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Bruce W. Turnbull works on survival analysis and the statistical monitoring of clinical trials. His Cornell profile connects research on statistical methods with the practical problem of deciding what accumulating trial data justify.^[ref]

His 1974 paper studied nonparametric estimation when event times are subject to both left censoring and right censoring. A time known only to precede a threshold and one known only to exceed a threshold convey different information about the same survival function. Retaining those inequalities makes incomplete observations useful without substituting an invented event time.^[ref]

Competing end-of-life criteria are two or more conditions, any one of which can terminate useful operation under the stated specification. They define a remaining useful life by the earliest relevant crossing. If the prospective crossing times, measured from the present, are $T_1,\dots ,T_n$, thenThis is the event structure studied by competing risks. A condition ending one operating cycle must not inadvertently be treated as the component's aging endpoint: NASA's battery experiment separately defines a terminal-potential discharge cutoff and a30% rated-capacity-loss aging criterion.^[ref]

For an original mechanical example, suppose a component can reach either an unacceptable vibration level in12h or an unacceptable temperature in20h under a specified operating scenario. If either condition stops operation, its predicted useful life is12h. Requiring both conditions would instead define another stopping rule. An observed early crossing also does not establish when the other criterion would have been reached if the test had continued.

For uncertain crossing times, the survival function of useful life isOnly under independence does this equal the product of the individual survival functions. For an explicitly hypothetical independent pair of exponential distributions with rates $0.01{\mathrm{h}}^{-1}$ and $0.02{\mathrm{h}}^{-1}$, the product is $e^{-0.03t}$ when $t$ is measured in hours. The expected first crossing is $1/0.03\approx 33.33\mathrm{h}$, smaller than either individual mean,100h or50h.

The independence assumption is substantive. Shared temperature, load or damage can correlate crossing times, so multiplying marginal survival functions can give the wrong answer. Moreover, an intervention triggered by the first crossing can change the subsequent trajectory. The physical end-of-life criterion, dependence model and assumed future operation therefore belong with any numerical lifetime prediction.

A Cox–Snell residual is the fitted cumulative hazard function evaluated at an observed event or censoring time: $r_i={\hat{H}}_i(Y_i)$. Its reference probability distribution comes from the probability integral transform. If the true event time $T_i$ has a proper continuous survival function $S_i$, then $H_i=-\log S_i$ satisfiesFor example, an exponential distribution with rate $\lambda$ has $H(t)=\lambda t$, so rescaling its event time by $\lambda$ produces a unit-rate exponential distribution.

With right censoring, $Y_i=\min (T_i,C_i)$ and the transformed observation is $H_i(Y_i)$ together with its original event indicator ${\delta }_i$. Under independent censoring, a Kaplan–Meier estimator applied to these transformed observations should approximately follow $e^{-r}$; its logarithm should approximately follow a straight line of slope minus one. Censored residuals must remain censored in this comparison.

The exact exponential identity uses the true model, whereas ${\hat{H}}_i$ is estimated from data: fitted residuals are approximate diagnostics and need not be independent exact exponentials. Atoms in a discrete event-time probability distribution, or a cure fraction with positive probability of no finite event, invalidate the unmodified proper-continuous reference law. Such models require diagnostics respecting their jumps or surviving mass.

Suppose $T\ge 0$ has an absolutely continuous probability distribution, with survival function $S(0)=1$. On an interval where $S(t)>0$, its cumulative hazard isHere $h$ is the hazard function; its integral is dimensionless because the units of rate and time cancel. The identity follows by integrating the derivative of $\log S$, so mere continuity of the survival function is not enough: the absolute continuity assumption supplies a probability density function. If $S$ reaches zero, interpret $-\log S$ as $+\infty$. The normalization $S(0)=1$ also matters; an initial probability mass at zero would require an initial term $-\log S(0)$.

Accumulating hazard functions adds on the $H$ scale and multiplies survival factors on the $S$ scale. For $S(a)>0$ and $b>a$,For instance, a cumulative hazard increment of $2$ gives event probability $1-e^{-2}\approx 0.864665$, rather than a probability of $2$. With $h(t)=0.2$ per year, $H(5)=1$ and $S(5)=e^{-1}\approx 0.367879$. With the increasing hazard function $h(t)=0.04t$ per year when $t$ is expressed in years, $H(t)=0.02t^2$, so $H(5)=0.5$ gives $S(5)=e^{-0.5}\approx 0.606531$. Its coefficient has units of inverse years squared.

The constant hazard function produces an exponential distribution; the increasing example is a Weibull distribution. Their survival functions both decrease, although one hazard function stays constant and the other increases. At year ten both have $H(10)=2$ and $S(10)=e^{-2}$ despite different earlier histories. The accumulation identity is described in Lu Tian's Stanford notes, Section 1.1.

