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For independent and identically distributed random variables with distribution function $F$, let $M_n={\max }_{1\le i\le n}{X}_i$ be the sample maximum. In extreme value theory, $F$ belongs to the maximum domain of attraction $D(G)$ of a nondegenerate distribution function $G$ when there exist $a_n>0$ and $b_n\in \mathbb{R}$ such thatat every continuity point of $G$. Thus the normalized sample maxima have convergence in distribution to $G$.

A positive measurable function $q$ has regular variation at infinity with index $\rho$, written $q\in {\mathrm{RV}}_{\rho }$, ifThe case $\rho =0$ defines a slowly varying function; equivalently a regularly varying function is $q(t)=t^{\rho }\ell (t)$ with $\ell$ a slowly varying function.

Write $\overline{F}=1-F$ for the survival function and $x_F=\sup \{x:F(x)<1\}$ for the right endpoint of a distribution. For $\alpha >0$, use the following standard extreme value theory normalization conventions:They are respectively the Fréchet distribution, the negative Weibull distribution, and the Gumbel distribution for maxima. The negative Weibull distribution is supported to the left of its finite endpoint; it is not the usual positive Weibull distribution.

The necessary and sufficient maximum domain of attraction criteria are:

For the corresponding sufficient hazard function conditions, assume $F$ is absolutely continuous near $x_F$, its survival function tends to zero there, and its probability density function $f$ is positive there. Write $\eta (t)=f(t)/\overline{F}(t)$. The Von Mises conditions for extreme values giveThe last condition additionally assumes that the reciprocal hazard function is continuously differentiable near the endpoint. It concerns the derivative of the reciprocal hazard function, not the derivative of the hazard function itself. Equivalently, when $f$ is differentiable, it is $\overline{F}(t)f^{\prime }(t)/f(t{)}^2\to -1$.

To see the connection, taking an integral of the hazard function givesThe first two limits make the logarithmic tail ratios over multiplicative or endpoint-distance intervals tend to $-\alpha \log u$ or $\alpha \log u$, giving the required regular variation. For the Gumbel distribution, set $a(t)=1/\eta (t)$. The derivative condition makes $a(t+a(t)v)/a(t)\to 1$ uniformly on bounded $v$-intervals, soThe endpoint assumptions ensure these shifts are admissible: at an infinite endpoint $a(t)/t\to 0$, and at a finite endpoint $a(t)/(x_F-t)\to 0$. These are sufficient smooth-tail conditions; the tail-ratio characterizations above are necessary and sufficient without differentiability assumptions. For a source using the same reciprocal-hazard convention, see Smith and Weissman's university lecture notes, equations (2.18)–(2.24).

For the Gamma distribution with integer shape $m$ and rate one, let ${\overline{F}}_m(t)={\int }_t^{\infty }{u}^{m-1}{e}^{-u}du/(m-1)!$. The base case is ${\overline{F}}_1(t)=e^{-t}$, and integration by parts givesInduction provesThis survival function has the asymptotic form ${\overline{F}}_m(t)\sim e^{-t}{t}^{m-1}/(m-1)!$; for $m=1$ the relation is exact. Put $t_n={\beta }_n+x$. For every fixed real $x$, $t_n\to \infty$ and $t_n/\log n\to 1$. The proposed centering satisfiessoOnly the highest-degree term contributes to the limit, with the same reasoning covering $m=1$. The Taylor expansion $\log (1-u)=-u+O(u^2)$ now givesbecause $n{\overline{F}}_m({\beta }_n+x{)}^2\to 0$. Therefore the Gumbel limit for gamma maxima isEquivalently, $M_n-{\beta }_n$ has convergence in distribution to the standard Gumbel distribution, with scale $a_n=1$.

Writing $S_i(t)=1-F_i(t)$ for the survival function, the definition of the hazard function givesSince $S_i(0)=1$, integration yieldsThereforeThe exponent is the negative cumulative hazard.

For $g(u)=e^{-u}$,Therefore the population hazard function is

The quantityis the observed-minus-expected group-1 event count at time $a_j$. A positive value is local evidence that group 1 has the greater hazard function; a negative value points toward group 0.

Let $Y(t)$ be the at-risk process and $N(t)$ the counting process for observed events. Over a short interval, the multiplicative-intensity model giveswhere $h$ is the hazard function and $H$ the cumulative hazard function. Solving this relation for the infinitesimal hazard increment suggests $d\hat{H}(t)=dN(t)/Y(t)$. Summing over distinct event times gives the Nelson–Aalen estimatorwhere $d_j$ events occur among $r_j$ individuals at risk. Here there are no ties, so $d_j=1$.

Writing the two functions in the question as survivor functions, $S_2(t)=S_1(t{)}^{\lambda }$. Since $H_k(t)=-\log S_k(t)$,Differentiating at times where the hazard functions exist gives $h_2(t)=\lambda h_1(t)$. Their hazard ratio is therefore the constant $\lambda$, so they form a proportional hazards family.

A proportional hazards family has hazard functions related bywhere the hazard ratio $c(z)$ is positive and independent of time. Equivalently, its cumulative hazard functions satisfy $H(t\mid z)=c(z)H_0(t)$ and its survivor functions satisfy $S(t\mid z)=S_0(t{)}^{c(z)}$.

