From a Time-Varying Rate to Count Distributions
In a homogeneous Poisson process, the event rate is constant, so the expected number of events over an interval is the rate multiplied by the interval length. When the rate changes with time, that product no longer applies. Instead, the expected count over an interval is the accumulated rate over that interval. This accumulated quantity, called the cumulative intensity, determines the Poisson count distributions.
Let \((\Omega,\mathcal{F},\mathbb{P})\) be a probability space. A counting process \((N(t))_{t\ge0}\) has state space \(S=\{0,1,2,\ldots\}\), with time index \(t\ge0\). Its natural filtration is \(\mathcal{F}_t=\sigma(N(u):0\le u\le t)\). We consider processes with \(N(0)=0\), nondecreasing right-continuous paths, and independent increments: counts on disjoint time intervals are independent.
The function \(\lambda(t)\) describes the instantaneous event rate at time \(t\), when a pointwise rate is appropriate—for example, when \(\lambda\) is continuous. The cumulative intensity \(\Lambda(t)\) is the expected number of events by time \(t\). More generally, the expected count on \((s,t]\) is the integral of the rate over that interval, not the rate at its endpoint and not usually the rate at its midpoint multiplied by the interval length.
What the Cumulative Intensity Determines
Proof. By the definition of the process, \(N(t)-N(s)\) is Poisson with parameter \(a=\Lambda(t)-\Lambda(s)\), which is finite and nonnegative because \(\lambda\) is nonnegative and locally integrable. The Poisson probability formula gives the displayed probability. For \(a>0\), the probability mass function is \(p_k=e^{-a}a^k/k!\). Summing its first factorial moment gives \(\mathbb{E}X=\sum_{k\ge1}k p_k=a\sum_{r\ge0}e^{-a}a^r/r!=a\). Its second factorial moment is \(\mathbb{E}[X(X-1)]=a^2\), by the same index shift. Since \(X^2=X(X-1)+X\), \(\operatorname{Var}(X)=\mathbb{E}X^2-(\mathbb{E}X)^2=a^2+a-a^2=a\). If \(a=0\), the Poisson law is concentrated at zero, so the mean and variance are both zero and the formulas still hold. Taking \(s=0\), using \(N(0)=0\) and \(\Lambda(0)=0\), gives the final statement. \(\square\)
Independent increments add an important part of the description. If \((s_1,t_1],\ldots,(s_m,t_m]\) are pairwise disjoint intervals, then their counts are independent Poisson variables with respective means \(\int_{s_r}^{t_r}\lambda(u)\,du\). Thus the cumulative intensity specifies the distribution of each interval count, while independent increments specifies how counts on disjoint intervals fit together. For example, the probability of no events in \((s,t]\) is \(e^{-[\Lambda(t)-\Lambda(s)]}\), obtained by taking \(k=0\) in the count formula.
When \(\lambda\) is continuous, the expected cumulative count has derivative $$ \frac{d}{dt}\mathbb{E}N(t)=\frac{d}{dt}\Lambda(t)=\lambda(t). $$ For a merely measurable, locally integrable rate, \(\Lambda\) is absolutely continuous and this derivative equality holds for almost every \(t\), not necessarily at every point. Also, changing \(\lambda\) at a single time does not change any integral or count distribution. At a discontinuity, the assigned value \(\lambda(t)\) at that one point therefore does not alter the process law; one-sided rates or an almost-everywhere rate are the more informative descriptions.
Worked Example: A Rate That Increases Linearly
Suppose \(\lambda(t)=1+t\) for \(t\ge0\). Compute the expected count and the probability of exactly two events in \((1,3]\), and compare the instantaneous rate at time \(2\) with the average rate on that interval.
Integrating from zero gives \(\Lambda(t)=\int_0^t(1+u)\,du=t+t^2/2\). The endpoints give \(\Lambda(3)=3+9/2=15/2\) and \(\Lambda(1)=1+1/2=3/2\). Thus the interval mean is \(\Lambda(3)-\Lambda(1)=15/2-3/2=6\). The count on \((1,3]\) is Poisson with parameter \(6\), so its expected value and variance are both \(6\).
The probability of exactly two events is $$ \mathbb{P}(N(3)-N(1)=2)=e^{-6}\frac{6^2}{2!}=18e^{-6}. $$ The instantaneous rate at time \(2\) is \(\lambda(2)=1+2=3\). The average rate across the interval of length \(3-1=2\) is \(6/2=3\). In this example the average happens to equal the midpoint rate because the rate is linear; the interval mean itself is \(6\), not \(3\).
Check: The derivative of \(\Lambda(t)=t+t^2/2\) is \(1+t\), the specified rate. Also, \(\Lambda(3)-\Lambda(1)=6\), so the expected interval count from the formula agrees with the Poisson parameter used in the probability calculation.
Worked Example: A Rate With a Change at Time One
Let \(\lambda(t)=2\) for \(0\le t<1\), and \(\lambda(t)=1/2\) for \(t\ge1\). Find the cumulative intensity, the laws of the counts on \((0,1]\) and \((1,3]\), and the expected count by time \(3\).
For \(0\le t\le1\), integration gives \(\Lambda(t)=2t\). For \(t\ge1\), the contribution from \([0,1]\) is \(2\), and the contribution from \([1,t]\) is \((t-1)/2\). Therefore \(\Lambda(t)=2+(t-1)/2=(t+3)/2\) for \(t\ge1\). The two expressions agree at \(t=1\), since \(2(1)=2\) and \((1+3)/2=2\).
