Why Exponentiate the Negative Nelson-Aalen Estimate?
The Kaplan-Meier estimator combines conditional survival factors by multiplying them. The Nelson-Aalen estimator instead adds estimated hazard increments. These two ways of combining information are related: when each increment is small, the logarithm of a Kaplan-Meier factor is close to the negative of that increment. This motivates approximating Kaplan-Meier survival by exponentiating the negative Nelson-Aalen estimate.
The approximation is useful because it translates a sum of hazard increments into a survival-scale quantity. It is not an identity: the discrepancy depends on the sizes of the increments and on how their errors accumulate. In particular, “each increment is small” is not by itself enough to guarantee high accuracy if there are many increments.
The event-time calculation uses the risk set immediately before each failure time. If failures are tied, \(d_j\) counts all failures at that time, so the increment is still \(d_j/n_j\). Censoring affects which observations remain in later risk sets; it is not itself a failure and does not add a hazard increment. As with Kaplan-Meier and Nelson-Aalen estimation, interpreting the estimates as population survival and cumulative hazard relies on appropriate independent, noninformative censoring.
Deriving the Approximation and Its Direction
For an increment \(h\) between zero and one, the logarithm series gives \(\log(1-h)=-h-h^2/2-h^3/3-\cdots\). Thus the logarithm of each Kaplan-Meier factor is slightly more negative than \(-h\). Adding these identities over event times explains both the approximation and its direction.
The correction has the convergent expansion \(C(t)=\sum_j(h_j^2/2+h_j^3/3+\cdots)\). The first omitted term is quadratic, which is why small increments tend to make the approximation accurate. A useful rigorous bound follows from this same series. If \(h_{\max}=\max_j h_j<1\), then \(1/r\leq1/2\) for every integer \(r\geq2\), so \(\sum_{r=2}^{\infty}h_j^r/r\leq \frac12\sum_{r=2}^{\infty}h_j^r=h_j^2/[2(1-h_j)]\). Consequently,
This bounds the difference on the logarithmic scale. The absolute difference between the two survival estimates is \(\exp[-\widehat H_{\mathrm{NA}}(t)]-\widehat S_{\mathrm{KM}}(t) =\exp[-\widehat H_{\mathrm{NA}}(t)](1-\exp[-C(t)])\). Therefore a small bound for \(C(t)\) also controls the survival-scale difference, but it is not itself that difference.
Worked Calculations
Worked Example: Three Small Increments
At three successive event times, suppose the estimated hazard increments are \(0.02\), \(0.03\), and \(0.04\). Calculate Kaplan-Meier survival, the exponentiated Nelson-Aalen approximation, and a bound on the log-scale correction. Explain whether the approximation is close.
Method and common mistake avoided. Multiply the Kaplan-Meier factors, add the increments for Nelson-Aalen, and compare on both the log and survival scales. Do not treat the sum of increments as an exact substitute for the product.
First, the Nelson-Aalen estimate is \(\widehat H_{\mathrm{NA}}=0.02+0.03+0.04=0.09\). Thus the exponentiated estimate is \(\exp(-0.09)=0.9139312\), or \(0.9139\) to four decimal places.
The Kaplan-Meier estimate is \((1-0.02)(1-0.03)(1-0.04)=0.98(0.97)(0.96)\). The first two factors give \(0.98(0.97)=0.9506\), and \(0.9506(0.96)=0.912576\). Therefore Kaplan-Meier survival is \(0.9126\) to four decimal places. The exponentiated Nelson-Aalen value exceeds it by \(0.9139312-0.912576=0.0013552\), or \(0.0014\) to four decimal places.
For the bound, \(h_{\max}=0.04\) and \(\sum h_j^2=0.02^2+0.03^2+0.04^2=0.0004+0.0009+0.0016=0.0029\). Hence \(C\leq0.0029/[2(1-0.04)]=0.0029/1.92=0.0015104167\). Directly, \(C=-\log(0.912576)-0.09\approx0.0014839\), which is below the bound. The close estimates and small correction support using the approximation here.
