1. Lecture 3: Parametric Survival Modeling
Parametric models
Example and nuances in R

2. Parametric Distributions
We’ve discussed a variety of parametric distributions
Exponential, Weibull, log-normal, log-logistic, gamma, ….
But… how do we “fit” a model
Model parameterizations
Inclusion of coefficients

3. Modeling Homogeneous Population
Relatively “simple”
Once we’ve determined the distribution we need to estimate the parameters
For example, exponential

4. Covariates Frequently want to adjust survival for covariates
Two main approaches
Accelerated Failure Time model
Multiplicative model

5. Accelerated Failure Time
Under AFT model for two populations
expected survival time
median survival time
Survival at time t
for Population 1 are c times that of population 2, where c is constant.

6. Accelerated Failure Time
Data include
Failure time T > 0
Vector of covariates Z’=(Z1, Z2, …, Zp)
Quantitative
Qualitative
Log transform T for linear model approach

7. Accelerated Failure Time
When Z = 0, So(t) is survival function of em+sW

8. Accelerated Failure Time
First consider 2 populations that only differ by 1 unit in zk

9. Accelerated Failure Time
First consider 2 populations that only differ by 1 unit in zk

10. Exponential Models in R
Recall:
Parameterization is the same for exponential in R
rexp(n, rate)

11. Exponential Models in R
We can run an expontial survival model in R using
survreg(formula, data, dist)
R gives us:
But, we can find:
In a model with no covariates,

12. Exponential Models in R
The distribution of any T is exponential with constant hazard rate:
We can interpret as the hazard ratio corresponding to a 1 unit increase in the covariate

13. Weibull Models in R Recall: Now our scale parameter is no longer 1
Unlike exponential, the parameterization for Weibull is different in R…
Random weibull generation…
rweibull(n, shape, scale)

14. Weibull Models in R
Again we can run a Weibull model in R but parameterization different here too…
survreg(formula, data, dist)
R gives us:
But, we can find:

15. AML Example In R Survival in patients with Acute Myelogenous Leukemia.
Data
23 Subjects
Time to death
Censoring indicator
Treatment
Standard course of chemotherapy
Chemo extended ('maintainance') for additional cycles.

16. AML Dataset
> library(survival) > aml time status x Maintained Maintained Maintained Maintained Maintained Maintained Maintained Maintained Maintained Maintained Maintained Nonmaintained Nonmaintained Nonmaintained Nonmaintained Nonmaintained … Nonmaintained

17. AML model in R: exponential (no covariates)
>library(MASS) >library(survival) >sdat<-Surv(aml$time, aml$status) >exp_fit<-survreg(sdat~1, dist=“exponential“) >#exp_fit<-survreg(sdat~1, dist="weibull", scale=1)  alternative >summary(exp_fit) Call: survreg(formula = sdat ~ 1, dist = "exponential") Value Std. Error z p (Intercept) e-53 Scale fixed at 1 Exponential distribution Loglik(model)= Loglik(intercept only)= Number of Newton-Raphson Iterations: 4 n= 23

18. Checking Exponential Model Fit

19. Model Checks: Exponential
###Model checks for exponential par(mfrow=c(1,3)) lam_hat<-exp(-exp_fit$coefficient) logHt<-log(-log(emp_fit$surv)) logt<-log(emp_fit$time) # Plot log cumulative hazard vs. log time plot(logt, logHt, lwd=2, type="l", xlab="log(t)", ylab="log(H(t))") points(logt, logHt, pch=16) abline(log(lam_hat), 1, lwd=2, col="red") # Second model check: Plot of H(t) vs. t Ht<--log(emp_fit$surv) t<-emp_fit$time plot(t, Ht, lwd=2, type="l", xlab="time", ylab="H(t)") points(t, Ht, pch=16) abline(0,lam_hat, lwd=2, col="red") #Third model check fit.dat<-exp(-lam_hat*c(0:150)) plot(emp_fit, xlab="Time", ylab="Survival Fraction") lines(c(0:150), fit.dat, lwd=2, col=2)

20. Lets Look at some Specifics for Exponential
Exponential Model… 1st estimate lambda
12 month survival ?
Median survival ?
Mean survival ?

21. AML model in R: Weibull (no covariates)
>weib_fit<-survreg(sdat~1, dist="weibull", scale=0) > summary(weib_fit) Call: survreg(formula = sdat ~ 1, dist = "weibull", scale = 0) Value Std. Error z p (Intercept) e-63 Log(scale) e-01 Scale= Weibull distribution Loglik(model)= Loglik(intercept only)= Number of Newton-Raphson Iterations: 5 n= 23

22. Model Checks: Weibull

23. Model Checks: Weibull
###Model checks for weibull alp_hat<-1/exp(weib_fit$scale) lam_hat<-exp(-weib_fit$coefficient[1]/exp(weib_fit$scale)) logHt<-log(-log(emp_fit$surv)) logt<-log(emp_fit$time) # Plot log cumulative hazard vs. log time plot(logt, logHt, lwd=2, type="l", xlab="log(t)", ylab="log(H(t))") points(logt, logHt, pch=16) abline(log(lam_hat), alp_hat, lwd=2, col="red") # Plot of survival function vs. empircal fit.dat<-exp(-lam_hat*c(0:150)^alp_hat) plot(emp_fit, xlab="Time", ylab="Survival Fraction") lines(c(0:150), fit.dat, lwd=2, col=2)

24. Lets Look at some Specifics for Weibull
1st estimate lambda and alpha
12 month survival ?
Median survival ?
Mean survival ?

