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Chapter 10: Problem 19

Two individuals with cumulative hazard functions \(u H_{1}\left(y_{1}\right)\) and \(u H_{2}\left(y_{2}\right)\) are independent conditional on the value \(u\) of a frailty \(U\) whose density is \(f(u)\) (a) For this shared frailty model, show that $$ \mathcal{F}\left(y_{1}, y_{2}\right)=\operatorname{Pr}\left(Y_{1}>y_{1}, Y_{2}>y_{2}\right)=\int_{0}^{\infty} \exp \left\\{-u H_{1}\left(y_{1}\right)-u H_{2}\left(y_{2}\right)\right\\} f(u) d u $$ If \(f(u)=\lambda^{\alpha} u^{\alpha-1} \exp (-\lambda u) / \Gamma(\alpha)\), for \(u>0\) is a gamma density, then show that $$ \mathcal{F}\left(y_{1}, y_{2}\right)=\frac{\lambda^{\alpha}}{\left\\{\lambda+H_{1}\left(y_{1}\right)+H_{2}\left(y_{2}\right)\right\\}^{\alpha}}, \quad y_{1}, y_{2}>0 $$ and deduce that in terms of the marginal survivor functions \(\mathcal{F}_{1}\left(y_{1}\right)\) and \(\mathcal{F}_{2}\left(y_{2}\right)\) of \(Y_{1}\) and \(Y_{2}\) $$ \mathcal{F}\left(y_{1}, y_{2}\right)=\left\\{\mathcal{F}_{1}\left(y_{1}\right)^{-1 / \alpha}+\mathcal{F}_{2}\left(y_{2}\right)^{-1 / \alpha}-1\right\\}^{-\alpha}, \quad y_{1}, y_{2}>0 $$ What happens to this joint survivor function as \(\alpha \rightarrow \infty\) ? (b) Find the likelihood contributions when both individuals are observed to fail, when one is censored, and when both are censored. (c) Extend this to \(k\) individuals with parametric regression models for survival.

Short Answer

Expert verified
The joint survivor function approaches the product of marginal survivor functions, indicating independence, as \(\alpha \to \infty\).

Step by step solution

01

Express Probability Using Cumulative Hazard Functions

The probability of both individuals surviving beyond their respective times is given by \[ \mathcal{F}(y_1, y_2) = \operatorname{Pr}(Y_1 > y_1, Y_2 > y_2) = \int_{0}^{\infty} \exp \{-u H_1(y_1) - u H_2(y_2)\} f(u) du \]. This equation uses the property that survival times of individuals are conditionally independent given the frailty \(u\).
02

Substitute the Gamma Density Function

Substitute the given gamma density function for \(f(u)\) into the integral. The gamma density function is \(f(u) = \frac{\lambda^{\alpha} u^{\alpha-1} \exp(-\lambda u)}{\Gamma(\alpha)}\). Substituting this in, we have: \[ \mathcal{F}(y_1, y_2) = \int_{0}^{\infty} \frac{\lambda^{\alpha} u^{\alpha-1} \exp(-u(H_1(y_1) + H_2(y_2) + \lambda))}{\Gamma(\alpha)} du \].
03

Solve the Integral

The integral \( \int_{0}^{\infty} u^{\alpha-1} \exp(-u(H_1(y_1) + H_2(y_2) + \lambda)) du \) can be expressed as a gamma function \(\Gamma(\alpha)\) when factoring: \[ \frac{1}{(H_1(y_1) + H_2(y_2) + \lambda)^{\alpha}} \]. Simplifying the entire expression gives \[ \mathcal{F}(y_1, y_2) = \frac{\lambda^{\alpha}}{(H_1(y_1) + H_2(y_2) + \lambda)^{\alpha}} \].
04

Relate to Marginal Survivor Functions

Relate the joint survivor function \( \mathcal{F}(y_1, y_2) \) to the marginal survivor functions \( \mathcal{F}_1(y_1) \) and \( \mathcal{F}_2(y_2) \). These are found by recognizing that the marginal survivor functions relate to the cumulative hazard functions as: \( \mathcal{F}_1(y_1) = \exp(-H_1(y_1)) \) and \( \mathcal{F}_2(y_2) = \exp(-H_2(y_2)) \). Thus, \[ \mathcal{F}(y_1, y_2) = \left(\mathcal{F}_1(y_1)^{-1/\alpha} + \mathcal{F}_2(y_2)^{-1/\alpha} - 1\right)^{-\alpha} \].
05

