Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

In statistical engineering the "hazard rate" of a distribution is defined as:

$$r(x)=\frac{f(x)}{1-F(x)}$$

where $f(x)$ and $F(x)$ are the PDF and CDF. Basically $r(x)$ is the odds that, having reached a certain point on the abscissa (usually time), you won't get any further. In the study of mechanical failure, the relevant distributions are those with an $r(x)$ that is everywhere increasing for positive x, like the Weibull distribution.

My question is how would you interpret an $r(x)$ whose absolute value is everywhere increasing for positive x, but which is negative? Does the sign matter?

share|improve this question
add comment

1 Answer

up vote 2 down vote accepted

If your $f(x)$ and $F(x)$ have the necessary properties1 then it should not be possible to get a negative value of $r(x)$.

A common place to get confused here using a different range for the PDF and the evaluation of the failure rate.


1

  • $f(x)$ strictly non-negative over $[x_{min},x_{max})$
  • $F(x) = \int_{x_{min}}^x \mathrm{d}u f(u)$ for $ x \in [x_{min},x_{max}) $
  • $F(x)$ strictly non-negative over $[x_{min},x_{max})$ and monotonically increasing with $F(x_{min})=0$ (these are implied by the earlier conditions) and $F(x_{max}) \le 1$. Note that this implies that you express $f$ and $F$ in fractional quantities and never in percent.
share|improve this answer
    
Yeah, that's what I would expect, but I can't see where I'm going wrong. Take the lognormal distribution for example. For mean=0, standard dev>1.3 I get negative $r(x)$ for all positive $x$ (Acknowledge that the LN dist. would not be used in failure rate studies, still $r(x)$ should have some intelligible meaning for any distribution) –  ben Oct 17 '12 at 17:28
    
Wikipedia shows the PDF and CDF as well behaved for those values... Any chance that you are messing up your evaluation of one or the other at zero? –  dmckee Oct 17 '12 at 17:36
1  
Aha, it's that pesky constant of integration again. –  ben Oct 17 '12 at 18:09
    
(as is so often the case) –  Emilio Pisanty Oct 17 '12 at 22:10
    
We have certainly all been there. –  dmckee Oct 17 '12 at 23:24
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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