Counts on Disjoint Intervals
A count observed over a time window can be broken into counts on smaller, disjoint intervals. For a Poisson process, this partition is especially useful: the interval counts are independent, and the parameter for each count is the rate multiplied by that interval’s length. Once the interval counts are understood, their sum gives the count on the full window, while conditioning on that sum describes how a fixed total is distributed among the intervals.
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 \(S=\{0,1,2,\ldots\}\). Use the natural filtration \(\mathcal{F}_t=\sigma(N_s:0\le s\le t)\). The finite-dimensional laws of a Poisson process imply that increments over disjoint time intervals are independent and that \(N_b-N_a\) has a Poisson distribution with parameter \(\lambda(b-a)\) when \(0\le a<b\).
Fix \(T>0\) and a partition \(0=t_0<t_1<\cdots<t_m=T\). For \(r=1,\ldots,m\), define the interval length \(\ell_r=t_r-t_{r-1}>0\) and the interval count \(A_r=N_{t_r}-N_{t_{r-1}}\). Thus \(A_r\) counts events in \((t_{r-1},t_r]\), and the intervals are disjoint. In particular, \(N_T=\sum_{r=1}^m A_r\).
Proof. Each \(A_r\) is an increment over \((t_{r-1},t_r]\). The intervals are disjoint, so the independent-increments property gives independence of the \(A_r\). The increment distribution gives \(\mathbb{P}(A_r=k_r)=e^{-\lambda\ell_r}(\lambda\ell_r)^{k_r}/k_r!\). Multiplying these probabilities proves the joint formula.
For the sum, the event \(N_T=n\) is the disjoint union of the events \(\{A_1=k_1,\ldots,A_m=k_m\}\) over all nonnegative integer vectors with \(\sum_r k_r=n\). Since \(\sum_r\ell_r=T\), summing the joint formula gives $$ \mathbb{P}(N_T=n) =e^{-\lambda T}\lambda^n \sum_{k_1+\cdots+k_m=n}\prod_{r=1}^m\frac{\ell_r^{k_r}}{k_r!} =e^{-\lambda T}\lambda^n\frac{T^n}{n!}. $$ For the last equality, expand \((\ell_1+\cdots+\ell_m)^n\) by the multinomial theorem: the coefficient of \(\prod_r\ell_r^{k_r}\) is \(n!/\prod_r k_r!\). The resulting expression is the Poisson probability with parameter \(\lambda T\).
When \(\sum_r k_r=n\), the conditional probability is the joint probability divided by \(\mathbb{P}(N_T=n)\). The exponential factors cancel, as do the factors \(\lambda^n\), leaving $$ \frac{\prod_r\ell_r^{k_r}/k_r!}{T^n/n!} =\frac{n!}{k_1!\cdots k_m!}\prod_{r=1}^m\left(\frac{\ell_r}{T}\right)^{k_r}. $$ The denominator is positive because \(\lambda T>0\). If the counts do not sum to \(n\), their event is incompatible with \(N_T=n\), so its conditional probability is zero. \(\square\)
The conditional formula is a multinomial distribution: given the total \(n\), the interval counts have cell probabilities \(\ell_r/T\). This does not say that the counts remain independent after conditioning. Their sum is fixed, so knowing the count in one interval constrains the counts that can occur in the others.
Computing Joint and Aggregate Probabilities
Worked Example: A Specific Vector of Interval Counts
Let the rate be \(\lambda=2\), and partition \([0,4]\) at times \(0,1,3,4\). Find the probability that the three interval counts are \(1,2,0\), respectively. Then find the probability that the total count on \([0,4]\) is \(3\), and the conditional probability of this particular allocation given that total.
The interval lengths are \(1-0=1\), \(3-1=2\), and \(4-3=1\). The corresponding Poisson parameters are \(2\cdot1=2\), \(2\cdot2=4\), and \(2\cdot1=2\). Independence gives $$ \begin{aligned} \mathbb{P}(A_1=1,A_2=2,A_3=0) &=\left(e^{-2}\frac{2^1}{1!}\right) \left(e^{-4}\frac{4^2}{2!}\right) \left(e^{-2}\frac{2^0}{0!}\right)\\ &=(2e^{-2})(8e^{-4})(e^{-2})\\ &=16e^{-8}. \end{aligned} $$ Here \(4^2/2!=16/2=8\), and \(2^0/0!=1\).
The full interval has length \(4\), so \(N_4\) has parameter \(2\cdot4=8\). Therefore $$ \mathbb{P}(N_4=3)=e^{-8}\frac{8^3}{3!} =e^{-8}\frac{512}{6} =\frac{256}{3}e^{-8}. $$ Dividing gives the conditional probability $$ \mathbb{P}(A_1=1,A_2=2,A_3=0\mid N_4=3) =\frac{16e^{-8}}{(256/3)e^{-8}} =\frac{3}{16}. $$ The multinomial formula checks this: the interval probabilities are \(1/4,2/4,1/4\), so the allocation probability is $$ \frac{3!}{1!2!0!}\left(\frac14\right)^1 \left(\frac24\right)^2\left(\frac14\right)^0 =3\cdot\frac14\cdot\frac14 =\frac{3}{16}. $$ The two calculations agree.
There are two useful ways to answer a partition-count question. If the requested counts in each interval are specified, multiply their Poisson probabilities. If only the total is specified, use the Poisson distribution with parameter equal to the rate times the full window length. If both the total and the allocation are specified, use the conditional multinomial formula.
