# The Statistical Analysis of Recurrent Events (Statistics for Biology and Health)

## Richard J. Cook

Language: English

Pages: 404

ISBN: 0387698094

Format: PDF / Kindle (mobi) / ePub

This book presents models and statistical methods for the analysis of recurrent event data. The authors provide broad, detailed coverage of the major approaches to analysis, while emphasizing the modeling assumptions that they are based on. More general intensity-based models are also considered, as well as simpler models that focus on rate or mean functions. Parametric, nonparametric and semiparametric methodologies are all covered, with procedures for estimation, testing and model checking.

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φ−1 ) n = 0, 1, 2, . . . = Γ (φ−1 ) [1 + φµi (s, t)]n+φ−1 Pr(Ni (s, t) = n|zi ) = which is of negative binomial form. Note that the limit as φ → 0 gives the Poisson distribution (2.15). The unconditional mean and variance of Ni (s, t) can be obtained from (2.29) or by noting that E{Ni (s, t)} = E{E[Ni (s, t)|ui ]} = E{ui µi (s, t)} = µi (s, t) , (2.30) 2.2 Poisson Processes and Models for Event Counts 37 and var{Ni (s, t)} = E{var[Ni (s, t)|ui ]} + var{E[Ni (s, t)|ui ]} = E{ui µi (s, t)} +

Simple graphical displays based on nonparametric estimates of mean functions can provide insight into the diﬀerences in two groups. For example, plots of the log cumulative mean functions for the two groups should be roughly parallel if the simple multiplicative model is appropriate, whereas plots of the untransformed mean functions should be roughly parallel if an additive model is appropriate. Score tests also have appeal in this setting because of their simplicity. Consider two groups of

. aj,r−1 Show that (4.51) approaches (4.13) under the limit described above. [Section 4.2; Lawless et al., 2001; Lawless, 2003a, Section 7.4] 156 4 Analysis of Gap Times 4.4. Consider the random eﬀects model (4.16) in the case where the ui have gamma distributions with mean 1 and variance φ, with density function (2.28). Show that the likelihood contribution for an individual with ni observed gap times wij (j = 1, . . . , ni ) and a ﬁnal censored gap time wi,ni +1 is ⎧ ⎫ ⎨ ni ⎬ Γ (n + φ−1 )

likelihood (4.25). Show that this gives the test statistic ⎧ ⎫ m n +1 ⎪ ⎪ ⎪ ⎪ I(w k ≥ wij )z k ⎪ m ni ⎪ ⎨ ⎬ =1 k=1 zij − m n +1 , U (0) = ⎪ ⎪ i=1 j=1 ⎪ ⎪ ⎪ I(w k ≥ wij ) ⎪ ⎩ ⎭ =1 k=1 and obtain a variance estimate for U (0). b. The preceding test is not valid when there is heterogeneity across individuals. Consider the model where (4.53) is generalized to give hij (w) = h0i (w) exp(zij β) i = 1, . . . , m , 158 4 Analysis of Gap Times where zij is deﬁned as in part (a). In this case each

βtj−1 )}β −1 if j ≥ 2 and tj − tj−1 ≤ 20 Ej = {exp(α + βtj ) − [1 + eγ−20β − eγ ] exp[α + β(tj−1 + 20)]}β −1 if j ≥ 2 and tj − tj−1 > 20. Figure 5.2 gives plots of the ordered E(j) against the standard exponential quantiles, deﬁned as − log{1 − (j − 0.5)/30}, for j = 1, . . . , 30. The top panel, for Model 3, has a departure from linearity that might reﬂect a mixture of two distributions for the Ej instead of a single standard exponential. The bottom panel, for Model 2, is more nearly linear and