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With recovery rate one, the number of infected sites in a nearest-neighbour contact process on ${\mathbb{Z}}^d$ has death rate equal to its size and birth rate at most $2d\lambda$ times its size. It is dominated by a linear birth-death branching process with these per-particle rates. In particular, from one infection its expected size is at most $e^{(2d\lambda -1)t}$, so its Survival probability is zero when $2d\lambda <1$. This proves the displayed lower bound on the critical rate.

A Branching process with mean offspring number one. In a continuous-time birth–death example with equal per-cell birth and death rates $\beta$, one initial individual has Survival probability $1/(1+\beta t)$ and mean size $1+\beta t$ conditional on survival.

For a piecewise-deterministic Markov process, conditional on no jump before $u$, the state is $z(u)=\phi (s,x,u)$ and the survival probability is $S(u)=\exp [-{\int }_s^u\alpha (z(r))dr]$. The first jump has probability density function $\alpha (z(u))S(u)$; conditioning on its time and its destination gives $P_{s,t}f(x)=S(t)f(z(t))+{\int }_s^t\alpha (z(u))S(u)\int P_{u,t}f(y)K_u(z(u),dy)du$. This proves the identity for bounded Borel measurable functions under nonexplosion.

An unstable particle of rest lifetime ${\tau }_0$ traversing distance $H$ at constant speed $v$ has survival probability $p=\exp [-H/(\gamma v{\tau }_0)]$. Requiring a flight time no greater than one dilated mean lifetime selects $p\ge e^{-1}$. In the ultrarelativistic approximation $v\simeq c$, this gives $\gamma \gtrsim H/(c{\tau }_0)$ and $E\gtrsim m{c}^{2}H/(c{\tau }_0)$. It is a specified survival-probability criterion, not a sharp cutoff for individual random decays. Continuous energy loss requires integrating the decay probability along the path.

At a distinct event time $a_m$, estimate the conditional probability of surviving that time by $1-d_m/r_m$. Multiplication of these conditional survival probabilities gives the Kaplan–Meier estimator, and taking minus its logarithm gives an integrated hazard estimate:The conditional failure count has a binomial likelihood whose maximizing event fraction is $d_m/r_m$. Tied failures therefore enter directly as one count. Even ordering $d_m$ failures arbitrarily, with no intervening censoring, produces the same survivor function factor becauseThe failure-versus-censoring ordering convention still matters when both have the same recorded time. If $d_m=r_m$, the survivor function becomes zero and the logarithmic integrated hazard is infinite.

In the classical risk model, let $N(t)$ be the Poisson process of claim arrivals, independent of the independent claim sizes. The reserve starts at $u$ and earns premiums continuously at rate $c$, soHere $\psi$ is the ultimate ruin probability and $\varphi =1-\psi$ is the survival probability. The relative safety loading is $\theta =(c-\lambda \mu )/(\lambda \mu )>0$.

To derive the survival integro-differential equation, condition on the first claim time $T$, which has the exponential distribution of rate $\lambda$. Immediately before that claim the available capital is $u+cT$. A claim $x\le u+cT$ leaves survival probability $\varphi (u+cT-x)$, since the future claim process restarts with the same independent law; a larger claim causes ruin immediately. The first-claim decomposition for survival probability is thereforePut $r=\lambda /c$ and $H(v)={\int }_0^v\varphi (v-x)f(x)dx$. Changing variables to $v=u+ct$ yieldsThis representation justifies differentiation. The function $H$ is continuous: extend $\varphi$ by zero to negative arguments, then use continuity of translations of the integrable density $f$ in $L^1$ to control its convolution with the bounded function $\varphi$. Differentiating the lower endpoint of the displayed integral gives ${\varphi }^{\prime }=r\varphi -rH$. Since $c=(1+\theta )\lambda \mu$,This deduction uses the first-arrival law rather than taking the desired differential equation as an assumption.

