TITLE: Empirical Distribution Function.
QUESTION [5 upvotes]: Suppose $T$ denotes a nonnegative random variable representing the lifetimes of individuals in some population. Let $t_i,i=1,2,...,n,$ denotes an ordered observed value. Then the empirical survivor function(esf) is defined by 
$$S(t)=\frac{\text{number of observations}>t}{n}.$$
Consider the observed valus of $T$: $9,13,13,18,23,28,31,34,45,48,161$.
According to the formula, The values of esf for this data are given below:
$$
\begin{array}{c|cccccccccc}
t & 9&13&18&23&28&31&34&45&48&161 \\
\hline
S(t)&\frac{10}{11}&\frac{8}{11}&\frac{7}{11}&\frac{6}{11}&\frac{5}{11}&\frac{4}{11}&\frac{3}{11}&\frac{2}{11}&\frac{1}{11}&0
\end{array}.
$$
My question is: at $t=13$, the first individual with $t=9$ is died. That is, at $t=13$, I have actually $n=10$ individuals remaining. Then why doesn't the denominator (number of individuals) reduces with time in the formula of $S(t)$? Why is the calculation of esf  not like the following?
$$
\begin{array}{c|cccccccccc}
t & 9&13&18&23&28&31&34&45&48&161 \\
\hline
S(t)&\frac{10}{11}&\frac{8}{10}&\frac{7}{8}&\frac{6}{7}&\frac{5}{6}&\frac{4}{5}&\frac{3}{4}&\frac{2}{3}&\frac{1}{2}&0
\end{array}?
$$

REPLY [4 votes]: Comment. The usual 'empirical CDF' (ECDF) denoted $\hat F_n(x)$ starts at 0 to the left of the smallest observation,
jumps by $1/n$ at each sorted observation, and ends at 1 to the right of the largest observation. The 'emirical reliability' or 'empirical survival' function
that I recall seeing is $1 - \hat F_n(x).$
You indicate your text uses S, which is similar to R (which came later). In R, one has the following:
x = c(9, 13, 13, 18, 23, 28, 31, 34, 45, 48, 161)
plot(ecdf(x), lwd=2)


Notice the 'double jump' at repeated observation $13.$
This method refers to all $n$ subjects subjects throughout. Your method
seems to look only at the remaining survivors at each step. You might ask
your instructor whether your method is in use. 
I don't have my survival
analysis book at hand and don't want to risk "telling you more than I know"
without a review of this topic. Maybe someone else has insights to offer.