Analyze as 伪 Approaches Infinity

When \( \alpha \to \infty \), the term \( \left(\mathcal{F}_1(y_1)^{-1/\alpha} + \mathcal{F}_2(y_2)^{-1/\alpha} - 1\right)^{-\alpha} \) approaches \( \mathcal{F}_1(y_1) \cdot \mathcal{F}_2(y_2) \). Thus, the joint survival function becomes the product of marginal survivors, indicating independence.
06

Likelihood Contributions

- **Both Fail:** Given failures at \(y_1\) and \(y_2\), the likelihood contribution includes the density function \( f(U) \) and terms from the frailty-compounded hazard densities.- **One Censored:** The likelihood accounts for one survival term and the other's hazards.- **Both Censored:** The contribution only involves the survival distribution powered with frailty.
07

Extend to k Individuals

For \(k\) individuals with regression models, express the joint survival using Covariate-driven hazards: \(H_i(y) = \exp(X_i^T \beta) H_0(y)\). The joint survival function modifies to accommodate individual-specific covariates and a shared frailty.

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Key Concepts

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

Cumulative Hazard Function
The cumulative hazard function is a fundamental concept in survival analysis. It provides a way to characterize the time until an event happens, such as death or failure, in a population. The cumulative hazard function at a particular time is the integral of the hazard rate over time up to that point. More simply, it is a running total of the instantaneous risk. Key points to remember about cumulative hazard functions:
  • They accumulate hazard rates over time, reflecting increasing risk.
  • Higher values of the cumulative hazard function indicate a greater accumulated risk up to that point in time.
  • The cumulative hazard function is closely linked to the survival function, as the survival function can be expressed in terms of the cumulative hazard function: \( S(y) = e^{-H(y)} \) where \( S(y) \) is the survival probability at time \( y \), and \( H(y) \) is the cumulative hazard function.
By understanding the cumulative hazard, one gains insights into the dynamics of risk over time.
Frailty Model
Frailty models are widely used in survival analysis to account for unobserved heterogeneity among individuals. When studying time-to-event data, not all sources of risk may be directly observable. Frailty models introduce a random effect鈥攖ermed 'frailty'鈥攖o model this unobserved variability. The basics of frailty models include:
  • Frailty acts multiplicatively on the hazard function, capturing the effect of unobserved risk factors.
  • Individuals with high frailty values have higher hazard rates and thus are more likely to experience the event sooner.
  • In the shared frailty model, a common frailty is assumed to affect grouped individuals similarly, allowing for correlated event times within a group.
In the exercise, frailty is modeled as a gamma-distributed random variable, which allows for convenient mathematical properties that simplify analysis and interpretation.
Gamma Distribution
The gamma distribution is a continuous probability distribution often used in survival analysis, particularly in frailty modeling. It is characterized by two parameters, \(\alpha\) (shape) and \(\lambda\) (rate), which influence its form and properties. Key features of the gamma distribution:
  • The gamma distribution is flexible and can model various shapes, from exponential-type decay to more flexible curves, making it versatile for different data patterns.
  • In survival analysis, it is commonly used to model frailty. The choice of gamma distribution for frailty is advantageous because it mathematically simplifies the integration needed to derive the joint survival function.
  • The gamma function, \( \Gamma(\alpha) \), appears in its probability density function, simplifying the representation of combined hazards in the shared frailty model.
By utilizing a gamma frailty, complex dependencies in survival data can be accounted for more effectively, leading to more accurate models.
Regression Models for Survival
Regression models for survival analysis are instrumental in analyzing time-to-event data while considering the impact of covariates. These models help understand how factors like age, gender, or baseline health status affect survival times. Essentials of regression models in survival analysis:
  • They incorporate covariates, which are explanatory variables that potentially influence the hazard rate.
  • A common approach is to assume that these covariates affect the hazard multiplicatively \( H_i(y) = \exp(X_i^T \beta) H_0(y) \) where \( X_i \) are the covariates, \( \beta \) are the coefficients which need to be estimated, and \( H_0(y) \) is the baseline hazard function.
  • Commonly used regression models for survival include the Cox proportional hazards model and parametric models like the Weibull and exponential models.
These models extend survival analysis beyond simple descriptive statistics to powerful inferential techniques, allowing for more nuanced insights into how and why events occur.