For $f(x)=4x{e}^{-2x}$, the claim law is an Erlang distribution of shape two and rate two. Its mean is $\mu =4{\int }_0^{\infty }{x}^2{e}^{-2x}dx=1$. With $\theta =2$, we have $r=1/3$. Changing the convolution variable to $t=u-x$ givesHence the first requested specialization isLet $J(u)={\int }_0^u\varphi (t)e^{2t}dt$. Differentiation under the finite integral gives $I^{\prime }=J$ and $J^{\prime }=\varphi e^{2u}$. Differentiating the preceding equation and using $4e^{-2u}I=\varphi -3{\varphi }^{\prime }$ then givesThus the second identity isDifferentiate once more:The preceding second-derivative equation gives $8e^{-2u}J=-6{\varphi }^{\prime \prime }-10{\varphi }^{\prime }+4\varphi$. Substitution cancels the undifferentiated terms and provesThis is the Erlang claim-size differential equation for survival probability with $\beta =2$ and $r=1/3$.

Its characteristic polynomial factors as $z(3z+8)(z+1)$. The limit $\varphi (u)\to 1$ therefore makes the general bounded solutionThe supplied zero-capital survival probability is $\varphi (0)=2/3$, so $A+B=-1/3$. We also need the initial derivative carried by the original integral equation: $I(0)=0$ implies ${\varphi }^{\prime }(0)=\varphi (0)/3=2/9$, giving $-A-(8/3)B=2/9$. Solving gives $A=-2/5$ and $B=1/15$. Therefore the survival and ruin probabilities areThe extra initial derivative is essential: $\varphi (0)$ and the limit at infinity alone would leave one free parameter in the third-order equation. The result has $\psi (0)=1/3$ and tends to zero. It is positive and decreasing, since ${\psi }^{\prime }(u)=-(2/5)e^{-u}+(8/45)e^{-8u/3}<0$ for $u\ge 0$. These signs agree with its interpretation as an ultimate ruin probability.

Use the classical risk model surplus $U_t=u+ct-{\sum }_{j=1}^{N_t}{X}_j$, with $c>0$, and write $\phi$ for its ultimate survival probability. The first-claim decomposition for survival probability follows by conditioning on the first arrival time of the Poisson process. That time has exponential distribution of rate $\lambda$. A claim at time $t$ leaves capital $u+ct-x$ if $x\le u+ct$, after which the Markov property restarts the same risk model. ConsequentlyPut $C(v)={\int }_0^v\phi (v-x)f(x)dx$ and change variable $v=u+ct$:The convolution $C$ is continuous since $\phi$ is bounded and $f$ is integrable. Differentiating proves the survival integro-differential equation

For the specified claim law, expand its probability density function as $f(x)=\frac{1}{2}{e}^{-x}+\frac{1}{4}{e}^{-x/2}$. This is an equal-weight mixture distribution of exponential distributions with rates $1$ and $1/2$. In the convolution, set $z=u-x$ to obtainSince $\lambda /c=10/21$, the requested coefficients are

For the exponential-mixture differential equation for survival probability, define $A(u)=e^{-u}{I}_1(u)$, $B(u)=e^{-u/2}{I}_2(u)$ and $r=\lambda /c$. Differentiating the integrals yieldsApply $(D+1)(D+\frac{1}{2})$ to the last equation, where $D=d/du$. The first two equations eliminate $A,B$, givingThusThe characteristic polynomial factors asThe ordinary differential equation therefore has solution $C_0+C_1{e}^{-u/6}+C_2{e}^{-6u/7}$.

The claim expected value is $\frac{1}{2}\cdot 1+\frac{1}{2}\cdot 2=3/2$. The relative safety loading is $\rho =2.1/(3/2)-1=2/5$, and the given zero-capital survival probability is $\phi (0)=2/7$. The limit at infinity sets $C_0=1$. A third condition comes from the original survival integro-differential equation: its convolution vanishes at zero, so ${\phi }^{\prime }(0)=(10/21)(2/7)=20/147$. Thereforewhich gives $C_1=-20/29$ and $C_2=-5/203$. The final survival probability and ultimate ruin probability areBoth exponential coefficients of $\psi$ are positive. Hence $\phi$ increases from $2/7$ to one, and $\psi$ decreases from $5/7$ to zero. The initial slope is essential: the two stated boundary values alone do not determine all three constants of the eliminated differential equation.

With no censoring or new entry within the interval, all $r$ individuals can be classified at its end: $d$ have experienced the event and $r-d$ remain event-free. Thus the empirical conditional survival probability isThis assumes $r>0$ and that the observed individuals represent the group whose conditional survival probability is being estimated.

