# Estimate for the standard error of the probability of a residual lifetime

Suppose that we estimate the survival function using the Kaplan-Meier estimator. Based on that KM-curve $$\hat{S}(\cdot)$$, one can then estimate the probability that the residual lifetime is larger than 300 days (when the patient already survived 500 days) as $$\hat{S}(800)/\hat{S}(500)$$.

Question: How can we obtain (using the Delta method) an estimate of this probability?

Note: One is allowed to assume that the KM-estimates at different time points can be considered as independent.

Approach: Maybe, I could write $$\text{Var}\left(\log\left(\frac{\hat{S}(800)}{S(500)}\right)\right) = \text{Var}\left(\log \hat{S}(800)\right)+\text{Var}\left(\log \hat{S}(500)\right),$$ because of the assumption we made (see Note), but I don't know whether this is the correct way to go?

## 1 Answer

The most popular formula is the Greenwood one. (edit: by the way you have the good way to go)

The estimator of the KM (log version)

where di/ni is the proportion of events for the population at risk in ti and 1 - di/ni the proportion of non events for the populaiton at risk in ti.

The idea is to compute the variance of the sum that has only orthogonal elements and so its variance is just equal of the sum of the variance of each elements, then we compute the variance of log(St) to have with delta method Var(St)

By using the property of the variance of the maximum likelihood estimator, we have:

So the second derivative is easily compute

And the following result:

So if i /= j, the covariance matrix is diagonal and evaluated in (1-di/ni), we have:

Now, by using delta method, we have very simply the following result;

And so:

Since we have Var(log(St)), Var(St) = Var(exp(log(St))) and so with the delta method where g() = exp(), you have the following result: