Finite-Dimensional Laws of a Poisson Process

From Counts to Increments

A Poisson process records cumulative arrivals. Its increments over disjoint time intervals are independent, but its cumulative counts are not: a count at an earlier time is included in every later count. To find probabilities involving several observation times, first rewrite the cumulative counts as counts on the intervals between those times. Each such increment has a Poisson distribution whose parameter is the rate multiplied by the interval length.

Let $(\Omega,\mathcal{F},\mathbb{P})$ be a probability space, and let $(N_t)_{t\ge0}$ be a Poisson process with rate $\lambda>0$, taking values in $S=\{0,1,2,\ldots\}$ and indexed by continuous time $t\ge0$. We use the defining properties that $N_0=0$, increments on disjoint intervals are independent, and for $0\le s<t$, $N_t-N_s$ has a Poisson distribution with mean $\lambda(t-s)$. In particular, for $r=0,1,2,\ldots$,

\mathbb{P}(N_t-N_s=r) =e^{-\lambda(t-s)}\frac{(\lambda(t-s))^r}{r!}.

Fix strictly increasing observation times $0<t_1<\cdots<t_m$. Set $t_0=0$, and define the increments $Y_r=N_{t_r}-N_{t_{r-1}}$ for $r=1,\ldots,m$. The interval lengths are $\Delta t_r=t_r-t_{r-1}$. The variables $Y_1,\ldots,Y_m$ are independent, with $Y_r$ Poisson with mean $\lambda\Delta t_r$. The observed counts are cumulative sums: $N_{t_\ell}=Y_1+\cdots+Y_\ell$.

Theorem (Finite-dimensional joint probabilities): For integers $n_1,\ldots,n_m$, let $n_0=0$. If $0\le n_1\le n_2\le\cdots\le n_m$, then $$ \mathbb{P}(N_{t_1}=n_1,\ldots,N_{t_m}=n_m) =\prod_{r=1}^{m}e^{-\lambda\Delta t_r} \frac{(\lambda\Delta t_r)^{n_r-n_{r-1}}}{(n_r-n_{r-1})!}. $$ If any $n_r$ is negative or the sequence is not nondecreasing, the probability is zero.

This gives the joint law at any finite set of ordered times: the successive differences $n_r-n_{r-1}$ specify how many arrivals occurred on each intervening interval. The formula includes zero increments; if $n_r=n_{r-1}$, that interval contributes the probability of no arrivals.

Proof. Since $N_0=0$, the event $N_{t_1}=n_1,\ldots,N_{t_m}=n_m$ is exactly the event $Y_r=n_r-n_{r-1}$ for every $r=1,\ldots,m$. Each $Y_r$ is a nonnegative integer, so if any count is negative or a difference $n_r-n_{r-1}$ is negative, the event is impossible. Otherwise independence of the increments gives the probability of their joint event as the product of their individual probabilities. Substituting the Poisson probability with mean $\lambda\Delta t_r$ for each $Y_r$ gives the stated formula. $\square$

The statement is for strictly increasing observation times. If a list of observation times repeats a time, the corresponding counts must be equal; one can remove repeated times and apply the formula to the remaining strictly increasing list. The initial time $t=0$ is also deterministic: $N_0=0$.

Calculating Joint Probabilities

Worked Example: Two Observation Times

Let the rate be $\lambda=2$, and observe the process at $t_1=1$ and $t_2=3$. Find the probability that $N_1=2$ and $N_3=5$.

The first interval has length $\Delta t_1=1-0=1$, and the second has length $\Delta t_2=3-1=2$. The required increments are $n_1=2$ and $n_2-n_1=5-2=3$. Their Poisson means are $2\cdot1=2$ and $2\cdot2=4$, respectively. Independence gives

\begin{aligned} \mathbb{P}(N_1=2,N_3=5) &=\left(e^{-2}\frac{2^2}{2!}\right) \left(e^{-4}\frac{4^3}{3!}\right)\\ &=\left(e^{-2}\frac{4}{2}\right) \left(e^{-4}\frac{64}{6}\right)\\ &=2e^{-2}\cdot\frac{32}{3}e^{-4} =\frac{64}{3}e^{-6}. \end{aligned}

Check: the two increments are $2$ and $3$, which sum to $5$, the specified value of $N_3$. Their means are $2$ and $4$; the displayed factors are the Poisson probabilities for exactly those increments, and their product has exponential factor $e^{-2}e^{-4}=e^{-6}$.

Worked Example: An Interval With No Arrivals

Let $\lambda=1$, and observe at $t_1=\tfrac12$, $t_2=1$, and $t_3=2$. Find the probability of the count vector $(N_{1/2},N_1,N_2)=(0,2,3)$.

The interval lengths are $\Delta t_1=\tfrac12$, $\Delta t_2=1-\tfrac12=\tfrac12$, and $\Delta t_3=2-1=1$. The required increments are $0$, $2-0=2$, and $3-2=1$. The corresponding means are $\tfrac12,\tfrac12,1$. Therefore

\begin{aligned} \mathbb{P}(N_{1/2}=0,N_1=2,N_2=3) &=\left(e^{-1/2}\frac{(1/2)^0}{0!}\right) \left(e^{-1/2}\frac{(1/2)^2}{2!}\right) \left(e^{-1}\frac{1^1}{1!}\right)\\ &=e^{-1/2}\cdot\frac{e^{-1/2}}{8}\cdot e^{-1} =\frac{1}{8}e^{-2}. \end{aligned}

Check: the first factor is the probability of zero arrivals in a half-unit interval, the second is the probability of two arrivals in the next half-unit, and the third is the probability of one arrival in the final one-unit interval. The increments $0,2,1$ add successively to the specified cumulative counts $0,2,3$.