Check. The exact relation gives \(\exp(-0.09-0.0014839)\approx0.912576\), agreeing with the product calculation.
Worked Example: A Small Correction With Exact Product Arithmetic
For an algebraic three-step discrete survival model, take conditional failure probabilities \(h_1=0.02\), \(h_2=1/49\), and \(h_3=1/96\). Find the two survival estimates and bound the log-scale correction. This example specifies conditional probabilities directly; it does not assert that the three values arise from one observed risk table.
Method and common mistake avoided. Keep the product factors and hazard sum distinct. In particular, the bound uses \(\sum h_j^2\) multiplied by \(1/[2(1-h_{\max})]\); using an incorrect squared-increment sum can invalidate the conclusion.
The Nelson-Aalen sum is \(\widehat H_{\mathrm{NA}}=0.02+1/49+1/96 =0.02+0.0204081633+0.0104166667=0.0508248299\). Therefore the exponentiated value is \(\exp(-0.0508248299)\approx0.9504451\).
For Kaplan-Meier, the product simplifies exactly: \((1-0.02)(1-1/49)(1-1/96) =(49/50)(48/49)(95/96)\). Canceling \(49\) and \(48\) against the corresponding denominators gives \((48/50)(95/96)=(24/25)(95/96)=0.95\). The exponentiated estimate exceeds Kaplan-Meier survival by approximately \(0.9504451-0.9500000=0.0004451\).
The largest increment is \(h_{\max}=1/49\), which is greater than \(0.02\) and \(1/96\). The exact sum of squares is \[ 0.02^2+(1/49)^2+(1/96)^2 =0.0004+0.0004164931279+0.0001085069444 \approx0.0009250000723. \] The coefficient in the bound is \(1/[2(1-1/49)]=1/[2(48/49)]=49/96\). Thus \(C\leq0.0009250000723(49/96)\approx0.0004721355\).
The actual correction is \(C=-\log(0.95)-0.0508248299 \approx0.0512932944-0.0508248299=0.0004684645\), which is below the corrected bound. The values \(0.9504451\) and \(0.9500\) are close, consistent with this small log-scale correction.
Check. Substituting the correction into the exact relationship gives \(\exp[-0.0508248299-0.0004684645]\approx0.9500000\), matching the exact product.
Worked Example: Why a Moderate Increment Matters
Suppose three successive conditional failure probabilities are \(0.10\), \(0.20\), and \(0.05\). Calculate both survival values, the exact log correction, and its upper bound. Assess whether the approximation is as close as in the previous example.
Method and common mistake avoided. Use the same log-correction identity, but assess the combined error rather than judging accuracy from one increment in isolation.
The cumulative hazard estimate is \(0.10+0.20+0.05=0.35\), so the exponentiated Nelson-Aalen value is \(\exp(-0.35)=0.7046881\). The Kaplan-Meier product is \(0.90(0.80)(0.95)=0.72(0.95)=0.684\). The approximation therefore exceeds Kaplan-Meier survival by \(0.7046881-0.684=0.0206881\).
For the exact correction, evaluate each term: \(-\log(0.90)-0.10=0.0053605157\), \(-\log(0.80)-0.20=0.0231435513\), and \(-\log(0.95)-0.05=0.0012932944\). Their sum is \(C=0.0297973614\).
Here \(h_{\max}=0.20\) and \(\sum h_j^2=0.10^2+0.20^2+0.05^2=0.01+0.04+0.0025=0.0525\). The bound is \(C\leq0.0525/[2(1-0.20)]=0.0525/1.6=0.0328125\). The actual correction is below this bound, but it is much larger than in the small-increment examples.
