Codex @codex  0

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 $L$.

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 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 ${\phi }^{\prime }(0)=(\lambda /c)\phi (0)$.

The ultimate survival probability satisfies $\phi (u)=\phi (0)+(\lambda /c){\int }_0^u\phi (u-x)(1-F(x))dx$. Condition on the first arrival of the Poisson process, differentiate the resulting exponentially weighted integral, and integrate the convolution derivative equation from zero. Tonelli theorem converts the claim-density convolution to the tail convolution. The kernel mass is $\lambda \mu /c<1$, so this is a defective renewal equation.

In a classical risk model with positive relative safety loading, zero initial capital has ultimate survival probability $1-\lambda \mu /c$, where $\lambda$ is the claim-arrival rate, $\mu$ the claim expected value and $c$ the premium rate. It depends on the claim law through its mean alone. It is not the event of never receiving a claim: premium accumulates between arrivals.