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A reinsurance layer with attachment $M$ and payment limit $A$ pays nothing below $M$, the excess between $M$ and $M+A$, and $A$ above that. Its expected recovery is ${\int }_M^{M+A}\mathbb{P}(X>x)dx$, by the tail integral formula for moments. For continuous claims, conditioning on penetration of the layer divides this expectation by $\mathbb{P}(X>M)$.

Suppose everyone enters observation at zero and has one event or censoring time. At distinct recorded time $t_j$, let $n_j$ observations terminate, of which $d_j$ are events, and use the pre-time risk set size $r_j={\sum }_{k\ge j}{n}_k$. For the grouped Nelson–Aalen estimator, $\hat{H}(t_j)={\sum }_{k\le j}{d}_k/r_k$. Exchanging sums gives ${\sum }_j{n}_j\hat{H}(t_j)={\sum }_k(d_k/r_k){\sum }_{j\ge k}{n}_j={\sum }_k{d}_k$. The unweighted sum over distinct times generally does not have this property. Processing each observation separately after breaking ties also restores the corresponding unweighted identity on the resulting artificial time grid.

For events $A_n$ in a probability space, the infinitely-often event isIts complement of an event is ${\bigcup }_{m\ge 1}{\bigcap }_{n\ge m}{A}_n^c$: after some outcome-dependent index, no more events occur.^[ref]

Deleting finitely many events leaves the limsup of events unchanged. It is therefore a tail event of the sequence of indicator random variables $1_{A_n}$, whether or not those random variables are independent. For independent events, the Kolmogorov zero-one law restricts its probability to zero or one; the Borel-Cantelli lemmas determine which alternative holds from the series of probabilities. Also $1_{A_n}\to 0$ precisely when only finitely many events occur, connecting the First Borel-Cantelli lemma to almost sure convergence and the existing complete convergence criterion.

If the bank account $B$ is positive and nondecreasing and $S/B$ is a martingale under an equivalent martingale measure $Q$, then for $K\ge 0$ the European call price $C(T,K)$ is nondecreasing in $T$. Put $M_t=S_t/B_t$. For $u\ge t$, $K/B_u\le K/B_t$ and hence $(M_u-K/B_u{)}^{+}\ge (M_u-K/B_t{)}^{+}$. Conditional convexity gives ${\mathbb{E}}_Q[(M_u-K/B_t{)}^{+}\mid {\mathcal{F}}_t]\ge (M_t-K/B_t{)}^{+}$. Taking expectations and multiplying by $B_0$ proves the claim. The result is not generally true for negative strikes: with $S_t=B_t=e^{rt}$ and $r>0$, a negative-strike call costs $1-K{e}^{-rT}$, decreasing in maturity.

For any fixed maturity, a European call option price is convex in its strike because $K\mapsto (s-K{)}^{+}$ is convex for every terminal stock value $s$. Specifically, for $0\le a\le 1$, $(s-a{K}_1-(1-a)K_2{)}^{+}\le a(s-K_1{)}^{+}+(1-a)(s-K_2{)}^{+}$. Multiplication by the positive discount factor and expectation under the pricing measure preserve this inequality. This also explains why a violation of convexity creates a butterfly-spread arbitrage.

A down-and-out European call pays the usual European call option payoff only if a lower barrier has not been hit before maturity. In the zero-interest, zero-dividend Black-Scholes model, for $0<B\le K<S$, its value is $C(S,K)-(S/B)C(B^2/S,K)$. This follows from reflection of the drifted logarithmic price killed at the barrier.

For a sufficiently regular homogeneous function $f$ of degree $n$ on a three-dimensional region, Euler theorem for homogeneous functions gives $x\cdot \nabla f=nf$. Hence $\nabla \cdot (fx)=(n+3)f$, and the divergence theorem gives the displayed reduction when $n\ne -3$. At degree $-3$ only the zero-flux identity follows, so division by $n+3$ is unavailable. Singular functions additionally require an excision or limiting argument; the surface contribution at a removed singularity must not be silently dropped. On the sloping side of a cone with its apex at zero, $x$ is tangent to the side and $x\cdot n=0$, so only the cap contributes to this formula.

