# Why do we need second rate of failure even though we have already hazard function?

In reliability theory we have $h(x)=\frac{p(x)}{s(x)}$ where $p(x)$ and $s(x)$ are the probability mass function and the survival function, respectively.

Again $$SRF(x)=\log\left(\frac{s(x)}{s(x+1)}\right)$$ gives the same failure rate function.

Why do we use second one if we have already $h(x)=\frac{p(x)}{s(x)}$?

What is the need to calculate the second rate of failure function?

• Could you provide a reference for this second function $SRF(x)$. The units of the two are not the same. $h(x)$ has units of $[x]^{-1}$ while the 2nd function is unitless. Also, if failure times, $x$, are in the ms or $\mu$s range, it's not clear to me that adding "1" would be an incremental change in $x$. – user137329 Oct 17 '17 at 14:41