The set must contain every observed event time. Otherwise a discrete survival likelihood would assign probability zero to at least one exact event observation.

The set can include additional times without observed events. Include time zero when events at zero are present, and allow remaining tail probability beyond the fitted range. Estimation of a conditional event mass requires a nonempty risk set; once follow-up ends, later probability masses are not identified by the data. The usual Kaplan–Meier estimator retains the final estimated survival probability as its tail convention.

For the survivor functions, $F_A^{\prime }(u)=-f_A(u)$ and $F_B^{\prime }(u)=-f_B(u)$. The product rule givesIntegrating from 0 to $t$, using $F_A(0)=F_B(0)=1$, yieldsThe three terms partition the possible histories: A occurs first by time $t$, B occurs first by time $t$, or neither has occurred by $t$. The first two are the cumulative incidence functions for the two competing risks, while the third is the survival probability of their minimum. The initial value assumes nonnegative event times with no atom at zero.

The backward generator of the drift--diffusion is $\mathcal{L}=\alpha {\partial }_y+D{\partial }_y^2$. The survival probability satisfies the Kolmogorov backward equationwithThe target is absorbing, while reflection gives the Neumann boundary condition at $L$.

An unstable isolated level has survival amplitude $a(t)=\theta (t)e^{-i{E}_{0}t/\hslash }{e}^{-\Gamma t/(2\hslash )}$, because its survival probability falls as $e^{-\Gamma t/\hslash }$. Its energy-domain amplitude is proportional toTaking the modulus squared gives the denominator of the Breit-Wigner resonance cross sectionAt $E-E_0=\pm \Gamma /2$ the resonant factor is half its peak, so $\Gamma$ is the full width at half maximum. exponential decay is the isolated-level approximation; energy-dependent thresholds or overlapping levels need a more general lineshape.

The table gives the Xi baryon mean lifetime ${\tau }_{{\Xi }^0}=2.90\times 10^{-10}\mathrm{s}$. At $t_1=10^{-12}\mathrm{s}$, an initially resting particle has survival probabilityThus the most probable contents are one undecayed ${\Xi }^0$. The initial momentum is unspecified. For a moving parent, time dilation replaces the denominator by ${\gamma }_{\Xi }{\tau }_{{\Xi }^0}$, making the survival probability at least as large. The times below are characteristic proper mean lifetimes; laboratory times depend on the particle velocities.

The principal first step isIt typically occurs on the $10^{-10}\mathrm{s}$ scale. The neutral pion almost immediately converts to two photons, with ${\tau }_{{\pi }^0}=8.43\times 10^{-17}\mathrm{s}$ and $B_{\gamma \gamma }=0.98823$. The Lambda baryon has ${\tau }_{\Lambda }=2.632\times 10^{-10}\mathrm{s}$ and two dominant routes.

In the charged-pion route,The muon lifetime is a standard additional input, not present in the supplied five table pages.^[ref] The successive contents are approximately $\Lambda +2\gamma$, then $p+{\pi }^{-}+2\gamma$, then $p+{\mu }^{-}+{\bar{\nu }}_{\mu }+2\gamma$. After a few muon lifetimes the dominant stable endpoint isThe leading-path probability including the tabulated two-photon pion branch isHere the last factor is the tabulated dominant muonic charged-pion fraction; rare radiative corrections are being neglected in describing a fixed photon count. Before selecting pion decay channels, the probability of reaching the $p{\pi }^{-}{\pi }^0$ history is $0.99524\times 0.639=0.63596$.

In the neutral-pion route,This produces $n+4\gamma$ by roughly the nanosecond scale. It is a long-lived intermediate state, not the final endpoint: the free neutron has ${\tau }_n=879.4\mathrm{s}$ in the original table and eventually undergoes beta-minus decay,Thus after several thousand seconds the dominant stable endpoint isBefore selecting the two ${\pi }^0\to 2\gamma$ decays, the $n{\pi }^0{\pi }^0$ history has probability $0.99524\times 0.358=0.35630$. The familiar rough description is therefore a $64\%$ charged-pion route and a $36\%$ neutron route; including the specified leading pion branches gives approximately $62.8\%$ and $34.8\%$ for the two displayed endpoints. The remainder includes Dalitz decays and other rare channels, not a failure of probability conservation.

