Solved – confirm or validate underlining distribution association with survival analysis

distributionsepidemiologyrsurvivalweibull distribution

The following two question outline how one can plot the results from a survival analysis using R. Q1 and Q2

But both of the examples assume, or more directly specify a weibull distribution fitted to the survival model.

Refering to survreg() which is within the survival package in R, the following are possible fitting assumptions:

weibull
exponential
gaussian
logistic
lognormal
loglogistic

So if we take the example from the survreg() help.

library(survival)

data(ovarian)

head(ovarian)

survival.weibull <- survreg(Surv(time, status) ~ ph.ecog + age + strata(sex), 
                          dist='weibull', lung)

survival.logn <- survreg(Surv(time, status) ~ ph.ecog + age + strata(sex), 
                            dist='lognormal', lung)

survival.logl <- survreg(Surv(time, status) ~ ph.ecog + age + strata(sex), 
                         dist='loglogistic', lung)

How can one verify the best appropriate distribution to fit. From the summary statistic we get the Loglik(model) value. Is this the best indicator? Or is there a graphical method to visualise the best fit – I was suggested a QQ-plot may be of help.

Thanks in advance for an advice

Best Answer

The use of log-likelihoods are problematic because not all the survival models are nested within one another. But you can use something like AIC or BIC, which can be obtained using extractAIC(). For example:

> extractAIC(survival.weibull)
[1]    5.000 2284.504
> extractAIC(survival.logn)
[1]    5.000 2315.142
> extractAIC(survival.logl) 
[1]    5.000 2297.171

This would suggest, based on AIC, that the Weibull model fits best. In terms of visualizing the best fit, you could plot all three parametric fits atop a non-parameteric Kaplan-Meier curve to visualize the fit of the parametric forms to the underlying data. The means to do that is in your first question.

Related Solutions

Solved – fitting a log-logistic distribution

I am not sure I understand your problem, but this can be done directly with fitdistr. See the following code:

library(fitdistrplus)
library(actuar)

test <- actuar::rllogis(500, 3.3, 1)
fitdist(test, "llogis")
Fitting of the distribution ' llogis ' by maximum likelihood 
Parameters:
      estimate Std. Error
shape 3.374404 0.12585423
scale 0.998631 0.02296057

The fitdistr function do not have a set of predefined distribution models, it just assumes that pdqr... functions corresponding to basename llogis exist somewhere on the search path.

Solved – parametric survival regression and discrete time survival regression

First, with at most one bankruptcy event per company, you don't have data for a panel survival model as described on the page linked in your question. Quoting from that page:

Obviously, in survival data, we have repeated observations on the same person because we observed them over a period of time, from onset of risk until failure or the calling off of the data collection effort. Sometimes the multiple observations on a person are explicit; the data themselves contain multiple observations for some or all the individuals. That happens when covariates change over time. Other times, the multiple observations on the individuals are implicit; there is only one physical observation for each, but still that observation records a span of time.

Those kinds of repeated observations have nothing to do with panel data. Panel data arises, for instance, when individuals are from different countries and it was believed that country affects survival. In that case, in a panel-data model, there would be a random effect or, if you prefer, an unobserved latent effect for each country.

We can, however, write models in which the random effect occurs at the individual level if we have repeated failure events for them. (Emphasis added)

For example, if you were studying how quickly different people caught a common cold, which can happen often, you might include as a random effect the different tendencies for individuals to catch a cold, given the covariate values. But with only one event per individual you can't do that. What you have are data arranged in a standard format for survival analysis with time-dependent covariates. From the structure of your data you don't seem to have any random effects.

Second, unless you have a really strong reason to suspect that your survival data take a particular parametric form, you will typically be better off trying the semi-parametric approach of Cox proportional hazards regression rather than a parametric model. With Cox regression you don't need to know how the underlying hazard changes over time, removing an important assumption that you would have to make with a parametric model. Yes, you only have data for one time point per year, but that's no more incompatible with a Cox model than it would be with parametric models.

Third, you do have to be careful in the organization of the data for your model. The covariate values for any time point should represent their status just before the event. So it would be wrong to use year-end covariate values on the same row as bankruptcy events that could have occurred earlier in the year. Depending on the nature of your data you might need to reorganize the data so that covariates best represent predictors of events at the noted times.

Fourth, and perhaps most important, you might want to consider whether and how a survival model is appropriate for your study. For example, the Cox model assumes a basic shared shape of the hazard as a function of time starting from time 0. In clinical studies, for example, time 0 might be the time a patient received a particular treatment. In your case, the assumptions and implications of the assumption of a shared hazard-function shape could be quite different in 2 different scenarios: (a) if time 0 represents the time of formation of each company, or (b) time 0 represents, say, the calendar year 2000. You need to think carefully about how such assumptions correspond to what you know about the subject matter.

Best Answer

Related Solutions

Solved – fitting a log-logistic distribution

Solved – parametric survival regression and discrete time survival regression

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