Why Conditioning Changes the Question
For a Poisson process, fixing the number of arrivals in a time window gives useful information about where those arrivals occurred. It also changes the laws and dependence relations of counts in subintervals. Errors often arise from carrying an unconditional statement into a conditional calculation, or from confusing an unordered collection of locations with the ordered arrival times.
Let \((\Omega,\mathcal{F},\mathbb{P})\) be a probability space, and let \((N_t)_{t\ge0}\) be a homogeneous Poisson process with rate \(\lambda>0\), state space \(S=\{0,1,2,\ldots\}\), and continuous time \(t\ge0\). Thus \(N_0=0\), increments over disjoint time intervals are independent, and \(N_t-N_s\) has a Poisson distribution with mean \(\lambda(t-s)\) for \(0\le s<t\). Fix a horizon \(T>0\). Write \(S_1<\cdots<S_{N_T}\) for the arrival times in \((0,T]\) when \(N_T\ge1\).
The event \(N_T=n\) has positive probability for every integer \(n\ge0\), since \(\mathbb{P}(N_T=n)=e^{-\lambda T}(\lambda T)^n/n!>0\). Thus conditioning on this event is well-defined. When \(n=0\), there are no arrival locations to describe. For \(n\ge1\), the locations before ordering and the ordered arrival times must be treated differently.
Proof. The interarrival times of a rate-\(\lambda\) Poisson process are independent exponential random variables with density \(x\mapsto\lambda e^{-\lambda x}\) for \(x>0\). The transformation from the first \(n\) interarrival times to the arrival times \(s_1,\ldots,s_n\) has determinant \(1\). Consequently, the joint density of the first \(n\) arrival times at \(0<s_1<\cdots<s_n\), together with the event that the next arrival occurs after \(T\), is \(\lambda^n e^{-\lambda s_n}e^{-\lambda(T-s_n)}=\lambda^n e^{-\lambda T}\). Dividing this density by \(\mathbb{P}(N_T=n)=e^{-\lambda T}(\lambda T)^n/n!\) gives the conditional density \(n!/T^n\) on the ordered region. To compare with independent Uniform\((0,T)\) variables, their joint density before ordering is \(1/T^n\) on the cube \((0,T)^n\). Except on ties, an event of probability zero, the cube is split into \(n!\) possible strict orderings. Each ordering has the same density; sorting therefore gives density \(n!/T^n\) on the ordered region. This proves both claims. \(\square\)
This result does not say that \(S_1,\ldots,S_n\) are independent. It says that if the locations are viewed without their arrival order, they have the law of an independent uniform sample. Sorting that sample creates dependent order statistics.
Three Distinctions to Keep Straight
Worked Example: Ordered Times Are Not Each Uniform
Suppose \(N_T=2\). Let \(U_1,U_2\) denote the two locations viewed as an unordered sample of independent Uniform\((0,T)\) variables. The ordered arrival times are \(S_1=\min(U_1,U_2)\) and \(S_2=\max(U_1,U_2)\). Find the probabilities that the first arrival is before \(T/3\), and that both arrivals are before \(T/3\).
The event \(S_1<T/3\) means at least one of \(U_1,U_2\) lies in \((0,T/3)\). Its complement requires both uniform locations to be at least \(T/3\), which has probability \((2/3)^2=4/9\). Hence
The event \(S_2<T/3\) means both locations are before \(T/3\), so
Check: The probability exactly one location is before \(T/3\) is \(2(1/3)(2/3)=4/9\). The probabilities for zero, exactly one, and two such locations sum to \(4/9+4/9+1/9=1\). A common wrong calculation treats \(S_1\) as Uniform\((0,T)\) and gives probability \(1/3\) for \(S_1<T/3\). The correction is that the first ordered time is the minimum of two uniform locations, so the correct probability is \(5/9\).
A second frequent error is to use an unconditional Poisson law after the total count has been fixed. Counts in a subinterval are Poisson before conditioning, but conditional on the total they have a different distribution.
Proof. Before conditioning, the counts in the partition intervals are independent Poisson random variables with respective means \(\lambda\ell_1,\ldots,\lambda\ell_r\). For counts summing to \(n\), their joint probability is \(e^{-\lambda T}\prod_{a=1}^r(\lambda\ell_a)^{c_a}/c_a!\). The probability of the conditioning event is \(e^{-\lambda T}(\lambda T)^n/n!\). Dividing the joint probability by this positive probability cancels \(e^{-\lambda T}\); because \(\sum_a c_a=n\), the powers of \(\lambda\) also cancel. The result is exactly the displayed formula. Counts that do not sum to \(n\) cannot occur when \(N_T=n\). \(\square\)
Worked Example: Conditional and Unconditional Interval Counts
Take \(T=2/\lambda\), so \(\lambda T=2\), and let \(A=(0,T/2]\). Compare the probability that \(A\) contains exactly one arrival given \(N_T=3\) with its unconditional probability.
Conditional on \(N_T=3\), the interval \(A\) has length fraction \((T/2)/T=1/2\). The conditional interval-count formula gives \(\mathbb{P}(N(A)=1\mid N_T=3)=\binom31(1/2)^1(1/2)^2=3/8\).
Unconditionally, \(N(A)\) is Poisson with mean \(\lambda(T/2)=1\), so \(\mathbb{P}(N(A)=1)=e^{-1}1^1/1!=e^{-1}\). These probabilities answer different questions. For a direct check of the conditional calculation, let \(B=(T/2,T]\). The joint probability of \(N(A)=1,N(B)=2\) is \(e^{-2}(1)^1/1!\,(1)^2/2!=e^{-2}/2\). The probability \(N_T=3\) is \(e^{-2}2^3/3!=4e^{-2}/3\). Their ratio is \((e^{-2}/2)/(4e^{-2}/3)=3/8\), as above.
