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Synchronous radius and its limits. Kepler's third law for a circumplanetary orbit givesHere $G$ is the gravitational constant. A planet-synchronous orbit repeats after one spin period. A planet-stationary orbit additionally requires a circular orbit, zero orbital inclination to the equator and prograde motion. Then an antenna fixed on the ground points continuously at the same satellite; merely matching the orbital period does not give this property.

Write the planetary radius as $R_p=(3M/(4\pi \rho ){)}^{1/3}$. Requiring the semi-major axis to exceed $R_p$ gives $P^2>3\pi /(G\rho )$. The stellar tidal force restricts the circumplanetary orbit to the Hill sphere, of radius $r_H=a(M/(3M_{\star }){)}^{1/3}$. ThusThese are the surface and Hill-radius constraints in the idealized spherical, small-$M/M_{\star }$ model. Long-term prograde stability generally requires a radius appreciably inside the Hill sphere; for nonzero orbital eccentricity, the surface constraint applies to the periapsis, not just to the semi-major axis.

Collision time and the launch population. A phase-mixed isotropic swarm occupies a shell of radial scale $e{a}_s$. Its number density scales as $N_s/(e{a}_s^3)$, its relative speed as $a_s/P$, and its geometric collision cross-section as $r^2$. Consequently the total collision rate scales as $N_s^2{r}^2/(Pe{a}_s^2)$. To give the numerical normalization used here, adopt an effective shell volume $V=4\pi e{a}_s^3$. At a fixed position the velocity directions are uniformly distributed in the tangent plane, so their mean relative speed is $4v_K/\pi =8a_s/P$. For an unordered pair, the collision cross-section is $4\pi r^2$, givingThus the mean interval between collisions is $A/N_s^2$ in this large-population kinetic model. The effective shell width fixes an order-one coefficient: small orbital eccentricity and random planes alone do not specify a unique radial probability density. Phase mixing, negligible gravitational focusing, $r\ll e{a}_s$, and uncorrelated encounters are implicit in this estimate. Exactly identical orbital periods with perfectly fixed phases do not themselves produce a memoryless collision process.

For nearly planet-stationary orbits with $e\ll I\ll 1$, the swarm volume is smaller by a factor of order $I$, while the relative speed is smaller by the same factor, since vertical motion of scale $I{v}_K$ dominates the eccentric motion. The two changes cancel in the rate $n\sigma v$: there is no parametric factor $I$ in the collision time within the same phase-mixed kinetic approximation. Numerical factors and phase correlations can differ. This is not an argument that bringing all satellites into one nearly circular plane makes their phases random.

With $N_s=Rt$, an Inhomogeneous Poisson process has cumulative hazard functionThe survival function of the first collision is $\exp [-H(t)]$. Setting the expected number of collisions to one givesThis is a characteristic first-event population, with probability $1-e^{-1}$ of an earlier event. It is not a median: the median has an additional factor $(\log 2{)}^{1/3}$.

Which population collides next? Immediately after the first disruption let $S=N_{s0}-2$ and $F=2x^{-3}$. The latter follows from mass conservation for equal-density spherical pieces. Using the same unordered-pair counting and geometric collision cross-sections as above, define $b=(1+x{)}^2/2$; thenHere $S^2,F^2$ stand for the usual large-population approximations to $S(S-1),F(F-1)$. The factor two distinguishing identical and different species is essential. The probability that the next collision is fragment–fragment isIt is the most likely type if $S<4/[x(1+x{)}^2]$ and $S<2/x^2$. It has probability greater than one half ifFor $x\ll 1$, a strongly fragment-dominated next event therefore requires $Sx\ll 4$; the fragment–satellite comparison is more restrictive than the satellite–satellite comparison. If $S=0$, the next event must be fragment–fragment, provided fragments remain.

