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Parametric Survival Models (ch. 7). notations… Y = survival variable; S=survival function, f=density (pdf); F=cdf=1-S. All these functios involve the unknown parameter (or vector of parameters) theta ( . The 2 examples we’ll consider are: exponential with Weibull with .
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Parametric Survival Models (ch. 7) • notations… • Y = survival variable; S=survival function, f=density (pdf); F=cdf=1-S. All these functios involve the unknown parameter (or vector of parameters) theta (. The 2 examples we’ll consider are: • exponential with • Weibull with
The maximum likelihood estimator of and is a function of the observed Y’s which maximizes the likelihood, L(a constant multiple of the joint distribution of the observed data. So theta-hat maximizes L over all possible values of theta… • the mathematical method is discussed at the top of page 120 - note that usually the log of the likelihood is maximized…
I’ll list some of the important results given in this section: • define the score function as the derivative of the log-likelihood • define Fisher’s information as • then • at the bottom of p. 120 and top of p.121, the method of actually finding maximum likelihood estimators is discussed and it is shown that
Then hypothesis tests about can be based on any of the 3 statistics below. To test use one of : • Wald test: • LR test: • Score test:
Let’s use the LR test to compare models (see p. 179-180) - use the notation there: • The likelihood ratio test of the current model is • The likelihood ratio test of the full model is • Their difference (subtract the full minus the current likelihoods) is asymptotically chi-square with q d.f. and may be used to test whether the additional q parameters in the full model are zero. • This difference is called the deviance
Now go to p.127, the exponential model • Def. 7.2: Y ~ ( if the pdf of Y is here is the gamma function. Since 1)=1,, the exponential model is a special case of the gamma for 1. • The chi-square distribution is also a special case of the gamma.
here are some other relationships…(Lemma 7.1 on page 128): • now go to page 131… Maximum likelihood under Type I censoring . Suppose we have a survival variable Y ~exp(), subject to Type I censoring. Beta is the MTTF and we may use maximum likelihood methods to estimate it…(p.131 and 132)
Maximizing the likelihood equation gives the • MLE of as • so we may use this to construct a 95% CI for the MTTF
In the Weibull model we assume Y ~ W( subject to Type I censoring. Or we may work with the logs of the survival data, X=loge(Y). • The X’s have the 2-parameter (here u and b) extreme value distribution: • The survival function is given by • The original Weibull parameters can be estimated if u and b are estimated by u-hat and b-hat:
So, now go over example 7.4. Use SAS to read in the switch failure time data on p.135 (see website). Then get estimates of the the Weibull parameters for both the log-transformed and non-transformed data - use PROC LIFEREG - don’t forget: • Notice that we may use the NOLOG option in the MODEL statement to not take logs of the data… proc lifereg data=switch; model u*censor(0)= /nolog dist=weibull; title 'Modeling u=log(Y) w/ NOLOG option'; proc lifereg data=switch; model y*censor(0)= /dist=weibull; title 'Modeling non-transformed Y'; run; quit; • Return to Section 4.4 (p. 61-62) and use SAS to get a probability plot (formula 4.4) of this data…
Read section 8.5 on Diagnostics for choosing between models • Recall the empirical survivor plot: • Diagnostic 1: Plot the empirical survivor plots for each of the groups in a classification variable on the same plot. They should be “location shifted versions of the baseline survival function when an accelerated lifetimes model is appropriate” • Diagnostic 2: “Compare the standard deviations of the data in each group (or when enough data points permits, the s.d.s at each covariate level). This allows us to check the constancy of sigma in the AFT model.” • Diagnostic 3: “Plot to see if they are parallel - if so, the proportional hazards is correct”
Diagnostic 3 (continued); “These curves will be straight lines when log(y) is plotted as the x-coordinate (rather than y) and the Weibull model fits the data well.” • Go over Example 8.3 and use either R or SAS (or both!) to reproduce the diagnostic plots for this data.