The mean count on \((0,1]\) is \(\Lambda(1)-\Lambda(0)=2-0=2\), so \(N(1)-N(0)\) is Poisson with parameter \(2\). The mean count on \((1,3]\) is \(\Lambda(3)-\Lambda(1)=(3+3)/2-2=3-2=1\), so \(N(3)-N(1)\) is Poisson with parameter \(1\). These two counts are independent because the intervals are disjoint. The expected count by time \(3\) is \(\Lambda(3)=(3+3)/2=3\), which also equals \(2+1\), the sum of the two interval means.
At time \(3\), the rate is \(\lambda(3)=1/2\), whereas the expected cumulative count is \(\mathbb{E}N(3)=3\). These quantities answer different questions: \(1/2\) is the instantaneous rate at that time, while \(3\) is the expected total number of events accumulated since time zero.
Check: The integral over \((0,3]\) is \(2(1)+(1/2)(3-1)=2+1=3\), matching \(\Lambda(3)\). The interval means \(2\) and \(1\) sum to that same total.
Worked Example: A Rate That Rises Quadratically
Let \(\lambda(t)=3t^2\) for \(t\ge0\). Find the expected count by time \(2\), the mean count on \((1,2]\), and the probability of exactly one event in that interval.
The cumulative intensity is \(\Lambda(t)=\int_0^t3u^2\,du=t^3\). Hence \(\mathbb{E}N(2)=\Lambda(2)=2^3=8\). On \((1,2]\), the mean is \(\Lambda(2)-\Lambda(1)=8-1=7\), so the count in this interval is Poisson with parameter \(7\). Its probability of exactly one event is $$ \mathbb{P}(N(2)-N(1)=1)=e^{-7}\frac{7^1}{1!}=7e^{-7}. $$ The instantaneous rate at time \(2\) is \(\lambda(2)=3(2^2)=12\). It is not the expected count on \((1,2]\), which is \(7\), or the expected cumulative count by time \(2\), which is \(8\).
Check: Differentiating \(\Lambda(t)=t^3\) gives \(3t^2=\lambda(t)\). Also, integrating directly over the interval gives \(\int_1^2 3u^2\,du=2^3-1^3=7\), confirming the interval mean.
Interpreting Rates and Avoiding a Common Error
The rate \(\lambda(t)\) has units of events per unit time; the cumulative intensity \(\Lambda(t)\) and an interval mean have units of events. If the rate is constant, \(\lambda(t)=c\), then \(\Lambda(t)=ct\), and the familiar homogeneous-process formula is recovered: the count on \((s,t]\) has mean \(c(t-s)\). For a changing rate, the integral is essential. A useful average rate on an interval of positive length is $$ \frac{\Lambda(t)-\Lambda(s)}{t-s} =\frac{1}{t-s}\int_s^t\lambda(u)\,du. $$ This interval average need not equal the rate at either endpoint.
Common mistake. For the rate \(\lambda(t)=3t^2\), one might incorrectly use \(\lambda(2)=12\) as the mean count on \((1,2]\), and then claim the probability of no events there is \(e^{-12}\). This confuses the instantaneous rate at the endpoint with an interval count mean. The correct mean is \(\int_1^2 3u^2\,du=7\), so the probability of no events is \(e^{-7}\). The correction is to integrate the rate over the interval before applying a Poisson probability formula.
Local integrability is the condition that keeps the expected number of events finite on every bounded time interval. If \(\int_0^T\lambda(u)\,du=\infty\), the finite-parameter Poisson count formula on \((0,T]\) is not available. Conversely, if \(\lambda\) is locally integrable, every bounded interval has a finite Poisson mean, and its count is finite with probability one. The function \(\Lambda\) is nondecreasing because \(\lambda\ge0\), and it is additive across adjacent intervals: \(\Lambda(t)-\Lambda(s)=[\Lambda(r)-\Lambda(s)]+[\Lambda(t)-\Lambda(r)]\) whenever \(s\le r\le t\). This matches the decomposition of the count on \((s,t]\) into counts on \((s,r]\) and \((r,t]\).
The practical sequence is therefore: specify a nonnegative locally integrable rate, integrate it to obtain \(\Lambda\), take differences of \(\Lambda\) to get interval means, and then use the Poisson law for each count. For several disjoint intervals, use independent increments as well. This keeps a local rate, a cumulative expectation, and a random count distinct throughout the calculation.
Check Your Understanding
For each question, distinguish the rate function from the integrated mean and state which interval is being counted.
- If \(\lambda(t)=4\) for \(t\ge0\), what are \(\Lambda(5)\) and the distribution of the count on \((2,5]\)?
- For \(\lambda(t)=2t\), compute the expected count on \((1,2]\) and the instantaneous rate at time \(2\). Why are they different quantities?
- Suppose \(\Lambda(1)=3\) and \(\Lambda(4)=8\). Give the distribution, mean, and variance of the count on \((1,4]\).
- If a rate function is changed at exactly one time but is unchanged elsewhere, do its cumulative intensity and interval count laws change? Explain using the integral definition.
- For a piecewise constant rate, explain why counts on two adjacent intervals are independent and why their means add to the mean on the union.