These are particle decay cascades, treating the separated final proton, electron, photons and neutrinos as stable asymptotic particles and neglecting subsequent collisions and rare extra radiation. An infinitely large box prevents escaping particles from being removed from the inventory. Neutrino labels specify their production flavours. Each lifetime describes a stochastic waiting time, so there is no finite sharp instant at which every unstable particle has certainly decayed.

In standard configuration the frames are inertial, their corresponding axes are parallel, S${}^{\prime }$ moves at $v$ along the positive $x$ axis of S, and their origins coincide at $t=t^{\prime }=0$. Here $(t,x,y,z)$ and $(t^{\prime },x^{\prime },y^{\prime },z^{\prime })$ label the same event, $c$ is the speed of light and $\gamma =(1-v^2/c^2{)}^{-1/2}$.

A clock fixed in S${}^{\prime }$ has $\Delta x^{\prime }=0$. The inverse time transformation givesso the moving clock accumulates less proper time between its two ticks than the elapsed S-coordinate time.

Without time dilation the flight time from the given altitude would be $66.7\mu \mathrm{s}$, about $30.3$ rest-frame mean lifetimes, giving survival fraction only $e^{-30.3}\simeq 6.8\times 10^{-14}$ even at speed $c$. Relativistic muons can instead reach the ground with appreciable survival. Momentum-selected decay measurements test this clock-rate dependence experimentally.^[ref]

A mean-lifetime survival energy estimate requires a mean flight time no greater than the dilated mean lifetime:This corresponds to survival probability at least about $e^{-1}$. There is no sharp minimum energy for an individual random decay. For a required probability $p$, use $\exp [-H/(\gamma v{\tau }_0)]\ge p$. Energy loss in the atmosphere is neglected in this estimate. In the muon's frame the atmospheric path is length-contracted to $H/\gamma \simeq 660\mathrm{m}$; the ground reaches it in about ${\tau }_0$, giving the same survival probability.

For the two-body decay, use the rest frame of A. Momentum conservation makes B and the photon oppositely directed with equal momentum magnitude $p$. Energy conservation and the energy–momentum relation giveSubstitution and cancellation of $E_B^2$ giveA nonzero emitted photon requires $m_A>m_B$.

An ATP-binding step with pseudo-first-order rate $k_{1}c$ has survival probability $S_1=e^{-k_{1}ct}$ and waiting density $P_1=-S_1^{\prime }=k_{1}c{e}^{-k_{1}ct}$. It decreases from a finite value at zero, integrates to one and has mean $1/(k_{1}c)$. Its concentration dependence suggests binding-limited waiting before the large 80-degree substep.

Two independent sequential exponential steps of rates $k_a,k_b$ give the convolutionIt begins at zero, rises to a maximum at $\log (k_a/k_b)/(k_a-k_b)$ and decays; its mean is $k_a^{-1}+k_b^{-1}$. At equal rates it becomes $k^{2}t{e}^{-kt}$. This supports at least two kinetic events during the waiting associated with the 40-degree substep, plausibly chemical transformation/product release and conformational changes. It does not uniquely identify which rate belongs to hydrolysis or a particular product: different hidden-state networks can have the same waiting distribution. Three 120-degree catalytic cycles per revolution, partitioned into 80 and 40 degrees in the assay, are consistent with three catalytic sites.

The displayed identity requires a correction to the process construction in the source. In a piecewise-deterministic Markov process, the instantaneous hazard function is evaluated at the current flowing state $X_{u-}$, and the destination is sampled from $K_u(X_{u-},dy)$. Keeping the jump times and destinations of the original Markov jump process $Y$ instead evaluates both quantities at its last reset state $Y_{u-}$, which generally differs from $X_{u-}$. The source PDF confirms that these original jump times and destinations were retained. Consequently its definition does not in general produce the asserted transition kernel, or even a Markov process in the state variable $X$ alone.

