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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.

Short Answer

Expert verified
The mean excess life is expressed as proposed and verified with exponential density; survival function is derivable, showing decreasing hazard for lifetime data.

Step by step solution

01

Understanding Mean Excess Life Function

The mean excess life function is defined as \(e(y)= \mathrm{E}(Y-y \mid Y>y)\). It represents the expected remaining "life" given survival beyond a point \(y\). Our goal is to show that \(e(y)=\mathcal{F}(y)^{-1} \int_{y}^{\infty} \mathcal{F}(u) \, du\).
02

Expressing Conditional Expectation

Given \(Y > y\), the mean excess life function can be written as \(e(y) = \frac{\int_{y}^{\infty} (u-y)f(u) \, du}{\mathcal{F}(y)}\), where \(f(u)\) is the probability density function and \(\mathcal{F}(y) = 1-F(y) = \int_{y}^{\infty} f(u) \, du\) is the survival function.
03

Transform the Condition

By using the substitution \(v = u-y\), it transforms the integral \(\int_{y}^{\infty} (u-y)f(u) \, du = \int_{0}^{\infty} v f(v+y) \, dv\). Hence, integrating with respect to \(v\) yields the transformation needed to express \(e(y)\).
04

Derive Integral Form for \(e(y)\)

Recognizing \(v\) as the excess and applying the integral transformation, the form \(e(y) = \frac{1}{\mathcal{F}(y)} \int_{y}^{\infty} \mathcal{F}(u) \, du \) emerges after integrating by parts. This relates \(e(y)\) directly with the survival function \(\mathcal{F}(y)\).
05

Differential Equation for \(Q(y)\)

We can define \(Q(y) = \exp\left\{\int_{0}^{y} \frac{1}{e(u)} du\right\}\). Differentiating gives \(Q'(y) = Q(y) \frac{1}{e(y)}\), leading to the equation \(e(y)Q'(y) + Q(y) = 0\) which stems from the relationship with \(e(y)\).
06

Continuous Density and \(\mathcal{F}(y)\) Formula

Integrate the equation \(e(y)Q'(y) + Q(y) = 0\) to derive \(\mathcal{F}(y) = \frac{e(0)}{e(y)} \exp\left\{-\int_{0}^{y} \frac{1}{e(u)} du\right\}\), assuming continuity of the underlying density function.
07

Example with Exponential Density

For an exponential distribution \(f(u) = \lambda e^{-\lambda u}\), \(e(y) = \frac{1}{\lambda} = e(0)\), hence \(\mathcal{F}(y) = e^{-\lambda y}\), verifying the consistency of our derived result with known distributions.
08

Survival and Hazard Functions for Human Lifetime Data

Given \(e(y) = \gamma (1-y/\theta)^\beta\), the survival function \(\mathcal{F}(y) = e^{-\gamma(1-y/\theta)^{\beta}}\) and hazard function \(h(y) = \frac{\gamma \beta}{\theta} (1-y/\theta)^{\beta-1}\) can be derived. This model reflects a decreasing hazard function, typical in survival analyses of human lifetimes.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Mean Excess Life Function
The mean excess life function, denoted as \(e(y)\), serves as a cornerstone of survival analysis. It calculates the expected duration of time remaining, given that a certain threshold \(y\) has been surpassed. Mathematically, this is expressed as the conditional expectation \(\mathrm{E}(Y-y \mid Y>y)\). This simplifies to \(e(y) = \frac{1}{\mathcal{F}(y)} \int_{y}^{\infty} \mathcal{F}(u) \, du\). Understanding this function is crucial, as it provides insights into how one might anticipate future outcomes based solely on survival past a certain point. By substituting into the integral transformation with \(v = u - y\), it gives us a powerful tool to study how the probability of survival changes over time.
Probability Density Function
The probability density function (PDF), \(f(u)\), is fundamental in understanding the distribution of continuous random variables. It serves as the foundation for calculating probabilities in a continuous space. Essentially, integrating the PDF over a range gives the probability that the variable falls within that range. By using the PDF, we derive concepts such as the mean excess life function and the survival function. For instance, the PDF is integral when expressing the survival function \(\mathcal{F}(y) = \int_{y}^{\infty} f(u) \, du\). This not only reflects the likelihood of surviving beyond \(y\) but also sets the stage for further calculations involving the survival and hazard functions. In practice, understanding the PDF allows us to model complex systems like human life expectancy or the reliability of machinery.
Survival Function
The survival function, \(\mathcal{F}(y)\), represents the probability that a variable exceeds a certain value \(y\). It's determined as the complement of the cumulative distribution function: \(\mathcal{F}(y) = 1 - F(y)\). This function is crucial in many fields, especially in reliability engineering and medical statistics.It is linked intimately with the mean excess life function and establishes the basis for the hazard function. For continuous distributions, \(\mathcal{F}(y)\) can be precisely computed using the PDF. For example, in the exercise's context, we explored how \(\mathcal{F}(y)\) could be expressed in terms of the mean excess life function, making it possible to connect across different elements of survival analysis seamlessly. The survival function not only serves theoretical purposes but is also applied in determining the lifespan of products or analyzing patient lifetime data.
Hazard Function
The hazard function, \(h(y)\), is vital in understanding the risk of failure or event occurrence at a particular time, given survival up to that time. It's derived from the survival function by taking the ratio of the PDF to the survival function: \(h(y) = \frac{f(y)}{\mathcal{F}(y)}\).This function provides a moment to moment assessment of risk and is particularly useful in fields such as epidemiology and finance. It aids in making timely interventions or decisions. In the exercise, we derived the hazard function under the model of human lifetime data, which allowed us to observe a decreasing hazard function, a characteristic commonly seen in human and biological studies.By analyzing how \(e(y)\) and \(\mathcal{F}(y)\) interrelate, we could further simplify our understanding of \(h(y)\). This simplification is crucial for interpreting real-world data and for informing policies or strategies related to risk management.

