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The survival function of a right-censored sample.

public static KaplanMeierCurve Estimate(ReadOnlySpan<double> durations, ReadOnlySpan<bool> eventObserved, double confidenceLevel = 0.95)

Parameters — durations holds one non-negative time per subject. eventObserved is true where that time ends in the event and false where the subject was censored at it; the two spans must be the same length. confidenceLevel lies strictly inside (0, 1).

Returns — a KaplanMeierCurve whose Steps, Survival, Lower and Upper share one index.

Exceptions — ArgumentException when the spans differ in length, the sample is empty, or a duration is negative or NaN. ArgumentOutOfRangeException when confidenceLevel is not strictly inside (0, 1).

Example — a small sample, read step by step.

using Lodestar.Survival;

KaplanMeierCurve curve = KaplanMeier.Estimate([1, 2, 3], [true, false, true]);

double opening = curve.Survival[0];  // => 1
double afterFirst = curve.Survival[1];  // => 0.666…
int stillAtRisk = curve.Steps[2].AtRisk;  // => 2

Remarks — the second step is 1 - 1/3. The third carries the censoring at time 2, so the estimate does not move there while AtRisk falls from 3 to 2 — and the fourth then divides by that smaller risk set, which is the whole mechanism by which a censored subject still counts.

The confidence level is a level, not a multiplier. 0.95 asks for the two-sided 95% interval, so the critical value taken is the normal quantile at 0.975. That quantile is Distributions.NormalQuantile rather than a Student one at a large degrees of freedom: the substitute's accuracy peaks around 1e-8 and the log-log transform amplifies that into the seventh digit of a bound — decision 0003 has the measurement, as docs/guides/performance.md amended it, and this corpus is what caught it.

Greenwood's sum is accumulated, not the variance. The variance of the estimate is S² times that sum, but the log-log interval needs the sum alone, so it is what the loop carries. When the last subjects all have the event the increment is not finite; the sum becomes infinite, the estimate is zero, and both bounds collapse there. Each increment's denominator n(n - d) is taken in double: in int it overflows past 46,340 at risk.

Applies to — net10.0, netstandard2.0.

See also — KaplanMeier, NelsonAalen.Estimate.