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First consider unit requirements, $A_{jr}\in \{0,1\}$. The Erlang loss formula for a resource with capacity $C$ and offered traffic $a$ isThe Erlang fixed point approximation assumes that resources block independently and that the traffic retained after screening at other resources can be treated as a Poisson process. If $B_j$ is the approximate blocking probability and $q_j=1-B_j$, the reduced-load approximation givesThe link whose load is being calculated is excluded from the screening product. Otherwise one would confuse its offered load with its carried load.

For existence, these equations define a continuous map from $[0,1{]}^J$ into itself. The Brouwer fixed-point theorem gives a fixed point. Assume $C_j\ge 1$; a zero-capacity resource forces rejection of every call needing it and can be removed together with those call types.

For uniqueness, let $K$ have probabilities proportional to $a^k/k!$, $0\le k\le C$, and write $m_C(a)=\mathbb{E}K$. Direct differentiation givesThese identities hold for $a>0$, and $m_C(0)=0$. Also $E(a,C)$ increases from zero to one and $m_C(a)$ increases from zero to $C$.

Put $p_j=-\log q_j$. Define $a_C(p)$ by $E(a_C(p),C)=1-e^{-p}$, and define $h_C(p)=m_C(a_C(p))$. This is a continuous, strictly increasing function on $[0,\infty )$, starting at zero and tending to $C$. Multiplying each Erlang fixed point approximation load equation by $q_j$ transforms it intoThese are exactly the zero-gradient conditions ofEach integral is a strictly convex function; the exponential terms are convex functions. Thus $\Phi$ is a strictly convex function. It is also a coercive function, since $h_{C_j}(p)\to C_j>0$, so it has a unique minimizer. At $p_j=0$, its partial derivative is negative if some positive-traffic route uses $j$, so that coordinate of the minimizer is positive. If no route uses $j$, the unique minimizing coordinate is zero. The minimizer therefore satisfies the equations in every coordinate. Conversely any fixed point has $B_j<1$, finite $p$, and these zero-gradient equations, so must equal that unique minimizer. Hence the Erlang fixed point approximation exists and is unique for fixed routing.

For integer requirements, the common generalized Erlang fixed point approximation usesMultiplication by $q_j$ gives the same equations for $\Phi$, so the existence and uniqueness argument also covers this generalized approximation. It remains an approximation, rather than the exact blocking law for a call requesting several units.

A loss network models calls that require several resources simultaneously and are rejected if any required capacity is unavailable; rejected calls do not queue. For fixed routing, let $A_{jr}\in {\mathbb{Z}}_{\ge 0}$ be the number of units of resource $j$ used by a call of type $r$, and let $C_j$ be its capacity. Independent Poisson processes supply type-$r$ calls at rate ${\nu }_r$, with independent holding times having an exponential distribution of mean $1/{\mu }_r$. Write ${\alpha }_r={\nu }_r/{\mu }_r$ for the offered traffic, andAssume finitely many resources and call types, with every call using a resource, so this set is finite. The occupancy continuous-time Markov chain has ratesIts stationary distribution isIndeed, for each feasible upward transition,which is detailed balance for a continuous-time Markov chain. Thus this is a reversible Markov chain. With positive arrival rates it is irreducible on $\mathcal{N}(C)$: departures reach the empty state, and any feasible state can be assembled by arrivals. Its stationary distribution is consequently unique.

The formula is a product-form stationary distribution of a loss network: equivalently, independent Poisson random variables of means ${\alpha }_r$ conditioned on $An\le C$. The conditioning couples the occupancies, so the resources are generally not independent. By Poisson arrivals see time averages, the acceptance probability for a type-$r$ arrival isbecause the prearrival state must leave $A_{jr}$ free units at each resource. Set the numerator to zero if its capacity vector has a negative entry. This exact formula is often expensive to evaluate, which motivates the Erlang fixed point approximation.

The same occupancy stationary distribution extends to independent general holding-time distributions with these means by insensitivity of loss networks; the exponential assumption above makes the occupancy process itself a continuous-time Markov chain and permits the direct detailed balance proof.

