Conditioning Changes the Event-Time Distribution
For a homogeneous Poisson process, conditioning on the number of events in an interval makes the event locations uniform on that interval. With a nonhomogeneous rate, the same conditioning principle applies, but locations are no longer generally uniform: regions with higher intensity receive more events. The conditional law is governed by the intensity accumulated across the interval, not just by its length.
Let \((\Omega,\mathcal{F},\mathbb{P})\) be a probability space, and let \((N(t))_{t\ge0}\) be a nonhomogeneous Poisson process with state space \(S=\{0,1,2,\ldots\}\) and continuous time \(t\ge0\). Assume its deterministic intensity function \(\lambda(t)\) is nonnegative and integrable on \([0,T]\). Write \(\Lambda(t)=\int_0^t\lambda(u)\,du\). Thus, for \(0\le a<b\), \(N(b)-N(a)\) is Poisson with mean \(\Lambda(b)-\Lambda(a)\), and counts on disjoint intervals are independent. Let \(T_1<T_2<\cdots\) denote the event times. We condition on \(N(T)=n\), which has positive probability when \(\Lambda(T)>0\); the case \(n=0\) has no event times to describe.
The density expresses intensity-weighted sampling: a small interval with twice the intensity has twice the probability of receiving a sampled event, when its length is unchanged. If \(\lambda\) is constant and positive, \(f(t)=1/T\), recovering the uniform order-statistic representation. The theorem does not require \(\lambda\) to be positive everywhere; where \(\lambda=0\), the conditional density is zero.
Proof. For a finite partition of \([0,T]\) into disjoint intervals \(I_1,\ldots,I_r\), let \(C_j\) be the number of events in \(I_j\), and let \(\Lambda_j=\int_{I_j}\lambda(u)\,du\). The \(C_j\) are independent Poisson variables with means \(\Lambda_j\), and their sum is \(N(T)\). For nonnegative integers \(c_1,\ldots,c_r\) summing to \(n\), independence gives $$ \mathbb{P}(C_1=c_1,\ldots,C_r=c_r) =e^{-\Lambda(T)}\prod_{j=1}^r\frac{\Lambda_j^{c_j}}{c_j!}. $$ Also \(\mathbb{P}(N(T)=n)=e^{-\Lambda(T)}\Lambda(T)^n/n!>0\). Dividing the first probability by the second yields $$ \mathbb{P}(C_1=c_1,\ldots,C_r=c_r\mid N(T)=n) =\frac{n!}{\prod_{j=1}^r c_j!} \prod_{j=1}^r\left(\frac{\Lambda_j}{\Lambda(T)}\right)^{c_j}. $$ This is the multinomial allocation law of \(n\) independent samples, each falling in interval \(I_j\) with probability \(\Lambda_j/\Lambda(T)\). Those probabilities are exactly the interval probabilities of the density \(f(t)=\lambda(t)/\Lambda(T)\). Refining interval partitions determines the distribution of the unordered event locations, so conditionally those locations have the law of \(n\) independent samples with density \(f\). Since \(f\) is a density with respect to length, the probability of a tie between two samples is zero. For distinct sample values \(t_1<\cdots<t_n\), any of the \(n!\) permutations of the independent samples can produce that ordered list; each permutation contributes the product \(\prod_k f(t_k)\). Thus the ordered density is \(n!\prod_k f(t_k)\), which is the stated formula. \(\square\)
Proof. The independent Poisson probabilities for the interval counts and the positive probability of \(N(T)=n\) were computed in the theorem’s proof. Dividing by \(\mathbb{P}(N(T)=n)\) gives the displayed multinomial mass function for every count vector summing to \(n\). The cell probabilities sum to \(1\), because the intervals partition \([0,T]\) and \(\sum_j\Lambda_j=\Lambda(T)\). \(\square\)
Computing with Piecewise Rates
For piecewise-constant rates, the joint density is constant on each region where every event time lies in a specified rate segment. A reliable calculation has two steps: multiply the rate values to obtain the product in the density, and multiply by the volume of the ordered region. Alternatively, integrate the product of the rates over that region and then multiply by \(n!/\Lambda(T)^n\). These methods agree, but the rate factor must appear only once.
Worked Example: Two Events Under a Two-Level Rate
On \([0,4]\), let \(\lambda(t)=1\) for \(0\le t<2\) and \(\lambda(t)=3\) for \(2\le t\le4\). Find the conditional joint density given \(N(4)=2\), and compute the probabilities that both events are early, one is in each segment, or both are late.
The total cumulative intensity is \(\Lambda(4)=1\cdot2+3\cdot2=8\), so the conditioning event has positive probability. The joint density coefficient is \(2!/8^2=2/64=1/32\). Hence, for \(0<s<t<4\), the density is \(\lambda(s)\lambda(t)/32\). It equals \(1/32\) when both times are in \([0,2)\), \(3/32\) when \(s<2\le t\), and \(9/32\) when both times are in \([2,4]\).
For both events early, the ordered region \(0<s<t<2\) has area \(2\cdot2/2=2\). Its probability is \(2(1/32)=1/16\). For one event in each segment, the region \(0<s<2\le t<4\) has area \(2\cdot2=4\), so its probability is \(4(3/32)=12/32=3/8\). Equivalently, the product-rate integral over this mixed region is \(1\cdot3\cdot4=12\), and multiplying by the density coefficient \(1/32\) gives \(12(1/32)=3/8\). For both events late, the ordered region \(2\le s<t\le4\) has area \(2\), and its probability is \(2(9/32)=18/32=9/16\).
