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Suppose two independent background streams follow Poisson processes with rates $r_1$ and $r_2$. Define a coincidence by $|t_2-t_1|<w/2$: $w$ is the full accepted time-difference width. Around each first-counter signal, the expected number of second-counter signals is $r_{2}w$. The expected pair rate is therefore $r_1{r}_{2}w$. A circuit producing at most one output per first signal instead givesThe distinction matters when multiple detections occur in one window. If a different convention accepts $|t_2-t_1|<\tau$, its full width is $w=2\tau$ and the dilute expression is $2r_1{r}_2\tau$.

For a chosen numerical model, take background rates $r_1=100{\mathrm{s}}^{-1}$, $r_2=80{\mathrm{s}}^{-1}$ and $w=20\mu \mathrm{s}$. The dilute pair rate is $0.16{\mathrm{s}}^{-1}$. These are stipulated teaching inputs, not measurements of the historical chamber.

Making $w$ smaller also rejects some genuine events. Suppose true pairs arrive at rate $S_0$ before the timing cut, and their time difference has a centered normal distribution with standard deviation ${\sigma }_t$. Its accepted fraction, using the error function, isFor chosen $S_0=2.0{\mathrm{s}}^{-1}$ and ${\sigma }_t=5\mu \mathrm{s}$, the $20\mu \mathrm{s}$ window retains about $95.45\%$ of true pairs. In the dilute background model the fraction of accepted events that are true is $S/(S+R_{\mathrm{acc}})\simeq 92.27\%$. A narrow window trades detection efficiency against background; it does not make either quantity equal to one. Delaying one counter's signal far beyond genuine event correlations gives a proposed background check under stationarity, without assuming that the zero-delay excess itself is accidental.

Alternately routing a rate-$r$ Poisson process produces non-Poisson streams with common interarrival law Erlang$(2,r)$. The odd-numbered stream has an initial exponential delay, while the even-numbered stream is an ordinary renewal process. Their equilibrium delay survivor is $e^{-rt}(1+rt/2)$. The probability of no odd-stream point in $(s,s+t]$ is $e^{-rt}[1+rt(1-e^{-2rs})/2]$; for the even stream replace the minus by a plus. Conditioning on the parity of $N(s)$ proves these formulas.

With finitely many nonzero marks $z_j$ and independent Poisson processes $N^j$ of rates ${\lambda }_j$, the displayed process is a Lévy process with no Gaussian component. Its Lévy measure is ${\sum }_j{\lambda }_j{\delta }_{z_j}$, its mean is $t(\gamma +{\sum }_j{\lambda }_j{z}_j)$, and its variance is $t{\sum }_j{\lambda }_j{z}_j^2$. Its paths have finitely many jumps on every bounded interval and deterministic slope $\gamma$ between jumps. Coincident mark sizes combine their rates; zero marks have no effect. The drift here uses the uncompensated finite-sum convention, which differs from the drift parameter in a compensated Lévy–Khintchine formula.

A stable equilibrium M-M-s queue has a Poisson process of departures with the same rate as its input. Its present population is also independent of past departures, a component of quasireversibility. The birth-death process satisfies detailed balance; reversed upward jumps have a constant arrival rate, so they are generated by an independent Poisson process.

In stationarity, the departure process of an $M/M/\infty$ queue is a Poisson process of rate $\lambda$. By time reversibility, the departures before any fixed time are independent of the occupancy at that time.

The classical risk model has surplus $U_t=u+ct-{\sum }_{j=1}^{N_t}{X}_j$, where $N_t$ is a Poisson process of rate $\lambda$, independent claim sizes are positive, and premiums flow in at constant rate $c$. The relative safety loading is $\rho =c/(\lambda \mathbb{E}{X}_1)-1$. Ruin is the first time surplus becomes negative.

A compound Poisson process has the formwhere $N$ is a Poisson process and the independent identically distributed marks $Y_n$ are independent of $N$. It is a Lévy process with a finite Lévy measure.

At unit jump rate, this is $X_t={\sum }_{k=1}^{N_t}{\xi }_k$, where $N$ is a rate-one Poisson process and the independent marks are uniform signs. Equivalently it is the difference of two independent rate-$1/2$ Poisson processes. It is a Lévy process with characteristic function $\exp \{t(\cos \theta -1)\}$, mean zero, variance $t$, and Lévy characteristic exponent $1-\cos \theta$ in the negative-exponent convention. A general total jump rate rescales time.