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Most popular questions from this chapter

Consider independent exponential variables \(Y_{j}\) with densities \(\lambda_{j} \exp \left(-\lambda_{j} y_{j}\right)\), where \(\lambda_{j}=\) \(\exp \left(\beta_{0}+\beta_{1} x_{j}\right), j=1, \ldots, n\), where \(x_{j}\) is scalar and \(\sum x_{j}=0\) without loss of generality. (a) Find the expected information for \(\beta_{0}, \beta_{1}\) and show that the maximum likelihood estimator \(\widehat{\beta}_{1}\) has asymptotic variance \(\left(n m_{2}\right)^{-1}\), where \(m_{2}=n^{-1} \sum x_{j}^{2}\) (b) Under no censoring, show that the partial log likelihood for \(\beta_{1}\) equals $$ -\sum_{j=1}^{n} \log \left\\{\sum_{i=j}^{n} \exp \left(\beta_{1} x_{(i)}\right)\right\\} $$ where the elements of the rank statistic \(R=\\{(1), \ldots,(n)\\}\) are determined by the ordering on the failure times, \(y_{(1)}<\cdots

The rate of growth of an epidemic such as AIDS for a large population can be estimated fairly accurately and treated as a known function \(g(t)\) of time \(t\). In a smaller area where few cases have been observed the rate is hard to estimate because data are scarce. However predictions of the numbers of future cases in such an area must be made in order to allocate resources such as hospital beds. A simple assumption is that cases in the area arise in a non- homogeneous Poisson process with rate \(\lambda g(t)\), for which the mean number of cases in period \(\left(t_{1}, t_{2}\right)\) is \(\lambda \int_{t_{1}}^{t_{2}} g(t) d t\). Suppose that \(N_{1}=n_{1}\) individuals with the disease have been observed in the period \((-\infty, 0)\), and that predictions are required for the number \(N_{2}\), of cases to be observed in a future period \(\left(t_{1}, t_{2}\right)\). (a) Find the conditional distribution of \(N_{2}\) given \(N_{1}+N_{2}\), and show it to be free of \(\lambda\). Deduce that a \((1-2 \alpha)\) prediction interval \(\left(n_{-}, n_{+}\right)\)for \(N_{2}\) is found by solving approximately the equations $$ \begin{aligned} &\alpha=\operatorname{Pr}\left(N_{2} \leq n_{-} \mid N_{1}+N_{2}=n_{1}+n_{-}\right) \\ &\alpha=\operatorname{Pr}\left(N_{2} \geq n_{+} \mid N_{1}+N_{2}=n_{1}+n_{+}\right) \end{aligned} $$ (b) Use a normal approximation to the conditional distribution in (a) to show that for moderate to large \(n_{1}, n_{-}\)and \(n_{+}\)are the solutions to the quadratic equation $$ (1-p)^{2} n^{2}+p(p-1)\left(2 n_{1}+z_{\alpha}^{2}\right) n+n_{1} p\left\\{n_{1} p-(1-p) z_{\alpha}^{2}\right\\}=0 $$ where \(\Phi\left(z_{\alpha}\right)=\alpha\) and $$ p=\int_{t_{1}}^{t_{2}} g(t) d t /\left\\{\int_{t_{1}}^{t_{2}} g(t) d t+\int_{-\infty}^{0} g(t) d t\right\\} $$ (c) Find approximate \(0.90\) prediction intervals for the special case where \(g(t)=2^{t / 2}\), so that the doubling time for the epidemic is two years, \(n_{1}=10\) cases have been observed until time 0 , and \(t_{1}=0, t_{2}=1\) (next year) (Cox and Davison, 1989).

