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The continuous failure-time probability density function is the negative derivative of the survival function:The minus sign is essential because survivor probability decreases with time.

Writing $F$ for the survival function here, the relation $F^{\prime }(t)=-h(t)F(t)$ with $F(0)=1$ givesThis notation uses $F$ for survival rather than for the cumulative distribution function.

Assume independent individuals and independent censoring conditional on their explanatory variables, with the censoring mechanism carrying no parameter $\theta$. A failure contributes its failure-time probability density function $f_i(x_i;\theta )$, while a censored observation contributes the survival function $F_i(x_i;\theta )$ because its failure time exceeds $x_i$. Terms from the censoring distribution then factor out of the survival-parameter likelihood. Thus the survival likelihood and its logarithm areThe individual subscripts allow the same parameter vector to act through different covariate values.

For individual $i$, let ${\pi }_i={\pi }_i(\psi )$, $q_i(t)$ be the known baseline hazard function, and $r_i(t;\varphi )$ the extra hazard. PutThe class-specific survival functions are $e^{-Q_i(t)}$ and $e^{-Q_i(t)-R_i(t)}$. Consequently the unconditional survivor and failure density areMixture probabilities must be summed before taking a logarithm; a weighted average of class log-likelihoods would be a different, complete-data calculation. Substitution into the right-censored survival likelihood givesEquivalently, this additive excess-hazard mixture likelihood has contributionHere $q_i$ and $q_i+r_i$ must be nonnegative hazards. The factor $-Q_i(x_i)$ can be dropped when maximizing over $\varphi ,\psi$ because $q_i$ is known. Factoring out censoring presumes a common parameter-free censoring mechanism independent of failure and latent class, conditional on the observed covariates; unmodelled class-specific censoring would change these mixture contributions.

Use the intended model of independent random variables $A,B$ with exponential distributions of rates $\alpha ,\beta >0$. For $M=\min (A,B)$, independence gives the survival functionThusThe variance is $1/(\alpha +\beta {)}^2$, obtained from the standard first two exponential moments, $EM=1/(\alpha +\beta )$ and $E{M}^2=2/(\alpha +\beta {)}^2$.

For these competing exponential clocks, ties have probability zero. Integrating over Alice's firing time givesLet $H$ denote the event that the next shot hits, and use the stated success probability as $P(H\mid A<B)=1/2$. Bill's corresponding conditional success probability is one. The law of total probability givesThe Bayes' theorem therefore yields

The printed assumptions specify the two marginal exponential distributions but do not explicitly specify their independence. The answers above require that extra assumption; marginals alone do not determine the race. For a concrete counterexample, take one $Z\sim \mathrm{Exp}(1)$ and set $A=Z/\alpha$, $B=Z/\beta$. These have the required marginals, but $M\sim \mathrm{Exp}(\max (\alpha ,\beta ))$; if $\alpha >\beta$, Alice always fires first and $P(A<B\mid H)=1$. Thus the independence qualification changes actual answers, rather than merely their derivation.

Normalize the probability density function using $u=t^c$:The corresponding Weibull distribution has survival functionSince the distribution is continuous, the choices $>$ and $\ge$ have the same probabilities. The conditional probability is consequentlyFor $s,t>0$ and $c>1$, the supplied inequality, applied to $s/t$, gives $(s+t{)}^c>t^c+s^c$. Taking exponentials yieldsThis expresses the increasing hazard function $h_c(t)=c{t}^{c-1}$: surviving to a later age shortens the remaining lifetime in this conditional distribution. At $c=1$ equality holds, as expected from the memoryless property of the exponential distribution.

Statistician B uses the more appropriate analysis. Its start–stop representation includes left truncation as well as right censoring. Both entry and exit are ages since opening, so a business belongs to the risk set at age $t$ only when $\mathrm{ageatentry}<t\le \mathrm{obsage}$. The event indicator distinguishes closure from a censored end of observation. The counting-process representation does not mean that multiple closures per business have been observed.

