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Let the survival time $T$ have an exponential distribution with rate $\lambda =0.2$ per year. Its hazard function is $h(t)=0.2$ per year, its cumulative hazard is $H(t)=0.2t$ for $t$ in years, and its survival function is $S(t)=e^{-0.2t}$. For a subject still event-free after five years, the conditional probability of an occurrence within the next two years isThe original population's probability of an occurrence between those same endpoints is instead $S(5)-S(7)=e^{-1}-e^{-1.4}\approx 0.121282$. Dividing by $S(5)\approx 0.367879$ recovers the conditional answer. The memoryless property makes the two-year conditional answer independent of the preceding five years.

For just $\Delta =0.01$ year, the exact conditional event probability is $1-e^{-0.002}\approx 0.001998$, close to the instantaneous approximation $h(5)\Delta =0.002$. For two years that approximation gives $0.4$, whereas the exact answer is about $0.329680$. At the single exact time five, the event probability is zero. These are three different quantities: an instantaneous hazard function with inverse-time units, a short-interval approximation, and an exact finite-interval probability.

For independent survival times $T_1,T_2$ with cumulative distribution functions $F_1,F_2$, the probability both components have failed by time $t$ isThe probability at least one has failed isThe first equality uses independence of random variables; the second applies it to the complementary event that both survive. These are the distributions of the two extreme order statistics.

Marginal distributions alone do not determine these probabilities. Without independence, the Fréchet inequalities give $\max (0,F_1+F_2-1)\le P(\text{both})\le \min (F_1,F_2)$, and inclusion-exclusion gives $P(\text{at least one})=F_1+F_2-P(\text{both})$.

For a survival time with an absolutely continuous probability distribution, probability density function $f$, and survival function $S$, the hazard function iswhere $S(t)>0$, with the derivative identity holding almost everywhere. At points where the local limit exists, equivalentlyThe hazard function measures an instantaneous conditional rate. If time is in years, its units are inverse years, and its value can exceed one. A probability for a short interval is approximately $h(t)\Delta$, not $h(t)$; for an absolutely continuous probability distribution, the probability of an occurrence at the single exact time $t$ is zero. The discrete hazard, by contrast, is a conditional probability at a possible event time.

The factorization $f(t)=h(t)S(t)$ distinguishes a probability density function for the original population from a hazard function for those still event-free. With an exponential distribution of rate $0.2$ per year, $h(5)=0.2$ per year but $f(5)=0.2e^{-1}\approx 0.073576$ per year: the latter also accounts for the probability of reaching year five. The exponential survival risk over an interval example computes finite-horizon probabilities from this rate.

A simple sufficient independent censoring assumption is that the potential observation-ending time $C$ and survival time $T$ are independent random variables. More generally, appropriate conditional independence given the covariates used in the analysis can suffice for analysis within those conditions. The requirement concerns the potential times, rather than independence of the observed duration $Y=\min (T,C)$ and event indicator $\delta$: those recorded quantities ordinarily depend on each other.

For an ordinary pooled Kaplan–Meier estimator, the observed risk set must represent the future event experience of everyone still event-free at each time. Conditional independence given a covariate does not alone justify ignoring that covariate in a pooled calculation. Also, there must be a positive probability of remaining observed through the horizons being estimated. If follow-up ends at a common closing time, it supplies no observations to identify the later survival function. Independent subjects with a common survival distribution and independent censoring form the basic model in Lu Tian's STAT331 notes, Unit 3, pages 1–4.

For example, subjects who stop attending because their condition is deteriorating can produce informative censoring. A predetermined common closing date can instead give administrative censoring, provided calendar entry and other relevant conditions do not select future survival times. The reason for stopping observation must be justified from study design and context; the recorded event indicators alone cannot verify independent censoring.

At distinct observed event times $t_1<t_2<\cdots$, let $n_j$ be the risk set size immediately before $t_j$, and let $d_j$ count events at $t_j$. The Kaplan–Meier estimator of the survival function isThe empty product is one. Each factor estimates a conditional survival probability among those still observed and event-free; their product joins the successive steps. This is nonparametric statistics: no specified shape for the survival distribution is imposed. Use independent censoring and the appropriate risk sets to interpret the result as an estimate for the intended population.

The Kaplan–Meier estimator is a right-continuous function with downward jumps at event times. Right censoring changes later denominators without causing a jump. A tick on the curve marks right censoring, rather than a failure. For a tied group of $d_j$ events, use the one factor $(n_j-d_j)/n_j$; subjects right-censored at that recorded time remain in $n_j$ for this factor. The fraction $d_j/n_j$ estimates a discrete hazard, not a continuous hazard function in inverse-time units. With no censoring, the factors telescope to the fraction of the original cohort whose observed survival times exceed $t$, the complement of its empirical distribution function.

