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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 real random variable $X$, an upper-tail probability has the form $\mathbb{P}(X\ge t)$ and a lower-tail probability has the form $\mathbb{P}(X\le t)$. Each is the probability of an event lying beyond a chosen threshold in one direction. A tail probability need not be small: the threshold determines how much of the probability distribution it includes. A concentration inequality bounds such probabilities, often for deviations from an expected value.

Endpoint conventions matter. Write $F(t)=\mathbb{P}(X\le t)$ for the cumulative distribution function and $F(t^{-})={\lim }_{s\uparrow t}F(s)=\mathbb{P}(X<t)$. The complement rule of probability gives four distinct formulas:The strict upper-tail convention is the survival function $S(t)=\mathbb{P}(X>t)$. Its value differs from the inclusive upper tail probability by exactly the probability mass at the threshold:A singleton with positive probability mass is an atom of a measure. Thus $1-F(t)$ excludes an atom of a measure at $t$, whereas $1-F(t^{-})$ includes it. These endpoint identities follow directly from the cumulative distribution function properties in Kyle Siegrist, Distribution and Quantile Functions: endpoint probabilities and right tails.

As the threshold increases, both upper tail probabilities are nonincreasing functions. The strict survival function is a right-continuous function; the inclusive upper tail probability takes the upper endpoint value at a downward jump, equal to the left limit of the survival function. Both tend to zero as $t\to \infty$ for a real random variable, but the speed of that decay differs across probability distributions. The cumulative distribution function properties and their proof from continuity of probability measures are given in MIT 6.436J, Lecture 4: random variables and cumulative distribution function properties.

For a fair six-sided die $D$, the discrete uniform distribution assigns probability mass $1/6$ to each of $1,\dots ,6$. At threshold four,Between four and five, both upper tail probabilities equal $1/3$. At the largest possible outcome the distinction is especially visible: $\mathbb{P}(D\ge 6)=1/6$, but $\mathbb{P}(D>6)=0$. The diagram shows exact probabilities for this mathematical model; its dots specify which endpoints belong to each step function.

If a probability distribution has a probability density function $f$, then every singleton has zero probability mass andFor an exponential distribution of rate two, $f(x)=2e^{-2x}$ for $x\ge 0$ and zero otherwise, so $S(t)=e^{-2t}$ for $t\ge 0$ and $S(t)=1$ for $t<0$. Hence $\mathbb{P}(X\ge 1)=e^{-2}\approx 0.135335$. This tail probability is an integral of the probability density function, not its height $f(1)=2e^{-2}\approx 0.270671$. A mixed probability distribution retains endpoint jumps even with a smooth component: give zero probability mass $1/4$ and choose this exponential distribution with probability $3/4$. Then $\mathbb{P}(X\ge 0)=1$, $\mathbb{P}(X>0)=3/4$, and $S(t)=(3/4)e^{-2t}$ for $t\ge 0$.

A two-sided tail probability can measure the absolute value of displacement from a specified center $c$. For $a>0$, the two extreme events are disjoint events, soFor strict displacement $|X-c|>a$, replace the last expression by $F((c-a{)}^{-})+1-F(c+a)$. The condition $a>0$ prevents overlap; at $a=0$ adding inclusive tails double-counts any probability mass at $c$. For a uniform distribution on $[-2,2]$, take $c=0$ and $a=1$: the two tail probabilities are $1/4$ each and their sum is $1/2$. Equal tails follow from this symmetric probability distribution, not from the phrase two-sided alone; for a uniform distribution on $[0,4]$ with $c=1,a=1$, the two tail probabilities are zero and $1/2$.

The tail-sum formula for expectation turns a whole curve of tail probabilities into an expected value. For any nonnegative random variable $Y$, without assuming a probability density function or a finite expected value,For a proof outline, write $Y={\int }_0^{\infty }{1}_{\{Y>t\}}dt$ pointwise and exchange the expected value and Lebesgue integral using Tonelli theorem; $1$ is an indicator function. The inclusive version has the same Lebesgue integral: positive probability masses occur at only a countable set of thresholds, which has Lebesgue measure zero. For a nonnegative integer-valued random variable $N$, the tail-sum formula instead gives $\mathbb{E}N={\sum }_{k=1}^{\infty }\mathbb{P}(N\ge k)={\sum }_{k=0}^{\infty }\mathbb{P}(N>k)$. The integer hypothesis is essential for these sums. The integral identity, its indicator function proof, and the integer version appear in Kyle Siegrist, Additional Properties of Expected Value: integration and summation of right tails.

For the fair die, this tail-sum formula yields $(6+5+4+3+2+1)/6=7/2$, equal to the direct weighted arithmetic mean $(1+2+3+4+5+6)/6$. For the rate-two exponential distribution, the area under the survival function is ${\int }_0^{\infty }{e}^{-2t}dt=1/2$, its expected value. The first panel below shades an integral of the probability density function for a single threshold; the second shades an integral of the survival function across all nonnegative thresholds. These are analytical illustrations, with the displayed curves truncated at three; their exact total areas use an infinite integration range.

Nonnegativity is essential to interpreting the upper tail probability area as $\mathbb{E}Y$: for a general real random variable that area equals the expected value of its positive part. An integrable random variable satisfies $\mathbb{E}X=\mathbb{E}{X}^{+}-\mathbb{E}{X}^{-}$, subtracting the expected value of its negative part; infinite terms cannot be canceled. Even when $Y$ is finite almost surely, its expected value can be infinite. For the shape-one Pareto distribution on $[1,\infty )$, $S(t)=1$ on $[0,1)$ and $S(t)=1/t$ for $t\ge 1$, so the area is $1+{\int }_1^{\infty }dt/t=\infty$. The finite versus infinite expected value distinction is illustrated in Kyle Siegrist, Definitions and Basic Properties: exponential and Pareto expected values.

