From Event Counts to Waiting Times
A Poisson process describes events occurring over time when counts on disjoint intervals are independent and the distribution of a count depends only on the interval's length. Examples include arrivals to a service system or events recorded by a detector under a constant-rate model. Its defining increment properties lead both to a precise distribution for the number of events and to a simple distribution for the time between events.
Work on a filtered probability space \((\Omega,\mathcal{F},(\mathcal{F}_t)_{t\ge0},\mathbb{P})\) carrying a standard Brownian motion \(B_t\) and a counting process \(N_t\). Brownian motion is not used in this tutorial. Let \(\mathcal{F}_t^N=\sigma(N_s:0\le s\le t)\) denote the natural filtration of \(N\). Statements conditioning on the past of the counting process will use \(\mathcal{F}_t^N\). No stochastic integrals occur here.
The short-time conditions encode a rate \(\lambda\): over a small interval of length \(h\), one event has probability approximately \(\lambda h\), while two or more events have smaller-order probability. The increment assumptions then determine the count distribution at every time. The case \(\lambda=0\) is the degenerate process \(N_t=0\) for all \(t\); exponential waiting-time statements below concern \(\lambda>0\).
Deriving the Counting Probabilities
Proof. For \(0\le z\le1\), define the probability generating function \(G_t(z)=\mathbb{E}[z^{N_t}]\), using \(z^0=1\), including when \(z=0\). By independent stationary increments, \(N_{t+h}=N_t+(N_{t+h}-N_t)\) is a sum of independent variables, and the second summand has the same distribution as \(N_h\). Thus \(G_{t+h}(z)=G_t(z)G_h(z)\). The short-time assumptions give $$ G_h(z)=\mathbb{P}(N_h=0)+z\mathbb{P}(N_h=1)+\sum_{j\ge2}z^j\mathbb{P}(N_h=j) =1+\lambda h(z-1)+o(h). $$ Indeed, \(\mathbb{P}(N_h=0)=1-\lambda h+o(h)\), and the absolute value of the last sum is at most \(\mathbb{P}(N_h\ge2)=o(h)\). It follows that the right derivative of \(G_t(z)\) is \(\lambda(z-1)G_t(z)\), with \(G_0(z)=1\). Solving this scalar differential equation gives \(G_t(z)=\exp(\lambda t(z-1))\). Expanding, $$ G_t(z)=e^{-\lambda t}\sum_{k=0}^{\infty}\frac{(\lambda t)^kz^k}{k!}. $$ A probability generating function has coefficient \(\mathbb{P}(N_t=k)\) at \(z^k\), so comparison gives the asserted probabilities. For \(\lambda t>0\), differentiating the generating function at \(z=1\) gives \(\mathbb{E}[N_t]=\lambda t\) and \(\mathbb{E}[N_t(N_t-1)]=(\lambda t)^2\); therefore \(\operatorname{Var}(N_t)=(\lambda t)^2+\lambda t-(\lambda t)^2=\lambda t\). At \(t=0\), \(N_0=0\), so both moment statements hold there too. \(\square\)
Worked Example: Counts and a Tail Probability
Let \(N\) have rate \(\lambda=2\). Find the probabilities of zero, one, two, and three events by time \(t=2\), and then find the probability of at least four events.
Here \(\lambda t=2\cdot2=4\). Substituting this value into the count formula gives $$ \mathbb{P}(N_2=0)=e^{-4},\quad \mathbb{P}(N_2=1)=4e^{-4},\quad \mathbb{P}(N_2=2)=\frac{4^2}{2!}e^{-4}=8e^{-4}, $$ and $$ \mathbb{P}(N_2=3)=\frac{4^3}{3!}e^{-4}=\frac{64}{6}e^{-4}=\frac{32}{3}e^{-4}. $$ The event of at least four events is the complement of \(N_2\le3\), so the four probabilities just computed must be summed. Their coefficient sum is $$ 1+4+8+\frac{32}{3}=\frac{3+12+24+32}{3}=\frac{71}{3}. $$ Consequently \(\mathbb{P}(N_2\ge4)=1-\frac{71}{3}e^{-4}\). Check: the probabilities for counts zero through three sum to \(\frac{71}{3}e^{-4}\), so the displayed tail probability makes their total exactly one. The tail uses the inclusive event \(N_2\ge4\); its complement is precisely \(N_2\le3\).
