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For a single resource, assume unit-capacity calls arrive as a Poisson process of rate $a$, holding times have independent exponential distributions with mean $h$, and a call is rejected when all $C$ units are occupied. Let $\nu =ah$. Its occupancy is a birth-death process on $\{0,\dots ,C\}$, with birth rate $a$ below capacity and death rate $k/h$ in state $k$. Detailed balance for a birth-death process givesBy Poisson arrivals see time averages, the Erlang loss formula isThis derivation assumes exponential holding times; the standard insensitivity result extends the same formula to independent holding times of the same mean, but is not needed here.

A loss network with fixed routing has finitely many resources $j$, capacities $C_j$, and call types $r$ with a fixed nonempty set of resources. Type $r$ has Poisson arrivals of rate $a_r$, independent holding times of mean $h_r$, and offered load ${\nu }_r=a_r{h}_r$. It is accepted only if every resource on its route has a free unit, and otherwise is lost. In the displayed equations use the unit-incidence matrix $A_{jr}=1$ for $j\in r$ and zero otherwise. The equations concern this unit-demand model, rather than arbitrary multi-unit requirements.

The Erlang fixed point approximation replaces the correlated resource availabilities by independent ones. A type-$r$ call survives blocking at resources other than $j$ with approximate probability ${\prod }_{i\in r\setminus \{j\}}(1-B_i)$. Treating this filtering as Poisson thinning gives the reduced offered loadApply the single-resource formula to obtain $B_j=E({\rho }_j,C_j)$. Independence and Poisson thinning here are approximations; no such approximation is claimed for the exact network's joint stationary distribution.

Assume first $C_j\ge 1$. The right side is a continuous map from $[0,1{]}^J$ into itself, so the Brouwer fixed-point theorem gives existence. To prove uniqueness, let $m_C(\rho )=\rho [1-E(\rho ,C)]$ be the carried load of an Erlang loss resource, also the mean of the truncated Poisson occupancy. Differentiation yieldsThus $E$ increases continuously from zero to one and $m_C$ increases from zero to $C$.

Write $x_j=-\log (1-B_j)\ge 0$, and define ${\rho }_C(x)=E(\cdot ,C{)}^{-1}(1-e^{-x})$ and ${\ell }_C(x)=m_C({\rho }_C(x))$, with ${\ell }_C(0)=0$. Each ${\ell }_C$ is continuous and strictly increasing. A fixed point satisfiesThese are precisely the stationary equations for the convex potential for the Erlang fixed pointThe first term is convex; the second is strictly convex because each ${\ell }_{C_j}$ strictly increases. Hence $F$ is strictly convex. Its gradient cannot vanish at two distinct points: the difference of gradients dotted with the difference of points is strictly positive, with a positive contribution from every changed coordinate in the integral term. This also covers boundary coordinates $x_j=0$, using the right derivative there. Every fixed point has $B_j<1$, since its offered load is finite, so the change of variables is valid. There is exactly one fixed point. If a capacity is zero, its blocking probability is one; delete it and all routes using it before applying the argument to the remaining positive capacities. Unused resources have blocking probability zero.

Interpret the second process as an independent copy of the unit-rate Poisson process. First, the sums are finite almost surely. On $[0,1]$ there are finitely many arrivals, none at $0$, so their inverse-square contributions are finite. On $[1,\infty )$,For completeness, this expectation identity follows first for nonnegative simple functions from $\mathbb{E}N(A)=|A|$, and then for all nonnegative measurable functions by monotone convergence. Thus no unproved Poisson-integral formula is needed, and the tail sum is finite almost surely.

By (b), merging the two processes produces a rate-$2$ Poisson process with arrival times $V_i$. Its rescaled arrivals $2V_i$ form a unit-rate Poisson process: for a measurable set $A$, their count is the merged count on $A/2$, with parameter $2|A/2|=|A|$, and disjoint-set independence is preserved. ConsequentlyTherefore $\boxed{c=4}$. This is the scaling of an inverse-power sum over Poisson arrivals with exponent $2$.

The PDF does not explicitly repeat the second process's rate. If that rate were $\rho$ rather than $1$, the same count argument would give $X+Y\stackrel{d}{=}(1+\rho {)}^{2}X$. The numerical value $4$ uses the intended independent-unit-rate-copy interpretation.

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.