Statistician A records only exit and status and treats every sampled business as at risk from opening. That incorrectly adds already-established businesses to risk sets at ages before they entered observation, ignoring their necessary survival to entry. Its first risk count is $60$, whereas B's is $25$. The extra non-events dilute early estimated closure hazards and here bias the estimated survival upward. Correcting left truncation makes later risk sets capable of increasing as businesses enter at their respective ages.

For the Kaplan–Meier estimator with delayed entry, at event ages $t_j$ with $d_j$ closures and $r_j$ businesses at risk,The standard error follows from the Greenwood formula. Between event times this survival function estimate is constant. At age five the last preceding event is at $4.52$, and at age ten it is at $9.90$. Thus the requested estimates, including their precision, areandThese correspond to estimated proportions 46.2% and 28.0%, not to the probability of surviving five or ten additional years conditional on being sampled at an arbitrary age. A's corresponding estimates $0.650$ and $0.407$ use the inappropriate risk sets.

This interpretation assumes appropriately independent entry and censoring and a suitable common lifetime distribution across entry cohorts. A delayed-entry correction does not remove arbitrary informative sampling or calendar effects. More generally, if no observation covers the earliest ages, absolute survival from opening requires additional information about survival before the first observable age; the reported product-limit normalization is the one used in the supplied analysis.

For large $t$, write the survival function asA positive function $\ell$ is slowly varying if $\ell (tu)/\ell (t)\to 1$ for every fixed $u>0$. Heresince $\log (tu)=\log t+\log u$ and $\log \log (tu)-\log \log t\to 0$. Thus $\overline{F}$ has regular variation of index $-2$.

The relevant extreme-value criterion is that an infinite right endpoint and a survival function regularly varying with index $-r$, $r>0$, give attraction to the standard Fréchet distribution of shape $r$. Its distribution function is $G_r(x)=e^{-x^{-r}}$ for $x>0$ and zero for $x\le 0$. Here is a direct verification, so no additional theorem is needed to establish the criterion in this example. Choose $a_n\to \infty$ satisfying $n\overline{F}(a_n)\to 1$ and set $b_n=0$. For $x>0$,If $u_n\to 0$ and $n{u}_n\to c$, then $n\log (1-u_n)=-n{u}_n+O(n{u}_n^2)\to -c$. Applying this with $u_n=\overline{F}(a_{n}x)$ gives $F(a_{n}x{)}^n\to e^{-x^{-2}}$. For $x\le 0$, the distribution function is zero because the observations are supported on $[10,\infty )$. ThereforeFor example, elementary asymptotic normalizers arefor sufficiently large $n$, with arbitrary positive definitions at smaller indices. Indeed $\log a_n\sim \frac{1}{2}\log n$ and $\log \log a_n\sim \log \log n$, so $a_n^2\log a_n\log \log a_n\sim Cn$. The precise valid choice of $C$ at the lower endpoint does not change the limiting shape.

For $t\ge 0$, independence factors the survival function of the minimum:The minimum is nonnegative, and this is exactly the survival function of an exponential distribution. HenceThe parameter here is the rate, rather than the mean. The calculation is the simplest example of competing exponential clocks.

For a patient still in hospital, sum over the unobserved eventual outcome. Its likelihood contribution is the mixture survival functionThe probability of eventual death among those still hospitalized is generally different from $\theta$, because the two conditional time distributions can have different tails.

Let $f_j=\mathbb{P}(T=t_j)$ and $h_j=\mathbb{P}(T=t_j\mid T>t_{j-1})$. Since $S(t_{j-1})=\mathbb{P}(T>t_{j-1})$, the discrete hazard satisfies $h_j=f_j/S(t_{j-1})$. HenceStarting from $S(0)=1$ and iterating, the survival function isUnder independent censoring, the risk set just before $t_j$ gives the binomial conditional failure likelihood $h_j^{d_j}(1-h_j{)}^{n_j-d_j}$, whose maximizer is ${\hat{h}}_j=d_j/n_j$. This gives the Kaplan–Meier estimatorFailures at exactly $t$ are included because the target is $\mathbb{P}(T>t)$. Censored subjects contribute to earlier risk sets and leave before later ones; they do not create a failure factor. The usual convention includes a subject censored at a tied failure time in the risk set for that failure. Estimation beyond the last observed failure is limited by the available follow-up.

