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A counting process which, conditional on a random nonnegative locally integrable intensity $\Lambda$, is an Inhomogeneous Poisson process. For $L_t={\int }_0^t\Lambda (s)ds$, conditioning gives $\mathbb{E}N(t)=\mathbb{E}{L}_t$ and $\mathrm{Var}N(t)=\mathbb{E}{L}_t+\mathrm{Var}{L}_t$. Thus finite-variance counts are never underdispersed. Randomness of the intensity may induce dependence between increments which are independent conditionally.

A stimulus passes through a linear temporal or spatial filter, then a nonnegative nonlinear rate function. A conditionally Inhomogeneous Poisson process generates spikes at that rate. The filter and nonlinearity can be fitted from data and tested on held-out trials. Refractoriness and spike-history correlations require extensions beyond this basic cascade.

Fit a stimulus-conditioned linear-nonlinear-Poisson cascade model. Choose a causal stimulus-history window, subtract the stimulus mean, and estimate a temporal filter $k$. A useful initial estimate is the spike-triggered average, $\hat{a}(u)=N^{-1}{\sum }_i[s(t_i-u)-\bar{s}]$. For white Gaussian stimulation and a one-filter response, this estimates the filter direction. With correlated Gaussian stimulation it estimates a direction proportional to $Ck$, where $C$ is the history covariance matrix; use stimulus-correlated spike-triggered averaging to whiten or regularize this correction. An arbitrary stimulus or an even response nonlinearity does not guarantee that this average identifies the correct model.

A general fitting approach is to expand a causal filter in smooth basis functions, compute $a(t)=b+{\int }_0^{U}k(u)s(t-u)du$, and take a nonnegative rate $\lambda (t)=f(a(t))$. With $f(a)=e^a$, fit the coefficients by Poisson spike-train likelihood fitting:The first term rewards rate at observed spikes and the second penalizes overprediction elsewhere. Include a filter-size or smoothness penalty, and choose complexity on held-out data. Another option estimates $f$ by stimulus-projection histograms: divide the density of projection values at spikes by their occupancy density, multiplied by the mean firing rate. Avoid interpreting poorly sampled bins as reliable rates. Temporal train/test splits should avoid leakage through the filter-history window.

Given the fitted rate and the same stimulus, generate an Inhomogeneous Poisson process. For small bins of width $\Delta t$, spike with probability $1-e^{-\lambda (t)\Delta t}$, choosing the bins sufficiently short that multiple events are negligible. More exactly, advance integrated intensity until it increases by an independent unit-exponential random variable, then place the next event. Repeating this generates different synthetic spike trains with the same fitted stimulus-dependent rate, not copies of the original event times.

Compare held-out log-likelihood, stimulus-conditioned firing-rate predictions, count distributions, interspike-interval distributions and autocorrelation. With repeated identical stimulus presentations, compare trial-averaged time courses and trial-to-trial count variance, including the Fano factor. A single recorded train does not by itself supply a reliable repeated-trial average or variance; use held-out temporal data, model-based uncertainty and further recordings instead. Time-rescaled spike-train diagnostics test whether integrated-intensity intervals look independently exponential, rather than comparing raw intervals to an exponential law under a changing rate.

A successful model reproduces specified statistical features on new data; identical spike times are not the goal. The basic Poisson model can miss refractoriness, bursting, adaptation, history dependence or stimulus-unexplained trial variability. If needed, add a spike-history filter to the conditional rate, including an absolute dead time, and fit the resulting history-dependent model. A complex cell may also require multiple filters or quadratic stimulus features: its spike-triggered mean can vanish despite strong tuning. These failures identify model limitations rather than evidence that a random spike generator should match every observed interval.

For a constant-rate Poisson process, let $T$ be the waiting time from a spike to the next spike. By independent increments, there is no spike in an interval of length $u$ with probability $e^{-ru}$. Hence, for $r>0$,Its density is $r{e}^{-ru}$, so the interspike intervals are independent exponential distributions with mean $1/r$. If $U$ is uniform on $(0,1)$, inverse transform sampling gives $T=-\log U/r$. Repeatedly add independent draws to the current spike time to simulate a constant-rate spike train.

