a frailty model for censored family survival data
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A FRAILTY MODEL FOR CENSORED FAMILY SURVIVAL DATA, applied to the - PowerPoint PPT Presentation

Marianne Jonker VU University Amsterdam, The Netherlands. A FRAILTY MODEL FOR CENSORED FAMILY SURVIVAL DATA, applied to the age at onset of mental problems Joint work with: D.I Boomsma Motivation Observations from the Netherlands Twin


  1. Marianne Jonker VU University Amsterdam, The Netherlands. A FRAILTY MODEL FOR CENSORED FAMILY SURVIVAL DATA, applied to the age at onset of mental problems Joint work with: D.I Boomsma

  2. Motivation Observations from the Netherlands Twin Register: For a study on health and personality, longitudinal data were collected of twins and their sibs at different time points between 1991 and 2002. At every timepoint it is observed whether the individual ever had contacted social service before.

  3. Motivation Observations from the Netherlands Twin Register: For a study on health and personality, longitudinal data were collected of twins and their sibs at different time points between 1991 and 2002. At every timepoint it is observed whether the individual ever had contacted social service before. Aims: • to estimate the degree of heritability and environmental effects of the age at which individuals contact social service for non-physical problems for the first time, • to test if the survival functions for twins and non-twins differ.

  4. Heritability Heritability is the proportion of phenotypic variation in a population that is attributable to genetic variation among the individuals in that population. Phenotypic variation may be due to genetic and environmental factors. If a trait has a heritability of 0.5, this means that the phenotypic variation due to genetic variation is 50%. It does not imply that the trait is caused by genetics for 50%. Example: body height. Heritability of “height” in Western society is approximately ..%. Nowadays, environmental effects are very homogenous among these populations. Decades ago, the heriatabilty of height was probably much lower.

  5. Heritability Heritability is the proportion of phenotypic variation in a population that is attributable to genetic variation among the individuals in that population. Phenotypic variation may be due to genetic and environmental factors. If a trait has a heritability of 0.5, this means that the phenotypic variation due to genetic variation is 50%. It does not imply that the trait is caused by genetics for 50%. Example: body height Heritability of “height” in Western society is approximately 80%. Nowadays, environmental effects are very homogenous among these populations. Decades ago, the heriatabilty of height was probably much lower.

  6. Body height Heritability is implied if mono-zygotic twins are more similar to each other than dizygotic twins for the trait under study. 180 195 195 185 175 190 190 185 170 175 185 180 165 180 175 165 160 175 170 155 170 165 155 170 180 190 155 160 165 170 175 180 170 175 180 185 190 195 160 165 170 175 180 185

  7. Estimation of Heritability Suppose we are interested in a trait that has been observed for every individual. We model the trait value as: T = µ + G + C + E. Then for a monozygotic twin: T MZ = µ + G + C + E 1 1 T MZ = µ + G + C + E 2 2 and for a dizygotic twin T DZ = µ + G 1 + C + E 1 1 T DZ = µ + G 2 + C + E 2 . 2

  8. Estimation of Heritability Heritability is defined as h 2 = Var G Var T . It can be computed that ρ MZ = Var G + Var C ρ DZ = 0 . 5Var G + Var C . Var T Var T So, h 2 = Var G Var T = 2( ρ MZ − ρ DZ ) . The term at the right can be easily estimated by inserting sample correlations.

  9. Interval Censored Data The age at which an individual visists social service for the first time is interval censored; we do not observe the exact age, but an interval in which this age falls. An interval may start at zero or end at infinity. Heritability as defined before can not be estimated.

  10. The model, hazards Suppose T is the age at which an individual contacted social service for the first time with distribution function F and density f , then the hazard is defined as f ( t ) λ ( t ) = 1 − F ( t ) � t e − Λ( t ) , 1 − F ( t ) = Λ( t ) = λ ( s )d s 0 Interpretation If f ( t ) dt is interpreted as the probability that an individual visits social service for the first time in the interval [ t, t + dt ), then λ ( t ) dt ≈ P( t ≤ T < t + dt ) = P( t ≤ T < t + dt | T ≥ t ) P( T ≥ t )

  11. The model, hazards Suppose T is the age at which an individual contacted social service for the first time with distribution function F and density f , then the hazard is defined as f ( t ) λ ( t ) = 1 − F ( t ) � t e − Λ( t ) , 1 − F ( t ) = Λ( t ) = λ ( s )d s 0 Interpretation If f ( t ) dt is interpreted as the probability that an individual visits social service for the first time in the interval [ t, t + dt ), then λ ( t ) dt ≈ P( t ≤ T < t + dt ) = P( t ≤ T < t + dt | T ≥ t ) P( T ≥ t )

  12. Statistical Model (Frailty Model) A frailty model is a random effects proportional hazards model. The hazard function is defined as: λ i ( t | Z i ) = Z i λ ( t ). Family survival data can be modelled with a correlated frailty model where the random effects (“frailties”) account for the dependence between family members. Let: T 1 , T 2 , T 3 : ages at onset of mental help for a twin and a sib Z 1 , Z 2 , Z 3 : corresponding frailty variables • T 1 , T 2 , T 3 are independent given ( Z 1 , Z 2 , Z 3 ) • T 1 , T 2 , T 3 have hazard functions t → λ i ( t | Z i ) = Z i λ ( t ) , i = 1 , 2 , 3 Equivalently P( T 1 > t 1 , T 2 > t 2 , T 3 > t 3 | Z 1 , Z 2 , Z 3 ) = e − Z 1 Λ( t 1 ) − Z 2 Λ( t 2 ) − Z 3 Λ( t 3 ) .