Why not accept every delayed pulse? A longer window captures more genuine neutron capture signals, but also admits more accidental coincidences. Consider an original model with a capture waiting time following an exponential distribution of mean $\tau >0$. Assume detection after a prompt signal accepts $0<t<W$, with perfect time measurement and no lower cut. Integrating the normalized density givesThis assumes that moderation, competing capture channels, boundaries and electronic recovery can be represented or neglected in a single effective waiting-time model. It is not a fitted historical capture law.

Let independent prompt-like and delayed-like background streams have Poisson process rates $R_p$ and $R_d$. The accidental coincidence pair rate in a directed window of full width $W$ is $R_p{R}_{d}W$. Electronics producing at most one accepted delayed match per prompt instead give $R_p(1-e^{-R_{d}W})$. These definitions coincide to first order when $R_{d}W\ll 1$. Correlated backgrounds, such as one external neutron producing both a prompt recoil and a later capture, require separate controls; an independent-stream estimate cannot remove them.

For chosen $\tau =20\mu \mathrm{s}$ and $W=50\mu \mathrm{s}$, the accepted fraction is $1-e^{-2.5}\simeq 0.918$. With chosen $R_p=40{\mathrm{s}}^{-1}$ and $R_d=30{\mathrm{s}}^{-1}$, the pair rate is $0.0600{\mathrm{s}}^{-1}$, while the at-most-one rate is approximately $0.05996{\mathrm{s}}^{-1}$. These are teaching parameters, not reported Savannah River rates. A measured uncorrelated background can be tested using displaced time windows; a reactor-off measurement addresses a different question and cannot by itself characterize every reactor-correlated background.

Independent marking is the hypothesis that guarantees the Poisson thinning theorem. Equal marginal retention probabilities do not guarantee it. Consider one Bernoulli random variable $B$ with success probability $p\in (0,1)$, independent of a Poisson process, and retain every arrival when $B=1$ and none when $B=0$. Each arrival is retained with probability $p$, but the marks share the same coin. In an interval whose original count has Poisson distribution of mean $m>0$, the retained count is $R=BN$ and the rejected count is $D=(1-B)N$, soThe extra variance shows that $R$ is not Poisson distributed; the negative covariance proves dependence. Its no-arrival probability is $1-p+p{e}^{-m}$, rather than $e^{-pm}$. For the toy values $p=1/2,m=2$, both rules have retained mean $1$, but the common-coin rule has variance $2$ and no-arrival probability $(1+e^{-2})/2\approx 0.567668$, compared with variance $1$ and no-arrival probability $e^{-1}\approx 0.367879$ for independent marking. These formulas follow by conditioning on $B$ and using the law of total variance.

An arrival-dependent or adversarial retention rule can change the law just as strongly. Keeping every second arrival gives $R_t=\lfloor N_t/2\rfloor$. Its successive retained gaps are sums of two independent exponential random variables, hence have the Erlang distribution of shape two and rate $\lambda$, not an exponential distribution. The resulting renewal process and its count variance are analyzed in Every second Poisson arrival has underdispersed counts, and the two dependent streams in alternating split of a Poisson process. As an even more restrictive rule, keeping only the first arrival ever gives $R_t=1_{\{N_t\ge 1\}}$, a Bernoulli distribution at each $t>0$, so it cannot have a positive-mean Poisson distribution. A thinning rule must specify how decisions depend on arrivals and on one another; the independent-marking conclusion cannot be applied automatically to arbitrary deletion.

The interval during which a detector or readout cannot register another event while recovering from a response. A detector trigger may be fast while the camera, chamber or data acquisition is still busy.

In a simple nonextending model, every accepted event starts a fixed busy interval ${\tau }_d$; arrivals during it are lost and do not extend it. With Poisson process input rate $R$, the average cycle lasts ${\tau }_d+1/R$, giving observed rateThis is a specified mathematical model, not a universal correction. An instrument whose busy interval is extended by further arrivals behaves differently. Correcting the total rate also does not remove the multiplicity bias from a detector trigger.

