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In the context of Survival Analysis, I have seen that the standard (non-parametric) estimate of the Hazard Function is through the Nelson-Aalen Estimator :

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As seen from this formula, the Nelson-Aalen Estimator provides an estimate of the cumulative hazard (hazard can not go down with time):

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However, I have often heard one of the main advantages of the Hazard Function to be non-monotonic (i.e. can go up and down). This allows the Hazard Function to model fluctuating risk in real life concepts such as human mortality (e.g. human mortality is high when someone is born, then decreases as the years go by, and finally increases again in old age). However, based on the estimate of the Hazard Function from the Nelson-Aalen Estimator, I don't understand how the Hazard Function can be anything but monotonic.

While researching the SAS documentation on Survival Analysis, I came across the following entry on (Kernel) Smoothed Estimates of the Hazard Function:

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Supposedly, if you follow the code they provide, this apparently yields "non-cumulative" hazard estimates (note that the estimates below are clearly non-monotonic):

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My Question: Is my understanding of the Nelson-Aalen Estimator correct? Even though that the Nelson-Aalen Estimator by definition is an estimate of the cumulative hazard function - by taking weighted averages of consecutive differences of the Nelson-Aalen estimates, we arrive at the "instantaneous" estimate of the hazard function which can now fluctuate across time? I.e.

  1. Evaluate the Nelson-Aalen Estimates at each required point

  2. Make a new column where you subtract consecutive values of the Nelson-Aalen Estimates (e.g. row_2 - row_1, row_3 - row_2, etc.).

  3. Now, pass these differences through some Kernel Smoothing Function

Is this correct? Does anyone know a standard way of doing this in R?

Thanks!

References:

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  • $\begingroup$ If you wanna apply differencing you'd get a bunch of delta functions. It boils down to smoothing those deltas by fitting cubic splines or other basis functions, or KDE if you will. Or with good assumptions, it becomes traditional parametric modelling, which defeats the purpose of non-parametric estimator. $\endgroup$ Nov 23 '21 at 4:41
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You need to distinguish between the instantaneous hazard as a function of time, $h(t)$, and the corresponding cumulative hazard $H(t)$. The latter is the integral of the former up through time $t$, $H(t)= \int_0^t h(\tau) d \tau$. As instantaneous hazard is necessarily non-negative its integral, the cumulative hazard, is necessarily non-decreasing. That restriction, however, puts no restriction on the shape of the instantaneous hazard $h(t)$ as a function of time except that it cannot go below 0.

Steps 1 and 2 in your question, without the smoothing in Step 3, is how hazard functions can be defined in non- and semi-parametric models. Without the smoothing and with non-parametric models, the instantaneous hazard is 0 between events; at event times, it's the ratio of cases having events at that time to the total number still at risk. With modeled covariates in Cox regressions, at each event time one weights the number of cases at risk by their relative hazards as estimated by the model to get a baseline hazard estimate.

Smoothing of such hazard and survival functions is certainly possible, but seldom is done with non- and semi-parametric models. For such models, I think that audiences prefer to see plots that more directly represent the data. Parametric survival models typically involve smooth survival/hazard functions directly.

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I think your logic here is correct. If you model the event times parametrically you can produce a smooth hazard and cumulative hazard function. When the event times are modeled non-parametrically the hazard function is a blip function. You could try plotting this with vertical bars rather than a smoothing function.

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  • $\begingroup$ @ Geoffery Johnson : Thank you for your reply! I guess now the challenge will be to choose a suitable kernel function with a appropriate bandwidth parameter. different choices of the above will inevitably produce different estimates! :) $\endgroup$
    – stats555
    Nov 23 '21 at 3:38

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