  13. Statistical Model (Frailty Model) A frailty model is a random effects proportional hazards model. The hazard function is defined as: λ i ( t | Z i ) = Z i λ ( t ). Family survival data can be modelled with a correlated frailty model where the random effects (“frailties”) account for the dependence between family members. Let: T 1 , T 2 , T 3 : ages at onset of mental help for a twin and a sib Z 1 , Z 2 , Z 3 : corresponding frailty variables • T 1 , T 2 , T 3 are independent given ( Z 1 , Z 2 , Z 3 ) • T 1 , T 2 , T 3 have hazard functions t → λ i ( t | Z i ) = Z i λ ( t ) , i = 1 , 2 , 3 Equivalently P( T 1 > t 1 , T 2 > t 2 , T 3 > t 3 | Z 1 , Z 2 , Z 3 ) = e − Z 1 Λ( t 1 ) − Z 2 Λ( t 2 ) − Z 3 Λ( t 3 ) .

  14. Statistical Model (Frailty Variable) The coordinates in the frailty vector ( Z 1 , Z 2 , Z 3 ) are decomposed as Z i = A i + C + B i + E i i = 1 , 2 , 3 with • A i : genetic effects for individual i • C : common environmental effects for twin and sib • B i : specific twin environmental effects • E i : non-shared, specific environmental and genetic effects

  15. Model Assumptions Z i = A i + C + B i + E i i = 1 , 2 , 3 • A i : cor( A 1 , A 2 ) = 1 if MZ-twin-pair, otherwise cor( A i , A j ) = 1 / 2 • B i : cor( B 1 , B 2 ) = 1, otherwise cor( B i , B j ) = 0 , i � = j • E i : E 1 , E 2 , E 3 are independent; cor( E i , E j ) = 0 , i � = j

  16. Model Assumptions Z i = A i + C + B i + E i i = 1 , 2 , 3 • A i : cor( A 1 , A 2 ) = 1 if MZ-twin-pair, otherwise cor( A i , A j ) = 1 / 2 • B i : cor( B 1 , B 2 ) = 1, otherwise cor( B i , B j ) = 0 , i � = j • E i : E 1 , E 2 , E 3 are independent; cor( E i , E j ) = 0 , i � = j Furthermore: • ( A 1 , A 2 , A 3 ) , C, ( B 1 , B 2 , B 3 ) and ( E 1 , E 2 , E 3 ) are independent. • A i , C, B i , E i have gamma-distributions with inverse scale parameter η and shape parameter ν, ν c , ν b , ν e , respectively. So, Z i has a gamma distribution with parameters η and ν + ν c + ν b + ν e Assume: η = ν + ν c + ν b + ν e , then E Z 1 = E Z 2 = E Z 3 = 1.

  17. Heritability, Environmental and Twin effect Heritability, and environmental and twin effect are defined as the proportions of variance of the frailty associated with genetic, environmental and twin effects: h 2 = Var A i Var Z = ν c 2 = Var C Var Z = ν c b 2 = ν b η η η (see also Yashin et al, 1999 and Jonker et al, 2009). These definitions are as usual, except that the frailties are viewed as the latent phenotypes. Then, the correlations between the frailty variables of two individuals equal h 2 + c 2 + b 2 , ρ MZ = 0 . 5 h 2 + c 2 + b 2 , ρ DZ = 0 . 5 h 2 + c 2 . ρ S,S =

  18. Simultaneous Survival function Because the vector of frailties has a multivariate gamma distribution, the simultaneous survival functions can be written in terms of the marginal survival function S . For instance, S MZ,S ( t 1 , t 2 , t 3 ) = P( T 1 > t 1 , T 2 > t 2 , T 3 > t 3 ) S ( t 1 ) − σ 2 + S ( t 2 ) − σ 2 + S ( t 3 ) − σ 2 − 2 � − ν/ 2 − ν c � = S ( t 1 ) − σ 2 + S ( t 2 ) − σ 2 − 1 � − ν/ 2 − ν b � × × S ( t 3 ) ( ν/ 2+ ν b ) σ 2 ( S ( t 1 ) S ( t 2 ) S ( t 3 )) ν e σ 2 (1) with σ 2 = 1 /η and with S the marginal survival function. S can be taken gender and/or twin - non twin specific.

  19. Aims • to estimate the degree of heritability ( h 2 ), environmental effects ( c 2 ), and twin effects ( b 2 ), • to investigate whether the survival functions differ for twins and non-twins. Steps: • Compute the likelihood (interval censored data) • Estimate the unknown parameters in the model (maximum likelihood estimators) • Estimate h 2 , c 2 , b 2 and test whether they are positive (likelihood ratio test) • Test whether the survival fuctions for twins and non-twins differ (likelihood ratio test)

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