Counts on disjoint intervals are independent, not merely counts starting at time zero. For \(0\le s<t\), the increment \(N_t-N_s\) has the Poisson distribution with parameter \(\lambda(t-s)\). Its mean and variance are both \(\lambda(t-s)\), by the theorem applied to the increment, whose distribution equals that of \(N_{t-s}\).
Worked Example: Counts on Two Disjoint Intervals
Let \(N\) have rate \(\lambda=\tfrac12\). Find the probability of exactly one event in \([0,1]\) and no events in \((1,2]\).
The first interval has length \(1\), so its count is Poisson with parameter \(\tfrac12\cdot1=\tfrac12\). The second interval also has length \(1\), so its count has the same parameter. The increments are independent; hence $$ \begin{aligned} \mathbb{P}(N_1=1,\ N_2-N_1=0) &=\mathbb{P}(N_1=1)\mathbb{P}(N_2-N_1=0)\\ &=\left(\frac{(1/2)^1}{1!}e^{-1/2}\right) \left(\frac{(1/2)^0}{0!}e^{-1/2}\right) =\frac12e^{-1}. \end{aligned} $$ Check: the event requires one event in the first interval and zero in the second; the product contains exactly those two probabilities. As a second calculation, it is the event that the total count by time \(2\) is one and that this event lies in the first interval. The independent-increment calculation gives its probability directly, without treating the total count alone as sufficient to identify the event's location.
Interarrival Times
For each integer \(n\ge1\), define the \(n\)th arrival time by \(T_n=\inf\{t\ge0:N_t\ge n\}\), with the infimum of the empty set taken to be infinity. Set \(T_0=0\) and \(X_n=T_n-T_{n-1}\) when the arrival times are finite. Unit-sized jumps ensure that two arrivals cannot occur at the same time.
Proof. First, the event \(T_1>x\) is exactly the event \(N_x=0\). The count formula therefore gives \(\mathbb{P}(T_1>x)=e^{-\lambda x}\). To obtain the joint law of later arrival times, consider ordered times \(0<t_1<\cdots<t_n\). Put small disjoint intervals of lengths \(h_1,\ldots,h_n\) around these times, with their endpoints ordered and with all \(h_i>0\). The event that the first \(n\) arrivals fall one in each such interval requires no events in the intervening gaps and exactly one event in each small interval. Independent stationary increments give its probability as the product of the corresponding count probabilities. For a gap of length \(a\), the no-event probability is \(e^{-\lambda a}\); for a small interval of length \(h_i\), the one-event probability is \(\lambda h_i+o(h_i)\). As the interval lengths tend to zero, the gap lengths sum to \(t_n+o(1)\), so the product, divided by \(h_1\cdots h_n\), tends to \(\lambda^n e^{-\lambda t_n}\). Thus the joint density of the ordered arrival times is $$ f_{T_1,\ldots,T_n}(t_1,\ldots,t_n)=\lambda^n e^{-\lambda t_n}, \qquad 0<t_1<\cdots<t_n. $$ Transform to gaps \(x_1=t_1\) and \(x_j=t_j-t_{j-1}\) for \(2\le j\le n\). This triangular transformation has Jacobian determinant \(1\), and \(t_n=x_1+\cdots+x_n\). The joint density of the gaps is therefore $$ \lambda^n e^{-\lambda(x_1+\cdots+x_n)} =\prod_{j=1}^n\big(\lambda e^{-\lambda x_j}\big), \qquad x_1,\ldots,x_n>0. $$ Each factor integrates to \(1\), since \(\int_0^\infty\lambda e^{-\lambda x}\,dx=1\). The factorization proves that the gaps are independent and each has density \(\lambda e^{-\lambda x}\) for \(x>0\), whose survival probability is \(e^{-\lambda x}\). In particular, \(T_n\) is finite almost surely: the joint density integrates to one over finite positive gaps, so \(X_1,\ldots,X_n\) are finite with probability one. \(\square\)
There is also a useful direct relation between arrival times and counts. For \(t\ge0\), \(T_n>t\) holds exactly when fewer than \(n\) events have occurred by time \(t\), or \(N_t\le n-1\). Hence $$ \mathbb{P}(T_n>t)=e^{-\lambda t}\sum_{k=0}^{n-1}\frac{(\lambda t)^k}{k!}. $$ This is the survival probability of the \(n\)th arrival time. It follows from the count formula and does not require separately integrating the joint density.
