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Independent loss alone leaves normalized factorial moments invariant: if every input photon is retained with probability $\eta$, binomial thinning conditional on a total gives $⟨K⟩=\eta \mu$ and $⟨K(K-1)⟩={\eta }^2⟨N(N-1)⟩$. Both powers cancel from $g^{(2)}(0)$. Fewer coincidences still mean poorer statistical precision.

Now let each channel contain a signal rate $s_i$ and an independent Poisson process background rate $b_i$, independent also of the signal. Put ${\rho }_i=s_i/(s_i+b_i)$. The pair density is $s_1{s}_2{g}_{\mathrm{sig}}^{(2)}+s_1{b}_2+b_1{s}_2+b_1{b}_2$. Dividing by $(s_1+b_1)(s_2+b_2)$ proves the displayed formula, for the same bin and timing response. For chosen equal signal fractions ${\rho }_1={\rho }_2=4/5$, ideal signal values $0,1,2$ become $9/25,1,41/25$. Background pulls both dips and peaks toward one.

This calculation assumes additive, mutually independent background counts. Scattered coherent light that interferes with the signal, correlated electronic noise, or recovery effects require a different model. Correcting a dip without independently estimating the backgrounds cannot establish photon antibunching.

Poisson arrivals see time averages says that an exogenous, homogeneous Poisson process samples prearrival states with the same long-run frequencies as observation over time, provided the observed system cannot anticipate future arrivals and the relevant time averages exist. The system may depend on all past arrivals: an ordinary queue does. The required independence is between future increments of the Poisson process and the history available to the system, including its service information. Assume measurable state observations with ordinary piecewise continuous functions as sample paths; write $X(t-)$ for the state immediately before time $t$.

For a state set $B$, let $A(T)$ count offered arrivals by time $T$ and let $t_k$ be their times. With a positive finite constant arrival rate, the PASTA equality isalmost surely when the time limit exists. The indicator functions count visits on the left and time spent on the right. For a stationary process with ergodicity, the right side is the stationary probability of $B$. It matters that the arriving customer has not yet changed the observed state.

In the M/M/1 utilization and delay example, server utilization is $0.8$. Thus PASTA says that 80 percent of offered arrivals find the server busy and 20 percent find the whole queue empty. If an offered arrival seeing population $n$ is admitted with conditional probability $p(n)$, using fresh randomness independent of that state, the accepted fraction is ${\sum }_n{\pi }_{n}p(n)$, where $\pi$ is the stationary distribution. Multiplying this fraction by the offered rate gives the admitted queue throughput when every admitted customer eventually completes service. Accepted arrivals are selected by state and need not themselves form a Poisson process or see the unconditioned time frequencies.

Arbitrary arrivals need not see time averages. Suppose a customer arrives every two minutes and its service time is exactly one minute, with one server and no other customers. The server is busy half the time, but every arrival finds it idle. A uniformly random initial phase can even make the occupancy a stationary process, without repairing the disagreement. Nevertheless Little's law still gives $L=(1/2)\cdot 1=1/2$, with rate measured in customers per minute. This separates Little's law from PASTA.

Ronald W. Wolff's 1982 paper gave a general PASTA theorem under nonanticipation, rather than requiring the observed system to be independent of its driving Poisson process. Its Theorem 1 compares the two long-run fractions; stationarity is a way to establish convergence, not the central sampling condition. ^[ref]. ^[ref].

The time of the $n$-th event of a Poisson process. At positive rate $\lambda$, the Poisson interarrival times $T_j-T_{j-1}$, with $T_0=0$, are independent and identically distributed random variables with the rate-$\lambda$ exponential distribution. In particular, $T_n$ has expected value $n/\lambda$ and variance $n/{\lambda }^2$. The inverse-count identity is $\{T_n\le t\}=\{N_t\ge n\}$.

For a rate-one Poisson process with Poisson arrival times $T_n$, suppose $t_n\to \infty$ and $t_n/n\to \lambda \in [0,\infty ]$. Thenwhere $1/\sqrt{\infty }=0$ and $\mathrm{BN}(\rho )$ denotes the bivariate normal distribution with standard normal marginal distributions, including the singular cases. In particular, equal scales give $\rho =-1$ and separated scales give $\rho =0$.

