No-Arrival Conditioning and the Poisson Clock

A Quiet Interval and the Future Poisson Clock

A quiet interval can prompt two different questions. If no arrivals occurred during an interval that has already been observed, does that change the distribution of arrivals after the observation time? If we instead learn that there will be no arrivals during an interval still to come, how does that affect the wait after that interval ends? Independent increments answer both questions, but the conditioning events are different. Separating them prevents the common error of treating a past gap as though it were extra waiting time still to come.

Let \((\Omega,\mathcal{F},\mathbb{P})\) be a probability space, and let \((N_t)_{t\ge 0}\) be a Poisson process of rate \(\lambda\ge 0\), with state space \(\{0,1,2,\ldots\}\) and continuous time \(t\ge 0\). Thus \(N_0=0\), and for \(0\le u<v\), the increment \(N_v-N_u\) has a Poisson distribution with mean \(\lambda(v-u)\); increments on disjoint time intervals are independent. We count arrivals in \((u,v]\) by \(N_v-N_u\). For \(\lambda>0\), the probability of an arrival at any fixed time is zero, so using open or closed endpoints at a specified time does not change the probabilities below.

Fix \(0\le a<s\), and let \(A_{a,s}\) be the event that no arrivals occur in the already observed interval \((a,s]\). For \(\lambda>0\), this is a positive-probability event, since \(\mathbb{P}(A_{a,s})=e^{-\lambda(s-a)}>0\). The future count over an interval after \(s\) is independent of this event: the two intervals are disjoint. The same applies jointly to any finite collection of disjoint future intervals.

Proof. Write \(C_r=\{N_{u_r}-N_{u_{r-1}}=k_r\}\). The event \(A_{a,s}\) is determined by the increment on \((a,s]\). Each \(C_r\) is determined by an increment on a disjoint interval after \(s\). Independent increments therefore give \(\mathbb{P}(A_{a,s}\cap C_1\cap\cdots\cap C_m)=\mathbb{P}(A_{a,s})\prod_{r=1}^m\mathbb{P}(C_r)\). Since \(\mathbb{P}(A_{a,s})=e^{-\lambda(s-a)}>0\), divide both sides by it. Each increment has a Poisson distribution with mean \(\lambda(u_r-u_{r-1})\), giving the displayed formula. \(\square\)

Residual Waiting Time After a Quiet Past

For a fixed observation time \(s\), define the residual wait \(R_s=\inf\{x>0:N_{s+x}>N_s\}\), with the infimum of the empty set taken as \(\infty\). For \(\lambda>0\), the event \(R_s>x\) is exactly the event of no arrivals in \((s,s+x]\). Thus, for \(x\ge0\), the theorem gives the conditional survival probability after observing no arrivals in \((a,s]\):

Consequently, \(R_s\) has an exponential distribution with rate \(\lambda\) even after conditioning on that quiet past. The observation does not start a new process: it reveals a gap in the past, while independent increments say that the future increment has not been altered by that gap. The same calculation with \(x=0\) gives probability \(1\), as required for a positive-rate process with no arrival exactly at the fixed time \(s\).

Conditioning on a Quiet Future

A different condition is that no arrivals occur during an interval that begins at the observation time. Let \(h>0\), and condition on \(B_{s,h}=\{N_{s+h}-N_s=0\}\). This event has probability \(e^{-\lambda h}>0\) when \(\lambda>0\). It says something about the next \(h\) units of time, not about the history before \(s\). Independent increments imply that arrivals after \(s+h\) are unaffected by this condition.

Proof. The event \(\{R_s>h+x\}\) requires zero arrivals throughout \((s,s+h+x]\). Its probability is \(e^{-\lambda(h+x)}\). Since \(B_{s,h}\) has probability \(e^{-\lambda h}>0\), the conditional probability is \(e^{-\lambda(h+x)}/e^{-\lambda h}=e^{-\lambda x}\). This is the exponential survival function for the additional wait after the quiet interval ends. \(\square\)

The kth Future Arrival and a Common Mistake

The same count argument gives the distribution of the time until the \(k\)th future arrival, for any positive integer \(k\). Define \(T_k(s)=\inf\{x>0:N_{s+x}-N_s\ge k\}\). For \(x\ge0\), the event \(T_k(s)>x\) means that there have been at most \(k-1\) arrivals in \((s,s+x]\). Thus, conditional on a quiet past interval \(A_{a,s}\),

This follows by summing the conditional Poisson probabilities for counts \(r=0,\ldots,k-1\); the quiet-past theorem says each has its usual Poisson probability. For \(k=1\), the sum has just the term \(e^{-\lambda x}\), the exponential residual survival probability. No separate assumption about an arrival at time \(s\) is needed: for \(\lambda>0\), the probability of an arrival at any fixed time is zero.

Common mistake: adding the past quiet time to the residual wait. For example, let \(\lambda=\tfrac13\), observe no arrivals in \((s-3,s]\), and ask for the probability of waiting more than \(2\) further units after \(s\). An incorrect calculation treats the total quiet time as \(3+2=5\) and reports \(e^{-(1/3)5}=e^{-5/3}\) as the conditional probability. That is the joint probability of no arrivals in the past \(3\) units and the next \(2\) units, not the conditional probability. The conditioning event has probability \(e^{-(1/3)3}=e^{-1}\), so division gives \(e^{-5/3}/e^{-1}=e^{-2/3}\). Equivalently, independent increments give \(e^{-(1/3)2}=e^{-2/3}\) directly. The quiet past does not lengthen the wait still ahead.

If \(\lambda=0\), the process has no arrivals almost surely. The no-arrival conditions considered here then have probability \(1\), but the residual wait is \(\infty\), not a finite exponential random variable. The exponential formulas for residual waits therefore require \(\lambda>0\); this degenerate case should not be obtained by treating a zero-rate exponential distribution as an ordinary waiting-time distribution.