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Chapter 5: Problem 3

Use the relation \(\mathcal{F}(y)=\exp \left\\{-\int_{0}^{y} h(u) d u\right\\}\) between the survivor and hazard functions to find the survivor functions corresponding to the following hazards: (a) \(h(y)=\lambda\); (b) \(h(y)=\lambda y^{\alpha}\); (c) \(h(y)=\alpha y^{\kappa-1} /\left(\beta+y^{k}\right) .\) In each case state what the distribution is. Show that \(\mathrm{E}\\{1 / h(Y)\\}=\mathrm{E}(Y)\) and hence find the means in (a), (b), and (c).

Short Answer

Expert verified
The survivor functions are exponential, Weibull, and complex (inverse Gaussian-like), with means 1/位, 螕(1 + 1/伪)/位^(1/伪), and complex for (a), (b), (c) respectively.

Step by step solution

01

Survivor Function for Constant Hazard

For part (a), the hazard function is constant: \( h(y) = \lambda \). The survivor function, \( \mathcal{F}(y) \), is given by:\[\mathcal{F}(y) = \exp\left(-\int_{0}^{y} h(u) \, du\right) = \exp\left(-\int_{0}^{y} \lambda \, du\right) = \exp(-\lambda y)\]This matches the survivor function of the Exponential distribution with parameter \( \lambda \).
02

Survivor Function for Power Hazard

For part (b), the hazard function is \( h(y) = \lambda y^{\alpha} \). The survivor function is given by:\[\mathcal{F}(y) = \exp\left(-\int_{0}^{y} \lambda u^{\alpha} \, du\right) = \exp\left(-\lambda \frac{y^{\alpha+1}}{\alpha+1}\right)\]This matches the Weibull distribution with shape parameter \( \alpha \) and scale parameter \( \lambda \).
03

Survivor Function for Complicated Hazard

For part (c), the hazard function is \( h(y) = \frac{\alpha y^{\kappa-1}}{\beta + y^{k}} \). The survivor function becomes:\[\mathcal{F}(y) = \exp\left(-\int_{0}^{y} \frac{\alpha u^{\kappa-1}}{\beta + u^{k}} \, du\right)\]This integral doesn't have a simple closed form but defines a generalized version of the inverse Gaussian or similar complex distributions.
04

Relationship Between Expected Values

Using \( \mathrm{E}\{1 / h(Y)\} = \mathrm{E}(Y) \), we relate the expectation of the survivor function integral to that of the hazard function. This property illustrates that the mean of the survival times can be derived from integrating the reciprocal hazard.
05

Find Means for Each Case

For (a), since \( h(y) = \lambda \), \( \mathrm{E}(1/h(Y)) = 1/\lambda \), so \( \mathrm{E}(Y) = 1/\lambda \). For (b), as it relates to Weibull, \( \mathrm{E}(Y) = \Gamma(1 + 1/\alpha)/\lambda^{1/\alpha}\). For (c), solving \( \mathrm{E}\{1/h(Y)\} \) involves integrating a complex function with no simple closed form.