On an open domain invariant under positive scaling, $D{f}_x(x)=cf(x)$ implies $f(\lambda x)={\lambda }^{c}f(x)$. Indeed, $\frac{d}{d\lambda }[{\lambda }^{-c}f(\lambda x)]={\lambda }^{-c-1}[D{f}_{\lambda x}(\lambda x)-cf(\lambda x)]=0$. The converse follows by differentiating the scaling identity at $\lambda =1$. Connectedness of the whole domain is unnecessary: the argument works along each positive ray.

For a connected cellular spherical planar graph with vertex degrees at least three and face sizes at least three, degree/face double counting and the Euler formula for a connected planar graph give ${\sum }_n(6-n)F_n=12+{\sum }_m(2m-6)V_m\ge 12$. The left side is at most $3(F_3+F_4+F_5)$, proving the displayed bound. In particular a convex polyhedron has at least four faces of size at most five.

Hierholzer's algorithm constructs an Euler circuit by growing closed trails along unused edges and splicing them at shared vertices, as in the Euler circuit criterion. Each edge must have a distinct record. An implementation using adjacency lists, a stack, and a used flag for each edge runs in $O(|V|+|E|)$ time, including the connectivity and vertex degree checks. Store both incidences of a self-loop, while marking its single edge record used only once.

With fixed positive scale estimates $s_j$, this is Euclidean distance in standardized coordinates. It removes marginal units but ignores correlations. Small estimated variances can greatly amplify measurement noise; a zero scale must be handled separately.

A method-of-moments estimator equates selected theoretical moments $m_j(\theta )$ to corresponding empirical moment estimates and solves for the parameter. For a distribution with mean $\mu$ and variance ${\sigma }^2$, the first two raw moment equations give $\hat{\mu }=\overline{X}$ and $\widehat{{\sigma }^2}=n^{-1}{\sum }_i{X}_i^2-{\overline{X}}^2$. Existence, uniqueness and consistency require appropriate identifiability and moment assumptions; unbiased empirical moments do not make nonlinear parameter estimates unbiased. This statistical estimation procedure differs from the method of moments in probability for proving convergence in distribution.

For a sphere of radius $r_0$ and uniform mass density ${\rho }_0$, with no surrounding mass and gravitational potential zero at infinity, the spherical shell theorem gives $V_c^2=4\pi G{\rho }_0{r}^2/3$ inside. Matching to the exterior point-mass potential gives $\Phi (r)=-(2\pi G{\rho }_0/3)(3r_0^2-r^2)$. Since $v_e^2=-2\Phi$, the displayed relation follows for $0<r\le r_0$. Exterior shells leave the interior circular speed unchanged but deepen the potential and change the escape velocity; a uniform core alone is insufficient to determine escape.

A spacecraft trajectory is the path traced by a spacecraft's position over time. An unpowered exterior arc can approximate a Kepler orbit; other bodies' gravity, radiation pressure, and propulsion can change it.

Operational evidence comes from reconstructing a spacecraft's motion. A ground antenna transmits a coded signal; the spacecraft returns it. Corrected round-trip time constrains distance, while the Doppler effect of a coherent carrier constrains the line-of-sight relative velocity. NASA's navigation account describes calibration of onboard turnaround delays before launch, calibration of ground delays before use, and corrections for the terrestrial atmosphere. Measurements over multiple stations and epochs constrain the orbit; range and Doppler effect alone do not directly give all transverse motion. Navigators compare a fitted spacecraft trajectory with later tracking and remeasure after propulsion to assess the maneuver. The account identifies radiation pressure and imperfect propulsion as disturbances. This provides practical checks of a dynamical model and its corrections, rather than a laboratory measurement of a universal escape speed. The simple inverse-square potential predicts an unpowered exterior arc; accurate navigation includes additional forces and measurement delays.^[ref]

For a circular orbit, gravity supplies centripetal acceleration, giving $v_c^2/r=\mu /r^2$. ThusA body in a circular orbit is bound despite its continuous free fall. At the same $r$, escape needs twice its kinetic energy per unit mass. An instantaneous forward tangential velocity increment from the circular orbit requires $\Delta v=(\sqrt{2}-1)v_c$, rather than the full escape speed from rest. In the synthetic example this is exactly $2(\sqrt{2}-1)m/s\approx 0.828427m/s$.^[ref]

Escape speed is a coasting threshold, not a required continuously maintained speed or a measure of rocket propellant. A powered vehicle can start at rest and acquire energy gradually. The rocket equation relates an ideal velocity increment to rocket exhaust speed and rocket mass ratio; rocket thrust, drag, and the timing and direction of propulsion affect a real ascent. Rotation also changes the relation between surface-relative velocity and the inertial frame used in the calculation.

