Question

Suppose that X is a random variable having a continuous distribution with p.d.f.\(f\left( x \right)\)and c.d.f.\(F\left( x \right)\)and for which\({\rm P}\left( {X > 0} \right) = 1\)Let the failure rate\(h\left( x \right)\) be as defined in Exercise 18 of Sec. 5.7. Show that\(\exp \left[ { - \int\limits_0^x {h\left( t \right)dt} } \right] = 1 - F\left( x \right)\)

Short answer

\(\exp \left[ { - \int\limits_0^x {h\left( t \right)dt} } \right] = 1 - F\left( x \right)\)

Step-by-step solution

Given information

X be the random variable with \(f\left( x \right)\) as the pdf and \(F\left( x \right)\)as the cdf.

Verifying \(\exp \left[ { - \int\limits_0^x {h\left( t \right)dt} } \right] = 1 - F\left( x \right)\)

Let us consider the integral

\(\int\limits_0^x {h\left( t \right)dt = \int\limits_0^x {\frac{{f\left( t \right)}}{{1 - F\left( t \right)}}dt} } \)

F is the cdf of X, hence

\(\begin{array}{l}F\left( x \right) = \int\limits_0^x {f\left( t \right)dt} \\\frac{d}{{dt}}\left( {1 - F\left( t \right)} \right) = - f\left( t \right)\end{array}\)

Hence by using the integration by parts

\(\begin{aligned}{}\int\limits_0^x {h\left( t \right)dt} &= - \left[ {\log \left( {1 - F\left( t \right)} \right)} \right]_0^x\\ &= - \left[ {\log \left( {1 - F\left( x \right)} \right) - \log \left( {1 - F\left( 0 \right)} \right)} \right]\\ &= - \log \left( {1 - F\left( x \right)} \right)\end{aligned}\)

Substitute in the left-hand side of equation (1).

\(\begin{array}{l}\exp \left[ { - \int\limits_0^x {h\left( t \right)dt} } \right] = \exp \left[ {\left( { - \log \left( {1 - f\left( x \right)} \right)} \right)} \right]\\\exp \left[ { - \int\limits_0^x {h\left( t \right)dt} } \right] = 1 - F\left( x \right)\end{array}\)

Hence, we have proved the required equality.