Not every proposed vector of counts can occur. Since a counting process is nondecreasing, later counts cannot be smaller than earlier ones. For example, if $t_1<t_2$, the event $N_{t_1}=3,N_{t_2}=2$ has probability zero, not the product of two Poisson probabilities. This restriction is built into the joint formula through the requirement that each increment be nonnegative.

Worked Example: Checking an Impossible Count Vector

For any rate $\lambda>0$ and times $0<t_1<t_2$, consider the proposed values $N_{t_1}=2$ and $N_{t_2}=1$. The first count requires two arrivals by $t_1$, while the second would require a total of only one by the later time $t_2$. In increment form, the second interval would have count $1-2=-1$, which is impossible. Thus

\mathbb{P}(N_{t_1}=2,N_{t_2}=1)=0.

Check: every sample path of a counting process is nondecreasing, so $N_{t_2}\ge N_{t_1}$ whenever $t_2>t_1$. The proposed values violate this pathwise condition, independently of the rate and the lengths of the intervals.

Joint Generating Function and Covariance

The finite-dimensional law can also be summarized by a joint probability generating function. For $0\le z_1,\ldots,z_m\le1$, define it as $\mathbb{E}\!\left[\prod_{\ell=1}^m z_\ell^{N_{t_\ell}}\right]$, with the convention $z^0=1$, including when $z=0$. This function encodes the joint probabilities, and the increment representation gives a compact expression.

Result (Joint probability generating function): For $0<t_1<\cdots<t_m$ and $0\le z_1,\ldots,z_m\le1$, $$ \mathbb{E}\!\left[\prod_{\ell=1}^m z_\ell^{N_{t_\ell}}\right] =\prod_{r=1}^m \exp\!\left\{\lambda\Delta t_r \left(\prod_{\ell=r}^m z_\ell-1\right)\right\}. $$

Proof. Write $N_{t_\ell}=\sum_{r=1}^{\ell}Y_r$. In the product $\prod_{\ell=1}^m z_\ell^{N_{t_\ell}}$, the increment $Y_r$ appears in the counts at times $t_r,t_{r+1},\ldots,t_m$. Hence $$ \prod_{\ell=1}^m z_\ell^{N_{t_\ell}} =\prod_{r=1}^m\left(\prod_{\ell=r}^m z_\ell\right)^{Y_r}. $$ The $Y_r$ are independent. For a Poisson variable $Y$ with mean $\theta$, summing its probabilities gives $\mathbb{E}[w^Y]=\sum_{q=0}^{\infty}w^q e^{-\theta}\theta^q/q!=e^{\theta(w-1)}$ for $0\le w\le1$. Apply this identity to each $Y_r$, with $\theta=\lambda\Delta t_r$ and $w=\prod_{\ell=r}^m z_\ell$, and multiply the resulting expectations. This yields the displayed expression. $\square$

A related consequence describes dependence between two cumulative counts. The Poisson distribution with mean $\theta$ has variance $\theta$. If $0\le s\le t$, write $N_t=N_s+(N_t-N_s)$. The increment $N_t-N_s$ is independent of $N_s$, so the covariance is

\operatorname{Cov}(N_s,N_t) =\operatorname{Cov}(N_s,N_s)+\operatorname{Cov}(N_s,N_t-N_s) =\operatorname{Var}(N_s)+0 =\lambda s.

Interchanging the two times when $t<s$ gives the compact expression $\operatorname{Cov}(N_s,N_t)=\lambda\min(s,t)$. This shared part of the observation intervals explains why cumulative counts at different times are generally dependent, even though increments on disjoint intervals are independent.

A Common Mistake: Multiplying Marginal Probabilities

Independence applies to disjoint-interval increments, not to cumulative counts at distinct times. To see the difference, take $\lambda=1$, $s=1$, and $t=2$, and consider $N_1=1,N_2=2$. A tempting but incorrect calculation is to multiply the two marginal probabilities: $$ \mathbb{P}(N_1=1)\mathbb{P}(N_2=2) =\left(e^{-1}\frac{1^1}{1!}\right) \left(e^{-2}\frac{2^2}{2!}\right) =2e^{-3}. $$ This treats $N_1$ and $N_2$ as independent.

The correct calculation uses $N_1=1$ and $N_2-N_1=1$. The increments on $[0,1]$ and $(1,2]$ are independent, and each has mean $1$. Thus $$ \mathbb{P}(N_1=1,N_2=2) =\left(e^{-1}\frac{1^1}{1!}\right) \left(e^{-1}\frac{1^1}{1!}\right) =e^{-2}. $$ The difference is not a matter of approximation: the later count includes the earlier count. Converting to interval increments is the essential step in a joint-probability calculation.

Check Your Understanding

For each question, begin by identifying the increments on the disjoint intervals.

For rate $\lambda=1$, find $\mathbb{P}(N_1=0,N_3=2)$. State the two increment values and their means.
For times $0<t_1<t_2<t_3$, write the increment values corresponding to $N_{t_1}=2,N_{t_2}=2,N_{t_3}=5$. Which interval has no arrivals?
Explain why the joint probability of $N_{t_1}=4,N_{t_2}=3$ is zero when $t_1<t_2$.
Use the joint generating-function formula with $m=2$ to express $\mathbb{E}[z_1^{N_s}z_2^{N_t}]$ for $0<s<t$.
For $0\le s\le t$, explain why the independent increment $N_t-N_s$ contributes zero to $\operatorname{Cov}(N_s,N_t)$.

Finite-Dimensional Laws of a Poisson Process

What You'll Learn