To prove this, fix real $x,y$ and put $s_n=n+y\sqrt{n}>0$ for large $n$. The event $(T_n-n)/\sqrt{n}\le y$ is exactly $N_{s_n}\ge n$. Counts at the two deterministic times $t_n,s_n$ are sums of a shared count on $[0,\min (t_n,s_n)]$ and an independent count on the remaining interval. The central limit theorem for a centered Poisson random variable of diverging parameter follows by writing a count at a real time $u$ as its count at $\lfloor u\rfloor$, a sum of independent unit increments, plus an independent remainder with expected value at most one. Markov's inequality makes that remainder negligible after division by $\sqrt{u}$.

Apply this central limit theorem to the two independent count blocks whenever their lengths diverge. A block with bounded length, or length negligible compared with the scale of its coordinate, contributes a term tending to zero in probability. The standardized pair of counts therefore has a joint normal limit whose correlation coefficient isThis includes $\lambda =1$: the difference block then has negligible variance after normalization. In the arrival event the standardized count at $s_n$ must be at least $(n-s_n)/\sqrt{s_n}\to -y$. Reversing this inequality changes the sign of the second limiting normal random variable, giving the asserted negative correlation coefficient. Continuity of the marginal distributions permits the moving thresholds and the weak inequalities, including at the singular limits.

Every $\rho \in [-1,0)$ is realized by $t_n=\lambda n$ with $\lambda ={\rho }^{-2}\ge 1$, and zero by $t_n=n^4$. Conversely, even when $t_n/n$ has no limit, it has a subsequence converging in the extended interval $[0,\infty ]$. If the original pair has a joint bivariate normal distribution limit, that subsequence has the same limit, forcing its correlation coefficient into $[-1,0]$.

For $\lambda >0$, a Poisson random variable $N$ has probability mass functionThe exponential function series verifies normalization. At $\lambda =0$, define $N=0$ with certainty. The equality of mean and variance is a distinguishing model property, though it does not by itself characterize the law.

When counting events in a homogeneous Poisson process, distinguish its rate $r$ from the distribution parameter: an interval of duration $t$ has $\lambda =rt$. A rate measured per minute times a duration measured in minutes gives a dimensionless expected count. This process assumes independent increments, a constant rate, and a negligible chance of multiple events relative to interval length in a sufficiently short interval. A rate alone does not establish a Poisson distribution. See Bertsekas and Tsitsiklis, chapter 6 summary, pages 46-47.

For a constructed arrival model with $r=2$ events per minute over $t=1.5$ minutes, $\lambda =3$. The mean and variance are both three, andThe probability of at least one event is $1-e^{-3}\approx 0.9502129$. These results describe the specified process, not every arrival mechanism with the same average rate.

The Poisson limit theorem, often called the law of rare events, connects this law to the binomial distribution. Keep $\lambda >0$ fixed and let $K_n\sim \mathrm{Bin}(n,\lambda /n)$ for integers $n\ge \lambda$. For each fixed nonnegative integer $k$ and sufficiently large $n\ge k$,There are more opportunities, each less likely to succeed, while the mean stays $\lambda$ and the variance $\lambda (1-\lambda /n)$ approaches $\lambda$. For $\lambda =3$, the probability of three successes is $0.2668279$ at $n=10$, $0.2310569$ at $n=50$, and $0.2257398$ at $n=200$, approaching the Poisson distribution value $0.2240418$.

The panels show the same count scale and the same Poisson distribution reference. At $n=10$, $p=0.3$ is not especially small and the difference is visible. No universal trial-count cutoff guarantees an adequate approximation for every event.

The independence assumption matters even when every success is rare. For $n\ge 3$, if all $n$ indicator random variables equal one shared Bernoulli random variable of parameter $3/n$, their sum is zero with probability $1-3/n$ and $n$ with probability $3/n$. Its mean is three, yet its zero-count probability tends to one, not $e^{-3}$. Some dependent counts still admit approximation through the Stein-Chen method, but this requires additional control of dependence.

For historical orientation, the BnF records Poisson's original probability treatise, Recherches sur la probabilité des jugements, as published in Paris in 1837. This bibliographic fact alone does not establish when the modern distribution terminology arose. See BnF record of Poisson's 1837 treatise.

For a unit-rate Poisson process and real $\theta$, the displayed process is a nonnegative martingale with initial value one. Independent increments give $\mathbb{E}{e}^{\theta (N_t-N_s)}=\exp ((t-s)(e^{\theta }-1))$, proving its conditional mean property. The Ville inequality turns this exponential moment into a maximal tail bound.

Poissonization replaces a fixed number of independent trials by a Poisson process clock. Independent exponential arrival clocks can then simplify occupancy and collection problems, after which the original counting quantity is recovered from the clock rate.