Take two independent rate-one Poisson processes $N^{+}$ and $N^{-}$, and letThe difference of independent Poisson processes starts at zero and has stationary increments and independent increments. Its paths are càdlàg. For an interval of length $h$, the probability of any jump is $1-e^{-2h}\to 0$, proving stochastic continuity. Thus $X$ is a Lévy process. The characteristic function of a Poisson distribution with mean $t$ is $\exp (t(e^{iu}-1))$, so independence givesThe sample paths are integer-valued step functions with jumps $+1$ or $-1$. On every bounded interval there are only finitely many jumps, and independent Poisson arrival times coincide with probability zero. The combined arrival rate is two: holding times are independent exponentials of rate two, and each jump direction has probability $1/2$, independently of the holding times. This is equivalently a Compound Poisson process of rate two with Rademacher distribution jump sizes. Its paths have finite variation on compact time intervals, although there are infinitely many jumps over the whole half-line almost surely.

In the classical risk model write $U(t)=u+ct-{\sum }_{j=1}^{N(t)}{X}_j$, where the Poisson process $N(t)$ has rate $\lambda$ and is independent of the claim sizes. Define the ruin time and ultimate ultimate ruin probability byThe relative safety loading is $\rho =c/(\lambda \mu )-1>0$, so $c=(1+\rho )\lambda \mu$. The adjustment coefficient is the nonzero positive solutionExistence and uniqueness follow from the secant-slope existence criterion for an adjustment coefficient. Explicitly, $\lambda (M(r)-1)-cr$ has derivative $\lambda \mu -c<0$ at zero and is strictly convex for positive claims. The assumed divergence of $M$ makes it cross zero once at a positive $R<r_{\infty }$. When $r_{\infty }=\infty$, an exponential lower bound from any positive tail event shows that $M(r)/r\to \infty$. In particular $R$ is inside the finite-transform domain, not at its endpoint.

Put $\overline{F}=1-F$. Replacing $\phi =1-\psi$ in the survival renewal equation for a classical risk model yieldsBy the tail integral formula for moments, $\mu ={\int }_0^{\infty }\overline{F}(x)dx$. Therefore this is the defective renewal equationThe original kernel has mass $\lambda \mu /c=1/(1+\rho )<1$. For the exponential tilt of the ruin renewal kernel, setMultiplying by $e^{Ru}$ gives the required proper renewal equationTo verify that this is a probability renewal kernel, Tonelli theorem gives, for $r>0$ in the finite-transform domain,The adjustment coefficient equation consequently implies ${\int }_0^{\infty }{k}_R(x)dx=1$. Its expected value isIt is finite because $R$ is interior to the finite-transform domain and is positive because the strict convex crossing has derivative $\lambda M^{\prime }(R)-c>0$.

We quote the key renewal theorem in the following form: for a nonarithmetic distribution $K$ of positive increments with finite positive mean $m$, and a directly Riemann integrable nonnegative function $h$, the locally bounded solution of $z=h+z\ast K$ satisfies $z(u)\to m^{-1}{\int }_0^{\infty }h$. It has renewal representation $z=h\ast {\sum }_{j\ge 0}{K}^{\ast j}$; this follows by iterating the equation, since the probability that arbitrarily many positive increments have sum at most a fixed $u$ tends to zero.

Here $K_R(dx)=k_R(x)dx$ is absolutely continuous, and hence nonarithmetic distribution, even if the original claim law has atoms. The function $h_R$ is continuous. Choose $\delta >0$ with $M(R+\delta )<\infty$. The Markov inequality gives $\overline{F}(x)\le M(R+\delta )e^{-(R+\delta )x}$, and hence $h_R(u)\le C_{\delta }{e}^{-\delta u}$. Continuity on compact intervals and this exponential bound make the upper Riemann sums finite with uniformly vanishing tails, proving direct Riemann integrability.

A further application of Tonelli theorem evaluates the forcing integral:All hypotheses of the key renewal theorem are now checked, so the interior adjustment coefficient ruin prefactor isThis proves the requested Cramér–Lundberg ruin asymptotic, including its constant.

For the two-component hyperexponential distribution, conditioning on the chosen exponential component givesIts moment-generating function and derivative areThe moment-generating function diverges at $1/2$, while the derivative of the adjustment equation at zero is negative. Strict convexity therefore places its unique positive adjustment coefficient inThe adjustment equation, divided by $R$, becomesUsing this identity to subtract $(1+\rho )\mu$ from $M^{\prime }(R)$ givesThus the asymptotic constant in terms of $\rho$ and $R$ isThe denominator is positive on the identified domain. If desired, $R$ is the smaller root of $6(1+\rho )R^2-(5+9\rho )R+3\rho =0$; the other algebraic root lies outside the positive finite-transform interval and is not an adjustment coefficient.

