# Mean Residual Life of Weibull Distribution closed form

Weibull distribution function is parameterized as $$F(x) = 1 - e^{(-(x/\sigma)^a)}$$ and survival function is $$S(x)= e^{(-(x/\sigma)^a)}$$ where $$a$$ is shape parameter and $$\sigma$$ is the scale parameter.

Mean Residual Function is given by:

$$mrl(x) = \frac{\int_{x}^{\infty} S(t) dt}{S(x)} = \frac{\int_{x}^{\infty} e^{(-(t/\sigma)^a)}dt}{e^{(-(x/\sigma)^a)}}$$.

I can evaluate the above integral numerically. My question is there a closed form or any analytical form for above mean residual life $$mrl(x)$$ of Weibull distribution?

• No, not in the usual meaning of closed form at least, since it involves an incomplete gamma. Commented Nov 4, 2022 at 2:36
• @Glen_b, thanks. I have edited my question. Any analytical form would be good as well. Commented Nov 4, 2022 at 3:25

Use a substitution $$u = (\frac{t}{\sigma})^a$$. Then $$du = \frac{a}{\sigma} (\frac{t}{\sigma})^{a-1}dt$$.
Since $$\frac{t}{\sigma} = u^{1/a}$$, this gives us $$du = \frac{a}{\sigma} (u^{1/a})^{a-1}dt = \frac{a}{\sigma} u^{1-1/a}dt$$, so that $$dt = \frac{\sigma}{a} u^{1/a-1} du$$. Then
\begin{align*}mrl(x) &= e^{{(x/\sigma)}^a} \int_{(x/\sigma)^a}^{\infty} \frac{\sigma}{a} u^{1/a-1} e^{-u} \, du \\ &= \frac{\sigma}{a} e^{{(x/\sigma)}^a} \Gamma\left(\frac{1}{a}, \left(\frac{x}{\sigma}\right)^a \right), \end{align*} where $$\Gamma(s,x)$$ is the upper incomplete gamma function.
This can be rewritten slightly by using a recurrence property of the upper incomplete gamma function: \begin{align*} mrl(x) &= \sigma e^{{(x/\sigma)}^a} \left(\Gamma\left(1 + \frac{1}{a}, \left(\frac{x}{\sigma}\right)^a \right) - \left(\left(\frac{x}{\sigma}\right)^a\right)^{1/a} e^{-{(x/\sigma)}^a} \right) \\ &= \sigma e^{{(x/\sigma)}^a} \Gamma\left(1 + \frac{1}{a}, \left(\frac{x}{\sigma}\right)^a \right) - x. \end{align*} The reader can decide which is simpler.