For a discrete hazard $q_j$ at each possible event time $t_j$, the exact counterpart is $S(t)={\prod }_{t_j\le t}(1-q_j)$. Taking its negative logarithm sums $-\log (1-q_j)$, rather than $q_j$. Therefore exponentiating a Nelson–Aalen estimator does not give exactly the Kaplan–Meier estimator; small-jump comparison of cumulative hazard estimators quantifies the difference.

At ordered possible failure times $t_j$, the discrete hazard is $h_j=\mathbb{P}(T=t_j\mid T>t_{j-1})$. Its survival function satisfies $S(t_j)=S(t_{j-1})(1-h_j)$, and hence $S(t)={\prod }_{t_j\le t}(1-h_j)$. Estimating $h_j$ by the observed number of failures divided by the risk set size under independent censoring yields the Kaplan–Meier estimator. Unlike a continuous-time hazard rate, a discrete hazard is a probability between zero and one.

In the usual left-and-right-censoring sense, doubly censored data contain some left-censored and some right-censored observations, possibly alongside exact values. The record identifies which side is censored. This is the convention used for datasets mixing both sides of censoring.^[ref].

For known observation limits $L<R$, one can record $Y=\min \{\max (T,L),R\}$ with an indicator specifying $T\le L$, an exact interior event, or $T\ge R$. For a continuous event-time probability distribution with cumulative distribution function $F$, probability density function $f$ and survival function $S$, the respective likelihood contributions are $F(L)$, $f(T)$ and $S(R)$. A measurement instrument that reports “below its lower detection limit” or “above its upper detection limit” produces precisely this distinction.

If the record only says that $T$ lies outside $(L,R)$ without identifying the side, its contribution would be $F(L)+S(R)$. That is different recorded information and is not the defining convention above. Terminology also needs context when both the origin and endpoint of a duration are uncertain; an interval-censored origin together with an interval-censored endpoint describes a different observation scheme from simple left-and-right censoring of one event time.

For inter-renewal times with an Erlang distribution of integer shape $r$ and rate $\lambda$, the mean is $r/\lambda$ and survival function is $e^{-\lambda t}{\sum }_{j=0}^{r-1}(\lambda t{)}^j/j!$. The equilibrium first-delay density is survival divided by the mean,This is the equal mixture of Erlang distributions of shapes $1,\dots ,r$. It can be constructed by observing a Poisson clock with uniformly randomized phase modulo $r$, whose every $r$th event is a renewal.

The equilibrium residual-life distribution describes the remaining wait to a renewal when observing a stationary renewal process at an ordinary time. For positive interarrivals $T$ with finite mean $\mu$, it has survival function and densityThe density integrates to one by the Tail-sum formula for expectation. It exists even when the original lifetime law has atoms. A deterministic lifetime $T=c$ gives a uniform residual on $(0,c)$; an exponential lifetime gives the same exponential distribution as the original lifetime.

For the discrete chain convention in residual lifetime Markov chain, the invariant probabilities are instead ${\pi }_i=\mathbb{P}(T>i)/\mathbb{E}T$ for $i\ge 0$. A deterministic integer lifetime $m$ gives a uniform law on $\{0,\dots ,m-1\}$. These formulas describe observation at equilibrium; they do not make the first lifetime of an ordinary renewal process already have that residual law.

For a nonnegative random variable $Y$ and a fixed cap $a\ge 0$, the tail-sum formula givesBelow the cap, the events $\{\min (Y,a)>t\}$ and $\{Y>t\}$ coincide; at and above the cap, the former event is empty. For the second identity, the positive part $(Y-a{)}_{+}$ exceeds $t\ge 0$ precisely when $Y>a+t$. Thus the two integrals divide the area under the survival function, just as $Y=\min (Y,a)+(Y-a{)}_{+}$ divides each outcome. The capped expected value is always finite, while the excess expected value may be infinite.

For the uniform distribution on the interval $[0,4]$ and cap $a=2$, the capped expected value is ${\int }_0^2(1-t/4)dt=3/2$ and the excess expected value is ${\int }_2^4(1-t/4)dt=1/2$. Directly averaging the capped values also gives ${\int }_0^{2}y(1/4)dy+2\mathbb{P}(Y>2)=1/2+1=3/2$. Capping therefore measures a different average from the uncapped expected value two, even when both are finite.

For independent and identically distributed random variables $U_i$ with the continuous uniform distribution on $[0,1]$, let $M_n={\max }_{1\le i\le n}{U}_i$. Its cumulative distribution function is $x^n$ on $[0,1]$, so $M_n\stackrel{\mathrm{d}}{\to }1$. The endpoint error has a nonconstant limiting probability distribution: for fixed $z\ge 0$ and $n\ge z$,This survival function gives the displayed exponential distribution limit. More generally, for observations uniform on $[a,b]$, $n(b-M_n)/(b-a)$ has the same limit. The scaling identifies an error of order $1/n$, in contrast to the $1/\sqrt{n}$ fluctuations of a typical sample mean with nonzero finite variance.

