# general question survival models: Difference time component in survival and classification model

I have a general question about survival models. Survival models take the time dependence into account in a natural way. But what is exactly the difference in inference if we would model the surival as a binary variable with time as the independent variable instead of a survival model? Are there any gains in modelling data as a survival model instead of some classification model? Thanks in advance!

Below is my attemp to answer this question. Please feel free to delete it if it's not correct

The purpose of survival analysis is to infer on the distribution of a time-to-event random variable $$T$$.
One important feature of experiments based on time-to-event processes is the presence of right censoring: for some subjects we don't know $$T$$, only that it is larger than their total follow-up time. Usual survival models (Kaplan-Meier, Cox regression model,...) are designed in order to take into account this censoring. But in the end the purpose is always to infer about the distribution of $$T$$.

For your suggestion, say the censoring time is denoted by $$C$$ (it is the time the subject will stay under observation). The follow-up is then $$Y = \min(T,C)$$ and the observable variables are $$(Y, \delta)$$ where $$\delta$$ indicates if $$T \leq C$$ i.e if $$Y=T$$ (that is, if the follow-up ends by the event of interest).
Most survival models infer on $$T$$ from $$(Y,\delta)$$ (and possibly by including some set of covariates for regression models, like the Cox model).

What you aks (if I understand correctly) is if we can replace common survival models by a regression model taking $$\delta$$ as response and $$Y$$ as a covariate. For example a logistic model: $${\rm logit} \big( \mathbb P(\delta=1 \mid Y)\big) = \beta Y \qquad (1)$$ and $$\mathbb P(\delta=1 \mid Y) = \mathbb P \big ( T \leq C \mid \min(T,C) \big )$$ In fact if we assume $$T \perp C$$ we can show that $$\delta \perp Y$$. \begin{align*} \mathbb P( T \leq C, \min (T,C) \in [l,l+dl]) &= \mathbb P (T \leq C, T \in[l,l+dl]) \\ &= \mathbb E \Big ( \delta I_{T \in [l,l+dl]} \Big) \end{align*} One interesting property of this last expectation is that, under $$T \perp C$$, for any (bounded) function $$h$$, $$\mathbb E \Big ( \delta h(T) \Big) = \mathbb E \Big ( S_C(T) h(T) \Big)$$ where $$S_C$$ is the survival function of $$C$$. The last expectation is thus only with respect to $$T$$! Taking $$h$$ as $$h(T) = I_{T \in [l,l+dl]}$$ we have: \begin{align*} \mathbb P( T \leq C, \min (T,C) \in [l,l+dl]) &= \mathbb E \Big ( S_C(T) I_{T \in [l,l+dl]} \Big) \\ &= \int_l^{l+dl} S_C(t) f_T(t) dt \end{align*}

And when $$dl \to 0$$ :

$$\int_l^{l+dl} S_C(t) f_T(t) dt \xrightarrow[dl \to 0]{} S_C(l) f_T(l)$$ Hence, $$\mathbb P ( T \leq C , \min (T,C) = l) = \mathbb P (T \leq C, T=l) = \lim_{dl \to 0} \mathbb P(C \leq l, T \in [l,l+dl] ) = S_C(l) f_T(l)$$ The last part show that the probability of both events can be expressed as a product of the probability of each event, hence the independence.

The conclusion of all this computation is that, under independent censoring $$(T \perp C)$$, we have $$\delta \perp Y$$ and the logit model defined above is then:

$${\rm logit} \big( \mathbb P(\delta=1 \mid Y)\big) = {\rm logit} \big( \mathbb P(\delta=1)\big)$$ So including $$Y$$ in the model would not be useful. Hence this approach and the model associated cannot be used to learn something about $$T$$. And finally, survival analysis and this second are thus not equivalent.

Here is an R code that run some simulations and computes the model $$\mathbb P(\delta=1 \mid Y) = \mathbb P \big ( T \leq C \mid \min(T,C) \big )$$ where you can see that the estimator of $$\beta$$ (in equation $$(1)$$ above) is close to $$0$$.