25. Compare Weibull/Exponential Fits to the Empirical Distribution (no covariates)

26. Empirical Distribution: What about specific times (no covariates)?
12 month survival = 0.74
Median survival = 27

27. What about relative to the empirical distribution (no covariates)?
12 month survival = 74%
Median survival = 27 months
Exponential Model:
12 month survival = 73%
Median survival = 26.1 months
Mean survival = 37.7 months
Weibull model:
12 month survival = 75.5%
Median survival = 27.3 months
Mean survival = 36.9 months

28. What about covariates….

29. AML model in R: exponential (with covariate)
> exp_fit2<-survreg(sdat~x, dist="exponential", data=aml) > summary(exp_fit2) Call: survreg(formula = sdat ~ x, data = aml, dist = "exponential") Value Std. Error z p (Intercept) e-27 xNonmaintained e-02 Scale fixed at 1 Exponential distribution Loglik(model)= Loglik(intercept only)= Chisq= 4.06 on 1 degrees of freedom, p= Number of Newton-Raphson Iterations: 4 n= 23

30. Exponential Fit by Group

31. What about estimates by group?
Maintiained?
Non-maintained?

32. AML model in R: Weibull (with covariates)
> weib_fit2<-survreg(sdat~x, dist="weibull", data=aml, scale=0) > summary(weib_fit2) Call: survreg(formula = sdat ~ x, data = aml, dist = "weibull", scale = 0) Value Std. Error z p (Intercept) e-43 xNonmaintained e-02 Log(scale) e-01 Scale= Weibull distribution Loglik(model)= Loglik(intercept only)= Chisq= 5.31 on 1 degrees of freedom, p= Number of Newton-Raphson Iterations: 5 n= 23

33. Weibull fit

34. What about estimates by group?
Maintiained?
Non-maintained?

35. Exponential and Weibull Fits for Maintained vs. Non-maintained
Red = weibull
Blue = exponetial

36. Empirical Distribution: What about specific survival times (with covariate)?
Maintained:
12 month survival = 91%
Median survival = 31 months
Non-Maintained:
12 Month survival = 58%
Median survival = 23 months

37. Comparisons Maintained Empirical: Exponential Model: Weibull model:
Non-maintained
Empirical:
12 month survival = 91%
Median survival = 31 months
Exponential Model:
12 month survival = 82%
Median survival = 41.9 months
Weibull model:
12 month survival = 88%
Median survival = 45.9 months
Empirical:
12 month survival = 58%
Median survival = 23 months
Exponential Model:
12 month survival = 60%
Median survival = 16.1 months
Weibull model:
12 month survival = 66%
Median survival = 18 months

38. Compare Exponential & Empirical Distribution (with covariates)

39. Compare Weibull & Empirical Distribution (with covariates)

41. Multiplicative Hazard Rate Models
Hazard rate of individual with covariate vector z is:
In these models ho(t) may be parametric or arbitrary non-negative function
Most common link function proposed by Cox

42. Multiplicative Hazard Rate Models
Key feature is proportional hazards

43. Multiplicative Hazard Rate Model
These parametric models are very similar to semi-parametric Cox proportional hazard models we will discuss later…
The AFT models using the exponential/Weibull are also classified as multiplicative models due to their proportional hazards property
This is not true for any other parametric distribution
Since Cox models are so commonly used, it is rare to see a parametric implementation of these models

44. Advantages of Parametric Models
If we correctly characterize the underlying distribution, our estimates will be more precise than semi- and non-parametric estimates.
This means we may have greater power to identify relationships between our outcome and predictors
However…

45. Disadvantages of Parametric Models
If we use the wrong distribution problems can arise
Distribution often chosen based on the shape of the model without covariates,
This can/will change as covariates are added\
Alternatively use intuition/theory about what the dependency is expected to be
BUT the time-dependency is what is left over after conditioning on covariates so we are also likely to fail here.

46. Brief SAS Code  /************************************/
/* Accelerated Failure Time Models */
/*Exponential models: 1st is intercept only, second is with the covariate*/
proc lifereg data=aml;
model time*status(0) = /dist=exponential;
run;
class x;
model time*status(0) = x/dist=exponential;

47. Brief SAS Code 
/************************************/ /* Accelerated Failure Time Models */ /*Weibull models: 1st is intercept only, second is with the covariate*/ proc lifereg data=aml; model time*status(0) = /dist=weibull; run; class x; model time*status(0) = x/dist=weibull;

48. Example of SAS Output
Analysis of Maximum Likelihood Parameter Estimates
Parameter DF
Estimate
Standard Error
95% Confidence Limits
Chi-Square
Pr > ChiSq
Intercept 1
3.6288
0.2357
3.1668
4.0907
237.02
<.0001
Scale
1.0000
0.0000
Weibull Scale
8.8781
Weibull Shape
Lagrange Multiplier Statistics
Parameter
Chi-Square
Pr > ChiSq
Scale
0.3305
0.5654

49. Next Time
Likelihoods!!!