Confidence intervals and functions of an estimator?

Question

I'm studying non-parametric estimators for survival functions and for the Nelson-Aalen estimator we have that the estimator for the survival function is $\exp(-\hat Y)$ (i.e. exponential of negative $\hat Y$ which is the estimator for the cumulative hazard rate).

Now my questions is, since the Survival function is just the negative exponential of the cumulative hazard, if we find a confidence interval for the cumulative hazard can I just take the negative exponential of that confidence interval to get an (approximate) confidence interval of the SURVIVAL function, or is there a better method (but not like using the delta method I've read about).

So lets say the 95% confidence interval for a cumulative hazard is [0.04,0.13] then could I say that the 95% CI for the corresponding survival function is: [exp(-0.13), exp(-0.04)] since Survival function = $\exp( - \text{cumulative hazard})$.

This actually is what one of the examples in my textbook does but I encountered a similar problem before when i was doing linear regression and I had something like log(y)= a + bx + error and to construct a confidence interval for y I would need to do: $y= \exp (a + bx + \text{error})$ so there was that exponential relationship and again if we had confidence intervals for the regression would there be any special allowance we have to make for the fact that we are exponentiating.

I know this may seem kinda broad and that I'm asking more for a detailed explanation then a simple answer so feel free just pour out your thoughts and examples, the more the merrier.

Answer 1

The issue raised in this question probably arose from the distinction between confidence intervals and prediction intervals and how they take into account the "error" terms. Confidence intervals represent error in the expected value. A prediction interval adds in the residual "error" to illustrate the uncertainty in predicting an individual case.

In a linear regression of the form $y=a+bx+\epsilon$, the confidence interval for a point estimate $\hat{y}_i = \hat a +\hat b x_i$, based on the estimates of the regression coefficients $\hat a$ and $\hat b$ and a covariate value $x_i$, doesn't explicitly involve $\epsilon$, as $\hat y$ is an estimate of the expected value (at $\epsilon=0$ under the assumptions of ordinary least squares). The variance in $\epsilon$ helps determine the (co)variances of $\hat a$ and $\hat b$ and thus the confidence interval for $\hat y$, but doesn't further enter the calculation.

In that situation, as the page linked by Kjetil B Halvorsen in a comment indicates, you are welcome to use any strictly monotone invertible transformation of $\hat y$ along with its confidence interval and still end up with a valid confidence interval for the transformed $\hat y$. Non-parametric survival models, as in the current question, are modeling expected survival, so that's what's done when transforming the confidence interval for an estimated cumulative hazard $\hat \Lambda(t)$ to the corresponding expected survival $\hat S(t) = \exp\left( -\hat \Lambda(t)\right)$.*

If you want to illustrate the error in making a prediction about an individual case, you must take into account the further variance arising from $\epsilon$. That's where you need to consider just how $\epsilon$ was incorporated into the model and how, in individual cases, it adds to the predicted expected value. That's the issue raised in the next to last paragraph of this question, and would differ, for example, between modeling individual observations in the form $\log(y) = a + bx +\epsilon$ and modeling the mean value $\bar y$ of the response with a generalized linear model, as explained on this page.

Parametric survival models that assume particular distributions of survival times around their expected values can provide prediction intervals in addition to confidence intervals. This document is a convenient reference to the underlying principles.

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*This relationship between cumulative hazard and survival is general, not restricted to the Nelson-Aalen estimate mentioned in this question.