How to interpret a negative failure rate?

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?

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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.
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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
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