Assume independent censoring: the censoring mechanism contributes no factor involving the lifetime rate $\theta$. The exponential density and survival function are $f_{\theta }(x)=\theta e^{-\theta x}$ and $S_{\theta }(x)=e^{-\theta x}$. An observed event contributes the density; a right-censored lifetime contributes the probability of surviving its censoring time. Thus the likelihood for $\theta$, up to censoring factors independent of it, isFor $D>0$ and $T>0$, ${\ell }^{\prime }(\theta )=D/\theta -T$ vanishes at $D/T$, and ${\ell }^{\prime \prime }(\theta )=-D/{\theta }^2<0$ proves the maximum:Censored individuals add follow-up time to the denominator but no event to the numerator. If $D=0$ and $T>0$, the likelihood decreases for $\theta >0$ and has only a supremum as $\theta \downarrow 0$; zero is an extended boundary estimate, not a positive-rate exponential MLE. The derivation is for independent individuals entering at time zero; delayed entry would require conditional survival contributions and exposure measured from entry.

For a proper continuous event time, the survival function is $S(t)=e^{-H(t)}$. The assumed invertibility of the cumulative hazard function gives, for $u\ge 0$,ThereforeThis is the cumulative hazard probability transformation; it applies also conditionally on a subject's covariates, using that subject's correctly specified cumulative hazard function.

The fitted transformed times are Cox–Snell residuals. They retain their event/right censoring indicators, so a right-censored residual represents an exponential observation known only to exceed its displayed value. Under the fitted model and independent right censoring, calculate the Kaplan–Meier estimator of residual survival and compare it with $e^{-y}$, or calculate the residual Nelson–Aalen estimator and compare its cumulative hazard with the diagonal $H(y)=y$. Systematic departures reveal model inadequacy; sparse extreme residual risk sets and parameter estimation require caution. Treating all censored residuals as observed event times would invalidate this diagnostic.

Without right censoring, the residual mean should be approximately one. With right censoring, use the modified Cox–Snell residualFor a true unit-rate exponential distribution, the memoryless property gives $E(U\mid U>c)=c+1$. Thus an event keeps its known transformed time, while a censored observation is replaced by the conditional expected event time. Under independent right censoring, iterated expectation makes the mean of these adjusted values one when the true hazards are used, and approximately one when fitted hazards are used. These mean-imputed values do not themselves have an exponential distribution, so the survival-curve diagnostic should still use the original censored residual dataset. Moreover fitting equations can force the adjusted sample mean to one, making its mean alone a weak diagnostic.

For the proposed mixture of a finite right censoring time $c\ge 0$ and no right censoring, $P(C<U)=\pi e^{-c}$ andConsequentlyWhen right censoring has positive probability this choice is unique; if $\pi e^{-c}=0$, no correction is needed and any $k$ has the same effect. Its independence from $c$ is the content of exponential memorylessness: the expected extra lifetime after any right censoring time is one. Conditioning on an arbitrary independent right censoring time proves the same correction beyond this special two-point mixture.

Write $r={\theta }_A+{\theta }_B$, with ${\theta }_A\ge 0$, ${\theta }_B>0$ and a finite $\tau \ge 0$. The competing risks model with transient surgical mortality has survival functionIntegrating the disease cause-specific hazard against this survival function givesThe first term accounts for disease deaths while both causes act; the second includes survivors of that period who then face only disease mortality. The probability of eventually dying from surgery isTaking the limit in $F_B$ gives $F_B(\infty )={\theta }_B(1-e^{-r\tau })/r+e^{-r\tau }$, and thereforeThe ultimate-death conclusion uses the positive continuing disease hazard. If ${\theta }_B=0$, the surviving fraction $e^{-{\theta }_A\tau }$ would instead live indefinitely in this model; the asserted conclusion is not valid in that boundary case.

Use the Aalen–Johansen estimator of the cumulative incidence function, updating the overall Kaplan–Meier estimator for either event:Right censoring changes subsequent risk sets, not the survival function by itself. Starting with the given estimates at $a_k$, the complete calculation isThe middle event is of the competing type: it decreases overall survival while leaving the disease cumulative incidence function unchanged at that instant. The event-free intervals leave every estimate and risk set unchanged.

