A short note on Nelson-Aalen

Hi All

Last time I made a post deriving the variance of the Kaplan Meier survival function. I had intended to also give the Nelson Aalen estimate as well buts its variance is much harder to get. This post just highlights the difficulties that I had.

The Nelson Aalen estimate is \hat{S}(t) = \prod_{j=1}^{k} \exp(-d_j/r_j). It feels a bit random to me, however David Collett points out that \exp(-x) = 1 - x + x^2/2 - x^3/6 ... so for x << 1 \exp(-x) \approx  1-x \hat{S}(t) = \prod_{j=1}^{k} \exp(-d_j/r_j) \approx \prod_{j=1}^{k} )(1-d_j/r_j)  \approx \prod_{j=1}^{k} )(r_j-d_j)/r_j which is the Kaplan Meier estimate.

Collett give the variance of the Nelson Aalen estimate as Var[\hat{S}(t)] = \hat{S}(t)^2 \sum_{j=1}^k (d_j/r_j^2). I have been unable to verify the formula. If the same logic as my perfidious post is applied the variance of var(\log[\hat{S}(t)]) can be found to be var(\log[\hat{S}(t)]) = \sum_{j=1}^{k} var[-d_j/r_j] and using everyone’s favorite identity (var[g(x)] \approx [(d/dx) g(x)]^2 var(x)) var(\log[\hat{S}(t)]) can also be found to be var(\log[\hat{S}(t)]) =\hat{S}(t)^{-2} var[\hat{S}(t)]. Putting to two sides together \hat{S}(t)^{-2} var[\hat{S}(t)]  = \sum_{j=1}^{k} var[-d_j/r_j] which is solved for var[\hat{S}(t)] to give var[\hat{S}(t)]  = \hat{S}(t)^{2} \sum_{j=1}^{k} var[-d_j/r_j]. So for Var[\hat{S}(t)] = \hat{S}(t)^2 \sum_{j=1}^k (d_j/r_j^2) to be true \sum_{j=1}^{k} var[-d_j/r_j] = \sum_{j=1}^k (d_j/r_j^2). I don’t see any clear reason why this is the case.

That is it for now. Tune in next time for hopefully some answers instead of just expositions on my inadequacies.

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