How to interpret a negative failure rate?

Question

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?

Answer 1

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.

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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.