Population equations and the normalization discrepancy. Every satellite–satellite event produces $2x^{-3}$ fragments and destroys two satellites. Every satellite–fragment event produces a net $x^{-3}-1$ fragments and destroys one satellite; every fragment–fragment event destroys two fragments. Consistent collision counting therefore givesFor $x\ll 1$, $b\simeq 1/2$ and $x^{-3}-1\simeq x^{-3}$. The last two terms in $A{\dot{N}}_f$ are then twice those in the printed equation. This is a genuine factor-of-two inconsistency: equal-size fragment pairs have a collision cross-section smaller by $x^2$, so their event rate must be $x^2{N}_f^2/A$ if the satellite event rate is $N_s^2/A$; destroying both fragments necessarily gives the sink $-2x^2{N}_f^2/A$.

If the printed equation is taken as a prescribed approximate rate model instead, its implicit event rates are $N_s^2/A$, $N_f{N}_s/(4A)$ and $x^2{N}_f^2/(2A)$. Under precisely that mixed normalization, its corresponding satellite equation isThe printed source term also neglects the consumed fragment in a satellite–fragment event, a legitimate relative $O(x^3)$ approximation. That approximation does not repair the pair-counting discrepancy.

The ensuing cascade. This is a two-size fragmentation cascade. First use the printed approximate model, dropping satellite–satellite events as requested. Put $S=N_s$, $F=N_f$, $k=4x^2$ and take $x<1/2$. ThenIntegrating this linear differential equation, with $F_0=2x^{-3}$ and $S_0=N_{s0}-2>0$, givesInitially the collisional cascade grows if $S_0>8x^2$, with approximate early exponential growth time $4A{x}^3/S_0$ when the satellite population is nearly fixed. The fragment population reaches its maximum atAfterwards satellites are depleted and fragment–fragment losses dominate. For $S_0>0$ the continuum solution has $S\to 0$, $F\to 0$, withWhen no satellites remain initially, $F(t)=F_0/[1+x^2{F}_{0}t/A]$ directly. Small integer populations eventually invalidate these deterministic differential equations.

The consistently counted model has, to leading order in $x$, exactly the same curve $F(S)$ and peak, but both retained time derivatives are twice as large. Its growth time is $2A{x}^3/S_0$, and $F(t)\sim A/(2x^{2}t)$. Keeping $(1+x{)}^2$ and the consumed fragment replaces $k$ by $4x^2/(1+x{)}^2$ and $x^{-3}$ in the curve by $x^{-3}-1$. This explicitly separates the physical collision bookkeeping from the printed normalization while giving the evolution under both conventions.

For a nonnegative random variable, $T={\int }_0^{\infty }1\{T>t\}dt$. Tonelli theorem permits interchanging expectation and this nonnegative integral, giving the tail integral formula for moments:The equality also holds when the mean is infinite. Thus mean completion time is the total area under the survival function, while a restricted mean survival time integrates only to a specified follow-up horizon.

The Kaplan–Meier estimator is $\hat{S}(t)={\prod }_{t_j\le t}(1-d_j/r_j)$, with event count $d_j$ and risk set size $r_j$. Its median is the first event time with estimated survival at most $1/2$. Since $\hat{S}(2)=0.5749$ and $\hat{S}(3)=0.4286$, the estimated median completion time is $3$ minutes.

The original last event exhausts the final risk set, so the fitted survival function drops to zero at $18$ and its conventional full area is finite. Integrate the right-continuous step function: its height is one before time one, and after each event the height is the newly computed survival value. There are three-minute gaps from $12$ to $15$ and from $15$ to $18$. ThusUsing the exact risk-set products rather than rounded printed survival values gives $4.187856$. This is not the sample mean of the observed event-or-censoring times; right censoring changes the estimated survival weights.

If the final observation at $18$ is instead censored, every earlier Kaplan–Meier estimator value is unchanged and no event factor is inserted at $18$. The median remains $3$ minutes, but the curve stays at $\hat{S}(15)\simeq 0.0107$ at the end of follow-up. By terminal censoring and survival-mean identifiability, a full unrestricted mean is no longer determined nonparametrically by these observations. Extending the final positive step forever would give an infinite integral, but that plotting convention does not establish an infinite population mean. Extra assumptions about the unobserved tail would be needed for a finite full-mean estimate.