A loss network models calls that need several resources simultaneously and are rejected, rather than queued, when insufficient capacity remains. Take finite resource and route sets. Let $A_{jr}\ge 0$ be the integer amount of resource $j$ required by a route-$r$ call, $C_j$ its capacity, andUnder fixed routing, each arriving call has a predetermined resource requirement. Take independent Poisson processes of rates ${\nu }_r$ and independent holding times with exponential distribution of rates ${\mu }_r$. A feasible arrival changes $n$ to $n+e_r$ at rate ${\nu }_r$; a departure changes it to $n-e_r$ at rate ${\mu }_r{n}_r$. Put ${\rho }_r={\nu }_r/{\mu }_r$ and assume each route uses a finite positive-capacity resource, so the state space is finite.

The product-form stationary distribution of a loss network isFor any feasible adjacent pair,These detailed balance equations prove stationarity and make the process a reversible Markov chain. Equivalently, Independent random variables with Poisson distributions of means ${\rho }_r$ are conditioned on satisfying the joint capacity constraints. The conditioning makes resource occupancies dependent even though the unconstrained counts are independent.

By Poisson arrivals see time averages, a route-$r$ arrival sees acceptance probability $G(C-A_r)/G(C)$, interpreting the numerator as zero for a negative capacity. Hence its blocking probability is $1-G(C-A_r)/G(C)$ and the expected value of its number in service is ${\rho }_{r}G(C-A_r)/G(C)$. This connects a stationary occupancy law to observable rejection and carried traffic.

The insensitivity of loss networks extends this occupancy formula to independent general holding-time distributions with the same expected values, under the usual fixed resource requirements and admission rule. Counts alone then need not be a Markov chain; residual holding times belong in a Markov description. The invariant occupancy formula survives. This is useful because detailed call-duration distributions can be difficult to estimate, while their means are much easier to measure.

Let $x_r$ be the rate of each individual flow on route $r$, so its aggregate rate is $w_r=n_r{x}_r$. The feasible allocations satisfy ${\sum }_{r:j\in r}{n}_r{x}_r\le C_j$. An allocation has proportional fairness if, for every feasible $\tilde{x}$,Equivalently it maximizes ${\sum }_{r:n_r>0}{n}_r\log x_r$ over positive active-route rates. This is the first-order optimality condition for a concave objective on a convex set. Rates for absent flows can be set to zero.

For the linear flow network, put $N={\sum }_{r=0}^I{n}_r$ and $M={\sum }_{i=1}^I{n}_i$. If $n_i>0$, unused capacity on link $i$ could increase $x_i$, soWhen $n_0>0$, write $w_0=n_0{x}_0$. For active local routes their aggregate rates are $1-w_0$, and the objective, up to constants independent of $w_0$, isFor $M>0$, differentiation gives $n_0/w_0=M/(1-w_0)$, hence $w_0=n_0/N$; when $M=0$ the maximizing endpoint is $w_0=1$, giving the same answer. Thus the proportionally fair allocation on a linear flow network isIf $n_0=0$ and $N>0$, this gives $x_i=1/n_i$ on every active local route, as expected. No departure occurs at the empty state.

Independent Poisson processes of flow arrivals and independent exponential distributions of document sizes make the count vector a continuous-time Markov chain. By the memoryless property, a surviving residual size with exponential distribution still has rate ${\mu }_r$ per unit of transferred data, so each route-$r$ flow completes at instantaneous rate ${\mu }_r{x}_r(n)$. The transition intensities areHence the through-route departure rate is ${\mu }_0{n}_0/N$ and each active local route's rate is ${\mu }_{i}M/N$.

To obtain the stationary law of a linear flow network, defineFor a through departure $H(n-e_0)/H(n)=n_0/N$; for an active local departure $H(n-e_i)/H(n)=M/N$. These are exactly the aggregate service rates, so the detailed balance identities $\pi (n-e_r){\nu }_r=\pi (n){\mu }_r{n}_r{x}_r(n)$ hold. This also shows why the binomial coefficient is essential; a factor $N$ would not have the required ratios.

For fixed local counts with total $M$, the negative binomial series givesSum the remaining independent geometric series to findThe given load conditions make every series converge. Setting $K=1/Z$ provesThe chain has bounded total departure rate, finite arrival rate and no explosion. With positive arrival rates it is irreducible, and this normalized invariant law gives the positive recurrent Markov chain property; zero arrival rates restrict the stationary support to the corresponding reachable class.

Summing out $n_0$ shows local-count independence in a linear flow network: the local counts have independent geometric distributions on ${\mathbb{Z}}_{+}$ with ratios $q_i={\rho }_i/(1-{\rho }_0)$. ThereforeThey are independent of one another under the stationary marginal, despite sharing the random through-flow population. The through count is generally dependent on them.