Edward L. Kaplan and Paul Meier published ^[ref]. Their paper addresses lifetime estimation when observation ends before failure and explicitly calls attention to the assumption of independence between lifetime and potential observation loss. The date 1958 is the original publication year; the publisher's later online-posting date is not the discovery date.

In his ^[ref], Kaplan traces Meier's interest to cancer-duration data in 1952 and his own to the lifetimes of vacuum tubes in undersea telephone-cable equipment a year later. The journal encouraged them to combine their manuscripts. The shared mathematical question was how to estimate a survival function when some lifetimes are still running or disappear from observation.

Consider six subjects observed from time zero. Recorded survival data are $A:(1,1)$, $B:(2,0)$, $C:(3,1)$, $D:(3,1)$, $E:(3,0)$, and $F:(5,0)$, with each pair giving $(Y,\delta )$ in years. Thus $A,C,D$ have observed events, while $B,E,F$ are right-censored. The letters label subjects, not treatment groups. Here are all exit times, including rows without events:

At year one, $\hat{S}(1)=1(1-1/6)=5/6$. Removing $B$ at year two leaves the Kaplan–Meier estimator unchanged. Immediately before year three, $C,D,E,F$ are still in the risk set: the two events give $\hat{S}(3)=(5/6)(1-2/4)=5/12\approx 0.416667$. Only afterwards do $C,D,E$ leave, leaving $F$ alone. Removing $E$ before counting the tied events would incorrectly use $n=3$ and give $(5/6)(1-2/3)=5/18$. The event-first convention is part of the recorded-time interpretation; coarsely rounded timestamps do not reveal a hidden ordering within that time bin.

The exact Kaplan–Meier estimator is $1$ on $[0,1)$, $5/6$ on $[1,3)$, and $5/12$ on $[3,5]$. The three right censoring ticks are $(2,5/6)$, $(3,5/12)$, and $(5,5/12)$; at a shared event time the tick is on the post-event step. Beyond year five the risk set is empty, so the horizontal endpoint does not determine later survival. The Kaplan–Meier median survival time is three years, since this is the first step at or below $1/2$.

Now separate the tied events by changing only $D$'s recorded event time from year three to year four. At year three the risk set still has four subjects, but $d=1$, so $\hat{S}(3)=(5/6)(3/4)=5/8$. After the event and $E$'s right censoring, only $D,F$ remain. At year four $n=2,d=1$, giving $\hat{S}(4)=(5/8)(1/2)=5/16=0.3125$. The intervening right censoring prevents the factors from telescoping to the original tied-event factor. If $E$ instead remained observed past year four, the factors would be $(3/4)(2/3)=1/2$, giving the original $5/12$ after both events. This is Kaplan–Meier telescoping across a censor-free interval in numbers.

Incorrectly counting every right-censored exit as an event estimates the distribution of observation-ending times $Y$, rather than survival times $T$. For the original tied cohort, the spurious curve is $5/6$ at year one, $(5/6)(4/5)=2/3$ at year two, $(2/3)(1-3/4)=1/6$ at year three, and zero at year five. That zero asserts a failure for $F$, whose only known information is $T_F>5$.

The computation uses the risk set and event-before-censoring convention in Lu Tian's STAT331 notes, Unit 3, pages 2–6; this six-subject example supplies its own exact counts.

These are separate data, so results from the randomized trial must not be carried over. First inspect the construction of the survival response: a command creating a survival time object must pair follow-up in months with an event indicator, whose death/censoring coding should be checked explicitly. A censoring time is not an observed death time. Stratified product-limit fits would then summarize Kaplan–Meier estimators for the dose groups before adjusted regression.

If the fitted command is a Cox proportional-hazards model with dose, sex and age, its specification isThere is no separate intercept estimate because an intercept is absorbed into the unspecified baseline hazard $h_0$. With the stated coding, $e^{{\beta }_G}$ compares high-dose with low-dose death hazards at equal sex and age; $e^{{\beta }_F}$ compares female with male hazards at equal dose and age; $e^{{\beta }_A}$ is the hazard multiplier for a one-year age increase. A negative treatment coefficient would correspond to a lower fitted hazard under high dose. A hazard ratio is not a ratio of survival probabilities, nor directly a ratio of survival times. Under this model the survivor function is $S(t\mid x)=S_0(t{)}^{\exp (x^T\beta )}$.

The coefficient standard errors, Wald ratios and exponentiated confidence intervals should be read together. The common large-sample interval for a coefficient's hazard ratio isGlobal likelihood-ratio, score and Wald tests assess the included slopes jointly; for exactly these three one-coefficient terms their reference degrees of freedom are three. They need not coincide in a small sample. The Cox partial likelihood uses the event risk sets; the fitted log-likelihood is not a full likelihood for an estimated baseline survival curve. The method for tied event times should be recorded.