Suppose that the cumulant-generating function of \(X\) can be written in the form \(m\\{b(\theta+\) \(t)-b(\theta)\\}\). Let \(\mathrm{E}(X)=\mu=m b^{\prime}(\theta)\) and let \(\kappa_{2}(\mu)\) and \(\kappa_{3}(\mu)\) be the variance and third cumulant respectively of \(X\), expressed in terms of \(\mu ; \kappa_{2}(\mu)\) is the variance function \(V(\mu)\). (a) Show that $$ \kappa_{3}(\mu)=\kappa_{2}(\mu) \kappa_{2}^{\prime}(\mu) \quad \text { and } \quad \frac{\kappa_{3}}{\kappa_{2}^{2}}=\frac{d}{d \mu} \log \kappa_{2}(\mu) $$ Verify that the binomial cumulants have this form with \(b(\theta)=\log \left(1+e^{\theta}\right)\). (b) Show that if the derivatives of \(b(\theta)\) are all \(O(1)\), then \(Y=g(X)\) is approximately symmetrically distributed if \(g\) satisfies the second-order differential equation $$ 3 \kappa_{2}^{2}(\mu) g^{\prime \prime}(\mu)+g^{\prime}(\mu) \kappa_{3}(\mu)=0 $$ Show that if \(\kappa_{2}(\mu)\) and \(\kappa_{3}(\mu)\) are related as in (a), then $$ g(x)=\int^{x} \kappa_{2}^{-1 / 3}(\mu) d \mu $$ (c) Hence find symmetrizing transformations for Poisson and binomial variables. (McCullagh and Nelder, 1989 , Section 4.8)

If \(X\) is a Poisson variable with mean \(\mu=\exp \left(x^{\mathrm{T}} \beta\right)\) and \(Y\) is a binary variable indicating the event \(X>0\), find the link function between \(\mathrm{E}(Y)\) and \(x^{\mathrm{T}} \beta\).

For a generalized linear model with known dispersion parameter \(\phi\) and canonical link function, write the deviance as \(\sum_{j=1}^{n} d_{j}^{2}\), where \(d_{j}^{2}\) is the contribution from the \(j\) th observation. Also let $$ u_{j}(\beta)=\partial \log f\left(y_{j} ; \eta_{j}, \phi\right) / \partial \eta_{j}, \quad w_{j}=-\partial^{2} \log f\left(y_{j} ; \eta_{j}, \phi\right) / \partial \eta_{j}^{2} $$ denote the elements of the score vector and observed information, let \(X\) denote the \(n \times p\) matrix whose \(j\) th row is \(x_{j}^{\mathrm{T}}\), where \(\eta_{j}=\beta^{\mathrm{T}} x_{j}\), and let \(H\) denote the matrix \(W^{1 / 2} X\left(X^{\mathrm{T}} W X\right)^{-1} X^{\mathrm{T}} W^{1 / 2}\), where \(W=\operatorname{diag}\left\\{w_{1}, \ldots, w_{n}\right\\}\) (a) Let \(\widehat{\beta}_{(k)}\) be the solution of the likelihood equation when case \(k\) is deleted, $$ \sum_{j \neq k} x_{j} u_{j}\left(\widehat{\beta}_{(k)}\right)=0 $$ and let \(\widehat{\beta}\) be the maximum likelihood estimate based on all \(n\) observations. Use first-order Taylor series expansion of \((10.65)\) about \(\widehat{\beta}\) to show that $$ \widehat{\beta}_{(k)} \doteq \widehat{\beta}-\left(X^{\mathrm{T}} W X\right)^{-1} x_{k} \frac{u_{k}(\widehat{\beta})}{1-h_{k k}} $$ Express \(\widehat{\beta}_{(k)}\) in terms of the standardized Pearson residual \(r_{P k}=u_{k} /\left\\{w_{k}\left(1-h_{k k}\right)\right\\}^{1 / 2}\) (b) Use a second order Taylor series expansion of the deviance to show that the change in the deviance when the \(k\) th case is deleted is approximately $$ r_{G k}^{2}=\left(1-h_{k k}\right) r_{D k}^{2}+h_{k k} r_{P k}^{2} $$ where \(r_{D k}\) is the standardized deviance residual \(d_{k} /\left(1-h_{k k}\right)^{1 / 2}\). (c) Suppose models \(A\) and \(B\) have deviances \(D_{A}\) and \(D_{B}\). Use (b) to find an expression for the change in the likelihood ratio statistic \(D_{A}-D_{B}\), when the \(k\) th case is deleted. (d) Show that your results (a)-(c) are exact in models with normal errors.
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