11
$\begingroup$

If X is a nonnegative random variable representing the life of a component having distribution function F,the mean residual life is defined by

$$ m(t) = E(X-t | X >t) = \frac{1}{\bar F(t)} \int_t^\infty (x-t) d\nu(x), t\geq 0 $$ In papaer R. C. Gupta and D. M. Bradley (2003)" Representing the Mean Residual Life in Terms of the Failure Rate"mentioned that by writing $$x - t = \int_{t}^{x} du$$ and employing Tonelli's theorem yields the equivalent formula $$ m(t) = \frac{1}{\bar F(t)}\int_t^\infty \int_t^x du d\nu(x) = \frac{1}{\bar F(t)}\int_t^\infty \int_u^\infty d\nu(x) du = \frac{1}{\bar F(t)}\int_t^\infty \bar{F}(u) du $$ How can we get this result by substituting the above integral and using Tonelli's theorem?

$\endgroup$

2 Answers 2

14
$\begingroup$

You have a nonnegative random variable $X\newcommand{\E}{\mathbb{E}}$ with distribution function $F$. The mean residual life is defined as $$ m(t) = \E\left[X - t\mid X >t\right] = \frac{\E\left[ (X-t) I_{\{X>t\}}\right]}{P\{X>t\}} = \frac{1}{1-F(t)} \int_t^\infty (x-t)\,dF(x) \, , $$ for $t>0$. But $$ \int_t^\infty (x-t)\,dF(x) = \int_t^\infty \left(\int_t^x du\right)dF(x) \, . \quad (*) $$ Tonellis' Theorem says that you can change the order of integration in $(*)$, but you have to be careful about the integration limits. Look at the following figure.

Integration domain

The original domain of integration is interpreted like this: $x$ varies from $t$ to $\infty$. For some fixed $x$, $u$ varies from $t$ to $x$, determining the filled region in the figure. Now, reverse the order: $u$ varies from $t$ to $\infty$. For a fixed $u$, $x$ varies from $u$ to $\infty$. Hence, $$ \int_t^\infty \left(\int_t^x du\right)dF(x) = \int_t^\infty \left(\int_u^\infty dF(x)\right)du $$ $$ = \int_t^\infty P\{X > u\}\,du = \int_t^\infty \left(1 - F(u)\right)\,du \, . $$ Therefore, $$ m(t) = \frac{1}{1-F(t)} \int_t^\infty \left(1 - F(u)\right)\,du \, . $$

$\endgroup$
1
  • $\begingroup$ How I can find $E(X|X<t)$? $\endgroup$
    – Alex Pozo
    Apr 14, 2020 at 12:59
2
$\begingroup$

Here is another way you can think about the problem. \begin{align*} & \mathbb{E}[X - t \, | \, X> t] = \int_0^{\infty} \mathbb{P}(X - t > s \, | \, X > t)ds = \int_0^{\infty} \dfrac{\mathbb{P}(X - t > s, \, X > t)}{\mathbb{P}(X > t)}ds \\ = & \dfrac{1}{\mathbb{P}(X > t)}\int_0^{\infty} \mathbb{P}(X > t + s, \, X > t)ds = \dfrac{1}{1 - F(t)} \int_0^{\infty} (1 - F(t+s))ds \\= & \dfrac{1}{1 - F(t)} \int_t^{\infty} (1 - F(s))ds \end{align*}

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.