Check the proportional hazards assumption using scaled Schoenfeld residuals and time trends; check the functional form of age, influential subjects and independent censoring. Dose-by-age or dose-by-sex interaction terms are scientifically possible but require enough information and a stated model comparison. Adjustment can alter a crude dose association, and the allocation mechanism here is not specified as randomized. No coefficients, confidence limits, significance results, exact fitted calls or diagnostics for this dataset are recoverable without the output sheet. The displayed model is an interpretation template, not a claim that it was the exact missing command.

Write $X=\min (T,c)$ and $\Delta =1_{\{T\le c\}}$ for the right-censored survival time and observed-failure indicator. Since the exponential distribution is continuous, including or excluding equality at $c$ makes no difference. Using the tail-integral expression for an expectation,Consequently the expected integrated hazard at the observed endpoint satisfiesAdministrative censoring therefore stops both the accumulated exposure and the opportunity to record a failure.

The paper uses $F$ for the survivor function, rather than for the cumulative distribution function. For a nonnegative survival time $T$ with an absolutely continuous probability distribution, defineWhere $F(t)>0$, the hazard function is the instantaneous event rate conditional on survival:The integrated hazard is $H(t)={\int }_0^{t}h(u)du$. Since $h=-F^{\prime }/F$, integration gives $H(t)=\log F(0)-\log F(t)$. With the usual time origin and no event at zero, $F(0)=1$, soConversely, differentiating $F=e^{-H}$ gives $F^{\prime }=-hF$, so these relationships determine each of the survivor function, hazard function and integrated hazard from either of the others. A positive limiting survivor function is possible if the integrated hazard remains finite; a proper finite event time instead has $F(t)\to 0$.

Assume independent individuals and independent censoring, with the censoring law not involving the event-rate parameters. If $T_j$ is the survival time and $C_j$ its censoring time, we observe $x_j=\min (T_j,C_j)$ and $v_j=1_{\{T_j\le C_j\}}$. The exponential distribution givesAn observed event contributes $f_j(x_j)$, while a right-censored observation contributes $S_j(x_j)$, because it only says the event has not occurred by $x_j$. Multiplying these survival likelihood contributions and omitting censoring factors independent of the parameters givesWithout independent censoring the censoring factors cannot generally be discarded in this way.

For a continuous positive survival time, $S^{\prime }(t)=-f(t)$ and the hazard function is $h(t)=f(t)/S(t)$ wherever $S(t)>0$. ThereforeSince $S(0)=1$, integrating gives the cumulative hazard function identityAt an endpoint where survival reaches zero, the corresponding cumulative hazard is interpreted as $+\infty$ and the exponential as zero.

An unobserved survival time is right-censored when it is known to exceed the last event-free follow-up time. For a potential survival time $T$ and potential observation-ending time $C$, recordWhen $\delta =1$, the occurrence was observed at $Y$; when $\delta =0$, only $T>Y=C$ is known. An intact machine at the study's closing date and an event-free subject who stops attending follow-up can both yield right censoring, but the reasons for stopping matter for independent censoring. A subject right-censored at year two has contributed two years to person-time at risk and belongs to earlier risk sets. Neither recording a failure at year two nor discarding that subject preserves this information.

Under independent censoring and separate parameters for the observation mechanism, the part of the survival likelihood involving an absolutely continuous probability distribution for $T$ is $f(Y{)}^{\delta }S(Y{)}^{1-\delta }$. Thus a right-censored observation contributes the survival function $S(Y)$, rather than the event probability density function $f(Y)$. See Lu Tian's Stanford notes, Section 2.1, equations (2.1)–(2.2). The Kaplan–Meier six-subject cohort shows exactly how this distinction changes the calculation.

Survival analysis studies the probability distribution of a survival time: how long from a specified starting point until a specified occurrence. The occurrence need not be death. It might be a machine's first breakdown, a customer's first renewal, or a person's recovery. The meaning of "survival" is simply that the chosen occurrence has not happened yet. Both the starting point and the outcome must be defined before interpreting a survival function.

The distinctive difficulty is incomplete observation. A machine still working when observation stops supplies a lower bound on its survival time, rather than an observed breakdown time. Right censoring lets us retain that information. The Kaplan–Meier estimator estimates a survival function using the changing risk sets, without choosing an exponential distribution or another particular survival distribution. A hazard function instead describes the current event rate among those still event-free; its cumulative hazard connects those local rates to the whole survival function.

This framework turns different follow-up lengths into usable evidence about waiting times. It supports questions such as the chance of remaining event-free for another year, or the restricted mean survival time within an observed horizon. Lu Tian's Stanford teaching materials introduce these interpretations in Survival Distributions, Hazard Functions, Cumulative Hazards, Section 1.1.