For a fixed occupancy $n$, let $x_r>0$ be the per-flow rate on every active route, so feasibility means ${\sum }_{r:j\in r}{n}_r{x}_r\le C_j$. A proportionally fair allocation is a feasible allocation satisfying, for every other feasible $\tilde{x}$,The sum counts the proportional changes over individual flows. By the first-order condition for a concave function, it is exactly the allocation solvingThis is weighted logarithmic utility optimization, and its negative objective is a strictly convex function, which gives unique rates on active routes. Routes with $n_r=0$ have no per-flow rate to determine.

Index the four routes by $1=\{1,2\}$, $2=\{2,3\}$, $3=\{3,4\}$, $4=\{4,1\}$, and let ${\Lambda }_r=n_r{x}_r$ be each route's total service. WriteThe four capacity constraints say ${\Lambda }_1+{\Lambda }_2\le 1$, ${\Lambda }_2+{\Lambda }_3\le 1$, ${\Lambda }_3+{\Lambda }_4\le 1$, and ${\Lambda }_4+{\Lambda }_1\le 1$. Equivalently,For fixed maxima $u,v$, every active route in the first group optimally takes total service $u$, and every active route in the second takes $v$, because its logarithmic objective is increasing. Up to the constant $-{\sum }_r{n}_r\log n_r$, the objective therefore becomes $P\log u+Q\log v$ with $u+v\le 1$. Its maximizer is $u=P/N$, $v=Q/N$ when $N>0$, with the evident endpoint choices when one group is empty. Hence the proportionally fair allocation on a four-cycle isSet ${\Lambda }_r=0$ on inactive routes. If $N=0$, all service totals are zero. In particular, for $n_1>0$, $n_1{x}_1=P/N$, as required. Two disjoint active routes in the same group can each receive the same total service; service totals need not sum to one over the entire network.

With independent document arrivals given by Poisson processes of rates ${\rho }_r$ and independent unit-mean exponential distributions for document sizes, each of the $n_r$ active documents has completion rate $x_r$. The flow-level network model is consequently a continuous-time Markov chain withAt $n=0$ all downward rates are zero. The total departure rate is at most four, so the bounded total jump rate ensures no explosion.

The original PDF's displayed factor is a binomial coefficient, not the plain quotient produced by the TeX extraction. The candidate stationary distribution for this reversible four-cycle flow model isUse $\left(\genfrac{}{}{0ex}{}{0}{0}\right)=1$. For $r\in \{1,3\}$ and $n_r>0$,so $\pi (n)q(n,n-e_r)=\pi (n-e_r){\rho }_r$. For $r\in \{2,4\}$ the corresponding ratio is ${\rho }_{r}N/Q$. Thus detailed balance for a continuous-time Markov chain holds for every adjacent pair, proving the displayed stationary distribution whenever $B<\infty$.

For completeness, its existence condition can be made explicit. Put $\alpha =\max ({\rho }_1,{\rho }_3)$ and $\beta =\max ({\rho }_2,{\rho }_4)$. At fixed $P,Q$,Therefore the sum over states with $N=m$ is at most $(m+1{)}^2(\alpha +\beta {)}^m$, using the binomial theorem. This is summable if $\alpha +\beta <1$. Conversely, restricting to the single maximizing route in each group leaves the series ${\sum }_{m\ge 0}(\alpha +\beta {)}^m$, which diverges if $\alpha +\beta \ge 1$. HenceThese are exactly the four strict resource-load inequalities. With positive arrival rates the Markov chain is irreducible, so the normalized stationary distribution is unique.

A coplanar target sweeps a strip of width $2R_p$ through the comet's local position probability. Hence the relevant geometric collision cross-section is a length, and the rare collision rate per comet is ${\Gamma }_{\mathrm{col}}=2R_p{\Sigma }_c(a_p)v_{\mathrm{rel}}$.

