Counts in Several Disjoint Windows
A Poisson process can be observed over several time windows, and a probability question may combine the counts from all of them: perhaps two windows contain the same number of events, one window is empty, or the total count is at least a specified value. The reliable starting point is to identify disjoint increments. Their counts are independent, and each has a Poisson distribution whose mean is the rate multiplied by the window length. Once the joint distribution is written down, even compound event conditions become explicit sums or products.
Let \((\Omega,\mathcal{F},\mathbb{P})\) be a probability space, and let \((N_t)_{t\ge0}\) be a Poisson process of rate \(\lambda\ge0\), with state space \(\mathbb{N}_0=\{0,1,2,\ldots\}\) and continuous time \(t\ge0\). For \(0\le a\le b\), write \(N_b-N_a\) for the number of events in the window \((a,b]\). The Poisson process properties used here are that this increment has a Poisson distribution with mean \(\lambda(b-a)\), and that increments on disjoint time intervals are independent. A window of zero length has count zero almost surely.
Proof. By the independent-increments property, \(X_1,\ldots,X_m\) are mutually independent. For each \(r\), the count \(X_r\) has a Poisson distribution with mean \(\theta_r\), so its probability of equaling \(k_r\) is \(e^{-\theta_r}\theta_r^{k_r}/k_r!\) when \(\theta_r>0\). Independence makes the probability of the simultaneous event the product of these probabilities. When \(\theta_r=0\), the count \(X_r\) is zero almost surely, so the stated factor gives precisely its probability at each possible \(k_r\). This proves the formula, including zero-length windows and the zero-rate case. \(\square\)
This joint formula is the basic tool for event conditions involving several windows. For example, an event requiring two counts to be equal is a disjoint union over their possible common values. A condition that the total count is at least a threshold can be handled by summing over count vectors meeting that threshold, or by first finding the distribution of the total.
Proof. If \(\Theta=0\), every \(\theta_r=0\), so each \(X_r=0\) almost surely and \(X=0\) almost surely. This is a Poisson distribution with mean zero and gives the claimed tail formula. Now suppose \(\Theta>0\). For any integer \(n\ge0\), the joint count formula gives $$ \mathbb{P}(X=n)=e^{-\Theta} \sum_{\substack{k_1+\cdots+k_m=n\\ k_1,\ldots,k_m\ge0}} \prod_{r=1}^m\frac{\theta_r^{k_r}}{k_r!}. $$ The multinomial theorem, applied to \((\theta_1+\cdots+\theta_m)^n\), says that the sum in this expression is \(\Theta^n/n!\). Thus \(\mathbb{P}(X=n)=e^{-\Theta}\Theta^n/n!\). Summing the probabilities for \(n=0,\ldots,q-1\) and taking the complement proves the tail formula. When \(q=0\), the event \(X\ge0\) has probability \(1\), as the empty-sum convention also gives. \(\square\)
Worked Probabilities
Worked Example: Equal Counts in Two Windows
Let \(\lambda=\tfrac12\), and take the disjoint windows \((0,1]\) and \((1,2]\). Let \(X=N_1-N_0\) and \(Y=N_2-N_1\). Find the probability that the counts agree and are at most \(1\), and write the exact probability that they agree with no upper restriction.
Each window has length \(1\), so both means are \(\tfrac12\). The joint formula gives, for \(k,\ell\ge0\), $$ \mathbb{P}(X=k,Y=\ell) =\left(e^{-1/2}\frac{(1/2)^k}{k!}\right) \left(e^{-1/2}\frac{(1/2)^\ell}{\ell!}\right) =e^{-1}\frac{(1/2)^{k+\ell}}{k!\ell!}. $$ The event \(X=Y\le1\) consists of exactly the two possibilities \((X,Y)=(0,0)\) and \((X,Y)=(1,1)\). Their probabilities are \(e^{-1}\) and \(e^{-1}(1/2)^2=e^{-1}/4\), respectively. Therefore, $$ \mathbb{P}(X=Y\le1)=e^{-1}+\frac{e^{-1}}4=\frac{5}{4e}. $$ For equality with no upper restriction, sum over all possible common counts: $$ \mathbb{P}(X=Y)=\sum_{k=0}^{\infty}\mathbb{P}(X=k,Y=k) =e^{-1}\sum_{k=0}^{\infty}\frac{(1/4)^k}{(k!)^2}. $$ This convergent series is an exact expression; the restricted event above is its \(k=0\) and \(k=1\) contribution.
Check: Substituting \(k=0\) into the summand gives \(e^{-1}\), and substituting \(k=1\) gives \(e^{-1}(1/4)\); their sum is \(5/(4e)\), as calculated directly. The unrestricted event contains these two cases and all cases \(k\ge2\), so its probability is at least \(5/(4e)\).
Worked Example: No Events in One Window and at Least Two in Another
Let \(\lambda=2\). Consider a first window of length \(\tfrac12\) and a disjoint second window of length \(1\). Find the probability that the first count is zero while the second count is at least \(2\).