For a nonnegative survival time $T$, the survival function iswhere $F$ is the cumulative distribution function. Thus $S(t)$ is the probability that the chosen occurrence is still in the future at time $t$. We consistently use the strict inequality $T>t$: a probability mass at $t$ has already left the survival function at $t$ itself. With this convention $S$ is a right-continuous function and a nonincreasing function taking values in $[0,1]$. If $\mathbb{P}(T=0)=0$, then $S(0)=1$; if $T$ is finite with probability one, then $S(t)\to 0$ as $t\to \infty$.

For $S(a)>0$ and $b>a$, conditional probability givesFor example, $S(2)=0.8$ and $S(5)=0.6$ mean that $0.6/0.8=0.75$ of those still event-free at year two remain event-free through year five. Their intervening event probability is $0.25$, whereas the original population's probability of an occurrence in $(2,5]$ is $0.8-0.6=0.2$. Conditioning changes the denominator.

The area under the survival function up to a horizon gives the restricted mean survival time. A positive terminal plateau in a plotted Kaplan–Meier estimator does not establish that the occurrence will never happen: terminal censoring and survival-mean identifiability explains why the unobserved tail remains undetermined.

The tail-sum formula expresses an expected value as the area under a survival function. For every nonnegative random variable $Y$,This Lebesgue integral and the expected value may both be $+\infty$; a probability density function is unnecessary. For a random variable $N$ taking values in the nonnegative integers, the corresponding sum isThe events $\{N>k\}$ and $\{N\ge k+1\}$ coincide because $N$ takes integer values. Thus the strict sum starts at zero, while the inclusive sum starts at one.^[ref]

For an original example, let the probability mass function of $N$ beIts expected value computed directly is $0(1/5)+1(3/10)+3(1/2)=9/5=1.8$. The strict tail probabilities at thresholds $0,1,2,3$ are $4/5,1/2,1/2,0$: their sum is again $1.8$. There is no probability mass at two, yet the event $\{N>2\}$ still has probability $1/2$. The survival function has heights $4/5,1/2,1/2$ on the successive intervals $[0,1),[1,2),[2,3)$; each interval has length one, so its integral is the same sum.

The indexing matters numerically. Starting the inclusive sum at zero adds $\mathbb{P}(N\ge 0)=1$, giving $2.8$ for this example. Starting the strict sum at one omits $\mathbb{P}(N>0)=4/5$, giving $1$. For a general nonnegative random variable, replacing $>$ with $\ge$ inside the integral is harmless: the difference is $\mathbb{P}(Y=t)$, nonzero at only a countable set of thresholds. For each positive integer $m$, at most $m$ thresholds can have probability mass at least $1/m$; their countable union contains all thresholds with positive probability mass and has Lebesgue measure zero. This reasoning concerns an integral, not equality of the two survival function conventions at every point.

For a continuous original example, take the uniform distribution on the interval $[0,4]$. Its survival function is $S(t)=1-t/4$ for $0\le t\le 4$ and zero for $t>4$. The area is a triangle of base four and height one:Directly integrating against the probability density function gives the same expected value, ${\int }_0^{4}y(1/4)dy=2$. The integral of the probability density function alone is one; it is the survival function whose area is the expected value.

The proof counts layers of each outcome. For a realized value $y\ge 0$, the indicator function $1_{\{y>t\}}$ equals one over an interval of length $y$, soTonelli theorem permits the interchange because the indicator function is nonnegative; no finite expected value is assumed. Similarly, an outcome $N=n$ contributes one indicator random variable for each of the $n$ events $\{N\ge k\}$ that occur:Again Tonelli theorem applies to the nonnegative sum, or equivalently the monotone convergence theorem applies to its partial sums. This exchanges counting layers within each outcome with averaging across outcomes. The existing integer-valued hypothesis in the tail-sum formula explains what changes when the values are not integers; the tail integral formula for moments weights the layers to obtain higher moments.

For an integrable random variable $X$ that can have either sign, apply the nonnegative formula separately to its positive part $X_{+}=\max (X,0)$ and negative part $X_{-}=\max (-X,0)$:Integrability makes both integrals finite. More generally the difference defines an extended expected value when at least one is finite; if both are infinite, subtraction does not define an expected value.^[ref]

Tail probabilities are often easier to measure or bound than a whole probability density function. For a device's survival time, the survival function records the probability of still operating at each threshold; its integral gives the mean operating time. A finite observation window naturally gives the expected value of a capped nonnegative random variable. For a mathematical bound, $\mathbb{P}(Y>t)\le e^{-t/2}$ for all $t\ge 0$ implies $\mathbb{E}[Y]\le {\int }_0^{\infty }{e}^{-t/2}dt=2$. The tail-sum formula turns control of a whole family of events into control of an average.