The independent, spatially homogeneous random-placement idealization gives a Poisson process of intersections along the sampling line. With line intensity $\mu$, an interval of length $s$ contains no ridge with probability $e^{-\mu s}$. The spacing $X$ thus has exponential distributionRandom orientation changes the intersection intensity, which has already been represented by the measured $\mu$. Merely saying that positions are random does not prove a Poisson process: independence and homogeneity are additional assumptions. Finite segments, clustering or excluded widths can invalidate them.

For a shifted lognormal distribution, write $Z=\log (X-\theta )\sim N(m,{\sigma }^2)$ with $\sigma >0$. Applying the change of variables formula, $dZ/dX=1/(X-\theta )$, givesThe threshold $\theta$ is a minimum observable separation. Finite keel widths impose geometric exclusion, and the sonar's footprint and ridge-identification criterion can suppress a shallow peak near a deeper peak. This sonar ridge shadowing or resolution effect means that $\theta$ need not be a universal physical distance between every pair of ridges; it can partly reflect how the profiles were processed.

A lognormal distribution is compatible with products of many positive factors: taking logarithms turns multiplicative changes into sums, which can approach a normal distribution. Repeated deformation, breakup, convergence and merging across scales are plausible contributors. The observed form therefore suggests correlated or multistage ridging rather than the simplest independent-intersection model. It does not identify a unique ridging mechanism. For example, directly generating $X=\theta +\exp Z$ with a normal distribution for $Z$ produces the same spacing law without specifying any particular mechanics. Distributional agreement must be supplemented by dynamical and spatial evidence.

For $t>0$, use the continuous logarithm of the characteristic function, normalized to zero at $u=0$, to writeThus the deterministic drift is $+c$. In particular the last contribution is not the negative drift of a compensated unit-jump process.

Take a Poisson random measure $N$ on $(0,\infty )\times \{+,-,p\}$ with intensityDefine the jump marks $j(+)=b$, $j(-)=-b$, and $j(p)=1$. A realization with the required process law is the Poisson stochastic integral with finite intensitywhere the three counts are independent Poisson processes of rates $a/2,a/2,c$. Their characteristic functions multiply towhich is exactly the given expression. The constructed process is a Lévy process, and stationary independent increments make its entire finite-dimensional law determined by these one-time characteristic functions. This gives a representation in law of the specified process.

Equivalently, using its nonzero-jump measure, the Lévy measure isand the unmarked representation is $X_t=ct+{\int }_{(0,t]\times (\mathbb{R}\setminus \{0\})}zN(ds,dz)$ with intensity $ds\nu (dz)$. The finite-jump case of the Lévy–Itô decomposition realizes this pathwise using the jump measure of a version of $X$; the integral is an uncompensated finite sum.

The atomic compound Poisson process with drift has càdlàg paths with finitely many nonzero jumps on every bounded time interval, linear slope $c$ between jumps, and no Brownian component. When $b>0$, jumps of sizes $b,-b,1$ occur at the stated rates; when $b=1$, positive unit-jump rates combine to $a/2+c$. When $b=0$, the two symmetric marks have zero effect and are omitted from the Lévy measure; the process reduces to $ct+N_t^p$, so $a$ has no effect. If $c=0$ the paths are piecewise constant, and if also $ab=0$ they are identically zero. All paths have finite variation on bounded intervals. As a check on the drift sign,

The likelihood function for the ordered arrival times of a Poisson process, including the absence of further arrivals before $T$, isThe arrival-time constraint does not depend on $\lambda$. Equivalently, the Poisson distribution of the total count gives $L(\lambda )=e^{-\lambda T}(\lambda T{)}^n/n!$, which differs by a parameter-independent factor. Conditional on the count, the Poisson process conditional arrival times have density $n!/T^n$, independent of $\lambda$. Thus the total count contains all the information about the rate when the exposure is known; it is a sufficient statistic.

A conjugate prior is a family of prior distributions whose members remain in that family after updating by the likelihood function. Here multiplying a shape-rate gamma distribution density by the Poisson process likelihood changes its power of $\lambda$ and its exponential rate, leaving a gamma distribution. The parameters change with the observations; conjugate prior does not mean that the Bayesian posterior equals the prior distribution.

Independent increments of the Poisson process give $M\mid \lambda \sim \mathrm{Poisson}(\lambda )$ for a one-hour interval. The Poisson distribution therefore givesThis conditional probability still depends on the unknown rate.