Why are tail probabilities useful? They answer threshold questions directly, and a concentration inequality can bound them without knowing the full probability distribution. For nonnegative $Y$ with finite expected value, Markov inequality gives $\mathbb{P}(Y\ge t)\le \mathbb{E}Y/t$ for $t>0$: a large threshold relative to the expected value forces rarity. The quantile function reverses the question by choosing a threshold for a prescribed cumulative level. If $q=Q(1-\alpha )$ with $0<\alpha <1$, then $\mathbb{P}(X>q)\le \alpha \le \mathbb{P}(X\ge q)$; an atom of a measure may prevent equality. For the die, $Q(0.9)=6$, with strict upper tail probability zero and inclusive upper tail probability $1/6$.

A tail probability describes rarity under a specified probability distribution; it is not automatically a conditional probability after extra information. For $u\ge t$ with $\mathbb{P}(X\ge t)>0$,because the threshold events are nested. Thus the die has $\mathbb{P}(D\ge 5)=1/3$ but $\mathbb{P}(D\ge 5\mid D\ge 4)=2/3$. Conditioning changes the sample space and renormalizes its probability measure; the same extreme outcomes can become common in that restricted population.

Condition an exponential distribution of rate $d>0$ on falling below $t>0$. Normalizing its probability density function gives the displayed density. Integration gives expectation $1/d-t/(e^{dt}-1)$. As $d\downarrow 0$, the density tends to the uniform distribution on $(0,t)$ and the expectation tends to $t/2$.

// Target: probability-and-statistics.bigb

For rate-$\alpha$ exponential distribution claims and $c>\lambda /\alpha$, the ruin integro-differential equation reduces to ${\psi }^{\prime \prime }+(\alpha -\lambda /c){\psi }^{\prime }=0$. Differentiate the convolution and use its equation $A^{\prime }=\alpha \psi -\alpha A$ to eliminate it. Positive relative safety loading implies $\psi (u)\to 0$ as $u\to \infty$, and the zero-capital survival probability gives $\psi (0)=\lambda /(\alpha c)$. These two conditions determine the displayed solution. The adjustment coefficient is $\alpha -\lambda /c$.

For an independent statistical sample from a gamma distribution with known shape $\lambda >0$ and unknown rate $\theta >0$, test $\theta ={\theta }_0$ against $\theta >{\theta }_0$. The alternative-to-null likelihood ratio for any larger rate ${\theta }_1$ is $({\theta }_1/{\theta }_0{)}^{n\lambda }\exp \{-({\theta }_1-{\theta }_0)S\}$, where $S={\sum }_i{X}_i$. This decreases in $S$, so the Neyman-Pearson lemma gives one common lower-tail ordering for all alternatives.

Let $G_a$ be the unit-rate gamma distribution cumulative distribution function with shape $a$. The uniformly most powerful test of exact test size $\alpha \in (0,1)$ rejects when $S<c_{\alpha }$, where $c_{\alpha }=G_{n\lambda }^{-1}(\alpha )/{\theta }_0$. Its power function is displayed above. If $n\lambda =1$, the sum has an exponential distribution, so $c_{\alpha }=-\log (1-\alpha )/{\theta }_0$ and $\pi (\theta )=1-(1-\alpha {)}^{\theta /{\theta }_0}$. Monotonicity of $G_a(\theta c_{\alpha })$ also makes this test valid and uniformly most powerful for the larger null hypothesis $\theta \le {\theta }_0$.

For independent samples from an exponential distribution, a test of rate at most ${\lambda }_0$ against rate greater than ${\lambda }_0$ rejects for small sample sums. All simple-alternative likelihood ratios against ${\lambda }_0$ decrease with that sum; its gamma distribution calibrates the size at ${\lambda }_0$. Monotonicity of the rejection probability in the rate proves that this is a uniformly most powerful test for the composite hypotheses.

For exponential distribution claims of mean $\mu$ and compound Poisson distribution count parameter $\lambda$, the total party variance under excess of loss reinsurance is $g(M)=2\lambda {\mu }^2(1-(M/\mu )e^{-M/\mu })$. Its derivative is $2\lambda e^{-M/\mu }(M-\mu )$, so $M=\mu$ is the unique global minimum, with value $2\lambda {\mu }^2(1-e^{-1})$.

The probability distribution of a nonnegative duration until an event. For a continuous duration with survival function $S(t)=P(T>t)$, its probability density function is $p(t)=-S^{\prime }(t)$. A constant hazard gives an exponential distribution; successive independent first-order waits give a convolution, as in the two-step catalytic waiting time.

If independent summands have exponential distribution of mean $\mu$ and the independent count has a zero-based geometric distribution of parameter $p$, their random sum of independent claims has an atom $p$ at zero and an exponential positive component of rate $p/\mu$. Its Laplace transform of a nonnegative random variable is $p/[1-(1-p)/(1+\mu s)]=p+(1-p)p/(p+\mu s)$. Thus $\mathbb{P}(S>x)=(1-p)e^{-px/\mu }$ for $x>0$. Unlike the geometric sum of exponential variables with a positive count, this law is not purely exponential.

For an independent exponential distribution default time of rate $\lambda$ and a continuous nonnegative martingale $S$, this process is a martingale. The compensating pre-default factor $e^{\lambda t}$ balances the survival probability $e^{-\lambda t}$. The strict survival inequality gives a càdlàg price process.