \(2/8^2=2/64=1/32=0.03125\).
Totals. The cumulative hazard estimate is \(0.25+0.25=0.50\). The variance estimate is \(1/48+1/32=2/96+3/96=5/96=0.0520833\ldots\). The estimated standard error is \(\sqrt{5/96}\approx0.228218\), or 0.2282 to four decimal places.
Check and trap avoided. The stated censoring correctly reduces the risk set from 9 to 8 before time 7. The three tied failures at time 4 contribute \(3/12^2\) as one grouped term, not as three sequential failures with different denominators.
From Risk-Set Data to the Final Estimate
In a life table or event-history summary, risk-set sizes may already be supplied. If they are not, determine them from the study records: start with the number under observation, remove failures and censorings that occur before the event time being considered, and retain individuals whose censoring occurs after that event time. Then calculate the variance contribution only for rows with failures.
The following sequence helps keep the calculation organized, especially when there are censorings between event times or tied failures.
Include event times at or before the requested time \(t\); do not include later failures.
For every included event time, identify \(Y_j\) immediately before the time and the full tied count \(d_j\).
Compute \(d_j/Y_j^2\), keeping enough digits to avoid premature rounding.
The sum is the variance estimate. Its square root is the estimated standard error, not another variance estimate.
Worked Example: Accounting for Censoring Before Later Events
In a hypothetical study, 30 individuals begin follow-up. At time 2, 1 failure occurs. Five other individuals are censored after time 2 but before time 5, when 4 failures occur. Eight individuals are then censored before time 8, when 2 failures occur. Find the cumulative hazard and variance estimates through time 8, along with the estimated standard error.
Determine the risk sets. At time 2, all 30 are still under observation, so \(Y_1=30\), with \(d_1=1\). After that failure and the five subsequent censorings, \(30-1-5=24\) individuals are at risk at time 5. Thus \(Y_2=24\), with \(d_2=4\). After the four failures at time 5 and the eight subsequent censorings, \(24-4-8=12\) individuals remain at risk at time 8. Thus \(Y_3=12\), with \(d_3=2\).
Cumulative hazard estimate. The increments are \(1/30\), \(4/24=1/6\), and \(2/12=1/6\). Their sum is \(1/30+1/6+1/6=1/30+10/30=11/30=0.366666\ldots\).
Variance estimate. The contributions are \(1/30^2=1/900\), \(4/24^2=4/576=1/144\), and \(2/12^2=2/144=1/72\). Using denominator 3600, their sum is \(4/3600+25/3600+50/3600=79/3600=0.0219444\ldots\).
Estimated standard error. Since \(\sqrt{79/3600}=\sqrt{79}/60\), the standard error is approximately \(8.88819/60=0.1481365\), or 0.1481 to four decimal places.
Check and trap avoided. The risk set for each time removes only earlier failures and censorings. The five censorings before time 5 and the eight before time 8 affect later risk sets, but they do not create variance terms of their own.
Common Calculation Errors
- Using the hazard increment as the variance contribution. The hazard increment is \(d_j/Y_j\); the variance contribution is \(d_j/Y_j^2\). For \(Y_j=20\) and \(d_j=2\), these are 0.10 and 0.005, respectively.
- Squaring the failure count. The contribution is \(d_j/Y_j^2\), not \(d_j^2/Y_j^2\). For three tied failures among 12 at risk, use \(3/144\), not \(9/144\).
- Sequentially updating the risk set within a tie. If three failures are tied with 12 at risk, the grouped contribution is \(3/12^2\). The calculation \(1/12^2+1/11^2+1/10^2\) imposes an unobserved ordering and is not the grouped Nelson-Aalen variance calculation.
- Letting censorings contribute as if they were failures. Censoring affects the risk sets at later event times, but does not enter the numerator \(d_j\) at its own time.
- Calling the variance the standard error. First sum \(d_j/Y_j^2\); then take the square root to obtain the standard error.
- Rounding each term too early. Preserve exact fractions or additional decimal places through the sum and square root. Round only the reported final result.
Check Your Understanding
For each hypothetical study, calculate the requested quantity using the risk sets immediately before the event times. Keep tied failures grouped. Answers are provided at the end of each problem.
- Two event times have \((Y_1,d_1)=(18,2)\) and \((Y_2,d_2)=(12,1)\). Calculate the variance estimate through the second time. (Answer: \(2/18^2+1/12^2=17/1296\approx0.0131\).)
- A study has \((Y_1,d_1)=(25,5)\) and \((Y_2,d_2)=(16,2)\). Calculate the cumulative hazard and variance estimates through time 2. (Answer: \(\widehat H=5/25+2/16=0.325\); variance \(=5/625+2/256=253/16000=0.0158125\).)
- At time 3, three failures are tied among 14 at risk; at time 6, one failure occurs among 9 at risk. Calculate the variance estimate. (Answer: \(3/14^2+1/9^2=439/15876\approx0.027651\).)
- Three event-time records are \((40,4)\), \((27,3)\), and \((15,5)\), where each pair is \((Y_j,d_j)\). Find the variance estimate and estimated standard error, rounding both to six decimal places. (Answer: variance \(=4/1600+3/729+5/225\approx0.028837\); standard error \(\approx0.169816\).)
- A hypothetical cohort has 50 at risk at time 1, with 2 failures. Before time 3, 7 additional individuals are censored; at time 3, 3 failures occur. Before time 6, 8 additional individuals are censored; at time 6, 4 failures occur. Calculate the variance estimate through time 6. (Answer: \(2/50^2+3/41^2+4/30^2\approx0.0070291\).)