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The Vandermonde product ${\prod }_{i<j}(r_i-r_j)$ is alternating, and its square is the discriminant of the monic polynomial with roots $r_i$.

A primitive polynomial has unit content.

Gauge fixing worldsheet diffeomorphism and Weyl symmetry introduces an anticommuting vector ghost $c^{\alpha }$ and a symmetric traceless antighost $b_{\alpha \beta }$. In complex coordinates they form holomorphic and antiholomorphic $bc$ systems.

Conformal gauge uses worldsheet diffeomorphisms and Weyl transformations to set $g_{\alpha \beta }=e^{2\varphi }{\eta }_{\alpha \beta }$ locally.

A Weyl transformation locally rescales a metric tensor, $g\mapsto e^{2\omega }g$, while retaining the same points and coordinates. Applied to a worldsheet metric, it leaves the embedding $X^{\mu }$ unchanged, and the classical Polyakov action is invariant under these Weyl transformations. Applied to a spacetime metric, a Weyl rescaling can relate different frame conventions, such as a string metric and an Einstein metric; the accompanying field normalizations must be transformed as well. It is not itself a coordinate transformation.

A worldsheet diffeomorphism is a smooth reparameterization of the worldsheet coordinates under which $X^{\mu }$ is a scalar and $g_{\alpha \beta }$ is a rank-two tensor.

With $J_0=0$, the age $A(t)=t-J_{N_t}$ of a rate-$\lambda$ Poisson process has meanGiven $N_t=n$, the last arrival is the maximum of $n$ uniform points and $\mathbb{E}[A(t)\mid N_t=n]=t/(n+1)$, including $n=0$.

Let $J_i$ be the points of a rate-$\lambda$ Poisson process and let independent identically distributed marks $X_i$ be independent of the process. ThenCondition on $N_t$, use uniform order statistics, and sum the resulting exponential series.

For independent Poisson processes of rates ${\lambda }_1$ and ${\lambda }_2$, the number $K$ of type-two arrivals before the first type-one arrival satisfiesThis follows by marking the superposed process by its arrival type.

Conditioned on $N_t=n$, the arrival times are distributed as the order statistics of $n$ independent uniform points on $[0,t]$. Their conditional joint density onis $n!/t^n$, obtained by dividing the Poisson-process density ${\lambda }^n{e}^{-\lambda t}$ by $\mathbb{P}(N_t=n)$.

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.

The upper-half-space Poisson kernel gives the harmonic extension of boundary data.

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.