For a unit-rate Poisson process, $h(\eta )=(1+\eta )\log (1+\eta )-\eta$ gives the displayed maximal deviation bound for $L,\eta >0$. Poisson exponential martingales and the Ville inequality control upper and lower deviations; the upper-tail rate $h$ also bounds the lower-tail probability. Since $h(0)=0$ and $h^{\prime }(\eta )=\log (1+\eta )>0$, the rate is strictly positive for every positive deviation level.

If no directed routing path returns from $k$ to $j$, the set of colonies reachable from $k$ has no outgoing edge to its complement. The complement is an autonomous open migration process, and the marked $j\to k$ transition is an exit from it. In time reversal of an open migration process, that exit is a state-independent Poisson immigration channel of rate ${\alpha }_j{p}_{jk}$. Reversing a stationary two-sided Poisson process preserves its law, proving the forward stream is Poisson. This argument does not assert Poisson internal streams for networks with feedback.

A Poisson process observed for a fixed duration $H$ has likelihood proportional to ${\lambda }^{N_H}{e}^{-\lambda H}$. Complete independent exponential distribution gaps observed until a fixed event count $n$ have likelihood ${\lambda }^n{e}^{-\lambda T_n}$. Both give count divided by exposure, but the exposure is random in the second design. Under fixed exposure the total count is sufficient: conditional on it, the event locations are independent of the rate, so exact timestamps add no rate information to interval counts. Gap estimates and fixed-horizon estimates should not be compared without distinguishing their stopping rules and any final censored gap.

A Homogeneous Poisson process models arrivals at a constant rate $\lambda >0$. Its counting process $N_t=N(t)$ counts arrivals in $(0,t]$, with $N_0=0$. Its paths are right-continuous nondecreasing functions, take nonnegative integer values, and have unit jumps almost surely. The defining Poisson distribution of each increment isIt has stationary increments: the probability distribution above depends on the duration $t$, not the starting time $s$. It also has independent increments: counts in any finite collection of disjoint time intervals are independent random variables. Together these conditions specify the joint count laws, not merely the Poisson distribution at a single time.^[ref]

The rate has units of arrivals per unit time; $\lambda t$ is dimensionless. The Poisson distribution givesFor a toy rate $\lambda =2$ per minute and a duration of $1.5$ minutes, the expected value and variance of the count are both $3$. The probability of no arrival is $e^{-3}\approx 0.049787$, and the probability of exactly three is $3^3{e}^{-3}/3!=\frac{9}{2}{e}^{-3}\approx 0.224042$. In two consecutive one-minute intervals, the probability of exactly two arrivals in the first and exactly one in the second is $(2e^{-2})(2e^{-2})=4e^{-4}\approx 0.073263$, using independent increments. These are exact model calculations with rounded decimals, not measured frequencies.

The diagram's nine hand-chosen arrivals illustrate a path; the line $\lambda t$ is an expected value, not a prescribed trajectory. Constant rate does not imply evenly spaced arrivals. Poisson interarrival times connect the count description to waiting, while Poisson thinning explains how independently routing arrivals produces separate streams. This makes the Poisson process useful as a simple arrival model for a queue, provided constant rate and independent increments are plausible. For a deterministic changing rate, use an Inhomogeneous Poisson process; bunching or history-dependent arrivals need additional assumptions. Conditional on $N_t=n$, the arrival times are the order statistics of $n$ independent uniform distributions on $[0,t]$, as developed in Poisson process conditional arrival times.

Sample a finite-state continuous-time Markov chain with generator $Q$ at the independent arrival times of a rate-$\lambda$ Poisson process. The sampled states form a discrete-time Markov chain with transition matrixIts invariant distributions are exactly those of the continuous-time chain.

Independent claim Poisson processes of rates ${\lambda }_i$ merge into a Poisson process of rate $\lambda ={\sum }_i{\lambda }_i$, by the Superposition theorem for Poisson point processes. The merged claim law is a mixture distribution of the individual claim laws with weights ${\lambda }_i/\lambda$. Equivalently, multiplication of the individual compound-Poisson transforms produces $\exp (\lambda [{\sum }_i({\lambda }_i/\lambda )M_i(r)-1])$.

