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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.

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.

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.

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.

For a homogeneous Poisson process of rate $\lambda >0$, $N(t)/t\to \lambda$ almost surely: unit-time increments are independent Poisson variables and the strong law of large numbers applies, then monotonicity fills the intervals between integers. At its $n$th point $Y_n$, simplicity gives $N(Y_n)=n$. Consequently $Y_n/n\to 1/\lambda$. Rate and mean interarrival time are reciprocals.

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.

Let $T_n=\inf \{t\ge 0:N_t\ge n\}$ be the $n$th arrival time, and set $T_0=0$. In a Homogeneous Poisson process of rate $\lambda >0$, the gaps $W_n=T_n-T_{n-1}$ are independent and identically distributed random variables with an exponential distribution:Here $\mathrm{Exp}(\lambda )$ uses the rate convention. Under the scale convention $\theta =1/\lambda$, the same probability density function is ${\theta }^{-1}{e}^{-w/\theta }$; a simulation function accepting a scale must receive $1/\lambda$. Conversely, summing independent rate-$\lambda$ exponential random variables to form $T_n$ and counting those sums constructs a Homogeneous Poisson process.^[ref]

The first-gap formula follows immediately from $\{W_1>w\}=\{N_w=0\}$ and the Poisson distribution. Independence of all successive gaps is the additional interval-theorem assertion; deterministic-time independent increments alone should not simply be substituted at random arrival times without justification. The memorylessness of the exponential distribution givesAt the toy rate of two per minute, the mean gap is half a minute, the median is $(\log 2)/2\approx 0.346574$ minutes, and the probability of waiting longer than half a minute is $e^{-1}\approx 0.367879$. Here $\log$ is the natural logarithm. Already having waited a minute without an arrival leaves this same conditional probability of waiting another half minute. The exponential distribution describes a model with no aging of the remaining wait; a scheduled arrival with a fixed gap has no such property.

Let $\Pi$ be a Poisson point process with intensity measure $\mu$ and let $f$ be measurable. If the pushforwardis locally finite, then the image counting measure is Poisson with intensity $f_{\ast }\mu$. If one requires a simple point process, the pushforward must also be diffuse, so distinct source points do not produce coincident image points with positive probability.

Let $N$ be a Poisson point process with intensity measure $\mu$, and let $g$ be a real-valued measurable function. For the ordinary Poisson sum $S_g=\int gdN$, the condition $\int |g|d\mu <\infty$ givesIf also $\int g^{2}d\mu <\infty$, thenExpand the square: equal-index terms contribute $\int g^{2}d\mu$, while the Poisson factorial moment measure makes distinct-index terms contribute $(\int gd\mu {)}^2$. Subtracting the squared expected value proves the variance formula. The Campbell first-moment formula also holds for general point processes; this variance identity uses the Poisson assumption.^[ref]

Square integrability alone is insufficient for the ordinary sum on an infinite-intensity space. Take intensity $dx$ on $[1,\infty )$ and $g(x)=1/x$. Then $\int g^{2}dx=1$, but the sum is infinite almost surely: the ordered points satisfy $Z_i/i\to 1$ by the strong law for Poisson arrival times, so comparison with the harmonic series proves divergence.

For this same $g$, the compensated Poisson integral exists as a mean-zero limit under convergence in Lp with $p=2$ and has variance $1$. Its definition is the L2 construction of compensated Poisson integrals, rather than subtraction of two infinite ordinary integrals. When the total intensity is finite, Cauchy-Schwarz inequality makes L2 imply L1, explaining why the shorter variance statement works in that special case.^[ref]

A Poisson germ-grain model assigns a measurable region to each point of a Poisson point process and takes the union of those regions. A location remains uncovered exactly when no Poisson point has a region containing it, so its coverage probability is determined by the intensity of the corresponding set of germs and marks.

Place equal masses at points of a homogeneous Poisson point process of intensity $\rho$ on the line, and truncate their signed inverse-power forces symmetrically at radius $R$. For force $m\mathrm{sign}(x)/|x{|}^p$, with $\alpha =1/p\in (0,2)$, the characteristic exponent tends to $-2\rho {\int }_0^{\infty }(1-\cos (mt/x^p))dx$. The cosine integral for a symmetric stable exponent evaluates it as displayed. Thus the weak limit is a symmetric stable distribution. Symmetric truncation is important when the far-field force series is not absolutely summable.

For a measurable map $f$ and a Poisson random measure of intensity $\nu$, image counts in disjoint sets equal original counts in their disjoint preimages. They are therefore independent Poisson variables with the pushforward intensity. The result is a Poisson random measure even if images collide; diffuse pushforward intensity gives a simple point process.