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Key Concepts

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

Survival Analysis
Survival Analysis is a branch of statistics used to predict the time until an event of interest, or endpoint, such as death or failure, occurs. It is widely applied in various fields like medicine, engineering, and sociology.
One of the core elements of survival analysis is the hazard function, which describes the instant rate of occurrence of the event given that it has not yet happened. Understanding the relationship between hazard and survivor functions is crucial for modeling survival data.
The survivor function, often denoted as \(\(\mathcal{F}(y)\)\), represents the probability that the time of event occurrence is later than a specified time point. It is related to the hazard function through the equation \( \(\mathcal{F}(y)=\exp \left\{-\int_{0}^{y} h(u) d u\right\}\)\). This relationship allows statisticians to convert hazard functions into survivor functions to analyze survival data effectively.
Exponential Distribution
The Exponential Distribution is a simple yet powerful probability distribution primarily used to model the time between events in a Poisson process.
It is characterized by a constant hazard rate, denoted by lambda (\(\lambda\)), indicating that the event has a consistent probability of occurring over time. This makes it memoryless, meaning the process does not depend on how much time has already passed.
In survival analysis, if the hazard function is \( h(y) = \lambda\), then the survivor function becomes \( \mathcal{F}(y) = \exp(-\lambda y)\). This characteristic shows that the Exponential is the only continuous distribution with a constant hazard rate, and it is frequently used to model simple life expectations in reliability engineering and other domains.
Weibull Distribution
The Weibull Distribution is a versatile distribution used for modeling lifetimes and is particularly popular because its hazard function can express both increasing and decreasing failure rates. This adaptability makes it suitable for analyzing a wide range of survival data.
The hazard function for a Weibull distributed random variable is \( h(y) = \lambda y^{\alpha}\), where \( \alpha\) is the shape parameter and \( \lambda\) is the scale parameter. The resulting survivor function is \( \mathcal{F}(y) = \exp\left(-\lambda \frac{y^{\alpha+1}}{\alpha+1}\right)\).
Depending on the value of the shape parameter:
  • If \( \alpha\) < 1, the failure rate decreases over time.
  • If \( \alpha\) = 1, the distribution reduces to the Exponential distribution with a constant hazard.
  • If \( \alpha\) > 1, the failure rate increases over time.

These versatile characteristics make the Weibull Distribution highly useful in reliability analysis and quality control.
Inverse Gaussian Distribution
The Inverse Gaussian Distribution is primarily used in survival analysis to model positively skewed response times and is especially potent in situations where the arrival of an event follows a normal process over time.
This distribution is defined through a complex hazard function. Example (c) involved a non-standard hazard expressed as: \(h(y) = \frac{\alpha y^{\kappa-1}}{\beta + y^{k}}\). A closed-form survivor function isn't easily attainable for such a complex hazard, often requiring numerical methods for solutions.
This distribution is useful in various fields, including finance for modeling stock returns and biology for modeling lifetimes, where the hazard rate might decrease initially and increase later in time.
Understanding its properties assists in analyzing real-world situations not conforming to the exponential or Weibull standards.
Expected Value Calculation
Calculating the Expected Value (or mean) is a crucial step in survival analysis as it indicates the average outcome of the modeled random variable. In survival contexts, it provides insights into the mean time until an event occurs.
In the context of hazard functions, the expected value calculation leverages the relationship \(\mathrm{E}\{1 / h(Y)\}\ = \mathrm{E}(Y)\), which signifies that integrating the reciprocal of the hazard can yield the expected lifetime.
For distributions like the Exponential, the expected value is straightforward: \( 1/\lambda\). For the Weibull, it is computed as \( \Gamma(1 + 1/\alpha)/\lambda^{1/\alpha}\), involving the gamma function. For more complex distributions like the Inverse Gaussian, the expectation might involve intricate integration without a simple closed form.
Grasping these calculations allows students to predict the average lifetime or failure time under different types of distribution models.

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Most popular questions from this chapter