The GRW model takes a normalized wave function in $L^2({\mathbb{R}}^{3N})$. Between jumps it obeys the Schrodinger equation, $i\hslash {\partial }_t\Psi =H\Psi$, with the usual many-particle Hamiltonian operator, for example $H={\sum }_i[-{\hslash }^2{\nabla }_i^2/(2m_i)]+V({\mathbf{x}}_1,\dots ,{\mathbf{x}}_N)$. Each particle has an independent Poisson process of collapse times with rate $1/\tau$. Thus the total jump rate is $N/\tau$, the mean wait is $\tau /N$, and the jumping label is uniform on $\{1,\dots ,N\}$.

At a jump of particle $i$, use the GRW localization operatorIts centre has probability density function $p_i(\mathbf{c})=\Vert L_i(\mathbf{c})\Psi {\Vert }^2$. The post-jump quantum state is $L_i(\mathbf{c})\Psi /\Vert L_i(\mathbf{c})\Psi \Vert$. The normalization is consistent because $\int d^3\mathbf{c}{L}_i(\mathbf{c}{)}^2=I$. The collapse times and labels have these state-independent rates; the centres depend on the wave function through the Born rule-like squared norm.

For precision about the stated spatial event, let $B=Y_{10a}=\{\mathbf{c}:\mathrm{dist}(\mathbf{c},Y)\le 10a\}$. With $H=0$, the GRW collapse-centre event probabilities are exactlyThe last expression is defined only for $p_1(B)>0$. It includes the possibility that the next jump hits the same particle.

Write $f_i^Y(\mathbf{x})=|\varphi (\mathbf{x}-{\mathbf{y}}_i){|}^2$, and similarly $f_i^Z$. Putwith the same definitions for $Z$. The macroscopic separation and negligible packet tails make the two branches essentially orthogonal, so their interference contribution can be neglected. The first-event probability is thereforeThe $Z$ contribution is extremely small: its actual positions are within $r$ of $Z$, whereas centres in $B$ are within $10a$ of $Y$, leaving a distance much larger than $a$ in the Gaussian. Packet-tail errors add to this Gaussian-tail error.

For the next-event calculation, two distinct labels have independent position densities within either product branch, giving factors ${\eta }_i{\eta }_j$. A repeated label instead gives $R_i$, since both localization Gaussians multiply the same position variable. ConsequentlyThese formulas, with $p_{2\mid 1}=p_{12}/p_1$, answer the spatial questions without an extra containment assumption. If the first event is dominated by the $Y$ branch, they simplify toNeglecting the $Z$ term in this conditional expression requires it to be small relative to the retained first-event probability, not merely small in absolute value. The exact operator formulas above cover rare-event exceptions as well.

The intended GRW branch persistence versus geometric containment approximation is obtained if $Y$ and $Z$ contain essentially all the position mass of their corresponding branches. Then positions in $Y$ have a Gaussian-centre margin of $10a$ inside $B$, so ${\eta }_i^Y\simeq R_i^Y\simeq 1$, while ${\eta }_i^Z,R_i^Z\simeq 0$. This givesFor the Gaussian convention used here, the three-dimensional probability of a displacement exceeding $10a$ is $\mathrm{erfc}(10)+20e^{-100}/\sqrt{\pi }$, an extremely small number. A first centre near $Y$ also suppresses the $Z$ branch relative to the $Y$ branch by the very small ratio $|\beta {|}^2{q}_i^Z(\mathbf{c})/(|\alpha {|}^2{q}_i^Y(\mathbf{c}))$ when the latter branch has appreciable weight. All particles are correlated with the same pointer alternative, so one localization selects the macroscopic branch even when the next particle label differs.

The literal PDF, however, only places the packet centres ${\mathbf{y}}_i$ in $Y$; it does not say that the packets themselves fit inside $Y$. Since it permits $r>a$, the two numerical approximations just displayed do not follow for every allowed packet and region. For example, take all ${\mathbf{y}}_i=0$, let $Y$ be a very small ball, and let $|\varphi {|}^2$ be uniform on a ball of radius $r\gg 10a$. Translate the other branch to a similarly small region $Z$ at distance much larger than $100r$. ThenIf these equal packet probabilities are denoted by $\eta$ and their repeated-label integral by $R$, the next probability iswhich approaches $\eta$, not one, for large $N$. Thus packet containment is needed for the stated near-region event to have the usual branch probabilities. The literal general answer is given by the integrals above.

The physical GRW amplification mechanism remains intact: with properly chosen macroscopic branch regions, spontaneous localization selects the alternatives with approximately the squared branch amplitudes and stabilizes the selected record, on a time scale $\tau /N$. Gaussian tails are not exactly eliminated, so the persistence is approximate. The containment qualification concerns how the region is defined, rather than whether the dynamics selects a macroscopically separated branch.