For a Poisson point process of intensity measure $\mu$, its $k$th Poisson factorial moment measure is ${\alpha }^{(k)}={\mu }^{\otimes k}$. Writing its counting measure as $N={\sum }_i{\delta }_{Z_i}$, the identity means that for nonnegative measurable function $g$,The indices select distinct point occurrences. Their locations need not be different when $\mu$ has an atom of a measure. For a simple point process, the locations are different as well.^[ref]

For example, if all intensity is concentrated at one point $a$, with $\mu (\{a\})=\lambda$, then $N(\{a\})$ has Poisson distribution. There are $N(\{a\})(N(\{a\})-1)$ ordered pairs of distinct occurrences there, with expected value ${\lambda }^2$, even though both locations equal $a$. Counting only pairs of different locations would give zero and fail the identity.

For $k=2$, separating equal-index terms from distinct-index terms gives the second-moment calculation in Campbell's theorem. Dividing the distinct-pair sum by two instead counts unordered pairs, as in the intersection point process of a Poisson line process.

Identify an unoriented line not through the origin by $L(p,\theta )=\{z:z\cdot (\cos \theta ,\sin \theta )=p\}$, with $p>0$. A Poisson point process on this parameter space is a random collection of lines. The measure $cdpd\theta$ is invariant under rotations and translations; signed coordinates $t\in \mathbb{R}$, $0\le \theta <\pi$ give the same measure $cdtd\theta$. The number of lines meeting a disk of radius $r$ is Poisson with mean $2\pi cr$.

For the Poisson equation $\Delta u={\sum }_{n\ge 1}{b}_n\sin (n\theta )$ on the unit disk with zero Dirichlet boundary condition, separation of variables in polar coordinates gives $r^2{R}_n^{\prime \prime }+r{R}_n^{\prime }-n^2{R}_n=r^2$. For $n\ne 2$, boundedness at $r=0$ and $R_n(1)=0$ select $R_n=(r^n-r^2)/(n^2-4)$. The resonant mode is $R_2=r^2\log r/4$. For discontinuous angular data, the Poisson equation is interpreted in its weak formulation; its Dirichlet energy gives uniqueness. Orthogonality of the angular Fourier modes gives $\int ufdA=-{\sum }_n\pi b_n^2/[4(n+2{)}^2]$, with the resonant case obtained directly or by the limit $n\to 2$.

The full conditional distributions are $\lambda \mid m,x\sim \mathrm{Gamma}(\alpha +S_m,\beta +m)$ and $\varphi \mid m,x\sim \mathrm{Gamma}(\gamma +S_N-S_m,\delta +N-m)$, in shape-rate notation. Conditional on the two rates, $m$ has weights $w_j=\exp [j(\varphi -\lambda )+S_j\log (\lambda /\varphi )]$. A Gibbs sampler cycles through these three draws. Each update preserves the joint posterior distribution; subtracting the largest log weight before exponentiation stabilizes the categorical draw. Averages of $1_{\{m=N\}}$ estimate the posterior probability of no observed change.

Integrating the two rates out of the Poisson count change-point model with gamma priors gives these weights up to a common factor, by the defining Gamma function integral. Normalizing across the finitely many possible indices gives the exact marginal posterior distribution of $m$, providing a check on Markov chain Monte Carlo estimates. One can also draw $m$ from these weights and then draw both rates from their conditional Gamma distributions, producing independent exact joint posterior samples.

Start with $Y_0\sim \mathrm{Poisson}(\theta )$ and conditionally let $Y_i\sim \mathrm{Poisson}(\theta Y_{i-1})$. The mean generation size is ${\theta }^{i+1}$. The log-likelihood is $S_n\log \theta -\theta T_n$ plus a parameter-free term, where $S_n={\sum }_{i=0}^n{Y}_i$ and $T_n=1+{\sum }_{i=0}^{n-1}{Y}_i$. Expected negative curvature gives the displayed Fisher information, bounded as the horizon grows. Extinction makes the observations eventually all zero, so one longer observed family does not supply indefinitely increasing information.

// Target: convergence-of-random-variables.bigb

Let $h$ be central and a non-zero-divisor in an associative algebra, with commutative quotient modulo $h$. Dividing commutators by $h$ and reducing modulo $h$ is independent of the lifts, since changing a lift by $hc$ changes the quotient by a commutator, zero modulo $h$. Antisymmetry, the Leibniz rule and the Jacobi identity descend from the corresponding commutator identities. For the order filtration of a Weyl algebra, the convention is $\{{\xi }_i,x_j\}={\delta }_{ij}$, with the opposite sign for $\{x_i,{\xi }_j\}$.

The optical transfer function is the Fourier transform of the normalized intensity point spread function for incoherent imaging with translation invariance, using spatial frequency in cycles per unit length. It has $H(0)=1$. For an ideal circular pupil it is the normalized pupil autocorrelation, with cutoff $2\mathrm{NA}/{\lambda }_0$ in object coordinates: no overlap remains when two pupil copies are shifted apart by more than one pupil diameter.^[ref]