Suppose an independent superposition of a positive-rate Poisson process and a renewal process with strictly positive nonarithmetic finite-mean interarrivals is itself a renewal process. Its first interarrival must be exponential; in fact the other process's interarrival law must also be exponential.

Let $\lambda >0$ be the Poisson rate, $\overline{F}$ the other interarrival survival function and $\mu$ its finite mean. The first superposed interarrival has survival $\overline{H}(t)=e^{-\lambda t}\overline{F}(t)$ and mean $h$. Adding the two long-run counting rates gives $h^{-1}=\lambda +{\mu }^{-1}$. Computing the limiting excess in two ways, using the renewal excess limit theorem for the superposition and independence of the two excesses, givesSet $A(t)={\int }_t^{\infty }\overline{F}(u)du$. Differentiating the displayed identity almost everywhere yields $\overline{F}(t)=\lambda hA(t)/(\mu -h)=A(t)/\mu$. Thus $A^{\prime }=-A/\mu$ with $A(0)=\mu$, forcing $\overline{F}(t)=e^{-t/\mu }$. It follows that $\overline{H}(t)=e^{-(\lambda +1/\mu )t}$. The renewal assumption on the superposition is essential; independent superpositions of general renewal processes need not be renewal processes.

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.

Extreme value theory studies limiting probability distributions of normalized sample maxima and related extremes. The standard limits for maxima include the Fréchet distribution, negative Weibull distribution, and Gumbel distribution.

The question is often practical: if a batch finishes only when its slowest task finishes, its completion time is a sample maximum. A typical individual delay or a sample mean does not answer that question. The central limit theorem describes suitably normalized sums and sample means under its assumptions; it does not give a limit law for sample maxima. Extreme value theory instead describes how the largest observation changes as the number of opportunities for a large observation grows.

For independent and identically distributed random variables $X_1,X_2,\dots$ with distribution function $F$, define the sample maximumThe event $M_n\le x$ means that every observation is at most $x$, so independence givesExtreme value theory asks whether there are constants $a_n>0$ and $c_n\in \mathbb{R}$ for which $(M_n-c_n)/a_n$ has convergence in distribution to a non-degenerate probability distribution. Equivalently, $F(c_n+a_{n}z{)}^n$ must tend to a limiting distribution function at its continuity points. The extremal types theorem says that any such limit belongs, up to a change of location and positive scale, to the Gumbel distribution, Fréchet distribution, or negative Weibull distribution. It does not guarantee that a limit exists for every $F$. ^[ref].

The maximum of independent exponential random variables gives a particularly transparent worked example: its center grows as $(\log n)/\lambda$, and its fluctuations have a Gumbel distribution limit on the scale $1/\lambda$. Here and below $\log$ denotes the natural logarithm. This normalization depends on the exponential distribution. Other survival functions can lead to the Fréchet distribution or negative Weibull distribution, with different constants $a_n,c_n$; the Gumbel distribution and the exponential example's normalization are not universal.

Independence is essential to the product calculation. If instead $X_1=X_2=\cdots =X$ for one rate-$\lambda$ exponential random variable, then $M_n=X$ for every $n$. Its distribution function stays $1-e^{-\lambda x}$ for $x\ge 0$ and zero below zero, and its median stays $(\log 2)/\lambda$. Perfect dependence creates no logarithmic growth: repeating the same random variable supplies no new chance of a larger value.

A historical starting point is R. A. Fisher and L. H. C. Tippett's 1928 paper on limiting laws of the largest or smallest sample member, published in volume 24, issue 2, pages 180–190 of the Proceedings of the Cambridge Philosophical Society. Its abstract studies the possible limiting forms and emphasizes that approach to a limit can be slow. ^[ref].

Let $k$ be a positive integer, $0<p<1$, and $\lambda >0$. For independent rate-$\lambda$ exponential distribution claims and a zero-based negative binomial distribution count of shape $k$ and parameter $p$, the random-sum transform identity gives $M_S(t)=[p(\lambda -t)/(p\lambda -t){]}^k$. This equals the transform of a compound binomial distribution with count $\mathrm{Bin}(k,1-p)$ and rate-$p\lambda$ claim sizes. Conditional on its positive count $n$, the total has Erlang distribution, whose survival function is the inner exponential-polynomial sum. Conditioning yields the displayed finite mixture distribution formula for $x>0$. It replaces an infinite mixture and includes the zero-total atom $p^k$ through the omitted $n=0$ term.