At the final time, both the event subject and the subject censored at that time are in the just-before risk set, so its denominator is eight. This is the usual event-before-censoring convention for recorded ties. After the event and right censoring, six subjects remain at risk. ConsequentlyThe nine numbered source items are the inputs to this single calculation, not nine further questions.

A non-degenerate probability distribution is not concentrated at a single point. In extreme value theory, a non-degenerate distribution function $F$ is max-stable if, for each integer $n\ge 1$, constants $a_n>0,b_n\in \mathbb{R}$ satisfy $F(a_{n}x+b_n{)}^n=F(x)$ for every $x$. This says that a normalized sample maximum has the same law as one observation. A distribution function $F$ belongs to the maximum domain of attraction of a non-degenerate $G$ if there are $a_n>0,b_n$ such thatat every continuity point of $G$. Equivalently, $(M_n-b_n)/a_n$ has convergence in distribution to $G$.

The extremal types theorem says that any such non-degenerate limit is max-stable and, up to a positive affine change of variable, is exactly one of the following:These are respectively the Gumbel distribution, Fréchet distribution, and negative Weibull distribution. The theorem is also known as the Fisher–Tippett–Gnedenko theorem.

Useful sufficient conditions can be expressed using the survival function $\overline{F}=1-F$ and the right endpoint of a distribution $x_F$. An infinite endpoint with $\overline{F}(tx)/\overline{F}(t)\to x^{-\alpha }$ for every $x>0$ gives the Fréchet distribution domain. A finite endpoint with $\overline{F}(x_F-sx)/\overline{F}(x_F-s)\to x^{\alpha }$ as $s\downarrow 0$ gives the negative Weibull distribution domain. For the Gumbel distribution, it suffices that a positive Gumbel auxiliary function $a(t)$ gives $\overline{F}(t+a(t)x)/\overline{F}(t)\to e^{-x}$ for every real $x$ as $t\uparrow x_F$. These are regular variation and exponential tail-ratio conditions; they need no proof here.

In case (i), ${\overline{F}}_1(x)=(1-x{)}^2$ on $0<x<1$, so the finite-endpoint ratio is exactly $x^2$. Therefore the domain is negative Weibull with shape $2$. With $a_n=n^{-1/2}$ and $b_n=1$,and the limit is one for $x\ge 0$.

In case (ii), ${\overline{F}}_2(t)=e^{-\lambda t}$ has exponential tail ratio with the constant Gumbel auxiliary function $1/\lambda$. Therefore the domain is Gumbel. Choosing $a_n=1/\lambda$ and $b_n=\log n/\lambda$ givesIn case (iii), ${\overline{F}}_3(t)=(1+t{)}^{-\lambda }$ has regular variation of index $-\lambda$, so the domain is Fréchet with shape $\lambda$. The convenient choices $a_n=n^{1/\lambda }$ and $b_n=-1$ givewith limit zero for $x\le 0$.

Finally, define the empirical distribution function $F_n(x)=n^{-1}{\sum }_{i=1}^n{1}_{\{X_i\le x\}}$. On a sample with positive threshold $t=X_{(n-k)}$ and exactly $k$ observations strictly above $t$, its survival function has $1-F_n(t)=k/n$. Using the Tonelli theorem to integrate the finite nonnegative sum givesThis is the Hill estimator as an empirical plug-in version of the tail-integral limit. It does not prove consistency for fixed $k$; that is a separate issue.

There is a finite-sample qualification because the question permits ties and does not assume positive observations. The formula requires $1\le k<n$ and $t>0$. If $r=\#\{i:X_i>t\}<k$ because the threshold is tied, direct substitution giveswhere terms equal to the threshold contribute zero. This is generally $k/r$ times the displayed Hill estimator, rather than the same estimator. If $r=0$, the empirical plug-in denominator is zero. For a continuous probability distribution the no-tie condition holds almost surely; for a Fréchet distribution domain and fixed $k$, the threshold is positive with probability tending to one. Thus the stated plug-in identity holds at a positive untied threshold, and the literal claim for all samples from an arbitrary distribution function needs this qualification.