For a prescribed time-dependent nonnegative rate $r(t)$, use an Inhomogeneous Poisson process. Conditional on the present event at $t_i$, its next waiting time has survival probabilityThus its conditional CDF is $1$ minus that expression for $u\ge 0$. Define the cumulative intensity $\Lambda (t)={\int }_0^{t}r(s)ds$. For each new independent uniform draw, set $E_i=-\log U_i$ and solveThis transforms independent unit-rate exponential intervals in integrated-intensity time into physical spike times. Use a generalized inverse when the rate vanishes on an interval: no spikes are placed in a flat part of $\Lambda$. On a finite simulation window, stop when the next intensity target exceeds the available $\Lambda$. If the total future intensity is finite, a target beyond it means that no further spike occurs.

Numerically, integrate the known rate and invert the accumulated integral, or integrate exactly across piecewise-constant rate bins. Drawing an exponential interval using only $r(t_i)$ is not exact when the rate changes during the interval. The Poisson model also omits the neuronal refractory period and other spike-history effects.

One useful starting point is reverse correlation, implemented as a spike-triggered average. Present a long stimulus record and retain the stimulus history before each observed spike. After subtracting the stimulus mean, estimatewhere $K$ is the number of spikes and $T_k$ is the history window. This identifies a candidate temporal receptive field. For white Gaussian stimulation and a one-filter encoding model, the average is proportional to the relevant filter, provided the response nonlinearity does not make that average vanish. For example an even response to the projection can have zero average despite strong stimulus dependence. Colored inputs require accounting for stimulus covariance, and more than one relevant stimulus direction may require a richer fitted model.

Use the candidate filter in an LNP model:Estimate the static nonlinearity $g$ by binning filter projections and dividing spike counts by exposure time in each bin. Equivalently, Bayes' rule gives $g(z)=\overline{r}p(z\mid \text{spike})/p(z)$ under this model. Refine the filter and nonlinearity by maximizing the conditional Poisson log-likelihoodwith suitable regularization. Simulate an Inhomogeneous Poisson process at the fitted rate using integrated-intensity inversion from the previous solution. This gives a statistical encoding model, not a claim to reconstruct the neuron's ion channels.

Use cross-validation: estimate parameters on one set of trials and test them on independent held-out trials. Under repeated identical stimuli, compare the observed and simulated trial-averaged firing rates, spike-count distributions, interspike-interval distributions and autocorrelations. Compare held-out likelihoods and predictive errors with simpler baseline models and with the variability between real repeats. Exact spike-by-spike agreement is not expected from two independent realizations of a stochastic model. Large rate agreement accompanied by incorrect short-interval or correlation statistics reveals missing dynamics. Adding a spike-history term, for example in a point-process generalized linear model, can account for refractoriness or bursting when the basic Poisson cascade fails.

A good model predicts held-out response distributions, not just its training spike train. Such a model summarizes what stimulus features affect the neuron, permits controlled simulated changes of those features, and tests whether the proposed computation is sufficient. Its failures can suggest further experiments or additional stimulus and history variables.

An Inhomogeneous Poisson process with deterministic, nonnegative locally integrable rate $\lambda (t)$ is a counting process $N$ with $N(0)=0$, independent increments, and increment distributionEquivalently, for continuous rates it is a simple counting process with independent increments and infinitesimal probabilities $\mathbb{P}(N(t+h)-N(t)=1)=\lambda (t)h+o(h)$ and $\mathbb{P}(N(t+h)-N(t)\ge 2)=o(h)$. For merely locally integrable rates use the integrated-rate formulation, which avoids a pointwise regularity assumption.

For an Inhomogeneous Poisson process, the likelihood function is the product of the intensities at arrivals times the exponential of minus the integrated intensity. The integrated intensity is $\theta +2(T-\theta )=2T-\theta$. There are $j(\theta )$ arrivals at rate one and $n-j(\theta )$ at rate two, soSet $t_0=0$ and $t_{n+1}=T$ to include changes before the first or after the last arrival. An arrival exactly at the change has probability zero, so the convention for $j$ at that point does not affect the Bayesian posterior.

