# Remaining useful life for Weibull distributed lifetimes

Suppose we model component lifetimes with a two-parameter Weibull distribution. With $$\alpha$$ as the scale parameter and $$\beta$$ as the shape parameter, the component's mean survival time is known to be: $$E(t) = \int^{\infty}_{0}S(t)dt = \int^{\infty}_{0}e^{-(\frac{t}{\alpha})^\beta}dt = \alpha \Gamma(\frac{1}{\beta}+1)$$ where $$\Gamma$$ is the Gamma function and the remaining useful life $$m$$ of a component aged $$t_0$$ is: $$m(t_0)=\frac{\int^{\infty}_{t_0}S(t)dt}{S(t_0)}=\frac{E(t) - \int^{t_0}_{0}S(t)dt}{S(t_0)}$$

My question is: why is the remaining useful time not simply $$m(t_0)=E(t)-t_0$$?

• Gut check: that quantity can be negative Apr 15, 2021 at 4:23
• @Cam.Davidson.Pilon Noted. Is there an intuitive explanation, though? Apr 15, 2021 at 15:37

All the information about lifetimes (whether Weibull distributed or not) is contained in the survival function. So let's start by deriving the survival function given the subject is age $$t_0$$.
Let $$T$$ denote the (random) lifetime of the subject. Recall that the survival function $$S(t)$$ is really just notation for $$P(T > t)$$. We'd like the survival function for $$T | T > t_0$$, that is, we want a formula for $$P(T > t \;|\; T > t_0)$$ where $$t > t_0$$
\begin{align*} S(t | T > t_0) = P(T > t \;|\; T > t_0) &= \frac{P( T > t \;\text{and}\; T > t_0)}{P(T > t_0)}\\ &= \frac{P( T > t)}{P(T > t_0)}\\ &=\frac{S(t)}{S(t_0)} \end{align*}
The numerator reduces in line 2 because if $$T > t$$, then $$T > t_0$$ because $$t > t_0$$.
From here, if we wanted the expected conditional lifetime (aka remaining lifetime), we just take the integral from $$t_0$$ to $$\inf$$, which is what your formula for $$m(t_0)$$ is.