Poisson process

A rate-$\lambda$ Poisson process has independent stationary increments. Conditional on $N_t=n$, arrival times are the order statistics of $n$ independent uniform points on $[0,t]$.

An inhomogeneous Poisson process with locally integrable intensity $\lambda (t)\ge 0$ has independent increments andfor $0\le s<t$. Its cumulative intensity is $\Lambda (t)={\int }_0^t\lambda (u)du$.

If $\lambda >0$ and $g={\Lambda }^{-1}$, then $M_t=N_{g(t)}$ is a rate-one Poisson process, becauseConversely, $N_t=M_{\Lambda (t)}$, so $N_t$ has a Poisson distribution of mean $\Lambda (t)$.

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 time $s$ with probability $p(s)$ gives a Poisson count of mean $\lambda \int p(s)ds$.

Mark each member of an independent Poisson infection cohort by its symptom-onset interval, using independent delays. Poisson thinning gives independent onset-bin counts, and sums over independent cohorts remain independent Poisson counts. Fixed cohort totals instead induce multinomial dependence. The marking assumptions are essential.

Conditioned on a total of $n$ Poisson points, the number falling in a region of relative intensity $p$ is $\mathrm{Binomial}(n,p)$.

With empty initial state, Poisson arrivals of rate $\lambda$ and exponential lifetimes of rate $\mu$ produce a Poisson occupancy of mean $\lambda (1-e^{-\mu t})/\mu$ at time $t$.

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 $\lambda t$.

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 $\mathbb{P}(N_t=n)$.

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.

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.

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 $n=0$.