(a) Show that when data \((X, Y)\) are available, but with values of \(Y\) missing at random, the log likelihood contribution can be written $$ \ell(\theta) \equiv I \log f(Y \mid X ; \theta)+\log f(X ; \theta) $$ and deduce that the expected information for \(\theta\) depends on the missingness mechanism but that the observed information does not. (b) Consider binary pairs \((X, Y)\) with indicator \(I\) equal to zero when \(Y\) is missing; \(X\) is always seen. Their joint distribution is given by $$ \operatorname{Pr}(Y=1 \mid X=0)=\theta_{0}, \quad \operatorname{Pr}(Y=1 \mid X=1)=\theta_{1}, \quad \operatorname{Pr}(X=1)=\lambda $$ while the missingness mechanism is $$ \operatorname{Pr}(I=1 \mid X=0)=\eta_{0}, \quad \operatorname{Pr}(I=1 \mid X=1)=\eta_{1} $$ (i) Show that the likelihood contribution from \((X, Y, I)\) is $$ \begin{aligned} &{\left[\left\\{\theta_{1}^{Y}\left(1-\theta_{1}\right)^{1-Y}\right\\}^{X}\left\\{\theta_{0}^{Y}\left(1-\theta_{0}\right)^{1-Y}\right\\}^{1-X}\right]^{I}} \\\ &\quad \times\left\\{\eta_{0}^{I}\left(1-\eta_{0}\right)^{1-I}\right\\}^{1-X}\left\\{\eta_{1}^{I}\left(1-\eta_{1}\right)^{1-1}\right\\}^{X} \times \lambda^{X}(1-\lambda)^{1-X} \end{aligned} $$ Deduce that the observed information for \(\theta_{1}\) based on a random sample of size \(n\) is $$ -\frac{\partial^{2} \ell\left(\theta_{0}, \theta_{1}\right)}{\partial \theta_{1}^{2}}=\sum_{j=1}^{n} I_{j} X_{j}\left\\{\frac{Y_{j}}{\theta_{1}^{2}}+\frac{1-Y_{j}}{\left(1-\theta_{1}\right)^{2}}\right\\} $$ Give corresponding expressions for \(\partial^{2} \ell\left(\theta_{0}, \theta_{1}\right) / \partial \theta_{0}^{2}\) and \(\partial^{2} \ell\left(\theta_{0}, \theta_{1}\right) / \partial \theta_{0} \partial \theta_{1}\). (ii) Statistician A calculates the expected information treating \(I_{1}, \ldots, I_{n}\) as fixed and thereby ignores the missing data mechanism. Show that he gets \(i_{A}\left(\theta_{1}, \theta_{1}\right)=M \lambda /\left\\{\theta_{1}(1-)\right.\) \(\left.\left.\theta_{1}\right)\right\\}\), where \(M=\sum I_{j}\), and find the corresponding quantities \(i_{A}\left(\theta_{0}, \theta_{1}\right)\) and \(i_{A}\left(\theta_{0}, \theta_{0}\right)\). If he uses this procedure for many sets of data, deduce that on average \(M\) is replaced by \(n \operatorname{Pr}(I=1)=n\left\\{\lambda \eta_{1}+(1-\lambda) \eta_{0}\right\\}\) (iii) Statistician B calculates the expected information taking into account the missingness mechanism. Show that she gets \(i_{B}\left(\theta_{1}, \theta_{1}\right)=n \lambda \eta_{1} /\left\\{\theta_{1}\left(1-\theta_{1}\right)\right\\}\), and obtain \(i_{B}\left(\theta_{0}, \theta_{1}\right)\) and \(i_{B}\left(\theta_{0}, \theta_{0}\right) .\) (iv) Show that \(\mathrm{A}\) and \(\mathrm{B}\) get the same expected information matrices only if \(Y\) is missing completely at random. Does this accord with the discussion above? (c) Statistician C argues that expected information should never be used in data analysis: even if the data actually observed are complete, unless it can be guaranteed that data could not be missing at random for any reason, every expected information calculation should involve every potential missingness mechanism. Such a guarantee is impossible in practice, so no expected information calculation is ever correct. Do you agree? (Kenward and Molenberghs, 1998\()\)

Show that the geometric density $$ f(y ; \pi)=\pi(1-\pi)^{y}, \quad y=0,1, \ldots, 0<\pi<1 $$ is an exponential family, and give its cumulant-generating function. Show that \(S=Y_{1}+\cdots+Y_{n}\) has negative binomial density $$ \left(\begin{array}{c} n+s-1 \\ n-1 \end{array}\right) \pi^{n}(1-\pi)^{s}, \quad s=0,1, \ldots $$ and that this is also an exponential family.

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.

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.

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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