Synchronous radius and its limits. Kepler's third law for a circumplanetary orbit givesHere $G$ is the gravitational constant. A planet-synchronous orbit repeats after one spin period. A planet-stationary orbit additionally requires a circular orbit, zero orbital inclination to the equator and prograde motion. Then an antenna fixed on the ground points continuously at the same satellite; merely matching the orbital period does not give this property.

Write the planetary radius as $R_p=(3M/(4\pi \rho ){)}^{1/3}$. Requiring the semi-major axis to exceed $R_p$ gives $P^2>3\pi /(G\rho )$. The stellar tidal force restricts the circumplanetary orbit to the Hill sphere, of radius $r_H=a(M/(3M_{\star }){)}^{1/3}$. ThusThese are the surface and Hill-radius constraints in the idealized spherical, small-$M/M_{\star }$ model. Long-term prograde stability generally requires a radius appreciably inside the Hill sphere; for nonzero orbital eccentricity, the surface constraint applies to the periapsis, not just to the semi-major axis.

Collision time and the launch population. A phase-mixed isotropic swarm occupies a shell of radial scale $e{a}_s$. Its number density scales as $N_s/(e{a}_s^3)$, its relative speed as $a_s/P$, and its geometric collision cross-section as $r^2$. Consequently the total collision rate scales as $N_s^2{r}^2/(Pe{a}_s^2)$. To give the numerical normalization used here, adopt an effective shell volume $V=4\pi e{a}_s^3$. At a fixed position the velocity directions are uniformly distributed in the tangent plane, so their mean relative speed is $4v_K/\pi =8a_s/P$. For an unordered pair, the collision cross-section is $4\pi r^2$, givingThus the mean interval between collisions is $A/N_s^2$ in this large-population kinetic model. The effective shell width fixes an order-one coefficient: small orbital eccentricity and random planes alone do not specify a unique radial probability density. Phase mixing, negligible gravitational focusing, $r\ll e{a}_s$, and uncorrelated encounters are implicit in this estimate. Exactly identical orbital periods with perfectly fixed phases do not themselves produce a memoryless collision process.

For nearly planet-stationary orbits with $e\ll I\ll 1$, the swarm volume is smaller by a factor of order $I$, while the relative speed is smaller by the same factor, since vertical motion of scale $I{v}_K$ dominates the eccentric motion. The two changes cancel in the rate $n\sigma v$: there is no parametric factor $I$ in the collision time within the same phase-mixed kinetic approximation. Numerical factors and phase correlations can differ. This is not an argument that bringing all satellites into one nearly circular plane makes their phases random.

With $N_s=Rt$, an Inhomogeneous Poisson process has cumulative hazard functionThe survival function of the first collision is $\exp [-H(t)]$. Setting the expected number of collisions to one givesThis is a characteristic first-event population, with probability $1-e^{-1}$ of an earlier event. It is not a median: the median has an additional factor $(\log 2{)}^{1/3}$.

Which population collides next? Immediately after the first disruption let $S=N_{s0}-2$ and $F=2x^{-3}$. The latter follows from mass conservation for equal-density spherical pieces. Using the same unordered-pair counting and geometric collision cross-sections as above, define $b=(1+x{)}^2/2$; thenHere $S^2,F^2$ stand for the usual large-population approximations to $S(S-1),F(F-1)$. The factor two distinguishing identical and different species is essential. The probability that the next collision is fragment–fragment isIt is the most likely type if $S<4/[x(1+x{)}^2]$ and $S<2/x^2$. It has probability greater than one half ifFor $x\ll 1$, a strongly fragment-dominated next event therefore requires $Sx\ll 4$; the fragment–satellite comparison is more restrictive than the satellite–satellite comparison. If $S=0$, the next event must be fragment–fragment, provided fragments remain.

Population equations and the normalization discrepancy. Every satellite–satellite event produces $2x^{-3}$ fragments and destroys two satellites. Every satellite–fragment event produces a net $x^{-3}-1$ fragments and destroys one satellite; every fragment–fragment event destroys two fragments. Consistent collision counting therefore givesFor $x\ll 1$, $b\simeq 1/2$ and $x^{-3}-1\simeq x^{-3}$. The last two terms in $A{\dot{N}}_f$ are then twice those in the printed equation. This is a genuine factor-of-two inconsistency: equal-size fragment pairs have a collision cross-section smaller by $x^2$, so their event rate must be $x^2{N}_f^2/A$ if the satellite event rate is $N_s^2/A$; destroying both fragments necessarily gives the sink $-2x^2{N}_f^2/A$.

