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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?

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Tonelli tells you that you can always switch the order of integration on a nonnegative function (and thereby allows you to use Fubini's theorem if you show one such ordering of integration is finite). In your example, everything is nonnegative so there's no need for Fubini in this case. The crux of the matter is integrating:

$$\int_{-\infty}^\infty\int_{-\infty}^\infty 1_{[t,\infty]}(x)1_{[t,x]}(u)dud\nu(x)$$

To see that $\int_t^\infty\int_t^xdud\nu(x)=\int_t^\infty\int_t^xd\nu(x)du$, notice that the left integral is integrating over the region $x\geq u$ and $u\geq t$, in other words a triangle that starts at $t$, defined by $(u,x)$. The left hand side is integrating along vertical slices, while the right integral is going along horizontal ones

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Actually I did not underestand what you say. I have two questions:1. what do you mena by a triangle that starts at t, defined by (u,x) ? 2- what do you mean by left hand side is integrating along vertical slices, while the right integral is going along horizontal ones? – ben Oct 8 '12 at 5:21
@ben: sorry for being vague. Check this out for a very similar example ("Example 2" in particular):… – Alex R. Oct 8 '12 at 5:31
I underestand the example 2, but I do not know how to draw my triangle to underestand the integration along vertical and horizontal slices? – ben Oct 8 '12 at 5:53
in $(u,x)$ coordinates, your triangle tip is at $(t,t)$, and when I say horizontal i mean keeping $x$ constant, and vice versa for vertical. – Alex R. Oct 8 '12 at 5:59
I can not draw this triangle. – ben Oct 8 '12 at 6:11

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