Poisson thinning splits a Poisson process by independently marking each arrival as retained or rejected. Let the original Homogeneous Poisson process have rate $\lambda >0$, and let $p(s)\in [0,1]$ be a deterministic measurable function of time. Attach independent uniform distributions $U_1,U_2,\dots$ on $(0,1)$, independent of the entire original Poisson process. At arrival time $T_i$, retain the arrival if $U_i\le p(T_i)$. Given all arrival times, these decisions are independent and have the specified retention probabilities. Define the two counting processes using an indicator function:The Poisson thinning theorem gives independent retained and rejected Inhomogeneous Poisson processes. Their rates are $\lambda p(s)$ and $\lambda (1-p(s))$, respectively, in the sense of the intensity function of a point process. In particular, for $0\le a<b$,The independence is of the two whole processes: every finite collection of retained counts is independent of every finite collection of rejected counts. It is not merely zero covariance for one interval. For constant $p$, the streams are Homogeneous Poisson processes of rates $\lambda p$ and $\lambda (1-p)$; the endpoints $p=0,1$ include the identically zero stream.^[ref]

For an exact toy calculation, take $\lambda =2$ per minute and retain with $p(s)=1/4$ on $(0,2]$ and $p(s)=3/4$ on $(2,4]$. In the first two minutes, the retained and rejected expected values are $2\cdot 2\cdot 1/4=1$ and $2\cdot 2\cdot 3/4=3$. In the next two minutes they are $3$ and $1$. Over all four minutes, $R_4,D_4$ are independent Poisson distributions with mean $4$ each, and their sum $N_4$ has mean $8$. ThusThese probabilities and expected values describe the model, not empirical measurements. In the diagram, the original nine synthetic arrivals split into five retained and four rejected arrivals; those realized counts need not equal their expected values.

The mechanism explains why mutually exclusive labels can yield independent counts: randomness of the original total offsets the dependence from allocating a fixed total. For constant $p$ and an interval with original mean $m$, the conditional retained count has a binomial distribution given $N=n$. The joint probability generating function isThis factorization proves the two independent Poisson distributions; disjoint original intervals give independent pairs. For varying $p(s)$, the Independent marking theorem for Poisson point processes gives the same conclusion with the two integrated intensities. The joint red and green Poisson count formula gives the corresponding direct calculation with probability mass functions.

Independence here is unconditional. After conditioning on $N=n$, the identity $D=n-R$ gives covariance $\mathrm{Cov}(R,D\mid N=n)=-np(1-p)$ for constant $p$. Thus the retained and rejected counts are dependent for $n>0$ and $0<p<1$, even though their unconditional covariance is zero. Poisson thinning can model independent routing of arrivals, or independent detection of arrivals with a known efficiency schedule; using only a retention fraction without checking the marking mechanism can give a wrong count law.

A jump process with event types $k$, increments ${\nu }_k$, and rates $a_k(X)$ can be represented as $X(t)=X(0)+{\sum }_k{\nu }_k{P}_k({\int }_0^t{a}_k(X(s-))ds)$, with independent unit-rate Poisson processes. This represents the Markov jump-process generator through time-changed counts.

A Markov chain $Z$ with transition kernel $K$, sampled at an independent Poisson process $N$ of rate $r$, is a pseudo-Poisson process. Its infinitesimal generator is $r(K-I)$ and its transition semigroup is $e^{-rt}{\sum }_{n\ge 0}(rt{)}^n{K}^n/n!$. Conditioning on $N_t$ proves the latter formula, and differentiation at zero gives the former. Its resolvent operator is $(\lambda +r{)}^{-1}{\sum }_{n\ge 0}[r/(\lambda +r){]}^n{K}^n$ for $\lambda >0$, by integrating the nonnegative scalar Poisson weights. The series converge in operator norm on bounded Borel measurable functions.

Queueing theory studies how jobs accumulate while waiting for limited service: customers at a desk, requests at a computer, or calls competing for a connection. A queue model specifies how jobs arrive, their service times, the number of servers, the buffer size of a queue, and the service discipline. Its practical questions include the queue population, queueing delay, customer sojourn time, queue throughput, and server utilization.

Average spare capacity does not guarantee short waits. Even when the service rate of a queue exceeds the arrival rate, bursts of arrivals and unusually long service times can leave work waiting. The M/M/1 utilization and delay example shows that an 80 percent server utilization can turn a 12-minute mean service time into a 60-minute mean customer sojourn time. As capacity is approached, congestion grows sharply; reserving spare capacity can be much more valuable than its idle fraction suggests.