Check. The exact relation gives \(\exp(-0.35-0.0297973614)=\exp(-0.3797973614)=0.684\), agreeing with the product. The survival-scale difference of about \(0.0207\) is large enough that the approximation should not be treated as interchangeable with Kaplan-Meier here.
When Is the Approximation Reliable?
The expansion shows that the leading contribution to the log correction is approximately \(\frac12\sum_j h_j^2\). A useful sufficient condition for a small correction is therefore that both the largest increment and the sum of squared increments are small. Since \(\sum_j h_j^2\leq h_{\max}\sum_j h_j=h_{\max}\widehat H_{\mathrm{NA}}\), the bound also implies that if \(h_{\max}\) is small and the cumulative hazard remains moderate, the correction will be controlled.
This qualification matters. A long sequence of individually small increments can still accumulate a non-negligible correction if their squared values sum to a material amount. Conversely, a moderate total hazard spread across many very small increments can yield a good approximation. The calculations should therefore use the actual increments rather than relying on a rule such as “all hazards are below a chosen percentage.”
A common exam trap is to reverse the direction of the approximation. Because \(\log(1-h)<-h\) for \(0<h<1\), the Kaplan-Meier product is smaller than \(\exp(-\sum h_j)\), not larger. Another trap is to report the log correction \(C\) as though it were the survival-scale difference. The exact survival difference includes the factor \(\exp(-\widehat H_{\mathrm{NA}})\) and \(1-\exp(-C)\).
For each event time, use \(h_j=d_j/n_j\), with the risk set immediately before that time.
Multiply \(1-h_j\) for Kaplan-Meier and add \(h_j\) before exponentiating the negative sum for the Nelson-Aalen approximation.
Use \(C=\sum_j[-\log(1-h_j)-h_j]\), or bound it using the maximum increment and sum of squares.
The exponentiated Nelson-Aalen value must be at least the Kaplan-Meier value; distinguish a log-scale correction from an absolute survival difference.
These relationships are aligned with the SOA Exam ASTAM (Advanced Short-Term Actuarial Mathematics) syllabus. In calculation problems, showing the product, sum, correction or bound, and direction of comparison makes the approximation argument transparent.
Check Your Understanding
For each problem, calculate the requested quantities before checking the answer.
- A discrete survival curve has increments \(0.01\) and \(0.02\). Calculate the Kaplan-Meier estimate and the exponentiated Nelson-Aalen approximation, both to four decimal places. State which is larger. (Answer: Kaplan-Meier \(=0.9702\); exponentiated Nelson-Aalen \(=0.9704\), which is larger.)
- At three event times the increments are \(0.04\) and \(0.05\) (two event times). Find both survival estimates and their difference, to four decimal places. (Answer: Kaplan-Meier \(=0.9120\); exponentiated Nelson-Aalen \(=\exp(-0.09)=0.9139\); difference \(=0.0019\).)
- For increments \(0.02\), \(0.04\), and \(0.01\), calculate \(h_{\max}\), \(\sum h_j^2\), and the stated upper bound \(\sum h_j^2/[2(1-h_{\max})]\) on \(C\). (Answer: \(h_{\max}=0.04\), \(\sum h_j^2=0.0021\), and the bound is \(0.0021/1.92=0.00109375\).)
- A model has increments \(0.10\), \(0.05\), and \(0.02\). Calculate its Kaplan-Meier survival and exponentiated Nelson-Aalen approximation, to four decimal places. (Answer: Kaplan-Meier \(=0.90(0.95)(0.98)=0.8379\); exponentiated Nelson-Aalen \(=\exp(-0.17)=0.8437\), an overestimate.)
- For increments \(0.02\), \(0.03\), and \(0.04\), determine the upper bound on the log correction and whether that bound is below \(0.0016\). (Answer: \(h_{\max}=0.04\), \(\sum h_j^2=0.0029\), and \(C\leq0.0029/1.92\approx0.0015104\), which is below \(0.0016\).)