For an explicit counterexample, take $a(t,x)=1$, $\alpha (x)=2+\tanh x$, $y_0=0$, and let every $K_t(x,dy)$ be the standard normal distribution, independently of $t,x$. These are smooth coefficients with bounded positive jump rate, so nonexplosion holds. In the literal construction, the first holding time has the exponential distribution of rate $2$. Thus the probability distribution of $X_t$ has an atom of mass $e^{-2t}$ at $t$. Paths with at least one jump contribute no atom there, because the reset has a normal density. The asserted identity instead assigns to this no-jump atom the masswhich differs for every $t>0$.

Here is the full derivation for the intended piecewise-deterministic Markov process. Assume nonexplosion; smoothness alone does not guarantee it. Start at $(s,x)$, let $\tau$ be the first jump after $s$, and setOne precise construction draws an independent unit-rate exponential random variable $E$ and takes $\tau =\inf \{u\ge s:H(u)\ge E\}$, with $\inf \varnothing =\infty$. This implements a hazard function along the flow of a vector field, rather than a fixed exponential holding time at the last reset. Its survival probability and jump-time probability density function areEquivalently, $S^{\prime }=-\alpha (z)S$ with $S(s)=1$. On $\{\tau >t\}$, the terminal state is $z(t)$. Conditional on $\tau =u$, draw the new state $y$ from $K_u(z(u),dy)$ and restart the same construction from $(u,y)$. The Markov property and the tower property of conditional expectation therefore give the first-jump decomposition of a piecewise-deterministic Markov processThis proves the requested identity under the corrected construction, for every bounded Borel measurable function $f$. The two contributions partition according to whether the first jump occurs before $t$; later jumps are already contained in $P_{u,t}$.

For comparison, the literal construction does have a different renewal identity when started at a reset with $X_s=Y_s=x$. If $Q_{s,t}$ denotes only these reset-start operators, the same conditional expectation argument givesThese $Q_{s,t}$ need not be the transition operators conditional on $X_s=x$ at an arbitrary time: that observation does not specify the hidden reset state. This distinction explains exactly why replacing only the between-jump paths is insufficient.

The identity for the killed deterministic flow follows directly from the chain rule and the fundamental theorem of calculus, independently of the construction issue in part (b). For a continuously differentiable $f$, put $z(t)=\phi (s,x,t)$ and $H(t)={\int }_s^t\alpha (\phi (s,x,u))du$. ThenHence, on such test functions,The time subscript on $L_t^{\alpha }$ is necessary because the vector field may depend on time; it is suppressed on the right-hand side of the source's displayed operator identity. The killed deterministic flow transports an observable along the flow of a vector field and multiplies it by the no-jump survival probability.

Multiply each time-sliced path weight by $\exp [-{\sum }_{j}V(x_j)\Delta t]$. A nonnegative finite $V$ is a local killing rate: the Feynman-Kac weight attenuates paths in proportion to the time spent in regions where $V>0$. Let that rate tend to infinity outside the allowed interval. Paths that exit are removed, leaving Brownian motion killed at its first boundary hit.

Consequently the required kernel is the Gaussian path measure restricted to surviving paths, equivalentlywith the infinite-barrier interpretation stated above. It solves ${\partial }_{t}K=D{\partial }_x^{2}K$ in the interior with absorbing Dirichlet boundary conditions at $\pm L$. A reflecting boundary condition for a diffusion would describe a different process.

Strictly, this is a transition probability density in the final position, jointly with survival; a probability for an interval of final positions includes an integral over $x_f$. Its integral over the whole allowed interval is at most one. It is not the normalized endpoint density conditional on survival, which would require division by the survival probability. In these diffusion units $V$ and the eigenvalues of $H$ have units of inverse time.

The survival probability from an absorbing transition density follows by integrating over final position. Sincethe exact survival probability for $t>0$ isThe slowest Dirichlet mode dominates at large time:The first relative correction is $-\frac{1}{3}{e}^{-8u}$. The approximation is a late-time result; extrapolating its prefactor to $t=0$ would incorrectly give a probability larger than one. The full alternating series tends to one there.

In the classical risk model with premium rate $c>0$ and claim Poisson process rate $\lambda$, the first-claim decomposition for survival probability implies the displayed equation for ultimate survival probability. Change the first-claim integral to an integral over available capital and differentiate its lower limit. The bounded survival probability function convolved with the integrable claim probability density function is continuous, so this also establishes the needed differentiability. In particular ${\phi }^{\prime }(0)=(\lambda /c)\phi (0)$.