For $k=1$, the formula reduces to $(1-p)e^{-p\lambda x}$ for $x>0$: there is an atom $p$ at zero and a rate-$p\lambda$ exponential distribution positive component. This checks both the omitted zero-count term and the parameter scaling.

When a first-passage time $\tau$ has an absolutely continuous distribution, its density is $f_{\tau }(t)=-S^{\prime }(t)$, where $S(t)=\mathbb{P}(\tau >t)$ is its survival function. In a continuous-time Markov chain, this density is the total Probability flux from surviving states into the absorbing set.

For $\alpha >0$, the standard Fréchet distribution has distribution function ${\Phi }_{\alpha }(x)=e^{-x^{-\alpha }}$ for $x>0$ and zero for $x\le 0$. Its maximum domain of attraction consists of distribution functions with an infinite right endpoint of a distribution and a survival function with regular variation of index $-\alpha$.

Consider successive independent Bernoulli trials with the same success probability $0<p\le 1$. A geometric distribution models waiting for their first success. Two conventions coexist, so always say what is counted. Let $T$ count all trials up to and including the successful trial, and let $F=T-1$ count preceding failures. Their probability mass functions areThe factors describe failures followed by one success; independence permits multiplication. Summing the geometric series verifies that each law has total mass one. Their expected values differ by one, while shifting by a constant leaves the variance unchanged:The positive convention and these moments are also given by Penn State STAT 414.^[ref]

For a constructed retry mechanism with independent attempts and $p=0.3$, success on attempt four has probability $\mathbb{P}(T=4)=0.7^3(0.3)=0.1029$. More than four attempts are needed precisely when the first four fail: $\mathbb{P}(T>4)=0.7^4=0.2401$. The expected values are $\mathbb{E}T=10/3$ and $\mathbb{E}F=7/3$, and the common variance is $70/9\approx 7.7778$. The large spread explains why the average waiting count is not a deadline.

The two probability mass functions have identical heights with different labels: $T=4$ and $F=3$ describe the same event. For $0<p<1$, consecutive masses have ratio $1-p$, explaining the name geometric distribution. The support of a probability distribution is the positive integers for $T$ and the nonnegative integers for $F$. At $p=1$, it collapses to $T=1$ and $F=0$, both with variance zero; interpret an empty product, including $(1-p{)}^0$, as one. These counts concern attempts; converting them into elapsed time requires specifying the time spent per attempt.

The survival function makes the counting convention especially visible. Put $q=1-p$. For nonnegative integers $n$, and for $k\ge 1$,The event $T>n$ requires $n$ failures, whereas $T\ge n$ requires only $n-1$ when $n\ge 1$. For the constructed $p=0.3$ mechanism, $\mathbb{P}(T\ge 4)=0.343$ differs from $\mathbb{P}(T>4)=0.2401$. The geometric success deadline uses the latter tail to control the chance of still waiting.

The striking memorylessness of the geometric distribution says that, for nonnegative integers $m,r$ with $\mathbb{P}(T>m)>0$,After five failures, the conditional probability of requiring more than two additional attempts is still $0.7^2=0.49$. The analogous zero-based identity uses $\mathbb{P}(F\ge m+r\mid F\ge m)=\mathbb{P}(F\ge r)$; keeping the inequalities straight avoids an off-by-one error. This property follows from fresh independent trials with unchanged $p$; learning or changing conditions can destroy it. At $p=0$, success never occurs, so there is no proper finite waiting-count law.

For independent positive-support geometric random variables with success probabilities $p_i$, all must exceed $k$ for their minimum to exceed $k$. The resulting survival function is ${\prod }_i(1-p_i{)}^k$. This is the positive-support geometric distribution with the displayed success probability. independence is needed for multiplying the survival probabilities.

Germán Rodríguez works on statistical methods for population research. His Princeton teaching and research site connects generalized linear models, survival analysis and demographic questions about event timing. The site's biographical note identifies his training in biostatistics and his work in population research.^[ref]

His Survival Models by Germán Rodríguez explains how survival functions and hazard functions encode the same event-time distribution in different ways. With right censoring, an event observed at $t$ contributes density $f(t)$, whereas an observation ending without an event contributes $S(t)$. Writing the contribution as $S(t)h(t{)}^{\delta }$, with event indicator $\delta \in \{0,1\}$, makes that distinction explicit. These likelihood conventions let incomplete follow-up inform estimation without treating a censored time as an observed failure.^[ref]

For the Gamma distribution with fixed positive integer shape $m$ and rate one, the survival function is $e^{-t}{\sum }_{j=0}^{m-1}{t}^j/j!$. Its sample maximum centered by $\log n+(m-1)\log \log n-\log ((m-1)!)$ has convergence in distribution to the standard Gumbel distribution, without an additional scale factor.