The restricted mean survival time through $18$ is still $4.187856$ minutes in either version: the two curves differ only at or after the endpoint, which does not affect the area up to it. All these interpretations require independent censoring. Giving up may be related to how long a person would have taken to finish, so that assumption needs scientific justification in this study.

The expected value and variance under quota share reinsurance follow by scaling the exponential distribution:The retained stop loss moments for an exponential aggregate follow from the payout $Y=\min (S,M)$, its survival function equals $e^{-x/{\mu }_S}$ for $0\le x<M$ and zero for $x\ge M$. The tail integral formula for moments gives, with $m=M/{\mu }_S$ and $r=e^{-m}$,Consequently the retained moments areMatching the two expected values forces $\alpha =1-r$, which lies strictly between zero and one. The difference of the variances simplifies toIndeed $h(m)=m-1+e^{-m}$ has $h(0)=0$ and $h^{\prime }(m)=1-e^{-m}\ge 0$ for $m\ge 0$. Because $M>0$, the difference is actually positive. At equal retained expected value, aggregate stop loss reinsurance reduces the variance more than quota share reinsurance.

Use the rate convention for the exponential distribution. By convolution of independent random variables, the probability density function of $S=X+Y$ is, for $s\ge 0$,It is zero for $s<0$. The expression is nonnegative whichever rate is larger, and defines a two-stage hypoexponential distribution. As a check, its limit as $\mu \to \lambda$ is ${\lambda }^{2}s{e}^{-\lambda s}$, the shape-two gamma distribution.

For $M=\min \{X,Y\}$, independence gives the survival function $\mathbb{P}(M>s)=e^{-\lambda s}{e}^{-\mu s}$ for $s\ge 0$. Differentiating its complement givesThus the minimum has an exponential distribution whose rate is the sum of the rates.

The survivor function of a nonnegative event time isHere the paper uses $F$ for the survival function, rather than for the cumulative distribution function; the latter is $1-F(t)$ for a continuous distribution. The survivor function is nonincreasing, starts at 1 when there is no atom at zero, and tends to zero for an almost surely finite event time.

Let $d_j$ be the number of events at potential time $a_j$ and let $r_j$ be the size of the risk set immediately before that time, including observations censored at $a_j$. Define the discrete conditional event probabilityUnder independent censoring, and ignoring the separate censoring distribution, the survival likelihood factorizes asThis follows by multiplying a survival factor for each individual at risk who does not fail at $a_j$, and an event factor for each failure. The log factor $d_j\log q_j+(r_j-d_j)\log (1-q_j)$ is maximized at ${\hat{q}}_j=d_j/r_j$, with the corresponding boundary values when $d_j=0$ or $d_j=r_j$.

The survival function is the product of successive conditional survival probabilities. ThereforeThis is the Kaplan–Meier estimator. A product over no event times is 1. If everyone remaining experiences an event, the product becomes zero and stays zero; subsequent empty risk sets supply no additional factor.

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.

Before the first product forms, the system is in either $E_0$ or $C_0$. Product formation is possible only through $C_0\to E_1$, so the first-passage-time density is the outgoing probability fluxEquivalently, the survival function is $E_0(\tau )+C_0(\tau )$ and its negative derivative is this density.

Conditioning on $T_1=t$ and using independent random variables giveswhere the notation in the paper uses $F_2$ for the survival function.

A proportional frailty model specifiesIts population survival function is the Laplace transform of the frailty density,If $m=\mathbb{E}U<\infty$, replacing $U$ by $U/m$ and $h_0$ by $m{h}_0$ leaves their product, and hence the model, unchanged. The normalization $\mathbb{E}U=1$ identifies this otherwise arbitrary scale and makes $h_0(0)$ the initial population hazard when $H_0(0)=0$.

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.

The right endpoint of a distribution with distribution function $F$ is $x_F=\sup \{x:F(x)<1\}$, possibly infinite. The survival function is positive below $x_F$.

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.