Check: The three disjoint cases cover the ordered region, and their probabilities sum to \(1/16+3/8+9/16=1/16+6/16+9/16=1\). The mixed-region density is \(3/32\) on area \(4\), giving \(4(3/32)=3/8\), in agreement with the product-rate integral calculation.
Worked Example: A Linear Rate and a Cross-Interval Event
Let \(\lambda(t)=1+t\) on \([0,2]\). Given \(N(2)=2\), find the probability that exactly one event occurs before time \(1\), and verify it by integrating the joint density.
The cumulative intensity is \(\Lambda(2)=\int_0^2(1+t)\,dt=2+2=4\). The intensity masses in the two unit intervals are \(\int_0^1(1+t)\,dt=1+1/2=3/2\) and \(\int_1^2(1+t)\,dt=[t+t^2/2]_1^2=4-3/2=5/2\). By the conditional count law, the probability of one event in each interval is $$ \frac{2!}{1!1!}\left(\frac{3/2}{4}\right)\left(\frac{5/2}{4}\right) =2\left(\frac38\right)\left(\frac58\right)=\frac{30}{64}=\frac{15}{32}. $$
For the density check, the ordered event times must satisfy \(0<s<1<t<2\). The joint density there is \(2(1+s)(1+t)/4^2=(1+s)(1+t)/8\). Therefore the probability is $$ \int_0^1\int_1^2\frac{(1+s)(1+t)}8\,dt\,ds =\frac18\left(\int_0^1(1+s)\,ds\right)\left(\int_1^2(1+t)\,dt\right) =\frac18\cdot\frac32\cdot\frac52=\frac{15}{32}. $$
Check: The density integrates over this region to the same value as the conditional multinomial calculation, \(15/32\). In particular, treating the event times as uniform would incorrectly use interval probabilities \(1/2\) and \(1/2\), rather than the intensity-weighted probabilities \(3/8\) and \(5/8\).
Worked Example: Three Events Across Two Rate Segments
Let \(\lambda(t)=1\) on \([0,1)\) and \(\lambda(t)=2\) on \([1,2]\). Given \(N(2)=3\), compute the probability that exactly two events occur before time \(1\) and one occurs after time \(1\), then verify by integrating the ordered density.
The total intensity is \(\Lambda(2)=1\cdot1+2\cdot1=3\). The two segment probabilities for an intensity-weighted sample are \(1/3\) and \(2/3\). The conditional count probability is $$ \frac{3!}{2!1!}\left(\frac13\right)^2\left(\frac23\right) =3\cdot\frac19\cdot\frac23=\frac{2}{9}. $$
For the density calculation, the relevant ordered region is \(0<t_1<t_2<1<t_3<2\). On it the joint density is $$ \frac{3!}{3^3}(1)(1)(2)=\frac{6}{27}\cdot2=\frac49. $$
The region for \((t_1,t_2)\) is a triangle of area \(1/2\), and the range for \(t_3\) has length \(1\). Its total volume is therefore \(1/2\), so integrating the constant density gives \((4/9)(1/2)=2/9\). The two calculations agree.
Check: The product of the three rates is \(2\), while the coefficient \(3!/3^3\) is \(2/9\), giving density \(4/9\). Multiplying that density by the ordered-region volume \(1/2\) gives \(2/9\), not \(4/9\).
Common Pitfall: Counting the Rate Factor Twice
When integrating over a region where the rate is constant, distinguish the joint-density coefficient \(n!/\Lambda(T)^n\) from the product of the rate values. For the mixed region in the first example, the product-rate integral is \(12\), and the coefficient is \(1/32\). The probability is \(12(1/32)=3/8\). Equivalently, the density is \(3/32\) throughout a region of area \(4\), giving \(4(3/32)=3/8\).
A wrong calculation is to multiply the product-rate integral \(12\) by the already rate-weighted density \(3/32\), obtaining \(12(3/32)=9/8\). This uses the rate factor twice and even produces a value greater than \(1\). Use either the coefficient times the product-rate integral, or the full density times the region’s ordinary volume—not a mixture of the two.
The order-statistic representation is useful both for probability calculations and for simulation: draw independent locations from the normalized intensity density, then sort them. It also clarifies what conditioning does and does not change. It fixes the total count, but it does not make the locations uniform unless the rate is constant. When rates vary, the conditional allocation to a subinterval is determined by that subinterval’s share of total cumulative intensity.
Check Your Understanding
Use the conditional density and the intensity masses to justify each answer.
- For a constant positive rate on \([0,T]\), what is the density of one event time given \(N(T)=1\)?
- If the intensity masses of two intervals are \(2\) and \(6\), what is the conditional probability that a single event lies in the first interval, given that the total count is \(1\)?
- For \(n=2\), explain the factor \(2!\) in the ordered joint density.
- With \(\lambda(t)=1\) on \([0,1)\) and \(\lambda(t)=2\) on \([1,2]\), compute the conditional probability that both events lie in the second interval, given \(N(2)=2\).
- In a piecewise-constant example, describe the two equivalent ways to integrate the joint density without counting the intensity factor twice.