Reaching another body is a finite-destination spacecraft trajectory problem. In the synthetic radial example, a launch at $2m/s$ reaches $r=4\mathrm{m}$ while remaining bound. A spacecraft can similarly encounter the Moon on a bound geocentric transfer; actual planning must include the Moon's gravity, target motion, arrival velocity, and any propulsion needed for capture. Escaping Earth alone does not imply escaping the Sun, because the relevant reference frame and attracting mass have changed. The isolated formula describes one gravitational well; a full mission requires more than its scalar threshold.

“Arrives at infinity with zero speed” is an asymptotic statement. For outward radial motion at zero specific orbital energy,Every finite distance takes finite time, but $t(r)\to \infty$ as $r\to \infty$. There is no instant of rest at an infinitely distant destination. Differentiating the displayed solution gives $\ddot{r}=-\mu /r^2$, independently checking Newton's second law. In the synthetic example, reaching $r=4\mathrm{m}$ takes exactly $(4-\sqrt{2})/3\mathrm{s}\approx 0.861929\mathrm{s}$. For nonzero specific angular momentum, the Barker equation gives the corresponding infinite-time limit along a parabolic trajectory. Positive-energy escape also takes infinite time to reach infinity; it retains nonzero asymptotic speed.

The same initial distance can lead to strikingly different orbits. Let $s=v_0/\sqrt{\mu /r_0}\ge 1$ and launch tangentially in one orbital plane. Then periapsis is the launch point, specific angular momentum is $h=r_0{v}_0$, andwhere $f$ is true anomaly. These follow by combining the initial specific orbital energy with the orbital eccentricity relation and the Kepler orbit equation. With $\mu =8{\mathrm{m}}^3/{\mathrm{s}}^2$ and $r_0=2\mathrm{m}$, the initial speeds $2$, $\sqrt{6}$, $2\sqrt{2}$, and $2\sqrt{3}m/s$ give $e=0,1/2,1,2$: a circular orbit, an elliptic orbit with apoapsis distance $6\mathrm{m}$, a parabolic trajectory, and a hyperbolic Kepler orbit. The two open paths have asymptotic speeds zero and $2m/s$ respectively. Their different directions at large distance are part of the Kepler orbit, not an extra scalar escape threshold.

Use the synthetic standard gravitational parameter $\mu =8{\mathrm{m}}^3/{\mathrm{s}}^2$ and launch distance $r_0=2\mathrm{m}$. Then escape speed is $2\sqrt{2}m/s$. Three outward radial launches give exact comparisons: $v_0=2m/s$ has specific orbital energy $-2J/kg$ and stops outward at $r=4\mathrm{m}$; $v_0=2\sqrt{2}m/s$ has zero specific orbital energy; $v_0=2\sqrt{3}m/s$ has specific orbital energy $+2J/kg$ and asymptotic speed $2m/s$. The second case has speed $2m/s$ at $r=4\mathrm{m}$. Doubling $r_0$ divides escape speed by $\sqrt{2}$, whereas doubling $M$ at fixed $r_0$ multiplies it by $\sqrt{2}$. Changing the test particle's mass changes $E$ but cancels from the escape speed.

An inviscid fluid with conservative body force, a materially conserved stratifying scalar $\alpha$, and an equation of state making $\nabla \rho ,\nabla p,\nabla \alpha$ coplanar conserves its Ertel potential vorticity. The stretching of absolute vorticity per unit density cancels the deformation of the scalar gradient; the remaining baroclinic triple product vanishes. In an incompressible density-stratified ideal fluid, choosing the advected density as $\alpha$ gives that cancellation directly.