Two complementary ideas make queueing theory especially useful. Little's law converts a measured queue population and queue throughput into a mean customer sojourn time, using a counting identity rather than a particular arrival law. Poisson arrivals see time averages answers a different question: when do arriving customers see the same state frequencies as an observer sampling over time? That requires an exogenous Poisson process and a system that cannot anticipate its future arrivals. Neither result by itself determines all queueing delays; detailed models such as the M/M/1 queue and M/G/1 queue supply the remaining information.

For independent Poisson process arrivals and exponential distribution claim sizes of mean $\mu$, minimizing the logarithm of the negative expected exponential utility gives $H^{\prime }(\alpha )=\lambda \beta \mu [(1-\beta \mu \alpha {)}^{-2}-(1+{\theta }_R)]$. Its strictly positive second derivative makes the displayed stationary point unique. On $0\le \alpha \le 1$, the optimum is $\min (1,{\alpha }_{\ast })$. On the strictly open interval $0<\alpha <1$, there is no maximizer when ${\alpha }_{\ast }\ge 1$, only a supremum at full retention. The stationary retention increases with reinsurer loading and is independent of the direct insurer's loading.

Three nodes have integer link capacities $C_{12},C_{13},C_{23}$. Calls between a node pair use either its direct link or the two-link alternative. With unrestricted instantaneous rearrangement of existing calls, counts $n_{12},n_{13},n_{23}$ are feasible exactly when $n_{12}+n_{13}\le C_{12}+C_{13}$, $n_{12}+n_{23}\le C_{12}+C_{23}$ and $n_{13}+n_{23}\le C_{13}+C_{23}$. Necessity follows by counting calls crossing each node cut. For sufficiency, the inequalities allow at most one count to exceed its direct-link capacity. Divert precisely that excess to the other two links; the same inequalities ensure capacity there. The counts thus have the same feasible set as a fixed routing loss network with one virtual resource at each node, capacity equal to the sum of its two incident physical-link capacities, and each call using its two endpoint resources. With independent Poisson processes for arrivals and independent routing-independent exponential distributions for holding times, this gives the usual product-form stationary distribution of a loss network and exact Poisson arrivals see time averages blocking probabilities.

If every interspike interval is at least $t_{\mathrm{abs}}>0$, a sustained spike train has mean firing rate at most $1/t_{\mathrm{abs}}$. Across a window of duration $T$, the event count satisfies $N(T)\le 1+T/t_{\mathrm{abs}}$; dividing by $T$ and taking the long-time limit gives the bound. The extra one accounts for a possible first event. The bound permits much slower firing and says nothing about whether stronger drive can maintain the limiting rate in a real neuron.

For the chosen phenomenological paired-pulse recovery, the hold alone gives $500\mathrm{Hz}$. Add the further assumption that every successful event resets precisely the same recovery curve, with no accumulation from earlier events. Then $0.35\mathrm{nA}$ onset-timed pulses can sustain at most $1000/4.942488\approx 202.327\mathrm{Hz}$ in that model. This repeatable reset is an extra assumption: a two-pulse result alone cannot establish it. A real neuron's sustained rate also depends on neural adaptation, input delivery and evolving ion-channel inactivation. The existing leaky integrate-and-fire model similarly adds integration time to its imposed hold.

A homogeneous Poisson process of rate $\lambda$ has interspike intervals with an exponential distribution, soWith the purely illustrative choices $\lambda =100\mathrm{Hz}$ and $t_{\mathrm{abs}}=0.002\mathrm{s}$, this is $1-e^{-0.2}\approx 0.181269$. Thus about $18.13\%$ of intervals predicted by that model violate this chosen hold. Matching a mean spike train count does not make a stimulus-only Poisson process respect recent firing.

A minimal alternative is a renewal process with independent and identically distributed intervals $S=t_{\mathrm{abs}}+X$, where $X$ has an exponential distribution of rate $\lambda$. Its hazard function is zero during the hold and $\lambda$ afterwards. Its mean interval is $t_{\mathrm{abs}}+1/\lambda$ and long-run firing rate isThe same $100{\mathrm{s}}^{-1}$ rate after a $2\mathrm{ms}$ hold gives $83.333\mathrm{Hz}$, not $100\mathrm{Hz}$. This separate stochastic example has no gradual relative recovery; it illustrates why an available-state rate differs from an observed mean firing rate. Smooth dependence on time since the last event can extend the hazard function, while longer histories require more than this simple renewal process. Time-rescaled spike-train diagnostics can test the resulting history-dependent predictions against recorded spike trains.