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Let $X_t=N_t^{+}-N_t^{-}$, where $N^{+}$ and $N^{-}$ are independent Poisson processes of rate $\lambda$. This symmetric Poisson difference process is centered, has jumps $\pm 1$, and has variance rate ${\sigma }^2=2\lambda$. If $a,b$ are positive integers, then $X_T\in \{-a,b\}$ exactly. Optional sampling of the martingale $X_t$ givesConsequentlyPart (d) now gives the mean exit time

For the Poisson process,Differentiating at zero gives

For ordered times $0<t_1<\cdots <t_n<t$, the joint density together with $N_t=n$ is $e^{-\lambda t}{\lambda }^n$. Dividing by $P(N_t=n)=e^{-\lambda t}(\lambda t{)}^n/n!$ gives $n!/t^n$, exactly the density of uniform order statistics. See Poisson process.

The Poisson arrivals see time averages property says that an external Poisson process independent of a stationary system sees the system-state distribution at its arrival times. Thus the fraction of arrivals that accept a state-dependent admission rule $p(n)$ is ${\sum }_n{\pi }_{n}p(n)$.

Sample a finite-state continuous-time Markov chain with generator $Q$ at the independent arrival times of a rate-$\lambda$ Poisson process. The sampled states form a discrete-time Markov chain with transition matrixIts invariant distributions are exactly those of the continuous-time chain.

Independent claim Poisson processes of rates ${\lambda }_i$ merge into a Poisson process of rate $\lambda ={\sum }_i{\lambda }_i$, by the Superposition theorem for Poisson point processes. The merged claim law is a mixture distribution of the individual claim laws with weights ${\lambda }_i/\lambda$. Equivalently, multiplication of the individual compound-Poisson transforms produces $\exp (\lambda [{\sum }_i({\lambda }_i/\lambda )M_i(r)-1])$.

A jump process with event types $k$, increments ${\nu }_k$, and rates $a_k(X)$ can be represented as $X(t)=X(0)+{\sum }_k{\nu }_k{P}_k({\int }_0^t{a}_k(X(s-))ds)$, with independent unit-rate Poisson processes. This represents the Markov jump-process generator through time-changed counts.

In the classical risk model with premium rate $c>0$ and claim Poisson process rate $\lambda$, the first-claim decomposition for survival probability implies the displayed equation for ultimate survival probability. Change the first-claim integral to an integral over available capital and differentiate its lower limit. The bounded survival probability function convolved with the integrable claim probability density function is continuous, so this also establishes the needed differentiability. In particular ${\phi }^{\prime }(0)=(\lambda /c)\phi (0)$.

The ultimate survival probability satisfies $\phi (u)=\phi (0)+(\lambda /c){\int }_0^u\phi (u-x)(1-F(x))dx$. Condition on the first arrival of the Poisson process, differentiate the resulting exponentially weighted integral, and integrate the convolution derivative equation from zero. Tonelli theorem converts the claim-density convolution to the tail convolution. The kernel mass is $\lambda \mu /c<1$, so this is a defective renewal equation.

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 $t$.

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 $\Lambda (t)$.