Worked Example: Cumulative Counts Are Not Independent
Let \(\lambda=2\). Compare the event \(\{N_1=1,N_2=1\}\) with the product of its marginal probabilities. This calculation illustrates why the independent objects are disjoint-interval counts, not cumulative counts.
Set \(A=N_1\), the count in \((0,1]\), and \(B=N_2-N_1\), the count in \((1,2]\). These are independent, and each has a Poisson distribution with parameter \(2\). The event \(N_1=1,N_2=1\) is exactly the event \(A=1,B=0\), because \(N_2=A+B\). Hence $$ \mathbb{P}(N_1=1,N_2=1) =\mathbb{P}(A=1)\mathbb{P}(B=0) =(2e^{-2})(e^{-2}) =2e^{-4}. $$
The marginal probability \(\mathbb{P}(N_1=1)\) is \(2e^{-2}\), since \(N_1\) has parameter \(2\). The count \(N_2\) has parameter \(2\cdot2=4\), so $$ \mathbb{P}(N_2=1)=e^{-4}\frac{4^1}{1!}=4e^{-4}. $$ If one incorrectly treated \(N_1\) and \(N_2\) as independent, the product of the marginals would be $$ \mathbb{P}(N_1=1)\mathbb{P}(N_2=1) =(2e^{-2})(4e^{-4}) =8e^{-6}, $$ which differs from the joint probability \(2e^{-4}\). Thus \(N_1\) and \(N_2\) are not independent. The distinction is that \(N_2\) includes all of \(N_1\), whereas \(A\) and \(B\) count events on disjoint intervals.
Allocation Given a Fixed Total
The conditional multinomial formula can be used even when the intervals have different lengths. An interval twice as long as another receives twice the conditional probability per event, but the counts are still constrained to add up to the fixed total. The formula gives a full probability for each allocation vector; it does not assign independent Poisson probabilities after conditioning.
Worked Example: Allocating Four Events Across Three Intervals
Let \(\lambda=3\) and partition \([0,6]\) into intervals of lengths \(1,2,3\). Given that \(N_6=4\), find the probability that the interval counts are \(1,1,2\).
The length proportions are \(1/6,2/6,3/6\), and the counts sum to \(1+1+2=4\), as required. The conditional probability is $$ \begin{aligned} \mathbb{P}(A_1=1,A_2=1,A_3=2\mid N_6=4) &=\frac{4!}{1!1!2!} \left(\frac16\right)^1 \left(\frac26\right)^1 \left(\frac36\right)^2\\ &=12\cdot\frac16\cdot\frac13\cdot\frac14\\ &=\frac16. \end{aligned} $$ Each factor follows from the specified count and interval length: \(4!/(1!1!2!)=24/2=12\), \(2/6=1/3\), and \((3/6)^2=(1/2)^2=1/4\).
Check by the unconditional joint probability. The three interval parameters are \(3\cdot1=3\), \(3\cdot2=6\), and \(3\cdot3=9\). Thus $$ \mathbb{P}(A_1=1,A_2=1,A_3=2) =e^{-18}\frac{3^1}{1!}\frac{6^1}{1!}\frac{9^2}{2!} =729e^{-18}. $$ The total count has parameter \(3\cdot6=18\), and $$ \mathbb{P}(N_6=4)=e^{-18}\frac{18^4}{4!} =e^{-18}\frac{104976}{24} =4374e^{-18}. $$ Their ratio is \(729/4374=1/6\), which confirms the conditional calculation.
Checks and Common Pitfalls
Before calculating, identify whether the requested variables count disjoint intervals or cumulative windows. For a partition, the interval counts are independent and their parameters are \(\lambda\ell_r\). Cumulative counts such as \(N_{t_1}\) and \(N_{t_2}\), where \(t_1<t_2\), overlap and generally are dependent. Also check that the proposed interval counts add to the stated total before using a conditional allocation formula.
Common mistake. For a rate-\(2\) process, it is incorrect to use \(2e^{-4}\) as \(\mathbb{P}(N_2=1)\). The parameter of \(N_2\) is \(\lambda t=2\cdot2=4\), so the correct probability is \(e^{-4}4^1/1!=4e^{-4}\). Consequently, the product of the marginal probabilities for \(N_1=1\) and \(N_2=1\) is \((2e^{-2})(4e^{-4})=8e^{-6}\). The joint probability is instead \(2e^{-4}\), as found by expressing the event through the disjoint increments \(N_1\) and \(N_2-N_1\). Both the parameter and the event being counted must be checked before multiplying probabilities.
Finally, the total-count formula is consistent with the partition: summing the interval counts telescopes to \(N_T-N_0=N_T\), since \(N_0=0\) almost surely. The lengths likewise add to \(T\), so the sum of their Poisson parameters is \(\sum_r\lambda\ell_r=\lambda T\). These checks help catch a missing interval or an incorrect endpoint difference.
Check Your Understanding
Use disjoint increments for joint probabilities, and verify that any conditional allocation counts sum to the specified total.
- A rate-\(2\) process is partitioned at times \(0,1,3\). What are the parameters of the two interval counts?
- For the partition in Question 1, find the probability that the interval counts are \(0\) and \(2\), respectively.
- For a rate-\(3\) process on \([0,2]\), what is the distribution of the total count?
- A window is split into intervals of lengths \(1\) and \(3\). Given a total of \(2\), find the conditional probability that both events fall in the longer interval.
- Explain why \(N_1\) and \(N_2-N_1\) are independent, while \(N_1\) and \(N_2\) are generally dependent.