Construct the two countries' period-specific risk sets with the same left truncation and right censoring rules, then compare their event counts by a log-rank test. This is a nonparametric comparison because it does not specify the shape of either country's hazard function.

For specified exposure periods with constant death hazard functions $h_k$, the integrated hazard through the endpoint is ${\sum }_k{h}_k\Delta t_k$. The displayed event probability follows from the survivor function identity. For different latent exposure schedules mixed with probabilities $w_j$, average the schedule-specific probabilities, ${\sum }_j{w}_j(1-e^{-H_j})$, rather than exponentiating the average integrated hazard. When the hazards are small, $1-e^{-H_j}\simeq H_j$ gives a person-time approximation. A comparison must specify the entire custody or treatment schedule; later changes to exposure cannot be inferred from an endpoint rate alone.

For an equally weighted mixture of individual survivor functions $F_i$, the population hazard function is $\overline{h}(t)={\sum }_i{F}_i(t)h_i(t)/{\sum }_i{F}_i(t)$. Thus the weights are the surviving proportions, and the cumulative hazard function is $-\log (n^{-1}{\sum }_i{e}^{-H_i(t)})$, not generally the mean individual integrated hazard.

If every interspike interval is at least $t_{\mathrm{abs}}>0$, a sustained spike train has mean firing rate at most $1/t_{\mathrm{abs}}$. Across a window of duration $T$, the event count satisfies $N(T)\le 1+T/t_{\mathrm{abs}}$; dividing by $T$ and taking the long-time limit gives the bound. The extra one accounts for a possible first event. The bound permits much slower firing and says nothing about whether stronger drive can maintain the limiting rate in a real neuron.

For the chosen phenomenological paired-pulse recovery, the hold alone gives $500\mathrm{Hz}$. Add the further assumption that every successful event resets precisely the same recovery curve, with no accumulation from earlier events. Then $0.35\mathrm{nA}$ onset-timed pulses can sustain at most $1000/4.942488\approx 202.327\mathrm{Hz}$ in that model. This repeatable reset is an extra assumption: a two-pulse result alone cannot establish it. A real neuron's sustained rate also depends on neural adaptation, input delivery and evolving ion-channel inactivation. The existing leaky integrate-and-fire model similarly adds integration time to its imposed hold.

A homogeneous Poisson process of rate $\lambda$ has interspike intervals with an exponential distribution, soWith the purely illustrative choices $\lambda =100\mathrm{Hz}$ and $t_{\mathrm{abs}}=0.002\mathrm{s}$, this is $1-e^{-0.2}\approx 0.181269$. Thus about $18.13\%$ of intervals predicted by that model violate this chosen hold. Matching a mean spike train count does not make a stimulus-only Poisson process respect recent firing.

A minimal alternative is a renewal process with independent and identically distributed intervals $S=t_{\mathrm{abs}}+X$, where $X$ has an exponential distribution of rate $\lambda$. Its hazard function is zero during the hold and $\lambda$ afterwards. Its mean interval is $t_{\mathrm{abs}}+1/\lambda$ and long-run firing rate isThe same $100{\mathrm{s}}^{-1}$ rate after a $2\mathrm{ms}$ hold gives $83.333\mathrm{Hz}$, not $100\mathrm{Hz}$. This separate stochastic example has no gradual relative recovery; it illustrates why an available-state rate differs from an observed mean firing rate. Smooth dependence on time since the last event can extend the hazard function, while longer histories require more than this simple renewal process. Time-rescaled spike-train diagnostics can test the resulting history-dependent predictions against recorded spike trains.

Under the additive hazard function model $h^{(i)}=h_B^{(i)}+h_E$, the survivor function factorization gives $F_E=F^{(i)}/F_B^{(i)}$, independently of $i$. If the excess hazard is nonnegative, this has a survival-probability interpretation for a hypothetical process with just that hazard. If excess hazards are negative, relative survival can exceed one and is a comparison ratio rather than an ordinary survivor function.

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.

A survival distribution is the probability distribution of an event or failure time, commonly summarized by its survivor function, hazard function or cumulative hazard function. It may contain atoms or a positive probability of an event never occurring; an ordinary density-based hazard therefore need not describe every survival distribution.

In a population mixture with initial component probabilities $p_i$, conditioning on survival to time $t$ changes them to the displayed probabilities by Bayes' theorem. Components with larger survivor functions receive greater relative weight among survivors. The population hazard function is the weighted mean of the component hazards with these time-dependent weights. This selection can change the population hazard even when all component hazards are constant, as shown by the hazard derivative for a mixture of exponential distributions.

A survival time is the elapsed time to an event of interest, which may be death, failure, recovery or completion of a task. Its survival function and hazard function describe the event-time law. Right censoring records only that the event time exceeds the observed follow-up time.

Let the frailty random variable be uniform on $[a,b]$ with $0<a<b$, and let the conditional individual hazard function be the constant $U\theta$, $\theta >0$. Averaging conditional exponential survivor functions gives the displayed population survivor function for $t>0$, with value one at zero. Its population hazard of a survival mixture isIt starts at $\theta (a+b)/2$ and tends to $a\theta$: the frailty distribution among survivors progressively concentrates near its smallest frailty. Although each individual's hazard is constant, the population hazard decreases through survival selection.