Remaining useful life for Weibull distributed lifetimes

Question

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 $m(t_0)=E(t)-t_0$?

Answer 1

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.