# 1000 simulations of a survival dataset with 200 subjects
simulations<-sapply(1:1000,function(j){
data<-data.frame(do.call(rbind,lapply(1:200,function(i){
time<-rexp(1,0.01)  #time-to-event
cens<-rexp(1,0.01) #independent censoring
fup<-min(time,cens) #follow-up
d<-1*(time<=cens) #delta
return(c(fup=fup,d=d))
})))
mod<-glm(d~fup,data,family = "binomial") #logit regression with fup (Y) as IV
return(mod\$coefficients[2]) #return the estimated value of beta
})
mean(simulations) #mean value of the estimated beta
#should be close to 0

If I understand your question correctly, what is at issue here is the difference between parametric survival models and the semi-parametric models used in Cox proportional hazards analysis.

It's possible to incorporate time directly into your model if you know the baseline shape of the relationship between survival probability and time. For example, if individuals are known to have a constant hazard of having an event over time, the survival probability over time follows an exponential curve and your model would determine how the covariates affect that hazard of an event and thus the steepness of the survival curve. There are several such parametric distributions that are used in survival analysis.

The advantage of the semi-parametric Cox proportional hazards model is that you don't need to know the baseline shape of the relationship between survival probability and time. The Cox model works with the assumption that the baseline shape is the same for all individuals, and determines how the covariates affect the steepness of the (empirically determined) survival curve.

One more thought:

When survival data are only collected at a handful of defined times, survival modeling can be performed as a set of binary analyses over each discrete time interval. That might be like your idea of time being an "independent variable," in that you effectively choose the times at which survival data are collected.

Other than that, though, I think you will get into trouble using the terminology "independent variable" for time in a survival model. In a fundamental sense the time of an event is the observation, the dependent variable, in survival analysis.

• A proportional odds model with a log-logistic baseline distribution is a special case of a logistic regression model with a log-time effect. However, I don't know how to fit this model with logistic regression without augmenting an infinitely fine grid of values, which is computationally impossible. See my answer for more details. Jan 20, 2020 at 17:25

First of all, I will say that there is an equivalence between logistic regression and the proportional odds model. In fact, with the special case of current status data, the proportional odds model with a log-logistic baseline can be fit in a straightforward manner. In addition, the semi parametric proportional odds can be perfectly fit with right censored data using logistic regression, although this will require a huge expansion of the data that will be incredibly computationally inefficient.

To explain the computational inefficiencies, let's first start with the wrong way to do things. A common first-pass idea is to simply just include time (or log-time) as a predictor in your dataset and fit a logistic regression model from that. This is not a good idea. A very simple way to demonstrate this is consider what would happen if we had survival data in which none of the data was right censored, i.e., we observed an event for all subjects of the dataset. In that case, if we tried the naive logistic regression approach, we now have a logistic regression dataset in which the outcome is all 1's. That's not good: it means all predictions for model will be 1, regardless of what's the input, and it also means all the estimates of the parameters are undefined. Yet survival models have no problem with valid estimation when there is no right censored data, so we must be doing something wrong!

To demonstrate the issue here, consider for now that we have two time points with $$T_1 < T_2$$. In a survival analysis setting, if we know that an event for subject $$i$$ occurred at $$T_2$$, this means that the event did not occur at $$T_1$$. So if we wanted to use logistic regression to fit our survival model, we could do this by augmenting our data: every time we have an event that occurred at $$T_2$$, we would have one data point in our logistic regression model that a dummy variable indicates time = $$T_1$$ and outcome = 0 and another data point with another dummy variable that indicates time = $$T_2$$ and outcome = 1 (all other covariates are the same between the two new data points). If we fit this model, we fully recreate the semi-parametric proportional odds model.

However, note that we had to make our dataset bigger, which means it runs slower. If our data set only has two unique times ($$T_1$$ and $$T_2$$), then this isn't so bad; our new dataset will be less than twice as big as the original (since events that occurred at $$T_1$$ don't need any augmentation). However, if we have three unique times, it gets worse: if an event happened at time $$T_3$$, we will need to augment one data point at $$T_1$$ and another at $$T_2$$, meaning three data points will be needed to represent this single record. In general, the size of the new dataset will be of order $$O(n n_{unq})$$, where $$n_{unq}$$ is the number of unique times in the dataset. If times are continuous, this means our new data is now of size $$O(n^2)$$, with $$n$$ new covariates. Computationally, that's a nightmare for even moderately sized datasets ($$n = 1000$$)!