Using the corrected phase-mixed comet surface density and the notation $s=q/a_p$, $\gamma =\sqrt{3-2\sqrt{2s}}$, the coplanar comet collision time isFor $q\ll a_p$, this becomes ${\pi }^2{Q}^{3/2}{a}_p/(2\sqrt{3}{R}_p\sqrt{G{M}_{\star }})$, so $t_{\mathrm{col}}\propto Q^{3/2}{M}_p^{-1/3}{a}_p$ at fixed stellar mass and planetary mass density. This is the mean waiting time for one orbiting comet; the mean interval between impacts from $N_c$ such independent comets is $t_{\mathrm{col}}/N_c$. It assumes rare encounters and orbital phase mixing, not a Poisson process derived from a single deterministic trajectory.

An Inhomogeneous Poisson process with intensity $\lambda$ is a counting process with $N_0=0$, independent increments, andThe continuity and positivity of $\lambda$ make $\Lambda$ strictly increasing.

For $M_t=N_{g(t)}$ and $0\le s<t$,has a Poisson distribution with meanDisjoint increments remain independent, so $M$ is a homogeneous rate-one Poisson process. Taking $g={\Lambda }^{-1}$ gives $N_t=M_{\Lambda (t)}$, and henceThis is the time change of an inhomogeneous Poisson process.

For the bicycle model, let an arrival occur at time $s\le t$ and have velocity $V$. At time $t$ its position is $V(t-s)$, so it lies in the first $x$ miles exactly whenConditional on its arrival time, independently marking the bicycle by whether this event occurs is Poisson thinning. Therefore the desired count is Poisson with meanThus

For the finite state space $S$, the generator or Q-matrix $G=(g_{ij})$ is defined byIts off-diagonal entries are nonnegative jump rates and $g_{ii}=-{\sum }_{j\ne i}{g}_{ij}$. The transition semigroup of a continuous-time Markov chain is $P(t)=e^{tG}$.

An invariant distribution of a continuous-time Markov chain is a probability row vector $\pi$ satisfying $\pi P(t)=\pi$ for every $t\ge 0$. Differentiation at zero gives $\pi G=0$; conversely, $\pi G=0$ implies $\pi e^{tG}=\pi$. ThusInvariant distributions need not be unique: each closed communicating class supports one, and their convex combinations are invariant. A finite irreducible chain has a unique invariant distribution.

Let $0=T_0<T_1<\cdots$ be the arrival times of the independent rate-$\lambda$ Poisson process. The increments $T_{n+1}-T_n$ are independent exponential variables of rate $\lambda$. Conditional on $Y_n=X_{T_n}=i$, the Markov property, time homogeneity, and independence of the next increment giveHence $(Y_n)$ is a discrete-time Markov chain. Its transition matrix is the Poisson-sampled continuous-time Markov chain resolvent

If $\pi G=0$, then $\pi e^{tG}=\pi$ and integration gives $\pi P_{\lambda }=\pi$. Conversely, if a probability row vector $\nu$ satisfies $\nu P_{\lambda }=\nu$, multiplyingon the right by $\lambda I-G$ yields $\nu G=0$. Therefore the sampled chain and the original chain have exactly the same invariant distributions.

Immediately after departure $n$, either $Q_n>0$ and one of those customers starts the next service, or $Q_n=0$ and the next arrival both initiates service and is not counted among those waiting after it. During that service $A_n$ further customers arrive. Thus the embedded departure chain of an M-G-1 queue obeys

Conditional on the next service time $S_{n+1}=s$, the Poisson process property givesThe law of total expectation and law of total variance therefore giveand

In equilibrium let $Q$ have the common distribution of $Q_n,Q_{n+1}$ and put $H=1_{\{Q>0\}}=h(Q)$. The fresh pair $(A_n,S_{n+1})$ is independent of $Q_n$. Taking expectations in the recursion givesBecause $QH=Q$ and $H^2=H$, stationarity of the second moment givesSolving,

In stationarity, level crossing together with Poisson arrivals see time averages identifies this as the mean number $L$ in the shop. The Little law gives $L=\lambda (W_q+\mathbb{E}S)$, where $W_q$ is waiting time before service. Since $\rho =\lambda \mathbb{E}S$,This is the Pollaczek-Khinchine mean formula.