For independent Poisson processes with rates $r_{+},r_{-}\ge 0$, their difference is a Lévy process with characteristic function $\exp (t[r_{+}(e^{iu}-1)+r_{-}(e^{-iu}-1)])$. When the combined rate is positive, its jumps occur at rate $r_{+}+r_{-}$, with direction probabilities proportional to the two rates. Equal unit rates give $\exp (2t(\cos u-1))$ and a rate-two Compound Poisson process with Rademacher distribution jumps. Its paths are integer-valued, càdlàg, and of finite variation on compact intervals.

For a rate-$\lambda$ Poisson process, keeping every second arrival gives a renewal process with independent interarrival times equal to sums of two exponential waiting times. Put $x=\lambda t$. Its count mean and variance are $(2x+e^{-2x}-1)/4$ and $(4x-8x{e}^{-2x}-e^{-4x}+1)/16$. The variance is strictly smaller than the mean for $x>0$: sixteen times their difference has derivative $4[-1+4x{e}^{-2x}+e^{-4x}]<0$ and is zero at zero, because $\sinh (2x)>2x$. Therefore this process is not a Cox process.

Let $Y_t={\sum }_{j=1}^{N_t}{X}_j$ for a rate-$\lambda$ Poisson process and independent and identically distributed nonnegative marks independent of that process, with $\varphi (q)=\mathbb{E}{e}^{-q{X}_1}$. The Laplace transform of a nonnegative random variable for the increment over $[s,t]$ is $\exp [\lambda (t-s)(\varphi (q)-1)]$. That increment is independent of the past of $Y$, so the displayed nonnegative process has mean one and satisfies $\mathbb{E}[Z_t\mid \sigma (Y_u:u\le s)]=Z_s$. No finite first moment of the marks is needed for this martingale when $q\ge 0$.

A flow-level model treats each document transfer as a flow whose rate depends on the current occupancy and resource capacities. With independent Poisson processes for arrivals and independent unit-mean exponential distributions for sizes, a route's total completion rate equals its allocated aggregate service.

A target galaxy with geometric encounter cross section $\pi R^2$ in a population of number density $N$ has encounter rate $\Gamma =\pi R^{2}N⟨v_{\mathrm{rel}}⟩$. A Poisson process gives merger probability $1-e^{-\Gamma T}$, approximated by $\Gamma T$ only for rare encounters. Counts of distinct pairs across the whole population additionally require a factor of one-half.

Place recovery marks at each vertex according to independent rate-$\mu$ Poisson processes and infection arrows on each oriented nearest-neighbour edge according to independent rate-$\lambda$ Poisson processes. A vertex is infected at time $t$ exactly when a time-directed path from an initially infected vertex reaches it while following arrows and avoiding recovery marks.

A queue with a Poisson process of arrivals and two independent service times with exponential distributions of different rates must distinguish the two possible single-customer states. If an arrival chooses either idle server with probability $1/2$, the continuous-time Markov chain is reversible. Write $a,b$ for the states with only server one or two busy, and $n\ge 2$ for total population. Its detailed balance equations give ${\pi }_a={\pi }_0\nu /(2{\mu }_1)$, ${\pi }_b={\pi }_0\nu /(2{\mu }_2)$ and ${\pi }_n={\pi }_0{\nu }^2(\nu /\mu {)}^{n-2}/(2{\mu }_1{\mu }_2)$. These weights normalize exactly when $\nu <\mu$. In equilibrium, reversed upward transitions have total rate $\nu$ in every state, so time reversal of a continuous-time Markov chain makes the departure stream a Poisson process of rate $\nu$. This output conclusion requires stationarity.

An increment of a real- or vector-valued stochastic process is the difference $X_t-X_s$ between its values at two times $s<t$. A random walk accumulates its one-step increments; a Poisson process has nonnegative integer increments counting arrivals in $(s,t]$. A process can have independent increments even though its values at different times are dependent.

A stochastic process has independent increments when, for every $t_0<t_1<\cdots <t_k$, the random variables $X_{t_1}-X_{t_0},\dots ,X_{t_k}-X_{t_{k-1}}$ are independent. Adjacent intervals are allowed: each increment represents the change after its left endpoint up to its right endpoint. A Poisson process, Brownian motion, and a random walk with independent steps have this property. Their values at different times need not be independent: for a fair simple random walk on the integer line, $\mathrm{Cov}(S_m,S_n)=\min (m,n)$ because both positions contain the same first $\min (m,n)$ increments. Independent increments need not be stationary increments; summing independent steps with time-dependent variances gives a counterexample.