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Most popular questions from this chapter

Show that the inverse Gaussian density $$ f(y ; \mu, \lambda)=\left(\frac{\lambda}{2 \pi y^{3}}\right)^{1 / 2} \exp \left\\{-\lambda(y-\mu)^{2} /\left(2 \mu^{2} y\right)\right\\}, \quad y>0, \lambda, \mu>0 $$ is an exponential family of order \(2 .\) Give a general form for its cumulants.

Show that \(\sum s\left(Y_{j}\right)\) is minimal sufficient for the parameter \(\omega\) of an exponential family of order \(p\) in a minimal representation.

Let \(Y_{1}, \ldots, Y_{n}\) be independent exponential variables with hazard \(\lambda\) subject to Type I censoring at time \(c\). Show that the observed information for \(\lambda\) is \(D / \lambda^{2}\), where \(D\) is the number of the \(Y_{j}\) that are uncensored, and deduce that the expected information is \(i(\lambda \mid c)=n\\{1-\exp (-\lambda c)\\} / \lambda^{2}\) conditional on \(c\) Now suppose that the censoring time \(c\) is a realization of a random variable \(C\), whose density is gamma with index \(v\) and parameter \(\lambda \alpha\) : $$ f(c)=\frac{(\lambda \alpha)^{v} c^{v-1}}{\Gamma(v)} \exp (-c \lambda \alpha), \quad c>0, \alpha, v>0 $$ Show that the expected information for \(\lambda\) after averaging over \(C\) is $$ i(\lambda)=n\left\\{1-(1+1 / \alpha)^{-v}\right\\} / \lambda^{2} $$ Consider what happens when (i) \(\alpha \rightarrow 0\), (ii) \(\alpha \rightarrow \infty\), (iii) \(\alpha=1, v=1\), (iv) \(v \rightarrow \infty\) but \(\mu=v / \alpha\) is held fixed. In each case explain qualitatively the behaviour of \(i(\lambda)\).

(a) Suppose that \(Y_{1}\) and \(Y_{2}\) have gamma densities (2.7) with parameters \(\lambda, \kappa_{1}\) and \(\lambda, \kappa_{2}\). Show that the conditional density of \(Y_{1}\) given \(Y_{1}+Y_{2}=s\) is $$ \frac{\Gamma\left(\kappa_{1}+\kappa_{2}\right)}{s^{\kappa_{1}+\kappa_{2}-1} \Gamma\left(\kappa_{1}\right) \Gamma\left(\kappa_{2}\right)} u^{\kappa_{1}-1}(s-u)^{\kappa_{2}-1}, \quad 00 $$ and establish that this is an exponential family. Give its mean and variance. (b) Show that \(Y_{1} /\left(Y_{1}+Y_{2}\right)\) has the beta density. (c) Discuss how you would use samples of form \(y_{1} /\left(y_{1}+y_{2}\right)\) to check the fit of this model with known \(v_{1}\) and \(v_{2}\).

In the linear model (5.3), suppose that \(n=2 r\) is an even integer and define \(W_{j}=Y_{n-j+1}-\) \(Y_{j}\) for \(j=1, \ldots, r\). Find the joint distribution of the \(W_{j}\) and hence show that $$ \tilde{\gamma}_{1}=\frac{\sum_{j=1}^{r}\left(x_{n-j+1}-x_{j}\right) W_{j}}{\sum_{j=1}^{r}\left(x_{n-j+1}-x_{j}\right)^{2}} $$ satisfies \(\mathrm{E}\left(\tilde{\gamma}_{1}\right)=\gamma_{1} .\) Show that $$ \operatorname{var}\left(\tilde{\gamma}_{1}\right)=\sigma^{2}\left\\{\sum_{j=1}^{n}\left(x_{j}-\bar{x}\right)^{2}-\frac{1}{2} \sum_{j=1}^{r}\left(x_{n-j+1}+x_{j}-2 \bar{x}\right)^{2}\right\\}^{-1} $$ Deduce that \(\operatorname{var}\left(\tilde{\gamma}_{1}\right) \geq \operatorname{var}\left(\widehat{\gamma_{1}}\right)\) with equality if and only if \(x_{n-j+1}+x_{j}=c\) for some \(c\) and all \(j=1 \ldots, r\)

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