The count $N_t=M(0,t]$ is a rate-$\lambda$ Poisson process. Over a time interval $(s,t]$, the incrementdepends only on the Poisson points and marks in that interval. Disjoint intervals give independent increments, and the distribution depends only on $t-s$. The paths are càdlàg step functions, $X_0=0$, andThus $X$ is stochastically continuous and

Ignoring gravitational focusing, a galaxy sweeps a cylinder of volume $\pi R^2{v}_{\mathrm{rel}}T$. Multiplying this by the number density $N$ of target galaxies and averaging the relative speed gives the mean encounter countFor independent encounters modeled by a Poisson process, the probability of at least one galaxy merger is $1-e^{-\mu }$. The printed expression $P=\mu$ is its rare-encounter approximation $\mu \ll 1$; it is not an exact probability for arbitrary $T$.

For an illustrative present-day field population take an effective merger impact parameter $R=20\mathrm{kpc}=0.02\mathrm{Mpc}$, $N=10^{-2}{\mathrm{Mpc}}^{-3}$, and $⟨v_{\mathrm{rel}}⟩=200\mathrm{km}{\mathrm{s}}^{-1}$. These are assumed order-of-magnitude inputs, not precise observational measurements. Taking a Hubble time of $T\simeq 1.4\times 10^{10}\mathrm{yr}\simeq 4.4\times 10^{17}\mathrm{s}$, the given distance conversion yieldsand henceThus the geometric field estimate is of order $10^{-5}$ to $10^{-4}$ per Hubble time. It scales as $R^{2}N⟨v_{\mathrm{rel}}⟩T$ and is very sensitive to environment and the adopted effective merger radius. Enhanced density in groups, gravitational focusing, and the evolution of the galaxy population are all omitted; this number is not a prediction of the full cosmological merger fraction.

Now use the stipulated rapid, distant encounter. The consistent rectilinear trajectory isWith the stated impact parameter along $x$ and velocity along $z$, this trajectory lies in the $xz$-plane. The original PDF's reference to the $xy$-plane is a typo; the local TeX also corrupts the impact-parameter direction. The original PDF fixes that direction as $x$.

Use the positive potential $G{m}_p/|\mathbf{R}-\mathbf{r}|$, whose gradient is the attractive acceleration under the question's sign convention. Its multipole expansion isThe first term has no force; the linear term accelerates the entire galaxy and disappears in the frame following its centre of mass. The leading internal tidal potential is thereforeThis is a quadrupole approximation valid for $r\ll p$, rather than an exact equality for every $r<p$. Its acceleration tidal tensor acts on $\mathbf{r}$ asIn the impulse approximation, each stellar position is held fixed during the flyby and the velocity kick is the integral of this acceleration. Put $s=vt$; terms odd in $s$ integrate to zero, whileThe integrated diagonal coefficients of the tidal tensor are consequently $(2,-2,0)/(v{p}^2)$, givingThe encounter stretches the galaxy along the impact parameter, compresses it along $y$, and gives no net leading kick along the flyby direction.

Denote the pre-encounter stellar velocity by $\mathbf{u}$, to distinguish it from the relative flyby speed $v$. The instantaneous change of specific kinetic energy isUncorrelated kicks with zero mean cross term imply $⟨\mathbf{u}\cdot \Delta \mathbf{v}⟩=0$, so the phase-averaged heating at a specified position isFor an individual star the cross term need not vanish: the formula is an ensemble or orbital-phase average. Integrating over a spherical galaxy of mass $m_g$, symmetry gives $⟨x^2⟩=⟨y^2⟩=⟨z^2⟩=⟨r^2⟩/3$. Therefore the total tidal heating isFor two identical galaxies, each receives this heating with $m_p=m_g$. Adding both contributions gives

In the centre of mass frame, equal masses approach with speeds $v/2$ if $v$ is their relative speed at infinity. The initial orbital energy iswhere $m_g/2$ is the reduced mass. The orbital potential energy vanishes at infinite separation. Within the weak-deflection approximation this asymptotic speed is also the nearly constant speed used in the flyby calculation.