Call the counts \(X\) and \(Y\). Their means are \(2(\tfrac12)=1\) and \(2(1)=2\), and the counts are independent. Hence \(\mathbb{P}(X=0)=e^{-1}\). For the second window, the excluded cases \(Y=0\) and \(Y=1\) have probabilities \(e^{-2}\) and \(e^{-2}\cdot 2\), respectively. Therefore, $$ \mathbb{P}(Y\ge2)=1-e^{-2}-2e^{-2}=1-3e^{-2}. $$ Multiplying the independent event probabilities gives $$ \mathbb{P}(X=0,\ Y\ge2) =e^{-1}(1-3e^{-2}) =e^{-1}-3e^{-3}. $$
Check by summing joint probabilities: For every integer \(y\ge2\), the joint formula gives $$ \mathbb{P}(X=0,Y=y) =e^{-1}\left(e^{-2}\frac{2^y}{y!}\right). $$ Summing over \(y\ge2\) yields \(e^{-1}\mathbb{P}(Y\ge2)=e^{-1}(1-3e^{-2})\), the same result. The first window’s zero-count condition has been included once as its probability \(e^{-1}\), not confused with a condition on the cumulative count.
Worked Example: A Minimum Total Across Three Windows
Let \(\lambda=2\), and consider three pairwise disjoint windows, each of length \(\tfrac12\). Find the probability that the total count across the three windows is at least \(2\).
Write the counts as \(X_1,X_2,X_3\). Each has mean \(2(\tfrac12)=1\), so the total mean in the theorem is \(\Theta=1+1+1=3\). The total \(X_1+X_2+X_3\) is Poisson with mean \(3\). Its probability of being less than \(2\) is the sum of its probabilities at \(0\) and \(1\): $$ \mathbb{P}(X_1+X_2+X_3<2) =e^{-3}\frac{3^0}{0!}+e^{-3}\frac{3^1}{1!} =e^{-3}+3e^{-3} =4e^{-3}. $$ Taking the complement gives $$ \mathbb{P}(X_1+X_2+X_3\ge2)=1-4e^{-3}. $$ The strict inequality in the complement is important: the excluded total counts are \(0\) and \(1\), while a total of \(2\) belongs to the requested event.
Check directly from the windows: All three counts are zero with probability \(e^{-1}e^{-1}e^{-1}=e^{-3}\). A total of exactly one means that one of the three counts is \(1\) and the other two are \(0\). Each of these three disjoint possibilities has probability \((e^{-1})(e^{-1})(e^{-1})=e^{-3}\), because the count of \(1\) in a mean-one window has probability \(e^{-1}\cdot1^1/1!=e^{-1}\). Thus the probability of a total below \(2\) is \(e^{-3}+3e^{-3}=4e^{-3}\), confirming the result.
Disjoint Increments Are Not Cumulative Counts
Independence applies to counts in disjoint windows, not in general to counts from time zero up to different endpoints. For \(0\le s<t\), write \(N_t=N_s+(N_t-N_s)\). The two terms on the right are counts on disjoint intervals and are independent. But \(N_s\) is itself part of \(N_t\). In the nondegenerate case \(\lambda>0\) and \(s>0\), this dependence can be verified from the covariance: $$ \operatorname{Cov}(N_s,N_t) =\operatorname{Cov}(N_s,N_s)+\operatorname{Cov}(N_s,N_t-N_s) =\operatorname{Var}(N_s)+0 =\lambda s>0. $$ Here the zero covariance term follows from independence of the increments, and a Poisson random variable with mean \(\lambda s\) has variance \(\lambda s\). When \(s=0\), \(N_s=0\) almost surely; when \(\lambda=0\), both counts are constant zero. In either degenerate case, \(N_s\) and \(N_t\) are independent.
Common mistake. Take rate \(\lambda=1\) and times \(s=1,t=2\). Treating \(N_1\) and \(N_2\) as independent would give the incorrect calculation $$ \mathbb{P}(N_1=0,N_2=0) \mathrel{\color{black}{=}} \mathbb{P}(N_1=0)\mathbb{P}(N_2=0) =e^{-1}e^{-2}=e^{-3}. $$ The correction is that the event \(N_2=0\) already implies \(N_1=0\), so the joint event is just \(\{N_2=0\}\), with probability \(e^{-2}\). Equivalently, the covariance is \(\operatorname{Cov}(N_1,N_2)=1\), so these two cumulative counts are not independent. Use independent-increment calculations for the disjoint counts \(N_1-N_0\) and \(N_2-N_1\), not for the overlapping cumulative counts.
For a practical calculation, list the windows, compute each mean \(\theta_r=\lambda(b_r-a_r)\), and then translate the event into count vectors. For specified counts, use the joint product formula. For equality, sum the joint probabilities over the allowed common values. For a minimum total, use the Poisson law of the sum and subtract the probabilities below the threshold. If a window has zero length or the rate is zero, its count is deterministically zero; the same formulas still apply with the stated zero-mean convention.
Check Your Understanding
In each question, identify whether the counts refer to disjoint windows or cumulative counts before using independence.
- Two disjoint windows each have mean \(1\). Write an exact series for the probability that their counts are equal.
- For independent window counts with means \(2\) and \(1\), find the probability that the first count is zero and the second is exactly \(2\).
- Three disjoint windows each have mean \(2\). Find the probability that their total count is at least \(1\), and verify it by identifying the complementary event.
- For a Poisson process of rate \(\lambda>0\), explain why \(N_s\) and \(N_t-N_s\) are independent but, when \(s>0\), \(N_s\) and \(N_t\) are not independent for \(t>s\).
- If \(\lambda=0\), what is the probability that the counts in any finite collection of windows are all zero? Which term in the joint formula gives this result?