Check: Given the total count of three, the probabilities of zero, one, two, and three arrivals in \(A\) are respectively \(1/8,3/8,3/8,1/8\), which sum to \(1\). The unconditional probability \(e^{-1}\) is not obtained by using a Poisson law with mean \(1\) after the total has been fixed.
Conditioning Can Create Dependence
Unconditionally, counts in disjoint intervals are independent by the independent-increments property. After conditioning on their sum, they cannot generally remain independent: if one interval receives more of the fixed total, fewer arrivals remain for the others.
Proof. By the conditional interval-count result, the \(n\) arrivals can be represented by \(n\) independent uniform locations before ordering. Let \(I_k\) be \(1\) if location \(k\) lies in \(A\), and \(0\) otherwise. Then \(N(A)=\sum_{k=1}^n I_k\), while \(N(B)=n-N(A)\) when the two intervals form the entire partition. In that case, \(\operatorname{Cov}(N(A),N(B)\mid N_T=n)=\operatorname{Cov}(N(A),n-N(A)\mid N_T=n)=-\operatorname{Var}(N(A)\mid N_T=n)\). The indicators \(I_k\) are independent, each with probability \(p\) of being \(1\), so their variances are \(p(1-p)\), and the variance of their sum is \(np(1-p)\). This proves the formula. More generally, if other intervals are also present, the same indicator representation gives \(N(A)=\sum I_k\) and \(N(B)=\sum J_k\), where for each \(k\), \(I_kJ_k=0\), \(\mathbb{E}I_k=p\), and \(\mathbb{E}J_k=q=|B|/T\). Thus \(\operatorname{Cov}(I_k,J_k)=-pq\); indicators from distinct locations are independent, giving covariance \(-npq\). \(\square\)
Worked Example: Two Half-Window Counts Are Dependent
Split \((0,T]\) into two halves and condition on \(N_T=2\). Let \(C_1\) and \(C_2\) be the counts in the first and second halves. Compute their joint probability of being \((1,1)\), compare it with the product of their conditional marginal probabilities, and calculate their covariance.
The conditional interval-count formula gives \(\mathbb{P}(C_1=1,C_2=1\mid N_T=2)=\frac{2!}{1!1!}(1/2)(1/2)=1/2\). Each marginal count has a Binomial\((2,1/2)\) distribution, so \(\mathbb{P}(C_1=1\mid N_T=2)=\mathbb{P}(C_2=1\mid N_T=2)=\binom21(1/2)(1/2)=1/2\). If they were conditionally independent, the joint probability would be \((1/2)(1/2)=1/4\), not \(1/2\).
For the covariance, \(C_2=2-C_1\), and \(C_1\) has variance \(2(1/2)(1/2)=1/2\). Therefore \(\operatorname{Cov}(C_1,C_2\mid N_T=2)=\operatorname{Cov}(C_1,2-C_1\mid N_T=2)=-1/2\).
Check: The possible count pairs are \((0,2),(1,1),(2,0)\), with probabilities \(1/4,1/2,1/4\); they sum to \(1\). The marginal probability \(C_1=1\) is the middle probability \(1/2\), and the negative covariance agrees with the fixed-sum relation.
A Practical Error Check
Before calculating with a fixed Poisson count, identify exactly what is being conditioned on and what is being counted. The following distinctions prevent the most common mistakes.
| Question | Correct conditional statement | Common error |
|---|---|---|
| Where are the \(n\) arrivals? | Unordered locations have the law of \(n\) independent Uniform\((0,T)\) variables. | Claiming the ordered arrival times are independent uniforms. |
| How many arrivals lie in a subinterval? | Given \(N_T=n\), use the binomial or multinomial allocation law. | Using the unconditional Poisson law without conditioning. |
| Are disjoint interval counts independent? | They are independent unconditionally, but generally dependent given their total. | Assuming independence survives conditioning on the sum. |
| Does the total count determine every location? | No; for \(n\ge1\), the conditional locations remain random. | Treating a fixed count as fixing the arrival times. |
At a fixed \(n\), the conditional law is a statement about arrivals inside the specified window \((0,T]\); it does not assert that counts or times outside the window have the same conditional distribution. Also, interval endpoints cause no ambiguity in these calculations: a fixed time has probability zero of being an arrival time, so open or closed endpoint choices do not change the probabilities.
Common mistake and correction. A tempting but incorrect argument is: “Counts in the two halves are independent Poisson variables, so they are still independent given their sum.” Independence before conditioning does not imply independence after conditioning. In the example above, the correct conditional joint probability is \(1/2\), while multiplying the conditional marginal probabilities gives \(1/4\). The fixed-total constraint \(C_1+C_2=2\) forces dependence; the covariance is \(-1/2\).
Check Your Understanding
For each question, specify whether the statement is conditional or unconditional before calculating.
- Given \(N_T=3\), what is the probability that exactly two arrivals lie in the first quarter of \((0,T]\)?
- For \(N_T=2\), explain why the probability that the first ordered arrival is before \(T/2\) differs from the probability that a particular unordered location is before \(T/2\).
- If two disjoint intervals have length fractions \(1/3\) and \(2/3\), what is the conditional covariance of their counts given \(N_T=4\)?
- For two equal halves and a fixed total of three arrivals, state the conditional distribution of the count in the first half.
- Why is it incorrect to use a Poisson distribution for a subinterval count after conditioning on the total count, without checking the conditional law?