Questions

1. Explaining the derivation of the hazard function

A bit of context

The hazard function is often found stated in brevity as:

h(t)=f(t)S(t)h(t)=\frac{f(t)}{S(t)}

where f(⋅)f(\cdot) is the probability density function, and S(⋅)S(\cdot) is the survival function. Throughout this question I will be referring the descriptions given by Rodrأguez and Tian.

Traditionally the survival and hazard functions come into play when the random variable TT is non-negative and continuous. In this sense, at least the concept of the survival function is remarkably straight forward being the probability that TT is greater than tt.

S(t)=1−F(t)=P(T>t)S(t)= 1-F(t) = P(T>t)

This is especially intuitive when put in context, e.g. the time following diagnosis of a disease until death.

From the definition of the hazard function above, it is clear that it is not a probability distribution as it allows for values greater than one.

My confusion comes in at Rodrأguez’s definition:

h(t)=limdt→0P(t≤T

Passing to the limit means taking limit (after some calculations). You need to learn the definition of limit of sequence / limit of function if you are not sure about the concept.

It depends on your fundamental definition of h(t)h(t):

h(t) = \frac {f(t)} {S(t)}

= \frac {1} {S(t)} \frac {d} {dt} F(t)

= \frac {1} {S(t)} \frac {d} {dt} [1 – S(t)]

= -\frac {1} {S(t)} \frac {d} {dt} S(t)

= – \frac {d} {dt} \ln S(t) h(t) = \frac {f(t)} {S(t)}

= \frac {1} {S(t)} \frac {d} {dt} F(t)

= \frac {1} {S(t)} \frac {d} {dt} [1 – S(t)]

= -\frac {1} {S(t)} \frac {d} {dt} S(t)

= – \frac {d} {dt} \ln S(t)

You see from definition it is unitless – survival function is just a probability, and pdf is the derivative of CDF.

Not sure about your last question. Hazard function is often used to in time modelling of survival analysis. Inherently there is nothing prohibiting hazard function to be used in other places.