Worked Example: The Second Arrival by a Given Time
For a rate-\(2\) process, find the probability that the second arrival occurs by time \(1\), and find the probability that the first waiting time exceeds \(\tfrac12\).
The second arrival has occurred by time \(1\) exactly when \(N_1\ge2\). The parameter for this count is \(\lambda t=2\cdot1=2\), so $$ \mathbb{P}(T_2\le1)=\mathbb{P}(N_1\ge2) =1-\mathbb{P}(N_1=0)-\mathbb{P}(N_1=1) =1-e^{-2}-2e^{-2}=1-3e^{-2}. $$ For the first waiting time, the exponential survival formula gives $$ \mathbb{P}(T_1>\tfrac12)=e^{-2(1/2)}=e^{-1}. $$ Check: the complement of \(T_2\le1\) is \(T_2>1\), which is the event \(N_1\le1\); its probability is \(e^{-2}+2e^{-2}=3e^{-2}\), agreeing with the first calculation. For the second result, \(T_1>\tfrac12\) is exactly \(N_{1/2}=0\), whose Poisson probability is \(e^{-2(1/2)}=e^{-1}\).
Memorylessness and a Common Pitfall
The exponential waiting-time law has the memoryless property. For \(x,y\ge0\), $$ \mathbb{P}(T_1>x+y\mid T_1>x) =\frac{e^{-\lambda(x+y)}}{e^{-\lambda x}} =e^{-\lambda y}. $$ The conditioning event has positive probability, since \(e^{-\lambda x}>0\). The result says that, conditional on no event by time \(x\), the probability of waiting at least another \(y\) is the same as the probability of waiting at least \(y\) from time zero. More generally, independent stationary increments give \(\mathbb{P}(N_{t+s}-N_t=0\mid\mathcal{F}_t^N)=e^{-\lambda s}\) almost surely: the future increment is independent of the past and has the same law as \(N_s\).
Common pitfall: confusing a count mean with a waiting-time mean. The expected number of events by time \(t\) is \(\lambda t\); it is not the expected time to the next event. For example, at rate \(\lambda=2\), the expected count by time \(1\) is \(2\), while the first waiting time has mean \(1/\lambda=1/2\). The latter follows by integrating its density: \(\int_0^\infty x\,2e^{-2x}\,dx=1/2\). Substituting \(t=1\) into the count mean to claim that the expected waiting time is \(2\) reverses the units and gives the wrong quantity.
The constant-rate assumption is essential to these formulas: stationary increments make the count law depend on interval length, and independent increments make counts on disjoint intervals factor. The results here concern the event counts and arrival times themselves; questions about compensating a counting process or integrating against its jumps require additional constructions.
Check Your Understanding
Use the count formula and arrival-time relationships to answer the following questions.
- For a rate-\(3\) process, compute the probability of exactly two events by time \(1\).
- For a rate-\(\tfrac12\) process, compute the probability of no events in an interval of length \(4\).
- Explain why the probability that the second arrival exceeds \(t\) is the probability that the count by time \(t\) is at most one.
- For a rate-\(2\) process, find the probability that the first waiting time is at most \(\tfrac12\).
- Show from the exponential survival function that, conditional on \(T_1>x\), the probability of waiting more than a further \(y\) is \(e^{-\lambda y}\).