Conservation of energy transfers the positive internal heating out of the relative orbit. Tidal capture of galaxies occurs in this model if $E_{\mathrm{orb}}-\Delta E_{\mathrm{int}}<0$, which givesThe equality is marginal capture. A bound pair still needs subsequent evolution to coalesce, so the result is an approximate capture criterion used here as a merger criterion. Its right-hand side has dimensions of length times speed, as required.

The impulse approximation requires the flyby duration $\tau \sim p/v$ to be short compared with the stellar dynamical time ${\Omega }^{-1}$, namely $v/p\gg \Omega$. If $v/p<\Omega$, a star moves substantially during the encounter. Its orbital adiabatic invariance of an orbital action suppresses net heating by a slowly varying tidal field: adiabatic shielding of tidal encounters replaces the frozen-position calculation. The divergent $v^{-2}$ heating predicted by extrapolating the impulsive formula to slow encounters is therefore spurious. Close passages with $p$ comparable to the galaxy size or strong gravitational focusing also lie outside the derivation.

Here $\Lambda$ is planar Lebesgue measure. For a Borel set $B\subseteq [0,\infty )$, polar coordinates giveThe pushforward is therefore locally finite and atomless. By the mapping theorem for Poisson point processes, $f(\Pi )$ is a homogeneous Poisson process on $[0,\infty )$ with constant intensity $\pi$. This is the two-dimensional radial volume transform of a homogeneous Poisson point process.

Put $S_k=R_k^2$. The variable $S_k$ is the $k$th arrival time of that rate-$\pi$ Poisson process, hence has the gamma distribution with shape $k$ and rate $\pi$:Since $s=r^2$ and $ds/dr=2r$, the change-of-variables formula for a probability density gives the Kth-nearest-neighbour distance in a homogeneous Poisson point process

The Poisson output theorem for a feedback queue, equivalently the Burke theorem applied to the effective queue-length process with service rate $\delta \mu$, shows that successful departures in equilibrium form a Poisson process of rate $\lambda$.

The process of all entries into the queue is different. Its long-run rate isImmediately after any queue-entry event the server is busy, so the conditional probability of another entry in the next interval of length $h$ isthe next entry can be external or can be a service completion followed by feedback. A Poisson process of the displayed long-run rate would instead give $(\lambda /\delta )h+o(h)$. These coefficients are unequal when $\lambda <\delta \mu$, sinceHence the combined arrival process is not Poissonian.

An M-M-1 queue has Poisson process arrivals of rate $\lambda$, independent exponential service times of rate $\mu$, and one server. The two letters $M$ indicate the memoryless property of the interarrival and service-time distributions. Its queue length is a birth-death process with birth rate $\lambda$ and death rate $\mu$ away from zero. When $\rho =\lambda /\mu <1$, its normalized stationary distribution isthe powers ${\rho }^n$ quoted in the question form the corresponding unnormalized invariant measure.

In equilibrium the chain is reversible by detailed balance. Reverse time. Forward departures become reverse accepted arrivals; reversibility identifies the reversed chain with the original one, whose attempted arrivals form a Poisson process. The accepted points are exactly those occurring while the state is below $N$. The standard quasi-reversibility argument therefore makes the equilibrium departure process Poisson, with rate equal to the mean departure rate

Independent Poisson processes have stationary independent increments, so their weighted sum does too and is stochastically continuous. Moreover,where

Let $q=({\mathbf{q}}_1,\dots ,{\mathbf{q}}_N)$ denote the particle positions. In the Ghirardi--Rimini--Weber model, the wavefunction obeys the ordinary Schrodinger equationbetween random collapses. Each particle $i$ has an independent Poisson process of collapse times with rate $\lambda$. At such a time the state jumps according towhere the random centre has probability density $\Vert L_i(\mathbf{X})\psi {\Vert }^2$. The normalization is arranged so that $\int d^{3}X{L}_i(\mathbf{X}{)}^2=I$.

