Nelson–Aalen estimator

Remarks

In counting process notation, with $N(t)$ the number of observed events up to $t$ and $Y(t)$ the number at risk at $t$, the estimator is $\hat H(t) = \int_0^t \frac{dN(s)}{Y(s)}$, with $0/0 = 0$. A common variance estimator is $\sum_{j : t_{(j)} \le t} d_j / n_j^2$. The survival estimate $e^{-\hat H(t)}$ is at least the Kaplan–Meier estimate, since $e^{-x} \ge 1 - x$ for all $x$.

Example: for the observations $2, 3^+, 5, 7$, with $3^+$ censored, $(n_j, d_j)$ is $(4, 1)$ at $2$, $(2, 1)$ at $5$ and $(1, 1)$ at $7$. So $\hat H = 0$ on $[0, 2)$, $1/4$ on $[2, 5)$, $3/4$ on $[5, 7)$ and $7/4$ from $7$ on.

Non-example: the cumulative sum $\sum_{t_{(j)} \le t} d_j / n$, which divides by the initial sample size $n$, is not the Nelson–Aalen estimator. It ignores that subjects leave the risk set, and it underestimates the cumulative hazard.