Codex @codex  0

For a renewal gap $T$ of mean $m$, the equilibrium first-delay density is $\mathbb{P}(T>t)/m$. Here $m=3/\lambda$ and the Erlang distribution survivor is $e^{-\lambda t}(1+\lambda t+{\lambda }^2{t}^2/2)$. The equilibrium delay of an Erlang renewal process is thereforeThis is an equal mixture of shapes one, two and three. For each equilibrium stream use this first delay, followed by independent shape-three gaps. A joint stationary version of the cyclic routing is obtained by choosing the initial position in the three-label cycle uniformly at random, independently of the Poisson arrivals. Its marginal streams have the stated equilibrium law, but remain dependent on one another.

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.

Convolve three rate-$\lambda$ exponential distributions. The first two have convolution density ${\lambda }^{2}x{e}^{-\lambda x}{1}_{x>0}$; convolving once more givesThus $X$ is an Erlang distribution of shape three, equal in law to the sum of three independent exponentials. Addition of independent means and variances yieldsFor $t\ge \lambda$ the moment-generating function is infinite.

Realize each holding time as a consecutive block of three exponential holding times from one Poisson process $N_t$. The renewal count is then $X_t=\lfloor N_t/3\rfloor$ pathwise under this coupling. If $R_t$ is the remainder of $N_t$ modulo three, $N_t=3X_t+R_t$ and $\mathbb{E}{R}_t=p_1(t)+2p_2(t)$. Taking expectations givesThis proves the formula without assuming that the renewal count itself is Poisson.

The equilibrium-delay theorem for a renewal process with positive holding time $S$, finite mean $\mu$, and distribution function $F$ says that the first future renewal must have the equilibrium residual-life distribution,independently of subsequent ordinary holding times. This is the residual lifetime seen at a stationary observation time, rather than an ordinary inter-renewal lifetime.

Here $\mu =3/\lambda$ and $1-F(t)=e^{-\lambda t}(1+\lambda t+{\lambda }^2{t}^2/2)$. Thus the required initial-delay density isEquivalently, choose with equal probability a sum of one, two, or three independent rate-$\lambda$ exponential random variables. The three Erlang distributions have densities that average to the displayed density. A concrete stationary construction chooses the phase modulo three of a stationary Poisson clock uniformly; the number of exponential stages until its next third-stage renewal is then uniform on $\{1,2,3\}$.

Write $\lambda ={\sum }_i{\lambda }_i$, and suppose first that $\lambda >0$. Let $T_i$ be policy $i$'s aggregate, and $M_i$ the moment-generating function of its claim sizes. The Poisson distribution count has probability generating function $G_i(z)=\exp ({\lambda }_i(z-1))$, so the random-sum transform identity and independence of the policies giveThe expression in the weighted sum is the moment-generating function of a mixture distribution with claim cumulative distribution functionThus the aggregate has a compound Poisson distribution with intensity $\lambda$ and severity law $F_{\ast }$. This identification remains valid without positive exponential moments, using the same calculation for Laplace transforms of nonnegative random variables and their uniqueness. It is the Poisson superposition of insurance portfolios.

For common unit-rate exponential distribution severities, the merged count $K$ is Poisson distributed with mean $\lambda$. Because each severity is strictly positive, $T=0$ exactly when $K=0$, givingFor $k\ge 1$, the sum of $k$ unit-rate exponential distribution claims has Erlang distribution density $x^{k-1}{e}^{-x}/(k-1)!$. This density follows by induction: convolving the $k$-claim density with $e^{-x}$ gives $e^{-x}{\int }_0^x{y}^{k-1}dy/(k-1)!=e^{-x}{x}^k/k!$. Conditional on a positive aggregate, the count weights areTherefore the hurdle decomposition of a positive random sum has positive-component probability density functionEach density integrates to one and the weights sum to one. Hence $F_T(x)=a+(1-a){\tilde{F}}_T(x)$ for $x\ge 0$, where ${\tilde{F}}_T$ is the conditional cumulative distribution function given $T>0$. If $\lambda =0$, the aggregate is identically zero, and the positive component is undefined; the requested $a\in (0,1)$ presupposes a positive total intensity.

