Chapter 5: Problem 4
The mean excess life function is defined as \(e(y)=\mathrm{E}(Y-y \mid Y>y)\). Show that $$ e(y)=\mathcal{F}(y)^{-1} \int_{y}^{\infty} \mathcal{F}(u) d u $$ and deduce that \(e(y)\) satisfies the equation \(e(y) Q^{\prime}(y)+Q(y)=0\) for a suitable \(Q(y)\). Hence show that provided the underlying density is continuous, $$ \mathcal{F}(y)=\frac{e(0)}{e(y)} \exp \left\\{-\int_{0}^{y} \frac{1}{e(u)} d u\right\\} $$ As a check on this, find \(e(y)\) and hence \(\mathcal{F}(y)\) for the exponential density. One approach to modelling survival is in terms of \(e(y)\). For human lifetime data, let \(e(y)=\gamma(1-y / \theta)^{\beta}\), where \(\theta\) is an upper endpoint and \(\beta, \gamma>0\). Find the corresponding survivor and hazard functions, and comment.
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