# Kaplan-Meier Estimate with no censoring

Let $T_i$ is the survival time for individual $i$ $(i=1,2,\ldots, n)$ and $C_i$ be the time to censoring. Let $U_i=\min(T_i,C_i)$. And $\hat S(U_i)$ is the Kaplan-Meier estimator for the censoring distribution. Suppose $R_i$ and $Z_i$ are two indicator functions. Also, $p$ is a probability. Consider the following estimator of cumulative distribution function:

$$\hat F(t)= \sum_{i=1}^{n}\frac{I(T_i<C_i)(1-R_i+R_iZ_i/p)I(U_i\le t)}{\hat S(U_i)},$$ where $I(.)$ is an indicator function.

Now it is written that, with no censoring $\hat F(t)$ becomes

$$\hat F(t)= \sum_{i=1}^{n}(1-R_i+R_iZ_i/p)I(T_i\le t).$$

I understand that if there is no censoring, then $I(T_i<C_i)=1$, that is, we will always observe the survival time. Also, with no censoring $U_i=T_i$ and hence $I(U_i\le t)=I(T_i\le t)$.

But I do not understand why does Kaplan-Meier estimator for the censoring distribution, $\hat S(U_i)$, which appears in the above first equation vanish in the second equation with no censoring?

• Recall, KM and most survival methods are based on unique failure times, as the informativeness for survival comes from failures only -- not censored. Also, each object (patient, mouse, etc) contributes to a failure variable (0-censored, 1-failed) and a time variable [$t$=time to failure or time to censoring (end of study or loss to follow-up)]. If you look at the product limit variant of KM, there's not a censoring time and a failure time, but rather bin counts for failures as a function of time interval. – user32398 Dec 18 '16 at 1:57
• Can you give a source to where you found this expression? I am not following the meaning of $R$ and $Z$. In Fine/Gray JASA 1999 Proportional Hazards Model for the Subdistribution of a Competing Risk, $Z$ is the modeled effect of covariates and $R$ is a cause-specific indicator matrix. You might want to start with a simpler case. – AdamO Jun 17 '18 at 13:05