The original GRW scales are approximatelyAn isolated microscopic particle is therefore exceedingly unlikely to collapse during a laboratory experiment. A macroscopic pointer containing about $N\sim 10^{23}$ relevant particles has total collapse rate $N\lambda \sim 10^7{\mathrm{s}}^{-1}$ and collapse time about $10^{-7}\mathrm{s}$. After a measurement interaction correlates different microscopic outcomes with pointer positions separated by much more than $r_C$, one constituent's localization suppresses all incompatible pointer branches. This GRW amplification mechanism produces one definite macroscopic outcome with probabilities given by the Born rule, while leaving ordinary microscopic unitary time evolution almost unchanged.

For one spatial coordinate, average over the random centre of one collapse. The resulting density operator has position-space kernelIts diagonal is unchanged, so $⟨x⟩$ and $⟨x^2⟩$ are unchanged. The first derivative of the Gaussian factor vanishes at $x=x^{\prime }$, so $⟨p⟩$ is also unchanged. Its second derivative does not vanish, and the position representation of the momentum operator givesIn three dimensions the increase in total ${\mathbf{p}}^2$ is $3{\hslash }^2/(2r_C^2)$, corresponding to kinetic-energy increase $3{\hslash }^2/(4m{r}_C^2)$ per collapse. These are ensemble statements: conditioning on one specified collapse centre can shift the position moments. Repetition at rate $\lambda$ predicts GRW spontaneous heating, so precision searches for anomalous bulk heating, spontaneous radiation, momentum diffusion, and loss of matter-wave interference test the model.

Let the Poisson process $X$ have rate $\lambda$, let $\overline{F}(t)=\mathbb{P}({\xi }_1>t)$ be the survival function of an inter-renewal time of $N$, and let $\overline{H}$ be the survival function of the first event time of $Y$. Independence givesPut

At a large time, the excess of $Y$ is the minimum of the excesses of $X$ and $N$. The Poisson excess is an independent rate-$\lambda$ exponential random variable. Applying the renewal excess limit theorem to $N$ and to the assumed renewal process $Y$ givesSubstituting $\overline{F}(y)=e^{\lambda y}\overline{H}(y)$ yields the integral equation

Set $A(x)={\int }_x^{\infty }\overline{H}(y)dy$ and $c={\mu }_H/{\mu }_F<1$. Differentiating the equation almost everywhere givesSince $A^{\prime }=-\overline{H}$, it follows thatUsing $\overline{H}(0)=1$ shows that $\overline{H}(x)=e^{-x/{\mu }_H}$. Thus the first event time of $Y$ is exponential. This is the exponential first interarrival in a renewal superposition containing a Poisson process theorem.

In Kendall notation, the first two letters $M$ mean Poisson arrivals and exponential service times. An M-M-1 queue has arrival rate $\lambda$, one server of rate $\mu$, and queue-length transitionsIt has a stationary distribution of an M-M-1 queue exactly whenand then

An M-M-infinity queue has infinitely many servers, so every customer begins service immediately. Its occupancy transitions areThe detailed balance for a birth-death process equations giveThus for every $\lambda ,\mu >0$ the stationary distribution of an M-M-infinity queue exists and isthe Poisson distribution of mean $\rho$. The Burke theorem for an M-M-infinity queue states that in this stationary regime the departure process is a Poisson process of rate $\lambda$; equivalently, time reversal turns departures into arrivals. The past departure process is also independent of the number in the system at the observation time.

For an M-M-1 queue, the first service lasts on average $1/\mu$. During it, the mean number of arrivals is $\lambda /\mu$, and each arrival initiates another statistically identical busy-period family. Thus, when $\lambda <\mu$,so the mean busy period of an M-M-1 queue isFor $\lambda =\mu$ the busy period is almost surely finite but has infinite mean, while for $\lambda >\mu$ it is infinite with positive probability; in either case its expectation is infinite.

For an M-M-infinity queue, stationary occupancy is Poisson with mean $\rho =\lambda /\mu$, so the long-run probability of an empty system isThe process alternates between idle periods and busy periods. An idle period waits for the next Poisson arrival, so its mean is $1/\lambda$. The renewal-reward theorem, with reward equal to time spent empty, givesSolving yields the mean busy period of an M-M-infinity queue: