The Poisson kernel on the circle is nonnegative, has normalized integral one, and its mass outside any fixed arc about zero tends to zero as $r\uparrow 1$. Split the convolution error into a short arc, controlled by the modulus of continuity, and its complement, controlled by the vanishing kernel mass. This proves uniform convergence for every continuous periodic function. Its Fourier multiplier is $r^{|n|}$, so the convolution is the Abel sum of the Fourier series.

For a meromorphic function near $0<|z|\le 1$, and a target $a$ off its image of the unit circle, let $u_a=-\log k(f,a)$ for the chordal metric, $M_a(s)$ its normalized angular mean, and $I(a)=M_a^{\prime }(1)$. The weighted spherical area $T(r)=(4\pi {)}^{-1}{\int }_{r<|z|<1}\log (|z|/r)(f_{\mathrm{round}}^{\#}{)}^{2}dA$ satisfies the displayed formula, where $N$ counts $a$-points with multiplicity and weight $\log (|z|/r)$ and $\overline{m}=M_a(r)-M_a(1)$. Indeed $\Delta u_a=(f_{\mathrm{round}}^{\#}{)}^2/2-2\pi \sum {\mathrm{deg}}_z(f){\delta }_z$, and Green's second identity with the radial logarithmic weight proves the formula. The angular mean must include $1/(2\pi )$.

If the positive baseline rate has a gamma distribution of shape $a$ and rate $b$, its logarithm has this probability density function with respect to $d\mu$. The extra factor $e^{\mu }$ is the Jacobian determinant of the transformation. After a Gibbs sampler draws the conditional baseline rate, taking its logarithm is an exact update of the intercept. Omitting the Jacobian changes the prior.

Let independent counts satisfy $Y_i\sim \mathrm{Poi}(\theta w_i)$, with positive known exposures $w_i$, $\theta >0$, and $a_n={\sum }_{i=1}^n{w}_i\uparrow a_{\infty }<\infty$. Their likelihood is proportional to ${\theta }^{S_n}{e}^{-a_n\theta }$, where $S_n={\sum }_{i=1}^n{Y}_i$. If $S_n>0$, the positive maximum-likelihood estimator is $S_n/a_n$; when $S_n=0$, no maximum is attained on the open parameter space, while the closed-space maximizer is zero. This extended estimate converges almost surely to $S_{\infty }/a_{\infty }$, where $S_{\infty }\sim \mathrm{Poi}(\theta a_{\infty })$. Finiteness follows from $\mathbb{E}{S}_{\infty }=\theta a_{\infty }$, and the limiting Poisson distribution follows from the finite-sum laws. In particular, the probability of a zero estimate tends to $e^{-\theta a_{\infty }}>0$, proving failure of statistical consistency. The total Fisher information tends to $a_{\infty }/\theta$: infinitely many observations need not provide infinite information.

A rate ratio compares event counts per unit exposure between two conditions. In a Poisson regression with a log link and binary indicator coefficient $\beta$, it equals $e^{\beta }$. Exponentiating an approximate Wald confidence interval for $\beta$ gives an interval for the rate ratio. A rate ratio is distinct from a relative risk or an odds ratio.

The Poisson equation for differential forms on a closed oriented Riemannian manifold is solvable exactly when its source is $L^2$-orthogonal to every harmonic differential form of the same degree. Necessity follows from the formal adjoint identity $⟨\Delta \alpha ,h⟩=⟨\alpha ,\Delta h⟩=0$. For sufficiency, the Green operator of the Hodge Laplacian gives $\alpha =G\beta$, since $\Delta G=I-P_{\mathcal{H}}$.

For $N$ with Poisson distribution of mean $\lambda$, the equality $n\mathbb{P}(N=n)=\lambda \mathbb{P}(N=n-1)$ proves the identity by summation. For an independent random sum of independent claims $S$ with this count, exchangeability of the summands gives $\mathbb{E}[Sh(S)]=\lambda \mathbb{E}[Xh(S+X)]$, where $X$ on the right is independent of $S$ and has the severity law. These identities hold for nonnegative measurable functions, and for integrable signed functions. They obtain moments without assuming that a positive exponential moment exists.

The Stein-Chen method compares an integer-valued count with a Poisson distribution by solving the Poisson Stein equation and estimating its expectation at the count. For sums of indicator random variables, dependency neighborhoods expose local overlap errors. This gives total variation distance bounds without having to compute the entire count distribution.

The information entropy $F(\lambda )$ of a Poisson distribution is strictly increasing for $\lambda >0$. Given ${\lambda }_2>{\lambda }_1$, take independent $U\sim \mathrm{Pois}({\lambda }_1)$ and $V\sim \mathrm{Pois}({\lambda }_2-{\lambda }_1)$. Their sum has Poisson distribution with mean ${\lambda }_2$, so entropy monotonicity under independent addition gives $F({\lambda }_2)\ge F({\lambda }_1)$. All these information entropies are finite, and $\mathrm{Cov}(U+V,V)=\mathrm{Var}(V)>0$ excludes equality.

// Target: probability-and-statistics.bigb

A mixture of Poisson distributions over a nonnegative random intensity. With exposure $n>0$ and prior density $\pi (\theta )$, its probability mass function is $p_n(s)={\int }_0^{\infty }{e}^{-n\theta }(n\theta {)}^s\pi (\theta )d\theta /s!$. A gamma mixing distribution gives a negative binomial distribution. Counts sharing the same latent intensity are conditionally independent, but generally dependent after mixing.

The distribution of $N_{+}-N_{-}$ for independent Poisson random variables with means ${\lambda }_{+},{\lambda }_{-}$ is the Skellam distribution. Its characteristic function is $\exp ({\lambda }_{+}(e^{iu}-1)+{\lambda }_{-}(e^{-iu}-1))$, its mean is ${\lambda }_{+}-{\lambda }_{-}$ and its variance is ${\lambda }_{+}+{\lambda }_{-}$. For equal means $\lambda >0$,This is the fixed-time law of a symmetric Poisson difference process.

// Target: markov-process.bigb

For one acoustic half-space and vacuum on the other side, a harmonic point force gives displacement amplitude $F\gamma /D$ and fluid acoustic velocity potential amplitude $i\omega F{e}^{-\gamma y}/D$. For time dependence $e^{-i\omega t}$, the outgoing acoustic square-root branch has $\gamma >0$ for $|k|>k_0$ and $\gamma =-i\sqrt{k_0^2-k^2}$ for $|k|<k_0$. The causal limit $\omega \mapsto \omega +i0$ places the positive guided-mode pole above the inversion contour and the negative pole below. The saddle point generates cylindrical radiation, whereas guided residues decay normally to the membrane.

For the Boussinesq top-hat plume in a stratified ambient with $N^2=C{z}^p$, the exact pure-power solution has the displayed radius and velocity, and $g^{\prime }=-2C{z}^{p+1}/(3p+8)$. With positive entrainment, radius, upward speed and reduced gravity, stable $C>0$ requires $-4<p<-8/3$, whereas unstable $C<0$ requires $p>-8/3$. The stable branch has divergent buoyancy flux at zero height, so it is not a finite-flux point-source solution. The unstable branch has zero limiting source buoyancy flux and extracts buoyancy from its maintained unstable ambient. The cases $p=-6,-4,-8/3$ are singular at nonzero $C$; $C=0$ instead admits the ordinary Boussinesq point-source plume.

For prescribed $N^2=-G^2<0$, the steady Boussinesq top-hat plume in a stratified ambient has kinematic fluxes $Q={\alpha }^{2}G{z}^3/9$, $M={\alpha }^2{G}^2{z}^4/36$, and $F={\alpha }^2{G}^3{z}^4/36$. Balancing powers in $Q_z=2\alpha \sqrt{M}$, $M_z=FQ/M$, and $F_z=G^{2}Q$ gives exponents $(3,4,4)$ and these positive amplitudes. The ambient's static instability is prescribed without allowing it to drive ambient motion; the Boussinesq approximation additionally requires small density differences over the modeled height range.

The intersection of a line in Plücker coordinates with a plane covector $F$ is the first displayed homogeneous vector: it lies in the span of the defining points and its contraction with $F$ is zero by antisymmetry. A zero vector means the whole line is contained in the plane. Dually, the plane through the line and a point or ideal direction $D$ is given by the second formula; it vanishes if $D$ is already on the line. These operations require no Euclidean lengths or angles.

For $\Phi =-\mu /r+{\alpha }^2\mu (3z^2-r^2)/r^5$, a nearly circular equatorial orbit has $n^2=\mu /R^3$ and azimuthal frequency ${\Omega }_c^2=n^2(1+3{\alpha }^2/R^2)$. The radial epicyclic frequency is ${\kappa }^2=Rd{\Omega }_c^2/dR+4{\Omega }_c^2$, and the vertical epicyclic frequency is ${\nu }^2={\partial }_z^2\Phi {|}_{z=0}$. Radial stability requires $R^2>3{\alpha }^2$. The nodal precession rate ${\Omega }_c-\nu$ is negative and the apsidal precession rate ${\Omega }_c-\kappa$ positive. These sign statements concern small eccentricity and inclination. For an arbitrary inclined orbit, the averaged argument of periapsis instead evolves as $(3/2)n{\alpha }^2[5{\cos }^{2}i-1]/[a^2(1-e^2{)}^2]$ and can regress.

For a sharp thermal front moving at fraction $\Gamma$ of the Darcy velocity $U$, a cold fluid of viscosity $\mu$ displacing its hot counterpart of viscosity $b\mu$ has the displayed growth rate in the local, unbounded planar approximation without thermal conduction. The harmonic pressure perturbations decay on either side. Pressure continuity and continuity of perturbed Darcy flux give the velocity response $\delta U=U|a|(b-1)\eta /(b+1)$; the thermal front's kinematic law is ${\eta }_t=\Gamma \delta U$. The prefactor is the thermal-front speed, not the material pore velocity. Increasing viscosity with heating therefore makes the trailing thermal interface unstable even when the leading displacement of the original fluid is stable.

The complete graph $K_5$ has $n=5$ and $m=\left(\genfrac{}{}{0ex}{}{5}{2}\right)=10$, exceeding the planar graph edge bound $3\cdot 5-6=9$. Therefore it is not a planar graph. The complete bipartite graph $K_{3,3}$ has $n=6$ and $m=3\cdot 3=9$. The general bound allows twelve edges, but the bipartite planar graph edge bound allows only eight, proving nonplanarity.

These are necessary bounds, not a complete planarity test. For an original counterexample, subdivide every edge of $K_{3,3}$ once. The resulting simple graph is a connected graph with $n=15$, $m=18$, and is bipartite: put all original vertices in one part and all nine new vertices in the other. Both inequalities hold, $18\le 39$ and $18\le 26$. Nevertheless, a crossing-free graph drawing would let us trace each two-edge graph path as one curve, producing an impossible embedding of $K_{3,3}$. The graph subdivision preserves this obstruction. Kuratowski's theorem supplies a stronger characterization than these counts.

If a finite simple graph is a connected graph, a bipartite graph, and a planar graph, with $n\ge 3$ vertices, then $m\le 2n-4$. Every facial boundary walk has even length and length at least three, hence at least four. Thus $4f\le 2m$; substitute the Euler formula for a connected planar graph.^[ref]

The square cycle graph attains equality: $4=2\cdot 4-4$. The three-vertex graph path also attains equality: $2=2\cdot 3-4$. Its repeated bridges in a graph illustrate why counting facial boundary walk length correctly matters. Neither bound applies unchanged to a multigraph: two vertices may support arbitrarily many parallel edges in a plane graph.

A planar Brownian motion range has zero two-dimensional Lebesgue measure almost surely. Assuming finite expected area, Brownian scaling gives half that expected area for each half-time range. Inclusion-exclusion then makes their expected overlap zero. Translate their ranges by the midpoint position; the independent Brownian arms at a deterministic time make the expected overlap the integral of the squared point-hitting probabilities for a half-time range. Those probabilities vanish almost everywhere, so the half-time expected area, and hence the full-time expected area, vanish. This argument does not assert zero hitting probability for every individual point solely from an almost-everywhere integral conclusion.