3
$\begingroup$

The hazard function is commonly used in models using distributions with positive support (gamma, weibull, lognormal, etcetera). However, I have not seen this concept (hazard) being used in the context of models using distributions with support on the entire real line, like the normal, Student-t, logistic, laplace, etcetera.

Is there any use of the hazard function associated with distributions with support on the real line?

EDIT: I just came across the concept of Inverse Mills ratio. It seems like the hazard function of distributions with support of the real line appears, implicitly, through this concept (en.wikipedia.org/wiki/Mills_ratio).

$\endgroup$

2 Answers 2

1
$\begingroup$

As @whuber notes in a comment on another answer, there is no reason to restrict hazard functions to distributions having only non-negative support. Although a Wikipedia page suggests that restriction, a survival function $S(x)$ can be taken in general to be the complement of a corresponding cumulative distribution $F(x)$, with $S(x)=1-F(x)$. For any value of $x$ for which $f(x)$ is defined and $S(x)$ isn't 0, there is a defined value of the hazard $h(x)=f(x)/S(x)$.

A survival function defined over the entire real line can be useful in evaluating a parametric survival model fit, as explained for example in Chapters 18 and 19 of Frank Harrell's Regression Modeling Strategies. A Kaplan-Meier survival plot of censored, standardized residuals over the real line can help evaluate whether the distribution of residuals matches that expected for a particular choice of parametric family (e.g., standard minimum extreme value, defined over the entire real line, for a Weibull model).

Furthermore, survival analysis with left-censored survival times can be thought of as allowing for negative survival times. In R, a left-censored observation at time $x$ can be coded as Surv(-Inf, x, event).

Survival functions defined over the entire real line thus can be useful.

The question, however, is how useful a hazard function is for distributions with support over the entire real line. The hazard function is perhaps most useful as describing the probability of an event at time $x$ given that there hasn't yet been an event. In practice for analysis of event times, it's useful to define a time origin such that $f(x)=0$ (and thus $F(x)=0$) for $x \le 0$. In that case, $h(x)=0$ for $x \le 0$. I suppose there might be circumstances in which a different time origin might make sense, but it's hard to think of one.

$\endgroup$
1
  • $\begingroup$ I just came across the concept of Inverse Mills ratio. It seems like the hazard function of distributions with support of the real line appears, implicitly, through this concept (en.wikipedia.org/wiki/Mills_ratio). $\endgroup$
    – Armindo
    Commented May 6, 2023 at 16:03
0
$\begingroup$

The hazard function is a function of time so a non-negative support is required.

In survival analysis, the hazard function is the probability of dying at time $t$ given that you survived until time $t$.

$\endgroup$
5
  • $\begingroup$ Your logic is unconvincing, because there's no mathematical obstacle to defining a hazard function for any (absolutely continuous) distribution, nor must people always measure time as positive values. $\endgroup$
    – whuber
    Commented May 5, 2023 at 17:51
  • $\begingroup$ @whuber The hazard function can be defined as $$h(t) = \frac{f(t)}{S(t)}$$ where $f(t)$ is a probability density function defined on the interval $[0,\infty]$ and $S(t)$ is the corresponding Survival function, i.e. $S(t)=1-F(t)$. It does not make sense to perform survival analysis with time being negative. $\endgroup$
    – 29703461
    Commented May 5, 2023 at 18:20
  • 1
    $\begingroup$ Sure it does: it makes perfect sense, and works perfectly well, to shift the time coordinate by any constant amount. Moreover, that formula for $h$ works provided $f$ exists and $S$ is nonzero; and even for positive random variables it's not defined when $S$ is zero. $\endgroup$
    – whuber
    Commented May 5, 2023 at 19:52
  • $\begingroup$ By definition $f(t)$ is on the interval $[0,\infty]$. Using something like a Gaussian would require truncation. Obviously when $S$ is 0 the survival analsyis is complete? $\endgroup$
    – 29703461
    Commented May 6, 2023 at 13:40
  • 2
    $\begingroup$ That's perfectly circular: by your definition $f$ has non-negative support, therefore "non-negative support is required"! $\endgroup$
    – whuber
    Commented May 6, 2023 at 14:35

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.