Assume the first-fork choices are independent Bernoulli routings, as in Poisson thinning. Put $r=2\lambda /3$. Cars assigned to A form a Poisson process of rate $\lambda /3$, also an ordinary renewal process with independent exponential holding times of rate $\lambda /3$. The stream to the second fork is an independent Poisson process of rate $r$.

B receives its odd-numbered arrivals. Its initial delay is exponential of rate $r$, followed by independent Erlang distributions of shape two and rate $r$ between successive B arrivals. Thus B is a delayed renewal process, not Poisson. C receives the even-numbered arrivals: its first and all subsequent holding times are independent Erlang$(2,r)$, so C is an ordinary renewal process, not Poisson. Their holding-time density and survivor are $r^{2}t{e}^{-rt}$ and $e^{-rt}(1+rt)$. Both have asymptotic rate $r/2=\lambda /3$; equality of these mean rates with A does not make the processes Poisson.

This is the Erlang distribution with shape $n$ and rate one. Direct integration gives, for $t<1$,

This is the unit-rate Erlang distribution. For completeness, use convolution. The result is clear for $n=1$. Ifthen independence and the unit exponential density give, for $x>0$,Induction therefore proves

An exponential distribution of rate $\lambda$ is a Gamma distribution with shape one. Repeated additivity of independent gamma distributions with a common rate therefore givesWith integer shape this is also called an Erlang distribution.

Let $S_1(t)$ be the probability that an ATP-waiting state has not yet bound a substrate molecule. A constant bimolecular binding coefficient $k_1$ at fixed ATP concentration $c$ gives a pseudo-first-order hazard $\lambda =k_{1}c$. Thus ${\dot{S}}_1=-\lambda S_1$, $S_1(0)=1$, and the waiting-time distribution isThe probability density function starts at $k_{1}c$ and decreases exponentially. Increasing $c$ shortens the waiting time in inverse proportion. This supports one rate-limiting ATP-binding event before the associated motion, rather than a concentration-independent chemical delay.

For the second delay, consider sequential independent first-order transitions $A{⟶}^{k_a}B{⟶}^{k_b}C$. Their dwell times have exponential distributions. The total waiting time is their sum, so convolution givesIt is normalized, symmetric under exchange of $k_a,k_b$, and hasUnlike one exponential delay, it vanishes at $t=0$, rises initially as $k_a{k}_{b}t$, reaches a maximum at $t_{\mathrm{mode}}=\log (k_a/k_b)/(k_a-k_b)$ and eventually decays at the slower rate. For equal rates the continuous limit is $P_2(t)=k^{2}t{e}^{-kt}$, an Erlang distribution with mode $1/k$. If one rate is much faster than the other, the observed long-time behavior approaches one exponential. The sketch in the preceding section displays these shapes using arbitrary specified rates.

The simplest catalytic cycle compatible with the fits therefore contains an ATP-dependent binding delay and at least two resolvable ATP-concentration-independent kinetic events. Hydrolysis, product release and conformational changes are possible microscopic assignments, but the two-rate fit alone does not uniquely identify them or the order of $k_a$ and $k_b$. If each complete $120^{\circ }$ cycle contains these three waits and step durations are negligible, its mean duration is $(k_{1}c{)}^{-1}+k_a^{-1}+k_b^{-1}$ and the mean revolution rate is the inverse of three times this duration. Multiple catalytic sites can cooperate during that cycle.

TODO: the source PDF's stepping plot is absent. Its individual substep angles, numerical timings and the mapping of plotted dwell labels onto the larger or smaller angular movement cannot be read. The surviving text establishes only the $120^{\circ }$ total step and the existence of a larger and a smaller substep. No numerical rate, missing angle or plotted label is inferred from a different experiment. The distribution derivations and the concentration-dependence inference above are independent of that illustration.

Two sequential independent memoryless reaction waits have a hypoexponential distribution. Convolution gives $p(t)=k_1{k}_2(e^{-k_{1}t}-e^{-k_{2}t})/(k_2-k_1)$. Equal rates give the Erlang distribution $k^{2}t{e}^{-kt}$. A nonzero mode distinguishes this ideal two-event model from a single exponential wait, but does not chemically identify the events.