# Kaplan-Meier Estimate

#### Cynderella

##### New Member
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?