If the printed equation is taken as a prescribed approximate rate model instead, its implicit event rates are $N_s^2/A$, $N_f{N}_s/(4A)$ and $x^2{N}_f^2/(2A)$. Under precisely that mixed normalization, its corresponding satellite equation isThe printed source term also neglects the consumed fragment in a satellite–fragment event, a legitimate relative $O(x^3)$ approximation. That approximation does not repair the pair-counting discrepancy.

The ensuing cascade. This is a two-size fragmentation cascade. First use the printed approximate model, dropping satellite–satellite events as requested. Put $S=N_s$, $F=N_f$, $k=4x^2$ and take $x<1/2$. ThenIntegrating this linear differential equation, with $F_0=2x^{-3}$ and $S_0=N_{s0}-2>0$, givesInitially the collisional cascade grows if $S_0>8x^2$, with approximate early exponential growth time $4A{x}^3/S_0$ when the satellite population is nearly fixed. The fragment population reaches its maximum atAfterwards satellites are depleted and fragment–fragment losses dominate. For $S_0>0$ the continuum solution has $S\to 0$, $F\to 0$, withWhen no satellites remain initially, $F(t)=F_0/[1+x^2{F}_{0}t/A]$ directly. Small integer populations eventually invalidate these deterministic differential equations.

The consistently counted model has, to leading order in $x$, exactly the same curve $F(S)$ and peak, but both retained time derivatives are twice as large. Its growth time is $2A{x}^3/S_0$, and $F(t)\sim A/(2x^{2}t)$. Keeping $(1+x{)}^2$ and the consumed fragment replaces $k$ by $4x^2/(1+x{)}^2$ and $x^{-3}$ in the curve by $x^{-3}-1$. This explicitly separates the physical collision bookkeeping from the printed normalization while giving the evolution under both conventions.

Assume individual incubation delays are independent of the Inhomogeneous Poisson process of infections and of one another. This marking assumption is needed in addition to specifying a marginal incubation density.

In the discrete cohort model let $N_i\sim \mathrm{Poisson}(h_i)$ be the independent infection counts. Mark each infection by its eventual onset interval. For cohort $i$, the categories have probabilities $q_{ik}$ for observed intervals, with the remaining probability assigned to onsets outside the window. Poisson thinning makes the counts $N_{ik}$ in these categories mutually independent Poisson random variables with means $h_i{q}_{ik}$. One direct proof is their probability generating function:Different cohorts are independent. Summing their counts therefore gives independent Poisson onset counts in disjoint intervals:The same argument applies exactly in continuous time by the Independent marking theorem for Poisson point processes and mapping each marked infection to its onset time. The endpoint model approximates its means; its independent-Poisson conclusion is exact within that discrete model. A fixed cohort size would instead induce negatively correlated onset-bin counts, so the Poisson process infection assumption matters.

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 an Inhomogeneous Poisson process whose known positive intensity changes from ${\lambda }_1$ to ${\lambda }_2$ at an unknown time $\theta \in (0,T)$, the ordered arrival-time likelihood function isUnder a uniform prior, each interval between successive arrivals has an exponential posterior density with rate in the exponent ${\lambda }_2-{\lambda }_1$. Its integrated weights allow exact interval selection followed by a truncated exponential draw; when the two rates coincide, the Bayesian posterior is uniform and the data contain no change-time information.

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.

For a stimulus-conditioned Inhomogeneous Poisson process, the event log-likelihood is the displayed expression. A causal filter expansion $\lambda (t)=\exp [b+{\sum }_a{w}_a(s\ast b_a)(t)]$ can be fitted by maximum likelihood estimation, with a smoothness or size penalty and held-out validation. The integral term penalizes predicted spikes that did not occur; fitting only stimulus values at observed spikes omits this essential part.

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.