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The state probabilities satisfy $p_0^{\prime }=-\lambda p_0$ and $p_n^{\prime }=\lambda p_{n-1}-\lambda p_n$, yielding the Poisson law of mean $\lambda t$.

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.

The difference of two independent rate-$\lambda$ Poisson processes is a continuous-time simple symmetric random walk with total jump rate $2\lambda$ and variance $2\lambda t$ at time $t$.

Independently retaining a point at $x$ with probability $p(x)$ thins a Poisson point process of intensity $\lambda (x)$ to one of intensity $p(x)\lambda (x)$; retained and rejected processes are independent.

Independently assigning a mark to each point of a Poisson point process, with mark probabilities allowed to depend measurably on the point, produces independent Poisson point processes for the different marks. If the original intensity is $\lambda (x)dx$ and mark $j$ has probability $p_j(x)$, its intensity is $\lambda (x)p_j(x)dx$.

Let $\lambda$ be a nonnegative measurable intensity function on a space with reference measure $dx$. A Non-homogeneous Poisson point process has independent counts on disjoint measurable sets andwhenever the integral is finite.

The union of independent Poisson point processes with diffuse intensity measures ${\mu }_1,\dots ,{\mu }_m$ is a Poisson point process with intensity measureOn every finite collection of disjoint measurable sets, this follows by adding independent Poisson counts and checking that the resulting counts remain independent.

In three dimensions the Poisson kernel for the upper half-space isIt gives the harmonic function $\varphi (x,y,z)={\int }_{{\mathbb{R}}^2}{P}_z(x-\xi ,y-\eta )f(\xi ,\eta )d\xi d\eta$ for compatible boundary data. The method of images constructs the zero-boundary Dirichlet Green function by subtracting the reflected-source Green function of the Laplacian from the source kernel. On the boundary plane the outward unit normal is $-e_z$, and $-{\partial }_{n}G=P_z$, giving the integral formula through Green's second identity.

The kernel is positive and has integral one: in polar coordinates this is $z{\int }_0^{\infty }r(r^2+z^2{)}^{-3/2}dr=1$. Its mass outside any fixed disk tends to zero as $z\downarrow 0$. It is therefore an approximate identity, recovering bounded boundary data at continuity points. At a jump across a smooth curve, the vertical limit gives the average of the two adjacent values; assigning a different value only on that curve does not change the harmonic extension.

The function $D=q\cdot p$ generates the canonical scaling $(q,p)\mapsto (e^{t}q,e^{-t}p)$. An observable $F$ is invariant under this scaling exactly when

Along a trajectory generated by the Hamiltonian $H$,Thus a time-independent observable is conserved exactly when its Poisson bracket with $H$ vanishes.

The Poisson bracket satisfies

A hyperbolic half-turn is the unique orientation-preserving isometry of order two fixing a given point. It exchanges every pair of points lying at equal opposite distances along a geodesic through its centre.

Reflection in a hyperbolic line is the unique orientation-reversing hyperbolic isometry fixing that line pointwise. The product of reflections in intersecting lines is a rotation through twice their angle.

PSL2(R) is $S{L}_2(\mathbb{R})/\{\pm I\}$. It acts on the upper half-plane by the Möbius transformationsand is the group of orientation-preserving hyperbolic isometries.

The geodesics in the upper half-plane are vertical lines and semicircles centred on the boundary axis. Indeed, the length integrand $\sqrt{1+y^{\prime 2}}/y$ has the Beltrami identitywhich integrates to $(x-c{)}^2+y^2=R^2$. Swapping the coordinate axes gives circles centred on the vertical boundary for the metric $(d{x}^2+d{y}^2)/x^2$.

If $P$ is at hyperbolic distance $\rho$ from the origin, the Poincare-disc geodesic through $P$ perpendicular to $OP$ is represented by a Euclidean circle of radius $\mathrm{csch}\rho$ whose centre is at Euclidean distance $\coth \rho$ from the origin.

The geodesics are diameters and arcs of circles orthogonal to the unit circle. A radial segment is length minimizing becausewith equality for constant angle and monotone radius. Hence the distance from the origin to $z$ is $2\mathrm{artanh}|z|$.

For a piecewise smooth curve $z(t)$ in the Poincare disc, its hyperbolic length is

An isometry sends the centers of a collinear family of equal hyperbolic discs to points on one hyperbolic geodesic. In the upper-half-plane model no geodesic contains three distinct points on a horizontal Euclidean line, since geodesics are vertical lines or semicircles orthogonal to the boundary.

In curvature minus one, a hyperbolic disc of radius $R$ has areaIn the Poincare disc this follows by integrating $4r(1-r^2{)}^{-2}drd\theta$ up to Euclidean radius $\tanh (R/2)$.